RD SHARMA SOLUTION CHAPTER-27 Direction Cosines and Direction Ratios I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 27 Direction Cosines and Direction Ratios Exercise Ex. 27.1

Question 1

If a line makes angles of 90°, 60° and 30° with the positive direction of x,y and z-axis respectively, find its direction cosines.Solution 1

Let l, m and n be the direction cosines of a line.

l = cos 90° = 0

begin mathsize 12px style text m end text equals text cos   60 end text degree equals 1 half end style
begin mathsize 12px style straight n equals cos text   end text 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction end style
begin mathsize 12px style therefore straight T text he   direction   cosines   of   the   line   are   0 , end text 1 half comma fraction numerator square root of 3 over denominator 2 end fraction. end style

Question 2

If a line has direction ratios 2, -1, -2, determine its direction cosines.Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text the text    end text direction text    end text cosines text    end text of text    end text the text    end text line text    end text be text    end text straight l comma straight m comma straight n. end cell row cell Here comma end cell row cell straight a equals 2 comma straight b equals negative 1 comma straight c equals negative 2 text    end text are text    end text the text    end text direction text    end text ratios text    end text of text    end text the text    end text line. end cell row cell straight l equals plus-or-minus fraction numerator straight a over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction comma straight m equals plus-or-minus fraction numerator straight b over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction comma straight n equals plus-or-minus fraction numerator straight c over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction end cell row cell straight l equals fraction numerator 2 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction comma straight m equals fraction numerator negative 1 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction comma straight n equals fraction numerator negative 2 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction end cell row cell straight l equals fraction numerator 2 over denominator square root of 9 end fraction comma straight m equals fraction numerator negative 1 over denominator square root of 9 end fraction comma straight n equals fraction numerator negative 2 over denominator square root of 9 end fraction end cell row cell straight l equals 2 over 3 comma straight m equals negative 1 third comma straight n equals negative 2 over 3 end cell row cell therefore The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text are text    end text 2 over 3 comma negative 1 third comma negative 2 over 3. end cell end table end style

Question 3

Find the direction cosines of the line passing through two points (-2, 4, -5) and (1, 2, 3).Solution 3

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text joining text    end text left parenthesis negative 2 comma 4 comma negative 5 right parenthesis text   end text and text   end text left parenthesis 1 comma text   end text 2 comma text   end text 3 right parenthesis text   are end text comma end cell row cell left parenthesis 1 plus 2 comma 2 minus 4 comma 3 plus 5 right parenthesis equals left parenthesis 3 comma negative 2 comma 8 right parenthesis end cell row cell Here comma straight a equals 3 comma straight b equals negative 2 comma straight c equals 8 end cell row cell Direction text    end text cosines text    end text are end cell row cell fraction numerator 3 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction comma fraction numerator negative 2 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction comma fraction numerator 8 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction end cell row cell equals fraction numerator 3 over denominator square root of 77 end fraction comma fraction numerator negative 2 over denominator square root of 77 end fraction comma fraction numerator 8 over denominator square root of 77 end fraction end cell end table end style

Question 4

Using direction ratios show that the points A (2, 3, -4), (1, -2, 3) and (3, 8, -11) are collinear.Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell Here text    end text straight A text   end text left parenthesis 2 comma 3 comma negative text 4 end text right parenthesis comma text   end text straight B text   end text left parenthesis 1 comma negative 2 comma 3 right parenthesis text   end text and text   end text straight C text   end text left parenthesis 3 comma 8 comma negative 11 right parenthesis. end cell row cell Direction text    end text ratios text    end text of text    end text AB equals left parenthesis 1 minus 2 comma negative 2 minus 3 comma 3 plus 4 right parenthesis equals left parenthesis negative 1 comma negative 5 comma 7 right parenthesis end cell row cell Direction text    end text ratios text    end text of text    end text BC equals left parenthesis 3 minus 1 comma 8 plus 2 comma negative 11 minus 3 right parenthesis equals left parenthesis 2 comma 10 comma negative 14 right parenthesis end cell row blank row cell Here comma text    end text the text    end text respective text    end text direction text    end text cosines text    end text of text    end text AB text    end text and text    end text AC comma end cell row cell fraction numerator negative 1 over denominator 2 end fraction equals fraction numerator negative 5 over denominator 10 end fraction equals fraction numerator 7 over denominator negative 14 end fraction text    end text are text    end text proportional. end cell row blank row cell Also comma text   end text straight B text    end text is text    end text the text    end text common text    end text point text    end text between text    end text the text    end text two text    end text lines comma end cell row cell therefore The text    end text points text     end text straight A text   end text left parenthesis 2 comma 3 comma negative text 4 end text right parenthesis comma text   end text straight B text   end text left parenthesis 1 comma negative 2 comma 3 right parenthesis text   end text and text   end text straight C text   end text left parenthesis 3 comma 8 comma negative 11 right parenthesis text    end text are text    end text collinear. text   end text end cell row blank end table end style

Question 5

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2)Solution 5

begin mathsize 12px style table attributes columnalign left end attributes row cell straight A left parenthesis 3 comma 5 comma negative 4 right parenthesis comma straight B left parenthesis negative 1 comma 1 comma 2 right parenthesis text    end text and text    end text straight C left parenthesis negative 5 comma negative 5 comma negative 2 right parenthesis end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text AB equals left parenthesis negative 1 minus 3 comma 1 minus 5 comma 2 plus 4 right parenthesis end cell row cell equals left parenthesis negative 4 comma negative 4 comma 6 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text AB text    end text will text    end text be end cell row cell fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction comma fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction comma fraction numerator 6 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator 6 over denominator square root of 68 end fraction end cell row cell equals fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator 3 over denominator square root of 17 end fraction end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text BC equals left parenthesis negative 5 plus 1 comma negative 5 minus 1 comma negative 2 minus 2 right parenthesis end cell row cell equals left parenthesis negative 4 comma negative 6 comma negative 4 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text BC text    end text will text    end text be end cell row cell fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction comma fraction numerator negative 6 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction comma fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator negative 6 over denominator square root of 68 end fraction comma fraction numerator negative 4 over denominator square root of 68 end fraction end cell row cell equals fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator negative 3 over denominator square root of 17 end fraction comma fraction numerator negative 2 over denominator square root of 17 end fraction end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text AC equals left parenthesis negative 5 minus 3 comma negative 5 minus 5 comma negative 2 plus 4 right parenthesis end cell row cell equals left parenthesis negative 8 comma negative 10 comma 2 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text AC text    end text will text    end text be end cell row cell fraction numerator negative 8 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction comma fraction numerator negative 10 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction comma fraction numerator 2 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction end cell row cell equals fraction numerator negative 8 over denominator square root of 168 end fraction comma fraction numerator negative 10 over denominator square root of 168 end fraction comma fraction numerator 2 over denominator square root of 168 end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 42 end fraction comma fraction numerator negative 5 over denominator square root of 42 end fraction comma fraction numerator 1 over denominator square root of 42 end fraction end cell end table end style

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the acute angle between the lines whose direction ratios are proportional to 2:3:6 and 1:2:2.Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Find the angle between the lines whose direction cosines are given by equations

2l + 2m – n = 0, mn + ln + lm = 0Solution 16(iv)

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RD SHARMA SOLUTION CHAPTER-26 Scalar Triple Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 26 Scalar Triple Product Exercise Ex. 26.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 6

Solution 6

Question 7

S h o w space t h a t space t h e space p o i n t s space A open parentheses minus 1 comma 4 comma minus 3 close parentheses comma space B open parentheses 3 comma 2 comma minus 5 close parentheses comma space C open parentheses minus 3 comma 8 comma minus 5 close parentheses space a n d space D open parentheses minus 3 comma 2 comma 1 close parentheses
a r e space c o minus p l a n a r.

Solution 7

A B equals p o s i t i o n space v e c t o r space o f space B minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals 4 i with hat on top minus 2 j with hat on top minus 2 k with hat on top
A C equals p o s i t i o n space v e c t o r space o f space C minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals minus 2 i with hat on top plus 4 j with hat on top minus 2 k with hat on top
A D equals p o s i t i o n space v e c t o r space o f space D minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals minus 2 i with hat on top minus 2 j with hat on top plus 4 k with hat on top
T h e space f o u r space p o i n t s space a r e space c o minus p l a n a r space i f space t h e space v e c t o r s space a r e space c o minus p l a n a r.
T h u s comma space open vertical bar table row 4 cell minus 2 end cell cell minus 2 end cell row cell minus 2 end cell 4 cell minus 2 end cell row cell minus 2 end cell cell minus 2 end cell 4 end table close vertical bar equals 4 open square brackets 16 minus 4 close square brackets plus 2 open square brackets minus 8 minus 4 close square brackets minus 2 open square brackets 4 plus 8 close square brackets equals 48 minus 24 minus 24 equals 0
H e n c e space p r o v e d.

Question 8

Show that the four points whose position vectors are begin mathsize 12px style 6 i with hat on top minus 7 j with hat on top comma space 16 i with hat on top space minus space 19 j with hat on top minus 4 k with hat on top comma space 3 i with hat on top space minus space 6 k with hat on top comma space 2 i with hat on top space minus space 5 j with hat on top space plus space 10 k with hat on top end style Coplanar.Solution 8

L e t space O A equals 6 i with hat on top minus 7 j with hat on top comma space O B equals 16 i with hat on top minus 19 j with hat on top minus 4 k with hat on top comma space O C equals 3 i with hat on top minus 6 k with hat on top comma space O D equals 2 i with hat on top minus 5 j with hat on top plus 10 k with hat on top
T h u s comma
A B equals O B minus O A equals 10 i with hat on top minus 12 j with hat on top minus 4 k with hat on top
A C equals O C minus O A equals minus 3 i with hat on top plus 7 j with hat on top minus 6 k with hat on top
A D equals O D minus O A equals minus 4 i with hat on top plus 2 j with hat on top plus 10 k with hat on top
T h e space f o u r space p o i n t s space a r e space c o minus p l a n a r space i f space v e c t o r s space A B comma space A C space a n d space A D space a r e space c o minus p l a n a r.
T h u s comma space w e space h a v e
open vertical bar table row 10 cell minus 12 end cell cell minus 4 end cell row cell minus 3 end cell 7 cell minus 6 end cell row cell minus 4 end cell 2 10 end table close vertical bar equals 10 open parentheses 70 plus 12 close parentheses plus 12 open parentheses minus 30 minus 24 close parentheses minus 4 open parentheses minus 6 plus 28 close parentheses equals 820 minus 648 minus 88

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12(i)

Solution 12(i)

Question 12(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell text let   end text straight a with rightwards arrow on top text  =  end text straight i with hat on top text + end text straight j with hat on top text + end text straight k with hat on top text ,  end text straight b with rightwards arrow on top text  =  end text straight i with hat on top text   and   end text stack text c end text with rightwards arrow on top text  =  end text straight c subscript 1 straight i with hat on top text + end text straight c subscript 2 straight j with hat on top text + end text straight c subscript 3 straight k with hat on top text .  end text end cell row cell text Then ,  If   end text straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1 comma text   show   that   no   value   end text end cell row cell text of   end text straight c subscript text 1 end text end subscript text   can   make   end text straight a with rightwards arrow on top comma text   end text straight b with rightwards arrow on top text   and   end text straight c with rightwards arrow on top text   coplanar. end text end cell end table end style

Solution 12(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell straight a with rightwards arrow on top text  =  end text straight i with hat on top text + end text straight j with hat on top text + end text straight k with hat on top text ,  end text straight b with rightwards arrow on top text  =  end text straight i with hat on top text   and   end text stack text c end text with rightwards arrow on top text  =  end text straight c subscript 1 straight i with hat on top text + end text straight c subscript 2 straight j with hat on top text + end text straight c subscript 3 straight k with hat on top text. end text end cell row cell straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1 comma end cell row cell text If   end text straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma text    and   end text stack text c end text with rightwards arrow on top text   are   coplanar ,  then   their   scalar   triple   product   is   zero. end text end cell row cell left square bracket table row cell straight a with rightwards arrow on top end cell cell straight b with rightwards arrow on top end cell cell stack text c end text with rightwards arrow on top end cell end table right square bracket equals vertical line table row 1 1 1 row 1 0 0 row cell straight c subscript 1 end cell cell straight c subscript 2 end cell cell straight c subscript 3 end cell end table vertical line equals 0 end cell row cell rightwards double arrow negative straight c subscript 3 plus straight c subscript 2 equals 0 end cell row cell rightwards double arrow straight c subscript 2 equals straight c subscript 3 end cell row cell text But   this   is   a   contradiction   as   it   is   given   that   end text straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1. end cell row cell text Hence ,  no   value   of   end text straight c subscript 1 text   can   make   the   vectors   coplanar. end text end cell end table end style

Question 13

Solution 13

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RD SHARMA SOLUTION CHAPTER-25 Vector or Cross Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 25 Vector or Cross Product Exercise Ex. 25.1

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27(i)

Solution 27(i)

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Using Vectors, find the area of the triangle with vertices: (i) A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5) (ii) A (1, 2, 3), B(2, -1, 4) and C (4, 5, -1).Solution 34

Question 35

Solution 35

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RD SHARMA SOLUTION CHAPTER-24 Scalar or Dot Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 24 Scalar Or Dot Product Exercise Ex. 24.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Solution 5 (iv)

Question 5 (v)

Solution 5 (v)

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell text Dot   products   of   a   vector   with   vectors   end text straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma text   2 end text space straight i with hat on top plus straight j with hat on top minus 3 straight k with hat on top text   end text end cell row cell text and   end text straight i with hat on top plus straight j with hat on top plus straight k with hat on top text   are   respectively   4 , 0   and   2 .  Find   the   vector. end text end cell end table end style

Solution 7(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text the text    end text unknown text    end text vector text    end text be text    end text apostrophe straight a with rightwards arrow on top equals straight a subscript 1 straight i with hat on top plus straight b subscript 1 straight j with hat on top plus straight c subscript 1 straight k with hat on top apostrophe end cell row cell straight b with rightwards arrow on top equals straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma text    end text straight c with rightwards arrow on top equals 2 straight i with hat on top plus straight j with hat on top minus 3 straight k with hat on top comma straight d with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top plus straight k with hat on top end cell row cell It text    end text is text    end text given text    end text that text    end text straight a with rightwards arrow on top. straight b with rightwards arrow on top equals 4 end cell row cell straight a subscript 1 minus straight b subscript 1 plus straight c subscript 1 equals 4..... left parenthesis straight i right parenthesis end cell row cell straight a with rightwards arrow on top. straight c with rightwards arrow on top equals 0 end cell row cell 2 straight a subscript 1 plus straight b subscript 1 minus 3 straight c subscript 1 equals 0....... left parenthesis ii right parenthesis end cell row cell straight a with rightwards arrow on top. straight d with rightwards arrow on top equals 2 end cell row cell straight a subscript 1 plus straight b subscript 1 plus straight c subscript 1 equals 2........ left parenthesis iii right parenthesis end cell row blank row cell Solving text    end text left parenthesis straight i right parenthesis comma left parenthesis ii right parenthesis text    end text and text    end text left parenthesis iii right parenthesis comma end cell row cell straight a subscript 1 equals 2 comma straight b subscript 1 equals negative 1 comma straight c subscript 1 equals 1 end cell row blank row cell therefore the text    end text vector text    end text straight a with rightwards arrow on top equals 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top end cell end table end style

Question 8 (i)

Solution 8 (i)

Question 8 (ii)

Solution 8 (ii)

Question 9

Solution 9

Question 10

Solution 10

Given that begin mathsize 12px style straight a with rightwards arrow on top comma stack straight b comma with rightwards arrow on top straight c with rightwards arrow on top end style are mutually perpendicular, so,

begin mathsize 12px style straight a with rightwards arrow on top straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight c with rightwards arrow on top equals straight c with rightwards arrow on top. straight a with rightwards arrow on top equals 0
and space straight a with rightwards arrow on top comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top are space unit space vectors comma space so
open vertical bar straight a with rightwards arrow on top close vertical bar equals open vertical bar straight b with rightwards arrow on top close vertical bar equals open vertical bar straight c with rightwards arrow on top close vertical bar equals 1
Now comma
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close parentheses squared
equals open parentheses straight a with rightwards arrow on top close parentheses plus open parentheses straight b with rightwards arrow on top close parentheses squared open parentheses straight c with rightwards arrow on top close parentheses squared plus open parentheses straight c with rightwards arrow on top close parentheses squared plus 2 straight a with rightwards arrow on top straight b with rightwards arrow on top plus 2 straight b with rightwards arrow on top straight c with rightwards arrow on top plus 2 straight b with rightwards arrow on top straight c with rightwards arrow on top plus 2 straight c with rightwards arrow on top straight a with rightwards arrow on top
equals open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus open vertical bar straight c with rightwards arrow on top close vertical bar squared plus 2 left parenthesis 0 right parenthesis plus 2 left parenthesis 0 right parenthesis plus 2 left parenthesis 0 right parenthesis
equals open parentheses 1 close parentheses squared plus open parentheses 1 close parentheses squared plus left parenthesis 1 right parenthesis squared plus 0
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals 1 plus 1 plus 1
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals 3
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar equals square root of 3 end style

Question 11

Solution 11

Question 12

Show that the vector begin mathsize 12px style i with hat on top thin space plus thin space j with hat on top space plus space k with hat on top end style is equally inclined with the coordinate axes.Solution 12

begin mathsize 12px style Let space straight theta space be space the space angle space between space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space and space straight i with hat on top
Then comma
cos space straight theta equals fraction numerator open parentheses space straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses. left parenthesis straight i with hat on top right parenthesis over denominator space open vertical bar straight i with hat on top plus straight j with hat on top plus straight k with hat on top close vertical bar open vertical bar left parenthesis straight i with hat on top right parenthesis close vertical bar end fraction
space space space space space space space space space space equals fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction
space space space space space space space space space space equals square root of 3
Similarly comma space if space straight alpha space and space straight gamma space are space angles space that space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space make space with space straight j with hat on top space and space straight k with hat on top
Then comma
space space space space space space space space space space space space space cos space straight alpha equals square root of 3
and space space space space space space space cos space straight gamma equals square root of 3
Therefore comma space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space is space equally space inclined space the space three space axes. end style

Question 13

Show that the vectors begin mathsize 12px style straight a with rightwards arrow on top equals 1 over 7 open parentheses 2 straight i with hat on top space plus thin space 3 straight j with hat on top space plus space 6 straight k with hat on top close parentheses comma straight b with rightwards arrow space on top equals 1 over 7 open parentheses 3 straight i with hat on top space minus thin space 6 straight j with hat on top space plus space 2 straight k with hat on top close parentheses comma space straight c with rightwards arrow on top equals 1 over 7 open parentheses 6 straight i with hat on top space plus thin space 2 straight j with hat on top space minus space 3 straight k with hat on top close parentheses end style are mutually perpendicualr unit vectors.Solution 13

begin mathsize 12px style we space have comma
straight a with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 2 straight i with overbrace on top plus 3 straight j with overbrace on top plus 6 straight k with overbrace on top close parentheses
straight b with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 3 straight i with overbrace on top plus 6 straight j with overbrace on top plus 2 straight k with overbrace on top close parentheses
straight c with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 6 straight i with overbrace on top plus 2 straight j with overbrace on top plus 3 straight k with overbrace on top close parentheses
Then comma
straight a with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 2 straight i with overbrace on top plus 3 straight j with overbrace on top plus 6 straight k with overbrace on top close parentheses cross times 1 over 7 open parentheses 3 straight i with overbrace on top plus 6 straight j with overbrace on top plus 2 straight k with overbrace on top close parentheses
space space space space space space space space space 1 over 49 open parentheses 6 minus 18 plus 12 close parentheses equals 0
Similary comma
straight b with rightwards harpoon with barb upwards on top. straight c with rightwards harpoon with barb upwards on top equals straight a with rightwards harpoon with barb upwards on top. straight c with rightwards harpoon with barb upwards on top equals straight c
therefore space space straight a with rightwards harpoon with barb upwards on top comma straight b with rightwards harpoon with barb upwards on top comma straight c with rightwards harpoon with barb upwards on top space are space mutually space perpendicular end style

Question 14

begin mathsize 12px style for space any space two space vectors space straight a with rightwards arrow on top and space straight b with rightwards arrow on top comma end style

Solution 14

Question 15

Solution 15

Question 16

begin mathsize 12px style table attributes columnalign left end attributes row cell If space straight p with rightwards arrow on top equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top space and space straight q with rightwards arrow on top  = straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top ,  then space find space the space value space of space straight lambda straight comma end cell row cell so space that space straight p with rightwards arrow on top plus straight q with rightwards arrow on top space and space straight p with rightwards arrow on top minus straight q with rightwards arrow on top space are space perpendicular space vectors. end cell end table end style

Solution 16

begin mathsize 12px style table attributes columnalign left end attributes row cell straight p with rightwards arrow on top equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top text   end text and text   end text straight q with rightwards arrow on top text   end text equals straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top end cell row cell straight p with rightwards arrow on top plus straight q with rightwards arrow on top text   end text end cell row cell equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top plus straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top end cell row cell equals 6 straight i with hat on top plus left parenthesis straight lambda plus 3 right parenthesis straight j with hat on top minus 8 straight k with hat on top end cell row blank row cell straight p with rightwards arrow on top minus straight q with rightwards arrow on top text   end text end cell row cell equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top minus straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top end cell row cell equals 4 straight i with hat on top plus left parenthesis straight lambda minus 3 right parenthesis straight j with hat on top plus 2 straight k with hat on top end cell row blank row cell left parenthesis straight p with rightwards arrow on top plus straight q with rightwards arrow on top right parenthesis. left parenthesis straight p with rightwards arrow on top minus straight q with rightwards arrow on top right parenthesis equals 0 end cell row cell rightwards double arrow left square bracket 6 straight i with hat on top plus left parenthesis straight lambda plus 3 right parenthesis straight j with hat on top minus 8 straight k with hat on top right square bracket. left square bracket 4 straight i with hat on top plus left parenthesis straight lambda minus 3 right parenthesis straight j with hat on top plus 2 straight k with hat on top right square bracket equals 0 end cell row cell rightwards double arrow 24 plus left parenthesis straight lambda squared minus 9 right parenthesis minus 16 equals 0 end cell row cell rightwards double arrow straight lambda squared minus 9 plus 8 equals 0 end cell row cell rightwards double arrow straight lambda squared minus 1 equals 0 end cell row cell therefore straight lambda equals plus-or-minus 1 end cell end table end style

