JEE Main 2024 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2024, Shift 1, Maths Q13

Question 13 of 90 in this shift, Maths question 13 of 30, Section A.

One of the points of intersection of the curves

y=1+3x-2x^2

and

y=\frac{1}{x}

\left(\frac{1}{2}, 2\right)

. Let the area of the region enclosed by these curves be

\frac{1}{24}(l\sqrt{5}+m)-n\log_e(1+\sqrt{5})

, where

l, m, n \in \mathbf{N}

. Then

l+m+n

is equal to

(1) $30$ Official answer
(2) $29$
(3) $31$
(4) $32$

Official answer

Option 1

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

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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
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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
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
8 Apr 2024, Shift 2 · Q12The area of the region in the first quadrant inside the circle $x^2 + y^2 = 8$ and outside the parabola $y^2 = 2x$ is equal to :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.