The value of $$\int _{ 3/2 }^{ 2 }{ { \left( \cfrac { x-1 }{ 3-x } \right) }^{ 1/2 } } dx$$ is

$$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 6 } $$

$$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 8 } $$

$$\cfrac { \sqrt { 3 } }{ 2 } +1-\cfrac { \pi }{ 6 } $$

The value of integral $$\int _{ a }^{ b }{ \frac { \left| x \right| }{ x } dx,\quad a<b }$$ is

$$\left| a \right| -\left| b \right|$$

$$\left| b \right| -\left| a \right|$$

Let $$ 1 _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { n } } } d x $$ where $$ n > 2 , $$ then

$$ I _ { n } > \frac { \pi } { 6 } $$

$$ \mathrm { I } _ { \mathrm { n } } $$

$$ I _ { n } > \frac { 1 } { 2 } $$

If $${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int } } (lnx)^{ m }dx,$$ where $$m\epsilon N,$$then $${ I }_{ 10 }+10{ I }_{ 9 }$$ is equal to-

$$\frac { { e }^{ 10 } }{ 10 } $$

MCQ Questions for Class 12 Commerce Maths Integrals Quiz 11

What is

\int_{1}^{3} | 1 - x^{4} | d x

$\int _ { 1 } ^ { 3 } \left| 1 - x ^ { 4 } \right| d x$ equal to

?

$?$

0%
$- 232 / 5$
0%
$- 116 / 5$
0%
$116 / 5$
0%
$232 / 5$

The value of

\int_{3 / 2}^{2} {(\frac{x - 1}{3 - x})}^{1 / 2} d x

$\int _{ 3/2 }^{ 2 }{ { \left( \cfrac { x-1 }{ 3-x } \right) }^{ 1/2 } } dx$ is

0%
$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 6 }$
0%
$\cfrac { \sqrt { 3 } }{ 2 } -1+\cfrac { \pi }{ 8 }$
0%
$\cfrac { \sqrt { 3 } }{ 2 } +1-\cfrac { \pi }{ 6 }$
0%
None of these

The value of integral

\int_{a}^{b} \frac{| x |}{x} d x, a < b

$\int _{ a }^{ b }{ \frac { \left| x \right| }{ x } dx,\quad a<b }$ is

0%
$\left| a \right| -\left| b \right|$
0%
$\left| b \right| -\left| a \right|$
0%
$\left| a \right| -b$
0%
$\left| b \right| -a$

The value of the definite integral

\int_{0}^{1} (1 + e^{- x^{2}}) d x

$\int _0^1 (1+e^{-x^2})dx$ is

0%
$-1$
0%
$2$
0%
$1+e^{-1}$
0%
None of the above

Let

J = \int_{0}^{\infty} \frac{I n x}{1 + x^{3}} d x

$J=\int _{ 0 }^{ \infty }{ \dfrac { Inx }{ 1+{ x }^{ 3 } } } dx$ and

K = \int_{0}^{\infty} \frac{x I n x}{1 + x^{3}} d x

$K=\int _{ 0 }^{ \infty }{ \dfrac { xInx }{ 1+{ x }^{ 3 } } } dx$ then

0%
$J+K=0$
0%
$J-K=0$
0%
$J+K<0$
0%
$J+K>0$

The integral

\int_{π / 6}^{π / 3} \frac{f (x)}{f (x) + f (\frac{π}{2} - x)} d x

$\int^{\pi/3}_{\pi/6} \dfrac{f(x)}{f(x) +f(\dfrac{\pi}{2} -x)} dx$ is equal to

0%
$\dfrac{\pi}{12}$
0%
$\dfrac{12}{\pi}$
0%
$\dfrac{\pi}{13}$
0%
$\dfrac{13}{\pi}$

