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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 7
$$\displaystyle \int _{ 0 }^{ \pi /4 }{ \log { \left( 1+\tan ^{ 2 }{ \theta } +2\tan { \theta } \right) d\theta = } }$$
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0%
$$\pi \log 2$$
0%
$$(\pi \log 2)/2$$
0%
$$(\pi \log 2)/4$$
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$$\log 2$$
If $$\displaystyle\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx = K{{\sin }^{ - 1}}} ({2^x}) + C,$$ then K is equal to
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0%
$$\ell n2$$
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$$\displaystyle{1 \over 2}\ell n2$$
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$$\displaystyle{1 \over 2}$$
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$$\displaystyle{1 \over {\ell n2}}$$
Explanation
$$I=\displaystyle\int \dfrac{2^x}{\sqrt{1-4^x}}dx$$
Let $$2^x=z$$
$$\Rightarrow 2^x\log 2dx=dz$$
$$\Rightarrow I=\displaystyle\int \dfrac{\dfrac{1}{\log2}}{\sqrt{1-z^2}}dz$$
$$=\dfrac{1}{\log 2}\cdot \sin^{-1}z+c$$
$$=\dfrac{1}{\log 2}\cdot \sin^{-1}(e^x)+c$$
$$=k\sin^{-1}(2^x)+c$$
then $$k=\dfrac{1}{\log 2}$$.
The value of the defined integral $$\displaystyle \int^{\pi/2}_{0}(\sin x+\cos x)\sqrt {\dfrac {e^{x}}{\sin x}}dx$$ equals
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$$2\sqrt {e^{\pi/2}}$$
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$$\sqrt {e^{\pi/2}}$$
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$$2\sqrt {e^{\pi/2}}.\cos 1$$
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$$\dfrac {1}{2}e^{\pi/4}$$
Evaluate $$\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$$
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{4}$$
0%
$$\pi$$
0%
None of these
Explanation
Let $$I=\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$$
Put $$\sin x=t$$ and $$\cos xdx=dt$$.
$$[x=0\Rightarrow t=0]$$ and $$\left[x=\dfrac{\pi}{2}\Rightarrow t=1\right]$$.
$$\therefore I=\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$$
$$=[\tan^{-1}t]^1_0$$ ....... $$\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C$$
$$=\dfrac{\pi}{4}-0\\$$.
$$=\dfrac{\pi}{4}$$.
Evaluate $$\displaystyle\int^{\sqrt{8}}_{\sqrt{3}}x\sqrt{1+x^2}dx$$
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0%
$$\dfrac{19}{3}$$
0%
$$\dfrac{19}{6}$$
0%
$$\dfrac{38}{3}$$
0%
$$\dfrac{9}{4}$$
Evaluate $$\displaystyle\int^{\pi/2}_0\cos^3xdx $$
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$$1$$
0%
$$\dfrac{3}{4}$$
0%
$$\dfrac{2}{3}$$
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None of these
Evaluate : $$\displaystyle\int^1_0\sqrt{\dfrac{1-x}{1+x}}dx$$
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0%
$$\dfrac{\pi}{2}$$
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$$\left(\dfrac{\pi}{2}-1\right)$$
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$$\left(\dfrac{\pi}{2}+1\right)$$
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None of these
Evaluate: $$\int _{ 1/3 }^{ 1 }{ \cfrac { { \left( x-{ x }^{ 3 } \right) }^{ 1/3 } }{ { x }^{ 4 } } } dx=$$
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$$3$$
0%
$$0$$
0%
$$6$$
0%
$$4$$
Explanation
$$I=\displaystyle\int^1_{1/3}\dfrac{x^{1/3}(1-x^2)^{1/3}}{x^4}dx$$
Let $$x=\sin\theta, dx=\cos\theta d\theta$$
$$I=\displaystyle\int^{\pi/2}_{\sin^{-1}1/3}\dfrac{(\sin\theta)^{1/3}(\cos^{2/3}\theta)}{\sin^4\theta}\times \cos\theta d\theta$$
$$=\displaystyle\int \dfrac{\cos^{5/3}\theta}{\sin^{11/3}\theta}d\theta \times \dfrac{\cos^2\theta}{\cos^2\theta}$$
$$=\displaystyle\int \dfrac{\cos^{11/3}\theta}{\sin^{11/3}\theta\cos^2\theta}d\theta$$
$$=\displaystyle\cot^{11/3}\theta .\sec^2\theta d\theta$$
Let $$\tan\theta =t$$
$$\Rightarrow \sec^2\theta d\theta =dt$$
$$=\displaystyle\int t^{-11/3}dt$$
$$=\dfrac{t^{-8/3}}{(-8/3)}=\dfrac{-3}{8}t^{-8/3}$$
$$=\left.\dfrac{-3}{8}(\tan\theta)^{-8/3}\right|^{\pi/2}_{\sin^{-11}1/3}$$
As $$\sin^{-11/3}=\tan^{-1}\dfrac{1}{2\sqrt{2}}$$
$$\Rightarrow I=\dfrac{-3}{8}\left(\dfrac{1}{(\infty)^{8/3}}-\left(\dfrac{1}{2\sqrt{2}}\right)^{-8/3}\right)$$
$$=\dfrac{3}{8}\times (2\sqrt{2})^{8/3}=\dfrac{3}{8}(2^{3/2})^{8/3}$$
$$=\dfrac{3}{8}\times 16=6\Rightarrow (C)$$.
