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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 12 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 12
If $$f(x)=\begin{vmatrix} \cos { x } & 1 & 0 \\ 1 & 2\cos { x } & 1 \\ 0 & 1 & 2\cos { x } \end{vmatrix}$$ then $$\displaystyle\int _{ 0 }^{ \pi /2 }{ f\left( x \right) } dx$$ is equal to
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0%
$$1/4$$
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$$-1/3$$
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$$1/2$$
0%
$$none of these $$
The value of $$\int_{0}^{[x]} (x-[x])dx$$, where $$[x]$$ is the greatest integer $$|le x$$ is equal to
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$$4[x]$$
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$$2[x]$$
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$$\dfrac{1}{2} [x]$$
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$$\dfrac{1}{5} [x]$$
The value of $$\int_{1}^{-1}\dfrac{dx}{(2-x)\sqrt{1-x^{2}}}$$ is
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$$0$$
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$$\dfrac{\pi}{\sqrt{3}}$$
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$$\dfrac{2\pi}{3\sqrt{3}}$$
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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
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0%
$$0$$
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$$n(n^{2}-1)$$
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$$\dfrac{n}{(n^{2}-1)}$$
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$$n^{2}$$
$$\displaystyle \int_{-3\pi}^{3\pi}{\sin^{2}\theta\sin^{2} 2\theta d \theta}$$ is equal to-
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$$\pi$$
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$$\dfrac{3\pi}{2}$$
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$$\dfrac{5\pi}{2}$$
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$$6\pi$$
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 $$F\left(e\right)=$$
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$$\dfrac{-1}{2}$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$-1$$
Evaluate:
$$\displaystyle\int_{-2}^{3}{\left|1-{x}^{2}\right|dx}$$
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$$\dfrac{28}{3}$$
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$$\dfrac{2}{3}$$
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$$\dfrac{8}{3}$$
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$$\dfrac{5}{3}$$
$$\displaystyle \int_{0}^{\pi/2}{(\sin x-\cos x).\log(\sin x+\cos x)dx}$$=
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{2}$$
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$$0$$
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$$2$$
$$\displaystyle \overset{2\pi}{\underset{0}{\int}} x \,log \left(\dfrac{3 + \cos x}{3 - \cos x}\right)dx$$
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$$\dfrac{\pi}{2} \,log \,3$$
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$$\dfrac{\pi}{12} \,log \,3$$
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$$\dfrac{\pi}{3} \,log \,3$$
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$$0$$
$$\int^2_0 (\sqrt{\dfrac{4-x}{x}} - \sqrt{\dfrac{x}{4-x}})dx$$ is equal to
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$$0$$
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$$8$$
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$$4$$
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$$16$$
If $$2\displaystyle \int_{0}^{1}\tan^{-1}xdx=\displaystyle \int_{0}^{1}\cot^{-1}(1-x+x^{2})dx$$, then $$\displaystyle \int_{0}^{1}\tan^{-1}(1-x+x^{2})dx$$ is equal to:
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$$\log 4$$
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$$\dfrac {\pi}{2}+\log 2$$
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$$\log 2$$
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$$\dfrac {\pi}{2}-\log 4$$
The value of $${\int }_{0}^{1}\dfrac{dx}{x+\sqrt{1-x^{2}}}$$ is
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{2}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{\pi}{4}$$
Let $$f: { (2,3)\rightarrow (0,1) }$$ be defined by $$f{ (x)=x-[x] }$$ then $$f^{ -1 }{ (x) }$$ equals
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$$2x+1$$
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$$x+1$$
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$$x-1$$
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$$x+2$$
If $$I=\displaystyle\int _{ 8 }^{ 15 }{ \dfrac { dx }{ \left( x-3 \right) \sqrt { x+1 } } } $$, then $$I$$ equals
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$$\dfrac{1}{2}\log\dfrac{5}{3}$$
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$$2\log\dfrac{1}{3}$$
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$$\dfrac{1}{2}\log\dfrac{1}{5}$$
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$$2\log\dfrac{5}{3}$$
$$\displaystyle \int_{2}^{8}\dfrac {\sqrt {10-x}}{\sqrt {x}+\sqrt {10-x}}dx$$ is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
$${\int}_{0}^{\pi}\dfrac{\sqrt{1-x}}{1+x}dx=$$
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{2}-1$$
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$$\dfrac{\pi}{2}+1$$
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$$\pi+1$$
$$\int_{1}^{\infty }(e^{x+1}+e^{3-x})^{-1}dx$$ is equal to:
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$$\frac{\pi }{4e^{2}}$$
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$$\frac{\pi }{4e}$$
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$$\frac{1}{e^{2}}(\frac{\pi }{2}-tan^{-1}\frac{1}{e})$$
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$$\frac{\pi }{2e^{2}}$$
$$\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^{2}}}dx$$ is equal to
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$$\frac{1}{2}+\frac{\pi }{2\sqrt{3}}$$
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$$\frac{1}{2}-\frac{\pi }{2\sqrt{3}}$$
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$$\frac{1}{2}+\frac{\pi }{4\sqrt{3}}$$
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$$\frac{1}{2}-\frac{\pi }{4\sqrt{3}}$$
The integral $$\displaystyle \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}{\dfrac{8\cos 2x}{(\tan x+\cot x)^{3}}dx}$$ equals :
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$$\dfrac{15}{128}$$
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$$\dfrac{15}{64}$$
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$$\dfrac{13}{32}$$
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$$\dfrac{13}{256}$$
If $$P=\displaystyle \lim _{ n\rightarrow \infty }{ \frac { { \left( \prod _{ r=1 }^{ n }{ \left( { n }^{ 3 }+{ r }^{ 3 } \right) } \right) }^{ 1/n } }{ { n }^{ 3 } } }$$ and $$\lambda =\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 3 } } } $$ then $$In P$$ is equal to
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$$In 2-1+\lambda$$
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$$In 2-3+3\lambda$$
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$$2In 2-\lambda$$
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$$In 4-3+3\lambda$$
$$\displaystyle \overset{3}{\underset{0}{\int}} \dfrac{dx}{\sqrt{5 - x^2}}$$
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{2}$$
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$$-\dfrac{\pi}{2}$$
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$$-\dfrac{\pi}{6}$$
$$\displaystyle\int _{ 0 }^{ \infty }{ \dfrac { dx }{ \left( x+\sqrt { { x }^{ 2 }+1 } \right) ^{ 3 } } } =$$
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$$\dfrac{3}{8}$$
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$$\dfrac{1}{8}$$
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$$-\dfrac{3}{8}$$
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$$None\ of\ these$$
If $${ I }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ \cos(\sin x)dx, { I }_{ 2 }= } \int _{ 0 }^{ \pi /2 }{ \sin(\cos x)dx } $$ and $${ I }_{ 3 }=\int _{ 0 }^{ \pi /2 }{ \cos xdx } $$, then
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$${ I }_{ 1 }>{ I }_{ 2 }>{ I }_{ 3 }$$
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$${ I }_{ 3 }>{ I }_{ 2 }>{ I }_{ 1 }$$
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$${ I }_{ 3 }>{ I }_{ 1 }>{ I }_{ 2 }$$
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$${ I }_{ 1 }>{ I }_{ 3 }>{ I }_{ 2 }$$
If I =$$\overset { 2 }{ \underset { -3 }{ \int } } (|x + 1| + |x + 2| +|x -1|) dx$$, then i equals
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$$\dfrac{31}{2}$$
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$$\dfrac{35}{2}$$
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$$\dfrac{47}{2}$$
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$$\dfrac{39}{2}$$
If $$f(x)\int _{ 1 }^{ x }{ \frac { { tan }^{ 1 }t }{ t } } dt(x>0)$$, then the value of $$f({ o }^{ 2 })-f(\frac { 1 }{ o^{ 2 } } )$$
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$$\pi $$
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$$2\pi $$
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$$\frac { \pi }{ 2 } $$
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0
The floor value of integral $$\displaystyle \int_\dfrac{\pi }{4}^{3\pi }\dfrac{x}{1+4x}dx$$ is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
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
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8
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$$200 / 3$$
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$$100 / 3$$
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None of these
Explanation
In $$I_{2},$$ 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.)
If $$I_1=\displaystyle\int^1_x\dfrac{dt}{1+t^2}$$ and $$I_2=\displaystyle\int^{1/x}_1\dfrac{dt}{1+t^2}$$ for $$x > 0$$, then?
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$$I_1 = I_2$$
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$$I_1 > I_2$$
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$$I_2 > I_1$$
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$$I_2=\cot^{-1}x-\pi/4$$
If $$z=x+3i$$ 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?
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$$3\sqrt{2}$$
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$$6\sqrt{3}$$
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$$\sqrt{6}$$
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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
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$$I_1=0$$
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$$I_2+I_3=0$$
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$$I_1+I_2+I_3=0$$
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$$I_2=I_3$$
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
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$$\dfrac{(b-a)^{m+n}.m!n!}{(m+n)!}$$
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$$\dfrac{(b-a)^{m+n+1}.m!n!}{(m+n+1)!}$$
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$$\dfrac{(b-a)^{m}.m!}{m!}$$
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None of these
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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