Definite Integration — Page 9 | JEE Mathematics

Q81

If the real part of the complex number

{(1 - \cos \theta + 2i\sin \theta )^{ - 1}}

is

{1 \over 5}

for

\theta \in (0,\pi )

, then the value of the integral

\int_0^\theta {\sin x} dx

is equal to:

A 1

B 2

C

-

1

D 0

Correct Answer

Option A

Solution

z = {1 \over {1 - \cos \theta + 2i\sin \theta }}

= {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}

= {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}

{\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}

\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}

1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}

3{\cos ^2}{\theta \over 2} = {1 \over 2}

{\theta \over 2} = n\pi \pm {\pi \over 4}

\theta = 2n\pi \pm {\pi \over 2}

\theta \in (0,\pi )

$\therefore$

\theta = {\pi \over 2}

\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}}

= - (0 - 1)

= 1

Q82

Let

g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx

, where

f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R

. Then which one of the following is correct?

A g(1) = g(0)

B

\sqrt 2 g(1) = g(0)

C

g(1) = \sqrt 2 g(0)

D g(1) + g(0) = 0

Correct Answer

Option B

Solution

$\because$

f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)

$\therefore$

f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) + \ln \left( {\sqrt {{x^2} + 1} - x} \right)

$\therefore$

f(x) + f( - x) = 0

.... (i) $\because$

g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx

= \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi \over 4}t + f(x)} \right) + \cos \left( {{\pi \over 4}t + f( - x)} \right)} \right\}} dx

= \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx}

g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx}

$\therefore$

g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx

and

g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx

$\therefore$

\sqrt 2 g(1) = g(0)

Q83

If

\int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^2}}},\alpha \in R}

where [x] is the greatest integer less than or equal to x, then the value of

\alpha

is :

A 200 (1

-

e

-

1)

B 100 (1

-

e)

C 50 (e

-

1)

D 150 (e

-

1

-

1)

Correct Answer

Option A

Solution

I = \int_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx}

$\because$ Integrand is periodic with period 1 $\therefore$

I = 100\int_0^\pi {{{{{\sin }^2}x} \over {{e^{\left\{ {{x \over \pi }} \right\}}}}}} dx

Let

{x \over \pi } = t \Rightarrow dx = \pi dt

= 100\pi \int_0^1 {{{{{\sin }^2}(\pi t)dt} \over {{e^t}}}}

= 50\pi \int_0^1 {{e^{ - t}}(1 - \cos 2\pi t)dt}

= 50\pi \int_0^1 {{e^{ - t}}dt - 50\pi \int_0^1 {{e^{ - t}}} \cos (2\pi t)dt}

= - 50\left[ {{e^{ - t}}} \right]_0^1 - 50\pi \left[ {{{{e^{ - t}}} \over {1 + 4{\pi ^2}}}( - \cos 2\pi t + 2\pi \sin 2\pi t)} \right]_0^1

= - 50\pi ({e^{ - 1}} - 1) - {{50\pi } \over {1 + 4{\pi ^2}}}({e^{ - 1}}( - 1 + 0) - ( - 1 + 0))

= - 50\pi ({e^{ - 1}} - 1) - {{50\pi (1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}}

= {{200{\pi ^3}(1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}} = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^3}}}

(Given) $\therefore$

\alpha = 200(1 - {e^{ - 1}})

Q84

The value of the definite integral

\int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \sqrt[3]{\tan 2x} }}}

is :

A

{\pi \over 3}

B

{\pi \over 6}

C

{\pi \over {12}}

D

{\pi \over {18}}

Correct Answer

Option C

Solution

Let

I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\cos 2x)}^{1/3}}} \over {{{(\cos 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx}

...... (i)

\Rightarrow I = \int\limits_{\pi /2}^{5\pi /24} {{{{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}} \over {{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}} + {{\left( {\sin \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}}}} dx\left\{ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right\}

So,

I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\sin 2x)}^{1/3}}} \over {{{(\sin 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx}

..... (ii) Hence,

2I = \int\limits_{\pi /24}^{5\pi /24} {dx}

[(i) + (ii)]

\Rightarrow 2I = {{4\pi } \over {24}} \Rightarrow I = {\pi \over {12}}

Q85

Let

f:[0,\infty ) \to [0,\infty )

be defined as

f(x) = \int_0^x {[y]dy}

where [x] is the greatest integer less than or equal to x. Which of the following is true?

A f is continuous at every point in

[0,\infty )

and differentiable except at the integer points.

B f is both continuous and differentiable except at the integer points in

[0,\infty )

.

C f is continuous everywhere except at the integer points in

[0,\infty )

.

D f is differentiable at every point in

[0,\infty )

.