Evaluate$$\displaystyle \int _ { 0 } ^ { a } \dfrac { x d x } { \sqrt { a ^ { 2 } + x ^ { 2 } } } $$
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$$a ( \sqrt { 2 } - 1 )$$
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$$a ( 1 - \sqrt { 2 } )$$
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$$a ( 1 + \sqrt { 2 } )$$
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$$2 a \sqrt { 3 }$$
Explanation
$$\displaystyle \int _{ 0 }^{ a }{ \cfrac { xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } $$
Let $$t=\left( { a }^{ 2 }+{ x }^{ 2 } \right) \Rightarrow dt=2xdx$$
when $$x=0,t={ a }^{ 2 }\quad $$
when $$x=a,t=2{ a }^{ 2 }$$
$$\displaystyle =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { 2xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { dt }{ \sqrt { t } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ { t }^{ -1/2 }dt } $$
$$=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { { t }^{ -1/2+1 } }{ -\cfrac { 1 }{ 2 } +1 } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { \sqrt { t } }{ \cfrac { 1 }{ 2 } } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 2 }{ 2 } \left[ \sqrt { 2{ a }^{ 2 } } -\sqrt { { a }^{ 2 } } \right] $$
$$=a\left( \sqrt { 2 } -1 \right) $$
The correct evaluation of $$\displaystyle \int _ { 0 } ^ { \pi / 2 } \sin x \sin 2 x$$ is
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$$\dfrac { 4 } { 3 }$$
0%
$$\dfrac { 1 } { 3 }$$
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$$\dfrac { 3 } { 4 }$$
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$$\dfrac { 2 } { 3 }$$
Explanation
$$I=\displaystyle \int_0^{\pi/2} \sin x \sin 2x \, dx$$
$$I= \displaystyle \int_0^{\pi/2} 2 \sin^2 x \, \cos x\ dx$$
$$u = \sin x \Rightarrow du = \cos x\ dx$$
$$= \left.\begin{matrix}\displaystyle \int_0^1 2u^2 du = 2 \dfrac{u^3}{3}\end{matrix}\right|^1_0= \dfrac{2}{3}$$
$$\int { { e }^{ x^{ 3 } }+{ x }^{ 2-1 }(3{ x }^{ 4 }+{ 2x }^{ 3 }+{ 2x }^{ 2 }\quad x=h(x)+c } $$ then the value of $$h(1)h(-1)$$.
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1
0%
-1
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2
0%
-2
$$\int _{ 0 }^{ \pi }{ \cfrac { { x }^{ 2 } }{ { \left( 1+sinx \right) }^{ 2 } } } dx$$ equals
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$$\pi (\pi -2)$$
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$$\pi ^2 (\pi -2)$$
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$$\pi (2-\pi )$$
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none of these
The value of the integral $$\displaystyle \int_{-\pi/2}^{\pi/2} \left(x^{2}+\log \dfrac{\pi-x}{\pi+x}\right) \cos x dx $$ is
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$$0$$
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$$\dfrac{\pi^{2}}{2}-4$$
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$$\dfrac{\pi^{2}}{2}+4$$
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$$\dfrac{\pi^{2}}{2}$$
Explanation
$$I=\displaystyle \int_{-\pi/2}^{\pi/2}\left[x^2+log\left( \dfrac{\pi-x}{\pi+x}\right)\right]cosxdx$$
As,$$\displaystyle \int_{-a}^{a}f(x)dx=0,$$when$$f(-x)=-f(x)$$
$$\therefore I=\displaystyle \int_{-\pi/2}^{\pi/2}x^2cosxdx+0=2\displaystyle \int_{0}^{\pi/2}(x^2cosx)dx$$
$$=2\left\{ (x^2sinx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}(x^3cosx) \right\}$$
$$2\left[ \dfrac{\pi^2}{4}-2\left\{(-xcosx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}1.(-cosx)dx \right\}\right]$$
$$=2\left[ \dfrac{\pi^2}{4}-2(sinx)_{0}^{\pi/2}\right]=2\left[\dfrac{\pi^2}{4}-2\right]=\left( \dfrac{\pi^2}{2}-4\right)$$
Let $$\theta$$ be the angle between the lines $${ L }_{ { 1 } }:\left[ \begin{array}{l} { { x }=2{ t }+{ 1 } } \\ { { y }={ t }+{ 1 } } \\ { { z }=3{ t }+{ 1 } } \end{array} \right. $$ and $${ L }_{ { 2 } }:\left[ \begin{array}{l} { { x }=3{ s }+2 } \\ { { y }=6{ s }-1 } \\ { { z }=4 } \end{array} \right. $$ where $$s , t \in { R }.$$ Then the value of $$\int _ { 0 } ^ { \theta } \dfrac { 1 } { 1 + \tan x } d x =$$
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$$\pi / 6$$
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$$\pi / 4$$
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$$\pi / 2$$
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$$\pi / 3$$
Let $${ I }_{ 1 }=\displaystyle \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 50 } \right) }^{ 100 }dx }$$ and $${ I }_{ 2 }=\displaystyle \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 50 } \right) }^{ 101 }dx }$$, then $$\dfrac { { I }_{ 1 } }{ { I }_{ 2 } } =$$
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$$\dfrac {5051}{5050}$$
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$$\dfrac {5051}{5049}$$
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$$\dfrac {51}{50}$$
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$$\dfrac {101}{100}$$
$$\displaystyle\int _{ -9 }^{ 9 }{ \log { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } dx } $$ equal
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$$2\log (9^{2}+1)$$
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$$2\log (\sqrt{9^{2}+1}-9)$$
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$$0$$
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$$2\log (9+\sqrt{9^{2}+1})$$
If $$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\tan \theta}{\sqrt{2k \sec \theta}}d\theta=1-\dfrac{1}{\sqrt{2}},(k>0),$$ then the value of k is :
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$$2$$
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$$\dfrac{1}{2}$$
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$$4$$
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$$1$$
Explanation
$$LHS= \dfrac{1}{\sqrt{2k}}\displaystyle\int_0^{\dfrac{\pi}{3}} \dfrac{\tan \theta}{\sqrt{\sec \theta}}d\theta$$
$$=\dfrac{1}{\sqrt{2k}}\displaystyle \int_0^{\dfrac{\pi}{3}}\dfrac{\sin \theta}{\sqrt{\cos \theta}}d\theta$$
$$=-\dfrac{1}{\sqrt{2k}}|2\sqrt{\cos \theta}|_0^{\dfrac{\pi}{3}}$$
$$=-\dfrac{\sqrt{2}}{\sqrt{k}}\left(\dfrac{1}{\sqrt{2}}-1\right)$$
$$=\dfrac{\sqrt 2}{\sqrt k}\left(1-\dfrac{1}{\sqrt{2}}\right)$$
$$\because$$ it is given that $$RHS = 1-\dfrac{1}{\sqrt{2}}$$
$$\therefore $$ by comparing we get $$k=2$$
The value of $$\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$$ is equal to?
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$$2\pi$$
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$$\pi^2$$
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$$2\pi^2$$
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$$4\pi$$
Explanation
Let $$I = \displaystyle \int_0^{2 \pi} \dfrac{x \sin^8 x}{\sin^8 x + \cos^8 x}$$ ___(1)
$$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 (2\pi - x)}{\sin^8 (2 \pi - x) + \cos^8 (2 \pi - x)}$$
$$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x}$$ ___(2)
Add (1) + (2) we get,
$$2I = \displaystyle \int_0^{2 \pi} \dfrac{2 \pi \sin^8 x}{\sin^8 x + \cos^8 x}$$
$$2I = 2 \times 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x}$$
$$I = 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$$
$$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$$ __(3)
$$I = 4\pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 \left(\dfrac{\pi}{2} - x \right)}{\sin^8 \left(\dfrac{\pi}{2} - x \right) + \cos^8 \left(\dfrac{\pi}{2} - x \right)}$$
$$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\cos^8 (x)}{\sin^8 x + \cos^8 x} $$ __(4)
add (3) + (4) we get
$$I = 2 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx = 2\pi \displaystyle \int_0^{\pi/2} 1 dx = 2 \pi \times \dfrac{\pi}{2} = \pi^2$$
If $$f(a-x)=-f(x)$$, then $$\displaystyle \int_{0}^{a}f(x)dx=0$$.
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0%
True
0%
False
Explanation
$$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$$
$$\displaystyle\int_{a}^{a}{f\left(x\right)dx}=I+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$$
Now $$I=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}$$
Put $$x=-t$$
$$\Rightarrow dx=-dt$$
When $$x=-a,\,t=a$$ and when $$x=0,\,t=0$$
$$I=\displaystyle\int_{a}^{0}{f\left(-t\right)\left(-dt\right)}$$
$$=-\displaystyle\int_{a}^{0}{f\left(-t\right)dt}$$
$$=\displaystyle\int_{0}^{a}{f\left(-t\right)dt}...........$$ since $$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(x\right)dx}$$
$$=\displaystyle\int_{0}^{a}{f\left(-x\right)dt}..........$$ since $$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(t\right)dt}$$
Thus,
$$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{f\left(-x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$$
$$=\displaystyle\int_{0}^{a}{\left[f\left(-x\right)+f\left(x\right)\right]dx}$$
$$=2\displaystyle\int_{0}^{a}{f\left(x\right)dx}$$
If $$f\left(x\right)$$ is an odd function, then $$f\left(-x\right)=-f\left(x\right)$$
$$\Rightarrow \displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{\left[-f\left(x\right)+f\left(x\right)\right]dx}=0$$
Hence proved.
Thus, the given statement is true.
$$\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos { 2x } } }$$
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$$200\sqrt 2$$
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$$400\sqrt 2$$
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$$800\sqrt 2$$
0%
$$none$$
Explanation
$$I=\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos x } dx } \\ =400\int _{ 0 }^{ \pi }{ \sqrt { 1-\cos 2x } dx } \\ =800\int _{ 0 }^{ \pi }{ \sqrt { 2 } \sqrt { { { \sin }^{ 2 } }x } dx } \\ =800\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin xdx } \\ =800\sqrt { 2 } \left[ { \cos x } \right] _{ \frac { \pi }{ 2 } }^{ 0 } \\ =800\sqrt { 2 }$$
$$ \\ Hence,\, the\, option\, C\, is\, the\, correct\, answer. $$
$$\displaystyle \int_{0}^{1}\sin^{-1}x dx=\dfrac {\pi}{2}-1$$
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0%
True
0%
False
Let a function $$f:R\rightarrow R$$ be defined as $$f\left( x \right) =x+\sin { x } $$. The value of $$\int _{ 0 }^{ 2\pi }{ { f }^{ -1 }(x) } dx$$ will
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0%
$$2{ \pi }^{ 2 }$$
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$$2{ \pi }^{ 2 }-2$$
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$$2{ \pi }^{ 2 }+2$$
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$${ \pi }^{ 2 }$$
Explanation
We have $$f(x) = x + \sin x $$
Let $$I = \displaystyle \int_0^{2\pi} f^{-1} (x) dx$$
let $$x = f(t) \Rightarrow dx = f'(t) dt$$
$$I = \displaystyle \int_{f^{-1} (0)}^{f^{-1} (2 \pi)} t . f'(t) dt = \int_0^{2\pi} t. f'(t) dt$$
$$= t f(t) \displaystyle |_0^{2\pi} - \int_0^{2\pi} f(t) dt$$
$$= 2 \pi . f(2 \pi) - \displaystyle \int_0^{2\pi} t + \sin t$$
$$4 \pi^2 - \dfrac{t^2}{2} \displaystyle |_0^{2\pi} + \cos t |_0^{2\pi}$$
$$= 4 \pi^2 - 2 \pi^2 + 0$$
$$= 2 \pi^2$$
The value of $$\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta=$$
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$$\dfrac{\pi}{4}-1$$
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$$\dfrac{\pi}{4}$$
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$$1-\dfrac{\pi}{4}$$
0%
none of these
Explanation
Now,
$$\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta$$
$$=\displaystyle\int\limits_{0}^{\frac{\pi}{4}} (\sec^2 \theta-1)\ d\theta$$ [ Since $$\sec^2 \theta-\tan^2 \theta=1$$]
$$=\left[\tan \theta-\theta\right]_0^{\frac{\pi}{4}}$$
$$=1-\dfrac{\pi}{4}$$.
