Let g(x) = f(x) + f(1 - x) and f ^ " (x) > 0, x ∈ (0, 1) . If g is...

Given,

$g\left(x\right) = f$\left(x\right) + f\left(1 - x\right)$and${f}^{"}\left(x\right) > 0, x \in \left(0, 1\right)$. If$g$is decreasing in the interval \left(0, \alpha\right)$ and increasing in the interval \left(\alpha, 1\right)$,</p><p>Now solving,</p><p>$g\left(x\right) = f\left(x\right) + f\left(1 - x\right)$</p><p>Now differentiating both side we get,</p><p>${g}^{'}\left(x\right) = {f}^{'}$\left(x\right) - {f}^{'}\left(1 - x\right)$

Differentiating again we get,

${g}^{"}\left(x\right) = {f}^{"}\left(x\right) + {f}^{"}\left(1 - x\right) > 0$

So, ${g}^{'}\left(x\right)$ is increasing as given ${f}^{"}\left(x\right) > 0$

$⇒ {g}^{'}\left(0\right) < {g}^{'}\left(1\right)$

$⇒ {f}^{'}\left(0\right) - {f}^{'}\left(1\right) < {f}^{'}\left(1\right) - {f}^{'}\left(0\right)$

$⇒ {f}^{'}\left(0\right) < {f}^{'}\left(1\right)$

Now finding ${g}^{'}\left(x\right) = 0$ we get,

$⇒ {f}^{'}\left(x\right) = {f}^{'}\left(1 - x\right)$

$⇒ x = 1 - x$

$⇒ x = \dfrac{1}{2}$

Now ${g}^{'}\left(x\right)$ is positive for $x \in \left(0, \dfrac{1}{2}\right)$

And ${g}^{'}\left(x\right)$ is negative for $x \in \left(\dfrac{1}{2}, 1\right)$

$∴ \alpha = \dfrac{1}{2}$

So,

$\tan^{- 1}2\alpha + \tan^{- 1}\left(\dfrac{1}{\alpha}\right) + \tan^{- 1}\left(\dfrac{\alpha + 1}{\alpha}\right)$

$= \tan^{- 1}1 + \tan^{- 1}2 + \tan^{- 1}3$

$= \dfrac{\pi}{4} + \tan^{- 1}\left(\dfrac{2 + 3}{1 - 6}\right)$

$= \dfrac{\pi}{4} + \tan^{- 1}\left(- 1\right)$

$= \dfrac{\pi}{4} + \dfrac{3\pi}{4} = \pi$