NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

IIT-JEE 2012 Paper 2 Offline

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

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

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

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12