JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 16

Region (first quadrant, as intended; see the note). For $x>0$: $1\le y\le\min(x^2,27/x)$, which needs $1\le x\le27$. The curves $y=x^2$ and $y=27/x$ meet at $x=3$.

Area.
$$\int_1^3(x^2-1)\,dx+\int_3^{27}\left(\frac{27}x-1\right)dx=\frac{20}3+(27\ln9-24)=54\ln3-\frac{52}3.$$

Note. For $x\le-1$, $xy<0\le27$ always holds, so the set as printed also contains the unbounded strip $\{x\le-1,\ 1\le y\le x^2\}$; the options show the first-quadrant region is meant.

(B)