Correct Answer

Option A

Solution

f:[0,\infty ) \to [0,\infty ),f(x) = \int_0^x {[y]dy}

Let

x = n + f,f \in (0,1)

So,

f(x) = 0 + 1 + 2 + ... + (n - 1) + \int\limits_n^{n + f} {n\,dy}

f(x) = {{n(n - 1)} \over 2} + nf

= {{[x]([x] - 1)} \over 2} + [x]\{ x\}

Note

\mathop {\lim }\limits_{x \to {n^ + }} f(x) = {{n(n - 1)} \over 2},\mathop {\lim }\limits_{x \to {n^ - }} f(x) = {{(n - 1)(n - 2)} \over 2} + (n - 1)

= {{n(n - 1)} \over 2}

f(x) = {{n(n - 1)} \over 2}(n \in {N_0})

so f(x) is cont. $\forall$ x $\ge$ 0 nd diff. except at integer points

Q86

Let f : (a, b)

\to

R be twice differentiable function such that

f(x) = \int_a^x {g(t)dt}

for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :

A twelve roots in (a, b)

B five roots in (a, b)

C seven roots in (a, b)

D three roots in (a, b)

Correct Answer

Option C

Solution

f(x) = \int_a^x {g(t)dt}

$\Rightarrow$ f′(x) = g(x) $\Rightarrow$ f′'(x) = g'(x) Given, g(x).g'(x) = 0 $\Rightarrow$ f′(x).f′'(x) = 0 Also given f(x) has exactly 5 roots.

So from Rolle's theorem we can say, f′(x) has 4 roots and f′'(x) has 3 roots. $\therefore$ f′(x).f′'(x) = 0 has 4 + 3 = 7 roots.

Q87

The integral

16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}}

is equal to

A

{{11} \over {12}} + {\log _e}4

B

{{11} \over 6} + {\log _e}4

C

{{11} \over {12}} - {\log _e}4

D

{{11} \over 6} - {\log _e}4

Correct Answer

Option D

Solution

$I=\int \dfrac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$ $=\dfrac{1}{4} \int \dfrac{x}{x^{2}+2} d x+\dfrac{1}{4} \int \dfrac{x}{\left(x^{2}+2\right)^{2}}-\dfrac{1}{4} \int \dfrac{d x}{x}+\dfrac{1}{4} \int \dfrac{d x}{x^{3}}$ $=\dfrac{1}{8} \ln \left(x^{2}+2\right)-\dfrac{\ln x}{4}-\dfrac{1}{8\left(x^{2}+2\right)}-\dfrac{1}{8 x^{3}}$ Now, $16 \int_{1}^{2} \dfrac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\dfrac{11}{6}$ $=\dfrac{11}{6}-\ln 4$

Q88

The value of the definite integral

\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

is equal to :

A

- {\pi \over 2}

B

{\pi \over {2\sqrt 2 }}

C

- {\pi \over 4}

D

{\pi \over {\sqrt 2 }}

Correct Answer

Option B

Solution

I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

.... (1) Using

\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} }

I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{ - x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

Add (1) and (2)

2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}}

2I = 2\int\limits_0^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}}

I = \int\limits_0^{{\pi \over 4}} {{{\left( {1 + {1 \over {{{\tan }^2}x}}} \right){{\sec }^2}x} \over {{{\left( {\tan - {1 \over {\tan x}}} \right)}^2} + 2}}} dx

\tan x - {1 \over {\tan x}} = t

\left( {1 + {1 \over {{{\tan }^2}x}}} \right){\sec ^2}xdx = dt

I = \int\limits_{ - \infty }^0 {{{dt} \over {{t^2} + 2}}} = \left[ {{1 \over {\sqrt 2 }}{{\tan }^{ - 1}}\left( {{t \over {\sqrt 2 }}} \right)} \right]_{ - \infty }^0

I = 0 - {1 \over {\sqrt 2 }}\left( { - {\pi \over 2}} \right) = {\pi \over {2\sqrt 2 }}

Q89

The value of

\int\limits_{{{ - 1} \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left( {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right)}^{{1 \over 2}}}dx}

is :

A loge 4

B loge 16

C 2loge 16

D 4loge (3 + 2

{\sqrt 2 }

)

Correct Answer

Option B

Solution

\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2}

= \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}}

= \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|

= \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|

= \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|

= \left| {{{4x} \over {{x^2} - 1}}} \right|

$\therefore$

I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx}

I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}

Let

f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|

$\therefore$

f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|

\Rightarrow f(x) = f( - x)

$\therefore$ f(x) is a even function. $\therefore$

I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}

Using property, If f(x) is an even function then,

\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} }

x > 0 when

x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]

$\Rightarrow$ x2 > 0 when

x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]

$\Rightarrow$ x2 $-$ 1

\Rightarrow {1 \over {{x^2} - 1}}

\Rightarrow {{4x} \over {{x^2} - 1}} $\therefore$

\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}

when

x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]

$\therefore$

I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx}

= - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx}

= - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}

= - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]

= - 4{\log _e}\left| { - {1 \over 2}} \right|

= - 4{\log _e}{1 \over 2}

= - 4\log _e^{{2^{ - 1}}}

= 4\log _e^2

= \log _e^{{2^4}}

= \log _e^{16}

Q90

The value of

\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}}

is :

A

{1 \over 2}{\tan ^{ - 1}}(2)

B

{1 \over 2}{\tan ^{ - 1}}(4)

C

{\tan ^{ - 1}}(4)

D

{1 \over 4}{\tan ^{ - 1}}(4)

Correct Answer

Option B

Solution

L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}}

\Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx}

\Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4