If the area of the region \ (x, y): 1-2 x leq y leq 4-x 2 . , x geq 0, y geq 0\ is α/β, α, β, in N, gcd (α, β)=1 , then the value of (α+β) is :

Mathematics · Area under the Curves

JEE Main 2026 — 21 January, Evening Shift — Question 11

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

Option A:
$73$
Correct
Option B:
$85$
Option C:
$91$
Option D:
$67$

Answer: A

Step-by-step solution

\text{Area} &= \int_{0}^{2} (4 - x^2) \, dx - \int_{0}^{1/2} (1 - 2x) \, dx \\ &= \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} - \left[ x - x^2 \right]_{0}^{1/2} \\ &= \left( 8 - \frac{8}{3} \right) - \left( \frac{1}{2} - \frac{1}{4} \right) \\ &= \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12} \\ \alpha &= 61, \beta = 12 \implies \alpha + \beta = 73 \end{aligned}$$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves