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

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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
where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction
Here
space space space space straight a equals 1 comma space straight b equals 3 space and space straight f left parenthesis straight x right parenthesis equals 2 straight x squared space plus space 5 straight x
Now
space space space space straight h space equals space 2 over straight n
space space space space nh equals 2
Thus comma space we space have
straight I equals integral subscript straight t superscript 3 left parenthesis 2 straight x squared space plus space 5 straight x right parenthesis dx
space space space equals limit as straight k rightwards arrow 0 of space straight h 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 space plus space 2 straight h right parenthesis plus...... plus straight f left parenthesis plus 1 left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets end style
begin mathsize 12px style space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses 2 plus 5 close parentheses plus open curly brackets 2 left parenthesis 1 plus straight h right parenthesis squared space plus space 5 left parenthesis 1 plus straight h right parenthesis close curly brackets plus open curly brackets 2 left parenthesis 1 plus 2 straight h right parenthesis squared plus 5 left parenthesis 1 plus 2 straight h right parenthesis close curly brackets plus.... close square brackets
space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses 7 straight n plus 9 straight h left parenthesis 1 plus 2 plus 3 plus..... close parentheses plus 2 straight h squared left parenthesis 1 plus 2 squared plus 3 cubed plus..... right parenthesis right parenthesis close square brackets
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 equals limit as straight k rightwards arrow 0 of space 2 over straight n open square brackets 7 straight n plus 18 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction plus 8 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
space space equals 112 over 3 end style

Question 33

Solution 33

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RD SHARMA SOLUTION CHAPTER-18 Maxima and Minima I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 18 Maxima and Minima Ex 18.1

Question 1

Solution 1

Question 2

Find the maximum and minimum values, if any, without using derivatives of the following function given by f(x) = -(x-1)2 + 2 on R.Solution 2

Question 3

Solution 3

Question 4

Find the maximum and minimum values, if any, without using derivatives of the following function given by h(x) = sin(2x) + 5 on R.Solution 4

Question 5

Find the maximum and minimum values, if any,  usingwithout derivatives of the following function given by begin mathsize 12px style f left parenthesis x right parenthesis equals open vertical bar sin space 4 straight x space plus space 3 close vertical bar end style on R.Solution 5

Question 6

Solution 6

Question 7

Find the maximum and minimum values, if any, without using derivatives of the following function given by begin mathsize 12px style g italic left parenthesis x italic right parenthesis space equals negative 1 open vertical bar straight x plus 1 close vertical bar plus 3 end style on R.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 18 Maxima and Minima Ex. 18.2

Question 1

Solution 1

Question 2

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) =x– 3xSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

f(x) = sinx – cos x, 0 < x < 2πSolution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

f open parentheses x close parentheses equals 2 sin x space minus space x comma space minus straight pi over 2 space less or equal than x space less or equal than straight pi over 2

F o r space c h e c k i n g space t h e space m i n i m a space a n d space m a x i m a comma space w e space h a v e
f apostrophe open parentheses x close parentheses equals 2 cos x space minus 1 space equals space 0

rightwards double arrow cos x equals 1 half equals cos straight pi over 3
rightwards double arrow x equals minus straight pi over 3 comma space straight pi over 3
A t space x equals minus straight pi over 3 comma space f open parentheses x close parentheses space c h a n g e s space f r o m space minus v e space t o space plus space v e
rightwards double arrow x equals minus straight pi over 3 space i s space p o i n t space o f space l o c a l space m i n i m a space w i t h space v a l u e space equals space minus square root of 3 minus straight pi over 3

A t space x equals straight pi over 3 comma space f open parentheses x close parentheses space c h a n g e s space f r o m space plus v e space t o space plus space v e
rightwards double arrow x equals straight pi over 3 space i s space p o i n t space o f space l o c a l space m a x i m a space w i t h space v a l u e space equals space square root of 3 minus straight pi over 3

Question 12

Find the local maxima and local minima, if any, of the following functions using first derivative test. Find also the local maximum and the local minimum values, as the case may be:

begin mathsize 12px style f begin italic style left parenthesis x right parenthesis end style equals straight x square root of 1 minus x end root comma space x greater than 0 end style

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 18 Maxima and Minima Ex. 18.3

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 1(viii)

Solution 1(viii)

L o c a l space M a x i m u m space v a l u e space equals space f left parenthesis 4 right parenthesis
space space space space space space space space space space space space space space space space space 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 4 square root of 32 minus 4 squared end root
space space space space space space space space space space space space space space space space 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 4 square root of 32 minus 16 end root
space space space space space space space space space space space space space space space space space 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 4 square root of 16
space space space space space space space space space space space space space space space space space 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 16
L o c a l space m i n i m u m space a t space x equals minus 4 ;
L o c a l space M i n i m u m space v a l u e space equals space f left parenthesis minus 4 right parenthesis
space space space space space space space space space space space space space space space space space 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 minus 4 square root of 32 minus open parentheses minus 4 close parentheses squared end root
space space space space space space space space space space space space space space space space 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 minus 4 square root of 32 minus 16 end root
space space space space space space space space space space space space space space space space space 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 minus 4 square root of 16
space space space space space space space space space space space space space space space space space 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 minus 16

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Question 4

Show that the function given by f(x)=begin mathsize 12px style fraction numerator log x over denominator x end fraction end style has maximum value at x = e.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

If f(x) = x3 + ax2 + bx + c has a maximum at x = -1 and minimum at x = 3. Determine ab and c.Solution 7

Question 8

Prove that   has maximum value at  Solution 8

Given: 

Differentiating w.r.t x, we get

Take f'(x) = 0

Differentiating f'(x) w.r.t x, we get

At 

Clearly, f”(x) < 0 at 

Thus,   is the maxima.

