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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
Let α,β\alpha, \beta and γ\gamma be three positive real numbers. Let f(x)=αx5+βx3+γx,x∈Rf(x)=\alpha x^5+\beta x^3+\gamma x, x \in \mathbb{R} and g:R→Rg: \mathbb{R} \rightarrow \mathbb{R} be such that g(f(x))=xg(f(x))=x for all x∈Rx \in \mathbb{R}. If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n be in arithmetic progression mean is zero, then the value of f(g(1n∑i=1nf(ai)))f\left(g\left(\frac{1}{n} \sum\limits_{i=1}^n f\left(a_i\right)\right)\right) is equal to
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Correct option: A
0
The correct answer is A (0) because the function f(x)f(x) is odd, and the sum of an arithmetic progression with a mean of zero is zero, leading to f(g(0))=0f(g(0)) = 0.

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