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

If two vector begin mathsize 12px style a with rightwards arrow on top space a n d space b with rightwards arrow on top end style are such that begin mathsize 12px style open vertical bar a with rightwards arrow on top close vertical bar equals 2 comma open vertical bar b with rightwards arrow on top close vertical bar space equals space 1 space a n d space a with rightwards arrow on top b with rightwards arrow on top equals 1 end style, then find the value of (3a – 5b) . (2a + 7b).Solution 29

Question 30(i)

Solution 30(i)

Question 30(ii)

Solution 30(ii)

Question 31(i)

Solution 31(i)

Question 31(ii)

Solution 31(ii)

Question 31(iii)

Solution 31(iii)

Question 32(i)

Solution 32(i)

Question 32(ii)

Solution 32(ii)

Question 32(iii)

Solution 32(iii)

Question 33(i)

Solution 33(i)

Question 33(ii)

Solution 33(ii)

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

begin mathsize 12px style table attributes columnalign left end attributes row cell text If   end text straight a with rightwards arrow on top comma text   end text straight b with rightwards arrow on top text   are   two   vectors   such   that   end text vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line equals vertical line straight b with rightwards arrow on top vertical line comma text   then   prove   end text end cell row cell text that   end text straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top text   is   perpendicular   to   end text straight a with rightwards arrow on top. end cell end table end style

Solution 49

begin mathsize 12px style table attributes columnalign left end attributes row cell vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line squared equals vertical line straight b with rightwards arrow on top vertical line squared end cell row cell rightwards double arrow left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight a with rightwards arrow on top plus straight b with rightwards arrow on top. straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top equals 0 end cell row cell rightwards double arrow straight a with rightwards arrow on top. left parenthesis straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top right parenthesis equals 0 end cell row cell therefore straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top text    end text is text    end text perpendicular text    end text to text   end text straight a with rightwards arrow on top. end cell end table end style

Chapter 24 Scalar Or Dot Product Exercise Ex. 24.2

Question 1

Solution 1

Question 2

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.Solution 2

Question 3

(Pythagoras’s Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.Solution 3

Question 4

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.Solution 4

Question 5

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.Solution 5

Question 6

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.Solution 6

Question 7

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.Solution 7

Question 8

If AD is the median of D ABC, using vectors, prove that

AB2 + AC2 = 2 (AD2 + CD2).Solution 8

Question 9

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.Solution 9

Question 10

In a quadrilateral ABCD, prove that AB2 + BC2 + CD2+ DA2 = AC2 + BD2 + 4 PQ2, where P and Q are middle points of diagonals AC and BD. Solution 10

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RD SHARMA SOLUTION CHAPTER-23 Algebra of Vectors I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 23 Algebra of Vectors Exercise Ex. 23.1

Question 1(i)

Represent graphically a dispacement of 40 km, 30° east of north.Solution 1(i)

Question 1(ii)

Represent graphically a displacement of 50 km, south-eastSolution 1(ii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 50 km, south-east.Question 1(iii)

Represent graphically a displacement of 70 km, 40° north of west.Solution 1(iii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 70 km, 40° north of west.Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

(ii) 20 kg weight

(iii) 45°

(iv) 10 metres south-east

(v) 50 m/s2Solution 2

(i) 15 kg is a scalar quantity because it involves only

(ii) 20 kg weight is a vector quantity as it involves both magnitude and direction.

(iii) 45° is a scalar quantity as it involves only magnitude.

(iv) 10 metres south-east is a vector quantity as it involve direction.

(v) 50 m/s2 is a scalar quantity as it involves magnitude of acceleration.Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 23 Algebra of Vectors Exercise Ex. 23.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.3

Question 1

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors

Solution 1

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

T h e space v e r t i c e s space A comma space B comma space C space o f space t r i a n g l e space A B C space h a v e space r e s p e c t i v e l y space p o s i t i o n space v e c t o r s space
a with rightwards arrow on top comma space b with rightwards arrow on top comma space c with rightwards arrow on top space w i t h space r e s p e c t space t o space a space g i v e n space o r i g i n space O. space S h o w space t h a t space t h e space p o i n t space D space w h e r e space t h e space b i s e c t o r space o f space
angle A space m e e t s space B C space h a s space p o s i t i o n space v e c t o r
d with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction comma space w h e r e space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar equals gamma equals open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar
H e n c e space d e d u c e space t h a t space t h e space i n c e n t r e space I space h a s space p o s i t i o n space v e c t o r space fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction comma space w h e r e
alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar

Solution 7

L e t space A B C space b e space a space t r i a n g l e.
L e t space t h e space p o s i t i o n space v e c t o r s space o f space A comma space B space a n d space C space w i t h space r e s p e c t space t o space s o m e space o r i g i n comma space O space b e
a with rightwards arrow on top comma space b with rightwards arrow on top space a n d space c with rightwards arrow on top space r e s p e c t i v e l y.
L e t space D space b e space t h e space p o i n t space o n space B C space w h e r e space t h e space b i s e c t o r space o f space angle A space m e e t s.
L e t space d with rightwards arrow on top space p o s i t i o n space v e c t o r space o f space D space w h i c h space d i v i d e s space B C space i n t e r n a l l y space i n space t h e space r a t i o space beta space a n d space gamma comma
w h e r e space beta equals open vertical bar stack A C with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar stack A B with rightwards arrow on top close vertical bar
T h u s comma space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar
T h u s comma space b y space s e c t i o n space f o r m u l a comma space t h e space p o s i t i o n space v e c t o r space o f space D space i s space g i v e n space b y
stack O D with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction
L e t space alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar
I n c e n t r e space i s space t h e space c o n c u r r e n t space p o i n t space o f space a n g l e space b i s e c t o r s.
T h u s comma space I n c e n t r e space d i v i d e s space t h e space l i n e space A D space i n space t h e space r a t i o space alpha : beta plus gamma
T h u s comma space t h e space p o s i t i o n space v e c t o r space o f space i n c e n t r e space i s
e q u a l space t o space fraction numerator alpha a with rightwards arrow on top plus fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator open parentheses beta plus gamma close parentheses end fraction cross times open parentheses beta plus gamma close parentheses over denominator alpha plus beta plus gamma end fraction equals fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction

Chapter 23 – Algebra of Vectors Exercise Ex. 23.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.Solution 6

Chapter 23 – Algebra of Vectors Exercise Ex. 23.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find |AB| in each case.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

rightwards double arrow b with hat on top equals 1 half i with hat on top plus fraction numerator square root of 3 over denominator 2 end fraction j with hat on top

Question 12

Solution 12

Chapter 23 Algebra of Vectors Exercise Ex. 23.6

Question 1

Solution 1

Question 2

Solution 2

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 3

Solution 3

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top space a n d space b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top

Solution 18

G i v e n space t h a t space
a with rightwards arrow on top equals 2 stack i space with hat on top plus 3 j with hat on top minus k with hat on top
space a n d
b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top
T h u s comma space F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top plus b with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top plus space i with hat on top minus 2 j with hat on top plus k with hat on top
rightwards double arrow a with rightwards arrow on top plus b with rightwards arrow on top equals 3 i with hat on top plus j with hat on top
rightwards double arrow open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 9 plus 1 end root equals square root of 10
T h u s comma space t h e space u n i t space v e c t o r space a l o n g space t h e space r e s u l tan t space v e c t o r space a with rightwards arrow on top plus b with rightwards arrow on top space i s space
fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction
T h e space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t
v e c t o r equals fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction cross times 5 equals square root of 5 over 2 end root open parentheses 3 i with hat on top plus j with hat on top close parentheses

Question 19

Solution 19

Chapter 23 Algebra of Vectors Exercise Ex. 23.7

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

U sin g space v e c t o r space s h o w space t h a t space t h e space p o i n t s space A open parentheses minus 2 comma 3 comma 5 close parentheses comma space B open parentheses 7 comma 0 comma minus 1 close parentheses space a n d space C open parentheses minus 3 comma minus 2 comma minus 5 close parentheses space
a n d space D open parentheses 3 comma 4 comma 7 close parentheses space a r e space s u c h space t h a t space A B space a n d space C D space i n t e r s e c t space a t space t h e space p o i n t space P open parentheses 1 comma 2 comma 3 close parentheses

Solution 12

W e space h a v e
stack A P with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space P minus P o s i t i o n space v e c t o r space o f space A
rightwards double arrow stack A P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 2 i with hat on top plus 3 j with hat on top plus 5 k with hat on top close parentheses equals 3 i with hat on top minus j with hat on top minus 2 k with hat on top
stack P B with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space B minus P o s i t i o n space v e c t o r space o f space P
rightwards double arrow stack P B with rightwards arrow on top equals 7 i with hat on top minus k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 6 i with hat on top minus 2 j with hat on top minus 4 k with hat on top
C l e a r l y comma space stack P B with rightwards arrow on top equals 2 stack A P with rightwards arrow on top
s o space v e c t o r s space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space p o i n t space c o m m o n space t o space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top. space
H e n c e space P comma space A comma space B space a r e space c o l l i n e a r space p o i n t s.
S i m i l a r l y comma space stack C P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 3 i with hat on top minus 2 j with hat on top minus 5 k with hat on top close parentheses equals 4 i with hat on top plus 4 j with hat on top plus 8 k with hat on top
a n d space stack P D with rightwards arrow on top equals 3 i with hat on top plus 4 j with hat on top plus 7 k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 2 i with hat on top plus 2 j with hat on top plus 4 k with hat on top
S o space v e c t o r s space stack C P with rightwards arrow on top space a n d space stack P D with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space c o m m o n space p o i n t space t o space stack C P with rightwards arrow on top space a n d space stack C D with rightwards arrow on top.
H e n c e comma space C comma space P comma space D space a r e space c o l l i n e a r space p o i n t s.
T h u s comma space A comma space B comma space C comma space D space a n d space P space a r e space p o i n t s space s u c h space t h a t space A comma space P comma space B space a n d space C comma space P comma space D space
a r e space t w o space s e t s space o f space c o l l i n e a r space p o i n t s. space H e n c e space A B space a n d space C D space i n t e r s e c t space a t space t h e
p o i n t space P

Question 13

Using vectors, find the value of λ such that the points

 (λ, – 10, 3), (1 -1, 3) and (3, 5, 3) are collinear.Solution 13

Chapter 23 Algebra of Vectors Exercise Ex. 23.8

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

If a unit vector begin mathsize 12px style straight a with rightwards arrow on top end style makes and angles begin mathsize 12px style straight pi over 3 space with space straight i with hat on top comma space straight pi over 4 space with space straight j with hat on top end style and an acute angle θ with begin mathsize 12px style straight k with hat on top end style, then find θ and hence, the components of begin mathsize 12px style straight a with rightwards arrow on top end style.Solution 10

Question 11

Solution 11

Question 12

Solution 12

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RD SHARMA SOLUTION CHAPTER – 22 Differential Equations I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 22 Differential Equations Exercise Ex. 22.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

D e t e r m i n e space t h e space o r d e r space a n d space d e g r e e space o f space t h e space f o l l o w i n g space d i f f e r e n t i a l space e q u a t i o n. space S t a t e
a l s o space w h e t h e r space i t space i s space l i n e a r space o r space n o n minus l i n e a r.
square root of 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared end root equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 1 third end exponent

Solution 4

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n comma space square root of 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared end root equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 1 third end exponent
S q u a r i n g space o n space b o t h space t h e space s i d e s comma space w e space h a v e
1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 2 over 3 end exponent
C u b i n g space o n space b o t h space t h e space s i d e s comma space w e space h a v e
open square brackets 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets cubed equals open curly brackets open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 2 over 3 end exponent close curly brackets cubed
rightwards double arrow 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 6 plus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 4 equals c squared open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared
rightwards double arrow c squared open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 6 minus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 4 minus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared minus 1 equals 0
T h e space h i g h e s t space o r d e r space d i f f e r e n t i a l space c o e f f i c i e n t space i n space t h i s space
e q u a t i o n space i s space fraction numerator d squared y over denominator d x squared end fraction space a n d space i t s space p o w e r space i s space 2.
T h e r e f o r e comma space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n space i s space a space
n o n minus l i n e a r space d i f f e r e n t i a l space e q u a t i o n space o f space s e c o n d space o r d e r space a n d space s e c o n d space d e g r e e.

Question 5

Solution 5

Question 6

D e t e r m i n e space t h e space o r d e r space a n d space d e g r e e space o f space t h e space f o l l o w i n g space d i f f e r e n t i a l space e q u a t i o n. space S t a t e
a l s o space w h e t h e r space i s space l i n e a r space o r space n o n minus l i n e a r.
3 root of fraction numerator d squared y over denominator d x squared end fraction end root equals square root of fraction numerator d y over denominator d x end fraction end root

Solution 6

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n comma
3 root of fraction numerator d squared y over denominator d x squared end fraction end root equals square root of fraction numerator d y over denominator d x end fraction end root
C u b i n g space o n space b o t h space t h e space s i d e s space o f space t h e space a b o v e space e q u a t i o n comma space w e space h a v e
fraction numerator d squared y over denominator d x squared end fraction equals open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 3 over 2 end exponent
S q u a r i n g space o n space b o t h space t h e space s i d e s space o f space t h e space a b o v e space e q u a t i o n comma space w e space h a v e
open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared equals open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 3 over 2 end exponent close square brackets squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared equals open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses close square brackets cubed
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared minus open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses close square brackets cubed equals 0
T h e space h i g h e s t space o r d e r space d i f f e r e n t i a l space c o e f f i c i e n t space i n space t h i s space e q u a t i o n space i s space fraction numerator d squared y over denominator d x squared end fraction
a n d space i t s space p o w e r space i s space 2.
T h e r e f o r e comma space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n space i s space a space n o n minus l i n e a r space d i f f e r e n t i a l
e q u a t i o n space o f space s e c o n d space o r d e r space a n d space s e c o n d space d e g r e e.

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Determine the order and degree of the following differential equations. State also whether they are linear or non-linear.

Solution 27

The order of a differential equation is the order of the highest order derivative appearing in the equation.

The degree of a differential equation is the degree of the highest order derivative.

Consider the given differential equation

In the above equation, the order of the highest order derivative is 1.

So the differential equation is of order 1.

In the above differential equation, the power of the highest order derivative is 3.

Hence, it is a differential equation of degree 3.

Since the degree of the above differential equation is 3, more than one, it is a non-linear differential equation.

Chapter 22 – Differential Equations Exercise Ex. 22.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Form the differential equation having y = (sin-1x)2 + A cos -1 x + B, where A and B are arbitrary constants, as its general solution.Solution 14

Question 15(i)

F o r m space t h e space d i f f e r e n t i a l space e q u a t i o n space o f space t h e space f a m i l y space o f space
c u r v e s space r e p r e s e n t e d space b y space t h e space e q u a t i o n space left parenthesis apostrophe a apostrophe space b e i n g space t h e space p a r a m e t e r right parenthesis.
open parentheses 2 x plus a close parentheses squared plus y squared equals a squared

Solution 15(i)

C o n s i d e r space t h e space g i v e n space e q u a t i o n. comma
open parentheses 2 x plus a close parentheses squared plus y squared equals a squared.... left parenthesis 1 right parenthesis
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
2 open parentheses 2 x plus a close parentheses plus 2 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses 2 x plus a close parentheses plus y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow 2 x plus a equals minus y fraction numerator d y over denominator d x end fraction
rightwards double arrow a equals minus 2 x minus y fraction numerator d y over denominator d x end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space a space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
open parentheses 2 x minus 2 x minus y fraction numerator d y over denominator d x end fraction close parentheses squared plus y squared equals open parentheses minus 2 x minus y fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses y fraction numerator d y over denominator d x end fraction close parentheses squared plus y squared equals open parentheses 4 x squared plus y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 4 x y fraction numerator d y over denominator d x end fraction close parentheses

rightwards double arrow y squared equals 4 x squared plus 4 x y fraction numerator d y over denominator d x end fraction
rightwards double arrow y squared minus 4 x squared minus 4 x y fraction numerator d y over denominator d x end fraction equals 0

Question 15(ii)

Solution 15(ii)

Question 15(iii)

F o r m space t h e space d i f f e r e n t i a l space e q u a t i o n space o f space t h e space f a m i l y space o f space c u r v e s space r e p r e s e n t e d space b y space t h e
e q u a t i o n space left parenthesis apostrophe a apostrophe space b e i n g space t h e space p a r a m e t e r right parenthesis :
open parentheses x minus a close parentheses squared plus 2 y squared equals a squared

Solution 15(iii)

C o n s i d e r space t h e space g i v e n space e q u a t i o n comma
open parentheses x minus a close parentheses squared plus 2 y squared equals a squared.... left parenthesis 1 right parenthesis
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e
2 open parentheses x minus a close parentheses plus 4 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses x minus a close parentheses plus 2 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses x minus a close parentheses equals minus 2 y fraction numerator d y over denominator d x end fraction
rightwards double arrow a equals x plus 2 y fraction numerator d y over denominator d x end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space a space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
open parentheses x minus x plus 2 y fraction numerator d y over denominator d x end fraction close parentheses squared plus 2 y squared equals open parentheses x plus 2 y fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow 4 y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 2 y squared equals x squared plus 4 y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 4 x y fraction numerator d y over denominator d x end fraction
rightwards double arrow 2 y squared minus x squared equals 4 x y fraction numerator d y over denominator d x end fraction

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Represent the following family of curves by forming the corresponding differential equation (a,b being parameters):

x+ (y – b)2 = 1Solution 16(iv)

begin mathsize 12px style straight x squared plus left parenthesis straight y minus straight b right parenthesis squared equals 1 space space space space space space space space space space space space space space space space space space space space space space space space ____ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
2 straight x space plus space 2 left parenthesis straight y minus straight b right parenthesis dy over dx equals 0
straight x space plus space left parenthesis straight y minus straight b right parenthesis dy over dx equals 0
left parenthesis straight y minus straight b right parenthesis dy over dx equals negative straight x
left parenthesis straight y minus straight b right parenthesis equals fraction numerator begin display style fraction numerator negative straight x over denominator dy end fraction end style over denominator dx end fraction
Put space the space value space of space left parenthesis straight y minus straight b right parenthesis space is space equation space left parenthesis straight i right parenthesis
straight x squared open parentheses fraction numerator negative straight x over denominator begin display style dy over dx end style end fraction close parentheses squared equals 1
straight x squared open parentheses dy over dx close parentheses squared plus straight x squared equals open parentheses dy over dx close parentheses squared
straight x squared open curly brackets open parentheses dy over dx close parentheses squared plus 1 close curly brackets equals open parentheses dy over dx close parentheses squared
end style

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 16(vii)

Solution 16(vii)

Question 16(viii)

Solution 16(viii)

Question 16(ix)

Solution 16(ix)

Question 16(x)

Solution 16(x)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Chapter 22 – Differential Equations Exercise Ex. 22.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

show that y = ae2x + be-x is a solution of the differential equation begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y equals 0. end styleSolution 3

begin mathsize 12px style straight y equals ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals 2 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent space space space space space space space space space space space space space ___ left parenthesis ii right parenthesis
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals 4 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent space space space space space space space space space space space space space ___ left parenthesis iii right parenthesis
Now comma
fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y
equals left parenthesis 4 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent right parenthesis minus left parenthesis 2 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent right parenthesis minus 2 left parenthesis ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent right parenthesis
equals 4 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent minus 2 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent minus 2 ae to the power of 2 straight x end exponent minus 2 be to the power of negative straight x end exponent
equals 4 ae to the power of 2 straight x end exponent minus 4 ae to the power of 2 straight x end exponent plus 2 be to the power of negative straight x end exponent minus 2 be to the power of negative straight x end exponent
equals 0
S o comma
fraction numerator d squared y over denominator d x squared end fraction minus fraction numerator d y over denominator d x end fraction minus 2 y equals 0 end style

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Verify that y = begin mathsize 12px style straight a over straight x end style + b is a solution of the differential equation begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 over straight x open parentheses dy over dx close parentheses equals 0 end styleSolution 7

begin mathsize 12px style straight y equals straight a over straight x plus straight b space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals negative straight a over straight x squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis ii right parenthesis
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator 2 straight a over denominator straight x cubed end fraction
equals negative 2 over straight x open parentheses negative straight a over straight x squared close parentheses
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative 2 over straight x open parentheses dy over dx close parentheses
fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 over straight x open parentheses dy over dx close parentheses equals 0 end style

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Show that y = ex(A cos x + B sin x) is the solution of the differential equation

