JEE Advanced 2025 — Paper 2 — MATH-II solution

JEE Advanced2025Paper 2MATH-II

Let $S$ denote the locus of the mid-points of those chords of the parabola $y^{2} = x$ , such that the area of the region enclosed between the parabola and the chord is $\frac{4}{3}$ . Let $R$ denote the region lying in the first quadrant, enclosed by the parabola $y^{2} = x$ , the curve $S$ , and the lines $x = 1$ and $x = 4$ .

Then which of the following statements is (are) TRUE?

A
$(4, 3) \in S$
B
$(5, 2) \in S$
C
Area of $R$ is $\frac{14}{3} - 23$
D
Area of $R$ is $\frac{14}{3} - 3$

Solution

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

For $y^{2} = x$ , the chord bisected at $(h, k)$ has equation (using $T = S_{1}$ ): $k y - \frac{x + h}{2} = k^{2} - h$ , i.e. $x - 2 k y + 2 k^{2} - h = 0$ .

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 $α, β$ satisfy $∣ α - β ∣ = 4 h - 4 k^{2}$ . The enclosed area is

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

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

Step 3: Test the candidate points.

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

$(5, 2)$ : $5 - 2 = 3 \neq = 1$ . Option (B) FALSE.

Step 4: Area of $R$ .

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

$A = \int_{1}^{4} (x - x - 1) d x = \frac{2}{3} [x^{3/2} - (x - 1)^{3/2}]_{1}^{4} = \frac{2}{3} (8 - 33 - 1) = \frac{14}{3} - 23 .$

Option (C) TRUE; (D) FALSE.

$∴$ The correct options are (A) and (C).

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