Let $\mathrm{g}(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x) > 0, x \in(0,1)$. If $\mathrm{g}$ is decreasing in the interval $(0, a)$ and increasing in the interval $(\alpha, 1)$, then $$\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$$ is equal to :

We have, $g(x)=f(x)+f(1-x)$ <br/><br/>Differentiating both side, we get <br/><br/>$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$ <br/><br/>As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function. <br/><br/>Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$ <br/><br/>So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$ <br/><br/>$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$. <br/><br/>Now, <br/><br/>$$ \begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned} $$