$$\int _{ 0 }^{ 1 }{ \frac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ \frac { 3 }{ 2 } } } dx } =........$$
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0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{2}{3}$$
0%
$$\dfrac{3}{2}$$
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$$1-\dfrac{1}{\sqrt{2}}$$
Explanation
$$\displaystyle\int _{ 0 }^{ 1 }{ \dfrac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ \frac { 3 }{ 2 } } } dx } =\dfrac{1}{2}\int ^{1}_0 \dfrac{d(x^2+1)}{(x^2+1)^{3/2}}=\dfrac{1}{2}\bigg[\dfrac{(x^2+1)^{-3/2+1}}{-3/2+1}\bigg]^{1}_{0}=\bigg[\dfrac{1}{2}\times \dfrac{(x^2+1)^{-1/2}}{-1/2}\bigg]^{1}_{0}=-[\dfrac{1}{\sqrt{2}}-1\bigg]=1-\dfrac{1}{\sqrt{2}}$$
$$\displaystyle\int^{1/2}_0\dfrac{dx}{(1+x^2)\sqrt{1-x^2}}$$ is equal to?
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0%
$$\dfrac{1}{\sqrt{2}}\tan^{-1}\sqrt{\dfrac{2}{3}}$$
0%
$$\dfrac{2}{\sqrt{2}}\tan^{-1}\left(\dfrac{3}{\sqrt{2}}\right)$$
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$$\dfrac{\sqrt{2}}{2}\tan^{-1}\left(\dfrac{3}{2}\right)$$
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$$\dfrac{\sqrt{2}}{2}\tan^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$
Explanation
Let $$x=\sin\theta$$
$$dx=\cos \theta d\theta$$
$$\displaystyle\int^{\pi/6}_0\dfrac{\cos\theta d\theta}{(1+\sin^2\theta)\cos \theta}=\displaystyle\int^{\pi/6}_0\dfrac{\sec^2\theta}{1+2\tan^2\theta}=d\theta$$.
Multiply numeration and denomination by $$\sec^2\theta$$
$$=\dfrac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\theta)^{\dfrac{\pi}{6}}_{0}$$
$$=\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{\dfrac{2}{3}}\right)$$.
$$\int _{ -1 }^{ 1/2 }{ \dfrac { { e }^{ x }\left( 2-{ x }^{ 2 } \right) dx }{ \left( 1-x \right) \sqrt { 1-{ x }^{ 2 } } } } $$ is equal to
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0%
$$\dfrac { \sqrt { e } }{ 2 } \left( \sqrt { 3 } +1 \right) $$
0%
$$\dfrac { \sqrt { 3e } }{ 2 } $$
0%
$$\sqrt { 3e } $$
0%
$$\sqrt { \dfrac { e }{ 3 } } $$
$$\int _{ 1 }^{ 3} \dfrac{4x}{x^2+1} \, dx$$
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0%
$$\log { 5 } $$
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$$\cfrac { 1 }{ 2 } \log { 5 } $$
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$$\log { 25 } $$
0%
$$\log { 100 } $$
Explanation
$$\int _{ 1 }^{ 3 }{ \left( \cfrac { 4x }{ { 1+x }^{ 2 } } \right) } dx$$
substitute $$x^2+1=t$$
$$x=1$$ then $$t=2$$
$$x=3$$ then $$t=10$$
$$\int_2^{10}\dfrac2tdt$$
$$=2{ \left[ \ln { \left( { 1+x }^{ 2 } \right) } \right] }_{ 1 }^{ 3 }$$
$$=2\left[ \ln { 10 } -\ln { 2 } \right]$$
$$ =\ln { 25 } $$
If $$\int _{ log2 }^{ x }{ \dfrac { dx }{ \sqrt { { e }^{ x }-1 } } } =\dfrac { \pi }{ 6 } ,$$then x is equal to _________.
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0%
4
0%
in 8
0%
in 4
0%
None of these
$$\displaystyle\int^1_0\dfrac{dx}{e^x+e^{-x}}$$ is equal to?
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0%
$$\dfrac{\pi}{4}-\tan^{-1}(e)$$
0%
$$\tan^{-1}(e)-\dfrac{\pi}{4}$$
0%
$$\tan^{-1}(e)+\dfrac{\pi}{4}$$
0%
$$\tan^{-1}(e)$$
Explanation
$$\displaystyle\int^1_0\dfrac{dx}{e^x+e^{-1}}=\displaystyle\int^1_0\dfrac{e^x}{(e^x)^2+1}dx$$
Put $$e^x=t\Rightarrow e^x dx=dt$$
$$x\rightarrow 0\Rightarrow t\rightarrow e^0=1$$
$$x\rightarrow 1\Rightarrow t\rightarrow e^1=e$$
$$\displaystyle \int^e_1 \dfrac{1}{1+t^2}dt=\tan^{-1}t+c$$
$$=(\tan^{-1}(e^x))^1_0+c=\tan^{-1}(e)-\dfrac{\pi}{4}$$.
