JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 14)
Let $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$$ and $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. If $$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$$ be in arithmetic progression with mean zero, then 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 equal to :
0
3
9
27
Explanation
$$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$$
$${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$$
$$\therefore$$ First and last term, second and second last and so on are equal in magnitude but opposite in sign.
$$f(x) = \alpha {x^5} + \beta {x^3} + \gamma x$$
$$\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)} $$
$$ = 0\alpha + 0\beta + 0\gamma $$
$$ = 0$$
$$\therefore$$ $$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0} $$
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