Question

Find the maximum possible integral value of $\frac{\beta -\alpha }{{\tan }^{-1}\beta -{\tan }^{-1}\alpha }$, where $0<\alpha <\beta <\sqrt{3}$, is

Answer 1

Let $\alpha =\tan \theta ,\beta =\tan \varphi ⟹0<\theta <\varphi <\frac{\pi }{3}$

We know that $\tan x$ is continouos on $[0,\frac{\pi }{4}]$ and differentiable on $(0,\frac{\pi }{3})$.

By cauchy's inequality there exists some $c\in (0,\frac{\pi }{3})$;

$f^{\prime }\left(c\right)=\frac{\tan \varphi -\tan \theta }{\varphi -\theta }⟹f^{\prime }\left(c\right)={\text{sec}}^{2}x⟹$ maximum possible integral value is $3$