a > 0

$a > 0$ and

A = \int_{0}^{a} \cos^{- 1} x d x

$A=\displaystyle \int^{a}_{0}\cos^{-1}xdx$ , then

\int_{- a}^{a} (\cos^{- 1} x - \sin^{1} \sqrt{1 - x^{2}}) d x = π a - λ A

$\int ^{a}_{-a}(\cos^{-1}x-\sin^{1}\sqrt {1-x^{2}})dx=\pi a-\lambda A$ . Then

λ

$\lambda$

0%
$0$
0%
$3$
0%
$2$
0%
$none\ of\ these$

\int_{π / 4}^{π / 2} \sqrt{2 + \sqrt{2 + 2 \cos 4 x}} d x

$\displaystyle \int_{\pi/4}^{\pi/2}{\sqrt{2+\sqrt{2+2\cos 4x}}dx}$ is equal to

0%
$\sqrt{2}$
0%
$\sqrt{2(\sqrt{2-1})}$
0%
$2$
0%
$none\ of\ these$

\int {(1 + x^{4})}^{\frac{1}{4}} d x = A I n (\frac{{(1 + x^{4})}^{\frac{1}{4}} + x}{{(1 + x^{4})}^{\frac{1}{4}} - x}) + B \tan^{- 1} (\frac{{(1 + x^{4})}^{\frac{1}{4}}}{x}) + C, t h e n A + B =

$\int {{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}dx = A\;In\;\left( {\dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}} + x}}{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}} - x}}} \right) + B\;\tan {\;^{ - 1}}\left( {\dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}}}{x}} \right) + C,\;then\;A + B = }$

0%
1/4
0%
1/2
0%
$- \dfrac{1}{4}$
0%
$- \dfrac{1}{2}$

\int_{4}^{5} e^{(x + 5)^{2}} d x + 3 \int_{1 / 3}^{2 / 3} e^{9 {(x \frac{2}{3})}^{2}} d x

$\displaystyle \int_{4}^{5}e^{(x + 5)^2} dx + 3\int_{1/3}^{2/3}e^{9\left ( x \dfrac{2}{3} \right )^2}dx$ is equal to

0%
1
0%
-1
0%
0
0%
2

Let

1_{n} = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{n}}} d x

$1 _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { n } } } d x$ where

n > 2,

$n > 2 ,$ then

0%
$I _ { n } < \frac { \pi } { 6 }$
0%
$I _ { n } > \frac { \pi } { 6 }$
0%
$\mathrm { I } _ { \mathrm { n } } < \frac { 1 } { 2 }$
0%
$I _ { n } > \frac { 1 } { 2 }$

\int_{0}^{b - c} f (x + c) d x = k \int_{b}^{c} f (x) d x

$\int _{ 0 }^{ b-c }{ f\left( x+c \right) dx=k\int _{ b }^{ c }{ f\left( x \right) dx } }$ then k=

0%
0
0%
1
0%
2
0%
-1

l_{1} - \int_{0}^{1} 2^{x^{2}}, l_{2} - \int_{0}^{1} 2^{x^{3}} d x, l_{3} - \int_{1}^{2} 2^{x^{2}} d x

$l_1-\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 2 } } } ,{ l }_{ 2 }-\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 3 } } } dx,{ l }_{ 3 }-\int _{ 1 }^{ 2 }{ { 2 }^{ { x }^{ 2 } } } dx$ and

l_{4} - \int_{1}^{2} 2^{x^{3}} d x

$l_4-\int _{ 1 }^{ 2 }{ { 2 }^{ { x }^{ 3 } } } dx$ then

0%
$l_2 > l_1$
0%
$l_1 > l_2$
0%
$l_3-l_2$
0%
None of these

Find the value of the equation :

\int_{\ln λ}^{\ln (\frac{1}{λ})} \frac{f (\frac{x^{2}}{3}) (f (x) + f (- x))}{g (3 x^{2}) (g (x) - g (- x))} d x = ?

$\int _ { \ln \lambda } ^ { \ln \left( \frac { 1 } { \lambda } \right) } \dfrac { f \left( \dfrac { x ^ { 2 } } { 3 } \right) ( f ( x ) + f ( - x ) ) } { g \left( 3 x ^ { 2 } \right) ( g ( x ) - g ( - x ) ) } d x =?$ where

λ > 1

$\lambda > 1$

0%
$0$
0%
$1$
0%
$\lambda$
0%
$1 / \lambda$

Find the value of the equation

\int_{0}^{\infty} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{3}} = ?

$\int _ { 0 } ^ { \infty } \dfrac { d x } { \left( x + \sqrt { x ^ { 2 } + 1 } \right) ^ { 3 } } =?$

0%
$3 / 8$
0%
$1/8$
0%
$-3/8$
0%
$-1/8$

7 (\int_{π}^{0} \frac{x^{4} (1 - x)^{4} d x}{1 + x^{2}} + π)

$7 \left( \int_{\pi}^{0}\dfrac{x^{4}(1-x)^{4} dx}{1+x^{2}}+\pi \right)$ is equal to

0%
$21$
0%
$22$
0%
$23$
0%
$none\ of\ these$

\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x

$\int _{ 0 }^{ 1 }{ \sqrt { \cfrac { 1-x }{ 1+x } } dx }$ , is equal to:

0%
$\cfrac { \pi } { 4 } - 1$
0%
$\cfrac { \pi } { 2 } - 1$
0%
$\frac { \pi } { 4 } + 1$
0%
$\cfrac { \pi } { 2 } + 1$

\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x,

$\int _ { 0 } ^ { 1 } \sqrt { \cfrac { 1 - x } { 1 + x } } d x ,$ is equal to

0%
$\cfrac { \pi } { 4 } - 1$
0%
$\cfrac { \pi } { 2 } - 1$
0%
$\cfrac { \pi } { 4 } + 1$
0%
$\cfrac { \pi } { 2 } + 1$

\frac{1}{π} \int_{- 2}^{2} \frac{1}{4 + x^{2}} d x =

$\frac { 1 }{ \pi } \int _{ -2 }^{ 2 }{ \frac { 1 }{ 4+{ x }^{ 2 } } dx= }$

0%
$0$
0%
$\frac { 1 }{ 4 }$
0%
$\frac { \pi }{ 4 }$
0%
$\frac { 1 }{ 2 }$

\int_{0}^{8} [\sqrt{t}] d t

$\int _{ 0 }^{ 8 }{ \left[ \sqrt { t } \right] } dt$ at equals to (where [.] greatest integer function.)

0%
28
0%
11
0%
2
0%
8

f (x) = \int_{0}^{1} (x f (t) + 1) d t, t h e n \int_{0}^{3} f (x) d x = 12

$f\left( x \right) =\int _{ 0 }^{ 1 }{ \left( xf\left( t \right) +1 \right) dt,then\int _{ 0 }^{ 3 }{ f\left( x \right) dx=12 } }$
because
Statement-2: f(x) = 3x + 1

0%
Statements-1 is true, statements-2 is true and statements-2 is correct explanation for statement-1.
0%
Statements-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statements-1.
0%
Statements-1 is true, statements-2 is false.
0%
Statements-1 is false, statements-2 is true.

I_{m} = \overset{e}{\int_{1}} (l n x)^{m} d x,

${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int } } (lnx)^{ m }dx,$ where

m ϵ N,

$m\epsilon N,$ then

I_{10} + 10 I_{9}

${ I }_{ 10 }+10{ I }_{ 9 }$ is equal to-

0%
${ e }^{ 10 }$
0%
$\frac { { e }^{ 10 } }{ 10 }$
0%
e
0%
e-1

If for every integer n,

\int_{n}^{n + 1} f (x) d x = n^{2}

$\int _{ n }^{ n+1 }{ f(x)dx={ n }^{ 2 } }$ , then the value of

\int_{- 2}^{4} f (x) d x

$\int _{ -2 }^{ 4 }{ f(x)dx }$ is -

0%
16
0%
14
0%
19
0%
None of these.

E = \int_{R}^{\infty} \frac{G M m}{x^{2}}

$\displaystyle E = \int_{R}^{\infty}\dfrac{GMm}{x^2}$ dx , (where

G

$G$ ,

M

$M$ ,

m

$m$ are constants ) equal to

0%
$-\dfrac{GMm}{R^2}$
0%
$+\dfrac{GMm}{R^2}$
0%
$-\dfrac{GMm}{R}$
0%
$+\dfrac{GMm}{R}$

\int_{0}^{4036} \frac{2^{x}}{2^{x} + 1^{4036 - x}} d x = . . . . . . . . . . . .

$\int _{ 0 }^{ 4036 }{ \dfrac { { 2 }^{ x } }{ { 2 }^{ x }+{ 1 }^{ 4036-x } } } dx=............$

0%
2018
0%
4035
0%
2017
0%
-2015

I_{1} = \int_{x}^{1} \frac{1}{1 + t^{2}} d t

${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$ and

I_{2} = \int_{1}^{1 / x} \frac{1}{1 + t^{2}} d t

${ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$ for x > 0, then

0%
${ I }_{ 1 }={ I }_{ 2 }$
0%
${ I }_{ 1 }>{ I }_{ 2 }$
0%
${ I }_{ 2 }>{ I }_{ 1 }$
0%
None of these.

The value of the integral

\int_{\frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} d x

$\int _{ \dfrac { 1 }{ 3 } }^{ 1 }{ \dfrac { \left( x-{ x }^{ 3 } \right) ^{ \dfrac { 1 }{ 3 } } }{ { x }^{ 4 } } }dx$ is

0%
6
0%
0
0%
3
0%
4

\overset{- 5}{\int_{- 4}} e^{(} x + 5)^{2} d x + 3 \overset{2 / 3}{\int_{1 / 3}} e^{9} (9 (x - 2 / 3)^{2}

$\overset { -5 }{ \underset { -4 }{ \int } } e^(x + 5)^2 dx + 3 \overset { 2/3}{ \underset { 1/3 }{ \int } } e^9(9(x-2/3)^2$ dx is equal toi

0%
$e^5$
0%
$e^4$
0%
$3e^2$
0%
0

z = x + 3 i

$z=x+3i$ then value of

\int_{2}^{4} [a r g | \frac{z - i}{z + i} |] d x

$\displaystyle\int^4_2\left[arg\left|\dfrac{z-i}{z+i}\right|\right]dx$ , where

[.]

$[.]$ denotes the greatest integer function, is?

0%
$3\sqrt{2}$
0%
$6\sqrt{3}$
0%
$\sqrt{6}$
0%
None

The area of the region bounded by the lines

x = 1, x = 2

$x = 1, x = 2$ , and the curves

x (y - e^{x}) = \sin x

$x(y - e^x) = \sin x$ and

2 x y = 2 \sin x + x^{3}

$2xy = 2 \sin x + x^3$ is

0%
$e^2 - e - \dfrac{1}{6}$
0%
$e^2 - e - \dfrac{7}{6}$
0%
$e^2 - e + \dfrac{1}{6}$
0%
$e^2 - e + \dfrac{7}{6}$

Explanation

Given:

x (y - e^{x}) = \sin x

$x\left(y-{e}^{x}\right)=\sin{x}$

\Rightarrow x y - x e^{x} = \sin x

$\Rightarrow\,xy-x{e}^{x}=\sin{x}$

\Rightarrow x y = \sin x + x e^{x}

$\Rightarrow\,xy=\sin{x}+x{e}^{x}$

\Rightarrow y = \frac{\sin x}{x} + e^{x}

$\Rightarrow\,y=\dfrac{\sin{x}}{x}+{e}^{x}$

Given:

2 x y = 2 \sin x + x^{3}

$2xy=2\sin{x}+{x}^{3}$

\Rightarrow y = \frac{\sin x}{x} + \frac{x^{2}}{2}

$\Rightarrow\,y=\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}$

A = \int_{1}^{2} {\frac{\sin x}{x} + e^{x} - (\frac{\sin x}{x} + \frac{x^{2}}{2})} d x

$A=\displaystyle\int_{1}^{2}{\left\{\dfrac{\sin{x}}{x}+{e}^{x}-\left(\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}\right)\right\}dx}$

