Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

Let $T={\left({\tan }^{-1}x\right)}^3+{\left({\cot }^{-1}x\right)}^3$

$={\left({\tan }^{-1}x+{\cot }^{-1}x\right)}^3-3{\tan }^{-1}x·{\cot }^{-1}x\left({\tan }^{-1}x+{\cot }^{-1}x\right)$

$=\frac{{\pi }^3}{8}-\frac{3\pi }{2}{\tan }^{-1}x\left(\frac{\pi }{2}-{\tan }^{-1}x\right)$ $\left(\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right)$

$=\frac{3\pi }{2}{\left({\tan }^{-1}x\right)}^2-\frac{3{\pi }^2}{4}{\tan }^{-1}x+\frac{{\pi }^3}{8}$

$T=\frac{3\pi }{2}{\left({\tan }^{-1}x-\frac{\pi }{4}\right)}^2+\frac{{\pi }^3}{32}$

We know than ${\tan }^{-1}x\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

Minimum value of $T$ is at ${\tan }^{-1}x=\frac{\pi }{4}$ and maximum value at ${\tan }^{-1}x=-\frac{\pi }{2}$

$\Rightarrow \frac{{\pi }^3}{32}\le T<\frac{7}{8}{\pi }^3$

$=\frac{{\pi }^3}{32}\le k{\pi }^3<\frac{7}{8}{\pi }^3$

$\Rightarrow \frac{1}{32}\le k<\frac{7}{8}$