RD SHARMA SOLUTION CHAPTER- 7 Trigonometric Ratios of Compound Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Ratios of Compound Angles Exercise Ex. 7.1

Question 34

I f space tan space left parenthesis A plus B right parenthesis equals p comma space tan space left parenthesis A minus B right parenthesis equals q comma space t h e n space s h o w space t h a t space tan space 2 A equals fraction numerator p plus q over denominator 1 minus p q end fraction

Solution 34

R H S comma
fraction numerator p plus q over denominator 1 minus p q end fraction
equals fraction numerator tan left parenthesis A plus B right parenthesis plus tan thin space left parenthesis A minus B right parenthesis over denominator 1 minus tan left parenthesis A plus B right parenthesis. tan left parenthesis A minus B right parenthesis end fraction
equals fraction numerator begin display style fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction end style plus begin display style fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end style over denominator 1 minus fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction. fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end fraction
equals fraction numerator begin display style fraction numerator left parenthesis tan thin space A plus tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis plus left parenthesis tan space A minus tan space B right parenthesis left parenthesis 1 minus tan space A. space tan thin space thin space B right parenthesis over denominator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style over denominator begin display style fraction numerator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis minus left parenthesis tan space A plus tan space B right parenthesis. left parenthesis tan space A minus tan space B right parenthesis over denominator left parenthesis 1 minus tan space A. space tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style end fraction
equals fraction numerator tan space A plus tan space B plus tan squared A. tan space B plus tan space A. tan squared B plus tan space A minus tan thin space B minus tan thin space squared A. tan space B plus tan space A. tan squared B over denominator 1 minus tan squared space A. tan squared space B minus tan squared space A plus tan squared space B end fraction
equals fraction numerator 2 space tan space A plus 2 space tan space A. tan squared B over denominator left parenthesis 1 minus tan squared A right parenthesis left parenthesis 1 plus tan squared B right parenthesis end fraction equals fraction numerator 2 space tan space A left parenthesis 1 plus tan squared space B right parenthesis over denominator left parenthesis 1 minus tan squared space A right parenthesis left parenthesis 1 plus tan squared space B right parenthesis end fraction equals fraction numerator 2 space tan space A over denominator 1 minus tan squared space A end fraction equals tan space 2 A equals L H S
H e n c e space P r o v e d

Question 19

I f space fraction numerator sin space left parenthesis x plus y right parenthesis over denominator sin space left parenthesis x minus y right parenthesis end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma space s h o w space t h a t space fraction numerator tan space x over denominator tan space y end fraction equals a over b

Solution 19

fraction numerator sin space x. cos space y plus sin space y. cos space x over denominator sin space x. cos space y minus sin space y. cos space x end fraction equals fraction numerator a plus b over denominator a minus b end fraction
rightwards double arrow fraction numerator sin space x. cos space y plus sin space y. cos space x plus sin space x. cos space y minus sin space y. cos space x over denominator sin space x. cos space y plus sin space y. cos space x minus sin space x. cos space y plus sin space y. cos space x end fraction equals fraction numerator a plus b plus a minus b over denominator a plus b minus a plus b end fraction left square bracket U sin g space C o m p o n e n d o space a n d space D i v i d e n d o right square bracket
rightwards double arrow fraction numerator 2 sin space x. cos space y over denominator 2 sin space y. cos space x end fraction equals fraction numerator 2 a over denominator 2 b end fraction
rightwards double arrow fraction numerator tan space x over denominator tan space y end fraction equals a over b
H e n c e space P r o v e d

Question 32

I f space a n g l e space theta space i s space d i v i d e d space i n t o space t w o space p a r t s space s u c h space t h a t space t h e space tan g e n t s space o f space o n e space p a r t space i s space lambda space p a r t
t i m e s space t h e space tan g e n t space o f space t h e space o t h e r comma space a n d space Ï• space i s space t h e i r space d i f f e r e n c e space t h e n space s h o w space t h a t space space
sin space theta equals fraction numerator lambda plus 1 over denominator lambda minus 1 end fraction sin space Ï•.

Solution 32

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

Question 33

I f space tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction comma space t h e n space s h o w space t h a t space sin space alpha plus cos space alpha equals square root of 2 cos space theta

Solution 33

tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction
rightwards double arrow tan space theta equals space fraction numerator tan space alpha minus 1 over denominator tan space alpha plus 1 end fraction left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r space b y space cos space alpha right square bracket
rightwards double arrow tan space theta equals fraction numerator tan space alpha minus tan space begin display style pi over 4 end style over denominator 1 plus tan space begin display style pi over 4 end style. tan space alpha end fraction
rightwards double arrow tan space theta equals tan space open parentheses alpha minus pi over 4 close parentheses
rightwards double arrow theta equals alpha minus pi over 4 space space space left square bracket R e m o v i n g space tan space f r o m space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos open parentheses alpha minus pi over 4 close parentheses space left square bracket T a k i n g space cos space o n space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos space alpha. cos space pi over 4 plus sin space alpha. sin space pi over 4
rightwards double arrow cos space theta equals cos space alpha. fraction numerator 1 over denominator square root of 2 end fraction plus sin space alpha. fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow cos space theta equals fraction numerator cos space alpha plus sin space alpha over denominator square root of 2 end fraction
rightwards double arrow square root of 2 space cos space theta equals sin space alpha plus cos space alpha
H e n c e space P r o v e d

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 2(a)

Solution 2(a)

Question 2(b)

Solution 2(b)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 12(i)

Prove that:

Sin (60o – q) cos (30o + q) + cos (60o – q) sin (30o + q) = 1Solution 12(i)

Question 12(iii)

Prove that:

Solution 12(iii)

Question 12(ii)

Prove that:

Solution 12(ii)

Question 13

Solution 13

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 17(i)

Solution 17(i)

Question 17(ii)

Solution 17(ii)

Question 17(iii)

Solution 17(iii)

Question 17(iv)

Solution 17(iv)

Question 18

Solution 18

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(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 29(i)

Solution 29(i)

Question 29(ii)

Solution 29(ii)

Question 29(iii)

Solution 29(iii)

Question 30

Solution 30

Question 31

Solution 31

Chapter 17 Trigonometric Ratios of Compound Angles Exercise Ex. 7.2

Question 4

Solution 4

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 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

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RD SHARMA SOLUTION CHAPTER-6 Graphs of Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.1

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)

Sketch the graph of the following pairs of functions on the same axes:

  1. y = sin x, y = sin begin mathsize 11px style open parentheses straight x plus straight pi over 4 close parentheses end style

Solution 2(i)

Question 2(ii)

Sketch the graph of the following pairs of functions on the same axes:

ii.  y = sin x, y = sin 3xSolution 2(ii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.2

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 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.3

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

Sketch the graph of the following functions on the same scale.

y = tan x, y = tan2 xSolution 9

Question 10

Solution 10

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RD SHARMA SOLUTION CHAPTER- 5 Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Functions Exercise Ex. 5.1

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

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

Prove that cosθ(tanθ+2)(2 tanθ+1)=2 secθ+5 sinθSolution 16

Question 17

I f space x equals fraction numerator 2 space sin space theta over denominator 1 plus cos space plus sin space theta end fraction comma space t h e n space p r o v e space t h a t space fraction numerator 1 minus cos space plus sin space theta over denominator 1 plus sin space theta end fraction i s space a l s o space e q u a l space t o space x.

Solution 17

fraction numerator 2 space sin space theta over denominator 1 plus cos space theta plus sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus cos space theta plus sin space theta right parenthesis left parenthesis 1 minus cos space theta plus sin space theta right parenthesis end fraction equals x space space space left square bracket R a t i o n a l i z i n g space t h e space d e n o m i n a t o r right square bracket
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus sin space theta right parenthesis squared minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta minus 2 space sin space theta cos space theta plus 2 space sin squared theta over denominator 1 plus sin squared space theta plus 2 sin space theta minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin squared space theta plus 2 space sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin space theta left parenthesis 1 plus sin space theta right parenthesis end fraction equals x
rightwards double arrow fraction numerator 1 plus cos space theta space minus sin space theta over denominator 1 plus sin space theta end fraction equals x space left square bracket C a n c e l l i n g space t h e space 2 space sin space theta space i n space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r right square bracket
H e n c e space P r o v e d

Question 18

Solution 18

Question 19

I f space tan space theta equals a over b 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 square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root.

