Mathematics · Area under the Curves

JEE Main 2026 — 21 January, Morning Shift — Question 2

The area of the region, inside the ellipse $x^{2}+4 y^{2}=4$ and outside the region bounded by the curves $\mathrm{y}=|\mathrm{x}|-1$ and $\mathrm{y}=1-|\mathrm{x}|$ , is :

Option A:
$2(\pi-1)$
Correct
Option B:
$2 \pi-\frac{1}{2}$
Option C:
$3(\pi-1)$
Option D:
$2 \pi-1$

Answer: A

Step-by-step solution

The ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$ , area = $\pi \times 2 \times 1 = 2\pi$ . The curves $y = |x| - 1$ and $y = 1 - |x|$ form a square with vertices at $(\pm 1, 0)$ and $(0, \pm 1)$ . Area of this square = $4 \times \frac{1}{2} \times 1 \times 1 = 2$ . Required area = area of ellipse - area of square = $2\pi - 2 = 2(\pi - 1)$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves