Verified Solution • 2025–26 Session moderate

QJEEM25Ma872c: JEE Main 2025 Mathematics - Class 12 JEEM25Ma872c Solution: Let the area enclosed between the curves and be . If are…

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:

Multiple Choice Options

  • A 15
  • B 18
  • C 33
  • D 27

Step-by-Step Verified Solution

amp;C1:∣y∣=1−x2amp;C2:x2+y2=1\begin{aligned}& C_1:|y|=1-x^2 \\& C_2: x^2+y^2=1\end{aligned}

JEE Main 2025 (Online) 29th January Evening Shift Mathematics - Area Under The Curves Question 10 English Explanation

∴\therefore Required Area

=α=4[=\alpha=4\left[\right. Area of circle in 1st 1^{\text {st }} quad. −∫01(1−x2)dx]\left.-\int_0^1\left(1-\mathrm{x}^2\right) \mathrm{dx}\right]

amp;=4[π4−[x−x33]01]amp;α=π−83amp;∴3α=3π−8amp;∴9α=9π−24amp;∴β=9,γ=−24amp;∴∣β−γ∣=33\begin{aligned}& =4\left[\frac{\pi}{4}-\left[\mathrm{x}-\frac{\mathrm{x}^3}{3}\right]_0^1\right] \\& \alpha=\pi-\frac{8}{3} \\& \therefore 3 \alpha=3 \pi-8 \\& \therefore 9 \alpha=9 \pi-24 \\& \therefore \beta=9, \gamma=-24 \\& \therefore|\beta-\gamma|=33\end{aligned}

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