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Monotonicity of g built from f when f'' > 0

This type of question, asked 3 times: 10 Apr 2023, 27 Jan 2024, 21 Jan 2026 S2. All 3 questions 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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  3. Question 3

    21 Jan 2026 S2

    52
    Let f:R→Rf:R \rightarrow R be a twice differentiable function such that f′′(x)>0f^{''}(x) > 0 for all x∈Rx \in R and f′(a−1)=0f^{'}(a - 1) = 0, where a is real number. Let g(x)=f(tan⁡2x−2tan⁡x+a)g(x) = f\left( \tan^{2}x - 2\tan x + a \right), 0<x<π20 < x <\frac{\pi}{2}. Consider the following two statements : (I) g is increasing in (0,π4)\left( 0,\frac{\pi}{4} \right) (II) g is decreasing in (π4,π2)\left( \frac{\pi}{4},\frac{\pi}{2} \right) Then,

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