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Self-inverse function with f(f(x)) = x: find the constant or a value

This type of question, asked 3 times: 2 May 2023 S1, 10 May 2023 S2, 16 Apr 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    2 May 2023 S1

    128
    For a suitable chosen real constant aa, let a function f:R∖{−a}→Rf: \mathbb{R} \setminus \{-a\} \to \mathbb{R} be defined by f(x)=a−xa+xf(x) = \dfrac{a-x}{a+x}. Further suppose that for any real number x≠−ax \ne -a and f(x)≠−af(x) \ne -a, (f∘f)(x)=x(f \circ f)(x) = x. Then f(−15)f\left(-\dfrac{1}{5}\right) is equal to

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  2. Question 2

    10 May 2023 S2

    147
    Let f(x)=αxx+1, x≠−1f(x) = \frac{\alpha x}{x+1},\, x \neq -1, then for α= ?\alpha = \,?, f(f(x))=xf(f(x)) = x.

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  3. Question 3

    16 Apr 2026 S1

    69
    If f(x)=4x+36x−4,x≠23f(x) = \dfrac{4x + 3}{6x - 4}, x \neq \dfrac{2}{3} and (fof)(x)=g(x)(x) = g(x) where g:R−{23}→R−{23}g : \mathbb{R} - \left\{\dfrac{2}{3}\right\} \to \mathbb{R} - \left\{\dfrac{2}{3}\right\}, then (g∘g∘g∘g∘g)(3)=(g \circ g \circ g \circ g \circ g)(3) =

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