Exam

JEE Main

Year

2025

Date

23 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Area Under Curves

If the area of the larger portion bounded between the curves...

If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b \pi+c), b, c \in N$, then $b+c$ is equal to

Step-by-step solution

$$\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2=5 \\ & \mathrm{x}^2+(\mathrm{x}-1)^2=25 \Rightarrow \mathrm{x}=4 \\ & \mathrm{x}^2+(-\mathrm{x}+1)^2=5 \Rightarrow \mathrm{x}=-3 \\ & \mathrm{~A}=25 \pi-\int_{-3}^4 \sqrt{25-\mathrm{x}^2} \mathrm{dx}+\frac{1}{2} \times 4 \times 4+\frac{1}{2} \times 3 \times 3 \\ & \mathrm{~A}=25 \pi+\frac{25}{2}-\left[\frac{\mathrm{x}}{2} \sqrt{25-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_{-3}^4 \\ & \mathrm{~A}=25 \pi+\frac{25}{2}-\left[6+\frac{25}{2} \sin ^{-1} \frac{4}{5}+6+\frac{25}{2} \sin ^{-1} \frac{3}{5}\right] \\ & \mathrm{A}=25 \pi+\frac{1}{2}-\frac{25}{2} \cdot \frac{\pi}{2} \\ & \mathrm{~A}=\frac{75 \pi}{4}+\frac{1}{2} \\ & \mathrm{~A}=\frac{1}{4}(75 \pi+2) \\ & \mathrm{b}=75, \mathrm{c}=2 \\ & \mathrm{~b}+\mathrm{c}=75+2=77\end{aligned}$$