JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 21)

Let

$${A_1} = \left\{ {(x,y):|x| \le {y^2},|x| + 2y \le 8} \right\}$$ and

$${A_2} = \left\{ {(x,y):|x| + |y| \le k} \right\}$$. If 27 (Area A₁) = 5 (Area A₂), then k is equal to :

Answer

JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Area Under The Curves Question 75 English Explanation 1

Required area (above x-axis)

$${A_1} = 2\int\limits_0^4 {\left( {{{8 - x} \over 2} - \sqrt x } \right)dx} $$

$$ = 2\left( {16 - {{16} \over 4} - {8 \over {3/2}}} \right) = {{40} \over 3}$$

and $${A_2} = 4\left( {{1 \over 2}\,.\,{k^2}} \right) = 2{k^2}$$

JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Area Under The Curves Question 75 English Explanation 2

$$\therefore$$ $$27\,.\,{{40} \over 3} = 5\,.\,(2{k^2})$$

$$\Rightarrow$$ k = 6