Let g(x)=f(x)+f(1-x) and f^''(x)0,x (0,1).

Question

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

Accepted Answer

We have, $g(x)=f(x)+f(1-x)$

Differentiating both side, we get

$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$

As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.

Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$

So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$

$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.

Now,

$$
\begin{aligned}
& \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\
& =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi
\end{aligned}
$$

Solutions · 1