JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

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

A
$2 (π - 1)$
B
$2 π - \frac{1}{2}$
C
$3 (π - 1)$
D
$2 π - 1$

Solution

The ellipse $\frac{x ^{2}}{4} + y^{2} = 1$ has semi-axes $a = 2$ , $b = 1$ , so its area is $π ab = 2 π$ .

The curves $y = ∣ x ∣ - 1$ and $y = 1 - ∣ x ∣$ enclose a square (rhombus) with vertices $(\pm 1, 0)$ and $(0, \pm 1)$ , which lies entirely inside the ellipse. Its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$ .

Required area $= 2 π - 2 = 2 (π - 1)$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →