Let $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x)0, x \in(0,1)$. If $\mathrm{g}$ is decreasing in the interval $(0, \alpha)$ 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: [JEE Main 2023, 10 Apr (Shift 2)]

Question

Let $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x)>0, x \in(0,1)$. If $\mathrm{g}$ is decreasing in the interval $(0, \alpha)$ 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: [JEE Main 2023, 10 Apr (Shift 2)]

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

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