JEE Main 2021 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 17 March 2021, Shift 2, Maths Q87

Question 87 of 90 in this shift, Maths question 27 of 30, Section B.

Let

f:[-3, 1]\to\mathbf{R}

be given as

f(x)=\begin{cases}\min\{(x+6), x^2\}, & -3\leq x\leq 0\\ \max\{\sqrt{x}, x^2\}, & 0\leq x\leq 1.\end{cases}

If the area bounded by

y=f(x)

and

x

-axis is A, then the value of

6A

is equal to __________.

Official answer

NTA final key.

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

Same idea in other shifts

Asked 8× 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
13 Apr 2023, Shift 1 · Q12The area of the region enclosed by the curve $f(x)=\max\{\sin x,\cos x\}$ , $-\pi\leq x\leq\pi$ and the $x$ -axis isMediumSingle correct
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

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 108 of the 174 JEE Main shifts held from 2017 to 2026.