Let α, β and γ be three positive real numbers.Let fx=αx5+βx3+γx, x∈R and g : R→Rbe such that gfx=x for all x∈R.If a1, a2, a3, ..... an be in arithmetic progression with mean zero, then the value of fg1n∑i=1nfai is equal to:

Consider a case when α=β=0 then fx=γx, and gfx=x⇒gγx=x ⇒gx=xγ 1n∑i=1nfai⇒yna1+a2+.....+an =0 ⇒fg0⇒f0 ⇒0

Let α, β and γ be three positive real numbers.Let fx=αx5+βx3+γx, x∈R and g : R→Rbe such that gfx=x for all x∈R.If a1, a2, a3, ..... an be in arithmetic progression with mean zero, then the value of fg1n∑i=1nfai is equal to:

Consider a case when α=β=0 then fx=γx, and gfx=x⇒gγx=x ⇒gx=xγ 1n∑i=1nfai⇒yna1+a2+.....+an =0 ⇒fg0⇒f0 ⇒0

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Let $α, β a n d γ$ be three positive real numbers.
$L e t f (x) = α x^{5} + β x^{3} + γ x, x \in R a n d g : R \to R$ be such that $g (f (x)) = x f o r a l l x \in R .$
$I f a_{1}, a_{2}, a_{3}, . . . . . a_{n}$ be in arithmetic progression with mean zero, then the value of
$f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))$ is equal to:

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Detailed Solution

$C o n s i d e r a c a s e w h e n α = β = 0 t h e n f (x) = γ x, a n d g (f (x)) = x \Rightarrow g (γ x) = x \Rightarrow g (x) = \frac{x}{γ} \frac{1}{n} \sum_{i = 1}^{n} f (a_{i}) \Rightarrow \frac{y}{n} (a_{1} + a_{2} + . . . . . + a_{n}) = 0 \Rightarrow f (g (0)) \Rightarrow f (0) \Rightarrow 0$