JEE Main 2020 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 6 September 2020, Shift 2, Maths Q62

Question 62 of 75 in this shift, Maths question 12 of 25.

The area (in sq. units) of the region enclosed by the curves

y=x^2-1

and

y=1-x^2

is equal to :

(1) $\frac{16}{3}$
(2) $\frac{7}{2}$
(3) $\frac{8}{3}$ Official answer
(4) $\frac{4}{3}$

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Application of Integrals
Topic: Area Under Curves
Idea tested: Area Between Curves

Same idea in other shifts

Asked 12× in all

2 Apr 2026, Shift 2 · Q24If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$ , then $3(A + 6\log_e(3))$ is equal toHardNumerical value
5 Apr 2026, Shift 2 · Q24Let $f:\mathbf{R}\to\mathbf{R}$ be a function such that $f(x)+3f\left(\frac{\pi}{2}-x\right)=\sin x$ , $x \in \mathbf{R}$ . Let the maximum…HardNumerical value
6 Apr 2026, Shift 1 · Q20Let $e$ be the base of natural logarithm and let $f:\{1,2,3,4\}\to\{1,e,e^2,e^3\}$ and…HardSingle correct
6 Apr 2026, Shift 2 · Q19The area of the region $\{(x,y): x^2-8x\le y\le -x\}$ is :EasySingle correct
8 Apr 2026, Shift 2 · Q18Let $f : (1, \infty) \to \mathbf{R}$ be a function defined as $f(x) = \frac{x-1}{x+1}$ . Let $f^{i+1}(x) = f\left(f^{i}(x)\right)$ ,…HardSingle correct
7 Apr 2025, Shift 1 · Q19If the area of the region bounded by the curves $y = 4 - \frac{x^2}{4}$ and $y = \frac{x-4}{2}$ is equal to $\alpha$ , then $6\alpha$ equalsEasySingle correct
31 Jan 2024, Shift 2 · Q11The area of the region enclosed by the parabolas $y=4x-x^2$ and $3y=(x-4)^2$ is equal toMediumSingle correct
29 Jun 2022, Shift 1 · Q6The area enclosed by $y^2 = 8x$ and $y = \sqrt{2}\,x$ that lies outside the triangle formed by $y = \sqrt{2}\,x$ , $x = 1$ ,…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 42 of the 174 JEE Main shifts held from 2017 to 2026.