Solution 19

square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root
equals square root of fraction numerator begin display style a over b end style plus 1 over denominator begin display style a over b end style minus 1 end fraction end root plus square root of fraction numerator begin display style a over b end style minus 1 over denominator begin display style a over b end style plus 1 end fraction end root space left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space d e n o m i n a t o r space b y space b right square bracket
equals square root of fraction numerator tan space theta plus 1 over denominator tan space theta minus 1 end fraction end root plus square root of fraction numerator tan space theta minus 1 over denominator tan space theta plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style end fraction end root
equals square root of fraction numerator sin space theta plus cos space theta over denominator sin space theta minus cos space theta end fraction end root plus square root of fraction numerator sin space theta minus cos space theta over denominator sin space theta plus cos space theta end fraction end root
equals fraction numerator sin space theta plus cos space theta plus sin space theta minus cos space theta over denominator square root of sin squared space theta minus cos squared space theta end root end fraction
equals fraction numerator 2 space sin space theta space over denominator square root of sin squared space theta space minus space cos squared space theta end root end fraction

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Syntax error from line 1 column 702 to line 1 column 707. Unexpected '</mi>'.

Solution 24

L H S space equals space a b plus a minus b plus 1
equals left parenthesis s e c space theta minus tan space theta right parenthesis left parenthesis cos e c space plus c o t space right parenthesis plus s e c space theta minus tan space theta minus cos e c space theta minus c o t space theta plus 1
equals open parentheses fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction close parentheses open parentheses fraction numerator 1 over denominator sin space theta end fraction plus fraction numerator cos space theta over denominator sin space theta end fraction close parentheses plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator cos space theta end fraction cross times fraction numerator cos space theta over denominator sin space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction cross times fraction numerator 1 over denominator sin space theta end fraction minus tan space theta cross times c o t space theta plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator 1 over denominator cos space theta end fraction minus 1 plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 minus sin squared space theta minus cos squared space theta over denominator sin space theta. cos space theta end fraction
equals fraction numerator 1 minus left parenthesis cos squared space theta plus sin squared space theta right parenthesis over denominator sin space theta. space cos space theta end fraction
equals fraction numerator 1 minus 1 over denominator sin space theta. cos space theta end fraction equals 0 equals R H S. space H e n c e space P r o v e d

Question 25

Solution 25

Question 26(i)

Solution 26(i)

Question 26(ii)

Solution 26(ii)

Question 26(iii)

Solution 26(iii)

Chapter 5 Trigonometric Functions Exercise Ex. 5.2

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 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 5 Trigonometric Functions Exercise Ex. 5.3

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)

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 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

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 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 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

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RD SHARMA SOLUTION CHAPTER- 4 Measurement of Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 4 Measurement of Angles Exercise Ex. 4.1

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

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

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RD SHARMA SOLUTION CHAPTER-3 Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 3 Functions Exercise Ex. 3.1

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

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

Express the function f: X ® R given by f (x) = x 3 + 1 as set of ordered pairs, where X = {-1, 0, 3, 9, 7}.Solution 18

Chapter 3 Functions Exercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

begin mathsize 12px style If space straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator 1 minus straight x end fraction comma space Show space that space straight f left square bracket straight f left curly bracket straight f left parenthesis straight x right parenthesis right curly bracket right square bracket equals straight x. end style

Solution 4

Question 5

Solution 5

Question 6

begin mathsize 11px style Find space colon left parenthesis straight i right parenthesis space straight f space left parenthesis 1 divided by 2 right parenthesis space space left parenthesis ii right parenthesis space straight f left parenthesis negative 2 right parenthesis space space left parenthesis iii right parenthesis space straight f left parenthesis 1 right parenthesis space left parenthesis iv right parenthesis space straight f left parenthesis square root of 3 right parenthesis space and space left parenthesis straight v right parenthesis space straight f left parenthesis square root of negative 3 end root right parenthesis. end style

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10

Solution 10

Question 11

Solution 11

Chapter 3 Functions Exercise Ex. 3.3

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 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(viii)

find the domain and range of f open parentheses x close parentheses space equals space square root of 9 minus x squared end rootSolution 3(viii)

Question 3(vii)

Find domain and range of f (x) = -|x|Solution 3(vii)

As |x|is defined for all real numbers, its domain is R and range is only negative numbers because, |x| is always positive real number for all real numbers and -|x| is always negative real numbers.

Chapter 3 Functions Exercise Ex. 3.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 4(vii)

Solution 4(vii)

Question 4(viii)

Solution 4(viii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals 3 x minus 2
f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals minus x plus 4
f : R minus open curly brackets 3 over 2 close curly brackets rightwards arrow R space d e f i n e d space b y space f over g left parenthesis x right parenthesis equals fraction numerator x plus 1 over denominator 2 x minus 3 end fraction

Question 9

Solution 9

f plus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals square root of x plus x ; space
f minus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals square root of x minus x ;
f g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals x to the power of 3 divided by 2 end exponent ;
f over g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of x end fraction ;

Question 10

Solution 10

left parenthesis f plus g right parenthesis : R rightwards arrow left square bracket 0 comma infinity right parenthesis space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals x squared plus 2 x plus 1 equals left parenthesis x plus 1 right parenthesis squared
left parenthesis f minus g right parenthesis : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals x squared minus 2 x minus 1
open parentheses f g close parentheses : R rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals 2 x cubed plus x squared
open parentheses f over g close parentheses : R rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator x squared over denominator 2 x plus 1 end fraction
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RD SHARMA SOLUTION CHAPTER- 2 Relations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 2 Relations Exercise Ex. 2.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

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(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Chapter 2 Relations Exercise Ex. 2.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Chapter 2 Relations Exercise Ex. 2.3

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 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

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RD SHARMA SOLUTION CHAPTER- 1 Sets I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 1 Sets Exercise Ex. 1.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

If A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then insert the appropriate symbol or  in each of the following blank spaces:

  1. 4…A
  2. -4 …A
  3. 12 ….A
  4. 9 …A
  5. 0 …..A
  6. -12 ….A

Solution 3

Chapter 1 Sets Exercise Ex. 1.2

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 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

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 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 1 Sets Exercise Ex. 1.3

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

Chapter 1 Sets Exercise Ex. 1.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

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

Chapter 1 Sets Exercise Ex. 1.5

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 1 Sets Exercise Ex. 1.6

Question 2(i)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)Solution 2(i)

Question 2(ii)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)Solution 2(ii)

Question 2(iii)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B – C) = (A ∩ B) – (A ∩ C)Solution 2(iii)

Question 2(iv)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A – (B ∪ C) = (A – B) ∩ (A – C)Solution 2(iv)

Question 2(v)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A – (B ∩ C) = (A – B) ∪ (A – C)Solution 2(v)

Question 2(vi)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B D C) = (A ∩ B) D (A ∩ C)Solution 2(vi)

Question 4(i)

For any two sets A and B, prove that

B ⊂ A ∪ BSolution 4(i)

Question 4(ii)

For any two sets A and B, prove that

A ∩ B ⊂ BSolution 4(ii)

Question 4(iii)

For any two sets A and B, prove that

A ⊂ B ⇒ A ∩ B = ASolution 4(iii)

Question 14(i)

Show that For any sets A and B,

A = (A ∩ B) ∩ (A – B)Solution 14(i)

Question 14(ii)

Show that For any sets A and B,

A ∪ (B – A) = A ∪ BSolution 14(ii)

Question 15

Each set X, contains 5 elements and each set Y, contains 2 elements and  each element of S belongs to exactly 10 of the X’rs and to exactly 4 of Y’rs, then find the value of n.Solution 15

Question 1

Solution 1

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12(i)

Solution 12(i)

Question 12(ii)

Solution 12(ii)

Question 13

Solution 13

Chapter 1 Sets Exercise Ex. 1.7

Question 4(i)

For any two sets A and B, prove that

(A ∪ B) – B = A – BSolution 4(i)

Question 4(ii)

For any two sets A and B, prove that

A- (A ∩ B) = A – BSolution 4(ii)

Question 4(iii)

For any two sets A and B, prove that

A – (A – B) = A ∩ BSolution 4(iii)

Question 4(iv)

