1

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

The area of the region

A = {(x, y) : 0 $\le$ y $\le$x |x| + 1  and  $-$1 $\le$ x $\le$1} in sq. units, is :
A
${2 \over 3}$
B
2
C
${4 \over 3}$
D
${1 \over 3}$

## Explanation

Required area

$= \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx$

$= 0 + \left( x \right)_{ - 1}^1 = 2$
2

### 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

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

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

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

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

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

$\Rightarrow \,\,k = 2$
3

### 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
4

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