JEE Main 2022 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 28 June 2022, Shift 2, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

The area of the bounded region enclosed by the curve

y=3-\left|x-\frac{1}{2}\right|-|x+1|

and the

x

-axis is :

(1) $\frac{9}{4}$
(2) $\frac{45}{16}$
(3) $\frac{27}{8}$ Official answer
(4) $\frac{63}{16}$

Official answer

Option 3

NTA final key.

Chapter: Application of Integrals
Topic: Area Under Curves
Idea tested: Area with Modulus

Same idea in other shifts

Asked 7× in all

2 Apr 2025, Shift 1 · Q21If the area of the region $\left\{(x, y): \left|4-x^2\right|\le y\le x^2,\ y\le4,\ x\ge0\right\}$ is…HardNumerical value
5 Apr 2024, Shift 2 · Q12The area enclosed between the curves $y = x|x|$ and $y = x - |x|$ is :MediumSingle correct
6 Apr 2023, Shift 2 · Q14The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal toEasySingle correct
10 Apr 2023, Shift 2 · Q26If the area of the region $\{(x, y) : |x^2 - 2| \leq y \leq x\}$ is A, then $6A + 16\sqrt{2}$ is equal to ______.MediumNumerical value
26 Jul 2022, Shift 2 · Q20The area bounded by the curves $y=|x^2-1|$ and $y=1$ isMediumSingle correct
29 Jul 2022, Shift 1 · Q16The area of the region $\{(x, y) : |x - 1| \leq y \leq \sqrt{5 - x^2}\}$ 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.