JEE Mathematics 2026 Question with Solution

Given: The region is $\{(x, y) : 1 - 2x \le y \le 4 - x^2, x \ge 0, y \ge 0\}$.

Find: The value of $(\alpha + \beta)$ when the area is $\frac{\alpha}{\beta}$ in lowest terms.

The region is bounded by the parabola $y = 4 - x^2$, the line $y = 1 - 2x$, and the coordinate axes in the first quadrant.

The parabola meets the $x$-axis at $x = 2$.

The line meets the $x$-axis at $x = \frac{1}{2}$.

For $0 \le x \le \frac{1}{2}$, the lower boundary is $y = 1 - 2x$.

For $\frac{1}{2} \le x \le 2$, the lower boundary is $y = 0$.

So the total area is

$$A = \int_0^2 (4 - x^2) \, dx - \int_0^{1/2} (1 - 2x) \, dx$$

Now,

$$\int_0^2 (4 - x^2) \, dx = \left[4x - \frac{x^3}{3}\right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$$

and

$$\int_0^{1/2} (1 - 2x) \, dx = \left[x - x^2\right]_0^{1/2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Therefore,

$$A = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$$

Hence $\alpha = 61$ and $\beta = 12$, with $\gcd(61,12)=1$.

Therefore,

$$\alpha + \beta = 61 + 12 = 73$$

So, the correct option is D.

Given: Upper curve $y = 4 - x^2$, lower condition $y \ge 1 - 2x$, and first-quadrant restrictions $x \ge 0, y \ge 0$.

Find: The area of the feasible region.

The first-quadrant restriction changes the lower boundary. Although the line is $y = 1 - 2x$, once it goes below the $x$-axis, the condition $y \ge 0$ becomes stronger.

• From $x = 0$ to $x = \frac{1}{2}$, the line lies above the $x$-axis, so the lower boundary is $y = 1 - 2x$.
• From $x = \frac{1}{2}$ to $x = 2$, the line lies below the $x$-axis, so the lower boundary is $y = 0$.

Thus,

$$A = \int_0^{1/2} \big[(4 - x^2) - (1 - 2x)\big] \, dx + \int_{1/2}^{2} (4 - x^2) \, dx$$

This is equivalent to

$$A = \int_0^2 (4 - x^2) \, dx - \int_0^{1/2} (1 - 2x) \, dx$$

Evaluating gives

$$A = \frac{16}{3} - \frac{1}{4} = \frac{61}{12}$$

Therefore, $\alpha + \beta = 73$, so the correct option is D.