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight m squared straight y equals 0
straight y equals straight e to the power of mcos to the power of negative 1 end exponent straight x end exponent
dy over dx equals fraction numerator me to the power of mcos to the power of negative 1 end exponent straight x end exponent over denominator negative square root of 1 minus straight x squared end root end fraction
dy over dx equals fraction numerator negative my over denominator square root of 1 minus straight x squared end root end fraction........ left parenthesis straight i right parenthesis space
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator square root of left parenthesis 1 minus straight x squared right parenthesis end root. open parentheses negative straight m begin display style dy over dx end style close parentheses minus left parenthesis negative my right parenthesis begin display style fraction numerator left parenthesis negative 2 straight x right parenthesis over denominator 2 square root of left parenthesis 1 minus straight x squared right parenthesis end root end fraction end style over denominator left parenthesis 1 minus straight x squared right parenthesis end fraction space left square bracket From space left parenthesis straight i right parenthesis right square bracket
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator left parenthesis negative straight m right parenthesis open parentheses begin display style negative my end style close parentheses minus straight x begin display style dy over dx end style over denominator left parenthesis 1 minus straight x squared right parenthesis end fraction left square bracket From space left parenthesis straight i right parenthesis right square bracket
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight m squared straight y minus straight x dy over dx
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx minus straight m squared straight y equals 0
Hence space Proved

Question 18

Solution 18

Question 19

Solution 19

Question 20

Show that y = e-x + ax + b is solution of the differential equation begin mathsize 12px style straight e to the power of straight x fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 end styleSolution 20

begin mathsize 12px style straight y equals straight e to the power of negative straight x end exponent plus ax plus straight b
Differentiating space it space with space respect space to space straight x comma
dy over dx equals negative straight e to the power of negative straight x end exponent plus straight a
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight e to the power of negative straight x end exponent
1 over straight e to the power of negative straight x end exponent fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1
straight e to the power of straight x fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 end style

Question 21(i)

For the following differential equation verify that the accompanying function is a solution in the mentioned domain (a, b are parameters) begin mathsize 12px style straight x dy over dx equals straight y space space space space space space space space space space space space straight y space equals space ax comma space straight x element of straight R minus left curly bracket 0 right curly bracket end styleSolution 21(i)

begin mathsize 12px style straight y equals ax space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals straight a
equals ax over straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because straight x element of straight R minus left curly bracket 0 right curly bracket right square bracket
dy over dx equals straight y over straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Using space equation space left parenthesis straight i right parenthesis right square bracket
straight x dy over dx equals straight y
So comma space straight y space equals space ax space is space the space solution space of space the space given space equation. end style

Question 21(ii)

Solution 21(ii)

Question 21(iii)

Solution 21(iii)

Question 21(iv)

Solution 21(iv)

Question 21(v)

Solution 21(v)

Chapter 22 – Differential Equations Exercise Ex. 22.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

For the following initial value problem verify that the accompanying function is a solution:

begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y equals 0 comma space straight y open parentheses 0 close parentheses space equals space 0 comma space straight y apostrophe open parentheses 0 close parentheses equals 1 space space space space space space space space space space space straight y equals sinx end style

Solution 3

begin mathsize 12px style Here comma space straight y space equals space sin space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals cos space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis ii right parenthesis
Again space differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative sin space straight x space
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative straight y
fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y equals 0
So comma space straight y equals sin space straight x space is space straight a space solution space of space the space equation.
Put space space space space space straight x equals 0 space in space equation space left parenthesis straight i right parenthesis comma
rightwards double arrow space space space space space straight y space equals space sin space 0
rightwards double arrow space space space space space straight y equals 0
rightwards double arrow space space space space space straight y left parenthesis 0 right parenthesis space equals 0
Put space straight x space equals 0 space in space equation space left parenthesis ii right parenthesis
straight y apostrophe space equals cos space 0
straight y apostrophe space equals 1
rightwards double arrow straight y apostrophe left parenthesis 0 right parenthesis space equals space 1 end style

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 22 – Differential Equations Exercise Ex. 22.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

w h e r e space x not equal to open parentheses 2 n plus 1 close parentheses pi comma space n element of Z

Question 6

Solution 6

Question 7

Solution 7

w h e r e space x element of R

Question 8

Solution 8

fraction numerator d y over denominator d x end fraction equals log x
rightwards double arrow d y equals log x cross times d x
rightwards double arrow integral d y equals integral log x d x
rightwards double arrow y equals log x cross times integral 1 d x minus integral open parentheses 1 over x integral 1 d x close parentheses d x plus C space space space space space open square brackets U sin g space i n t e g r a t i o n space b y space p a r t s close square brackets
rightwards double arrow y equals x log x minus integral d x plus C
rightwards double arrow y equals x log x minus x plus C
rightwards double arrow y equals x open parentheses log x minus 1 close parentheses plus C comma space w h e r e space x element of open parentheses 0 comma infinity close parentheses

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solve the following differential equation:

(sin x + cos x)dy + (cos x – sin x) dx = 0Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following differential equation:

begin mathsize 12px style cos space x fraction numerator d y over denominator d x end fraction minus cos space 2 x space equals space cos space 3 x end style

Solution 15

begin mathsize 12px style cos space straight x space dy over dx minus cos space 2 space straight x space equals space cos space 3 space straight x
cos space straight x space dy over dx minus cos space 3 space straight x space equals space cos space 2 space straight x
dy over dx equals fraction numerator 4 cos cubed space straight x minus 3 cos space straight x space plus 2 space cos squared straight x space minus 1 over denominator cos space straight x end fraction
dy over dx equals fraction numerator 4 cos cubed straight x over denominator cos space straight x end fraction minus fraction numerator 3 cos space straight x over denominator cos space straight x end fraction plus fraction numerator 2 cos squared straight x over denominator cos space straight x end fraction minus fraction numerator 1 over denominator cos space straight x end fraction
dy over dx equals 4 cos squared straight x minus 3 plus 2 space cos space straight x minus space sec space straight x
dy over dx equals 4 open parentheses fraction numerator cos space 2 straight x plus 1 over denominator 2 end fraction close parentheses minus 32 space cos space straight x space minus space sec space straight x
dy equals left parenthesis 2 space cos space 2 space straight x space plus space 2 minus 3 plus 2 space cos space straight x space minus sec space straight x right parenthesis dx
integral dy equals integral left parenthesis 2 cos space 2 straight x space minus 1 plus 2 space cos space straight x minus space sec space straight x right parenthesis dx
straight y space equals space sin space 2 straight x space minus straight x plus 2 space sin space straight x space minus log open vertical bar sec space straight x space plus space tan space straight x close vertical bar plus straight c end style

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solve the following differential equation

begin mathsize 12px style open parentheses 1 plus straight x squared close parentheses dy over dx minus straight x equals 2 space tan to the power of negative 1 end exponent straight x end style

Solution 18

begin mathsize 12px style left parenthesis 1 plus straight x squared right parenthesis dy over dx minus straight x equals 2 tan to the power of negative 1 end exponent straight x
left parenthesis 1 plus straight x squared right parenthesis dy over dx equals 2 tan to the power of negative 1 end exponent straight x plus straight x
dy equals open parentheses fraction numerator 2 tan to the power of negative 1 end exponent straight x plus straight x over denominator 1 plus straight x squared end fraction close parentheses dx
integral dy equals integral open parentheses fraction numerator 2 tan to the power of negative 1 end exponent straight x plus straight x over denominator 1 plus straight x squared end fraction close parentheses dx
straight y equals integral left parenthesis 2 straight t plus tant right parenthesis dt space space space space space space space space space space space space space space space space left square bracket tan to the power of negative 1 end exponent straight x equals straight t right square bracket
equals 1 half log open vertical bar 1 plus straight x squared close vertical bar plus left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared plus straight c end style

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

solve the following differential equation

begin mathsize 12px style straight x open parentheses straight x squared minus 1 close parentheses dy over dx equals 1 comma space straight y left parenthesis 2 right parenthesis space equals space 0 end style

Solution 26

begin mathsize 12px style straight x left parenthesis straight x squared minus 1 right parenthesis dy over dx equals 1 comma straight y left parenthesis 2 right parenthesis equals 0
dy over dx equals fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction
dy equals fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction dx
integral dy equals integral open parentheses fraction numerator 1 over denominator straight x open parentheses straight x squared minus 1 close parentheses end fraction close parentheses dx
straight y equals 1 half integral fraction numerator 1 over denominator straight x minus 1 end fraction dx minus integral 1 over straight x dx plus 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx
equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c
Putting space straight x equals 2 comma space straight y equals 0 comma space we space have
straight y equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c space
0 equals 1 half log open vertical bar 2 minus 1 close vertical bar minus log open vertical bar 2 close vertical bar plus 1 half log open vertical bar 2 plus 1 close vertical bar plus straight c space
straight c equals log open vertical bar 2 close vertical bar minus 1 half log open vertical bar 3 close vertical bar
Putting space the space value space ofc comma space we space have
straight y equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c
equals log 4 over 3 open parentheses fraction numerator straight x squared minus 1 over denominator straight x squared end fraction close parentheses end style

Chapter 22 – Differential Equations Exercise Ex. 22.6

Question 1

Solve the following differential equation:

begin mathsize 12px style dv over dx plus fraction numerator 1 plus straight y squared over denominator straight y end fraction equals 0 end style

Solution 1

Question 2

Solve the following differential equation:

begin mathsize 12px style dy over dx equals fraction numerator 1 plus straight y squared over denominator straight y cubed end fraction end style

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 22 – Differential Equations Exercise Ex. 22.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

begin mathsize 12px style cos space straight x space cos space straight y space dy over dx equals negative sin space straight x space sin space straight y
fraction numerator cos space straight y over denominator sin space straight y end fraction dy space equals space minus fraction numerator sin space straight x over denominator cos space straight x end fraction dx
integral cot space ydy space equals negative integral tan space xdx
log space sin space straight y space equals space log space cos space straight x space plus space log space straight c
sin space straight y space equals space straight c space cos space straight x end style

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

begin mathsize 12px style left parenthesis straight y space plus space xy right parenthesis dx space plus open parentheses straight x minus xy squared close parentheses dy space equals space 0
straight y left parenthesis 1 plus straight x right parenthesis dx equals open parentheses xy squared minus straight x close parentheses dy
straight y left parenthesis 1 plus straight x right parenthesis dx equals straight x open parentheses straight y squared minus 1 close parentheses dy
fraction numerator open parentheses straight y squared minus 1 close parentheses dy over denominator straight y end fraction equals fraction numerator 1 plus straight x over denominator straight x end fraction dx
integral open parentheses straight y minus 1 over straight y close parentheses dy equals integral open parentheses 1 over straight x plus 1 close parentheses dx
straight y squared over 2 minus log open vertical bar straight y close vertical bar equals log open vertical bar straight x close vertical bar plus straight x plus straight c subscript 1
straight y squared over 2 minus straight x minus log open vertical bar straight y close vertical bar minus log open vertical bar straight x close vertical bar equals straight c subscript 1
log open vertical bar straight x close vertical bar plus straight x plus log open vertical bar straight y close vertical bar minus straight y squared over 2 equals straight c end style

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solve the following differential equation:

begin mathsize 12px style fraction numerator d y over denominator d x end fraction equals e to the power of x plus y end exponent space plus space e to the power of negative x plus y end exponent end style

Solution 35

begin mathsize 12px style dy over dx equals straight e to the power of straight x plus straight y end exponent plus straight e to the power of negative straight x plus straight y end exponent
equals straight e to the power of straight x space xe to the power of straight y plus straight e to the power of negative straight x end exponent space xe to the power of straight y
dy over dx equals straight e to the power of straight y open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses
dy over straight e to the power of straight y equals open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses dx
integral straight e to the power of negative straight y end exponent dy equals integral open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses dx
minus straight e to the power of negative straight y end exponent equals straight e to the power of straight x minus straight e to the power of negative straight x end exponent plus straight c
straight e to the power of negative straight x end exponent minus straight e to the power of negative straight y end exponent equals straight e to the power of straight x plus straight c end style

Question 36

Solve the following differential equation:

begin mathsize 12px style fraction numerator d y over denominator d x end fraction equals open parentheses cos squared x space minus space sin squared x close parentheses cos squared space y end style

Solution 36

begin mathsize 12px style dy over dx equals open parentheses cos squared straight x minus sin squared straight x close parentheses cos squared straight y
fraction numerator dy over denominator cos squared straight y end fraction equals open parentheses cos squared straight x minus sin squared straight x close parentheses dx
integral sec squared ydy equals integral cos 2 xdx
tan space straight y equals fraction numerator sin 2 straight x over denominator 2 end fraction plus straight c end style

Question 37(i)

Solution 37(i)

Question 37(ii)

Solve the following differential equation:

Solution 37(ii)

Question 38(i)

Solution 38(i)

Question 38(ii)

Solution 38(ii)

Question 38(iii)

yex/y dx = (xex/y + y2) dy, y ¹ 0Solution 38(iii)

Question 38(iv)

(1 + y2) tan-1 x dx + 2y (1 + x2)dy = 0Solution 38(iv)

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

begin mathsize 12px style dy over dx equals 2 straight e to the power of straight x straight y cubed comma straight y left parenthesis 0 right parenthesis equals 1 half
integral dy over straight y cubed equals integral 2 straight e to the power of straight x dx
minus fraction numerator 1 over denominator 2 straight y squared end fraction equals 2 straight e to the power of straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight x space equals space 0 comma space straight y space equals space 1 half
minus 4 over 2 equals 2 straight e to the power of 0 plus straight c
minus 2 equals 2 plus straight c
straight c equals negative 4
Put space straight c equals negative 4 space in space equation space left parenthesis straight i right parenthesis
minus fraction numerator 1 over denominator 2 straight y squared end fraction equals 2 straight e to the power of straight x minus 4
minus 1 equals 4 straight e to the power of straight x straight y squared minus 8 straight y squared
minus 1 equals negative straight y squared left parenthesis 8 minus 4 straight e to the power of straight x right parenthesis
straight y squared left parenthesis 8 minus 4 straight e to the power of straight x right parenthesis equals 1 end style

Question 43

Solution 43

begin mathsize 12px style dr over dt equals negative rt comma space straight r left parenthesis 0 right parenthesis space equals space straight r subscript 0
integral dr over straight r equals negative integral tdt
log open vertical bar straight r close vertical bar equals negative straight t squared over 2 plus straight c space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative negative left parenthesis straight i right parenthesis
Put space straight t space equals space 0 comma space straight r space equals space straight r subscript 0 space inequation space left parenthesis straight i right parenthesis comma
log open vertical bar straight r subscript 0 close vertical bar equals 0 plus straight c
log open vertical bar straight r subscript 0 close vertical bar equals straight c
Now comma
log open vertical bar straight r close vertical bar equals negative straight t squared over 2 plus log open vertical bar straight r subscript 0 close vertical bar
straight r over straight r subscript 0 equals straight e to the power of negative straight t squared over 2 end exponent
straight r equals straight r subscript 0 straight e to the power of negative straight t squared over 2 end exponent end style

Question 44

Solution 44

Question 45(i)

Solution 45(i)

Question 45(ii)

Solution 45(ii)

Question 45(iii)

Solution 45(iii)

begin mathsize 12px style dy over dx equals 2 straight e to the power of 2 straight x end exponent straight y squared comma space straight y space left parenthesis 0 right parenthesis equals negative 1
integral dy over straight y squared equals integral 2 straight e to the power of 2 straight x end exponent dx
minus 1 over straight y equals fraction numerator 2 straight e to the power of 2 straight x end exponent over denominator 2 end fraction plus straight c
minus 1 over straight y equals straight e to the power of 2 straight x end exponent plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals space minus 1 comma space straight x space equals space 0
1 equals straight e to the power of 0 plus straight c
1 equals 1 plus straight c
straight c equals 0
Put space straight c equals 0 space in space equation space left parenthesis straight i right parenthesis comma
minus 1 over straight y equals straight e to the power of 2 straight x end exponent
straight y equals negative straight e to the power of negative 2 straight x end exponent end style

Question 45(iv)

Solution 45(iv)

begin mathsize 12px style cos space straight y dy over dx equals straight e to the power of straight x comma space straight y space left parenthesis 0 right parenthesis equals straight pi over 2
integral cos space ydy equals integral straight e to the power of straight x dx
sin space straight y equals straight e to the power of straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight x space equals 0 comma space straight y equals straight pi over 2
sin open parentheses straight pi over 2 close parentheses equals straight e to the power of 0 plus straight c
1 equals 1 plus straight c
straight c equals 0
Put space straight c equals 0 space in space equation space left parenthesis straight i right parenthesis comma
sin space straight y space equals space straight e to the power of straight x
straight y equals sin to the power of negative 1 end exponent open parentheses straight e to the power of straight x close parentheses end style

Question 45(v)

Solution 45(v)

Question 45(vi)

Solve the following initial value problem

begin mathsize 12px style dy over dx end style=1 + x2 + y2 + x2y2, y(0) = 1Solution 45(vi)

begin mathsize 12px style dy over dx equals 1 plus straight x squared plus straight y squared plus straight x squared straight y squared comma space straight y left parenthesis 0 right parenthesis equals 1
equals left parenthesis 1 plus straight x squared right parenthesis left parenthesis 1 plus straight y squared right parenthesis
integral fraction numerator dy over denominator 1 plus straight y squared end fraction equals integral left parenthesis 1 plus straight x squared right parenthesis dx
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space space straight x equals 0 comma space straight y equals 1
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight c
straight c equals straight pi over 4
Put space straight c equals straight pi over 4 space in space equation space left parenthesis straight i right parenthesis
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight pi over 4 end style

Question 45(vii)

Solve the following initial value problem

begin mathsize 12px style xy dy over dx equals open parentheses straight x plus 2 close parentheses open parentheses straight y plus 2 close parentheses comma space straight y open parentheses 1 close parentheses space equals space minus 1 end style

Solution 45(vii)

begin mathsize 12px style xy dy over dx equals open parentheses straight x plus 2 close parentheses open parentheses straight y plus 2 close parentheses comma space straight y open parentheses 1 close parentheses equals negative 1
fraction numerator ydy over denominator open parentheses straight y plus 2 close parentheses end fraction equals fraction numerator open parentheses straight x plus 2 close parentheses over denominator straight x end fraction dx
integral open parentheses 1 minus fraction numerator 2 over denominator straight y plus 2 end fraction close parentheses dy equals integral open parentheses 1 plus 2 over straight x close parentheses dx
straight y minus straight x minus 2 log left parenthesis straight y plus 2 right parenthesis minus 2 logx equals straight c
Put space straight x equals 1 comma space straight y equals negative 1
minus 1 minus 1 minus 2 log left parenthesis negative 1 plus 2 right parenthesis minus 2 log 1 equals straight c
rightwards double arrow negative 2 equals straight c
Thus comma space we space have
straight y minus straight x minus 2 log left parenthesis straight y plus 2 right parenthesis minus 2 logx equals negative 2
end style

Question 45(viii)

Solution 45(viii)

Question 45(ix)

Solution 45(ix)

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Find the particular solution of ebegin mathsize 12px style dy over dx end style= x + 1, given that y = 3 when x = 0.Solution 49

begin mathsize 12px style dy over straight e to the power of dx equals straight x plus 1
dy over dx equals log left parenthesis straight x plus 1 right parenthesis comma space straight y equals 3 space at space straight x equals 0
integral dy equals integral log left parenthesis straight x plus 1 right parenthesis dx
straight y equals log open vertical bar straight x plus 1 close vertical bar straight x integral 1 cross times dx minus integral open parentheses fraction numerator 1 over denominator straight x plus 1 end fraction cross times integral 1 dx close parentheses dx plus straight c
Using space in space tegration space by space parts
straight y equals straight x space log open vertical bar straight x space plus 1 close vertical bar minus integral fraction numerator straight x over denominator straight x plus 1 end fraction dx plus straight c
straight y equals straight x space log open vertical bar straight x plus 1 close vertical bar minus open parentheses integral open parentheses 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx close parentheses plus straight c
equals space straight x space log space open vertical bar straight x plus 1 close vertical bar minus open parentheses straight x minus log open vertical bar straight x plus 1 close vertical bar close parentheses plus straight c
straight y equals xlog open vertical bar straight x plus 1 close vertical bar minus straight x plus log open vertical bar straight x plus 1 close vertical bar plus straight c
straight y equals open parentheses straight x plus 1 close parentheses log open vertical bar straight x plus 1 close vertical bar minus straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals 3 space and space straight x space equals 0
3 equals 0 minus 0 plus straight c
straight c equals 3
Put space straight c space equals space 3 space in space equation space left parenthesis straight i right parenthesis comma
straight y equals left parenthesis straight x plus 1 right parenthesis log open vertical bar straight x plus 1 close vertical bar minus straight x plus 3 end style

Question 50

Solution 50

begin mathsize 12px style cos space ydy space plus space cos space straight x space sin space ydx equals 0
cos space ydy equals negative cos space straight x space sin space ydx
fraction numerator cos space straight y over denominator sin space straight y end fraction dy equals negative cos space xdx
integral cot space ydy equals negative integral cos space xdx
log open vertical bar sin space straight y close vertical bar equals negative sin space straight x space plus straight c space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
put space straight y space equals space straight pi over 2 and space straight x space equals straight pi over 2
log space open vertical bar sin straight pi over 2 close vertical bar equals negative sin straight pi over 2 plus straight c
0 equals negative 1 plus straight c
straight c equals 1
Put space straight c space equals 1 space in space equation space left parenthesis 1 right parenthesis comma
log open vertical bar sin space straight y close vertical bar equals 1 minus sin space straight x
log open vertical bar sin space straight y close vertical bar plus sin space straight x equals 1 end style

Question 51

Solution 51

begin mathsize 12px style dy over dx equals negative 4 xy squared comma space straight y equals 1 space when space straight x space equals 0
integral dy over straight y squared equals negative 4 integral xdx
minus 1 over straight y equals negative 4 straight x squared over 2 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals 1 space and space straight x space equals 0
minus 1 equals 0 plus straight c
straight c equals negative 1
Put space straight c equals negative 1 space in space equation space left parenthesis ii right parenthesis comma
minus 1 over straight y equals negative 2 straight x squared minus 1
1 over straight y equals 2 straight x squared plus 1
straight y equals fraction numerator 1 over denominator 2 straight x squared plus 1 end fraction end style

