Let the area enclosed between the curves $|\mathrm{y}|=1-\mathrm{x}^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ be $\alpha$. If $9 \alpha=\beta \pi+\gamma ; \beta, \gamma$ are integers, then the value of $|\beta-\gamma|$ equals

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Area Under Curves

Let the area enclosed between the curves

∣ y ∣ = 1 - x^{2}

and

x^{2} + y^{2} = 1

α

. If

9 α = β π + γ; β, γ

are integers, then the value of

∣ β - γ ∣

equals

Answer: (D)

∣ y ∣ = 1 - x^{2}

consists of the two parabolic arcs

y = 1 - x^{2}

and

y = x^{2} - 1

for

x \in [- 1, 1]

; both touch the unit circle

x^{2} + y^{2} = 1

x = 0, \pm 1

and lie inside the circle between these points (since for

u = x^{2} \in [0, 1]

u + (1 - u)^{2} = 1 - u + u^{2} \leq 1

).

Taking

α

as the area lying inside the circle but outside the region between the two arcs:

area between arcs = 2 \int_{- 1}^{1} (1 - x^{2}) d x = 2 [x - \frac{x ^{3}}{3}]_{- 1}^{1} = \frac{8}{3}

α = π - \frac{8}{3}

9 α = 9 π - 24 \Rightarrow β = 9, γ = - 24 \Rightarrow ∣ β - γ ∣ = 33

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option