For any two sets A and B, prove that

A ∪ (B – A) = A ∪ BSolution 4(iv)

Question 4(v)

For any two sets A and B, prove that

(A – B) ∪ (A ∩ B) = ASolution 4(v)

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Chapter 1 Sets Exercise Ex. 1.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 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

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RD SHARMA SOLUTION CHAPTER-19 Indefinite Integrals I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 19 Indefinite Integrals Ex. 19.1

Question 1

Integrate each of the following functions with respect to x:

(i) ∫ x4dx

(ii) ∫ x5/4dx

(iii) begin mathsize 12px style integral 1 over straight x to the power of 5 dx end style

(iv) begin mathsize 12px style integral 1 over x to the power of 3 divided by 2 end exponent d x end style

(v) ∫ 3xdx

(vi) begin mathsize 12px style integral fraction numerator 1 over denominator square root of x squared end root end fraction d x end style

(vii) ∫ 32log3x dx

(viii) ∫ logxx dx Solution 1

Question 2

Evaluate:

begin mathsize 12px style left parenthesis i right parenthesis space integral square root of fraction numerator 1 plus cos 2 x over denominator 2 end fraction end root d x
left parenthesis i i right parenthesis space integral square root of fraction numerator 1 minus cos 2 x over denominator 2 end fraction d x end root end style

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.2

Question 1

Evaluate begin mathsize 12px style integral open parentheses 3 cross times square root of x plus 4 square root of x plus 5 close parentheses d x end styleSolution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

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

Evaluate :integral fraction numerator cos x over denominator 1 plus cos x end fraction d xSolution 42

Question 43

Evalute :

integral fraction numerator 1 minus cos x over denominator 1 plus cos x end fraction d x

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 19 Indefinite Integrals Ex. 19.3

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

Integrate begin mathsize 12px style integral sin x square root of 1 plus cos 2 x end root space d x end styleSolution 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

Evaluate the following integrals:

begin mathsize 12px style integral fraction numerator 1 over denominator cos squared x left parenthesis 1 minus tan x right parenthesis squared end fraction end style

Solution 19

Chapter 19 – Indefinite Integrals Exercise Ex. 19.4

Question 1

Solution 1

Question 2

Integratebegin mathsize 12px style integral fraction numerator x cubed over denominator x minus 2 end fraction d x end styleSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.5

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

Chapter 19 Indefinite Integral Ex. 19.6

Question 1

Solution 1

Question 2

Evalute the following integral :integral sin cubed left parenthesis 2 x plus 1 right parenthesis d xSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Integrate begin mathsize 12px style integral sin x square root of 1 minus cos 2 x end root space d x end styleSolution 8

Chapter 19 Indefinite Integrals Ex. 19.7

Question 1

Integrate ∫ sin4x cos7x dxSolution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 19 Indefinite Integrals Ex. 19.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

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

Solution 44

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

Chapter 19 – Indefinite Integrals Exercise Ex. 19.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

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

begin mathsize 12px style Evaluate colon
integral fraction numerator 4 x plus 3 over denominator square root of 2 x squared plus 3 x plus 1 end root end fraction d x 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

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

Solution 53

Question 54

begin mathsize 12px style Integrate space the space function
fraction numerator e to the power of m tan to the power of negative 1 end exponent x end exponent over denominator 1 plus x squared end fraction end style

Solution 54

L e t space tan to the power of minus 1 end exponent x equals t
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 comma space w comma space w e space h a v e comma
fraction numerator 1 over denominator 1 plus x squared end fraction d x equals d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals integral e to the power of m t end exponent cross times d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m t end exponent over m
R e s u b s t i t u t i n g space t h e space v a l u e space o f space t space i n space t h e space a b o v e space s o l u t i o n comma space w e space h a v e comma
space space space space space space rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m tan to the power of minus 1 end exponent x end exponent over m plus C

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

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Chapter 19 Indefinite Integrals Exercise Ex. 19.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 19 – Indefinite Integrals Exercise Ex. 19.11

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

begin mathsize 12px style Let space iota equals integral cos e c to the power of 4 3 x d x. space T h e n
iota equals integral cos e c squared 3 x space cos e c squared 3 d x
equals integral open parentheses 1 plus c o t squared 3 x close parentheses cos e c squared 3 x d x
equals integral open parentheses cos e c squared 3 x space plus space c o t squared 3 x space cos e c squared 3 x close parentheses d x
rightwards double arrow iota equals integral cos e c squared 3 x d x space plus space integral c o t squared 3 x space cos e c squared 3 x d x
Substituting space c o t space 3 x space equals t space a n d space cos e c t squared 3 x d x space equals negative d t space i n space 2 nd space integral comma space we space get
iota equals integral cos e c squared 3 x d x space minus space integral t squared fraction numerator d t over denominator 3 end fraction
equals fraction numerator negative 1 over denominator 3 end fraction c o t 3 x minus t cubed over 9 plus C equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x minus 1 over 9 c o t cubed space 3 x plus c
therefore iota equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x space minus space fraction numerator c o t cubed 3 x over denominator 9 end fraction plus c
end style

Question 9

Solution 9

begin mathsize 12px style L e t space iota equals integral c o t to the power of n x space cos e c squared x d x. n not equal to negative 1......... open parentheses i close parentheses
L e t space c o t space x equals t. space T h e n
d open parentheses c o t x close parentheses equals d t
rightwards double arrow negative cos e c squared x d x equals d t
rightwards double arrow cos e c squared x d x equals negative d t
P u t t i n g space c o t x space equals t space a n d space c os e c squared x d x equals negative d t space i n space e q u a t i o n space open parentheses i close parentheses comma space w e space g e t
iota equals integral t to the power of n cross times open parentheses negative d t close parentheses
equals negative fraction numerator t to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c
rightwards double arrow equals negative fraction numerator open parentheses c o t x close parentheses to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c end style

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integral:

∫ sin5x cosx dxSolution 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

Chapter 19 Indefinite Integrals Exercise Ex. 19.14

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 19 Indefinite Integrals Exercise Ex. 19.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 19 – Indefinite Integrals Exercise Ex. 19.16

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.17

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 19 Indefinite Integrals Exercise Ex. 19.18

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.19

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

Evaluate the following integrals:

Solution 10

Question 11

Solution 11

Question 12

Evaluate the following integrals:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.20

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 19 – Indefinite Integrals Exercise Ex. 19.32

Question 3

Solution 3

Question 1

Solution 1

Question 2

Solution 2

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.21

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

Integrate the function

begin mathsize 12px style fraction numerator x minus 1 over denominator square root of x squared plus 1 end root end fraction end style

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Evaluate begin mathsize 12px style integral square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root d x end styleSolution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Evaluate the following integral

integral fraction numerator 5 x plus 3 over denominator square root of x squared plus 4 x plus 10 end root end fraction d x

Solution 17

Question 18

Evaluate the following integral:

Solution 18

Chapter 19 Indefinite Integrals Exercise Ex. 19.22

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.23

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.24

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.25

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find ∫xedxSolution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Evaluate the following integral:

Solution 27

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

begin mathsize 12px style Evaluate space the space integrals colon
integral 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 dx end style

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 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

begin mathsize 12px style Evaluate space the space integrals colon
integral tan to the power of negative 1 end exponent square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root dx end style

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Chapter 19 Indefinite Integrals Exercise Ex. 19.26

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

begin mathsize 12px style Evaluagte space the space following space integrals colon
integral straight e to the power of straight x open parentheses fraction numerator straight x minus 1 over denominator 2 straight x squared end fraction close parentheses dx end style

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

Evaluate the following integral:

Solution 24

Chapter 19 Indefinite Integrals Exercise Ex. 19.27

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 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.28

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.29

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

Evaluate the following integral:

Solution 11

Question 12

Evaluate the following integral:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.30

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 15

Solution 15

Question 16

Evaluate the following integral:

Solution 16

Question 17

Solution 17

Question 18

Evaluate the following integral:

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

Evaluate the following integral:

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

Evaluate the following integral:

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Evaluate the following integral:

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

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

Chapter 19 – Indefinite Integrals Exercise Ex. 19.31

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

Evaluate the following integral:

Solution 11

Chapter 19 – Indefinite Integrals Exercise MCQ

Question 1

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space is space equal space to

open parentheses straight a close parentheses space 1 fourth tan to the power of negative 1 end exponent straight x squared space plus space straight C
open parentheses straight b close parentheses space 1 fourth tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight c close parentheses space 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight d close parentheses space none space of space these end style

Solution 1

begin mathsize 12px style Correct space option colon thin space left parenthesis straight b right parenthesis
straight I equals integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space
Put space straight x squared equals straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral fraction numerator begin display style dt over 2 end style over denominator 4 plus straight t squared end fraction
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight t over 2 close parentheses plus straight C
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses plus straight C
end style

Question 2

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator 1 over denominator cos space straight x space plus square root of 3 space sin space straight x end fraction dx space is space equal space to

open parentheses straight a close parentheses space log space tan open parentheses straight pi over 3 plus straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space log space tan open parentheses straight x over 2 minus straight pi over 3 close parentheses plus straight C
open parentheses straight c close parentheses space 1 half space log space tan open parentheses straight x over 2 plus straight pi over 3 close parentheses plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 2

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator cosx plus square root of 3 sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator begin display style cosx over 2 end style plus begin display style fraction numerator square root of 3 over denominator 2 end fraction end style sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator cos open parentheses straight x minus begin display style straight pi over 6 end style close parentheses end fraction dx
straight I equals 1 half integral sec open parentheses straight x minus straight pi over 6 close parentheses dx
straight I equals 1 half ln open vertical bar tan open parentheses straight x over 2 plus straight pi over 3 close parentheses close vertical bar plus straight C end style

Question 3

Mark the correct alternative in each of the following:

begin mathsize 12px style integral straight x space sec space straight x squared space dx space is space equal space to

open parentheses straight a close parentheses space 1 half log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight b close parentheses space straight x squared over 2 log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight c close parentheses space 2 space log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C

open parentheses straight d close parentheses space none space of space these
end style

Solution 3

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral xsecx squared dx
Put space straight x squared equals straight t rightwards double arrow straight x equals square root of straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral sect dt over 2
straight I equals 1 half log open vertical bar sect plus tant close vertical bar plus straight C
straight I equals 1 half log open vertical bar sec straight x squared plus tan straight x squared close vertical bar plus straight C
end style

Question 4

begin mathsize 12px style If space integral fraction numerator 1 over denominator 5 plus 4 space sin space straight x end fraction dx equals straight A space tan to the power of negative 1 end exponent open parentheses straight B space tan straight x over 2 plus 4 over 3 close parentheses plus straight C comma space then

open parentheses straight a close parentheses space straight A equals 2 over 3 comma space straight B equals 5 over 3
open parentheses straight b close parentheses space straight A equals 1 third comma space straight B equals 2 over 3
open parentheses straight c close parentheses space straight A equals negative 2 over 3 comma space straight B equals 5 over 3
open parentheses straight d close parentheses space straight A equals 1 third comma space straight B equals negative 5 over 3 end style

Solution 4

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx equals Atan to the power of negative 1 end exponent open parentheses Btan straight x over 2 plus 4 over 3 close parentheses plus straight C
Put space tan straight x over 2 equals straight t rightwards double arrow straight x equals 2 tan to the power of negative 1 end exponent straight t
rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow sinx equals fraction numerator 2 tan begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx
equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 5 plus 4 cross times fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
equals integral fraction numerator 2 dt over denominator 5 straight t squared plus 8 straight t plus 5 end fraction
equals 2 over 5 integral fraction numerator dt over denominator straight t squared plus begin display style 8 over 5 end style straight t plus 1 end fraction
Using space completing space square space method
we space get
straight I equals 2 over 3 tan to the power of negative 1 end exponent open parentheses 5 over 3 tan straight x over 2 plus 4 over 3 close parentheses plus straight C
straight A equals 2 over 3 space and space straight B equals 5 over 3
end style

Question 5

begin mathsize 12px style integral straight x to the power of sin space straight x end exponent open parentheses fraction numerator sin space straight x over denominator straight x end fraction plus cos space straight x. log space straight x close parentheses dx space is space equal space to

open parentheses straight a close parentheses space straight x to the power of sin space straight x end exponent plus straight C

open parentheses straight b close parentheses space straight x to the power of sin space straight x end exponent space cos space straight x plus straight C
open parentheses straight c close parentheses space open parentheses straight x to the power of sin space straight x end exponent close parentheses squared over 2 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 5

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral straight x to the power of sinx open parentheses sinx over straight x plus cosxlogx close parentheses dx
Put space straight x to the power of sinx equals straight t
taking space log space on space both space sides comma
logt equals sinxlogx
1 over straight t dt equals sinx over straight x plus cosxlogx
rightwards double arrow straight I equals integral straight t cross times dt over straight t
straight I equals straight t plus straight C
straight I equals straight x to the power of sinx plus straight C
end style

Question 6

begin mathsize 12px style Integration space of space fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction with space respect space to space log subscript straight e straight x space is

open parentheses straight a close parentheses space fraction numerator tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses over denominator straight x end fraction plus straight C
open parentheses straight b close parentheses space tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x end fraction plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 6

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
integral fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction straight d open parentheses log subscript straight e straight x close parentheses
Put space log subscript straight e straight x equals straight t
integral fraction numerator dt over denominator 1 plus straight t squared end fraction equals tan to the power of negative 1 end exponent straight t plus straight C equals tan to the power of negative 1 end exponent open parentheses space log subscript straight e straight x close parentheses plus straight C
end style

Question 7

begin mathsize 12px style If space integral fraction numerator cos space 8 straight x plus 1 over denominator tan space 2 straight x space minus space cot space 2 straight x end fraction dx equals straight a space cos space 8 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 over 16
open parentheses straight b close parentheses space 1 over 8
open parentheses straight c close parentheses space 1 over 16
open parentheses straight d close parentheses space minus 1 over 8 end style

Solution 7

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
integral fraction numerator cos 8 straight x plus 1 over denominator tan 2 straight x minus cot 2 straight x end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator begin display style fraction numerator sin 2 straight x over denominator cos 2 straight x end fraction end style minus begin display style fraction numerator cos 2 straight x over denominator sin 2 straight x end fraction end style end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator sin squared 2 straight x minus cos squared 2 straight x end fraction cross times sin 2 xcos 2 xdx
equals integral negative fraction numerator cos squared 4 xsin 4 straight x over denominator cos 4 straight x end fraction dx
equals fraction numerator negative 1 over denominator 2 end fraction integral sin 8 xdx
equals fraction numerator cos 8 straight x over denominator 16 end fraction plus straight C
straight a equals 1 over 16 end style

Question 8

begin mathsize 12px style If space integral fraction numerator sin to the power of 8 space straight x space minus space cos to the power of 8 space straight x over denominator 1 minus 2 space sin squared space straight x space cos squared space straight x end fraction dx equals straight a space sin space 2 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 divided by 2

open parentheses straight b close parentheses space 1 divided by 2

open parentheses straight c close parentheses space minus 1

open parentheses straight d close parentheses space 1 end style

Solution 8

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral fraction numerator sin to the power of 8 straight x minus cos to the power of 8 straight x over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin to the power of 4 straight x minus cos to the power of 4 straight x close parentheses over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x plus cos squared straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin squared straight x plus cos squared straight x close parentheses squared minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses end fraction dx
integral negative cos 2 xdx
fraction numerator negative sin 2 straight x over denominator 2 end fraction plus straight C
rightwards double arrow straight a equals fraction numerator negative 1 over denominator 2 end fraction end style

Question 9

begin mathsize 11px style integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent space dx space is space equal space to

open parentheses straight a close parentheses space minus xe to the power of straight x plus straight C
open parentheses straight b close parentheses space xe to the power of straight x space plus space straight C
open parentheses straight c close parentheses space minus xe to the power of negative straight x end exponent plus straight C
open parentheses straight d close parentheses space xe to the power of negative straight x end exponent plus straight C
end style

Solution 9

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent dx
equals open parentheses straight x minus 1 close parentheses integral straight e to the power of negative straight x end exponent dx minus integral open parentheses open square brackets fraction numerator straight d open parentheses straight x minus 1 close parentheses over denominator dx end fraction close square brackets integral straight e to the power of negative straight x end exponent dx close parentheses dx
equals open parentheses straight x minus 1 close parentheses fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx
equals negative open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent plus fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction plus straight C
equals negative xe to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent minus straight e to the power of negative straight x end exponent plus straight C
equals negative xe to the power of negative straight x end exponent plus straight C
end style

Question 10

begin mathsize 11px style If space integral 2 to the power of 1 divided by straight x end exponent over straight x squared dx space equals space straight k space 2 to the power of 1 divided by straight x end exponent plus straight C comma space then space straight k space is space equal space to
open parentheses straight a close parentheses space minus fraction numerator 1 over denominator log subscript straight e 2 end fraction
open parentheses straight b close parentheses space minus log subscript straight e 2
open parentheses straight c close parentheses space minus 1
open parentheses straight d close parentheses space 1 half end style

Solution 10

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral 2 to the power of begin display style 1 over straight x end style end exponent over straight x squared dx
Put space 1 over straight x equals straight t
rightwards double arrow fraction numerator negative 1 over denominator straight x squared end fraction dx equals dt rightwards double arrow 1 over straight x squared dx equals negative dt
straight I equals integral 2 to the power of straight t open parentheses negative dt close parentheses
straight I equals fraction numerator negative 2 to the power of straight t over denominator log subscript straight e 2 end fraction plus straight C
rightwards double arrow straight I equals fraction numerator negative 2 to the power of begin display style 1 over straight x end style end exponent over denominator log subscript straight e 2 end fraction plus straight C
straight k equals fraction numerator negative 1 over denominator log subscript straight e 2 end fraction
end style

Question 11

begin mathsize 11px style integral fraction numerator 1 over denominator 1 plus tan space straight x end fraction dx equals

open parentheses straight a close parentheses space log subscript straight e open parentheses straight x plus sin space straight x close parentheses plus straight C
open parentheses straight b close parentheses space log subscript straight e open parentheses sin space straight x plus cos space straight x close parentheses plus straight C
open parentheses straight c close parentheses space 2 space sec squared straight x over 2 plus straight C
open parentheses straight d close parentheses space 1 half open curly brackets straight x space plus space log open parentheses sin space straight x space plus space cos space straight x close parentheses close curly brackets space plus space straight C end style

Solution 11

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator 1 over denominator 1 plus tanx end fraction dx
equals integral fraction numerator cosx over denominator sinx plus cosx end fraction dx
Numerator space can space be space written space as colon
cosx equals straight A open parentheses sinx plus cosx close parentheses plus straight B fraction numerator straight d open parentheses sinx plus cosx close parentheses over denominator dx end fraction
cosx equals open parentheses straight A minus straight B close parentheses sinx plus open parentheses straight A plus straight B close parentheses cosx
rightwards double arrow straight A minus straight B equals 0 space and space straight A plus straight B equals 1
rightwards double arrow straight A equals 1 half equals straight B
straight I equals integral fraction numerator open square brackets begin display style 1 half end style open parentheses sinx plus cosx close parentheses plus 1 half open parentheses cosx minus sinx close parentheses close square brackets dx over denominator sinx plus cosx end fraction
straight I equals 1 half integral open parentheses 1 plus fraction numerator cosx minus sinx over denominator sinx plus cosx end fraction close parentheses dx
straight I equals 1 half open square brackets 1 plus ln open parentheses sinx plus cosx close parentheses close square brackets plus straight C end style

Question 12

begin mathsize 12px style integral open vertical bar straight x close vertical bar cubed space dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator negative straight x to the power of 4 over denominator 4 end fraction plus straight C
open parentheses straight b close parentheses space open vertical bar straight x close vertical bar to the power of 4 over 4 plus straight C
open parentheses straight c close parentheses space straight x to the power of 4 over 4 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 12

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
integral open vertical bar straight x close vertical bar cubed dx
If space straight x greater than 0
rightwards double arrow integral straight x cubed dx
equals straight x to the power of 4 over 4 plus straight C
If space straight x less than 0
rightwards double arrow integral negative straight x cubed dx
equals negative straight x to the power of 4 over 4 plus straight C end style

Question 13

begin mathsize 12px style The space value space of space integral fraction numerator cos space square root of straight x over denominator square root of straight x end fraction dx space is
open parentheses straight a close parentheses space 2 space cos square root of straight x plus straight C
open parentheses straight b close parentheses space square root of fraction numerator cos space straight x over denominator straight x end fraction end root plus straight C
open parentheses straight c close parentheses space sin space square root of straight x plus straight C
open parentheses straight d close parentheses space 2 space sin space square root of straight x plus space straight C end style

Solution 13

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator cos square root of straight x over denominator square root of straight x end fraction dx
Put comma space
square root of straight x space equals straight t
fraction numerator 1 over denominator 2 square root of straight x end fraction dx equals dt
rightwards double arrow fraction numerator 1 over denominator square root of straight x end fraction dx equals 2 dt
straight I equals integral cost space 2 dt
straight I equals 2 sint plus straight C equals 2 sin square root of straight x plus straight C end style

Question 14

begin mathsize 12px style integral straight e to the power of straight x open parentheses 1 minus cot space straight x space plus space cot squared straight x close parentheses space dx equals

open parentheses straight a close parentheses space straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight c close parentheses space straight e to the power of straight x space cosec space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight e to the power of straight x space cosec space straight x space plus space straight C end style

Solution 14

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
straight I equals integral straight e to the power of straight x open parentheses 1 minus cotx plus cot squared straight x close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses 1 plus cot squared straight x minus cotx close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses cosec squared straight x minus cotx close parentheses dx
Here comma space straight f left parenthesis straight x right parenthesis equals negative cotx rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals cosec squared straight x
straight I equals negative straight e to the power of straight x cotx plus straight C end style

Question 15

Error converting from MathML to accessible text.

Solution 15

begin mathsize 12px style Correct space option colon thin space left parenthesis straight b right parenthesis
straight I equals integral fraction numerator sin to the power of 6 xdx over denominator cos to the power of 8 straight x end fraction
straight I equals integral tan to the power of 6 xsec squared xdx
Put space tanx equals straight t rightwards double arrow sec squared xdx equals dt
straight I equals integral straight t to the power of 6 dt
straight I equals straight t to the power of 7 over 7 plus straight C
straight I equals fraction numerator tan to the power of 7 straight x over denominator 7 end fraction plus straight C end style

Question 16

begin mathsize 12px style integral fraction numerator 1 over denominator 7 plus 5 cos space straight x end fraction dx equals

open parentheses straight a close parentheses space fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 6 end fraction tan straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 7 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C end style

Solution 16

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator dx over denominator 7 plus 5 cosx end fraction
put space tan straight x over 2 equals straight t rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow cosx equals fraction numerator 1 minus tan squared begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
straight I equals integral fraction numerator fraction numerator 2 dt over denominator 1 plus straight t squared end fraction over denominator 7 plus 5 cross times fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end fraction
rightwards double arrow straight I equals integral fraction numerator 1 over denominator straight t squared plus 6 end fraction dt
straight I equals fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator square root of 6 end fraction close parentheses plus straight C
rightwards double arrow fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tan begin display style straight x over 2 end style over denominator square root of 6 end fraction close parentheses plus straight C
end style

Question 17

begin mathsize 12px style integral fraction numerator 1 over denominator 1 minus cosx minus sinx end fraction dx equals
open parentheses straight a close parentheses space log space open vertical bar 1 plus cot straight x over 2 close vertical bar plus straight C
open parentheses straight b close parentheses space log space open vertical bar 1 minus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight c close parentheses space log space open vertical bar 1 minus cot fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight d close parentheses space log space open vertical bar 1 plus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C end style

Solution 17

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
integral fraction numerator dx over denominator 1 minus cosx minus sinx end fraction
Put space straight t equals space tan straight x over 2 rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
cosx equals fraction numerator 1 minus tan squared straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
sinx equals fraction numerator 2 tan straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
put space in space the space question
straight I equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 1 minus fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction minus fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t plus begin display style 1 fourth end style minus begin display style 1 fourth end style end fraction
rightwards double arrow straight I equals ln open vertical bar 1 minus cot straight x over 2 close vertical bar plus straight C

end style

Question 18

begin mathsize 12px style integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx equals
open parentheses straight a close parentheses space fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style straight x plus 3 end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C end style

