JEE Mathematics 2025 Question with Solution

Given: The region is $\{(x, y): |4 - x^2| \le y \le x^2, y \ge 0\}$.

Find: The value of $\alpha + \beta$ if the area is written in the form obtained from the integration.

We first determine where the region exists. From

$$|4-x^2| \le x^2$$

we get

$$4-x^2 \le x^2 \Rightarrow x^2 \ge 2.$$

So the relevant interval begins at $x = \sqrt{2}$.

Now split at $x = 2$ because the sign of $4-x^2$ changes there.

For $\sqrt{2} \le x \le 2$,

$$|4-x^2| = 4-x^2,$$

so the vertical strip length is

$$x^2-(4-x^2)=2x^2-4.$$

Hence

$$A_1 = \int_{\sqrt{2}}^{2} (2x^2-4) \, dx.$$

Evaluating,

$$A_1 = \left[\frac{2x^3}{3}-4x\right]_{\sqrt{2}}^{2}$$ $$= \left(\frac{16}{3}-8\right)-\left(\frac{4\sqrt{2}}{3}-4\sqrt{2}\right)$$ $$= \frac{8\sqrt{2}}{3}-\frac{8}{3}.$$

For $2 < x \le 2\sqrt{2}$,

$$|4-x^2| = x^2-4.$$

The upper bound becomes $y=4$ in the extracted solution, so the strip length is

$$4-(x^2-4)=8-x^2.$$

Thus

$$A_2 = \int_{2}^{2\sqrt{2}} (8-x^2) \, dx.$$

Evaluating,

$$A_2 = \left[8x-\frac{x^3}{3}\right]_{2}^{2\sqrt{2}}$$ $$= \left(16\sqrt{2}-\frac{16\sqrt{2}}{3}\right)-\left(16-\frac{8}{3}\right)$$ $$= \frac{32\sqrt{2}}{3}-\frac{40}{3}.$$

Therefore total area is

$$A = A_1 + A_2$$ $$= \left(\frac{8\sqrt{2}}{3}-\frac{8}{3}\right)+\left(\frac{32\sqrt{2}}{3}-\frac{40}{3}\right)$$ $$= \frac{40\sqrt{2}}{3}-16.$$

Now compare with the form used in the extracted solution,

$$\frac{80\sqrt{2}}{\alpha}-\beta = \frac{40\sqrt{2}}{3}-16.$$

So,

$$\alpha = 6, \quad \beta = 16.$$

Hence,

$$\alpha + \beta = 22.$$

Therefore, the correct option is C.

Note: The given question statement and the solution statement differ in form; the solution working clearly concludes that the required value is $22$, which matches option C.