1

### JEE Main 2019 (Online) 9th January Evening Slot

If   $\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$ then value of k is :
A
4
B
${1 \over 2}$
C
1
D
2

## Explanation

$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$

$\Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)$

$\Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}$

$\Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} \right) = 1 - {1 \over {\sqrt 2 }}$

$\Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }}$

$\Rightarrow \,\,k = 2$
2

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

Let  ${\rm I} = \int\limits_a^b {\left( {{x^4} - 2{x^2}} \right)} dx.$  If I is minimum then the ordered pair (a, b) is -
A
$\left( {\sqrt 2 , - \sqrt 2 } \right)$
B
$\left( {0,\sqrt 2 } \right)$
C
$\left( { - \sqrt 2 ,\sqrt 2 } \right)$
D
$\left( { - \sqrt 2 ,0} \right)$

## Explanation

Let f(x) = x2(x2 $-$ 2)

As long as f(x) lie below the x-axis, define integral will remain negative,
so correct value of (a, b) is ($-$ $\sqrt 2$, $\sqrt 2$) for minimum of I
3

### 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 $= {1 \over {\sqrt 3 }}$
4

### 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

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

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

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

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

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

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

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

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