1

JEE Main 2019 (Online) 10th January Morning Slot

If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is -
A
$\sqrt 3$
B
${{\sqrt 3 } \over 2}$
C
${2 \over {\sqrt 3 }}$
D
${1 \over {\sqrt 3 }}$

Explanation

Area bounded by

y2 = 4ax & x2 = 4by, a, b $\ne$ 0

is $\left| {{{16ab} \over 3}} \right|$

by using formula :

4a $=$ ${1 \over k} = 4b,k > 0$

Area $= \left| {{{16.{1 \over {4k}}.{1 \over {4k}}} \over 3}} \right| = 1$

$\Rightarrow$  k2 $= {1 \over 3}$

$\Rightarrow$  k $= {1 \over {\sqrt 3 }}$
2

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

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

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

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

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

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

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

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

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

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

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

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