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g(x) = f(x) + f(a - x) type with f'' > 0 decreases then increases at alpha: find alpha

The same question, asked 2 times: 10 Apr 2023*, 27 Jan 2024. Every wording below, oldest first.

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  1. Question 1

    10 Apr 2023

    159
    Let g(x)=f(x)+f(1−x)g(x) = f(x) + f(1 - x) and f′′(x)>0,x∈(0,1)f^{''}(x) > 0,x \in (0,1). If gg is decreasing in the interval (0,α)(0,\alpha) and increasing in the interval (α,1)(\alpha,1), then tan⁡1(2α)+tan⁡−1(1α)+tan⁡−1(α+1α)\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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  2. Question 2

    27 Jan 2024

    158
    Let g(x)=3f(x3)+f(3−x)g(x) = 3f\left( \frac{x}{3} \right)+f(3 -x) and f′′(x)>0f^{''}(x) > 0 for all x∈(0,3)x\in (0,3). If gg is decreasing in (0,α)(0,\alpha) and increasing in (α,3)(\alpha,3), then 8α8\alpha is

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