Question 52

Find the equation of a curve passing through the point (0,0) and whose differential equation is begin mathsize 12px style dy over dx equals straight e to the power of straight x space sin space straight x. end styleSolution 52

Question 53

Solution 53

Question 54

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after after t seconds.Solution 54

begin mathsize 12px style Let space the space rate space of space change space of space volume space of space the space balloon space be space straight k left parenthesis where space straight k space is space straight a space constant right parenthesis
rightwards double arrow dv over dt equals straight k
rightwards double arrow straight d over dt open parentheses 4 over 3 πr cubed close parentheses equals straight k space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets Volume space of space sphere equals 4 over 3 πr cubed close square brackets
rightwards double arrow 4 over 3 straight pi times 3 straight r squared times dr over dt equals straight k
rightwards double arrow 4 πr squared dr equals straight k space dt
Intrgrating space both space sides comma space we space get colon
4 straight pi integral straight r squared dr equals straight k integral dt
rightwards double arrow 4 straight pi times straight r squared over 3 equals kt plus straight c
rightwards double arrow 4 πr cubed equals 3 open parentheses kt plus straight c close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space........ left parenthesis straight i right parenthesis
Now comma space at space straight t space equals space 0 comma space straight r space equals space 3 colon
4 straight pi cross times 3 cubed equals 3 left parenthesis straight k cross times 0 plus straight c right parenthesis
108 straight pi equals 3 straight c
straight c equals 36 straight pi end style
begin mathsize 12px style At space straight t space equals 3 comma space straight r equals 6
4 straight pi cross times 6 cubed equals 3 left parenthesis straight k cross times 3 plus straight c right parenthesis
864 straight pi equals 3 left parenthesis 3 straight k plus 36 straight pi right parenthesis
3 straight k equals negative 288 straight pi minus 36 straight pi equals 252 straight pi
straight k equals 84 straight pi
Substituting space the space values space of space straight k space and space straight C space in space equation space left parenthesis 1 right parenthesis comma space we space get colon
4 πr cubed equals 3 open square brackets 84 πt plus 36 straight pi close square brackets
rightwards double arrow 4 πr cubed equals 4 straight pi left parenthesis 63 straight t plus 27 right parenthesis
rightwards double arrow straight r cubed equals 63 straight t plus 27
rightwards double arrow straight r equals left parenthesis 63 straight t plus 27 right parenthesis to the power of 1 third end exponent space Thus comma space the space radius space of space the space balloon space after space straight t space seconds space is space left parenthesis 63 straight t plus 27 right parenthesis to the power of 1 third end exponent. end style

Question 55

in a bank,principal increases continuously at the rate of r% per  year. Find The value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).Solution 55

Let p, t and represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

begin mathsize 12px style rightwards double arrow dp over dt equals open parentheses straight r over 100 close parentheses straight p
rightwards double arrow dp over straight p equals open parentheses straight r over 100 close parentheses dt end style

Integrating both side, we get:

begin mathsize 12px style integral dp over straight p equals straight r over 100 integral dt
rightwards double arrow log space straight p space equals space rt over 100 plus straight k
rightwards double arrow straight p equals straight e to the power of rt over 100 plus straight k end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space...... left parenthesis straight i right parenthesis
It space is space given space that space when space straight t space equals 0 comma space straight p space 100.
rightwards double arrow 100 space equals straight e to the power of straight k space space space space space space space space space space space space space space space space space space space space space..... left parenthesis 2 right parenthesis
Now comma space it space calligraphic l equals 10 comma space then space straight p space equals 2 space cross times 100 space equals 200.
200 equals straight e to the power of calligraphic l over 10 plus straight k end exponent
rightwards double arrow 200 space equals straight e to the power of calligraphic l over 10 end exponent. straight e to the power of straight k
rightwards double arrow 200 space equals straight e to the power of calligraphic l over 10 end exponent.100 space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis From left parenthesis 2 right parenthesis right parenthesis
rightwards double arrow straight e to the power of calligraphic l over 10 end exponent equals 2
rightwards double arrow straight r over 10 log subscript straight e 2
rightwards double arrow straight r over 10 equals 0.6931
rightwards double arrow straight r space equals 6.931
Hence comma space the space value space of space straight r space is space 6.93 percent sign end style

Question 56

Solution 56

Question 57

Solution 57

..Question 58

Solution 58

Chapter 22 – Differential Equations Exercise Ex. 22.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

2 v minus v fraction numerator d v over denominator d x end fraction equals fraction numerator d v over denominator d x end fraction
rightwards double arrow 2 v equals v fraction numerator d v over denominator d x end fraction plus fraction numerator d v over denominator d x end fraction
rightwards double arrow 2 v equals open parentheses v plus 1 close parentheses fraction numerator d v over denominator d x end fraction
rightwards double arrow fraction numerator open parentheses v plus 1 close parentheses over denominator v end fraction d v equals 2 d x

Question 10

Solution 10

Question 11

Solve the following differential equation.

Solution 11

Chapter 22 – Differential Equations Exercise Ex. 22.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solve the following differential equation:

begin mathsize 12px style straight x dy over dx equals straight x plus straight y end style

Solution 4

Question 5

Solve the following differential equation:

begin mathsize 12px style open parentheses straight x squared minus straight y squared close parentheses dx minus 2 xydy space equals 0 end style

Solution 5

Question 6

Solve the following initial value problem

begin mathsize 12px style dy over dx equals fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction end style

Solution 6

begin mathsize 12px style dy over dx equals fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction
Here space it space is space straight a space homogeneous space equation
put space straight y equals vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction
straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction minus straight v
straight x dv over dx equals fraction numerator 1 plus straight v squared over denominator 1 minus straight v end fraction
fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv equals dx over straight x
integral fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv equals integral dx over straight x
integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv minus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv equals integral dx over straight x
tan to the power of negative 1 end exponent straight v minus 1 half log left parenthesis 1 plus straight v squared right parenthesis equals log space straight x space plus straight c
tan to the power of negative 1 end exponent straight y over straight x equals 1 half log open parentheses straight x squared plus straight y squared close parentheses plus straight c end style

Question 7

Solution 7

Question 8

S o l v e space t h e space d i f f e r e n t i a l space e q u a t i o n space x squared fraction numerator d y over denominator d x end fraction equals x squared minus 2 y squared plus x y

Solution 8

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n
space x squared fraction numerator d y over denominator d x end fraction equals x squared minus 2 y squared plus x y
rightwards double arrow fraction numerator d y over denominator d x end fraction equals fraction numerator x squared minus 2 y squared plus x y over denominator x squared end fraction
T h i s space i s space a space h o m o g e n e o u s space d i f f e r e n t i a l space e q u a t i o n.
S u b s t i t u t i n g space y equals v x space a n d space fraction numerator d y over denominator d x end fraction equals v plus x fraction numerator d v over denominator d x end fraction comma space w e space h a v e
v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator x squared minus 2 v squared cross times x squared plus x cross times v cross times x over denominator x squared end fraction
rightwards double arrow v plus x fraction numerator d v over denominator d x end fraction equals 1 minus 2 v squared plus v
rightwards double arrow x fraction numerator d v over denominator d x end fraction equals 1 minus 2 v squared
rightwards double arrow fraction numerator d v over denominator 1 minus 2 v squared end fraction equals fraction numerator d x over denominator x end fraction
rightwards double arrow fraction numerator d v over denominator v squared minus begin display style 1 half end style end fraction equals minus 2 fraction numerator d x over denominator x end fraction
rightwards double arrow integral fraction numerator d v over denominator open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared minus v squared end fraction equals 2 integral fraction numerator d x over denominator x end fraction
rightwards double arrow integral fraction numerator d v over denominator open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared minus v squared end fraction equals 2 integral fraction numerator d x over denominator x end fraction
rightwards double arrow fraction numerator square root of 2 over denominator 2 end fraction log open parentheses fraction numerator fraction numerator 1 over denominator square root of 2 end fraction plus v over denominator fraction numerator 1 over denominator square root of 2 end fraction minus v end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator fraction numerator 1 over denominator square root of 2 end fraction plus begin display style y over x end style over denominator fraction numerator 1 over denominator square root of 2 end fraction minus y over x end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals log x squared plus log C
rightwards double arrow log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses to the power of fraction numerator 1 over denominator square root of 2 end fraction end exponent equals log C x squared
rightwards double arrow open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses to the power of fraction numerator 1 over denominator square root of 2 end fraction end exponent equals C x squared
rightwards double arrow open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals open parentheses C x squared close parentheses to the power of square root of 2 end exponent

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solve the following initial value problem

begin mathsize 12px style dy over dx equals straight y over straight x plus sin open parentheses straight y over straight x close parentheses end style

Solution 19

begin mathsize 12px style dy over dx equals straight y over straight x plus sin open parentheses straight y over straight x close parentheses
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space straight y space equals space vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals straight v plus sin space straight v
straight x dv over dx equals sin space straight v
cosecvdv space equals dx over straight x
integral cosecvdv equals integral dx over straight x
log space tan space straight v over 2 equals log space straight x plus log space straight c
tan straight v over 2 equals Cx
tan fraction numerator straight y over denominator 2 straight x end fraction equals Cx end style

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following initial value poblem

begin mathsize 12px style open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses dx plus straight e to the power of straight x over straight y end exponent open parentheses 1 minus straight x over straight y close parentheses dy equals 0 end style

Solution 25

begin mathsize 12px style open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses dx plus straight e to the power of straight x over straight y end exponent open parentheses 1 minus straight x over straight y close parentheses dy equals 0
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space space space space space space straight x equals vy
And
dx over dy equals straight v plus straight y dv over dy
So comma
dx over dy equals negative fraction numerator straight e to the power of begin display style straight x over straight y end style end exponent open parentheses 1 minus begin display style straight x over straight y end style close parentheses over denominator open parentheses 1 plus straight e to the power of begin display style straight x over straight y end style end exponent close parentheses end fraction
straight v plus straight y dv over dy equals negative fraction numerator straight e to the power of begin display style vy over straight y end style end exponent open parentheses 1 minus begin display style vy over straight y end style close parentheses over denominator blank end fraction
equals negative straight e to the power of straight v fraction numerator left parenthesis 1 minus straight v right parenthesis over denominator left parenthesis 1 plus straight e to the power of straight v right parenthesis end fraction
straight y dv over dy equals negative fraction numerator straight e to the power of straight v left parenthesis 1 minus straight v right parenthesis over denominator open parentheses 1 plus straight e to the power of straight v close parentheses end fraction
equals fraction numerator negative straight e to the power of straight v left parenthesis 1 minus straight v right parenthesis minus straight v left parenthesis 1 plus straight e to the power of straight v right parenthesis over denominator left parenthesis 1 plus straight e to the power of straight v right parenthesis end fraction
fraction numerator left parenthesis 1 plus straight e to the power of straight v right parenthesis over denominator negative straight e to the power of straight v open parentheses 1 minus straight v close parentheses minus straight v open parentheses 1 plus straight e to the power of straight v close parentheses end fraction dv equals dy over straight y
straight x plus ye to the power of straight x divided by straight y end exponent equals straight c end style

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solve the following differential equation:

begin mathsize 12px style ydx plus open curly brackets xlog space open parentheses straight y over straight x close parentheses dy minus 2 xdy equals 0 close curly brackets end style

Solution 35

Question 36(i)

Solution 36(i)

Question 36(ii)

Solution 36(ii)

Question 36(iii)

Solve the following initial value problem

begin mathsize 12px style dy over dx minus straight y over straight x plus cosec straight y over straight x equals 0 comma space straight y left parenthesis 1 right parenthesis equals 0 end style

Solution 36(iii)

begin mathsize 12px style dy over dx minus straight y over straight x plus cosec straight y over straight x equals 0 comma space straight y left parenthesis 1 right parenthesis equals 0
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space straight y space equals vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals vx over straight x minus cosec vx over straight x
straight x dv over dx equals straight v minus cosecv minus straight v
equals negative cosecv
dv over cosecv equals negative dx over straight x
sin space vdv equals negative dx over straight x
minus cos space straight v equals negative log open vertical bar straight x close vertical bar plus straight c
minus cos straight y over straight x equals negative log open vertical bar straight x close vertical bar plus straight c
Now space putting space straight y equals 0 comma space straight x equals 1 comma space we space have
straight c equals negative 1
Now
minus cos straight y over straight x plus 1 equals negative log open vertical bar straight x close vertical bar
log open vertical bar straight x close vertical bar equals cos straight y over straight x minus 1 end style

Question 36(iv)

Solution 36(iv)

Question 36(v)

Solution 36(v)

Question 36(vi)

Solution 36(vi)

Question 36(vii)

Solution 36(vii)

Question 36(viii)

Solution 36(viii)

Question 36(ix)

Solve the following initial value problem

Error converting from MathML to accessible text.

Solution 36(ix)

Error converting from MathML to accessible text.

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Chapter 22 – Differential Equations Exercise Ex. 22.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solve the following differential equation:

begin mathsize 12px style open parentheses sin space straight x close parentheses dy over dx plus ycos space straight x equals 2 space sin squared space straight x space cos space straight x end style

Solution 30

begin mathsize 12px style Here comma space open parentheses sin space straight x close parentheses dy over dx plus space straight y space cos space straight x space equals 2 space sin squared straight x space cos space straight x
dy over dx plus straight y space cot space straight x space equals space 2 space sin space straight x space cos space straight x
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight p equals cot space straight x comma space straight Q space equals 2 sin space straight x space cos space straight x
straight I. straight F. space space equals straight e to the power of integral pdx end exponent
space space space space space space space space space equals straight e to the power of integral cotxdx end exponent
space space space space space space space space space equals straight e to the power of log space sin space straight x end exponent
space space space space space space space space space equals space sin space straight x
solution space of space the space equation space is space given space by comma
straight y space cross times open parentheses straight I. straight F close parentheses equals integral straight Q cross times left parenthesis straight I. straight F right parenthesis dx plus straight c
straight y left parenthesis sin space straight x right parenthesis equals integral 2 sin space straight x space cos space straight x left parenthesis sin space straight x right parenthesis space dx space plus straight c
ysin space straight x equals space left parenthesis 2 divided by 3 right parenthesis sin to the power of logical and 3 straight x plus straight C
end style

Question 31

Solve the following differential equation:

begin mathsize 12px style open parentheses straight x squared minus 1 close parentheses dy over dx plus 2 left parenthesis straight x plus 2 right parenthesis straight y equals 2 left parenthesis straight x plus 1 right parenthesis end style

Solution 31

begin mathsize 12px style Here comma space open parentheses straight x squared minus 1 close parentheses dy over dx plus 2 open parentheses straight x plus 2 close parentheses straight y equals 2 left parenthesis straight x plus 1 right parenthesis
dy over dx plus fraction numerator 2 left parenthesis straight x plus 2 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction straight y equals fraction numerator 2 left parenthesis straight x plus 1 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight P equals fraction numerator 2 left parenthesis straight x plus 2 right parenthesis over denominator straight x squared minus 1 end fraction comma straight Q fraction numerator 2 left parenthesis straight x plus 1 right parenthesis over denominator left parenthesis straight x squared minus 1 right parenthesis end fraction
straight I. straight F. space equals straight e to the power of integral pdx end exponent
space space space space space space space space space space equals straight e to the power of 2 integral fraction numerator left parenthesis straight x plus 2 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction dx end exponent
space space space space space space space space space space equals straight e to the power of 2 integral fraction numerator 2 straight x over denominator straight x squared minus 1 end fraction dx plus 4 integral fraction numerator 1 over denominator straight x squared minus 1 end fraction dx end exponent
space space space space space space space space space space equals straight e to the power of log open vertical bar straight x squared minus 1 close vertical bar plus 4 straight x 1 half log open vertical bar fraction numerator straight x minus 1 over denominator straight x plus 1 end fraction close vertical bar end exponent
space space space space space space space space space space equals straight e to the power of log open vertical bar straight x squared minus 1 close vertical bar plus log open vertical bar fraction numerator straight x minus 1 over denominator straight x plus 1 end fraction close vertical bar squared end exponent
space space space space space space space space space space equals straight e to the power of log fraction numerator open parentheses straight x squared minus 1 close parentheses left parenthesis straight x minus 1 right parenthesis squared over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction end exponent
straight I. straight F space space space space space space equals fraction numerator open parentheses straight x plus 1 close parentheses left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 1 right parenthesis squared over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction
space space space space space space space space space space space space equals fraction numerator left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction
Solution space of space the space equation space is space given space by comma
4 cross times left parenthesis straight I. straight F right parenthesis equals integral straight Q cross times left parenthesis straight I. straight F right parenthesis dx plus straight c
fraction numerator straight y left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction equals 2 straight x squared over 2 minus 6 straight x minus 8 space log open vertical bar straight x plus 1 close vertical bar plus straight c
fraction numerator straight y left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction equals straight x squared minus 6 straight x plus 8 log open vertical bar straight x plus 1 close vertical bar plus straight c
straight y equals fraction numerator straight x plus 1 over denominator left parenthesis straight x minus 1 right parenthesis cubed end fraction open square brackets straight x squared minus 6 straight x minus 8 space log open vertical bar straight x plus 1 close vertical bar plus straight c close square brackets end style

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36(i)

Solution 36(i)

begin mathsize 12px style Here comma space dy over dx plus 3 straight y equals straight e to the power of mx
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight P equals 3 comma space straight Q equals straight e to the power of mx
straight I. straight F. equals straight e to the power of integral pdx end exponent
equals straight e to the power of integral 3 dx end exponent
equals straight e to the power of 3 straight x end exponent
Solution space of space the space equation space is space given space by comma
straight y cross times open parentheses straight I. straight F close parentheses equals integral space straight Q space cross times space open parentheses straight I. straight F. close parentheses dx space plus space straight c
straight y open parentheses straight e to the power of 3 straight x end exponent close parentheses space equals space integral straight e to the power of mx space straight e to the power of 3 straight x end exponent space dx space plus space straight c
equals space integral space straight e to the power of open parentheses straight m plus 3 close parentheses straight x end exponent space dx space plus space straight c
straight y open parentheses straight e to the power of 3 straight x end exponent close parentheses equals fraction numerator straight e to the power of open parentheses straight m plus 3 close parentheses straight x end exponent space dx over denominator open parentheses straight m plus 3 close parentheses end fraction plus straight c end style

Question 36(ii)

Solution 36(ii)

Question 36(iii)

Solution 36(iii)

Question 36(iv)

Solution 36(iv)

Question 36(v)

Solution 36(v)

Question 36(vi)

Solution 36(vi)

Question 36(vii)

Solution 36(vii)

Question 36(viii)

Solution 36(viii)

Question 36(ix)

Solution 36(ix)

Question 36(x)

Solution 36(x)

begin mathsize 12px style Here comma space e to the power of negative y space end exponent s e c squared space y d y equals d x space plus space x d y
e to the power of negative y space end exponent s e c squared space y equals fraction numerator d x over denominator d y end fraction plus x
fraction numerator d x over denominator d y end fraction plus x equals e to the power of negative y end exponent space s e c squared space y
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d x over denominator d y end fraction plus p x equals Q
P space equals space 1 comma space Q equals e to the power of negative y end exponent space s e c squared y
I. F. space equals space e to the power of integral p d y end exponent
space space space space space space space space equals space e to the power of integral d y end exponent
space space space space space space space space equals space e to the power of y
Solution space of space the space equation space is space given space by comma
x space cross times space open parentheses I. F close parentheses equals integral space Q space cross times space open parentheses I. F. close parentheses d y equals c
x open parentheses e to the power of y close parentheses equals integral e to the power of negative y end exponent space s e c squared y open parentheses e to the power of y close parentheses d y plus c
x e to the power of y equals t a n space y space plus c
x equals open parentheses t a n space y space plus space c close parentheses e to the power of negative y end exponent end style

Question 36(xi)

Solution 36(xi)

Question 36(xii)

Solution 36(xii)

Question 37(i)

Solution 37(i)

Question 37(ii)

Solution 37(ii)

Question 37(iii)

Solution 37(iii)

Question 37(iv)

Solution 37(iv)

Question 37(v)

Solve the following initial value problem:

begin mathsize 12px style open parentheses 1 plus straight y squared close parentheses dx space plus space open parentheses straight x minus straight e minus tan to the power of negative 1 end exponent straight t close parentheses dy equals 0 comma space straight y open parentheses 0 close parentheses equals 0 end style

Solution 37(v)

Question 37(vi)

Solution 37(vi)

Question 37(vii)

Solution 37(vii)

Question 37(viii)

Solve the following initial value problem

begin mathsize 12px style dy over dx plus ycotx equals 4 straight x space cosecx comma space straight y open parentheses straight pi over 2 close parentheses equals 0 end style

Solution 37(viii)

begin mathsize 12px style fraction numerator d y over denominator d x end fraction plus y space c o t space x space equals space 4 x space cos e c space x comma space y open parentheses straight pi over 2 close parentheses equals 0
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d y over denominator d x end fraction plus p y equals Q
p equals c o t x comma space Q equals 4 x space cos e c x
I. F.
equals e to the power of integral p d x end exponent space
equals e to the power of integral c o t space x d x end exponent
equals e to the power of log space sin space x end exponent
equals sin space x
Solution space of space the space equation space is space given space by comma
y cross times open parentheses I. F close parentheses equals integral Q cross times open parentheses I. F close parentheses d x plus c
y open parentheses sin space x close parentheses space equals space integral 4 x space cos e c space x space cross times open parentheses sin space x close parentheses d x space plus space c
space space space space space space space space space space space space space space equals space integral 4 x d x space plus space c
y space sin space x space equals 4 x squared over 2 plus c
space space space space space space space space space space space space space equals 2 x squared space plus thin space c
Put space y space equals space 0 comma space x equals straight pi over 2
space space space space space space space 0 equals straight pi squared over 2 plus c
space space space space space space space c equals negative straight pi squared over 2
Now comma space
y space sin space x space equals space 2 x squared minus straight pi squared over 2 end style

Question 37(ix)

Solution 37(ix)

Question 37(x)

Solution 37(x)

Question 37(xi)

Solution 37(xi)

Question 37(xii)

dy = cos x (2 – y cosec x) dxSolution 37(xii)

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solve the differential equation

begin mathsize 12px style open parentheses straight y plus 3 straight x squared close parentheses dx over dy equals straight x end style

Solution 40

begin mathsize 12px style open parentheses y plus 3 x squared close parentheses fraction numerator d x over denominator d y end fraction equals x
fraction numerator d x over denominator d y end fraction equals fraction numerator x over denominator y plus 3 x squared end fraction
fraction numerator d y over denominator d x end fraction equals fraction numerator y plus 3 x squared over denominator x end fraction
fraction numerator d y over denominator d x end fraction minus y over x equals 3 x
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d y over denominator d x end fraction plus p y equals Q
p equals 1 over x comma Q equals 3 x
I. F.
equals e to the power of integral p d z end exponent
equals e to the power of negative integral 1 over x d z end exponent
equals e to the power of negative l o g z end exponent
equals 1 over x
Solution space of space the space equation space is space given space by comma
y cross times open parentheses I. F close parentheses equals integral Q cross times open parentheses I. F close parentheses d x plus c
y open parentheses 1 over x close parentheses equals integral 3 x cross times open parentheses 1 over x close parentheses d x plus c
y over x equals 3 x plus c end style

Question 41

Solution 41

Chapter 22 – Differential Equations Exercise Ex. 22.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

I n space a space c u l t u r e comma space t h e space b a c t e r i a space c o u n t space i s space 100000. space T h e space n u m b e r space i s space i n c r e a s e d space b y space 10 %
i n space 2 space h o u r s. space I n space h o w space m a n y space h o u r s space w i l l space t h e space c o u n t space r e a c h space 200000 comma space i f space t h e space r a t e space o f space g r o w t h
o f space b a c t e r i a space i s space p r o p o r t i o n a l space t o space t h e space n u m b e r space p r e s e n t ?

