MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 12

$f(x)=\begin{vmatrix} \cos { x } & 1 & 0 \\ 1 & 2\cos { x } & 1 \\ 0 & 1 & 2\cos { x } \end{vmatrix}$ then

$\displaystyle\int _{ 0 }^{ \pi /2 }{ f\left( x \right) } dx$ is equal to

0%
$1/4$
0%
$-1/3$
0%
$1/2$
0%
$none of these$

The value of

$\int_{0}^{[x]} (x-[x])dx$ , where

$[x]$ is the greatest integer

$|le x$ is equal to

0%
$4[x]$
0%
$2[x]$
0%
$\dfrac{1}{2} [x]$
0%
$\dfrac{1}{5} [x]$

The value of

$\int_{1}^{-1}\dfrac{dx}{(2-x)\sqrt{1-x^{2}}}$ is

0%
$0$
0%
$\dfrac{\pi}{\sqrt{3}}$
0%
$\dfrac{2\pi}{3\sqrt{3}}$
0%
cannot be evaluated

The value of

$\displaystyle \int _{0}^{\infty}\dfrac{\sqrt{x^2+1}}{(x+\sqrt{x^{2}+1})^{n+1}} .dx \ \forall \ n\ \in \ N- \{ \pm 1 \}$ is

0%
$0$
0%
$n(n^{2}-1)$
0%
$\dfrac{n}{(n^{2}-1)}$
0%
$n^{2}$

$\displaystyle \int_{-3\pi}^{3\pi}{\sin^{2}\theta\sin^{2} 2\theta d \theta}$ is equal to-

0%
$\pi$
0%
$\dfrac{3\pi}{2}$
0%
$\dfrac{5\pi}{2}$
0%
$6\pi$

Let

$f\left(x\right)+f\left(\dfrac{1}{x}\right)=F\left(x\right)$ where

$f\left(x\right)=\displaystyle\int_{1}^{x}{\dfrac{\ln{t}}{1+t}dx}$ .Then

$F\left(e\right)=$

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

Evaluate:

$\displaystyle\int_{-2}^{3}{\left|1-{x}^{2}\right|dx}$

0%
$\dfrac{28}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{5}{3}$

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

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$0$
0%
$2$

$\displaystyle \overset{2\pi}{\underset{0}{\int}} x \,log \left(\dfrac{3 + \cos x}{3 - \cos x}\right)dx$

0%
$\dfrac{\pi}{2} \,log \,3$
0%
$\dfrac{\pi}{12} \,log \,3$
0%
$\dfrac{\pi}{3} \,log \,3$
0%
$0$

$\int^2_0 (\sqrt{\dfrac{4-x}{x}} - \sqrt{\dfrac{x}{4-x}})dx$ is equal to

0%
$0$
0%
$8$
0%
$4$
0%
$16$

$2\displaystyle \int_{0}^{1}\tan^{-1}xdx=\displaystyle \int_{0}^{1}\cot^{-1}(1-x+x^{2})dx$ , then

$\displaystyle \int_{0}^{1}\tan^{-1}(1-x+x^{2})dx$ is equal to:

0%
$\log 4$
0%
$\dfrac {\pi}{2}+\log 2$
0%
$\log 2$
0%
$\dfrac {\pi}{2}-\log 4$

The value of

${\int }_{0}^{1}\dfrac{dx}{x+\sqrt{1-x^{2}}}$ is

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}$

Let

$f: { (2,3)\rightarrow (0,1) }$ be defined by

$f{ (x)=x-[x] }$ then

$f^{ -1 }{ (x) }$ equals

0%
$2x+1$
0%
$x+1$
0%
$x-1$
0%
$x+2$

$I=\displaystyle\int _{ 8 }^{ 15 }{ \dfrac { dx }{ \left( x-3 \right) \sqrt { x+1 } } }$ , then

$I$ equals

0%
$\dfrac{1}{2}\log\dfrac{5}{3}$
0%
$2\log\dfrac{1}{3}$
0%
$\dfrac{1}{2}\log\dfrac{1}{5}$
0%
$2\log\dfrac{5}{3}$

$\displaystyle \int_{2}^{8}\dfrac {\sqrt {10-x}}{\sqrt {x}+\sqrt {10-x}}dx$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

${\int}_{0}^{\pi}\dfrac{\sqrt{1-x}}{1+x}dx=$

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{2}-1$
0%
$\dfrac{\pi}{2}+1$
0%
$\pi+1$

$\int_{1}^{\infty }(e^{x+1}+e^{3-x})^{-1}dx$ is equal to:

0%
$\frac{\pi }{4e^{2}}$
0%
$\frac{\pi }{4e}$
0%
$\frac{1}{e^{2}}(\frac{\pi }{2}-tan^{-1}\frac{1}{e})$
0%
$\frac{\pi }{2e^{2}}$

$\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^{2}}}dx$ is equal to

0%
$\frac{1}{2}+\frac{\pi }{2\sqrt{3}}$
0%
$\frac{1}{2}-\frac{\pi }{2\sqrt{3}}$
0%
$\frac{1}{2}+\frac{\pi }{4\sqrt{3}}$
0%
$\frac{1}{2}-\frac{\pi }{4\sqrt{3}}$

The integral

$\displaystyle \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}{\dfrac{8\cos 2x}{(\tan x+\cot x)^{3}}dx}$ equals :

0%
$\dfrac{15}{128}$
0%
$\dfrac{15}{64}$
0%
$\dfrac{13}{32}$
0%
$\dfrac{13}{256}$

$P=\displaystyle \lim _{ n\rightarrow \infty }{ \frac { { \left( \prod _{ r=1 }^{ n }{ \left( { n }^{ 3 }+{ r }^{ 3 } \right) } \right) }^{ 1/n } }{ { n }^{ 3 } } }$ and

$\lambda =\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 3 } } }$ then

$In P$ is equal to

0%
$In 2-1+\lambda$
0%
$In 2-3+3\lambda$
0%
$2In 2-\lambda$
0%
$In 4-3+3\lambda$

$\displaystyle \overset{3}{\underset{0}{\int}} \dfrac{dx}{\sqrt{5 - x^2}}$

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{2}$
0%
$-\dfrac{\pi}{2}$
0%
$-\dfrac{\pi}{6}$

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

0%
$\dfrac{3}{8}$
0%
$\dfrac{1}{8}$
0%
$-\dfrac{3}{8}$
0%
$None\ of\ these$

${ I }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ \cos(\sin x)dx, { I }_{ 2 }= } \int _{ 0 }^{ \pi /2 }{ \sin(\cos x)dx }$ and

${ I }_{ 3 }=\int _{ 0 }^{ \pi /2 }{ \cos xdx }$ , then

0%
${ I }_{ 1 }>{ I }_{ 2 }>{ I }_{ 3 }$
0%
${ I }_{ 3 }>{ I }_{ 2 }>{ I }_{ 1 }$
0%
${ I }_{ 3 }>{ I }_{ 1 }>{ I }_{ 2 }$
0%
${ I }_{ 1 }>{ I }_{ 3 }>{ I }_{ 2 }$

If I =

$\overset { 2 }{ \underset { -3 }{ \int } } (|x + 1| + |x + 2| +|x -1|) dx$ , then i equals

0%
$\dfrac{31}{2}$
0%
$\dfrac{35}{2}$
0%
$\dfrac{47}{2}$
0%
$\dfrac{39}{2}$

$f(x)\int _{ 1 }^{ x }{ \frac { { tan }^{ 1 }t }{ t } } dt(x>0)$ , then the value of

$f({ o }^{ 2 })-f(\frac { 1 }{ o^{ 2 } } )$

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

The floor value of integral

$\displaystyle \int_\dfrac{\pi }{4}^{3\pi }\dfrac{x}{1+4x}dx$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Let

$I_{1}=\int_{-2}^{2} \dfrac{x^{6}+3 x^{5}+7 x^{4}}{x^{4}+2} d x$ and

$I_{2}=\int_{-3}^{1} \dfrac{2(x+1)^{2}+11(x+1)+14}{(x+1)^{4}+2} d x,$ then the value of

$I_{1}+I_{2}$ is

0%
8
0%
$200 / 3$
0%
$100 / 3$
0%
None of these

Explanation

$I_{2},$ put

$x+1=t,$

then

$I_{2}=\int_{-2}^{2} \dfrac{2 t^{2}+11 t+14}{t^{4}+2} d t=\int_{-2}^{2} \dfrac{2 x^{2}+11 x+14}{x^{4}+2} d x\\$

$\therefore I_{1}+I_{2} = \\$

$=\int_{-2}^{2} \dfrac{\left(x^{2}+3 x+7\right)\left(x^{4}+2\right)+5 x}{x^{4}+2} d x \\$

$=\int_{-2}^{2}\left(x^{2}+3 x+7\right) d x+5 \int_{-2}^{2} \dfrac{x}{x^{4}+2} d x \\$

$=2 \int_{0}^{2}\left(x^{2}+7\right) d x=\dfrac{100}{3}$

(The other integrals are zero, being integrals of odd functions.)

$I_1=\displaystyle\int^1_x\dfrac{dt}{1+t^2}$ and

$I_2=\displaystyle\int^{1/x}_1\dfrac{dt}{1+t^2}$ for

$x > 0$ , then?

0%
$I_1 = I_2$
0%
$I_1 > I_2$
0%
$I_2 > I_1$
0%
$I_2=\cot^{-1}x-\pi/4$

$z=x+3i$ then value of

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

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

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

Suppose

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

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

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

using

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

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

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

On adding,

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

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

$\Rightarrow I_1=0$

Now,

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

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

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

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

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

$\Rightarrow I_2+I_3=0$

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

$\therefore I_2+I_3=0$

Hence,

$I_1+I_2+I_3=0$

If m,n

$\in$ N, then the value of

$\displaystyle \int_{a}^{b}(x-a)^m (b-x)^n dx$ is equal to

0%
$\dfrac{(b-a)^{m+n}.m!n!}{(m+n)!}$
0%
$\dfrac{(b-a)^{m+n+1}.m!n!}{(m+n+1)!}$
0%
$\dfrac{(b-a)^{m}.m!}{m!}$
0%
None of these

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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