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1

### IIT-JEE 2012 Paper 2 Offline

The value of the integral $$\int\limits_{ - \pi /2}^{\pi /2} {\left( {{x^2} + 1n{{\pi + x} \over {\pi - x}}} \right)\cos xdx}$$ is
A
$$0$$
B
$${{{\pi ^2}} \over 2} - 4$$
C
$${{{\pi ^2}} \over 2} + 4$$
D
$${{{\pi ^2}} \over 2}$$
2

### IIT-JEE 2011 Paper 2 Offline

Let f $$:$$$$\left[ { - 1,2} \right] \to \left[ {0,\infty } \right]$$ be a continuous function such that
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$

Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,}$$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then

A
$${R_1} = 2{R_2}$$
B
$${R_1} = 3{R_2}$$
C
$${2R_1} = {R_2}$$
D
$${3R_1} = {R_2}$$

## Explanation

$${R_1} = \int\limits_{ - 1}^2 {xf(x)dx = \int\limits_{ - 1}^2 {(2 - 1 - x)f(2 - 1 - x)dx} }$$

$$= \int\limits_{ - 1}^2 {(1 - x)f(1 - x)dx = \int\limits_{ - 1}^2 {(1 - x)f(x)dx} }$$

Hence, $$2{R_1} = \int\limits_{ - 1}^2 {f(x)dx = {R_2}}$$.

3

### IIT-JEE 2011 Paper 1 Offline

Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals
A
$${3 \over 4}$$
B
$${ 1\over 2}$$
C
$${1 \over 3}$$
D
$${1 \over 4}$$

## Explanation

We can write the integral

$$\int\limits_0^b {{{(1 - x)}^2}dx - \int\limits_0^1 {{{(1 - x)}^2}dx = {1 \over 4}} }$$

$$\Rightarrow \left. {{{{{(x - 1)}^3}} \over 3}} \right|_0^b - \left. {{{{{(x - 1)}^3}} \over 3}} \right|_b^1 = {1 \over 4}$$

$$\Rightarrow {{{{(b - 1)}^3}} \over 3} + {1 \over 3} - \left( {0 - {{{{(b - 1)}^3}} \over 3}} \right) = {1 \over 4}$$

$$\Rightarrow {{2{{(b - 1)}^3}} \over 3} = - {1 \over {12}} \Rightarrow {(b - 1)^3} = - {1 \over 8} \Rightarrow b = {1 \over 2}$$

4

### IIT-JEE 2011 Paper 1 Offline

The value of $$\,\int\limits_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {{{x\sin {x^2}} \over {\sin {x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}\,dx}$$ is
A
$${1 \over 4}\,\ell n{3 \over 2}$$
B
$$\,{1 \over 2}\,\ell n{3 \over 2}$$
C
$$\ell n{3 \over 2}$$
D
$$\,\,{1 \over 6}\,\ell n{3 \over 2}$$

## Explanation

$${x^2} = t \Rightarrow 2x\,dx = dt$$

$$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt}$$ and $$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt}$$

$$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}}$$.

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