JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 8

One variable. Let $t=\tan^{-1}x\in\left(-\frac\pi2,\frac\pi2\right)$; then $\cot^{-1}x=\frac\pi2-t$.

Sum of cubes with $a+b=\frac\pi2$:
$$t^3+\left(\tfrac\pi2-t\right)^3=\frac{\pi^3}8-\frac{3\pi}2\,t\left(\tfrac\pi2-t\right)=\frac{\pi^3}8-\frac{3\pi^2}4t+\frac{3\pi}2t^2.$$

Range. This quadratic in $t$ has its minimum at $t=\frac\pi4$: value $\frac{\pi^3}{32}$. As $t\to-\frac\pi2$ it tends to $\frac78\pi^3$ (not reached); as $t\to\frac\pi2$ it tends to $\frac{\pi^3}8$, which is smaller.

So $k\in\left[\frac1{32},\frac78\right)$.

(A)