Solution 18

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator straight x plus 4 minus 1 over denominator open parentheses straight x plus 4 close parentheses squared end fraction close parentheses straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator 1 over denominator straight x plus 4 end fraction minus 1 over open parentheses straight x plus 4 close parentheses squared close parentheses straight e to the power of straight x dx
straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator straight x plus 4 end fraction rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals negative 1 over open parentheses straight x plus 4 close parentheses squared
rightwards double arrow straight I equals fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
end style

Question 19

begin mathsize 12px style integral fraction numerator sin space straight x over denominator 3 space plus space 4 space cos squared straight x end fraction dx
open parentheses straight a close parentheses space log space open parentheses 3 plus 4 cos squared straight x close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator cos begin display style space end style begin display style straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight c close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C end style

Solution 19

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator sinx over denominator 3 plus 4 cos squared straight x end fraction dx
Put space cosx equals straight t
rightwards double arrow negative sinxdx equals dt
rightwards double arrow sinxdx equals negative dt
straight I equals integral fraction numerator negative dt over denominator 3 plus 4 straight t squared end fraction
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight t over denominator square root of 3 end fraction close parentheses plus straight C
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 cosx over denominator square root of 3 end fraction close parentheses plus straight C end style

Question 20

begin mathsize 12px style integral straight e to the power of straight x open parentheses fraction numerator 1 minus sin space straight x over denominator 1 minus cos space straight x end fraction close parentheses dx
open parentheses straight a close parentheses space minus straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x over 2 plus straight C
open parentheses straight c close parentheses space minus 1 half space straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight a close parentheses space minus 1 half space straight e to the power of straight x space cot space straight x over 2 plus straight C end style

Solution 20

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 minus sinx over denominator 1 minus cosx end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 over denominator 1 minus cosx end fraction minus fraction numerator sinx over denominator 1 minus cosx end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 over denominator 2 sin squared begin display style straight x over 2 end style end fraction minus fraction numerator 2 sin begin display style straight x over 2 end style cos straight x over 2 over denominator 2 sin squared straight x over 2 end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator cosec squared straight x over 2 over denominator 2 end fraction minus cot straight x over 2 close parentheses dx
straight f left parenthesis straight x right parenthesis equals negative cot straight x over 2 rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals fraction numerator cosec squared straight x over 2 over denominator 2 end fraction
straight I equals negative straight e to the power of straight x cot straight x over 2 plus straight C end style

Question 21

begin mathsize 12px style integral 2 over open parentheses straight e to the power of straight x space plus space straight e to the power of negative straight x end exponent close parentheses squared dx
open parentheses straight a close parentheses space fraction numerator negative straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction plus straight C
open parentheses straight b close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style negative 1 end style over denominator begin display style open parentheses straight e to the power of straight x plus space 1 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C end style

Solution 21

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Question 22

begin mathsize 12px style integral fraction numerator straight e to the power of straight x open parentheses 1 plus straight x close parentheses over denominator cos squared open parentheses xe to the power of straight x close parentheses end fraction dx equals

open parentheses straight a close parentheses space 2 space log subscript straight e cos open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight b close parentheses space sec open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight c close parentheses space tan open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight d close parentheses space tan open parentheses straight x plus straight e to the power of straight x close parentheses plus straight C end style

Solution 22

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator straight e to the power of straight x open parentheses 1 plus straight x close parentheses over denominator cos squared open parentheses xe to the power of straight x close parentheses end fraction dx
Put space xe to the power of straight x equals straight t
rightwards double arrow straight e to the power of straight x open parentheses 1 plus straight x close parentheses dx equals dt
straight I equals integral fraction numerator dt over denominator cos squared straight t end fraction
straight I equals integral sec squared tdt
straight I equals tant plus straight C
straight I equals tan open parentheses xe to the power of straight x close parentheses plus straight C end style

Question 23

begin mathsize 12px style integral fraction numerator sin squared straight x over denominator cos to the power of 4 straight x end fraction dx equals

open parentheses straight a close parentheses space 1 third space tan squared straight x space plus space straight C
open parentheses straight b close parentheses space 1 half space tan squared straight x space plus space straight C
open parentheses straight c close parentheses space 1 third space tan cubed straight x space plus space straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 23

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator sin squared straight x over denominator cos to the power of 4 straight x end fraction dx
straight I equals integral tan squared xsec squared xdx
Put space tanx equals straight t rightwards double arrow sec squared xdx equals dt
straight I equals integral straight t squared dt
straight I equals straight t cubed over 3 plus straight C
straight I equals fraction numerator tan cubed straight x over denominator 3 end fraction plus straight C end style

Question 24

begin mathsize 12px style The space primitive space of space the space function space straight f left parenthesis straight x right parenthesis space equals space open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent comma space straight a space greater than space 0 space is
open parentheses straight a close parentheses space fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction
open parentheses straight b close parentheses space log subscript straight e straight a times straight a to the power of straight x plus 1 over straight x end exponent
open parentheses straight c close parentheses space straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over straight x log subscript straight e straight a
open parentheses straight d close parentheses space straight x fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction end style

Solution 24

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight f left parenthesis straight x right parenthesis equals open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent
rightwards double arrow integral straight f left parenthesis straight x right parenthesis dx equals integral open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent dx
Put space straight x plus 1 over straight x equals straight t
rightwards double arrow open parentheses 1 minus 1 over straight x squared close parentheses dx equals dt
straight I equals integral straight a to the power of straight t dt
straight I equals fraction numerator straight a to the power of straight t over denominator log subscript straight e straight a end fraction plus straight C
straight I equals fraction numerator straight a to the power of straight x plus 1 over straight x end exponent over denominator log subscript straight e straight a end fraction plus straight C end style

Question 25

begin mathsize 12px style The space value space of space integral fraction numerator 1 over denominator straight x plus straight x space log space straight x end fraction dx space is
open parentheses straight a close parentheses space 1 space plus space log space straight x
open parentheses straight b close parentheses space straight x space plus space log space straight x
open parentheses straight c close parentheses space straight x space log space open parentheses 1 space plus space log space straight x close parentheses
open parentheses straight d close parentheses space log open parentheses 1 space plus space log space straight x close parentheses end style

Solution 25

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator 1 over denominator straight x plus xlogx end fraction dx
straight I equals integral fraction numerator dx over denominator straight x open parentheses 1 plus logx close parentheses end fraction
Put space 1 plus logx equals straight t
rightwards double arrow 1 over straight x dx equals dt
straight I equals integral 1 over straight t dt
rightwards double arrow straight I equals log open vertical bar straight t close vertical bar plus straight C
straight I equals log open parentheses 1 plus logx close parentheses plus straight C
end style

Question 26

begin mathsize 12px style integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx space is space equal space to
open parentheses straight a close parentheses space sin to the power of negative 1 end exponent square root of straight x plus straight C
open parentheses straight b close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight c close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight d close parentheses space sin to the power of negative 1 end exponent square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root plus straight C end style

Solution 26

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction cross times straight x over straight x end root dx
straight I equals integral fraction numerator xdx over denominator square root of straight x minus straight x squared end root end fraction
Consider comma
straight x equals straight A fraction numerator straight d open parentheses straight x minus straight x squared close parentheses over denominator dx end fraction plus straight B
straight x equals straight A open parentheses 1 minus 2 straight x close parentheses plus straight B
straight x equals negative 2 Ax plus straight A plus straight B
minus 2 straight A equals 1 rightwards double arrow straight A equals fraction numerator negative 1 over denominator 2 end fraction
rightwards double arrow straight A plus straight B equals 0 rightwards double arrow straight B equals 1 half
straight I equals integral fraction numerator fraction numerator negative 1 over denominator 2 end fraction open parentheses 1 minus 2 straight x close parentheses plus 1 half over denominator square root of straight x minus straight x squared end root end fraction dx
straight I equals integral open parentheses fraction numerator negative 1 over denominator 2 end fraction fraction numerator 1 minus 2 straight x over denominator square root of straight x minus straight x squared end root end fraction plus fraction numerator 1 over denominator 2 square root of straight x minus straight x squared end root end fraction close parentheses dx
straight I equals fraction numerator negative 1 over denominator 2 end fraction cross times 2 square root of straight x minus straight x squared end root plus 1 half integral fraction numerator 1 over denominator square root of straight x minus straight x squared end root end fraction dx
Second space term space after space completing space square space method space you space will space get space as
straight I equals negative square root of straight x minus straight x squared end root plus sin to the power of negative 1 end exponent square root of straight x plus straight C end style

