Key Concepts and Formulas

1. Inverse Functions: If $g$ is the inverse of $f$, then $g(f(x)) = x$ for all $x$ in the domain of $f$, and $f(g(y)) = y$ for all $y$ in the range of $f$.
2. Odd Functions: A function $h(x)$ is odd if $h(-x) = -h(x)$ for all $x$. A property of odd functions is that $h(0) = 0$ if $0$ is in the domain.
3. Arithmetic Progression (AP) with Mean Zero: If $a_1, a_2, \ldots, a_n$ are in an AP with mean zero, then their sum is zero, i.e., $\sum_{i=1}^n a_i = 0$. This implies a symmetry where for every term $a_i$, there is a corresponding term $-a_i$ (or the terms are symmetric around 0).

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Step-by-Step Solution

Step 1: Analyze the function $f(x)$ and its properties. The given function is $f(x)=\alpha x^{5}+\beta x^{3}+\gamma x$, where $\alpha, \beta, \gamma$ are positive real numbers. We check if $f(x)$ is an odd or even function. $f(-x) = \alpha (-x)^{5} + \beta (-x)^{3} + \gamma (-x)$ $f(-x) = -\alpha x^{5} - \beta x^{3} - \gamma x$ $f(-x) = -(\alpha x^{5} + \beta x^{3} + \gamma x)$ $f(-x) = -f(x)$ Since $f(-x) = -f(x)$, the function $f(x)$ is an odd function. For an odd function, $f(0) = 0$. Let's verify: $f(0) = \alpha (0)^{5} + \beta (0)^{3} + \gamma (0) = 0$ The derivative of $f(x)$ is $f'(x) = 5\alpha x^4 + 3\beta x^2 + \gamma$. Since $\alpha, \beta, \gamma > 0$ and $x^4 \ge 0$, $x^2 \ge 0$, $f'(x) > 0$ for all $x \in \mathbf{R}$. This means $f(x)$ is strictly increasing and thus invertible.

Step 2: Analyze the relationship between $f(x)$ and $g(x)$. We are given that $g(f(x)) = x$ for all $x \in \mathbf{R}$. This means $g$ is the inverse function of $f$. The problem asks for the value of $f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$. Using the property of inverse functions, $f(g(y)) = y$ for any $y$ in the range of $f$. Let $Y = \frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. Then the expression becomes $f(g(Y))$. If $Y$ is in the range of $f$, then $f(g(Y)) = Y$. Since $f(x)$ is a polynomial of odd degree with positive leading coefficients, its range is $\mathbf{R}$. Therefore, $Y$ will always be in the range of $f$. So, the expression simplifies to: $f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right) = \frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$

Step 3: Analyze the properties of the arithmetic progression $a_1, a_2, \ldots, a_n$. We are given that $a_1, a_2, \ldots, a_n$ are in an arithmetic progression with mean zero. The mean of the AP is given by $\frac{1}{n} \sum_{i=1}^n a_i$. We are given that this mean is zero: $\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} a_i = 0$ This implies that the sum of the terms is zero: $\sum\limits_{i=1}^{\mathrm{n}} a_i = 0$ This property is crucial. It means that the terms of the AP are symmetrically distributed around zero. For instance, if $n$ is even, say $n=2k$, then we can have pairs $a_i$ and $a_{n-i+1}$ such that $a_i + a_{n-i+1} = 0$. If $n$ is odd, the middle term is 0.

Step 4: Evaluate the sum $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. We need to evaluate $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. We know that $f(x)$ is an odd function, which means $f(-x) = -f(x)$. Since $\sum\limits_{i=1}^{\mathrm{n}} a_i = 0$, the terms $a_i$ are symmetrically distributed around 0. For every term $a_i$, there exists a term $a_j$ such that $a_j = -a_i$ (unless $a_i=0$). If the AP is $a_1, a_2, \ldots, a_n$, then the terms are of the form $a, a+d, a+2d, \ldots, a+(n-1)d$. The sum is $na + d\frac{(n-1)n}{2} = 0$. If $a_i$ is a term in the AP, then $-a_i$ is also effectively present in terms of their contribution to the sum. Consider the sum $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. Since $f$ is an odd function, for any term $a_i$, we have $f(-a_i) = -f(a_i)$. Because the AP has a mean of zero, the set of numbers $\{a_1, a_2, \ldots, a_n\}$ is symmetric about 0. This means that if $a_i$ is in the set, then $-a_i$ is also in the set (or if $a_i=0$, it is its own negative). Let's consider the sum of the $f(a_i)$ values: $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = f(a_1) + f(a_2) + \ldots + f(a_n)$ Due to the symmetry of $\{a_i\}$, we can pair up terms. For each $a_i$, there is an $a_j = -a_i$ (unless $a_i=0$). The sum can be written as: $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = \sum_{a_i \in \text{AP}} f(a_i)$ Since $f$ is odd, if $a_i$ is in the AP, then $-a_i$ is also in the AP (or $a_i=0$). The sum can be grouped as: $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = \sum_{a_i > 0} f(a_i) + \sum_{a_i < 0} f(a_i) + f(0) \quad (\text{if } 0 \text{ is in the AP})$ Since $f$ is odd, $f(a_i) + f(-a_i) = f(a_i) - f(a_i) = 0$. Therefore, the sum of $f(a_i)$ for all terms in the AP will be zero because each positive $f(a_i)$ term is cancelled by a corresponding negative $f(-a_i)$ term. $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = 0$

Step 5: Calculate the final expression. We established in Step 2 that the expression simplifies to $\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. From Step 4, we found that $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = 0$. Therefore, $\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right) = \frac{1}{\mathrm{n}} \times 0 = 0$ So, the value of $f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$ is $0$.

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Common Mistakes & Tips

• Assuming $a_i$ are symmetric: While $\sum a_i = 0$ implies symmetry, it's important to explicitly use the property that for every $a_i$, $-a_i$ is also in the set of terms (or $a_i=0$) when dealing with the sum of $f(a_i)$.
• Confusing $g(f(x))$ and $f(g(x))$: Remember that $g(f(x))=x$ and $f(g(y))=y$. Both are equal to the identity function on their respective domains.
• Ignoring the "positive real numbers" condition: The condition $\alpha, \beta, \gamma > 0$ ensures that $f(x)$ is strictly increasing, which guarantees the existence of a well-defined inverse function $g(x)$.

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Summary

The problem involves understanding the properties of inverse functions and odd functions, combined with the properties of an arithmetic progression with a mean of zero. We first identified that $f(x)$ is an odd function. Since $g$ is the inverse of $f$, the expression $f(g(Y))$ simplifies to $Y$. The core of the problem then reduces to evaluating $Y = \frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$. Given that $a_1, \ldots, a_n$ form an AP with mean zero, their sum is zero. Because $f(x)$ is an odd function, $f(a_i) + f(-a_i) = 0$. The symmetry of the AP terms around zero ensures that the sum $\sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)$ is zero. Consequently, $Y=0$, and thus $f(g(Y)) = Y = 0$.

The final answer is \boxed{0}. This corresponds to option (A).