= \int_{1}^{2} (e^{x} - \frac{x^{2}}{2}) d x

$=\displaystyle\int_{1}^{2}{\left({e}^{x}-\dfrac{{x}^{2}}{2}\right)dx}$

= {[e^{x} - \frac{x^{3}}{6}]}_{1}^{2}

$=\left[{e}^{x}-\dfrac{{x}^{3}}{6}\right]_{1}^{2}$

= e^{2} - \frac{8}{6} - (e - \frac{1}{6})

$={e}^{2}-\dfrac{8}{6}-\left(e-\dfrac{1}{6}\right)$

= e^{2} - e - \frac{7}{6}

$={e}^{2}-e-\dfrac{7}{6}$

Suppose

I_{1} = \int_{0}^{π / 2} \cos (π \sin^{2} x) d x; I_{2} = \int_{0}^{π / 2} \cos (2 π \sin^{2} x) d x

$I_1=\displaystyle \int_{0}^{\pi/2} \cos(\pi \sin^2 x)dx;I_2=\displaystyle \int_{0}^{\pi/2} \cos(2\pi \sin^2x)dx$ and

I_{3} = \int_{0}^{π / 2} \cos (π \sin x) d x

$I_3=\displaystyle \int_{0}^{\pi/2} \cos(\pi \sin x)dx$ then

0%
$I_1=0$
0%
$I_2+I_3=0$
0%
$I_1+I_2+I_3=0$
0%
$I_2=I_3$

Explanation

We have

I_{1} = \int_{0}^{π / 2} c o s (π s i n^{2} x) d x

$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi sin^2x)dx$

using

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

$\int_a^b f(x) \ dx=\int_a^b f(a+b-x) \ dx$

I_{1} = \int_{0}^{π / 2} c o s (π \sin^{2} (\frac{π}{2} - x)) d x

$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi \sin^2(\dfrac {\pi}2- x))dx$

I_{1} = \int_{0}^{π / 2} c o s (π c o s^{2} x) d x

$I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi cos^2x)dx$

On adding,

2 I_{1} = \int_{0}^{π / 2} c o s (π s i n^{2} x) + c o s (π c o s^{2} x) d x

$2I_1=\displaystyle \int_{0}^{\pi/2} cos(\pi sin^2x)+cos(\pi cos^2x )dx$

= \int_{0}^{π / 2} 2 c o s (\frac{π}{2}) . c o s (\frac{π}{2} c o s 2 x) d x

$= \displaystyle \int_{0}^{\pi/2} 2cos\left( \dfrac{\pi}{2}\right).cos\left( \dfrac{\pi}{2}cos2x\right)dx$

\Rightarrow I_{1} = 0

$\Rightarrow I_1=0$

Now,

I_{2} = \int_{0}^{π / 2} \cos (2 π \sin^{2} x) d x

$I_2=\displaystyle \int_{0}^{\pi/2} \cos(2\pi \sin^2x)dx$

I_{2} = \int_{0}^{π / 2} c o s {π (1 - \cos 2 x)} d x

$I_2=\displaystyle \int_{0}^{\pi/2} cos\left\{ \pi(1-\cos2x)\right\}dx$

= - \int_{0}^{π / 2} \cos (π \cos 2 x) d x

$=-\displaystyle \int_{0}^{\pi/2} \cos(\pi \cos2x)dx$

= - \frac{1}{2} \int_{0}^{π / 4} c o s (π c o s t) d t (p u t 2 x = t)

$=-\dfrac{1}{2}\displaystyle \int_{0}^{\pi/4} cos(\pi cos t)dt \qquad (put 2x=t)$

