1

### JEE Main 2019 (Online) 10th January Evening Slot

If  $\int\limits_0^x \,$f(t) dt = x2 + $\int\limits_x^1 \,$ t2f(t) dt then f '$\left( {{1 \over 2}} \right)$ is -
A
${{18} \over {25}}$
B
${{6} \over {25}}$
C
${{24} \over {25}}$
D
${{4} \over {5}}$

## Explanation

$\int\limits_0^x \,$f(t) dt = x2 + $\int\limits_x^1 \,$ t2f(t) dt           f '$\left( {{1 \over 2}} \right)$ = ?

Differentiate w.r.t. 'x'

f(x) = 2x + 0 $-$ x2 f(x)

f(x) = ${{2x} \over {1 + {x^2}}}$ $\Rightarrow$ f '(x) = ${{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}$

f '(x) = ${{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}$

f '$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$
2

### JEE Main 2019 (Online) 11th January Morning Slot

The value of the integral $\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$ (where [x] denotes the greatest integer less than or equal to x) is
A
0
B
4
C
4$-$ sin 4
D
sin 4

## Explanation

I $=$ $\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx}$

$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$   as   $\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0}$
3

### JEE Main 2019 (Online) 11th January Morning Slot

If  $\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}$ dx = A(x)${\left( {\sqrt {1 - {x^2}} } \right)^m}$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
A
${1 \over {27{x^6}}}$
B
${{ - 1} \over {27{x^9}}}$
C
${1 \over {9{x^4}}}$
D
${1 \over {3{x^3}}}$

## Explanation

$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}$ dx = A(x)${\left( {\sqrt {1 - {x^2}} } \right)^m}$ + C

$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$

Put  ${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$

Case-I   $x \ge 0$

$- {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$

$\Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$

$\Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$

$A(x) = - {1 \over {3{x^3}}}\,\,$ and $m = 3$

${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$

Case-II  $x \le 0$

We get  ${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$

$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$

${(A(x))^m} = {{ - 1} \over {27{x^9}}}$
4

### JEE Main 2019 (Online) 11th January Morning Slot

The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :
A
${3 \over 4}$
B
${5 \over 4}$
C
${7 \over 8}$
D
${9 \over 8}$

## Explanation

x = 4y $-$ 2 & x2 = 4y

$\Rightarrow$  x2 = x + 2 $\Rightarrow$ x2 $-$ x $-$ 2 = 0

x = 2, $-$ 1

So,   $\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}}$