JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 31 January 2024, Shift 2, Maths Q11

Question 11 of 90 in this shift, Maths question 11 of 30, Section A.

The area of the region enclosed by the parabolas

y=4x-x^2

and

3y=(x-4)^2

is equal to

(1) $\frac{14}{3}$
(2)6Official answer
(3) $\frac{32}{9}$
(4)4

Official answer

Option 2

NTA final key.

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

Same idea in other shifts

Asked 22× 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
4 Apr 2024, Shift 1 · Q13One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$ . Let the area of the…HardSingle correct
6 Apr 2024, Shift 1 · Q12Let the area of the region enclosed by the curves $y=3x$ , $2y=27-3x$ and $y=3x-x\sqrt{x}$ be $A$ . Then $10A$ is equal toMediumSingle 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.