= - \frac{2}{2} \int_{0}^{π / 2} c o s (π c o s t) d t = - I_{3}

$=-\dfrac{2}{2} \displaystyle \int_{0}^{\pi/2} cos(\pi cost)dt =-I_3$

\Rightarrow I_{2} + I_{3} = 0

$\Rightarrow I_2+I_3=0$

I_{3} = - \int_{0}^{π / 2} c o s (π s i n t) d t

$I_3=-\displaystyle \int_{0}^{\pi/2} cos(\pi sint)dt$

∴ I_{2} + I_{3} = 0

$\therefore I_2+I_3=0$

Hence,

I_{1} + I_{2} + I_{3} = 0

$I_1+I_2+I_3=0$

\int \frac{\cos x d x}{\sin^{3} x (1 + \sin^{6} x)^{2 / 3}} = f (x) (1 + \sin^{6} x)^{1 / α} + c

$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c$
Where

c

$c$ is a constant of integration, then

λ f (\frac{π}{3})

$\lambda f\left(\dfrac{\pi}{3}\right)$ is equal to:

0%
$\dfrac{9}{8}$
0%
$-2$
0%
$2$
0%
$-\dfrac{9}{8}$

Explanation

\int \frac{\cos x d x}{\sin^{3} x (1 + \sin^{6} x)^{2 / 3}} = f (x) (1 + \sin^{6} x)^{1 / α} + c . . . . . g i v e n

$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c.....given$

Let

\sin x = t

$\sin x = t$

\cot x d x = d t

$\cot x dx= dt$

I = \int \frac{d t}{t^{3} (1 + t^{6})^{2 / 3}}

$\displaystyle I= \int \dfrac{dt}{t^3(1+t^6)^{2/3}}$

I = \int \frac{d t}{t^{3} . t^{4} {(1 + \frac{1}{t^{6}})}^{2 / 3}}

$\displaystyle I=\int \dfrac{dt}{t^3.t^4\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$

I = \int \frac{d t}{t^{7} {(1 + \frac{1}{t^{6}})}^{2 / 3}}

$\displaystyle I=\int \dfrac{dt}{t^7\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$

Put

1 + \frac{1}{t^{6}} = r^{3}

$1+\dfrac{1}{t^6}=r^3$

\Rightarrow \frac{d t}{t^{7}} = \frac{- 1}{2} r^{2} d r

$\Rightarrow \dfrac{dt}{t^7} = \dfrac{-1}{2}r^2dr$

= \frac{- 1}{2} \int \frac{r^{2} d r}{r^{2}}

$= \dfrac{-1}{2}\displaystyle\int \dfrac{r^2dr}{r^2}$

= \frac{- 1}{2} r + c

$=\dfrac{-1}{2}r +c$

= \frac{- 1}{2} {(\frac{\sin^{6} x + 1}{\sin^{6} x})}^{1 / 3} + c . . . . . . .

$= \dfrac{-1}{2}\left(\dfrac{\sin^6x + 1}{\sin^6x}\right)^{1/3}+c .......$ As

r = {(1 + \frac{1}{t^{6}})}^{1 / 3}

$r=\left (1+\dfrac{1}{t^6}\right )^{1/3}$ and

t = \sin x

$t=\sin x$

= \frac{- 1}{2 \sin^{2} x} (1 + \sin^{6} x)^{1 / 3} + c

$=\dfrac{-1}{2\sin^2x} (1 + \sin^6x)^{1/3}+c$

f (x) = - \frac{1}{2} c o s e c^{2} x

$f(x) = -\dfrac{1}{2} cosec^{2}x$ and

λ = 3

$\lambda = 3$

Now for

x = \frac{π}{3}

$x=\dfrac{\pi}3$

c o s e c \frac{π}{3} = \frac{2}{\sqrt{3}}

$cosec \dfrac{\pi}3=\dfrac{2}{\sqrt3}$

λ f (\frac{π}{3}) = \frac{- 1}{2} \times {(\frac{2}{\sqrt{3}})}^{2} \times 3

$\lambda f \left(\dfrac{\pi}{3}\right) = \dfrac{-1}2\times \left (\dfrac{2}{\sqrt3}\right )^2\times 3$

λ f (\frac{π}{3}) = - 2 . . . . . A n s w e r

$\boxed{\lambda f \left(\dfrac{\pi}{3}\right) = -2}.....Answer$

Hence option

^{'} B^{'}

$'B'$ is the answer.

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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