JEE Mains · Maths · STD 12 - 8. Application and integration
Let the area enclosed between the curves \(|y|=1-x^2\) and \(x^2+y^2=1\) be \(\alpha\). If \(9 \alpha=\beta \pi+\gamma ; \beta, \gamma\) are integers, then the value of \(|\beta-\gamma|\) equals.
- A 27
- B 33
- C 15
- D 18
Answer & Solution
Correct Answer
(B) 33
Step-by-step Solution
Detailed explanation
Required area \(=\pi-4 \int_0^1\left(1-x^2\right) d x \) \( =\pi-4\left[x-\frac{x^3}{3}\right]_0^1 \) \( =\pi-4 \times \frac{2}{3}=\pi-\frac{8}{3} \) \( \therefore \alpha=\pi-\frac{8}{3} \) \( 9 \alpha=9 \pi-24 \rightarrow \beta=9, \gamma=-24 \) \( |\beta-\gamma|=|9+24|=33\)
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