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Definite Integration

JEE Mathematics · 230 questions · Page 5 of 23 · Click an option or "Show Solution" to reveal answer

Q41
If f(x)=2−xcos⁡x2+xcos⁡xf(x) = {{2 - x\cos x} \over {2 + x\cos x}} and g(x) = logex, (x > 0) then the value of integral ∫−π4π4g(f(x))dx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}} is
A loge3
B loge2
C loge1
D logee
Correct Answer
Option C
Solution
g(f(x))g\left( {f\left( x \right)} \right)

=

ln⁡(f(x))\ln \left( {f\left( x \right)} \right)

=

ln⁡(2−xcos⁡x2+xcos⁡x)\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)

∴\therefore I =

∫−π4π4ln⁡(2−xcos⁡x2+xcos⁡x)dx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx}

( Using property

∫−aaf(x)dx=∫0a(f(x)+f(−x))dx\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx

) I =

∫0π4(ln⁡(2−xcos⁡x2+xcos⁡x)+ln⁡(2+xcos⁡x2−xcos⁡x))dx\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx}

=

∫0π4(ln⁡(1))dx\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx}

= 0 = loge1

Q42
lim⁡n→∞[11+n+12+n+13+n + ... + 12n]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right] is equal to
A 0
B log⁡e2{\log _e}2
C log⁡e(23){\log _e}\left( {{2 \over 3}} \right)
D log⁡e(32){\log _e}\left( {{3 \over 2}} \right)
Correct Answer
Option B
Solution
lim⁡n→∞[11+n+12+n+13+n+……+12n]=lim⁡n→∞∑r=1n1r+n=lim⁡n→∞∑r=1n1n(1rn+1)=∫01dxx+1=log⁡e(1+x)∣01=log⁡e2\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\ & =\int_0^1 \frac{d x}{x+1} \\\\ & =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\ & =\log _e^2 \end{aligned}
Q43
∫−ππ∣π−∣x∣∣dx\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} is equal to :
A π2{\pi ^2}
B 2π2{\pi ^2}
C 2π2\sqrt 2 {\pi ^2}
D π22{{{\pi ^2}} \over 2}
Correct Answer
Option A
Solution
∫−ππ∣π−∣x∣∣dx\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx}

=

2∫0π∣π−∣x∣∣dx2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|} dx

[As it is even function] =

2∫0π(π−x)dx2\int\limits_0^\pi {\left( {\pi - x} \right)} dx

=

2[πx−x22]0π2\left[ {\pi x - {{{x^2}} \over 2}} \right]_0^\pi

=

π2{\pi ^2}
Q44
If the value of the integral ∫012x2(1−x2)32dx\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx is k6{k \over 6}, then k is equal to :
A 23+π2\sqrt 3 + \pi
B 32−π3\sqrt 2 - \pi
C 32+π3\sqrt 2 + \pi
D 23−π2\sqrt 3 - \pi
Correct Answer
Option D
Solution
I=∫012x2(1−x2)32dxI = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{(1 - {x^2})}^{{3 \over 2}}}}}} dx

Let

x=sin⁡θx = \sin \theta
⇒dx=cos⁡θdθ\Rightarrow dx = \cos \theta d\theta

When x = 0 then sinθ\theta = 0 ⇒\Rightarrow θ\theta = 0 When

x=12x = {1 \over 2}

then

sin⁡θ=12\sin \theta = {1 \over 2}

⇒\Rightarrow

θ=π6\theta = {\pi \over 6}

∴\therefore

I=∫0π6sin⁡2θ(cos⁡2θ)32cos⁡θdθI = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{({{\cos }^2}\theta )}^{{3 \over 2}}}}}\cos \theta d\theta }
=∫0π6sin⁡2θcos⁡2θdθ= \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}d\theta }
=∫0π6tan⁡2dθ=∫0π6(sec⁡2θ−1)dθ= \int\limits_0^{{\pi \over 6}} {{{\tan }^2}d\theta } = \int\limits_0^{{\pi \over 6}} {({{\sec }^2}\theta - 1)d\theta }
=[tan⁡θ−θ]0π6= \left[ {\tan \theta - \theta } \right]_0^{{\pi \over 6}}
=13−π6= {1 \over {\sqrt 3 }} - {\pi \over 6}

Given,

k6=13−π6=23−π6{k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}
⇒k=23−π\Rightarrow k = 2\sqrt 3 - \pi
Q45
Suppose f(x) is a polynomial of degree four, having critical points at –1, 0, 1. If T = {x ∈ \in R | f(x) = f(0)}, then the sum of squares of all the elements of T is :
A 6
B 2
C 8
D 4
Correct Answer
Option D
Solution

Critical points = −-1, 0, 1. ∴\therefore f'(x) = a(x −- 1)(x + 1)x ∴\therefore f(x) = a

(x44−x22)+C\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C

∵\because f(x) = f(0)

−a(x44−x22)+C=C- a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C
ax24(x2−2)=0a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0

∴\therefore x = 0,

2\sqrt 2

,

−2- \sqrt 2

∴\therefore T =

{0,2,−2}\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}

Sum of square of elements of

T=02+(2)2+(−2)2=4T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4
Q46
Let f(x)=∣x−2∣f(x) = \left| {x - 2} \right| and g(x) = f(f(x)), x∈[0,4]x \in \left[ {0,4} \right]. Then ∫03(g(x)−f(x))dx\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx is equal to:
A 1
B 0
C 12{1 \over 2}
D 32{3 \over 2}
Correct Answer
Option A
Solution
f(x)=∣x−2∣f(x) = |x - 2|

∴\therefore

f(f(x))=∣∣x−2∣−2∣=g(x)f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)
⇒g(x)=∣∣x−2∣−2∣={∣x−4∣if x≥2∣−x∣if x<2\Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ \begin{array}{ll}{|x - 4|} & {if\,x \ge 2} \\ {| - x|} & {if\,x < 2} \end{array} \right.

∴\therefore

∫03(g(x)−f(x))dx\int\limits_0^3 {(g(x) - f(x))dx}
=∫03g(x)dx−∫03f(x)dx= \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx}
=∫02∣−x∣dx+∫23−(x−4)dx−∫02−(x−2)dx−∫23(x−2)dx= \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx}
=∫02x dx−∫23(x−4)dx+∫02(x−2)dx−∫23(x−2)dx= \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx}
=[x22]02−[x22−4x]23+[x22−2x]02−[x22−2x]23= \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3
=2−{92−12−2+8}+{2−4}−{92−6−2+4}= 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}
=2−{92−6}−2−{92−4}= 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}
=10−92−92=20−9−92=22=1= 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1
Q47
The integral ∫π6π3tan⁡3x.sin⁡23x(2sec⁡2x.sin⁡23x+3tan⁡x.sin⁡6x)dx\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} is equal to:
A −19 - {1 \over {9}}
B −118 - {1 \over {18}}
C 718 {7 \over {18}}
D 92{9 \over 2}
Correct Answer
Option B
Solution

Given, I =

∫π6π3tan⁡3x.sin⁡23x(2sec⁡2x.sin⁡23x+3tan⁡x.sin⁡6x)dx\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx}
I=∫π/6π/3(2.tan⁡3xsec⁡2xsin⁡43x+3tan⁡4xsin⁡23x. 2sin⁡3xcos 3x )dxI = \int\limits_{\pi /6}^{\pi /3} ({2.{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3{\tan ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx
=12∫π/6π/3(4tan⁡3xsec⁡2xsin⁡43x+3.4tan⁡4xsin⁡33xcos 3x )dx= {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{\tan ^4}x{\sin ^3}3xcos\,3x\,)dx
=12∫π/6π/3ddx(tan⁡4xsin⁡43x)dx= {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{\tan }^4}x{{\sin }^4}3x} \right)} dx
=12[tan⁡4xsin⁡43x]π/6π/3= {1 \over 2}\left[ {{{\tan }^4}x{{\sin }^4}3x} \right]_{\pi /6}^{\pi /3}
=12[9.(0)−13.13(1)]=−118= {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} \right] = - {1 \over {18}}
Q48
The value of ∫−π2π211+esin⁡xdx\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} is:
A π\pi
B 3π2{{3\pi \over 2}}
C π2{{\pi \over 2}}
D π4{{\pi \over 4}}
Correct Answer
Option C
Solution

I =

∫−π2π211+esin⁡xdx\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx}

....(1) Replacing x with

(π2−π2+x)\left( {{\pi \over 2} - {\pi \over 2} + x} \right)

, we get I =

∫−π2π2dx1+e−sin⁡x\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{dx} \over {1 + {e^{ - \sin x}}}}}

=

∫−π2π2esin⁡xdx1+esin⁡x\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}}

.....(2) Adding (1) and (2), we get 2I =

∫−π2π21dx\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx}

=

(π2−(−π2))\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)

= π\pi ⇒\Rightarrow I =

π2{{\pi \over 2}}
Q49
The value of lim⁡n→∞1n∑j=1n(2j−1)+8n(2j−1)+4n\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{(2j - 1) + 8n} \over {(2j - 1) + 4n}}} is equal to :
A 5+log⁡e(32)5 + {\log _e}\left( {{3 \over 2}} \right)
B 2−log⁡e(23)2 - {\log _e}\left( {{2 \over 3}} \right)
C 3+2log⁡e(23)3 + 2{\log _e}\left( {{2 \over 3}} \right)
D 1+2log⁡e(32)1 + 2{\log _e}\left( {{3 \over 2}} \right)
Correct Answer
Option D
Solution
lim⁡n→∞1n∑j=1n(2jn−1n+8)(2jn−1n+4)\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{\left( {{{2j} \over n} - {1 \over n} + 8} \right)} \over {\left( {{{2j} \over n} - {1 \over n} + 4} \right)}}}
∫012x+82x+4dx=∫01dx+∫0142x+4dx\int\limits_0^1 {{{2x + 8} \over {2x + 4}}dx = \int\limits_0^1 {dx + \int\limits_0^1 {{4 \over {2x + 4}}} dx} }
=1+412(ln⁡∣2x+4∣)01= 1 + 4{1 \over 2}(\ln |2x + 4|)_0^1
=1+2ln⁡(32)= 1 + 2\ln \left( {{3 \over 2}} \right)
Q50
lim⁡x→1(∫0(x−1)2tcos⁡(t2)dt(x−1)sin⁡(x−1))\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)
A is equal to 0
B is equal to 12{1 \over 2}
C does not exist
D is equal to −12 - {1 \over 2}
Correct Answer
Option A
Solution
lim⁡x→1(∫0(x−1)2tcos⁡(t2)dt(x−1)sin⁡(x−1))\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)
(00)\left( {{0 \over 0}} \right)

Apply L Hospital Rule =

lim⁡x→12(x−1).(x−1)2cos⁡(x−1)4−0(x−1).cos⁡(x−1)+sin⁡(x−1)\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1} \right).{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4} - 0} \over {\left( {x - 1} \right).\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}}
(00)\left( {{0 \over 0}} \right)

=

lim⁡x→12(x−1)3cos⁡(x−1)4(x−1).[cos⁡(x−1)+sin⁡(x−1)(x−1)]\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^3}\cos {{\left( {x - 1} \right)}^4}} \over {\left( {x - 1} \right).\left[ {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right]}}

=

lim⁡x→12(x−1)2cos⁡(x−1)4cos⁡(x−1)+sin⁡(x−1)(x−1)\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4}} \over {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}}}

on taking limit =

01+1{0 \over {1 + 1}}

= 0

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