Question

Let $f(x)=(\arctan x)^3+(\operatorname{arccot} x)^3$. If the range of $f(x)$ is $[a, b)$ then find the value of $\frac{b}{a}$.

Answer 1

We have $f(x)=\left(\tan ^{-1} x\right)^3+\left(\cot ^{-1} x\right)^3=\left(\tan ^{-1} x+\cot ^{-1} x\right)\left(\left(\tan ^{-1} x\right)^2-\left(\tan ^{-1} x\right)\left(\cot ^{-1} x\right)+\left(\cot ^{-1} x\right)^2\right)$
$=\frac{\pi}{2}\left(\left(\tan ^{-1} x\right)^2-\left(\tan ^{-1} x\right)\left(\frac{\pi}{2}-\tan ^{-1} x\right)+\left(\frac{\pi}{2}-\tan ^{-1} x\right)^2\right) \left(U \operatorname{sing} \cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x\right)$
$=\frac{3 \pi}{2}\left(\left(\tan ^{-1} x-\frac{\pi}{4}\right)^2+\frac{\pi^2}{48}\right)$
Clearly, $f ( x )$ will be minimum when $\left(\tan ^{-1} x -\frac{\pi}{4}\right)^2=0$
and $f(x)$ will be maximum when $\left(\tan ^{-1} x-\frac{\pi}{4}\right)^2=\left(-\frac{\pi}{2}-\frac{\pi}{4}\right)^2$
$\therefore a = f ( x )_{\min }=\frac{3 \pi}{2}\left(0+\frac{\pi^2}{48}\right)=\frac{\pi^3}{32}$ and $ b=f(x)_{\max }=\frac{3 \pi}{2}\left(\left(\frac{-3 \pi}{4}\right)^2+\frac{\pi^2}{48}\right)=\frac{7 \pi^3}{8}$
Hence $\frac{ b }{ a }=\frac{\frac{7 \pi^3}{8}}{\frac{\pi^3}{32}}=28$