Solution 4

L e t space C space b e space t h e space c o u n t space o f space b a c t e r i a space a t space a n y space t i m e space t.
I t space i s space g i v e n space t h a t
fraction numerator d C over denominator d t end fraction infinity C
rightwards double arrow fraction numerator d C over denominator d t end fraction equals lambda C comma space w h e r e space lambda space i s space a space c o n s tan t space o f space p r o p o r t i o n a l i t y
rightwards double arrow fraction numerator d C over denominator C end fraction equals lambda d t
rightwards double arrow integral fraction numerator d C over denominator C end fraction equals lambda integral d t
rightwards double arrow log C equals lambda t plus log K.... left parenthesis 1 right parenthesis
I n i t i a l l y comma space a t space t equals 0 comma space C equals 100000
T h u s comma space w e space h a v e comma
log 100000 equals lambda cross times 0 plus log K.... left parenthesis 2 right parenthesis
rightwards double arrow log 100000 equals log K.... left parenthesis 3 right parenthesis
A t space t equals 2 comma space C equals 100000 plus 100000 cross times 10 over 100 equals 110000
T h u s comma space f r o m space left parenthesis 1 right parenthesis comma space w e space h a v e comma
log 110000 equals lambda cross times 2 plus log K.... left parenthesis 4 right parenthesis
S u b t r a c t i n g space e q u a t i o n space left parenthesis 2 right parenthesis space f r o m space left parenthesis 4 right parenthesis comma space w e space h a v e comma
log 110000 minus log 100000 equals 2 lambda
rightwards double arrow log 11 cross times 10000 minus log 10 cross times 10000 equals 2 lambda
rightwards double arrow log fraction numerator 11 cross times 10000 over denominator 10 cross times 10000 end fraction equals 2 lambda
rightwards double arrow log 11 over 10 equals 2 lambda
rightwards double arrow lambda equals 1 half log 11 over 10.... left parenthesis 5 right parenthesis
W e space n e e d space t o space f i n d space t h e space t i m e space apostrophe t apostrophe space i n space w h i c h space t h e space c o u n t space r e a c h e s space 200000.
S u b s t i t u t i n g space t h e space v a l u e s space o f space lambda space a n d space K space f r o m space e q u a t i o n s space left parenthesis 3 right parenthesis space a n d space left parenthesis 5 right parenthesis space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log 200000 equals 1 half log 11 over 10 t plus log 100000
rightwards double arrow 1 half log 11 over 10 t equals log 200000 minus log 100000
rightwards double arrow 1 half log 11 over 10 t equals log 200000 over 100000
rightwards double arrow 1 half log 11 over 10 t equals log 2
rightwards double arrow t equals fraction numerator 2 log 2 over denominator log 11 over 10 end fraction space h o u r s

Question 5

Solution 5

Question 6

Solution 6

Question 7

T h e space p o p u l a t i o n space o f space a space c i t y space i n c r e s e s space a t space a space r a t e space p r o p o r t i o n a l space t o space t h e space n u m b e r space o f space
i n h a b i tan t s space p r e s e n t space a t space a n y space t i m e space t. space I f space t h e space p o p u l a t i o n space o f space t h e space c i t y space w a s space 200000 space i n
1990 space a n d space 250000 space i n space 2000 comma space w h a t space w i l l space b e space t h e space p o p u l a t i o n space i n space 2010 ?

Solution 7


L e t space P space b e space t h e space p o p u l a t i o n space o f space t h e space c i t y space a t space a n y space t i m e space t.
I t space i s space g i v e n space t h a t
fraction numerator d P over denominator d t end fraction infinity P
rightwards double arrow fraction numerator d P over denominator d t end fraction equals lambda P comma space w h e r e space lambda space i s space a space c o n s tan t space o f space p r o p o r t i o n a l i t y
rightwards double arrow fraction numerator d P over denominator P end fraction equals lambda d t
rightwards double arrow integral fraction numerator d P over denominator P end fraction equals lambda integral d t
rightwards double arrow log P equals lambda t plus log K.... left parenthesis 1 right parenthesis
I n i t i a l l y comma space a t space t equals 1990 comma space P equals 200000
T h u s comma space w e space h a v e comma
log 200000 equals lambda cross times 1990 plus log K.... left parenthesis 2 right parenthesis
A t space t equals 2000 comma space P equals 250000
T h u s comma space f r o m space left parenthesis 1 right parenthesis comma space w e space h a v e comma
log 250000 equals lambda cross times 2000 plus log K.... left parenthesis 3 right parenthesis
S u b t r a c t i n g space e q u a t i o n space left parenthesis 2 right parenthesis space f r o m space left parenthesis 3 right parenthesis comma space w e space h a v e comma
log 250000 minus log 200000 equals 10 lambda
rightwards double arrow log 4 over 5 equals 10 lambda
rightwards double arrow lambda equals 1 over 10 log 4 over 5.... left parenthesis 4 right parenthesis
S u b s t i t u t i n g space t h e space v a l u e space o f space lambda space f r o m space e q u a t i o n space left parenthesis 4 right parenthesis space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log 200000 equals 1990 cross times 1 over 10 log 4 over 5 plus log K
rightwards double arrow log K equals log 200000 minus 199 cross times log 4 over 5.... left parenthesis 5 right parenthesis space space
S u b s t i t u t i n g space t h e space v a l u e space o f space lambda comma space log K space a n d space t equals 2010 space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log P equals open curly brackets 1 over 10 log 4 over 5 close curly brackets 2010 plus log 200000 minus 199 cross times log 4 over 5
rightwards double arrow log P equals log open curly brackets 4 over 5 close curly brackets to the power of 201 plus log open parentheses 200000 cross times open parentheses 5 over 4 close parentheses to the power of 199 close parentheses
rightwards double arrow P equals open curly brackets 4 over 5 close curly brackets to the power of 201 cross times 200000 cross times open parentheses 5 over 4 close parentheses to the power of 199
rightwards double arrow P equals open parentheses 5 over 4 close parentheses squared cross times 200000 equals 25 over 16 cross times 200000 equals 312500

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Chapter 22 – Differential Equations Exercise Ex. 22RE

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

F i n d space t h e space d i f f e r e n t i a l space e q u a t i o n space c o r r e s p o n d i n g space t o space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x comma space w h e r e
a comma space b comma space c space a r e space a r b i t r a r y space c o n s tan t s.

Solution 15

C o n s i d e r space t h e space g i v e n space e q u a t i o n space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d y over denominator d x end fraction equals 2 a e to the power of 2 x end exponent minus 3 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 1 right parenthesis
rightwards double arrow 7 fraction numerator d y over denominator d x end fraction equals 14 a e to the power of 2 x end exponent minus 21 b e to the power of minus 3 x end exponent plus 7 c e to the power of x.... left parenthesis 2 right parenthesis
D i f f e r e n t i a t i n g space e q u a t i o n space left parenthesis 1 right parenthesis space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d squared y over denominator d x squared end fraction equals 4 a e to the power of 2 x end exponent plus 9 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 3 right parenthesis
A g a i n space d i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d cubed y over denominator d x cubed end fraction equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 4 right parenthesis
N o w space c o n s i d e r space t h e space f o l l o w i n g space e x p r e s s i o n
fraction numerator d cubed y over denominator d x cubed end fraction minus 7 fraction numerator d y over denominator d x end fraction plus 6 y
equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x minus 14 a e to the power of 2 x end exponent plus 21 b e to the power of minus 3 x end exponent minus 7 c e to the power of x plus 6 open parentheses a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x close parentheses
equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x minus 14 a e to the power of 2 x end exponent plus 21 b e to the power of minus 3 x end exponent minus 7 c e to the power of x plus 6 a e to the power of 2 x end exponent plus 6 b e to the power of minus 3 x end exponent plus 6 c e to the power of x
equals 0
T h u s comma space t h e space r e q u i r e d space d i f f e r e n t i a l space e q u a t i o n space c o r r e s p o n d i n g space t o space
space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x space i s
fraction numerator d cubed y over denominator d x cubed end fraction minus 7 fraction numerator d y over denominator d x end fraction plus 6 y equals 0

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solve the following differential equation:

begin mathsize 12px style y space s e c squared space x space plus space open parentheses y plus 7 close parentheses space tan space x space fraction numerator d y over denominator d x end fraction equals 0 end style

Solution 46

Question 47

Solution 47

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64(i)

Solution 64(i)

Question 64(ii)

Solution 64(ii)

Question 64(iii)

Solution 64(iii)

Question 64(iv)

Solution 64(iv)

Question 64(v)

Solution 64(v)

Question 64(vi)

Solution 64(vi)

Question 65(i)

Solution 65(i)

Question 65(ii)

Solution 65(ii)

Question 65(iii)

Solution 65(iii)

Question 66(i)

Solution 66(i)

Question 66(ii)

Solution 66(ii)

Question 66(iii)

Solution 66(iii)

Question 66(iv)

Solution 66(iv)

Question 66(v)

Solution 66(v)

Question 66(vi)

Solution 66(vi)

Question 66(vii)

Solution 66(vii)

Question 66(viii)

Solution 66(viii)

Question 66(ix)

Solution 66(ix)

Question 66(x)

Solution 66(x)

Question 66(xi)

Solution 66(xi)

Question 66(xii)

Solution 66(xii)

Question 66(xiii)

Solution 66(xiii)

Question 66(xiv)

Solution 66(xiv)

Question 66(xv)

Solution 66(xv)

Question 67(i)

Solution 67(i)

Question 67(ii)

Solution 67(ii)

Question 67(iii)

Solution 67(iii)

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

Solution 76

Question 77

Solution 77

Question 78

Solution 78

Question 79

Solution 79

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RD SHARMA SOLUTION CHAPTER -21 Areas of Bounded Regions I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 21 Areas of bounded regions Exercise Ex. 21.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Thus, Required area =2 over 3 open parentheses 5 to the power of 3 over 2 end exponent minus 1 close parentheses square unitsQuestion 8

Solution 8

Question 9

Solution 9

Question 11

Sketch the region {(x, y):9x2 + 4y2 = 36} and find the area enclosed by it, using integration.Solution 11

9x2 + 4y2 = 36

Area of Sector OABCO =

Area of the whole figure = 4 × Ar. D OABCO

= 6p sq. unitsQuestion 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

S k e t c h space t h e space g r a p h space y equals open vertical bar x minus 5 close vertical bar. space E v a l u a t e space integral subscript 0 superscript 1 open vertical bar x minus 5 close vertical bar d x. space W h a t space d o e s space t h i s space v a l u e space o f space t h e
i n t e g r a l space r e p r e s e n t space o n space t h e space g r a p h ?

Solution 17

C o n s i d e r space t h e space s k e t c h space o f space t h e space g i v e n space g r a p h : y equals open vertical bar x minus 5 close vertical bar
T h e r e f o r e comma space
R e q u i r e d space a r e a equals integral subscript 0 superscript 1 y d x
equals integral subscript 0 superscript 1 open vertical bar x minus 5 close vertical bar d x
equals integral subscript 0 superscript 1 minus open parentheses x minus 5 close parentheses d x
equals open square brackets fraction numerator minus x squared over denominator 2 end fraction plus 5 x close square brackets subscript 0 superscript 1
equals open square brackets minus 1 half plus 5 close square brackets
equals 9 over 2 s q. space u n i t s
T h e r e f o r e comma space t h e space g i v e n space i n t e g r a l space r e p r e s e n t s space t h e space a r e a space b o u n d e d space b y space t h e space c u r v e s comma space
x equals 0 comma y equals 0 comma space x equals 1 space a n d space y equals minus open parentheses x minus 5 close parentheses.

Question 18

What dose this integral represent on the graph?.Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 10

and evaluate the area of the region under the curve and above the x-axis.Solution 10

Question 27

Find the area of the minor segment of the circle x2 + y2 = a2 cut off by the line x = Solution 27

Question 28

Find the area of the region bounded by the curve x = at2, y = 2at between the ordinates corresponding t = 1 and t = 2.Solution 28

Question 29

Find the area enclosed by the curve x = 3 cost,

y = 2 sin t.Solution 29

Chapter 21 Areas of Bounded Regions Exercise Ex. 21.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Find the area of the region bounded by x2 = 4ay and its latusrectum.Solution 3

Question 4

Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.Solution 4

Question 5

Find the area of the region bounded by the curve ay2 = x3, the y-axis and the lines y = a and y = 2a.Solution 5

Chapter 21 – Areas of Bounded Regions Exercise Ex. 21.3

Question 2

Find the area of the region common to the parabolas 4y2 = 9x and 3x2 = 16y.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find the area of the region between the circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.
Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 14

Solution 14

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23 (i)

Using Integration, find the area of the region bounded by the triangle whose vertices are (- 1, 2), (1, 5) and (3, 4).Solution 23 (i)

Equation of side AB,

Equation of side BC,

Equation of side AC,

Area of required region

= Area of EABFE + Area of BFGCB – Area of AEGCA

Question 25

F i n d space t h e space a r e a space o f space t h e space r e g i o n space i n space t h e space f i r s t space q u a d r a n t space e n c l o s e d space b y space x minus a x i s comma
t h e space l i n e space y equals square root of 3 x space a n d space t h e space c i r c l e space x squared plus y squared equals 16

Solution 25

C o n s i d e r space t h e space f o l l o w i n g space g r a p h.
W e space h a v e comma space y equals square root of 3 x
S u b s t i t u t i n g space t h i s space v a l u e space i n space x squared plus y squared equals 16 comma space
x squared plus open parentheses square root of 3 x close parentheses squared equals 16
rightwards double arrow x squared plus 3 x squared equals 16
rightwards double arrow 4 x squared equals 16
rightwards double arrow x squared equals 4
rightwards double arrow x equals plus-or-minus 2
S i n c e space t h e space s h a d e d space r e g i o n space i s space i n space t h e space f i r s t space q u a d r a n t comma space l e t space u s space t a k e space t h e space p o s i t i v e
v a l u e space o f space x.
T h e r e f o r e comma space x equals 2 space a n d space y equals 2 square root of 3 space a r e space t h e space c o o r d i n a t e s space
o f space t h e space i n t e r s e c t i o n space p o i n t space A.
T h u s comma space a r e a space o f space t h e space s h a d e d space r e g i o n space O A B equals A r e a space O A C plus A r e a space A C B
rightwards double arrow A r e a space O A B equals integral subscript 0 superscript 2 square root of 3 x d x plus integral subscript 2 superscript 4 square root of 16 minus x squared end root d x
rightwards double arrow A r e a space O A B equals open parentheses fraction numerator square root of 3 x squared over denominator 2 end fraction close parentheses subscript 0 superscript 2 plus 1 half open square brackets x square root of 16 minus x squared end root plus 16 sin to the power of minus 1 end exponent open parentheses x over 4 close parentheses close square brackets subscript 2 superscript 4
rightwards double arrow A r e a space O A B equals open parentheses fraction numerator square root of 3 cross times 4 over denominator 2 end fraction close parentheses plus 1 half open square brackets 16 sin to the power of minus 1 end exponent open parentheses 4 over 4 close parentheses close square brackets minus 1 half open square brackets 4 square root of 16 minus 12 end root plus 16 sin to the power of minus 1 end exponent open parentheses 2 over 4 close parentheses close square brackets
rightwards double arrow A r e a space O A B equals 2 square root of 3 plus 1 half open square brackets 16 cross times straight pi over 2 close square brackets minus 1 half open square brackets 4 square root of 3 plus 16 sin to the power of minus 1 end exponent open parentheses 1 half close parentheses close square brackets
rightwards double arrow A r e a space O A B equals 2 square root of 3 plus 4 straight pi minus 2 square root of 3 minus fraction numerator 4 straight pi over denominator 3 end fraction
rightwards double arrow A r e a space O A B equals 4 straight pi minus fraction numerator 4 straight pi over denominator 3 end fraction
rightwards double arrow A r e a space O A B equals fraction numerator 8 straight pi over denominator 3 end fraction s q. space u n i t s.

Question 26

Solution 26

Question 27

Solution 27

Question 29

Solution 29

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

C l e a r l y comma space A r e a space o f space capital delta A B C equals A r e a space A D B plus A r e a space B D C
A r e a thin space A D B : space T o space f i n d space t h e space a r e a space A D B comma space w e space s l i c e space i t space i n t o space v e r t i c a l space s t r i p s.
W e space o b s e r v e space t h a t space e a c h space v e r t i c a l space s t r i p space h a s space i t s space l o w e r space e n d space o n space s i d e space A C space a n d space t h e
u p p e r space e n d space o n space A B. space S o space t h e space a p p r o x i m a t i n g space r e c tan g l e space h a s space
L e n g t h equals y subscript 2 minus y subscript 1
W i d t h equals capital delta x space
A r e a equals open parentheses y subscript 2 minus y subscript 1 close parentheses capital delta x
S i n c e space t h e space a p p r o x i m a t i n g space r e c tan g l e space c a n space m o v e space f r o m space x equals 4 space t o space 6 comma space
t h e space a r e a space o f space t h e space t r i a n g l e space A D B equals integral subscript 4 superscript 6 open parentheses y subscript 2 minus y subscript 1 close parentheses d x
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals integral subscript 4 superscript 6 open square brackets open parentheses fraction numerator 5 x over denominator 2 end fraction minus 9 close parentheses minus open parentheses 3 over 4 x minus 2 close parentheses close square brackets d x
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals integral subscript 4 superscript 6 open parentheses fraction numerator 5 x over denominator 2 end fraction minus 9 minus 3 over 4 x plus 2 close parentheses d x
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals integral subscript 4 superscript 6 open parentheses fraction numerator 7 x over denominator 4 end fraction minus 7 close parentheses d x
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals open parentheses fraction numerator 7 x squared over denominator 4 cross times 2 end fraction minus 7 x close parentheses subscript 4 superscript 6
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals open parentheses fraction numerator 7 cross times 36 over denominator 8 end fraction minus 7 cross times 6 close parentheses minus open parentheses fraction numerator 7 cross times 16 over denominator 8 end fraction minus 7 cross times 4 close parentheses
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals open parentheses 63 over 2 minus 42 minus 14 plus 28 close parentheses
rightwards double arrow space a r e a space o f space t h e space t r i a n g l e space A D B equals open parentheses 63 over 2 minus 28 close parentheses
S i m i l a r l y comma space A r e a space B D C equals integral subscript 6 superscript 8 open parentheses y subscript 4 minus y subscript 3 close parentheses d x
rightwards double arrow A r e a space B D C equals integral subscript 6 superscript 8 open parentheses y subscript 4 minus y subscript 3 close parentheses d x
rightwards double arrow A r e a space B D C equals integral subscript 6 superscript 8 open square brackets open parentheses minus x plus 12 close parentheses minus open parentheses 3 over 4 x minus 2 close parentheses close square brackets d x
rightwards double arrow A r e a space B D C equals integral subscript 6 superscript 8 open square brackets fraction numerator minus 7 x over denominator 4 end fraction plus 14 close square brackets d x
rightwards double arrow A r e a space B D C equals open square brackets minus fraction numerator 7 x squared over denominator 8 end fraction plus 14 x close square brackets subscript 6 superscript 8
rightwards double arrow A r e a space B D C equals open square brackets minus fraction numerator 7 cross times 64 over denominator 8 end fraction plus 14 cross times 8 close square brackets minus open square brackets minus fraction numerator 7 cross times 36 over denominator 8 end fraction plus 14 cross times 6 close square brackets
rightwards double arrow A r e a space B D C equals open square brackets minus 56 plus 112 plus 63 over 2 minus 84 close square brackets
rightwards double arrow A r e a space B D C equals open parentheses 63 over 2 minus 28 close parentheses
T h u s comma space A r e a space A B C equals A r e a space A D B plus A r e a space B D C
rightwards double arrow A r e a space A B C equals open parentheses 63 over 2 minus 28 close parentheses plus open parentheses 63 over 2 minus 28 close parentheses
rightwards double arrow A r e a space A B C equals 63 minus 56
rightwards double arrow A r e a space A B C equals 7 space s q. space u n i t s

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

 Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

F i n d space t h e space a r e a space o f space t h e space r e g i o n comma space open curly brackets open parentheses x comma y close parentheses : x squared plus y squared less or equal than 4 comma space x plus y greater or equal than 2 close curly brackets

Solution 43

T h e space e q u a t i o n space o f space t h e space g i v e n space c u r v e s space a r e
x squared plus y squared equals 4.... left parenthesis 1 right parenthesis
x plus y equals 2....... left parenthesis 2 right parenthesis
C l e a r l y space x squared plus y squared equals 4 space r e p r e s e n t s space a space c i r c l e space a n d space x plus y equals 2 space i s space t h e space e q u a t i o n space o f space a
s t r a i g h t space l i n e space c u t t i n g space x space a n d space y space a x e s space a t space left parenthesis 0 comma 2 right parenthesis space a n d space left parenthesis 2 comma 0 right parenthesis space r e s p e c t i v e l y.
T h e space s m a l l e r space r e g i o n space b o u n d e d space b y space t h e s e space t w o space c u r v e s space i s space s h a d e d space i n space t h e space
f o l l o w i n g space f i g u r e.
L e n g t h space equals y subscript 2 minus y subscript 1
W i d t h equals capital delta x space a n d
A r e a equals open parentheses y subscript 2 minus y subscript 1 close parentheses capital delta x
S i n c e space t h e space a p p r o x i m a t i n g space r e c tan g l e space c a n space m o v e space f r o m space x equals 0 space t o space x equals 2 comma space t h e
r e q u i r e d space a r e a space i s space g i v e n space b y space
A equals integral subscript 0 superscript 2 open parentheses y subscript 2 minus y subscript 1 close parentheses d x
W e space h a v e space y subscript 1 equals 2 minus x space a n d space y subscript 2 equals square root of 4 minus x squared end root
T h u s comma
A equals integral subscript 0 superscript 2 open parentheses square root of 4 minus x squared end root minus 2 plus x close parentheses d x
rightwards double arrow A equals integral subscript 0 superscript 2 open parentheses square root of 4 minus x squared end root close parentheses d x minus 2 integral subscript 0 superscript 2 d x plus integral subscript 0 superscript 2 x d x
rightwards double arrow A equals open square brackets fraction numerator x square root of 4 minus x squared end root over denominator 2 end fraction plus a squared over 2 sin to the power of minus 1 end exponent open parentheses x over 2 close parentheses close square brackets subscript 0 superscript 2 minus 2 open parentheses x close parentheses subscript 0 superscript 2 plus open parentheses x squared over 2 close parentheses subscript 0 superscript 2
rightwards double arrow A equals 4 over 2 sin to the power of minus 1 end exponent open parentheses 2 over 2 close parentheses minus 4 plus 2
rightwards double arrow A equals 2 sin to the power of minus 1 end exponent open parentheses 1 close parentheses minus 2
rightwards double arrow A equals 2 cross times straight pi over 2 minus 2
rightwards double arrow A equals straight pi minus 2 space sq. units

Question 44

Solution 44

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 1

Calculate the area of the region bounded by the parabolas y2 = 6x and x2 = 6y.Solution 1

Question 18

Find the area of the region bounded by y = , x = 2y + 3 in the first quadrant and x-axis.Solution 18

Question 24

Find the area of the bounded by y = and y = x.Solution 24

Question 28

Find the area enclosed by the curve y = -x2 and the straight line x + y + 2 = 0.Solution 28

Question 30

Using the method of integration, find the area of the region bounded by the following lines: 3x – y – 3 = 0,

2x + y – 12 = 0, x – 2y – 1 = 0.Solution 30

Question 38

Find the area of the region enclosed by the parabola

 x2 = y and the line y = x + 2.Solution 38

Question 51

Solution 51

Question 52

Solution 52

Chapter 21 Areas of Bounded Regions Exercise Ex. 21.4

Question 1

Find the area of the region between the parabola x = 4y – y2 and the line x = 2y – 3.Solution 1

Question 2

Find the area bounded by the parabola x = 8 + 2y – y2; the y-axis and the lines y = -1 and y = 3.Solution 2

Question 3

Find the area bounded by the parabola y2 = 4x and the line

y = 2x – 4.

i. By using horizontal strips

ii. By using vertical stripsSolution 3

Question 4

Find the area of the region bounded the parabola y2 = 2x and straight line x – y = 4.Solution 4

Read More

RD SHARMA SOLUTION CHAPTER-20 Definite Integrals I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 20 Definite Integrals Exercise Ex. 20.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

T h e r e f o r e comma integral subscript 2 superscript 3 fraction numerator x over denominator x squared plus 1 end fraction equals 1 half log 2

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Evaluate the Integral in using substitution.

begin mathsize 12px style integral subscript negative 1 end subscript superscript 1 fraction numerator dx over denominator straight x squared space plus space 2 straight x space plus space 5 end fraction end style

Solution 44

begin mathsize 12px style integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator x squared plus 2 x plus 5 end fraction equals integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x squared plus 2 x plus 1 close parentheses plus 4 end fraction equals integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x plus 1 close parentheses squared plus open parentheses 2 close parentheses squared end fraction
Let space x space plus space 1 space equals space t space rightwards double arrow space d x space equals space d t
When space straight x space equals space minus 1 comma space t space equals space 0 space and space when space x space equals space 1 comma space t space equals space 2
therefore integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x plus 1 close parentheses squared plus open parentheses 2 close parentheses squared end fraction equals integral subscript 0 superscript 2 fraction numerator d t over denominator t squared plus 2 squared end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets 1 half tan to the power of negative 1 end exponent t over 2 close square brackets subscript 0 superscript 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half tan to the power of negative 1 end exponent space 1 minus 1 half tan to the power of negative 1 end exponent space 0
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open parentheses straight pi over 4 close parentheses equals straight pi over 8 end style

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 54

Solution 54

T h e r e f o r e comma space I equals 2 to the power of begin display style 5 over 2 end style end exponent over 3

Question 55

Solution 55

Question 56

Solution 56

Let cosx =u , Then

Hence

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Given :

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

We know , By reduction formula 

For n=2

For n=4

Hence

Note: Answer given at back is incorrect.Question 65

Solution 65

Using Integration By parts

Question 66

Solution 66

Question 67

Solution 67

Note: Answer given in the book is incorrect. Question 68

Solution 68

 =(1/4)log(2e)

Chapter 20 Definite Integrals Exercise Ex. 20.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Using Integration By parts

Hence

Question 25

Solution 25

Question 26

Solution 26

Question 27

Evaluate begin mathsize 11px style integral subscript 0 superscript straight pi fraction numerator 1 over denominator 5 plus 3 space cos space straight x end fraction dx end styleSolution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

?Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

begin mathsize 12px style Evaluate space integral subscript negative 1 end subscript superscript 1 space 5 straight x to the power of 4 space square root of straight x to the power of 5 plus 1 end root dx. end style

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Chapter 20 Definite Integrals Exercise Ex. 20.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

2x+3 is positive for x>-1.5 . Hence

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Evaluate the integral integral subscript 1 superscript 4 open curly brackets open vertical bar x minus 1 close vertical bar plus open vertical bar x minus 2 close vertical bar plus open vertical bar x minus 4 close vertical bar close curly brackets d xSolution 17

L e t space I equals integral subscript 1 superscript 4 open curly brackets open vertical bar x minus 1 close vertical bar plus open vertical bar x minus 2 close vertical bar plus open vertical bar x minus 4 close vertical bar close curly brackets d x
equals integral subscript 1 superscript 2 open curly brackets open parentheses x minus 1 close parentheses minus open parentheses x minus 2 close parentheses minus open parentheses x minus 4 close parentheses close curly brackets d x plus integral subscript 2 superscript 4 open curly brackets open parentheses x minus 1 close parentheses plus open parentheses x minus 2 close parentheses minus open parentheses x minus 4 close parentheses close curly brackets d x
equals integral subscript 1 superscript 2 open curly brackets open parentheses x minus 1 minus x plus 2 minus x plus 4 close parentheses close curly brackets d x plus integral subscript 2 superscript 4 open curly brackets open parentheses x minus 1 plus x minus 2 minus x plus 4 close parentheses close curly brackets d x
equals integral subscript 1 superscript 2 open parentheses 5 minus x close parentheses d x plus integral subscript 2 superscript 4 open parentheses x plus 1 close parentheses d x
equals open square brackets 5 x minus x squared over 2 close square brackets table row 2 row 1 end table plus open square brackets x squared over 2 plus x close square brackets table row 4 row 2 end table
equals open square brackets 10 minus 2 minus 5 plus 1 half close square brackets plus open square brackets 8 plus 4 minus 2 minus 2 close square brackets
equals 7 over 2 plus 8
I equals 23 over 2

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

For

Using Integration By parts

For

Using Integration By parts

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 27

Solution 27

[x]=0 for 0 < x

and [x]=1 for 1< x < 2

Hence

Question 28

Solution 28

Question 26

Evaluate the following integrals:

begin mathsize 12px style integral from negative straight pi divided by 2 to straight pi divided by 2 of fraction numerator negative straight pi divided by 2 over denominator square root of cosx space sin squared straight x end root end fraction dx end style

Solution 26

NOTE: Answer not matching with back answer.

Chapter 20 Definite Integrals Exercise Ex. 20.4A

Question 1

Solution 1

We know

Hence

We know

If

Then also

Hence

Question 2

Solution 2

We know

Hence

If

Then

Question 3

Solution 3

We know

Hence

If

Then

So

Question 4

Solution 4

We know

Hence

If

Then

Hence

Question 5

Solution 5

We know

Hence

If

Then

So

We know

If f(x) is even

If f(x) is odd

Here

f(x) is even, hence

Note: Answer given in the book is incorrect.Question 6

Solution 6

We know

Hence

If

Then

So

Question 7

Solution 7

We know

Hence

If

Then

So

Question 8

Solution 8

We know

Hence

If

Then

So

Note: Answer given in the book is incorrect. Question 9

Solution 9

If f(x) is even

If f(x) is odd

Here

  is odd and

  is even. Hence

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Chapter 20 Definite Integrals Exercise Ex. 20.4B

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

BQuestion 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

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Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Hence

Question 20

Solution 20

Question 21

Solution 21

Now

Let cosx=t

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25 (i)

Solution 25 (i)

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32 (i)

Solution 32 (i)

Question 33

Solution 33

Question 34

Evaluate the integral integral subscript 0 superscript 1 log open parentheses 1 over x minus 1 close parentheses d xSolution 34

L e t space I equals integral subscript 0 superscript 1 log open parentheses 1 over x minus 1 close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses fraction numerator 1 minus x over denominator x end fraction close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses 1 minus x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
A p p l y i n g space t h e space p r o p e r t y comma space integral subscript 0 superscript a f open parentheses x close parentheses d x equals integral subscript 0 superscript a f open parentheses a minus x close parentheses d x
T h u s comma space I equals integral subscript 0 superscript 1 log open parentheses 1 minus open parentheses 1 minus x close parentheses close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses 1 minus 1 plus x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals 0

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

We know

Also here

So

Hence

Question 39

Solution 39

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Chapter 20 Definite Integrals Exercise Ex. 20RE

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript 1 cos to the power of negative t end exponent x d x end style

Solution 4

begin mathsize 12px style integral subscript 0 superscript 1 cos to the power of negative 1 end exponent xdx equals integral subscript 0 superscript 1 cos to the power of negative 1 end exponent straight x times 1 dx
space space space space space space space space space space space space equals cos to the power of negative 1 end exponent straight x integral subscript 0 superscript 1 dx minus integral subscript 0 superscript 1 open curly brackets straight d over dx cos to the power of negative 1 end exponent straight x integral dx close curly brackets dx space space space space space space space open square brackets Using space Partial space Fraction close square brackets
space space space space space space space space space space space space equals xcos to the power of negative 1 end exponent straight x vertical line subscript 0 superscript 1 minus integral subscript 0 superscript 1 open curly brackets negative fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction close curly brackets dx
space space space space space space space space space space space space equals integral subscript 0 superscript 1 fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction dx
space space space space space space space space space space space space equals integral subscript 0 superscript 1 tdt over straight t dx space space space space space space space space space space space space space space space space space space space space open square brackets 1 minus straight x squared equals straight t squared close square brackets
space space space space space space space space space space space space equals 1 end style

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

begin mathsize 12px style We space have comma
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx over denominator square root of 1 plus cos end root end fraction dx
we space konw space that
space sin space straight x space equals space 2 sin straight x over 2 cos straight x over 2 space and space 1 plus cos space straight x equals 2 cos squared straight x over 2
therefore integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx over denominator square root of 1 space plus space cos end root end fraction
equals integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator square root of 2 cos squared begin display style straight x over 2 end style end root end fraction dx
equals fraction numerator 2 over denominator square root of 2 end fraction integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction dx
equals square root of 2 integral subscript 0 superscript straight pi over 2 end superscript sin space straight x over 2 space dx
equals square root of 2 space open square brackets negative 2 cos straight x over 2 close square brackets subscript 0 superscript straight pi over 2 end superscript
equals square root of 2 open square brackets 1 minus fraction numerator 1 over denominator square root of 2 end fraction close square brackets
equals 2 open parentheses square root of 2 minus 1 close parentheses
therefore integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator square root of 1 plus cosx end root end fraction dx equals 2 open parentheses square root of 2 minus 1 close parentheses end style

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

begin mathsize 12px style Evaluate space integral subscript 1 superscript 2 1 over straight x squared straight e to the power of fraction numerator negative 1 over denominator straight x end fraction end exponent dx end style

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Evaluate the integral integral subscript 0 superscript pi over 2 end superscript x squared cos 2 x d xSolution 21

begin mathsize 12px style integral subscript 0 superscript straight pi over 2 end superscript straight x squared cos space 2 xdx equals straight x squared integral subscript 0 superscript straight pi over 2 end superscript cos space 2 xdx minus integral subscript 0 superscript straight pi over 2 end superscript open curly brackets straight d over dx straight x squared integral cos space 2 xdx close curly brackets dx open square brackets Using space by space Parts close square brackets
space space space space space space equals straight x squared fraction numerator sin 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript minus integral subscript 0 superscript straight pi over 2 end superscript open curly brackets 2 straight x fraction numerator sin space 2 straight x over denominator 2 end fraction close curly brackets dx
space space space space space space equals integral subscript 0 superscript straight pi over 2 end superscript open curly brackets straight x space sin space 2 straight x close curly brackets dx
space space space space space space equals straight x fraction numerator cos 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript plus 1 half fraction numerator sin 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript space space space space space space space space space space space space space space space space space space space space space space space space open square brackets Using space by space Parts space again close square brackets
space space space space space space equals negative straight pi over 4 end style

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript straight pi over 4 end superscript tan to the power of 4 xdx end style

Solution 28

begin mathsize 12px style integral subscript 0 superscript straight pi over 4 end superscript tan to the power of 4 xdx equals integral subscript 0 superscript straight pi over 4 end superscript tan squared straight x space tan squared xdx
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript tan squared straight x open parentheses sec squared straight x minus 1 close parentheses dx open square brackets tan squared straight x equals sec squared space straight x minus 1 close square brackets
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript open parentheses tan squared xsec squared straight x minus tan squared straight x close parentheses dx
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript open parentheses tan squared xsec squared straight x minus sec squared straight x plus 1 close parentheses dx open square brackets tan squared straight x equals sec squared straight x minus 1 close square brackets
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript tan squared xsec squared xdx minus integral subscript 0 superscript straight pi over 4 end superscript sec squared xdx plus integral subscript 0 superscript straight pi over 4 end superscript dx
space space space space space space space space space space space space equals 1 third tan cubed straight x minus tanx plus straight x vertical line subscript 0 superscript straight pi over 4 end superscript
space space space space space space space space space space space space equals straight pi over 4 minus 2 over 3 end style

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx end style

Solution 38

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis equals cos to the power of 7 straight x. space Then space straight f left parenthesis 2 straight pi minus straight x right parenthesis equals open curly brackets cos left parenthesis 2 straight pi minus straight x right parenthesis close curly brackets to the power of 7 equals cos to the power of 7 straight x
space space space space space space space space space integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx equals 2 integral subscript 0 superscript straight pi cos to the power of 7 xdx
Now
space space space space space space space space space straight f left parenthesis straight pi minus straight x right parenthesis equals open curly brackets cos open parentheses straight pi minus straight x close parentheses close curly brackets to the power of 7 equals negative cos to the power of 7 straight x equals negative straight f left parenthesis straight x right parenthesis
Therefore
space space space space space space space space integral subscript 0 superscript straight pi cos to the power of 7 xdx equals 0
Hence
space space space space space space space space integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx equals 2 integral subscript 0 superscript straight pi cos to the power of 7 xdx equals 0 end style

Question 39

Solution 39

begin mathsize 12px style Let comma
straight I equals integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx space space space space space space space space space space space space space space space space space minus negative negative left parenthesis straight i right parenthesis
therefore straight l minus integral subscript 0 superscript straight a fraction numerator square root of straight a minus straight x end root over denominator square root of straight a minus straight x end root plus square root of straight x end fraction dx space space space space space space space space space space space space space minus negative negative left parenthesis ii right parenthesis open square brackets therefore integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis dx minus integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis dx close square brackets
Add space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
2 straight l equals integral subscript 0 superscript straight a fraction numerator square root of straight x plus square root of straight a minus straight x end root over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx
therefore 2 straight l equals integral subscript 0 superscript straight a dx space space space space space space space equals open square brackets straight x close square brackets subscript 0 superscript straight a equals straight a
rightwards double arrow straight l equals straight a over 2
therefore integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx equals straight a over 2 end style

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

begin mathsize 12px style integral from fraction numerator negative straight pi over denominator 4 end fraction to straight pi over 4 of open vertical bar tan space straight x close vertical bar dx equals negative integral from fraction numerator negative straight pi over denominator 4 end fraction to 0 of tan space straight x space dx space plus integral from 0 to straight pi over 4 of tan space straight x space dx space space space space space space open square brackets table row cell because tan space straight x greater or equal than 0 space space space end cell cell when space 0 less than straight x less than straight pi over 4 end cell row cell tan space straight x space less or equal than 0 end cell cell when space straight pi over 4 less than straight x less than 0 end cell end table close square brackets
equals open square brackets log space seg space straight x close square brackets subscript fraction numerator negative straight pi over denominator 4 end fraction end subscript superscript 0 minus open square brackets log space sec space straight x close square brackets subscript 0 superscript fraction numerator negative straight pi over denominator 4 end fraction end superscript
equals open square brackets 0 minus log fraction numerator 1 over denominator square root of 2 end fraction close square brackets minus open square brackets log fraction numerator 1 over denominator square root of 2 end fraction minus 0 close square brackets
equals negative 2 log space fraction numerator 1 over denominator square root of 2 end fraction
equals 2 space straight x space 1 half log 2
equals space log 2
therefore integral from fraction numerator negative straight pi over denominator 4 end fraction to straight pi over 4 of open vertical bar tan space straight x close vertical bar dx space equals space log 2 end style

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Chapter 20 Definite Integrals Exercise Ex. 20.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

begin mathsize 12px style Evaluate integral subscript 1 superscript 3 left parenthesis 3 straight x minus 2 right parenthesis dx end style

Solution 3

begin mathsize 12px style We space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a equals 1 comma space straight b equals 3 space and space straight f space left parenthesis straight x right parenthesis equals 3 straight x space minus 2
straight h equals 2 over straight n rightwards double arrow nh equals 2
Thus comma space we space have comma
straight l equals integral subscript 1 superscript 3 left parenthesis 3 straight x minus space 2 right parenthesis dx end style
begin mathsize 12px style rightwards double arrow straight l equals limit as straight h rightwards arrow 0 of open square brackets straight f left parenthesis 1 right parenthesis plus straight f left parenthesis 1 plus straight h right parenthesis plus straight f left parenthesis 1 plus 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis 1 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets 1 plus open curly brackets 3 left parenthesis 1 plus straight h right parenthesis minus 2 close curly brackets plus open curly brackets 3 left parenthesis 1 plus 2 straight h right parenthesis minus 2 close curly brackets plus negative negative negative negative plus open curly brackets 3 left parenthesis 1 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis minus 2 close curly brackets close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets straight n plus 3 straight h left parenthesis 1 plus 2 plus 3 plus negative negative negative negative left parenthesis straight n minus 1 right parenthesis right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets straight n plus 3 straight h fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
because straight h equals 2 over straight n space space space space space space space space space therefore if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
therefore limit as straight n rightwards arrow 0 of 2 over straight n open square brackets straight n plus 6 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
equals limit as straight n rightwards arrow 0 of space 2 plus 6 over straight n squared straight n squared open parentheses 1 minus 1 over straight n right parenthesis close parentheses
equals limit as straight n rightwards arrow 0 of space 2 space plus space 6 space equals space 8
therefore integral subscript 1 superscript 3 left parenthesis 3 straight x minus 2 right parenthesis dx equals space 8 end style