Question 27

begin mathsize 12px style integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses space plus space straight f apostrophe open parentheses straight x close parentheses close curly brackets dx space equals

open parentheses straight a close parentheses space straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight b close parentheses space straight e to the power of straight x space plus space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space straight e to the power of straight x space minus space straight f open parentheses straight x close parentheses space plus space straight C end style

Solution 27

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses plus straight f apostrophe open parentheses straight x close parentheses close curly brackets dx equals straight e to the power of straight x straight f open parentheses straight x close parentheses plus straight C
end style

Question 28

begin mathsize 12px style The space value space of space integral fraction numerator sin space straight x space plus space cos space straight x over denominator square root of 1 minus sin space 2 straight x end root end fraction dx space is space equal space to
open parentheses straight a close parentheses space square root of sin space 2 straight x space end root plus space straight C
open parentheses straight b close parentheses space square root of cos 2 straight x end root space plus space straight C
open parentheses straight c close parentheses space plus-or-minus open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space plus-or-minus space log space open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C end style

Solution 28

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator sinx plus cosx over denominator square root of 1 minus sin 2 straight x end root end fraction dx
straight I equals integral fraction numerator sinx plus cosx over denominator cosx minus sinx end fraction dx
Put space cosx minus sinx equals straight t rightwards double arrow open parentheses sinx plus cosx close parentheses dx equals plus-or-minus dt
rightwards double arrow open parentheses sinx plus cosx close parentheses dx equals plus-or-minus dt
straight I equals plus-or-minus integral dt over straight t
straight I equals plus-or-minus log open vertical bar straight t close vertical bar plus straight C
straight I equals plus-or-minus log open vertical bar sinx minus cosx close vertical bar plus straight C
end style

Question 29

begin mathsize 12px style If space integral straight x space sin space straight x space dx space equals space minus straight x space cos space straight x space plus space straight alpha comma space then space straight alpha space is space equal space to

open parentheses straight a close parentheses space sin space straight x space plus space straight C
open parentheses straight b close parentheses space cos space straight x space plus space straight C
open parentheses straight c close parentheses space straight C
open parentheses straight d close parentheses space none space fo space these end style

Solution 29

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral xsinxdx equals negative xcosx plus straight alpha
straight I equals integral xsinxdx
straight I equals straight x integral sinxdx minus integral open parentheses fraction numerator d straight x over denominator d straight x end fraction integral sinxdx close parentheses dx
straight I equals negative xcosx plus integral cosxdx
straight I equals xcosx plus sinx plus straight C
rightwards double arrow straight alpha equals sinx plus straight C
end style

Question 30

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space 1 over denominator cos space 2 straight x space plus space 1 end fraction dx equals

open parentheses straight a close parentheses space tan space straight x space minus space straight x space plus space straight C
open parentheses straight b close parentheses space straight x space plus space tan space straight x space plus space straight C
open parentheses straight c close parentheses space straight x space minus space tan space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight x space minus space cot space straight x space plus space straight C end style

Solution 30

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus 1 over denominator cos 2 straight x plus 1 end fraction dx
straight I equals negative integral fraction numerator 1 minus cos 2 straight x over denominator 1 plus cos 2 straight x end fraction dx
straight I equals negative integral fraction numerator 2 sin squared straight x over denominator 2 cos squared straight x over 2 end fraction dx
straight I equals negative integral tan squared xdx
straight I equals negative integral open parentheses sec squared straight x minus 1 close parentheses dx
straight I equals negative open parentheses tanx minus straight x close parentheses plus straight C
straight I equals straight x minus tanx plus straight C end style

Question 31

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space cos space 2 straight theta over denominator cos space straight x space minus space cos space straight theta end fraction dx space is space equal space to

open parentheses straight a close parentheses space 2 open parentheses sin space straight x space plus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight b close parentheses space 2 open parentheses sin space straight x space minus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 open parentheses sin space straight x space plus space 2 straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight d close parentheses space 2 open parentheses sin space straight x space minus space 2 straight x space cos space straight theta close parentheses space plus space straight C end style

Solution 31

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus cos 2 straight theta over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 cos squared straight x minus 1 minus open parentheses 2 cos squared straight theta minus 1 close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 open parentheses cos squared straight x minus cos squared straight theta close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals 2 integral open parentheses cosx plus cosθ close parentheses dx
straight I equals 2 open parentheses sinx plus xcosθ close parentheses plus straight C end style

Question 32

begin mathsize 12px style integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator 1 over denominator 5 straight x end fraction open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 5 end style end fraction open parentheses 4 plus fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 straight x end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
end style

Solution 32

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx
straight I equals integral fraction numerator straight x to the power of 9 over denominator straight x to the power of 12 open parentheses 4 plus begin display style 1 over straight x squared end style close parentheses to the power of 6 end fraction dx
straight I equals integral fraction numerator dx over denominator straight x cubed open parentheses 4 plus 1 over straight x squared close parentheses to the power of 6 end fraction
Put space open parentheses 4 plus 1 over straight x squared close parentheses equals straight t rightwards double arrow fraction numerator negative 2 over denominator straight x cubed end fraction dx equals dt
straight I equals 1 half integral negative straight t to the power of negative 6 end exponent dt
straight I equals 1 half open parentheses fraction numerator negative straight t to the power of negative 5 end exponent over denominator negative 5 end fraction close parentheses plus straight C
straight I equals 1 over 10 1 over straight t to the power of 5 plus straight C
straight I equals 1 over 10 open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C end style

Question 33

begin mathsize 12px style integral fraction numerator straight x cubed over denominator square root of 1 plus straight x squared end root end fraction dx equals straight a open parentheses 1 plus straight x squared close parentheses to the power of 3 divided by 2 end exponent space plus space straight b square root of 1 plus straight x squared end root plus straight C comma space then

open parentheses straight a close parentheses space straight a space equals space 1 third comma space straight b space equals space 1
open parentheses straight b close parentheses space straight a space equals space minus fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space 1
open parentheses straight c close parentheses space straight a space equals negative fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1
open parentheses straight d close parentheses space straight a space equals space fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1 end style

Solution 33

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator straight x cubed over denominator square root of 1 plus straight x squared end root end fraction dx
1 plus straight x squared equals straight t rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral fraction numerator straight x squared over denominator square root of 1 plus straight x squared end root end fraction xdx
straight I equals integral fraction numerator straight t minus 1 over denominator square root of straight t end fraction dt over 2
straight I equals 1 half integral open parentheses square root of straight t minus straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent close parentheses dt
straight I equals 1 half open parentheses 2 over 3 straight t to the power of 3 over 2 end exponent minus 2 square root of straight t close parentheses plus straight C
straight I equals 1 third open parentheses 1 plus straight x squared close parentheses to the power of 3 over 2 end exponent minus square root of 1 plus straight x squared end root plus straight C
straight a equals 1 third comma space straight b equals negative 1 end style

Question 34

begin mathsize 12px style integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx space is space equal space to

open parentheses straight a close parentheses space straight x space plus space straight x squared over 2 plus straight x cubed over 3 minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight b close parentheses space straight x space plus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight c close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
open parentheses straight d close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction plus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
end style

Solution 34

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator straight x cubed plus 1 minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator open parentheses straight x plus 1 close parentheses open parentheses straight x squared minus straight x plus 1 close parentheses minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral open parentheses straight x squared minus straight x plus 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx
straight I equals straight x cubed over 3 minus straight x squared over 2 plus straight x minus log open vertical bar straight x plus 1 close vertical bar plus straight C end style

Question 35

begin mathsize 12px style If space integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx space equals space straight a space log space open vertical bar 1 plus straight x squared close vertical bar space plus space straight b space tan to the power of negative 1 end exponent straight x space plus space 1 fifth log open vertical bar straight x space plus space 2 close vertical bar space plus space straight C comma space then

open parentheses straight a close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight b close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight c close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space 2 over 5
open parentheses straight d close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space 2 over 5 end style