If $$\displaystyle\int^{\dfrac{\pi}{2}}_0\dfrac{\cot x}{\cot x+cosec x}dx=m(\pi +n)$$, then $$mn$$ is equal to?
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0%
$$-1$$
0%
$$1$$
0%
$$\dfrac{1}{2}$$
0%
$$-\dfrac{1}{2}$$
Explanation
$$\displaystyle\int^{\pi /2}_0\dfrac{\cot xdx}{\cot x+cosec x}$$
$$\displaystyle\int^{\pi /2}_0\dfrac{\cos x}{\cos x+1}=\displaystyle\int \dfrac{2\cos^2\dfrac{x}{2}-1}{2\cos^2\dfrac{x}{2}}$$
$$\displaystyle\int^{\pi/2}_0\left(1-\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)dx$$
$$\left[x-\tan \dfrac{x}{2}\right]^{\dfrac{\pi}{2}}_0$$
$$\Rightarrow$$$$\dfrac{1}{2}[\pi -2]$$
$$\Rightarrow$$$$m=\dfrac{1}{2}, n=-2$$
$$mn=-1$$.
Evaluate : $$\displaystyle\int^1_0\dfrac{dx}{(1+x+x^2)}$$
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$$\dfrac{\pi}{\sqrt{3}}$$
0%
$$\dfrac{\pi}{3}$$
0%
$$\dfrac{\pi}{3\sqrt{3}}$$
0%
None of these
Evaluate$$\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$$
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0%
$$0$$
0%
$$\dfrac{\pi}{4}$$
0%
$$e^{\pi/2}$$
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$$(e^{\pi/2}-1)$$
Explanation
Let $$I=\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$$
$$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{(1+\cos x)}+\dfrac{\sin x}{(1+\cos x)}\right\}dx$$
$$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{2\cos^2(x/2)}+\dfrac{2\sin(x/2)\cos (x/2)}{2\cos^2(x/2)}\right\}dx$$
$$=\displaystyle\int^{\pi/2}_0e^x\left\{\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right\}dx$$
$$=\left[e^x\tan \dfrac{x}{2}\right]^{\pi/2}_0=e^{\pi/2}$$.
Evaluate : $$\displaystyle\int^{\sqrt{2}}_0\sqrt{2-x^2}dx$$
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$$\pi$$
0%
$$2\pi$$
0%
$$\dfrac{\pi}{2}$$
0%
None of these
Evaluate : $$\displaystyle\int^1_0\dfrac{(1-x)}{(1+x)}dx$$
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0%
$$(\log 2+1)$$
0%
$$(\log 2-1)$$
0%
$$(2 \log 2-1)$$
0%
$$(2 \log 2+1)$$
Evaluate : $$\displaystyle\int^9_0\dfrac{dx}{(1+\sqrt{x})}$$
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0%
$$(3-2 log 2)$$
0%
$$(3+2 log 2)$$
0%
$$(6-2 log 4)$$
0%
$$(6+2 log 4)$$
Evaluate : $$\displaystyle\int^2_{-2}|x|dx$$
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$$4$$
0%
$$3.5$$
0%
$$2$$
0%
$$0$$
Evaluate $$\displaystyle\int^{1}_0\dfrac{(1-x)}{(1+x)}dx$$
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0%
$$\dfrac{1}{2}log 2$$
0%
$$(2log 2+1)$$
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$$(2log 2-1)$$
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$$\left(\dfrac{1}{2}log 2-1\right)$$
Evaluate $$\displaystyle\int^{\pi/6}_0\cos x\cos 2xdx$$
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$$\dfrac{1}{4}$$
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$$\dfrac{5}{12}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{7}{12}$$
The value of $$\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx$$ is?
