JEE Advanced 2025 Paper 2, Mathematics Section 2 Q2: Area as Definite Integral

Step 1: Chord with midpoint $(h,k)$.

For $y^2=x$, the chord bisected at $(h,k)$ has equation (using $T=S_1$): $ky-\frac{x+h}{2}=k^2-h$, i.e. $x-2ky+2k^2-h=0$.

Step 2: Compute the enclosed area.

Swap roles of $x$ and $y$ to integrate against the standard parabola $y=x^2$. The two intersection abscissas $\alpha ,\beta$ satisfy $|\alpha -\beta |=\sqrt{4h-4k^2}$. The enclosed area is

$$A=\frac{(4h-4k^2{)}^{3/2}}{6}.$$

Setting $A=\frac{4}{3}$ gives $(4h-4k^2{)}^{3/2}=8$, so $4h-4k^2=4$, i.e. $h-k^2=1$.

Step 3: Test the candidate points.

$(4,\sqrt{3})$: $4-3=1$ ✓ so $(4,\sqrt{3})\in S$. Option (A) TRUE.

$(5,\sqrt{2})$: $5-2=3\ne 1$. Option (B) FALSE.

Step 4: Area of $R$.

$S$ is the parabola $x=y^2+1$, so in the first quadrant the region between $y=\sqrt{x}$ (upper) and $y=\sqrt{x-1}$ (lower) on $[1,4]$ has area

$$A={\int }_1^4\left(\sqrt{x}-\sqrt{x-1}\right)dx=\frac{2}{3}{\left[x^{3/2}-(x-1{)}^{3/2}\right]}_1^4=\frac{2}{3}(8-3\sqrt{3}-1)=\frac{14}{3}-2\sqrt{3}.$$

Option (C) TRUE; (D) FALSE.

$\therefore$ The correct options are (A) and (C).