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1

JEE Advanced 2016 Paper 2 Offline

MCQ (Single Correct Answer)
Area of the region

$$\left\{ {\left( {x,y} \right) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,5y \le x + 9 \le 15} \right\}$$

is equal to
A
$${1 \over 6}$$
B
$${4 \over 3}$$
C
$${3 \over 2}$$
D
$${5 \over 3}$$

Explanation

Here, $$\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $$

$$\therefore$$ $$y \ge \sqrt {x + 3} $$

$$ \Rightarrow y \ge \left\{ \matrix{ \sqrt {x + 3} ,\,when\,x \ge - 3 \hfill \cr \sqrt { - x - 3} ,\,when\,x \le - 3 \hfill \cr} \right.$$

or, $${y^2} \ge \left\{ \matrix{ x + 3,\,when\,x \ge - 3 \hfill \cr - 3 - x,\,when\,x \le - 3 \hfill \cr} \right.$$

Shown as

Also, $$5y \le (x + 9) \le 15$$

$$ \Rightarrow (x + 9) \ge 5y$$ and $$x \le 6$$

Shown as

$$\therefore$$ $$\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $$

$$\therefore$$ Required area = Area of trapezium ABCD $$-$$ Area of ABE under parabola $$-$$ Area of CDE under parabola

$$ = {1 \over 2}(1 + 2)(5) - \int_{ - 4}^{ - 3} {\sqrt { - (x + 3)} dx - \int_{ - 3}^1 {\sqrt {(x + 3)} dx} } $$

$$ = {{15} \over 2} - \left[ {{{{{( - 3 - x)}^{3/2}}} \over { - {3 \over 2}}}} \right]_{ - 4}^{ - 3} - \left[ {{{{{(x + 3)}^{3/2}}} \over {{3 \over 2}}}} \right]_{ - 3}^1$$

$$ = {{15} \over 2} + {2 \over 3}[0 - 1] - {2 \over 3}[8 - 0]$$

$$ = {{15} \over 2} - {2 \over 3} - {{16} \over 3} = {{15} \over 2} - {{18} \over 3} = {3 \over 2}$$

2

JEE Advanced 2016 Paper 2 Offline

MCQ (Single Correct Answer)
The value of $$\int\limits_{{\pi \over 2}}^{{\pi \over 2}} {{{{x^2}\cos x} \over {1 + {e^x}}}dx} $$ is equal to
A
$${{{\pi ^2}} \over 4} - 2$$
B
$${{{\pi ^2}} \over 4} + 2$$
C
$${\pi ^2} - {e^{{\pi \over 2}}}$$
D
$${\pi ^2} + {e^{{\pi \over 2}}}$$

Explanation

Let $$I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos x} \over {1 + {e^x}}}dx} $$ ...... (i) [$$\because$$ $$\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} } $$]

$$ \Rightarrow I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos ( - x)} \over {1 + {e^{ - x}}}}dx} $$ ...... (ii)

On adding Eqs. (i) and (ii), we get

$$2I = \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\left[ {{1 \over {1 + {e^x}}} + {1 \over {1 + {e^{ - x}}}}} \right]dx} $$

$$ = \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\,.\,(1)dx} $$ [$$\because$$ $$\int_{ - a}^a {f(x)dx} $$ $$ = 2\int_0^a {f(x)dx} $$, when $$f( - x) = f(x)$$]

$$ \Rightarrow 2I = 2\int_0^{\pi /2} {{x^2}\cos x\,dx} $$

Using integration by parts, we get

$$2I = 2[{x^2}(\sin x) - (2x)( - \cos x) + (2)( - \sin x)]_0^{\pi /2}$$

$$ \Rightarrow 2I = 2\left[ {{{{\pi ^2}} \over 4} - 2} \right]$$

$$\therefore$$ $$I = {{{\pi ^2}} \over 4} - 2$$

3

JEE Advanced 2015 Paper 2 Offline

MCQ (Single Correct Answer)
Let $$f'\left( x \right) = {{192{x^3}} \over {2 + {{\sin }^4}\,\pi x}}$$ for all $$x \in R\,\,$$ with $$f\left( {{1 \over 2}} \right) = 0$$.
If $$m \le \int\limits_{1/2}^1 {f\left( x \right)dx \le M,} $$ then the possible values of $$m$$ and $$M$$ are
A
$$m=13,$$ $$M=24$$
B
$$\,m = {1 \over 4},M = {1 \over 2}$$
C
$$m=-11,$$ $$M=0$$
D
$$m=1,$$ $$M=12$$

Explanation

We have, $$f'(x) = {{192{x^3}} \over {2 + {{\sin }^4}\pi x}}$$

Given, $$f\left( {{1 \over 2}} \right) = 0$$ and $$m \le \int\limits_{1/2}^1 {f(x)dx \le m} $$

$$\therefore$$ f(x) is increasing in $$\left( {{1 \over 2},1} \right)$$.

$$\therefore$$ $$f'{(x)_{\max }} = {{192} \over {24}} = 96$$

$$ \Rightarrow 96 = {{f(1) - f(1/2)} \over {1/2}} \Rightarrow f(1) = 96 \times {1 \over 2} = 48$$

$$M = {1 \over 2} \times {1 \over 2} \times 48 = 12$$

$$f'{(x)_{\min .}} = {{\left( {{{192} \over 8}} \right)} \over 3} = {{192} \over {24}}$$

$$ \Rightarrow {{192} \over {24}} = {{f(1) - 0} \over {(1/2)}} \Rightarrow f(1) = {{192} \over {24}} \times {1 \over 2}$$

$$\therefore$$ $$m = {1 \over 2} \times {1 \over 2} \times {{192} \over {24}} \times {1 \over 2} = 1$$

4

JEE Advanced 2014 Paper 2 Offline

MCQ (Single Correct Answer)
Given that for each $$a \in \left( {0,1} \right),\,\,\,\mathop {\lim }\limits_{h \to {0^ + }} \,\int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt} $$ exists. Let this limit be $$g(a).$$ In addition, it is given that the function $$g(a)$$ is differentiable on $$(0,1).$$

The value of $$g\left( {{1 \over 2}} \right)$$ is

A
$$\pi $$
B
$$2\pi $$
C
$${\pi \over 2}$$
D
$${\pi \over 4}$$

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