JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 10

This question was dropped in NTA's final key.

The regions. $|x^2-9|=3$ at $x^2=6$ and $x^2=12$. Two kinds of bounded region lie between the curve and $y=3$: the two pieces under $y=3$ for $\sqrt6\le|x|\le2\sqrt3$, and the cap above $y=3$ for $|x|\le\sqrt6$.

Under the line. $2\left[\int_{\sqrt6}^3(x^2-6)\,dx+\int_3^{2\sqrt3}(12-x^2)\,dx\right]=8\left(4\sqrt3+\sqrt6-9\right)\approx3.02$, which is option (D).

The cap. $\displaystyle\int_{-\sqrt6}^{\sqrt6}(6-x^2)\,dx=8\sqrt6$; counting it too gives a total that no option lists. The wording does not say which region is meant.