JEE Advanced 2024 · Paper 2, Mathematics question 2

Corners. The parabolas $y^2=4x$ and $y^2=12-2x$ meet where $4x=12-2x$, i.e. $x=2$, $y=2\sqrt2$.

The line $3y+2\sqrt2\,x=10\sqrt2$ passes through $(2,2\sqrt2)$ (since $6\sqrt2+4\sqrt2=10\sqrt2$) and meets the $x$-axis at $(5,0)$.

On $2\le x\le5$ the line lies below $y=\sqrt{12-2x}$; for example, at $x=5$ the line gives $y=0$ and the parabola gives $y=\sqrt2$. So the region is:

• under $y=2\sqrt x$ for $0\le x\le2$, plus
• the triangle with vertices $(2,0)$, $(5,0)$, $(2,2\sqrt2)$.

Area.
$$\int_0^2 2\sqrt x\,dx=\frac43\cdot2^{3/2}=\frac{8\sqrt2}{3},\qquad \text{triangle}=\frac12\cdot3\cdot2\sqrt2=3\sqrt2.$$
$$\text{Area}=\frac{8\sqrt2}{3}+\frac{9\sqrt2}{3}=\frac{17}{3}\sqrt2\implies\alpha=\frac{17}{3}.$$