JEE Main 2022 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 25 July 2022, Shift 1, Maths Q8

Question 8 of 90 in this shift, Maths question 8 of 30, Section A.

The area of the region given by

A=\{(x,y):x^2\le y\le\min\{x+2,\,4-3x\}\}

is :

(1) $\frac{31}{8}$
(2) $\frac{17}{6}$ Official answer
(3) $\frac{19}{6}$
(4) $\frac{27}{8}$

Official answer

Option 2

NTA final key (2022 Session 2).

Chapter: Application of Integrals
Topic: Area Under Curves
Idea tested: Area Involving Max/Min Functions

Same idea in other shifts

Asked 6× in all

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

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.