JEE Main Maths · Application of Integrals

Area Involving Max/Min Functions

Area Involving Max/Min Functions (Application of Integrals) was asked 6 times in 6 JEE Main shifts, 2022–2026, out of the 42 shifts Lumi analysed. 3 of the 6 had a numerical answer.

Times asked: 6; in 6 shifts
Years: 2022–2026
Usual format: Numerical value; Calculation
Pattern: Asked most years; repeats across shifts

How it is asked

Area bounded by parabola and min of lines
Area bounded by curves with modulus

Difficulty

Easy 0Medium 4Hard 2

Where it sits

Chapter: Application of Integrals. Topic: Area Under Curves, 41 questions over 15 ideas.

All 6 questions

Newest first, each with its options and official answer.

4 Apr 2026, Shift 1 · Q18The area of the region $\{(x,y): y\le\pi-|x|,\ y\le|x\sin x|,\ y\ge0\}$ is:HardSingle correct
3 Apr 2025, Shift 1 · Q25The area of the region bounded by the curve $y = \max\{|x|, x|x-2|\}$ , the $x$ -axis and the lines $x = -2$ and $x = 4$ is equal to ______MediumNumerical value
7 Apr 2025, Shift 2 · Q20If the area of the region $\{(x,y): 1+x^2\le y\le \min\{x+7,\ 11-3x\}\}$ is $A$ , then $3A$ is equal toMediumSingle correct
8 Apr 2024, Shift 1 · Q27Let the area of the region enclosed by the curve $y=\min\{\sin x,\cos x\}$ and the $x$ -axis between $x=-\pi$ to $x=\pi$ be $A$ . Then $A^2$ …MediumNumerical value
8 Apr 2023, Shift 2 · Q27Let the area enclosed by the lines $x + y = 2$ , $y = 0$ , $x = 0$ and the curve $f(x) = \min\left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$ …HardNumerical value
25 Jul 2022, Shift 1 · Q8The area of the region given by $A=\{(x,y):x^2\le y\le\min\{x+2,\,4-3x\}\}$ is :MediumSingle correct