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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Area Under Curves

Let the area enclosed between the curves |y|=1-x^2 and x^2+y^2=1 be...

Let the area enclosed between the curves ∣y∣=1−x2|y|=1-x^2 and x2+y2=1x^2+y^2=1 be α\alpha. If 9α=βπ+γ;β,γ9 \alpha=\beta \pi+\gamma ; \beta, \gamma are integers, then the value of ∣β−γ∣|\beta-\gamma| equals.

Options

  1. A27
  2. B33
  3. C15
  4. D18

Correct answer

B. 33

Step-by-step solution

 Required area =π−4∫01(1−x2)dx=π−4[x−x33]01=π−4×23=π−83∴α=π−839α=9π−24→β=9,γ=−24∣β−γ∣=∣9+24∣=33\begin{aligned} & \text{ Required area }=\pi-4 \int_0^1\left(1-x^2\right) d x \\ & =\pi-4\left[x-\frac{x^3}{3}\right]_0^1 \\ & =\pi-4 \times \frac{2}{3}=\pi-\frac{8}{3} \\ & \therefore \alpha=\pi-\frac{8}{3} \\ & 9 \alpha=9 \pi-24 \rightarrow \beta=9, \gamma=-24 \\ & |\beta-\gamma|=|^{9}+24|=33\end{aligned}
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