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$$50\sqrt{2}$$
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$$100\sqrt{2}$$
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$$150\sqrt{2}$$
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$$200\sqrt{2}$$
Evaluate : $$\displaystyle\int^2_1|x^2-3x+2|dx$$
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$$\dfrac{-1}{6}$$
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$$\dfrac{1}{6}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{2}{3}$$
Given $$I_{m}=\displaystyle \int_{1}^{e}(\log x)^{m} d x .$$ If $$\dfrac{I_{m}}{K}+\dfrac{I_{m-2}}{L}=e,$$ then the values of $$K$$ and $$L$$ are
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$$\dfrac{1}{1-m}, \dfrac{1}{m}$$
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$$(1-m), \dfrac{1}{m}$$
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$$\dfrac{1}{1-m}, \dfrac{m(m-2)}{m-1}$$
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$$\dfrac{m}{m-1}, m-2$$
Explanation
$$I_{m}= \displaystyle \int_{1}^{e}(\log x)^{m} d x \\$$
$$\therefore I_{m}=\left[x(\log x)^{m}\right]_{1}^{e}-\int_{1}^{e} x \dfrac{m(\log x)^{m-1}}{x} d x \text { (integrating by parts) }$$
$$\Rightarrow \dot{I}_{m}=e-m \displaystyle \int_{1}^{e}(\log x)^{m-1} d x=e-m I_{m-1}\\$$
Replacing $$m$$ by $$m-1\\$$
$$I_{m-1}=e-(m-1) I_{m-2}\\$$
From equations $$(1)$$ and $$(2),$$ we have
$$I_{m}=e-m\left[e-(m-1) I_{m-2}\right]\\$$
$$\Rightarrow I_{m}-m(m-1) I_{m-1}=e(1-m) \\$$
$$\Rightarrow \dfrac{I_{m}}{1-m}+m I_{m-2}=e \\$$
$$\Rightarrow K=1-m \text { and } L=\dfrac{1}{m}$$
$$\displaystyle\int^a_{-a}x|x|dx=?$$
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$$0$$
0%
$$2a$$
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$$\dfrac{2a^3}{3}$$
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None of these
Explanation
$$|x|=-x,x<0$$
$$x,x>0$$
$$I=\displaystyle\int^0_{-a}x(-x)dx+\displaystyle\int^a_0x\cdot xdx$$
$$=\displaystyle\int^0_{-a}-x^2dx+\displaystyle\int^a_0x^2dx$$
$$=\left[\dfrac{-x^3}{3}\right]^0_{-a}+\left[\dfrac{x^3}{3}\right]^a_0=+\dfrac{1}{3}(-a)^3+\dfrac{a^3}{3}=0$$.
$$\displaystyle\int^1_{-2}\dfrac{|x|}{2}dx=?$$
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$$3$$
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$$2.5$$
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$$1.5$$
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None of these
Explanation
$$|x|=-x,x<0$$
$$x,x>0$$
$$I=\displaystyle\int^0_{-2}\dfrac{-x}{x}dx+\displaystyle\int^1_0\dfrac{x}{x}dx=\displaystyle\int^0_{-2}-dx+\displaystyle\int^1_0dx=[-x]^0_{-2}+[x]^1_0=(-2+1)=-1$$.
Evaluate : $$\displaystyle\int^1_0|2x-1|dx$$
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$$2$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$0$$
Value of $$\displaystyle\int^3_2\dfrac{dx}{\sqrt{(1+x^3)}}$$ is?
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Less than $$1$$
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Greater than $$2$$
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Lies between $$3$$ and $$4$$
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None of these
Explanation
Given, $$ \int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}} $$
Let us consider the function
$$ f(x) = \int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}} $$
according to the question putting the lower and upper limit we get,
$$ f(2) = \dfrac{1}{\sqrt{1+2^3}} = \dfrac{1}{3}$$
$$ f(3) = \dfrac{1}{\sqrt{1+3^3}} = \dfrac{1}{\sqrt{28}}$$
$$ \therefore \dfrac{1}{3} > \dfrac{1}{\sqrt{28}} $$ we conclude that the Larger integrating value lies in the point 2
$$ \therefore \dfrac{1}{\sqrt{28}} < f(x) < \dfrac{1}{3} $$
Since, m < f(x) < M $$ \Rightarrow \int m dx < \int f(x) dx < \int M dx \, \, \, [ \, \forall \, m , M \in \mathbb{R} $$ and be the lower and upper limit of the f(n) ]
we can conclude that,
$$ \int_2^3 f(x) dx < \dfrac{1}{3} $$
Therefore the value is less than 1. (Ans)
The value of the definite integral $$\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x$$ is
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0
0%
$$\dfrac{\pi}{2}$$
0%
$$\pi$$
0%
$$2 \pi$$
Explanation
$$\sin n x-\sin (n-2) x=2 \cos (n-1) x \sin x\\$$
$$\Rightarrow \displaystyle \int \dfrac{\sin n x}{\sin x} d x=\int 2 \cos (n-1) d x+\int \dfrac{\sin (n-2) x}{\sin x} d x\\$$
$$\therefore \displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x=\int_{0}^{\pi / 2} 2 \cos 4 x d x+\int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\$$
$$=0+\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\=\displaystyle\int_{0}^{\pi / 2} 3-4sin^2x\ \ d x=\displaystyle\int_{0}^{\pi / 2} 1+2cos2x\ \ d x=\dfrac{\pi}{2}$$
Evaluate : $$\displaystyle\int^1_{-2}|2x+1|dx$$
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$$\dfrac{5}{2}$$
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$$\dfrac{7}{2}$$
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$$\dfrac{9}{2}$$
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$$4$$
The value of the definite integral $$\int_{2}^{4}(x(3-x)(4+x)(6-x)$$ $$(10-x)+\sin x) d x$$ equals
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$$\cos 2+\cos 4$$
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$$\cos 2-\cos 4$$
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$$\sin 2+\sin 4$$
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$$\sin 2-\sin 4$$
Explanation
Let $$\left.I=\int_{2}^{4}(x(3-x)(4+x)(6-x)(10-x)+\sin x) d x\right) $$ .........(i)
$$=\int_{2}^{4}((6-x)(3-(6-x))(4+(6-x))(6-(6-x)) $$
..........
$$\because \int_{a}^b f(x) \ dx= \int _{a}^{b} f(a+b-x) \ dx$$
$$=\int_{2}^{4}((6-x)(x-3)(10-x) x(4+x)+\sin (6-x)) d x$$ .......... (ii)
Adding equations (1) and $$(2),$$ we get
$$2 I=\int_{2}^{4}(\sin x+\sin (6-x)) d x \\$$
$$\quad=(-\cos x+\cos (6-x))_{2}^{4} \\$$
$$\quad=-\cos 4+\cos 2+\cos 2-\cos 4 \\$$
$$\quad=2(\cos 2-\cos 4) \\$$
$$\Rightarrow \quad I=\cos 2-\cos 4$$
Let $$f : R \rightarrow R$$ be a function as $$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$$. If $$g(x)$$ is a polynomial of degree $$\leq 3$$ such that $$\displaystyle \int \frac{g(x)}{f(x)} dx$$ does not contain any logarithm function and $$g(-2) = 10$$. Then
$$\displaystyle \int \frac{g(x)}{f(x)} dx$$, equals
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$$\tan^{-1} \left ( \frac{x - 2}{2}
\right ) + c$$
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$$\tan^{-1} \left ( \frac{x - 1}{1}
\right ) + c$$
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$$\tan^{-1} (x) + c$$
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None of these
Explanation
Here, $$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$$
$$ = (x^2 + 4x + 3) (x^2 - 4x - 12) - 100$$
$$ = (x^2 - 4x)^2 - 9(x^2 - 4x) - 136$$
$$ = (x^2 - 4x + 8)(x^2 - 4x + 17)$$
$$\displaystyle \int \frac{g(x)}{f(x)} = \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)}$$
$$ = \frac{Ax + B}{x^2 - 4x - 17} + \frac{Cx + D}{x^2 - 4x + 8}$$
Clearly, $$A, B$$ and $$C$$ must be zero.
$$\therefore \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{D}{x^2 - 4x + 8}$$
$$\therefore$$ $$g(x) = D (x^2 - 4x - 17)$$
$$g(-2) = D (4 + 8 - 17) = -10$$ [given]
$$\Rightarrow$$ $$\frac{g(x)}{f(x)} = \frac{2 (x^2 - 4x - 17)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{2}{x^2 - 4x + 8}$$
$$\therefore$$ $$\displaystyle \int \frac{g(x)}{f(x)} dx = \displaystyle \int \frac{2}{x^2 - 4x + 8}dx = 2 \displaystyle \int \frac{dx}{(x - 2)^2 + (2)^2}$$
$$= 2. \frac{1}{2} \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C = \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C$$
If $$\displaystyle \int_{-2}^{-1} (ax^2-5)dx $$ and
$$5+\displaystyle \int_{1}^{2} (bx+c)dx=0, $$ then
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$$ax^2-bx+c=0 $$ has atleast one root in (1,2)
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$$ax^2-bx+c=0 $$ has atleast one root in (-2,-1)
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$$ax^2+bx+c=0 $$ has atleast one root in (-2,-1)
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None of the above
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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