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1

### JEE Advanced 2013 Paper 1 Offline

Let $$f$$ $$:\,\,\left[ {{1 \over 2},1} \right] \to R$$ (the set of all real number) be a positive,
non-constant and differentiable function such that
$$f'\left( x \right) < 2f\left( x \right)$$ and $$f\left( {{1 \over 2}} \right) = 1.$$ Then the value of $$\int\limits_{1/2}^1 {f\left( x \right)} \,dx$$ lies in the interval
A
$$\left( {2e - 1,2e} \right)$$
B
$$\left( {e - 1,\,2e - 1} \right)$$
C
$$\left( {{{e - 1} \over 2},e - 1} \right)$$
D
$$\left( {0,{{e - 1} \over 2}} \right)$$
2

### JEE Advanced 2013 Paper 1 Offline

The area enclosed by the curves $$y = \sin x + {\mathop{\rm cosx}\nolimits}$$ and $$y = \left| {\cos x - \sin x} \right|$$ over the interval $$\left[ {0,{\pi \over 2}} \right]$$ is
A
$$4\left( {\sqrt 2 - 1} \right)$$
B
$$2\sqrt 2 \left( {\sqrt 2 - 1} \right)$$
C
$$2\left( {\sqrt 2 + 1} \right)$$
D
$$2\sqrt 2 \left( {\sqrt 2 + 1} \right)$$
3

### 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
$${{{\pi ^2}} \over 2}$$
4

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

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