QJEEM22Ma5228: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5228 Solution: Let and . If 27 (Area A 1 ) = 5 (Area A 2 ), then k is equal to :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted 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

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