The area of the region bounded by the curves $\mathrm{x}\left(1+\mathrm{y}^{2}\right)=1$ and $\mathrm{y}^{2}=2 \mathrm{x}$ is :

(C) $\frac{\pi}{2}-\frac{1}{3}$

(A) $2\left(\frac{\pi}{2}-\frac{1}{3}\right)$

(B) $\frac{\pi}{4}-\frac{1}{3}$

(D) $\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{3}\right)$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Area Under Curves

The area of the region bounded by the curves

x (1 + y^{2}) = 1

and

y^{2} = 2 x

is :

Answer: (C)

Rewrite the curves as

x = \frac{1}{1 + y ^{2}}

and

x = \frac{y ^{2}}{2}

. At the intersection,

\frac{1}{1 + y ^{2}} = \frac{y ^{2}}{2} ⟹ y^{4} + 2 y^{2} - 2 = 0

Setting

u = y^{2}

2 u^{2} + u - 1 = 0

, i.e.

(2 u - 1) (u + 1) = 0

, so

u = \frac{1}{2}

and

y = \pm 1

.

By symmetry in

y

, the area is

A = 2 \int_{0}^{1} (\frac{1}{1 + y ^{2}} - \frac{y ^{2}}{2}) d y = 2 [tan^{- 1} y - \frac{y ^{3}}{6}]_{0}^{1} = 2 (\frac{π}{4} - \frac{1}{6}) = \frac{π}{2} - \frac{1}{3}

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option