Question 4

Solution 4

Question 5

Solution 5

begin mathsize 12px style we space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f open parentheses straight a close parentheses plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a space equals space 0 comma space straight b space equals space 5 space and space straight f left parenthesis straight x right parenthesis equals left parenthesis straight x plus 1 right parenthesis
therefore straight h equals 5 over straight n rightwards double arrow nh equals 5
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 5 left parenthesis straight x plus 1 right parenthesis dx
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus negative negative negative negative negative straight f open curly brackets open parentheses straight n minus 1 close parentheses straight h close curly brackets close square brackets
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets 1 plus left parenthesis straight h plus 1 right parenthesis plus left parenthesis 2 straight h plus 1 right parenthesis plus negative negative negative negative negative left parenthesis left parenthesis straight n minus 1 right parenthesis straight h plus 1 right parenthesis close square brackets
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight n plus straight h left parenthesis 1 plus 2 plus 3 plus negative negative negative negative negative left parenthesis straight n minus 1 right parenthesis right parenthesis close square brackets
because straight h equals 5 over straight n space and space if space straight h space rightwards arrow 0 comma space straight n rightwards arrow infinity
straight l equals limit as straight h rightwards arrow 0 of space 5 over straight n open square brackets straight n plus 5 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
straight l equals limit as straight h rightwards arrow 0 of space 5 plus fraction numerator 25 over denominator 2 straight n squared end fraction straight n squared open parentheses 1 minus 1 over straight n close parentheses
equals 5 plus 25 over 2
therefore integral subscript 0 superscript 5 left parenthesis straight x plus 1 right parenthesis dx equals 35 over 2 end style

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a space plus space straight h right parenthesis plus straight f left parenthesis straight a space plus space 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis straight n minus straight a right parenthesis straight h close square brackets
where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a space equals 0 comma space straight b equals space 2 space and space straight f left parenthesis straight x right parenthesis space equals straight x squared plus space 4
therefore straight h equals 2 over straight n rightwards double arrow nh equals 2
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 2 open parentheses straight x squared plus space 4 close parentheses dx
equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis 0 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets 4 left parenthesis straight h squared plus 4 right parenthesis plus open curly brackets left parenthesis 2 straight h right parenthesis squared plus close curly brackets plus negative negative negative negative negative open curly brackets left parenthesis straight n minus 1 right parenthesis straight h squared plus 4 close curly brackets close square brackets
because straight h equals 2 over straight n space & space if space straight h space rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
equals limit as straight n rightwards arrow infinity of space 2 over straight n open square brackets 4 straight n plus 4 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction close square brackets
equals limit as straight n rightwards arrow infinity of space 8 space plus space fraction numerator 4 over denominator 3 straight n squared end fraction straight n cubed open parentheses 1 minus 1 over straight n close parentheses open parentheses 2 minus 1 over straight n close parentheses
equals 8 space plus space fraction numerator 4 space straight x space 2 over denominator 3 end fraction equals 32 over 3
therefore integral subscript 0 superscript 2 open parentheses straight x squared space plus space 4 close parentheses dx equals 32 over 3 space end style

Question 13

Solution 13

Question 14

Solution 14

Question 15

begin mathsize 12px style Evahuate space the space following space in space tegrals space as space limit space of space sums
integral subscript 0 superscript 2 straight e to the power of straight x dx end style

Solution 15

Question 16

Solution 16

Question 17

Solution 17

begin mathsize 12px style We space have comma
space space space space space space space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a space plus space straight h right parenthesis plus straight f left parenthesis straight a space plus space 2 straight h right parenthesis space plus....... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets comma space where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction.
Since space we space have space to space find space integral subscript straight a superscript straight b cosxdx
we space have comma space space space space space space space straight f left parenthesis straight x right parenthesis equals cos space straight x
therefore space space space space space space space straight l equals integral subscript straight a superscript straight b cosxdx
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets cos space straight a space plus cos left parenthesis straight a space plus space straight h right parenthesis space plus cos left parenthesis straight a plus 2 straight h right parenthesis plus..... plus cos left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis space straight h right parenthesis close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis begin display style straight h over 2 end style right parenthesis sin begin display style nh over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close square brackets equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos open parentheses straight a plus begin display style nh over 2 end style minus begin display style straight h over 2 end style space sin begin display style nh over 2 end style close parentheses over denominator sin begin display style straight h over 2 end style end fraction close square brackets
rightwards double arrow space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos open parentheses straight a plus begin display style fraction numerator straight b minus straight a over denominator 2 end fraction end style minus begin display style straight h over 2 end style close parentheses sin open parentheses begin display style fraction numerator straight b minus straight a over denominator 2 end fraction end style close parentheses over denominator sin begin display style straight h over 2 end style end fraction close square brackets space space space space space space space space space space space space space space space space space space space open square brackets because nh space equals space straight b minus straight a close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space open square brackets fraction numerator begin display style straight h over 2 end style over denominator sin begin display style straight h over 2 end style end fraction space straight x space 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction minus straight h over 2 close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space open square brackets fraction numerator begin display style straight h over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close square brackets straight x limit as straight h rightwards arrow 0 of space 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction minus straight h over 2 close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses equals 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses
rightwards double arrow space space space space space straight l equals sin space straight b space equals sin space straight alpha space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because 2 space cos space straight A space sin space straight B equals sin left parenthesis straight A minus straight B right parenthesis minus space sin space left parenthesis straight A space plus space straight B right parenthesis close square brackets end style

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

begin mathsize 12px style We space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h close square brackets
space space space space space space space space space space space where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 2 open parentheses straight x squared plus space straight x close parentheses dx
equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis 0 plus straight h right parenthesis plus straight f left parenthesis 0 plus 2 straight h right parenthesis plus negative negative negative negative negative straight f open curly brackets open parentheses straight n minus 1 close parentheses straight h close curly brackets close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets 0 space plus space left parenthesis straight h squared plus space straight h right parenthesis plus open curly brackets left parenthesis 2 straight h right parenthesis squared plus 2 straight h close curly brackets plus negative negative negative negative negative close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets open curly brackets straight h squared open parentheses 1 plus 2 squared plus 3 squared plus negative negative negative negative left parenthesis straight n minus 1 right parenthesis squared close parentheses close curly brackets plus straight h open curly brackets 1 plus 2 plus 3 minus negative negative negative left parenthesis straight n minus 1 right parenthesis close curly brackets close square brackets
because straight h equals 2 over straight n space & space if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
equals limit as straight n rightwards arrow infinity of space 2 over straight n open square brackets 4 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction plus 2 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
equals limit as straight h rightwards arrow infinity of space fraction numerator 4 over denominator 3 straight n cubed end fraction straight n cubed space open parentheses 1 minus 1 over straight n close parentheses open parentheses 2 minus 1 over straight n close parentheses plus 2 over straight n squared straight n squared open parentheses 1 minus 1 over straight n close parentheses
equals 8 over 3 plus 2 equals 14 over 3
therefore integral subscript 0 superscript 2 open parentheses straight x squared plus straight x close parentheses dx equals 14 over 3 end style

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Evaluate the following in tegrals as limit of sums

begin mathsize 12px style integral subscript 0 superscript 2 left parenthesis straight x squared minus straight x right parenthesis dx end style

Solution 31

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis equals dx limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus.... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Here
space space space straight a equals 0 comma space straight b equals 2 space and space straight f left parenthesis straight x right parenthesis equals straight x squared minus straight x
Now
space space space straight h equals 2 over straight n
space space space nh space equals space 2
Thus comma space we space have
straight l equals integral subscript 0 superscript 2 left parenthesis straight x squared minus straight x right parenthesis dx
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus.... plus straight f left parenthesis left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open curly brackets left parenthesis 0 right parenthesis squared minus left parenthesis 0 right parenthesis close curly brackets plus open curly brackets left parenthesis straight h right parenthesis squared minus left parenthesis straight h right parenthesis close curly brackets plus open curly brackets left parenthesis 2 straight h right parenthesis squared minus left parenthesis 2 straight h right parenthesis close curly brackets plus...... close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses open parentheses straight h close parentheses squared plus left parenthesis 2 straight h right parenthesis squared plus.... close parentheses minus open curly brackets left parenthesis straight h right parenthesis plus left parenthesis 2 straight h right parenthesis plus.... close curly brackets close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets straight h squared open parentheses 1 plus 2 squared plus 3 cubed plus..... plus left parenthesis straight n minus 1 right parenthesis squared close parentheses minus straight h open curly brackets 1 plus 2 plus 3 plus.... plus left parenthesis straight n minus 1 right parenthesis close curly brackets close square brackets
space space space because straight h equals 2 over straight n & if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
space space space equals limit as straight k rightwards arrow infinity of 2 over straight n open square brackets 9 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction minus 9 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
space space space equals 2 over 3 end style

Question 32

Evaluate the following in tegrals as limit of sums

begin mathsize 12px style integral subscript t superscript 3 left parenthesis 2 straight x squared plus 5 straight x right parenthesis dx end style

Solution 32

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis equals dx limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis space plus space straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus.... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
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Question 33

Solution 33

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RS Agarwal Solution | Class 11th | Chapter-7 |   Linear Equations in Two Variable | Edugrown

Exercise Ex. 7A

Question 1

Solve each of the following systems of equations graphically:

2x + 3y = 2

x – 2y = 8Solution 1

Question 3

Solve each of the following systems of equations graphically:

2x + 3y = 8

x – 2y + 3 = 0Solution 3

Question 4

Solve each of the following systems of equations graphically:

2x – 5y + 4 = 0

2x + y – 8 = 0Solution 4

Question 5

Solve each of the following systems of equations graphically:

3x + 2y = 12, 5x – 2y = 4.Solution 5

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.Question 6

Solve each of the following systems of equations graphically:

3x + y + 1 = 0

2x – 3y + 8 = 0Solution 6

Question 7

Solve each of the following systems of equations graphically:

2x + 3y + 5 = 0, 3x – 2y – 12 = 0.Solution 7

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.Question 8

Solve each of the following systems of equations graphically:

2x – 3y + 13 = 0, 3x – 2y + 12 = 0.Solution 8

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.Question 9

Solve each of the following systems of equations graphically:

2x + 3y – 4 = 0, 3x – y + 5 = 0.Solution 9

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.Question 10

Solve each of the following systems of equations graphically:

x + 2y + 2 = 0

3x + 2y – 2 = 0Solution 10

Question 11

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 3 = 0, 2x + 3y – 4 = 0.Solution 11

Question 12

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

2x – 3y + 4 = 0, x + 2y – 5 = 0.Solution 12

Question 13

Solve the following system of linear equations graphically:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Find the area of the region bounded by these lines and the x-axis.Solution 13

Question 14

Solve the following system of linear equation graphically:

x – y + 1 = 0, 3x + 2y – 12 = 0

Calculate the area bounded by these lines and x-axis.Solution 14

Question 15

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0.Solution 15

Question 16

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.Solution 16

Question 17

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

4x – y – 4 = 0, 3x + 2y – 14 = 0.Solution 17

Question 18

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

x – y – 5 = 0, 3x + 5y – 15 = 0.Solution 18

Question 19

Solve the following system of linear equations graphically:

2x – 5y + 4 = 0, 2x + y – 8 = 0

Find the point, where these lines meet the y-axisSolution 19

Question 20

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

5x – y – 7 = 0, x – y + 1 = 0.Solution 20

Question 21

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y = 12, x + 3y = 6.Solution 21

Question 22

Show graphically that each of the following given systems of equations has infinitely many solutions:

2x + 3y = 6, 4x + 6y = 12.Solution 22

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.Question 23

Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.Solution 23

Question 24

Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.Solution 24

Question 25

Show graphically that each of the following given systems of equations has infinitely many solutions:

x – 2y = 5, 3x – 6y = 15.Solution 25

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. Question 26

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

x – 2y = 6, 3x – 6y = 0Solution 26

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 27

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + 3y = 4, 4x + 6y = 12.Solution 27

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 28

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + y = 6, 6x + 3y = 20.Solution 28

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 29

Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.Solution 29

Exercise Ex. 7B

Question 1

Solve for x and y

x + y = 3

4x – 3y = 26Solution 1

Question 2

Solve for x and y:

Solution 2

Question 3

Solve for x and y

2x + 3y = 0

3x + 4y = 5Solution 3

Question 4

Solve for x and y

2x – 3y = 13

7x – 2y = 20Solution 4

Question 5

Solve for x and y

3x – 5y – 19 = 0

-7x + 3y + 1 = 0Solution 5

Question 6

Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0.Solution 6

Question 7

Solve for x and y:

Solution 7

Question 8

Solve for x and y

Solution 8

Question 9

Solve for x and y

Solution 9

Question 10

Solve for x and y

Solution 10

Question 11

Solve for x and y

Solution 11

Question 12

Solve for x and y

Solution 12

Question 13

Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.Solution 13

Question 14

Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74.Solution 14

Question 15

Solve for x and y

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2Solution 15

Question 16

Solve for x and y

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)Solution 16

Question 17

Solve for x and y

Solution 17

Question 18

×Solve for x and y

Solution 18

Putting the given equations become

5u + 6y = 13—(1)

3u + 4y = 7 —-(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52—(3)

18u + 24y = 42—(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 – 25

6y = -12

y = -2

Question 19

Solve for x and y

Solution 19

The given equations are and 

Putting 

x + 6v = 6 —-(1)

3x – 8v = 5—(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24—(3)

9x – 24v = 15 —(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 – 3 = 3

solution is x = 3, y = 2Question 20

Solve for x and y

Solution 20

Putting in the given equation

2x – 3v = 9 —(1)

3x + 7v = 2 —(2)

Multiplying (1) by7 and (2) by 3

14x – 21v = 63 —(3)

9x + 21v = 6 —(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 – 3v = 9

         -3v = 9 – 6

     -3v= 3

       v = -1

the solution is x = 3, y = -1Question 21

Solve for x and y

Solution 21

Question 22

Solve for x and y

Solution 22

Putting in the equation

9u – 4v = 8 —(1)

13u + 7v = 101—(2)

Multiplying (1) by 7 and (2) by 4, we get

63u – 28v = 56—(3)

52u + 28v = 404—(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 – 4v = 8

36 – 4v = 8

-4v = 8 – 36

-4v = -28

Question 23

Solve for x and y

Solution 23

Putting in the given equation, we get

5u – 3v = 1 —(1)

Multiplying (1) by 4 and (2) by 3, we get

20u – 12v = 4—-(3)

27u + 12v = 90—(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) – 3v = 1

10 – 3v = 1

-3v = 1 – 10 -3v = -9 

 v = 3

Question 24

Solve for x and y:

Solution 24

Question 25

Solve for x and y

4x + 6y = 3xy

8x + 9y = 5xy;Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3—(3)

8v + 9u = 5—(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 —(5)

48v + 54u = 30 —(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4Question 26

Solve for x and y:

Solution 26

Question 27

Solve for x and y:

Solution 27

Question 28

Solve for x and y

Solution 28

Putting 

3u + 2v = 2—-(1)

9u – 4v = 1—-(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4—-(3)

9u – 4v = 1—-(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is Question 29

Solve for x and y

Solution 29

The given equations are

Putting 

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5Question 30

Solve for x and y

Solution 30

Putting in the equation, we get

44u + 30v = 10—-(1)

55u + 40v = 13—-(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40—(3)

165u + 120v = 39—(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 – 8 = 3

the solution is x = 8, y = 3Question 31

Solve for x and y:

Solution 31

Question 32

Solve for x and y

71x + 37y = 253

37x + 71y = 287Solution 32

The given equations are

71x + 37y = 253—(1)

37x + 71y = 287—(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

solution is x = 2, y = 3Question 33

Solve for x and y

217x + 131y = 913

131x + 217y = 827Solution 33

217x + 131y = 913—(1)

131x + 217y = 827—(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5—-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

   x – y = 1—(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

solution is x = 3, y = 2Question 34

Solve for x and y:

23x – 29y = 98, 29x – 23y = 110.Solution 34

Question 35

Solve for x and y:

Solution 35

Question 36

Solve for x and y:

Solution 36

Question 37

Solve for x and y

where Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54—(5)

125u – 60v = —(6)

Adding (5) and (6), we get

Question 38

Solve for x and y:

Solution 38

Question 39

Solve for x and y:

Solution 39

Question 40

Solve for x and y:

x + y = a + b, ax – by = a2 – b2.Solution 40

Question 41

Solve for x and y

Solution 41

Question 42

Solve for x and y:

px + py = p – q, qx – py = p + q.Solution 42

Question 43

Solve for x and y:

Solution 43

Question 44

Solve for x and y

6(ax + by) = 3a + 2b

6(bx – ay) = 3b – 2aSolution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting  in(1), we get

Hence, the solution is  Question 45

Solve for x and y:

ax – by = a2 + b2, x + y = 2a.Solution 45

Question 46

Solve for x and y

bx – ay + 2ab = 0Solution 46

Question 47

Solve for x and y

x + y = 2abSolution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by 

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = abQuestion 48

Solve for x and y:

x + y = a + b, ax – by = a2 – b2.Solution 48

Question 49

Solve for x and y:

a2x + b2y = c2, b2x + a2y = d2.Solution 49

Question 50

Solve for x and y:

Solution 50

Exercise Ex. 7C

Question 1

Solve for x and y by method of cross multiplication:

x + 2y + 1 = 0

2x – 3y – 12 = 0Solution 1

x + 2y + 1 = 0 —(1)

2x – 3y – 12 = 0 —(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solutionQuestion 2

Solve for x and y by method of cross multiplication:

3x – 2y + 3 = 0

4x + 3y – 47 = 0Solution 2

3x – 2y + 3 = 0

4x + 3y – 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9Question 3

Solve for x and y by method of cross multiplication:

6x – 5y – 16 = 0

7x – 13y + 10 = 0Solution 3

6x – 5y – 16 = 0

7x – 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4Question 4

Solve for x and y by method of cross multiplication:

3x + 2y + 25 = 0

2x + y + 10 = 0Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20Question 5

Solve for x and y by method of cross multiplication:

2x +5y = 1

2x + 3y = 3Solution 5

2x + 5y – 1 = 0 —(1)

2x + 3y – 3 = 0—(2)

By cross multiplication we have

the solution is x = 3, y = -1Question 6

Solve for x and y by method of cross multiplication:

2x + y – 35 = 0

3x + 4y – 65 = 0Solution 6

2x + y – 35 = 0

3x + 4y – 65 = 0

By cross multiplication, we have

Question 7

Solve each of the following systems of equations by using the method of cross multiplication:

7x – 2y = 3, 22x – 3y = 16.Solution 7

Question 8

Solve for x and y by method of cross multiplication:

Solution 8

Question 9

Solve for x and y by method of cross multiplication:

Solution 9

Taking 

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have

the solution is Question 10

Solve for x and y by method of cross multiplication:

Solution 10

Let in the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

Question 11

Solve for x and y by method of cross multiplication:

Solution 11

Question 12

Solve for x and y by method of cross multiplication:

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0Solution 12

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0

By cross multiplication, we have

Question 13

Solve each of the following systems of equations by using the method of cross multiplication:

Solution 13

Exercise Ex. 7D

Question 1

Show that the following system of equations has a unique solution:

3x + 5y = 12, 5x + 3y = 4

Also, find the solution of the given system of equations.Solution 1

Question 2

Show that each of the following systems of equations has a unique solution and solve it:

2x – 3y = 17, 4x + y = 13.Solution 2

Question 3

Show that the following system of equations has a unique solution:

Also, find the solution of the given system of equationsSolution 3

Question 4

Find the value of k for which each of the following systems of equations has a unique solution:

2x + 3y – 5 = 0, kx – 6y – 8 = 0.Solution 4

Question 5

Find the value of k for which each of the following systems of equations has a unique solution:

x – ky = 2, 3x + 2y + 5 = 0.Solution 5

Question 6

Find the value of k for which each of the following systems of equations has a unique solution:

5x – 7y – 5 = 0, 2x + ky – 1 = 0.Solution 6

Question 7

Find the value of k for which each of the following systems of equations has a unique solution:

4x + ky + 8 = 0, x + y + 1 = 0.Solution 7

Question 8

Find the value of k for which each of the following systems of equations has a unique solution:

4x – 5y = k , 2x – 3y = 12Solution 8

4x – 5y – k = 0, 2x – 3y – 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solutionQuestion 9

Find the value of k for which each of the following systems of equations has a unique solution:

kx + 3y = (k – 3),12x + ky = kSolution 9

kx + 3y – (k – 3) = 0

12x + ky – k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solutionQuestion 10

Show that the system of equations

2x – 3y = 5, 6x – 9y = 15

has an infinite number of solutions.Solution 10

2x – 3y – 5 = 0, 6x – 9y – 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutionsQuestion 11

Show that the system of equations   Solution 11

Question 12

For what value of k, the system of equations

kx + 2y = 5, 3x – 4y = 10

has (i) a unique solution (ii) no solution?Solution 12

kx + 2y – 5 = 0

3x – 4y – 10 = 0

These equations are of the form

This happens when 

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have 

Hence, the given system of equations has no solution if Question 13

For what value of k, the system of equations

x + 2y = 5, 3x + ky + 15 = 0

has (i) a unique solution (ii) no solution?Solution 13

x + 2y – 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.Question 14

For what value of k, the system of equations

x + 2y = 3, 5x + ky + 7 = 0

has (i) a unique solution (ii) no solution?

