JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Area Under Curves

The constraints describe the region inside the parabola $x^2=2y$ (so $y\ge x^2/2$), outside the circle $x^2+y^2=2y$, and below the line $y=x$ (so $x\ge y$). The parabola meets $y=x$ at $(0,0)$ and $(2,2)$, while the circle meets $y=x$ at $(0,0)$ and $(1,1)$. The area is the triangle with vertices $(0,0),(2,0),(2,2)$ of area $\frac{1}{2}\cdot 2\cdot 2=2$, minus the area under the parabola ${\int }_0^2\frac{x^2}{2}dx=\frac{4}{3}$, minus the circular segment between the chord $y=x$ and the arc, which is $\frac{\pi }{4}-\frac{1}{2}$. This gives $2-\frac{4}{3}-\left(\frac{\pi }{4}-\frac{1}{2}\right)=\frac{7}{6}-\frac{\pi }{4}$. Matching $\frac{n+2}{n+1}-\frac{\pi }{n-1}=\frac{7}{6}-\frac{\pi }{4}$ gives $\frac{n+2}{n+1}=\frac{7}{6}$ and $\frac{1}{n-1}=\frac{1}{4}$, each yielding $n=5$.