1

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

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

The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :
A
${8 \over 3}$
B
${{14} \over 3}$
C
${{187} \over {24}}$
D
${{37} \over {24}}$

## Explanation

Area $= \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}$
3

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

The integral  $\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$  equals :
A
${\pi \over {40}}$
B
${1 \over {20}}{\tan ^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)$
C
${1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)} \right)$
D
${1 \over 5}\left( {{\pi \over 4}{{-\tan }^{ - 1}}\left( {{1 \over {3\sqrt 3 }}} \right)} \right)$

## Explanation

I $=$ $\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$

${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}}$   Put tan5x $=$ t

${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$
4

### JEE Main 2019 (Online) 12th January Morning Slot

The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is
A
${{15} \over 4}$
B
${{15} \over 2}$
C
${{21} \over 2}$
D
${{17} \over 4}$

## Explanation

Required area

$= \int\limits_0^3 {\left( {{x^2} + 2} \right)dx - {1 \over 2}.5.3 = 9 + 6 - {{15} \over 2}} {}$

$= {{15} \over 2}$