1

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

The value of   $\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,$  where [t] denotes the greatest integer less than or equal to t, is
A
${1 \over {12}}\left( {7\pi - 5} \right)$
B
${1 \over {12}}\left( {7\pi + 5} \right)$
C
${3 \over {10}}\left( {4\pi - 3} \right)$
D
${3 \over {20}}\left( {4\pi - 3} \right)$

## Explanation

${\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}}$

$= \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} }$

$\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}}$

$\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)$

$- 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}$

${{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}$

${{ - 9} \over {20}} + {{3\pi } \over 5}$
2

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

### 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   $\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0}$
4

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

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

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