Let g(x) = f(x) + f(1-x) and f''(x) 0 for x | vedclass

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(B) Given $g(x) = f(x) + f(1-x)$.Taking the derivative,$g'(x) = f'(x) - f'(1-x)$.Since $g$ is decreasing on $(0, \alpha)$ and increasing on $(\alpha,

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$\pi$

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(B) Given $g(x) = f(x) + f(1-x)$.Taking the derivative,$g'(x) = f'(x) - f'(1-x)$.Since $g$ is decreasing on $(0, \alpha)$ and increasing on $(\alpha, 1)$,$g'(x) = 0$ at $x = \alpha$.Thus,$f'(\alpha) = f'(1-\alpha)$.Since $f''(x) > 0$,$f'(x)$ is a strictly increasing function.Therefore,$f'(\alpha) = f'(1-\alpha)$ implies $\alpha = 1-\alpha$,which gives $\alpha = \frac{1}{2}$.Now,we evaluate the expression $	an^{-1}(2\alpha) + 	an^{-1}\left(\frac{1}{\alpha}ight) + 	an^{-1}\left(\frac{\alpha+1}{\alpha}ight)$ at $\alpha = \frac{1}{2}$.Substituting $\alpha = \frac{1}{2}$:$	an^{-1}(2 \cdot \frac{1}{2}) + 	an^{-1}\left(\frac{1}{1/2}ight) + 	an^{-1}\left(\frac{1/2+1}{1/2}ight)$$= 	an^{-1}(1) + 	an^{-1}(2) + 	an^{-1}(3)$.We know that $	an^{-1}(1) = \frac{\pi}{4}$.For $	an^{-1}(2) + 	an^{-1}(3)$,since $2 \cdot 3 > 1$,we use the formula $	an^{-1}(x) + 	an^{-1}(y) = \pi + 	an^{-1}\left(\frac{x+y}{1-xy}ight)$.$	an^{-1}(2) + 	an^{-1}(3) = \pi + 	an^{-1}\left(\frac{2+3}{1-6}ight) = \pi + 	an^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Thus,the total sum is $\frac{\pi}{4} + \frac{3\pi}{4} = \pi$.

Let $g(x) = f(x) + f(1-x)$ and $f''(x) > 0$ for $x \in (0, 1)$. If $g$ is decreasing in the interval $(0, \alpha)$ 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:

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