Hence, f(x) has maximum value at  .

Chapter 18 Maxima and Minima Ex. 18.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Find both the absolute maximum and absolute minimum of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 18 Maxima and Minima Ex. 18.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Divide 15 into two parts such that the square of one multiplied with the cube of the other is maximum.Solution 4

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

W e space h a v e comma space 4 l plus 3 a equals 20
rightwards double arrow 4 l equals 20 minus 3 a
rightwards double arrow l equals fraction numerator 20 minus 3 a over denominator 4 end fraction
F r o m space left parenthesis i right parenthesis comma space w e space h a v e comma
s equals open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses squared plus fraction numerator square root of 3 over denominator 4 end fraction a squared
fraction numerator d s over denominator d a end fraction equals 2 open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses open parentheses fraction numerator minus 3 over denominator 4 end fraction close parentheses plus 2 a cross times fraction numerator square root of 3 over denominator 4 end fraction
T o space f i n d space t h e space m a x i m u m space o r space m i n i m u m comma space fraction numerator d s over denominator d a end fraction equals 0
rightwards double arrow 2 open parentheses fraction numerator 20 minus 3 a over denominator 4 end fraction close parentheses open parentheses fraction numerator minus 3 over denominator 4 end fraction close parentheses plus 2 a cross times fraction numerator square root of 3 over denominator 4 end fraction equals 0
rightwards double arrow minus 3 open parentheses 20 minus 3 a close parentheses plus 4 a square root of 3 equals 0
rightwards double arrow minus 60 plus 9 a plus 4 a square root of 3 equals 0
rightwards double arrow 9 a plus 4 a square root of 3 equals 60
rightwards double arrow a open parentheses 9 plus 4 square root of 3 close parentheses equals 60
rightwards double arrow a equals fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction
D i f f e r e n t i a t i n g space o n c e space a g a i n comma space w e space h a v e comma
fraction numerator d squared s over denominator d a squared end fraction equals fraction numerator 9 plus 4 square root of 3 over denominator 8 end fraction greater than 0
T h u s comma space t h e space s u m space o f space t h e space a r e a s space o f space t h e space s q u a r e space a n d space t r i a n g l e space i s space m i n i m u m space w h e n space a equals fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction
W e space k n o w space t h a t comma space l equals fraction numerator 20 minus 3 a over denominator 4 end fraction
rightwards double arrow l equals fraction numerator 20 minus 3 open parentheses fraction numerator 60 over denominator 9 plus 4 square root of 3 end fraction close parentheses over denominator 4 end fraction
rightwards double arrow l equals fraction numerator 180 plus 80 square root of 3 minus 180 over denominator 4 open parentheses 9 plus 4 square root of 3 close parentheses end fraction
rightwards double arrow l equals fraction numerator 20 square root of 3 over denominator 9 plus 4 square root of 3 end fraction

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Also, find this maximum volume.Solution 12

M a x i m u m space v o l u m e space i s space V subscript x equals 3 end subscript equals 3 cross times open parentheses 18 minus 2 cross times 3 close parentheses squared
rightwards double arrow V equals 3 cross times 12 squared
rightwards double arrow V equals 3 cross times 144
rightwards double arrow V equals 432 space c m cubed

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

An isosceles triangle of vertical angle 2θ is inscribed in a circle radius a. show that the area of the triangle is maximum when  Solution 22

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

S i n c e space fraction numerator d squared S over denominator d x squared end fraction less than 0 comma space t h e space s u m space i s space l a r g e s t space w h e n space x equals y equals fraction numerator r over denominator square root of 2 end fraction

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

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its semi – circular ends is π : ( π + 2) Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 45

Solution 45

Question 5

Amongst all open (from the top) right circular cylindrical boxes of volume 125π cm3, find the dimensions of the box which has the least surface area.Solution 5

Let r and h be the radius and height of the cylinder.

Volume of cylinder   

  … (i)

Surface area of cylinder   

From (i), we get

Question 23

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is   Solution 23

Let   be an isosceles triangle with AB = AC.

Let   

Here, AO bisects 

Taking O as the centre of the circle, join OE, OF and OD such that

OE = OF = OD = r (radius)

Now, 

In 

Similarly, AF = r cot x

In 

As OB bisect   we have

In 

Similarly, BD = DC = CE = 

We have, perimeter of 

P = AB + BC + CA

 = AE + EC + BD + DC + AF + BF

Differentiating w.r.t x, we get

Taking 

As 

Therefore,   is an equilateral triangle.

Taking second derivative of P, we get

At 

Therefore, the perimeter is minimum when 

Least value of P 

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RD SHARMA SOLUTION CHAPTER-17 Increasing and Decreasing Functions I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 17 Increasing and Decreasing Functions Ex. 17.1

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

Question 8

Solution 8

Question 9

Solution 9

Chapter 17 Increasing and Decreasing Functions Ex. 17.2

Question 1(i)

Find the intervals in which the following functions are increasing or decreasing:

10 – 6x – 2x2Solution 1(i)

Question 1(ii)

Find the intervals in which the following functions are increasing or decreasing:

x2 + 2x – 5Solution 1(ii)

Question 1(iii)

Find the intervals in which the following functions are increasing or decreasing:

6 – 9x – x2Solution 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 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)

Solution 1(xiv)

Question 1(xv)

Solution 1(xv)

Question 1(xvi)

Solution 1(xvi)

Question 1(xvii)

Solution 1(xvii)

Question 1(xviii)

Solution 1(xviii)

Question 1(xix)

Solution 1(xix)

Question 1(xx)

Solution 1(xx)

Question 1(xxi)

Solution 1(xxi)

Question 1(xxii)

Solution 1(xxii)

Question 1(xxiii)

Solution 1(xxiii)

Question 1(xxiv)

Solution 1(xxiv)

Question 1(xxv)

Find the values of x for which the function y = [x(x – 2)]2 is increasing or decreasingSolution 1(xxv)

Question 1(xxvi)

Find the interval in which the following function is increasing or decreasing.

f(x) = 3x4– 4x3– 12x2 + 5Solution 1(xxvi)

Question 1(xxvii)

Find the interval in which the following function is increasing or decreasing.

