Let $ g(x)=f(x)+f(1-x) $ and $ f^{\prime \prime}(x)0, x \in(0,1) $. If $ g $ is decreasing in the interval $ (0, \alpha) $ and increasing in the interval $ (\alpha, 1) $, then the value of $ \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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Let $ g(x)=f(x)+f(1-x) $ and $ f^{\prime \prime}(x)>0, x \in(0,1) $. If $ g $ is decreasing in the interval $ (0, \alpha) $ and increasing in the interval $ (\alpha, 1) $, then the value of $ \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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Let \( g(x)=f(x)+f(1-x) \) and \( f^{\prime \prime}(x)>0, x \in(0,1) \). If \( g \) is decreasing in the interval \( …

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Let \( g(x)=f(x)+f(1-x) \) and \( f^{\prime \prime}(x)>0, x \in(0,1) \). If \( g \) is decreasing in the interval \( …