Is there any value of k for which the given system has an infinite number of solutions?Solution 14

x + 2y – 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if 

For infinite number of solutions we must have

This is never possible since 

There is no value of k for which system of equations has infinitely many solutionsQuestion 15

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3kSolution 15

2x + 3y – 7 = 0

(k – 1)x + (k + 2)y – 3k = 0

These are of the form

This hold only when

Now the following cases arises

Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equationsQuestion 16

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + (k – 2)y =k

6x + (2k – 1)y = (2k + 5)Solution 16

2x + (k – 2)y – k = 0

6x + (2k – 1)y – (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutionsQuestion 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

kx + 3y = (2k +1)

2(k + 1)x + 9y = (7k + 1)Solution 17

kx + 3y – (2k +1) = 0

2(k + 1)x + 9y – (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutionsQuestion 18

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

5x + 2y = 2k

2(k + 1)x + ky = (3k + 4)Solution 18

5x + 2y – 2k = 0

2(k +1)x + ky – (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.Question 19

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 1)x – y = 5

(k + 1)x + (1 – k)y = (3k + 1)Solution 19

(k – 1)x – y – 5 = 0

(k + 1)x + (1 – k)y – (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely manyQuestion 20

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k, kx + ky = 12.Solution 20

Question 21

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2

6x + (1 – 2b)y = 6Solution 21

(a – 1)x + 3y – 2 = 0

6x + (1 – 2b)y – 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4Question 22

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5

3x + (b – 1)y = 2Solution 22

(2a – 1)x + 3y – 5 = 0

3x + (b – 1)y – 2 = 0

These equations are of the form

These holds only when

Question 23

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x – 3y = 7

(a + b)x + (a + b – 3)y = (4a + b)Solution 23

2x – 3y – 7 = 0

(a + b)x + (a + b – 3)y – (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1Question 24

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y=7, (a + b + 1)x +(a + 2b + 2)y = 4(a + b)+ 1.Solution 24

Question 25

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, (a + b)x + (2a – b)y = 21.Solution 25

Question 26

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.Solution 26

Question 27

Find the value of k for which each of the following systems of equations no solution:

8x + 5y = 9, kx + 10y = 15.Solution 27

Question 28

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = 3, 12x + ky = 6.Solution 28

Question 29

Find the value of k for which each of the following systems of equations no solution:

Solution 29

Question 30

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = k – 3, 12x + ky = k.Solution 30

Question 31

Find the value of k for which the system of equations

5x – 3y = 0;2x + ky = 0

has a nonzero solution.Solution 31

We have 5x – 3y = 0 —(1)

2x + ky = 0—(2)

Comparing the equation with

These equations have a non – zero solution if 

Exercise Ex. 7E

Question 1

5 chairs and 4 tables together cost Rs.5600, while 4 chairs and 3 tables together cost Rs.4340. Find the cost of a chair and that of a table.Solution 1

Question 2

23 spoons and 17 forks together cost Rs.1770, while 17 spoons and 23 forks together cost Rs.1830. Find the cost of a spoon and that of a fork.Solution 2

Question 3

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all Rs.19.50, how many coins of each kind does she have?Solution 3

Question 4

The sum of two numbers is 137 and their difference is 43. Find the numbers.Solution 4

Let the two numbers be x and y respectively.

Given:

x + y = 137 —(1)

x – y = 43 —(2)

Adding (1) and (2), we get

2x = 180

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

  = 47

Hence, the two numbers are 90 and 47.Question 5

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 —(1)

4x – 7y = 2 —(2)

Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644 —(3)

12x – 21y = 6 —(4)

Adding (3) and (4), we get

Putting x = 25 in (1), we get

2 × 25 + 3y = 92

50 + 3y = 92

 3y = 92 – 50

y = 14

Hence, the first number is 25 and second is 14Question 6

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 —(1)

4x – y = 138 —(2)

Adding (1) and (2), we get

Putting x = 40 in (1), we get

3 × 40 + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.Question 7

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the number.Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x – 45 = y

2x – y = 45—(1)

and

2y – x = 21

 -x + 2y = 21—(2)

Multiplying (1) by 2 and (2) by 1

4x – 2y = 90—(3)

-x + 2y = 21 —(4)

Adding (3) and (4), we get

3x = 111

Putting x = 37 in (1), we get

2 × 37 – y = 45

 74 – y = 45

y = 29

Hence, the greater and the smaller numbers are 37 and 29.Question 8

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 —(1)

And

5y = x × 3 + 5

-3x + 5y = 5 —(2)

Adding (1) and (2), we get

y = 13

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.Question 9

If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.Solution 9

Let the required numbers be x and y respectively.

Then,

Therefore,

2x – y =-2—(1)

11x – 5y = 24 —(2)

Multiplying (1) by 5 and (2) by 1

10x – 5y = -10—(3)

11x – 5y = 24—(4)

Subtracting (3) and (4) we get

x = 34

putting x = 34 in (1), we get

 2 × 34 – y = -2

 68 – y = -2

-y = -2 – 68

y = 70

Hence, the required numbers are 34 and 70.Question 10

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.Solution 10

Let the numbers be x and y respectively.

According to the question:

x – y = 14 —(1)

From (1), we get

x = 14 + y —(3)

putting x = 14 + y in (2), we get

Putting y = 9 in (1), we get

x – 9 = 14

  x = 14 + 9 = 23

Hence the required numbers are 23 and 9Question 11

The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.Solution 11

Let the ten’s digit be x and units digit be y respectively.

Then,

x + y = 12—(1)

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x – (10x + y) = 18

10y + x – 10x – y = 18

9y – 9x = 18

y – x = 2 —-(2)

Adding (1) and (2), we get

Putting y = 7 in (1), we get

x + 7 = 12

x = 5

Number= 10x + y

                  = 10 × 5 + 7

                  = 50 + 7

                  = 57

Hence, the number is 57.
Question 12

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.Solution 12

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

 10x + y = 7(x + y)

     10x + y = 7x + 7y

       3x – 6y = 0—(1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

      10x – x + y – 10y = 27

        9x – 9y = 27

         (x – y) = 27

           x – y = 3—(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0—(3)

6x – 6y = 18 —(4)

Subtracting (3) from (4), we get

Putting x = 6 in(1), we get

3 × 6 – 6y = 0

18 – 6y = 0

Number = 10x + y

            = 10 × 6 + 3

            = 60 + 3

            = 63

Hence the number is 63.
Question 13

The sum of the digits of a two-digit number is 15. The number of obtained by interchanging the digits exceeds the given number by 9. Find the number.Solution 13

Let the ten’s digit and unit’s digits of required number be x and y respectively.

Then,

x + y = 15—(1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x – (10x + y) = 9

10y + x – 10x – y = 9

       9y – 9x = 9

Add (1) and (2), we get

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

                         = 10 × 7 + 8

                         = 70 + 8

                         = 78

Hence the required number is 78.
Question 14

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.Solution 14

Let the ten’s and unit’s of required number be x and y respectively.

Then,required number =10x + y

According to the given question:

 10x + y = 4(x + y) + 3

 10x + y = 4x + 4y + 3

6x – 3y = 3

  2x – y = 1 —(1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

 x – y = -2—(2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

                         = 10 × 3 + 5

                         = 30 + 5

                         = 35

Hence, required number is 35.
Question 15

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.Solution 15

Let the ten’s digit and unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder

According to the given questiion:

10x + y = 6 × (x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 —(1)

Number obtained by reversing the digits is 10y + x

 10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) =9

(x – y) = 1—(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 —(3)

5x – 5y = 5 —(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1), we get

x =5 and y= 4

Hence, required number is 54.Question 16

A two – digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchanged their places. Find the number.Solution 16

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y

Also,

(10x + y) + 18 = 10y + x

9x – 9y = -18

9(y – x) = 18—(1)

y – x = 2

Now, 

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5 × 10 + 7

                                        = 57Question 17

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.Solution 17

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

(10x + y) – 63 = (10y + x)

9x – 9y = 63

x – y = 7—(1)

Now,

Adding (1) and (2), we get

Putting x = 9 in (1), we get

9 – y = 7

y = 9 – 7

=2

x = 9, y = 2

Hence, the required number = 9 × 10 + 2

                                        = 92.Question 18

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.Solution 18

Question 19

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes Find the fraction.Solution 19

Let the numerator and denominator of fraction be x and y respectively.

According to the question:

x + y = 8—(1)

And

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24—(3)

4x – 3y = -3 —(4)

Add (3) and (4), we get

Putting x = 3 in (1), we get

3 + y= 8

y = 8 – 3

y = 5

x = 3, y = 5

Hence, the fraction is Question 20

If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fractionSolution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – 4

-y = -4 -6

y = 10

x = 3 and y = 10

Hence the fraction is Question 21

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes Find the fraction.Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is .

According to the given question:

y = x + 11

y- x = 11—(1)

and

-3y + 4x = -8 —(2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44—(3)

-3y + 4x = -8—(4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11 

  x = 25

x = 25, y = 36

Hence the fraction is Question 22

Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is 

Subtracting (1) from (2), we get

x = 15

Putting x = 15 in (1), we get

2 × 15 – y = 4

 30 – y = 4

 y = 26

x = 15 and y = 26

Hence the given fraction is Question 23

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.Solution 23

Question 24

The sum of two numbers is 16 and the sum of their reciprocals is Find the numbers.Solution 24

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16—(1)

And

—(2)

From (2),

xy = 48

We know,

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1),

y = 16 – x

   = 16 – 12

   = 4

The required numbers are 12 and 4Question 25

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.Solution 25

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

 x – y = 20—(1)

When 20 students are transferred from B to A:

 2(y – 20) = x + 20

 2y – 40 = x + 20

-x + 2y = 60—(2)

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

 x = 100

Hence, number of students of A and B are 100 and 80 respectively.Question 26

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays Rs.1330, and travelling 90 km, he pays Rs.1490. Find the fixed charges and rate per km.Solution 26

Question 27

A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs.4500, whereas a student B who takes food for 30 days, has to pay Rs.5200. Find the fixed charges per month and the cost of the food per day.Solution 27

Question 28

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received Rs. 1350 as an annual interest. Had he interchanged the amounts invested, he would have received Rs.45 less as interest. What amounts did he invest at different rates?Solution 28

Question 29

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves Rs.9000 per month, find the monthly income of eachSolution 29

Question 30

A man sold a chair and a table together for Rs.1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs.1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.Solution 30

Question 31

Points A and B are 70km apart on a highway. A car starts from A and another starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

Let the cars meet at point M.

AM = 7x km and BM = 7y km

AM – BM = AB

 7x – 7y = 70

7(x – y) = 70

x – y = 10 —-(1)

Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km


In this case let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

 x + y = 70—(2)

Adding (1) and (2), we get

2x = 80

 x = 40

Putting x = 40 in (1), we get

40 – y = 10

 y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.Question 32

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3hours less than the scheduled time. Find the length of the journey.Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

Case I:

Speed = (x + 5)km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5) (y – 3) = xy

xy + 5y -3x -15 = xy

5y – 3x = 15 —(1)

Case II:

Speed (x – 4)km/hr and time taken = (y + 3)hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy -4y + 3x -12 = xy

 3x – 4y = 12 —(2)

Multiplying (1) by 4 and (2) by 5, we get

20y – 12x = 60 —(3)

-20y + 15x = 60 —(4)

Adding (3) and (4), we get

3x = 120

or x = 40

Putting x = 40 in (1), we get

5y – 3 × 40 = 15 

  5y = 135  

  y = 27

Hence, length of the journey is (40 × 27) km = 1080 kmQuestion 33

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.Solution 33

Question 34

Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.Solution 34

Question 35

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.Solution 35

Question 36

A boat goes 12km upstream and 40km downstream in 8hours. It can go 16km upstream and 32 downstream in the same times. Find the speed of the boat in still water and the speed of the stream.Solution 36

Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.

Then,

Speed upstream = (x – y)km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream = 

Time taken to cover 40 km downstream = 

Total time taken = 8hrs

Again, time taken to cover 16 km upstream = 

Time taken to taken to cover 32 km downstream = 

Total time taken = 8hrs

Putting 

12u + 40v = 8 

   3u + 10v = 2 —(1)

and

16u + 32v = 8 

   2u + 4v = 1—(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8—(3)

20u + 40v = 10 —(4)

Subtracting (3) from (4), we get

Putting in (3), we get

On adding (5) and (6), we get

2x = 12 

  x = 6

Putting x = 6 in (6) we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speedof the stream = 2km/hrQuestion 37

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.Solution 37

Let man’s 1 day’s work be and 1 boy’s day’s work be 

Also let  and 

Multiplying (1) by 6 and (2) by 5 we get

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.Question 38

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

 x – y = 3 —-(1)

Also,

(x + 3)(y – 2) = xy

 3y – 2x = 6 —-(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 —(3)

3y – 2x = 6 —(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

 x= 15

x = 15, y = 12

Hence length = 15 metres and breadth = 12 metresQuestion 39

The area of a rectangle gets reduced by 8 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 . Find the length and breadth of the rectangle.Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy – [xy – 5y + 3x -15] = 8

xy – xy + 5y – 3x + 15 = 8 

 3x – 5y = 7 —(1)

And

(x + 3)(y + 2) – xy = 74

 xy + 3y +2x + 6 – xy = 74

2x + 3y = 68—(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21—(3)

10x + 15y = 340—(4)

Adding (3) and (4), we get

Putting x = 19 in (3) we get

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10mQuestion 40

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length. is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.Solution 40

Question 41

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs Rs.4150 while one full and one half reserved first class tickets cost Rs.6255. What is the basic first class full fare and what is the reservation charge?Solution 41

Question 42

Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.Solution 42

Question 43

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of man and his son.Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)


x – 2 = 5y – 10

  x – 5y = -8 —(1)

Two years later:

(x + 2) = 3(y + 2) + 8


x + 2 = 3y + 6 + 8

 x – 3y = 12 —(2)

Subtracting (2) from (1), we get

-2y = -20

  y = 10

Putting y = 10 in (1), we get

x – 5 10 = -8

x – 50 = -8

 x = 42

Hence the present ages of the man and the son are 42 years and 10 respectively.Question 44

If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95years. Find the age of the mother and that of the son.Solution 44

Let the present ages of  the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70—(1)

and

2x + y = 95—(2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 —(3)

4x + 2y = 190—(4)

Subtracting (3) from (4), we get

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

 y = 15

x = 40, y = 15

Hence, the ages of the mother and the son are 40 years and 15 years respectively.Question 45

The present age of a woman is 3 years more than three times the age of her daughter; three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3—(1)

Three years later:

 (x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

 x – 2y = 13—(2)

Subtracting (2) from (1), we get

 y = 10

Putting y = 10 in (1), we get

x – 3 × 10 = 3

 x = 33

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.Question 46

On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains Rs.7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains Rs.13. Find the actual price of each of the tea set and the lemon set.Solution 46

Question 47

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid Rs.27 for a book kept for 7 days, while Tanvy paid Rs. 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.Solution 47

Question 48

A chemist has one solution containing 50% acid and a second one containing 25% arid. How much of each should be used to make 10 litres of a 40% arid solution?Solution 48

Question 49

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given: Pure gold is 24-carat)Solution 49

Question 50

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of adds to be mixed to form the mixture.Solution 50

Question 51

The larger of the two supplementary angles exceeds the smaller by 18o.  Find them.Solution 51

Question 52

Solution 52

Question 58

Solution 58

Exercise Ex. 7F

Question 1

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.Solution 1

Question 2

Find the value of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.Solution 2

Question 3

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.Solution 3

Question 4

For what value of k will the following pair of linear equations have no ‘solution?

2x + 3y = 9, 6x + (k – 2)y = (3k – 2).Solution 4

Question 5

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.Solution 5

Question 6

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.Solution 6

Question 7

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.Solution 7

Question 8

The cost of 5 pens and 8 pencils is Rs.120, while the cost of 8 pens and 5 pencils is Rs.153. Find the cost of 1 pen and that of I pencil.Solution 8

Question 9

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.Solution 9

Question 10

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.Solution 10

Question 11

A man purchased 47 stamps of 20 p and 25 p for Rs.10. Find the number of each type of stamps.Solution 11

Question 12

A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x +y).Solution 15

Question 16

Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.Solution 16

Question 17

Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.Solution 17

Question 18

Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.Solution 18

Question 19

Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.Solution 19

Question 20

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.Solution 20

Question 21

Solution 21

Exercise MCQ

Question 1

If 2x + 3y = 12 and 3x – 2y = 5 then

(a) x = 2, y = 3

(b) x = 2, y = -3

(c) x = 3, y = 2

(d) x = 3, y = -2Solution 1

Question 2

(a) x = 4, y = 2

(b) x = 5, y = 3

(c) x = 6, y = 4

(d) x = 7, y = 5Solution 2

Question 3

(a) x = 2, y = 3

(b) x = -2, y = 3

(c) x = 2, y = -3

(d) x = -2, y = -3Solution 3

Question 4

Solution 4

Question 5

(a) x = 1, y = 1

(b) x = -1, y = -1

(c) x = 1, y = 2

(d) x = 2, y = 1Solution 5

Question 6

Solution 6

Question 7

If 4x+6y=3xy and 8x+9y=5xy then

(a) x=2, y=3

(b) x=1, y=2

(c) x=3, y=4

(d) x=1, y=-1Solution 7

Question 8

If 29x+37y=103 and 37x+29y=95 then

(a) x=1, y=2

(b) x=2, y=1

(c) x=3, y=2

(d) x=2, y=3Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The system kx – y = 2 and 6x – 2y = 3  has a unique solution only when

(a) k = 0

(b) k ≠ 0

(c) k = 3

(d) k ≠ 3Solution 11

Question 12

The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

(a) k = -6

(b) k ≠ -6

(c) k = 0

(d) k ≠ 0Solution 12

Question 13

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k=10

(b) k≠10

(c) 

(d) K=-21Solution 13

Question 14

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is

Solution 14

Question 15

For what value of k do the equations kx – 2y = 3 and

3x + y = 5 represent two lines intersecting at a unique point?

(a) k=3

(b) k=-3

(c) k=6

(d) all real values except -6Solution 15

Question 16

The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 16

We space know space that comma
the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0
has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2.
So comma space the space pair space of space equations space has space no space solution.

Question 17

The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 17

We space know space that comma
the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0
has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2.
So comma space the space pair space of space equations space has space no space solution.

Question 18

If a pair of linear equations is consistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 18

Question 19

If a pair of linear equations is inconsistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 19

Question 20

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠B = ?

(a) 20° 

(b) 40° 

(c) 60° 

(d) 80° Solution 20

Question 21

In a cyclic quadrilateral ABCD, it is being given that

∠A = (x + y + 10) °, ∠B = (y + 20) °,

∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?

(a) 70° 

(b) 80° 

(c) 100° 

(d) 110° Solution 21

Question 22

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96

(b) 69

(c) 87

(d) 78Solution 22

Question 23

Solution 23

Question 24

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(c) 47 years

(d) 40 yearsSolution 24

Question 25

The graphs of the equations 6x – 2y + 9 = 0 and

3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 25

Question 26

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 26

Question 27

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 27

Exercise FA

Question 1

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of

(a) parallel lines

(b) intersecting lines

(c) coincident lines

(d) none of theseSolution 1

Question 2

If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a+b have an infinite number of solutions then

(a) a= 5, b = 1

(b) a = -5, b = 1

(c) a = 5, b = -1

(d) a = -5, b = -1Solution 2

Question 3

The pair of equations 2x+y=5, 3x+2y=8 has

(a) a unique solution

(b) two solutions

(c) no solution

(d) infinitely many solutionsSolution 3

Question 4

If x = -y and y > 0, which of the following is wrong?

(a) x2y > 0

(b) x + y = 0

(c) xy < 0

(d)  Solution 4

Question 5

Solution 5

Question 6

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?Solution 6

Question 7

Solution 7

Question 8

Solve the system of equations x – 2y = 0, 3x + 4y = 20.Solution 8

Question 9

Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.Solution 9

Question 10

The difference between two numbers is 26 and one number is three times the other. Find the numbers.Solution 10

Question 11

Solve : 23x+29y=98, 29x+23y=110.Solution 11

Question 12

Solve : 6x+3y=7xy and 3x+9y = 11xy.Solution 12

Question 13

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.Solution 13

Question 14

In a ΔABC, ∠C =3∠B =2(∠A+∠B). Find the measure of each one of the ∠A, ∠B and ∠C. Solution 14

Question 15

5 pencils and 7 pens together cost Rs. 195 while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each one of the pencil and the pen.Solution 15

Question 16

Solve the following system of equations graphically :

2x-3y=1, 4x-3y+1=0.Solution 16

Since the intersection of the lines is the point with coordinates (-1, -1), x = -1 and y = -1.Question 17

Find the angles of a cyclic quadrilateral ABCD in which ∠A =(4x+20)°, ∠B=(3x-5)°, ∠C=(4y)° and ∠D=(7y+5)° Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

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RS Agarwal Solution | Class 11th | Chapter-2 |   Relations | Edugrown

Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Given colon space straight f open parentheses straight x close parentheses equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction
Let space straight y equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction
rightwards double arrow straight y plus yx squared equals straight x squared
rightwards double arrow straight y equals straight x squared minus yx squared
rightwards double arrow straight y equals straight x squared open parentheses 1 minus straight y close parentheses
rightwards double arrow fraction numerator straight y over denominator 1 minus straight y end fraction equals straight x squared
rightwards double arrow straight x equals square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root
straight x space is space defined space only space if space fraction numerator straight y over denominator 1 minus straight y end fraction greater or equal than 0
rightwards double arrow straight y greater or equal than 0 space and space 1 minus straight y greater or equal than 0
rightwards double arrow straight y greater or equal than 0 space and space 1 greater or equal than straight y
straight y not equal to 1
Therefore comma space straight y greater or equal than 0 space and space 1 greater than straight y
rightwards double arrow 0 less or equal than straight y less than 1
rightwards double arrow straight y element of left square bracket 0 comma space 1 right parenthesis
rightwards double arrow fraction numerator straight x squared over denominator 1 plus straight x squared end fraction element of left square bracket 0 comma space 1 right parenthesis
Hence comma space the space range space of space straight f space is space left square bracket 0 comma space 1 right parenthesis.

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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