Definite Integration — Page 7 | JEE Mathematics

Q61

\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}

is equal to :

A

{1 \over {15}}

B 0

C

{2 \over 3}

D

{3 \over 2}

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}

This is in

{0 \over 0}

form, so use L' Hospital rule

= \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \over {{d \over {dx}}\left( {{x^3}} \right)}}

= \mathop {\lim }\limits_{x \to {0^ + }} {{\sin x.2x - 0} \over {3{x^2}}}

(applying Leibnitz rule)

= {2 \over 3}\mathop {\lim }\limits_{x \to {0^ + }} {{\sin x} \over x}

= {2 \over 3}

Q62

The value of the integral,

\int\limits_1^3 {[{x^2} - 2x - 2]dx}

, where [x] denotes the greatest integer less than or equal to x, is :

A

-

5

B

- \sqrt 2 - \sqrt 3 + 1

C

-

4

D

- \sqrt 2 - \sqrt 3 - 1

Correct Answer

Option D

Solution

I = \int\limits_1^3 { - 3dx + \int\limits_1^3 {\left[ {{{(x - 1)}^2}} \right]dx} }

Put x $-$ 1 = t ; dx = dt

I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt

I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt}

I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3

I = - 1 - \sqrt 2 - \sqrt 3

Q63

Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2

-

x) for all x

\in

(0, 2), f(0) = 1 and f(2) = e2. Then the value of

\int\limits_0^2 {f(x)} dx

is :

A 1 + e2

B 2(1 + e2)

C 1

-

e2

D 2(1

-

e2)

Correct Answer

Option A

Solution

f'(x) = f'(2 $-$ x) On integrating both side f(x) = $-$ f(2 $-$ x) + c put x = 0 f(0) + f(2) = c $\Rightarrow$ c = 1 + e2 $\Rightarrow$ f(x) + f(2 $-$ x) = 1 + e2 ..... (i)

I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})}

Q64

Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x)

\ne

0 for all x

\in

R. If

\left| \begin{array}{ll}{f(x)} & {f'(x)} \\ {f'(x)} & {f''(x)} \end{array} \right|

= 0, for all x

\in

R, then the value of f(1) lies in the interval :

A (0, 3)

B (9, 12)

C (3, 6)

D (6, 9)

Correct Answer

Option D

Solution

\left| \begin{array}{ll}{f(x)} & {f'(x)} \\ {f'(x)} & {f''(x)} \end{array} \right| = 0

\Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0

Dividing by

{\left( {f(x)} \right)^2}

, we get

\Rightarrow {{f(x).f''(x) - {{\left( {f'(x)} \right)}^2}} \over {{{\left( {f(x)} \right)}^2}}} = 0

\Rightarrow {d \over {dx}}\left( {{{f'(x)} \over {f(x)}}} \right) = 0

Integrating both side,

{{f'(x)} \over {f(x)}} = c

(constant) At,

x = 0

,

{{f'(0)} \over {f(0)}} = c

\Rightarrow {2 \over 1} = c

\Rightarrow c = 2

$\therefore$

{{f'(x)} \over {f(x)}} = 2

\Rightarrow \int {{{f'(x)} \over {f(x)}}} dx = 2\int {dx}

\Rightarrow \ln |f(x)|\, = 2x + c'

at x = 0,

ln|f(0)|\, = 0 + c'

\Rightarrow 0 = 0 + c'

\Rightarrow c' = 0

$\therefore$

n|f(x)| = 2x

\Rightarrow f(x) = {e^{2x}}

f(1) = {e^2} = {(2.71)^2} = 7.34

So it lie between (6, 9).

Q65

The value of

\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx

, where [ t ] denotes the greatest integer

\le

t, is :

A

{{e + 1} \over 3}

B

{{e - 1} \over {3e}}

C

{1 \over {3e}}

D

{{e + 1} \over {3e}}

Correct Answer

Option D

Solution

I = \int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}dx}

Here -1 $\le$ x $\le$ 1 then -1 $\le$ x3 $\le$ 1 Integer between -1 and 1 is 0.

So integration will be divided into two parts, -1 to 0 and 0 to 1.

= \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx}

= \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx}

=

{1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1

= {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}

= {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}

Q66

If the value of the integral

\int\limits_{-1}^1 \dfrac{\cos \alpha x}{1+3^x} d x

is

\dfrac{2}{\pi}

.Then, a value of

\alpha

is

A

\dfrac{\pi}{2}

B

\dfrac{\pi}{4}

C

\dfrac{\pi}{3}

D

\dfrac{\pi}{6}

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Given, } \int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int\limits_0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int\limits_0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}

Q67

If

{I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx}

, then :

A

{1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}

are in A.P.

B I2 + I4, I3 + I5, I4 + I6 are in A.P.

C

{1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}

are in G.P.

D I2 + I4, (I3 + I5)2, I4 + I6 are in G.P.

Correct Answer

Option A

Solution

{I_n} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^n}xdx} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{n - 2}}x(\cos e{c^2}x - 1)dx}

=

\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx

= \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}

= {1 \over {n - 1}} - {I_{n - 2}}

\Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}

\Rightarrow {I_2} + {I_4} = {1 \over 3}

{I_3} + {I_5} = {1 \over 4}

{I_4} + {I_6} = {1 \over 5}

$\therefore$

{1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}

are in A.P.

Q68

\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]

is equal to :

A

{{1 \over 2}}

B

{{1 \over 3}}

C 1

D

{{1 \over 4}}

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]

\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\left( {n + 2} \right)}^2}}} + ... + {n \over {{{\left( {n + \left( {n - 1} \right)} \right)}^2}}}} \right]

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

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

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

= \int\limits_0^1 {{{dx} \over {{{(x + 1)}^2}}} = \left[ {{{ - 1} \over {(x + 1)}}} \right]_0^1 = {1 \over 2}}

Note :

\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{pn} {{1 \over n}f\left( {{r \over n}} \right)} = \int_\alpha ^\beta {f(x)} dx

where,

\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0

(as r = 1) and

\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = p

(as r = pn) Here

\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0

(as r = 0) and

\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over n}

= 1 (as r = n - 1)

Q69

The value of

\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx

is :

A

2\pi

B

{\pi \over 2}

C

4\pi

D

{\pi \over 4}

Correct Answer

Option D

Solution

Let

I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx

.... (1) Replace x with $-$ x, $\therefore$

I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}}

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

2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx

= \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx}

= \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx

[

{{{\cos }^2}x}

is a even function as

f(x) = f( - x)

for

{\cos ^2}x

]

= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx

= 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx

\Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx

= {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}

= {1 \over 2}\left[ {{\pi \over 2} - 0} \right]

= {\pi \over 4}

Q70

The value of

\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} }

, where [ x ] is the greatest integer

\le

x, is :

A 100e

B 100(e

-

1)

C 100(1 + e)

D 100(1

-

e)

Correct Answer

Option B

Solution

\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx}

Here,

n - 1 \le x < n

$\therefore$

[x] = n - 1

$\therefore$

\int\limits_{n - 1}^n {{e^{x - (n - 1)}}} dx

= \left[ {{e^{x - (n - 1)}}} \right]_{n - 1}^n

= {e^1} - {e^0}

= e - 1

Now,

\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)}