JEE Mathematics 2025 Question with Solution

Given: The curves are $x(1+y^2)=1$ and $y^2=2x$.

Find: The area of the region bounded by these two curves.

From $x(1+y^2)=1$, we get

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

From $y^2=2x$, we get

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

At the points of intersection,

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

So,

$$y^2(1+y^2)=2$$ $$y^4+y^2-2=0$$

Let $z=y^2$. Then

$$z^2+z-2=0$$ $$(z-1)(z+2)=0$$

Hence $z=1$, so $y=\pm 1$. Then

$$x=\frac{1}{1+1}=\frac{1}{2}$$

Thus, the curves intersect at $\left(\frac{1}{2},1\right)$ and $\left(\frac{1}{2},-1\right)$.

For $-1\le y\le 1$, the right curve is $x=\frac{1}{1+y^2}$ and the left curve is $x=\frac{y^2}{2}$. Therefore, area is

$$A=\int_{-1}^{1}\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy$$

Using symmetry,

$$A=2\int_{0}^{1}\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy$$

Now,

$$A=2\left[\tan^{-1}y-\frac{y^3}{6}\right]_{0}^{1}$$ $$A=2\left(\frac{\pi}{4}-\frac{1}{6}\right)$$ $$A=\frac{\pi}{2}-\frac{1}{3}$$

Therefore, the area of the bounded region is $\frac{\pi}{2}-\frac{1}{3}$ and the correct option is B.

Given: The equations are expressed conveniently as $x$ in terms of $y$.

Find: The enclosed area.

Since both curves are symmetric about the $x$-axis, integrate with respect to $y$ using horizontal strips.

• From $y^2=2x$, we write $x=\frac{y^2}{2}$.
• From $x(1+y^2)=1$, we write $x=\frac{1}{1+y^2}$.

The strip length is

$$\text{right} - \text{left} = \frac{1}{1+y^2}-\frac{y^2}{2}$$

for $y$ between the intersection values $-1$ and $1$.

So,

$$A=\int_{-1}^{1}\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right)dy$$

Evaluating,

$$\int \frac{1}{1+y^2}dy=\tan^{-1}y, \qquad \int \frac{y^2}{2}dy=\frac{y^3}{6}$$

Hence,

$$A=\left[\tan^{-1}y-\frac{y^3}{6}\right]_{-1}^{1}$$ $$A=\left(\frac{\pi}{4}-\frac{1}{6}\right)-\left(-\frac{\pi}{4}+\frac{1}{6}\right)$$ $$A=\frac{\pi}{2}-\frac{1}{3}$$

Therefore, the correct option is B.

The solution contains an incorrect intermediate area expression $2\left[\frac{\pi}{2}-\frac{1}{3}\right]$, but the final simplified result and the correct option are B, which agrees with the correct integral evaluation.