Solution 35

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx
Consider comma
fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction equals fraction numerator straight A over denominator straight x plus 2 end fraction plus fraction numerator Bx plus straight C over denominator open parentheses straight x squared plus 1 close parentheses end fraction
1 equals straight A open parentheses straight x squared plus 1 close parentheses plus open parentheses Bx plus straight C close parentheses open parentheses straight x plus 2 close parentheses
Comaring space coefficeints space and space solving space it space simultaneously space we space get
straight A equals 1 fifth comma space straight B equals negative 1 fifth comma space straight C equals 2 over 5
straight I equals integral open parentheses fraction numerator 1 over denominator 5 straight x plus 2 end fraction plus fraction numerator begin display style fraction numerator negative 1 over denominator 5 end fraction end style straight x plus begin display style 2 over 5 end style over denominator straight x squared plus 1 end fraction close parentheses dx
Integrating space we space get space as
1 fifth log open vertical bar straight x plus 2 close vertical bar minus 1 over 10 log open vertical bar straight x squared plus 1 close vertical bar plus 2 over 5 tan to the power of negative 1 end exponent straight x plus straight C
rightwards double arrow straight a equals negative 1 over 10 comma space straight b equals 2 over 5 end style
begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx
Consider comma
fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction equals fraction numerator straight A over denominator straight x plus 2 end fraction plus fraction numerator Bx plus straight C over denominator open parentheses straight x squared plus 1 close parentheses end fraction
1 equals straight A open parentheses straight x squared plus 1 close parentheses plus open parentheses Bx plus straight C close parentheses open parentheses straight x plus 2 close parentheses
Comaring space coefficeints space and space solving space it space simultaneously space we space get
straight A equals 1 fifth comma space straight B equals negative 1 fifth comma space straight C equals 2 over 5
straight I equals integral open parentheses fraction numerator 1 over denominator 5 straight x plus 2 end fraction plus fraction numerator begin display style fraction numerator negative 1 over denominator 5 end fraction end style straight x plus begin display style 2 over 5 end style over denominator straight x squared plus 1 end fraction close parentheses dx
Integrating space we space get space as
1 fifth log open vertical bar straight x plus 2 close vertical bar minus 1 over 10 log open vertical bar straight x squared plus 1 close vertical bar plus 2 over 5 tan to the power of negative 1 end exponent straight x plus straight C
rightwards double arrow straight a equals negative 1 over 10 comma space straight b equals 2 over 5 end style

Chapter 19 – Indefinite Integrals Exercise Ex. 19VSAQ

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

Write the value of ∫ e2x2 + In x dxSolution 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

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

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RD SHARMA SOLUTION CHAPTER- 33 Binomial Distribution I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 33 Binomial Distribution Exercise Ex. 33.1

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

Required Probability =4547 over 8192Question 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

Also, find the mean and variance of this distribution.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

                                = 0.0256Question 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

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.Solution 50

Question 51

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.Solution 51

Question 52

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?Solution 52

Question 53

A factory produces bulbs. The probability that one bulb is defective is   and they are packed in boxes of 10. From a single box, find the probability that

i. none of the bulbs is defective.

ii. exactly two bulls are defective.

iii. more than 8 bulbs work properly.Solution 53

Note: Answer given in the book is incorrect.

Chapter 33 Binomial Distribution Exercise Ex. 33.2

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

From a lot of 15 bulbs which include 5 defective, sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.Solution 22

Out of 15 bulbs 5 are defective.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Hence ,  the   probability   that   the   drawn   bulb   is   defective   is end text end cell row cell text P end text left parenthesis text Defective end text right parenthesis equals 5 over 15 equals 1 third end cell row cell text P end text left parenthesis text Not   defective end text right parenthesis equals 10 over 15 equals 2 over 3 end cell row cell text Let   X   denote   the   number   of   defective   bulbs   out   of   4. end text end cell row cell text Then ,  X   follows   binomial   distribution   with   end text end cell row cell straight n equals 4 comma text   end text straight p equals 1 third text   and   end text straight q equals 2 over 3 text   such   that end text end cell row cell straight P left parenthesis straight X equals straight r right parenthesis equals straight C presuperscript 4 subscript straight r open parentheses 1 third close parentheses to the power of straight r open parentheses 2 over 3 close parentheses to the power of 4 minus straight r end exponent semicolon straight r equals 0 comma 1 comma 2 comma 3 comma 4 end cell row cell text Mean end text equals sum from straight r equals 0 to 4 of rP left parenthesis straight r right parenthesis equals 1 cross times straight C presuperscript 4 subscript 1 open parentheses 1 third close parentheses open parentheses 2 over 3 close parentheses cubed plus 2 cross times straight C presuperscript 4 subscript 2 open parentheses 1 third close parentheses squared open parentheses 2 over 3 close parentheses squared end cell row cell plus 3 cross times straight C presuperscript 4 subscript 3 open parentheses 1 third close parentheses cubed open parentheses 2 over 3 close parentheses plus 4 cross times straight C presuperscript 4 subscript 4 open parentheses 1 third close parentheses to the power of 4 open parentheses 2 over 3 close parentheses to the power of 0 end cell row cell equals 32 over 81 plus 48 over 81 plus 24 over 81 plus 4 over 81 equals 108 over 81 equals 4 over 3 end cell end table end style

Question 23

A die is thrown three times. Let X be’ the number of twos seen’. Find the expectation of X.Solution 23

Question 24

A die is thrown twice. A ‘success’ is getting an even number on a toss. Find the variance of number of successes.Solution 24

Question 25

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of the number spades. Hence, find the mean of the distribution.Solution 25

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RD SHARMA SOLUTION CHAPTER- 32 Mean and Variance of a Random Variable I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.1

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

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

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, find the probability distribution of X.Solution 28

Question 29

The probability distribution of a random variable X is given below:

(i) Determine the value of k

(ii) Determine P (X  2) and P b(X > 2)

(iii) Find P (X  2) + P(X > 2)Solution 29

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.2

Question 1(i)

Find the mean and standard deviation of each of the following probability distributions:

xi : 2 3 4

pi : 2.2 0.5 0.3Solution 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)

Find the mean and standard deviation of each of the following probability distributions:

Solution 1(ix)

Question 2

A discrete random variable X has the probability distribution given below:

X : 0.5 1 1.5 2

P(X) : k k2 2k2 k

(i) Find the value of k.

(ii) Determine the mean of the distribution.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

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.Solution 17

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text apostrophe straight X apostrophe text    end text be text    end text the text    end text random text    end text variable text    end text which text    end text can text    end text assume end cell row cell values text    end text from text    end text 0 text    end text to text    end text 3. end cell row cell straight P left parenthesis straight X equals 0 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell straight P left parenthesis straight X equals 1 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 1 cross times to the power of 26 straight C subscript 2 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 2 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 2 cross times to the power of 26 straight C subscript 1 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 3 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell Probability text    end text distribution text    end text of text    end text straight X colon end cell row cell table row cell straight X equals straight x subscript straight i end cell 0 1 2 3 row cell straight p left parenthesis straight X equals straight x subscript straight i right parenthesis end cell cell 2 over 17 end cell cell 13 over 34 end cell cell 13 over 34 end cell cell 2 over 17 end cell end table end cell row blank row cell Mean equals sum from straight i equals 0 to 3 of left parenthesis straight x subscript straight i cross times straight p subscript straight i right parenthesis end cell row cell equals straight x subscript 0 straight p subscript 0 plus straight x subscript 1 straight p subscript 1 plus straight x subscript 2 straight p subscript 2 plus straight x subscript 3 straight p subscript 3 end cell row cell equals 0 cross times 2 over 17 plus 1 cross times 13 over 34 plus 2 cross times 13 over 34 plus 3 cross times 2 over 17 end cell row cell equals fraction numerator 13 plus 26 plus 12 over denominator 34 end fraction end cell row cell equals 51 over 34 end cell row cell equals 3 over 2 end cell row cell equals 1.5 end cell end table end style

Question 18

An urn contains 5 are 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.Solution 18

Question 19

Two numbers are selected at random (without replacement) from positive integers 2,3,4,5, 6 and 7. Let X denote the larger of the two number obtained. Find the mean and variance of the probability distribution of X.Solution 19

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