JEE Main 2022 — Inverse Trigonometric Functions Question with Solution

$$({\tan }^{-1}x{)}^3+({\cot }^{-1}x{)}^3=k{\pi }^3$$

Let $$f(t)=t^3+{\left(\frac{\pi }{2}-t\right)}^3$$

Where $$t={\tan }^{-1}x$$ ; $$x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$$

$$=t^3+{\left(\frac{\pi }{2}\right)}^3-\frac{3{\pi }^{2}t}{4}+\frac{3\pi }{2}{t}^2-t^3$$

$$f(t)=\frac{3\pi }{2}{t}^2-\frac{3{\pi }^2}{4}.t+\frac{{\pi }^3}{8}$$

This is a quadratic equation of t.

Here, coefficient of t² term is $$\frac{3\pi }{2}$$ which is > 0.

$$\therefore$$ It is a upward parabola.

Now, $$f^{\prime }(t)=3\pi t-\frac{3{\pi }^2}{4}$$

$$f^{\prime \prime }(t)=3\pi >0$$

$$\therefore$$ $$3\pi t-\frac{3{\pi }^2}{4}=0$$

$$\Rightarrow t=\frac{\pi }{4}$$ (minima)

$$\therefore$$ vertex of graph at $$\frac{\pi }{4}$$

$$\therefore$$ Minimum value at $$\frac{\pi }{4}$$ and maximum value at $$-$$$$\frac{\pi }{2}$$.

$$\therefore$$ $$f\left(\frac{\pi }{4}\right)=\frac{{\pi }^3}{64}+{\left(\frac{\pi }{2}-\frac{\pi }{4}\right)}^3=\frac{{\pi }^3}{32}$$

$$f\left(-\frac{\pi }{2}\right)=-\frac{{\pi }^3}{8}+{\pi }^3$$

$$=\frac{7{\pi }^3}{8}$$

$$\therefore$$ $$k{\pi }^3\in \left[\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}\right)$$

$$\Rightarrow k\in \left[\frac{1}{32},\frac{7}{8}\right)$$