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(A) Given $f(x) = \alpha x^5 + \beta x^3 + \gamma x$ where $\alpha, \beta, \gamma > 0$. Since $f'(x) = 5\alpha x^4 + 3\beta x^2 + \gamma > 0$ for all

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(A) Given $f(x) = \alpha x^5 + \beta x^3 + \gamma x$ where $\alpha, \beta, \gamma > 0$. Since $f'(x) = 5\alpha x^4 + 3\beta x^2 + \gamma > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function and thus invertible. Given $g(f(x)) = x$,we have $f(g(y)) = y$ for all $y \in \mathbb{R}$. We need to evaluate $f(g(\frac{1}{n} \sum_{i=1}^{n} f(a_i)))$. Since $f(g(y)) = y$,the expression simplifies to $\frac{1}{n} \sum_{i=1}^{n} f(a_i)$. Given $a_1, a_2, \dots, a_n$ are in arithmetic progression with mean zero,we have $\sum_{i=1}^{n} a_i = 0$. Since $f(x)$ is an odd function,$f(-x) = -f(x)$. For an arithmetic progression with mean zero,the terms are symmetric about zero. Thus,$\sum_{i=1}^{n} f(a_i) = 0$. Therefore,the value is $\frac{1}{n} 	imes 0 = 0$.

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