Question

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

Answer 1

Answer: 0028

We have $f(x)={\left({\tan }^{-1}x\right)}^3+{\left({\cot }^{-1}x\right)}^3=\left({\tan }^{-1}x+{\cot }^{-1}x\right)\left({\left({\tan }^{-1}x\right)}^2-\left({\tan }^{-1}x\right)\left({\cot }^{-1}x\right)+{\left({\cot }^{-1}x\right)}^2\right)$
$=\frac{\pi }{2}\left({\left({\tan }^{-1}x\right)}^2-\left({\tan }^{-1}x\right)\left(\frac{\pi }{2}-{\tan }^{-1}x\right)+{\left(\frac{\pi }{2}-{\tan }^{-1}x\right)}^2\right)\left(U\mathrm{sing}{\cot }^{-1}x=\frac{\pi }{2}-{\tan }^{-1}x\right)$
$=\frac{3\pi }{2}\left({\left({\tan }^{-1}x-\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{48}\right)$
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 $\frac{b}{a}=\frac{\frac{7{\pi }^3}{8}}{\frac{{\pi }^3}{32}}=28$