JEE Advanced 2023 · Paper 1, Mathematics question 6

The set $X$. $y^2<5x$ forces $x\ge1$, and $\frac{x^2}{8}<1$ forces $x\le2$.

• $x=1$: $y^2<5$ (the ellipse condition $y^2<17.5$ is weaker), so $y\in\{-2,\dots,2\}$: 5 points.
• $x=2$: $y^2<10$ from both conditions, so $y\in\{-3,\dots,3\}$: 7 points.

$|X|=12$, so there are $\binom{12}{3}=220$ choices.

Areas. Three points on one vertical line give area 0. Otherwise two points lie on one line and the third on the other, at horizontal distance 1, so
$$\text{area}=\tfrac12\,|y_1-y_2|,$$
where $y_1,y_2$ belong to the two points on the same line. It is a positive integer iff $y_1-y_2$ is even and non-zero.

Count.

• Two on $x=1$: same-parity pairs from $\{-2,\dots,2\}$ number $\binom32+\binom22=4$; times 7 choices on $x=2$: 28.
• Two on $x=2$: same-parity pairs from $\{-3,\dots,3\}$ number $\binom32+\binom42=9$; times 5 choices on $x=1$: 45.

$$P=\frac{28+45}{220}=\frac{73}{220}.$$