JEE Mathematics 2025 Question with Solution

Given: The region is defined by

Find: The area of the region.

The region is symmetric about the $y$-axis, so calculate the area for $x \geq 0$ and multiply by $2$.

For $x \geq 0$, the upper boundary is

Find the intersection points by equating the two curves:

Now compare the curves:

• For $0 \leq x \leq 2$, $x^2 + 1 \leq 2x + 1$
• For $2 \leq x \leq 3$, $2x + 1 \leq x^2 + 1$

So the first-quadrant area is

Evaluate the integrals:

Using symmetry, total area is

Therefore, the correct option is A and the area is $\frac{64}{3}$.

Given: The region lies below both $y = 2|x| + 1$ and $y = x^2 + 1$, above $y = 0$, with $|x| \leq 3$.

Find: The enclosed area.

Because both expressions involve $|x|$ or $x^2$, the region is symmetric about the $y$-axis. Hence,

In the first quadrant, $|x| = x$, so the two upper curves become $y = 2x + 1$ and $y = x^2 + 1$. The relevant upper boundary is the smaller of these two curves.

Their intersection is obtained from

Thus, they meet at $x = 0$ and $x = 2$.

So the area in the first quadrant is split into two parts:

The first integral is

The second integral is

Hence first-quadrant area is

Therefore, total area is

Therefore, the area of the region is $\frac{64}{3}$.