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QJEEM25Ma34a3: JEE Main 2025 Mathematics - Class 12 JEEM25Ma34a3 Solution: If the area of the larger portion bounded between the curves…

If the area of the larger portion bounded between the curves x2+y2=25x^2+y^2=25 and y=∣x−1∣\mathrm{y}=|\mathrm{x}-1| is 14( bπ+c),b,c∈N\frac{1}{4}(\mathrm{~b} \pi+\mathrm{c}), \mathrm{b}, \mathrm{c} \in N , then b+c\mathrm{b}+\mathrm{c} is equal to _________

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JEE Main 2025 (Online) 23rd January Morning Shift Mathematics - Area Under The Curves Question 5 English Explanation

amp;x2+y2=5amp;x2+(x−1)2=25⇒x=4amp;x2+(−x+1)2=5⇒x=−3amp; A=25π−∫−3425−x2dx+12×4×4+12×3×3amp; A=25π+252−[x225−x2+252sin⁡−1x5]−34amp; A=25π+252−[6+252sin⁡−145+6+252sin⁡−135]amp;A=25π+12−252⋅π2amp; A=75π4+12amp; A=14(75π+2)amp;b=75,c=2amp; b+c=75+2=77\begin{aligned}& \mathrm{x}^2+\mathrm{y}^2=5 \\& \mathrm{x}^2+(\mathrm{x}-1)^2=25 \Rightarrow \mathrm{x}=4 \\& \mathrm{x}^2+(-\mathrm{x}+1)^2=5 \Rightarrow \mathrm{x}=-3 \\& \mathrm{~A}=25 \pi-\int_{-3}^4 \sqrt{25-\mathrm{x}^2} \mathrm{dx}+\frac{1}{2} \times 4 \times 4+\frac{1}{2} \times 3 \times 3 \\& \mathrm{~A}=25 \pi+\frac{25}{2}-\left[\frac{\mathrm{x}}{2} \sqrt{25-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_{-3}^4 \\& \mathrm{~A}=25 \pi+\frac{25}{2}-\left[6+\frac{25}{2} \sin ^{-1} \frac{4}{5}+6+\frac{25}{2} \sin ^{-1} \frac{3}{5}\right] \\& \mathrm{A}=25 \pi+\frac{1}{2}-\frac{25}{2} \cdot \frac{\pi}{2} \\& \mathrm{~A}=\frac{75 \pi}{4}+\frac{1}{2} \\& \mathrm{~A}=\frac{1}{4}(75 \pi+2) \\& \mathrm{b}=75, \mathrm{c}=2 \\& \mathrm{~b}+\mathrm{c}=75+2=77\end{aligned}

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