Solution 1(xxvii)

Question 1(xxviii)

Find the interval in which the following function is increasing or decreasing.

Solution 1(xxviii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Show that the function given by f(x) = sin x is

(a) increasing in (0, π/2)

(b) decreasing in (π/2, π)

(c) neither increasing nor decreasing in (0, π)Solution 7

Question 8

Prove that the function f given by f(x) = log sin x is increasing on begin mathsize 12px style open parentheses 0 comma straight pi over 2 close parentheses end style and decreasing on begin mathsize 12px style open parentheses straight pi over 2 comma straight pi close parentheses end styleSolution 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(i)

Solution 30(i)

Question 31

begin mathsize 11px style Prove space that space the space function space straight f space given space by space straight f open parentheses straight x close parentheses space equals space log space cos space straight x space is space strictly space
increasing space open parentheses fraction numerator negative straight pi over denominator 2 end fraction comma 0 close parentheses space and space strictly space decreasing space on space open parentheses 0 comma straight pi over 2 close parentheses end style

Solution 31

Question 32

Solution 32

Question 33

Prove that the function f(x) = cos x is:

(i) strictly decreasing in (0, π)

(ii) strictly increasing in (π, 2π)

(iii) neither increasing nor decreasing in (0, 2π)Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39(i)

Find the interval in which f(x) is increasing or decreasing:

Solution 39(i)

Question 39(ii)

Find the interval in which f(x) is increasing or decreasing:

Solution 39(ii)

Question 39(iii)

Find the interval in which f(x) is increasing or decreasing:

Solution 39(iii)

Question 1(x)

Find the intervals in which the following functions are increasing or decreasing:

Solution 1(x)

Given: 

Differentiating w.r.t x, we get

Take f'(x) = 0

Clearly, f'(x) > 0 if x < -2 or x > -1

And, f'(x) < 0 if -2 < x < -1

Thus, f(x) increases on   and decreases on  Question 1(xxix)

Find the intervals in which the following functions are increasing or decreasing:

Solution 1(xxix)

Given: 

Differentiating w.r.t x, we get

Take f'(x) = 0

The points x = 2, 4 and -3 divide the number line into four disjoint intervals namely 

Consider the interval 

In this case, x – 2 < 0, x – 4 < 0 and x + 3 < 0

Therefore, f'(x) < 0 when 

Thus the function is decreasing in 

Consider the interval 

In this case, x – 2 < 0, x – 4 < 0 and x + 3 > 0

Therefore, f'(x) > 0 when 

Thus the function is increasing in 

Now, consider the interval 

In this case, x – 2 > 0, x – 4 < 0 and x + 3 > 0

Therefore, f'(x) < 0 when 

Thus the function is decreasing in 

And now, consider the interval 

In this case, x – 2 > 0, x – 4 > 0 and x + 3 > 0

Therefore, f'(x) < 0 when 

Thus the function is increasing in  Question 30(ii)

Prove that the following function is increasing on R:

Solution 30(ii)

Given: 

Differentiating w.r.t x, we get

Now, 

Hence, f(x) is an increasing function for all x.

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RD SHARMA SOLUTION CHAPTER-16 Tangents and Normals I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 16 Tangents and Normals Ex. 16.1

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)

Find the slopes of the tangent and the normal to the curve x = a(θ – sinθ), y =a(1 + cos θ) at θ = begin mathsize 12px style negative straight pi over 2 end style.Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

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

Chapter 16 Tangents and Normals Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Find the equations of the tangent and normal to the given curves at the indicated points:

y=x4 – 6x3 + 13x2 – 10x + 5 at (x = 1)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 3(vii)

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 3(ix)

Solution 3(ix)

Question 3(x)

Solution 3(x)

Question 3(xi)

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 3(xiii)

Solution 3(xiii)

Question 3(xiv)

Solution 3(xiv)

Question 3(xv)

Solution 3(xv)

Question 3(xvi)

Find the equation of the normal to curve y2 = 4x at the point (1, 2) and also find the tangent.Solution 3(xvi)

The equation of the given curve is y2 = 4x . Differentiating with respect to x, we have: 

begin mathsize 12px style 2 straight y dy over dx equals 4
rightwards double arrow dy over dx equals fraction numerator 4 over denominator 2 straight y end fraction equals 2 over straight y
therefore right enclose dy over dx end enclose subscript open parentheses 1 comma space 2 close parentheses end subscript equals 2 over 2 equals 1
Now comma space the space slope space at space point space left parenthesis 1 comma space 2 right parenthesis space is space 1 over right enclose begin display style dy over dx end style end enclose subscript open parentheses 1 comma space 2 close parentheses end subscript equals fraction numerator negative 1 over denominator 1 end fraction equals negative 1.
therefore Equation space of space the space tangent space at space left parenthesis 1 comma space 2 right parenthesis space is space straight y minus 2 space equals space minus 1 left parenthesis straight x minus 1 right parenthesis.
rightwards double arrow straight y minus 2 equals negative straight x plus 1
rightwards double arrow straight x plus straight y minus 3 equals 0
Equation space of space the space normal space is comma
straight y minus 2 equals negative left parenthesis negative 1 right parenthesis left parenthesis straight x minus 1 right parenthesis
straight y minus 2 equals straight x minus 1
straight x minus straight y plus 1 equals 0
end style

Question 3(xix)

Find the equations of the tangent and the normal to the given curves at the indicated points:

Solution 3(xix)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

From (A)

Equation of tangent is

begin mathsize 12px style open parentheses straight y minus straight a over 5 close parentheses equals 13 over 16 open parentheses straight x minus fraction numerator 2 straight a over denominator 5 end fraction close parentheses
16 straight y minus fraction numerator 16 straight a over denominator 5 end fraction equals 13 straight x minus fraction numerator 26 straight a over denominator 5 end fraction
13 straight x minus 16 straight y minus 2 straight a equals 0
Equation space of space normal space is comma
open parentheses straight y minus straight a over 5 close parentheses equals 16 over 13 open parentheses straight x minus fraction numerator 2 straight a over denominator 5 end fraction close parentheses
13 straight y minus fraction numerator 13 straight a over denominator 5 end fraction equals negative 16 straight x plus fraction numerator 32 straight a over denominator 5 end fraction
16 straight x plus 13 straight y minus 9 straight a equals 0 end style

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Find the equations of the tangent and the normal to the following curves at the indicated points:

X = 3 cosθ – cos3θ , y = 3 sinθ – sin3θSolution 5(vi)

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 21

Find the equation of the tangents to the curve 3x2 – y2 = 8, which passes through the point (4/3, 0).Solution 21

Question 3(xvii)

Find the equations of the tangent and the normal to the following curves at the indicated points:

Solution 3(xvii)

Given equation curve is 

Differentiating w.r.t x, we get

Slope of tangent at   is

Slope of normal will be

Equation of tangent at   will be

Equation of normal at   is

Question 3(xviii)

Find the equations of the tangent and the normal to the following curves at the indicated points:

Solution 3(xviii)

Given equation curve is 

Differentiating w.r.t x, we get

Slope of tangent at   is

Slope of normal will be

Equation of tangent at   will be

Equation of normal at   is

Question 20

At what points will tangents to the curve   be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.Solution 20

Given equation curve is 

Differentiating w.r.t x, we get

As tangent is parallel to x-axis, its slope will be m = 0

As this point lies on the curve, we can find y

Or

So, the points are (3, 6) and (2, 7).

Equation of tangent at (3, 6) is

y – 6 = 0 (x – 3)

y – 6 = 0

Equation of tangent at (2, 7) is

y – 7 = 0 (x – 2)

y – 7 = 0

Chapter 16 Tangents and Normals Ex. 16.3

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)

Find the angle of intersection of the folloing curves

begin mathsize 12px style straight x squared over straight a squared plus straight y squared over straight b squared equals 1 space and space straight x squared space plus space straight y squared space equals space ab end style

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1 (ix)

Find the angle of intersection of the following curves:

Y = 4 – x2 and y = x2Solution 1 (ix)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.Solution 6

Question 7

Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2)Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 9

Solution 9

Question 10

Solution 10

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RD SHARMA SOLUTION CHAPTER-15 Mean Value Theorems I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 15 Mean Value Theorems Ex 15.1

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 2(i)

Solution 2(i)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

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 3(vii)

Solution 3(vii)

Here,

f open parentheses x close parentheses equals fraction numerator sin x over denominator e to the power of x end fraction space o n space x space element of open square brackets 0 comma space straight pi close square brackets
W e space k n o w space t h a t comma space e x p o n e n t i a l space a n d space sin e space b o t h space f u n c t i o n s space a r e space c o n t i n u o u s space a n d space d i f f e r e n t i a b l e
e v e r y space w h e r e comma space s o space f open parentheses x close parentheses space i s space c o n t i n u o u s space i s space open square brackets 0 comma space straight pi close square brackets space a n d space d i f f e r e n t i a b l e space i s space open square brackets 0 comma space straight pi close square brackets

Now comma space

space space space space space space space space space space space space space straight f open parentheses 0 close parentheses equals fraction numerator sin space 0 over denominator straight e to the power of 0 end fraction equals 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 straight f open parentheses straight pi close parentheses equals fraction numerator sin space straight pi over denominator straight e to the power of straight pi end fraction equals 0

rightwards double arrow straight f open parentheses 0 close parentheses equals straight f open parentheses straight pi close parentheses
Since space Rolle apostrophe straight s space theorem space applicable comma space therefore space there space must space exist space straight a space point space straight c element of open square brackets 0 comma space straight pi close square brackets
such space that space straight f apostrophe open parentheses straight c close parentheses equals 0

Now comma
space space space space space space space space space space straight f open parentheses straight x close parentheses equals sinx over straight e to the power of straight x

space space space space space space space space space space rightwards double arrow straight f apostrophe open parentheses straight x close parentheses equals fraction numerator straight e to the power of straight x open parentheses cosx close parentheses minus straight e to the power of straight x open parentheses sinx close parentheses over denominator open parentheses straight e to the power of straight x close parentheses squared end fraction
Now comma
space space space space space space space space space space space space space space space space straight f apostrophe open parentheses straight c close parentheses equals 0
space space space space space space space space space space space rightwards double arrow straight e to the power of straight c open parentheses cosc minus sinc close parentheses equals 0
space space space space space space space space space space space rightwards double arrow space straight e to the power of straight c not equal to 0 space and space cosc minus sinc equals 0
space space space space space space space space space space space rightwards double arrow space tanc equals 1
space space space space space space space space space space space space space space space straight c equals straight pi over 4 element of open square brackets 0 comma straight pi close square brackets
Hence comma space Rolle apostrophe straight s space theorem space is space verified.

Question 3(viii)

Solution 3(viii)

Question 3(ix)

Solution 3(ix)

Question 3(x)

Solution 3(x)

Question 3(xi)

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 3(xiii)

Solution 3(xiii)

Question 3(xiv)

Solution 3(xiv)

Question 3(xv)

Solution 3(xv)

Question 3(xvi)

Solution 3(xvi)

Question 3(xvii)

Solution 3(xvii)

Question 3(xviii)

Verify Rolle’s theorem for function f(x) = sin x – sin 2x on [0, pi] on the indicated intervals.Solution 3(xviii)

Question 7

Solution 7

x = 0 then y = 16

Therefore, the point on the curve is (0, 16) Question 8(i)

Solution 8(i)

x = 0, then y = 0

Therefore, the point is (0, 0)Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

x = 1/2, then y = – 27

Therefore, the point is (1/2, – 27)Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 2(ii)

Verify Rolle’s theorem for each of the following functions on the indicated intervals:

Solution 2(ii)

Given function is 

As the given function is a polynomial, so it is continuous and differentiable everywhere.

Let’s find the extreme values

Therefore, f(2) = f(6).

So, Rolle’s theorem is applicable for f on [2, 6].

Let’s find the derivative of f(x)

Take f'(x) = 0

As 4 ∈ [2, 6] and f'(4) = 0.

Thus, Rolle’s theorem is verified.

Chapter 15 Mean Value Theorems Ex. 15.2

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 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)

Solution 1(xiv)

Question 1(xv)

Solution 1(xv)

Question 1(xvi)

Solution 1(xvi)

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

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RD SHARMA SOLUTION CHAPTER-14 Differentials, Errors and Approximations I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 14  Differentials, Errors and Approximations Ex 14.1

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

Question 8

Solution 8

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

Question 9(vi)

Solution 9(vi)

Question 9(vii)

Solution 9(vii)

Question 9(viii)

Solution 9(viii)

Question 9(ix)

Solution 9(ix)

Question 9(x)

Using differentials, find the approximate values of the following:

log1010.1, it being given that log10e = 0.4343.Solution 9(x)

Question 9(xi)

Solution 9(xi)

Question 9(xii)

Solution 9(xii)

Question 9(xiii)

Solution 9(xiii)

Question 9(xiv)

Solution 9(xiv)

Question 9(xv)

Solution 9(xv)

Question 9(xvi)

Solution 9(xvi)

Question 9(xvii)

Solution 9(xvii)

Question 9(xviii)

Solution 9(xviii)

Question 9(xix)

Solution 9(xix)

Question 9(xx)

Solution 9(xx)

Question 9(xxi)

Solution 9(xxi)

Question 9(xxii)

Solution 9(xxii)

Question 9(xxiii)

Solution 9(xxiii)

Question 9(xxiv)

Solution 9(xxiv)

Question 9(xxv)

Solution 9(xxv)

Question 9(xxvi)

Using differentials, find the approximate values of the following:

Solution 9(xxvi)

Question 9(xxvii)

Using differentials, find the approximate values of (3.968)3/2Solution 9(xxvii)

Question 9(xxviii)

Using differentials, find the approximate values of the following:

(1.999)5Solution 9(xxviii)

Question 9(xxix)

Using differentials, find the approximate values of the following:

Solution 9(xxix)

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 9(i)

Using differentials, find the approximate values of the following:

Solution 9(i)

Consider 

Let x = 25 and 

Also,

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RD SHARMA SOLUTION CHAPTER-13 Derivative as a Rate Measurer I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 13 Derivative as a Rate Measurer Ex 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Find the rate of change of the volume of a cone with respect to the radius of its base.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (Marginal revenue). If the total revenue (in rupees) received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5, find the marginal revenue, when x=5, and write which value does the question indicate.Solution 10

Chapter 13 Derivative as a Rate Measurer Ex. 13.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.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(i)

Find an angle θ, which increases twice as fast as its cosine.Solution 16(i)

Question 16(ii)

Find the angle θ

Whose rate of increase is twice the rate of decrease of its consine.Solution 16(ii)

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

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RD SHARMA SOLUTION CHAPTER-12 Higher Order Derivatives I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 12 Higher Order Derivatives Ex 12.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Find the second order derivative of log(sinx).Solution 1(iii)

L e t space y equals log left parenthesis sin x right parenthesis
D i f f e r e n t i a t i n g space w i t h space r e p e c t space t o space x comma space w e space g e t comma
fraction numerator d y over denominator d x end fraction equals fraction numerator cos x over denominator sin x end fraction
A g a i n space d i f f e r e n t i a t i n g space w i t h space r e s p e c t space t o space x comma space w e space g e t comma
fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator minus sin x cross times sin x minus cos x cross times cos x over denominator sin squared x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator minus sin squared x minus cos squared x over denominator sin squared x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator minus open parentheses sin squared x plus cos squared x close parentheses over denominator sin squared x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator minus 1 over denominator sin squared x end fraction
rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction equals minus cos e c squared x

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 1(viii)

Solution 1(viii)

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

Question 1(ix)

Solution 1(ix)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

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

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

If x equals a open parentheses theta minus sin theta close parentheses comma space y equals a open parentheses 1 plus cos theta close parentheses, find fraction numerator d squared y over denominator d x squared end fractionSolution 14

x equals a open parentheses theta minus sin theta close parentheses ; space y equals a open parentheses 1 plus cos theta close parentheses
D i i f e r e n t i a t i n g space t h e space a b o v e space f u n c t i o n s space w i t h space r e s p e c t space t o space theta comma space w e space g e t comma
fraction numerator d x over denominator d theta end fraction equals a open parentheses 1 minus cos theta close parentheses space space space... left parenthesis 1 right parenthesis
fraction numerator d y over denominator d theta end fraction equals a open parentheses minus sin theta close parentheses space space space space space space space... left parenthesis 2 right parenthesis
D i v i d i n g space e q u a t i o n space left parenthesis 2 right parenthesis space b y space left parenthesis 1 right parenthesis comma space w e space h a v e comma
fraction numerator d y over denominator d x end fraction equals fraction numerator a open parentheses minus sin theta close parentheses over denominator a open parentheses 1 minus cos theta close parentheses end fraction space equals fraction numerator minus sin theta over denominator 1 minus cos theta end fraction
D i f f e r e n t i a t i n g space w i t h space r e s p e c t space t o space theta comma space w e space h a v e comma
fraction numerator d open parentheses fraction numerator d y over denominator d x end fraction close parentheses over denominator d theta end fraction equals fraction numerator open parentheses 1 minus cos theta close parentheses open parentheses minus cos theta close parentheses plus sin theta open parentheses sin theta close parentheses over denominator open parentheses 1 minus cos theta close parentheses squared end fraction
equals fraction numerator minus cos theta plus cos squared theta plus sin squared theta over denominator open parentheses 1 minus cos theta close parentheses squared end fraction
equals fraction numerator 1 minus cos theta over denominator open parentheses 1 minus cos theta close parentheses squared end fraction
fraction numerator d open parentheses fraction numerator d y over denominator d x end fraction close parentheses over denominator d theta end fraction equals fraction numerator 1 over denominator 1 minus cos theta end fraction... left parenthesis 3 right parenthesis
D i v i d i n g space e q u a t i o n space left parenthesis 3 right parenthesis space b y space left parenthesis 1 right parenthesis comma space w e space h a v e comma
fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator 1 over denominator 1 minus cos theta end fraction cross times fraction numerator 1 over denominator a open parentheses 1 minus cos theta close parentheses end fraction
equals fraction numerator 1 over denominator a open parentheses 1 minus cos theta close parentheses squared end fraction
equals fraction numerator 1 over denominator a open parentheses 2 sin squared begin display style theta over 2 end style close parentheses squared end fraction
equals fraction numerator 1 over denominator 4 a sin to the power of 4 open parentheses theta over 2 close parentheses end fraction
equals fraction numerator 1 over denominator 4 a end fraction cos e c to the power of 4 open parentheses theta over 2 close parentheses

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

begin mathsize 12px style If space straight y equals straight e to the power of acos to the power of negative 1 end exponent straight x end exponent comma space show space that space open parentheses 1 minus straight x squared close parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0 end style

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

I f space y equals cos e c to the power of minus 1 end exponent x comma space x greater than 1 comma space t h e n space s h o w space t h a t space x open parentheses x squared minus 1 close parentheses fraction numerator d squared y over denominator d x squared end fraction plus open parentheses 2 x squared minus 1 close parentheses fraction numerator d y over denominator d x end fraction equals 0

Solution 42

W e space k n o w space t h a t comma space fraction numerator d over denominator d x end fraction open parentheses cos e c to the power of minus 1 end exponent x close parentheses equals fraction numerator minus 1 over denominator open vertical bar x close vertical bar square root of x squared minus 1 end root end fraction
L e t space y equals cos e c to the power of minus 1 end exponent x
fraction numerator d y over denominator d x end fraction equals fraction numerator minus 1 over denominator open vertical bar x close vertical bar square root of x squared minus 1 end root end fraction
S i n c e space x greater than 1 comma space open vertical bar x close vertical bar equals x
T h u s comma
fraction numerator d y over denominator d x end fraction equals fraction numerator minus 1 over denominator x square root of x squared minus 1 end root end fraction... 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 f u n c 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 squared y over denominator d x squared end fraction equals fraction numerator x begin display style fraction numerator 2 x over denominator 2 square root of x squared minus 1 end root end fraction end style plus square root of x squared minus 1 end root over denominator x squared open parentheses x squared minus 1 close parentheses end fraction
equals fraction numerator begin display style fraction numerator x squared over denominator square root of x squared minus 1 end root end fraction end style plus square root of x squared minus 1 end root over denominator x squared open parentheses x squared minus 1 close parentheses end fraction
equals fraction numerator x squared plus x squared minus 1 over denominator x squared open parentheses x squared minus 1 close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction
equals fraction numerator 2 x squared minus 1 over denominator x squared open parentheses x squared minus 1 close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction
T h u s comma space x open parentheses x squared minus 1 close parentheses fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator 2 x squared minus 1 over denominator x square root of x squared minus 1 end root end fraction... left parenthesis 2 right parenthesis
S i m i l a r l y comma space f r o m space left parenthesis 1 right parenthesis comma space w e space h a v e
open parentheses 2 x squared minus 1 close parentheses fraction numerator d y over denominator d x end fraction equals fraction numerator minus 2 x squared plus 1 over denominator x square root of x squared minus 1 end root end fraction... left parenthesis 3 right parenthesis
T h u s comma space f r o m space left parenthesis 2 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space h a v e comma
space x open parentheses x squared minus 1 close parentheses fraction numerator d squared y over denominator d x squared end fraction plus open parentheses 2 x squared minus 1 close parentheses fraction numerator d y over denominator d x end fraction equals fraction numerator 2 x squared minus 1 over denominator x square root of x squared minus 1 end root end fraction plus open parentheses fraction numerator minus 2 x squared plus 1 over denominator x square root of x squared minus 1 end root end fraction close parentheses equals 0
H e n c e space p r o v e d.

Question 43

I f space x equals cos t plus log tan t over 2 comma space y equals sin t comma space t h e n space f i n d space t h e space v a l u e space o f space fraction numerator d squared y over denominator d t squared end fraction space a n d space fraction numerator d squared y over denominator d x squared end fraction space a t space t equals straight pi over 4.

Solution 43

G i v e n space t h a t comma space x equals cos t plus log tan t over 2 comma space y equals sin t
D i f f e r e n t i a t i n g space w i t h space r e s p e c t space t o space t comma space w e space h a v e comma
fraction numerator d x over denominator d t end fraction equals minus sin t plus fraction numerator space 1 over denominator tan begin display style t over 2 end style end fraction cross times s e c squared t over 2 cross times 1 half
equals minus sin t plus fraction numerator space 1 over denominator begin display style fraction numerator sin begin display style t over 2 end style over denominator cos t over 2 end fraction end style end fraction cross times fraction numerator 1 over denominator cos squared begin display style t over 2 end style end fraction cross times 1 half
equals minus sin t plus fraction numerator space 1 over denominator begin display style fraction numerator sin begin display style t over 2 end style over denominator cos t over 2 end fraction end style end fraction cross times fraction numerator 1 over denominator cos squared begin display style t over 2 end style end fraction cross times 1 half
equals minus sin t plus fraction numerator space 1 over denominator 2 sin t over 2 cos t over 2 end fraction
equals minus sin t plus fraction numerator space 1 over denominator sin t end fraction
equals fraction numerator 1 minus sin squared t over denominator sin t end fraction
equals fraction numerator cos squared t over denominator sin t end fraction
equals cos t cross times c o t t
N o w space f i n d space t h e space v a l u e space o f space fraction numerator d y over denominator d t end fraction :
fraction numerator d y over denominator d t end fraction equals cos t
T h u s comma space fraction numerator d y over denominator d x end fraction equals fraction numerator d y over denominator d t end fraction cross times fraction numerator d t over denominator d x end fraction equals cos t cross times fraction numerator 1 over denominator cos t cross times c o t t end fraction
rightwards double arrow fraction numerator d y over denominator d x end fraction equals tan t
S i n c e space fraction numerator d y over denominator d t end fraction equals cos t comma space w e space h a v e space fraction numerator d squared y over denominator d t squared end fraction equals minus sin t
A t space t equals straight pi over 4 comma space open parentheses fraction numerator d squared y over denominator d t squared end fraction close parentheses subscript t equals straight pi over 4 end subscript equals minus sin open parentheses straight pi over 4 close parentheses equals fraction numerator minus 1 over denominator square root of 2 end fraction
fraction numerator d squared y over denominator d x squared end fraction equals fraction numerator begin display style fraction numerator d over denominator d t end fraction end style open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses over denominator begin display style fraction numerator d x over denominator d t end fraction end style end fraction
equals fraction numerator begin display style fraction numerator d over denominator d t end fraction end style open parentheses begin display style tan t end style close parentheses over denominator begin display style cos t cross times c o t t end style end fraction
equals fraction numerator s e c squared t over denominator begin display style cos t cross times c o t t end style end fraction
equals fraction numerator s e c squared t over denominator begin display style cos t cross times fraction numerator begin display style cos t end style over denominator sin t end fraction end style end fraction
equals fraction numerator s e c squared t over denominator begin display style cos squared t end style end fraction cross times sin t
equals s e c to the power of 4 t cross times sin t
T h u s comma space open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses subscript t equals straight pi over 4 end subscript equals s e c to the power of 4 open parentheses straight pi over 4 close parentheses cross times sin straight pi over 4 equals 2

Question 44

Solution 44

Question 45

Solution 45

Question 46

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 9

If   prove that   and   Solution 9

Given: 

Differentiating ‘x’ w.r.t   we get

Differentiating ‘y’ w.r.t   we get

Dividing (ii) by (i), we get

  … (iii)

Differentiating above equation w.r.t x, we get

Hence,   Question 48

If   find   Solution 48

Given: 

Differentiating ‘x’ w.r.t t, we get

Differentiating ‘y’ w.r.t t, we get

Dividing (ii) by (i), we get

Differentiating above equation w.r.t x, we get

Hence, 

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RD SHARMA SOLUTION CHAPTER-11 Differentiation I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 11 Differentiation Ex 11.1

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

Question 8

Differentiate f(x)=x2ex from first principles.Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 11 Differentiation Exercise Ex. 11.2

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

T h u s comma space fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator cos squared x end fraction plus fraction numerator sin x over denominator cos squared x end fraction
rightwards double arrow fraction numerator d y over denominator d x end fraction equals s e c squared x plus tan x s e c x
rightwards double arrow fraction numerator d y over denominator d x end fraction equals s e c x open square brackets tan x plus s e c x close square brackets

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

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

Differentiate the following functions with respect to x:

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 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

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 62

If   prove that  Solution 62

Given: 

Differentiating w.r.t x, we get

Hence,   Question 75

If   find  Solution 75

Given: 

Question 76

If   then find  Solution 76

Given: 

Chapter 11 Differentiation Exercise Ex. 11.3

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 13

Solution 13

Question 14

begin mathsize 12px style Differentiate space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus square root of 1 minus straight x squared end root over denominator square root of 2 end fraction close parentheses comma fraction numerator negative 1 over denominator square root of 2 end fraction less than straight x less than fraction numerator 1 over denominator square root of 2 end fraction end style

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

begin mathsize 12px style If space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses plus sec to the power of negative 1 end exponent open parentheses fraction numerator 1 plus straight x squared over denominator 1 minus straight x squared end fraction close parentheses comma space 0 less than straight x less than 1 comma space prove space that space dy over dx equals fraction numerator 4 over denominator 1 plus straight x squared end fraction. end style

Solution 35

Question 36

Solution 36

Question 37(i)

Solution 37(i)

Question 37(ii)

Solution 37(ii)

Question 38

show that dy/dx is independent of x.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

If y = tan-1 begin mathsize 12px style If space straight y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x end root minus space square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus space square root of 1 minus straight x end root end fraction close parentheses comma space find space dy over dx end styleSolution 45

Question 46

Solution 46

Question 47

Solution 47

Question 12

Differentiate the following function with respect to x:

Solution 12

Let 

Question 48

If   then find  Solution 48

Given:  ………. (i)

Let 

From (i), we get

Chapter 11 Differentiation Exercise Ex. 11.4

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

begin mathsize 12px style If space straight y square root of 1 minus straight x squared end root plus space straight x square root of 1 minus straight y squared end root equals 1 comma space prove space that space dy over dx equals negative square root of fraction numerator 1 minus straight y squared over denominator 1 minus straight x squared end fraction end root end style

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

begin mathsize 12px style dy over dx plus space straight e to the power of open curly brackets straight y minus straight x close curly brackets end exponent equals 0 end style.Solution 27

Question 28

Solution 28

Question 30

Solution 30

Question 31

Solution 31

Question 29

If  find   at x =1,  Solution 29

Given: 

Differentiating w.r.t x. we get

When x =1 and   we get

Chapter 11 Differentiation Exercise Ex. 11.5

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(i)

Solution 18(i)

Question 18(ii)

Solution 18(ii)

Question 18(iii)

Solution 18(iii)

Question 18(iv)

Solution 18(iv)

Question 18(v)

Solution 18(v)

Question 18(vi)

Solution 18(vi)

Question 18(vii)

Solution 18(vii)

Question 18(viii)

Solution 18(viii)

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 29(i)

Solution 29(i)

Question 29(ii)

Solution 29(ii)

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

begin mathsize 12px style If space straight x to the power of straight x plus straight y to the power of straight x equals 1 comma space prove space that space dy over dx equals negative open curly brackets fraction numerator straight x to the power of straight x open parentheses 1 plus logx close parentheses plus straight y to the power of straight x cross times space logy over denominator straight x space cross times space straight y to the power of open parentheses straight x minus 1 close parentheses end exponent end fraction close curly brackets end style

Solution 36

Question 37

begin mathsize 12px style If space straight x to the power of straight y space cross times space straight y to the power of straight x equals 1 comma space prove space that space dy over dx equals negative fraction numerator straight y open parentheses straight y plus xlogy close parentheses over denominator straight x open parentheses ylogx plus straight x close parentheses end fraction end style

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

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

begin mathsize 12px style If space straight y equals 1 plus fraction numerator straight alpha over denominator open parentheses begin display style 1 over straight x end style minus straight alpha close parentheses end fraction plus fraction numerator straight beta divided by straight x over denominator open parentheses begin display style 1 over straight x end style minus straight alpha close parentheses open parentheses begin display style 1 over straight x end style minus straight beta close parentheses end fraction plus fraction numerator straight gamma divided by straight x squared over denominator open parentheses begin display style 1 over straight x end style minus straight alpha close parentheses open parentheses begin display style 1 over straight x end style minus straight beta close parentheses open parentheses begin display style 1 over straight x end style minus straight gamma close parentheses end fraction comma space find space dy over dx. end style

Solution 61

Question 28

Find   when  Solution 28

Given: 

Let 

Differentiating ‘u’ w.r.t x, we get

Differentiating ‘v’ w.r.t x, we get

From (i), (ii) and (iii), we get

Question 62

If   find  Solution 62

Given: 

Let 

Taking log on both the sides of equation (i), we get

Taking log on both the sides of equation (ii), we get

Differentiating (iii) w.r.t x, we get

Using (iv) and (v), we have

Chapter 11 Differentiation Exercise Ex. 11.6

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

i f space y space equals space open parentheses cos x close parentheses to the power of open parentheses cos x close parentheses to the power of open parentheses cos x close parentheses to the power of negative y end exponent end exponent end exponent comma space p r o v e space t h a t space fraction numerator d y over denominator d x end fraction equals negative fraction numerator y squared tan x over denominator open parentheses 1 minus y log cos x close parentheses end fraction.

Solution 8

Chapter 11 Differentiation Exercise Ex. 11.7

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

begin mathsize 12px style If space straight x equals straight a open parentheses straight t plus 1 over straight t close parentheses space and space straight y equals straight a open parentheses straight t minus 1 over straight t close parentheses comma space prove space that space dy over dx equals straight x over straight y. end style

Solution 17

Question 18

begin mathsize 12px style If space straight x equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction close parentheses space and space straight y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction close parentheses. space minus 1 less than straight t less than 1 comma space prove space that space dy over dx equals 1. end style

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

begin mathsize 12px style Find space dy over dx comma space if space straight y equals 12 open parentheses 1 minus cost close parentheses comma straight x equals 10 open parentheses straight t minus sint close parentheses. end style

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   find   when   Solution 29

Given: 

Differentiate ‘x’ w.r.t  , we get

Differentiate ‘y’ w.r.t  , we get

Dividing (ii) by (i), we get

At 

Chapter 11 Differentiation Exercise Ex. 11.8

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

begin mathsize 12px style Differentiate space sin to the power of negative 1 end exponent square root of 1 minus straight x squared end root with space respect space to space cos to the power of negative 1 end exponent straight x comma space if
straight x space element of space open parentheses negative 1 comma space 0 close parentheses end style

Solution 4(ii)

Question 5(i)

begin mathsize 12px style Differentiate space sin to the power of negative 1 end exponent open parentheses 4 straight x square root of 1 minus 4 straight x squared end root close parentheses space space with space space respect space to space square root of 1 minus 4 straight x squared end root comma space if
straight x space element of open parentheses fraction numerator 1 over denominator negative 2 square root of 2 end fraction comma fraction numerator 1 over denominator 2 square root of 2 end fraction close parentheses end style

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

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 1

Differentiate   with respect to  Solution 1

We need to find 

Let 

So, we need to find 

Question 21

Differentiate   with respect to  Solution 21

We need to find 

Let 

Differentiating ‘u’ and ‘v’ w.r.t x, we get

Dividing (i) by (ii), we get

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