If the area of the region $\{(x, y): 1-2x \leq y \leq 4-x^2, x \geq 0, y \geq 0\}$ is $\frac{\alpha}{\beta}$, $\alpha, \beta \in \mathbf{N}$, $\gcd(\alpha, \beta)=1$, then the value of $(\alpha+\beta)$ is :

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Area Under Curves

If the area of the region

{(x, y) : 1 - 2 x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0}

\frac{α}{β}

α, β \in N

g cd (α, β) = 1

, then the value of

(α + β)

is :

Answer: (A)

The region lies below the parabola

y = 4 - x^{2}

(which meets the

x

-axis at

x = 2

) and above the larger of

y = 1 - 2 x

and

y = 0

, with

x \geq 0

. The line

y = 1 - 2 x

stays above the

x

-axis only for

0 \leq x \leq \frac{1}{2}

. Hence the area is the area under the parabola on

[0, 2]

minus the triangular area under the line on

[0, \frac{1}{2}]

\int_{0}^{2} (4 - x^{2}) d x - \int_{0}^{1/2} (1 - 2 x) d x = \frac{16}{3} - \frac{1}{4} = \frac{61}{12}

. With

α = 61, β = 12

and

g cd (61, 12) = 1

, the value of

α + β

73

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option