Let α, β and γ be three positive real numbers. Let f ( x )=α x 5+β x 3+γ x , x ∈ R and g: R arrow R be such that g(f(x))=x for all x ∈ R. If a1, a2, a3, ldots, an be in arithmetic progression with mean zero, then the value of f(g((1/n) displaystyle∑i=1n f(ai))) is equal to :

Question

Let α, β and &gamma; be three positive real numbers. Let f ( x )=α x 5+β x 3+&gamma; x , x ∈ R and g: R arrow R be such that g(f(x))=x for all x ∈ R. If a1, a2, a3, ldots, an be in arithmetic progression with mean zero, then the value of f(g((1/n) displaystyle∑i=1n f(ai))) is equal to : (A) 0 (B) 3 (C) 9 (D) 27

Tardigrade Admin · Accepted Answer

Consider a case when α=β=0 then f(x)=y x g(x)=(x/y) (1/ n ) displaystyle∑ i =1 n f ( a i ) ⇒ ( y / n )( a 1+ a 2+ ldots . .+ a n ) =0 ⇒ f ( g (0)) ⇒ f (0) ⇒ 0

Q. Let α,β and γ be three positive real numbers. Let f(x)=αx5+βx3+γx,x∈R and g:R→R be such that g(f(x))=x for all x∈R. If a1​,a2​,a3​,…,an​ be in arithmetic progression with mean zero, then the value of f(g(n1​i=1∑n​f(ai​))) is equal to :

0

3

9

27

Consider a case when α=β=0 then f(x)=yx g(x)=yx​ n1​i=1∑n​f(ai​)⇒ny​(a1​+a2​+…..+an​) =0 ⇒f(g(0))⇒f(0) ⇒0

Consider a case when $α = β = 0$ then
$f (x) = y x$
$g (x) = \frac{x}{y}$
$\frac{1}{n} i = 1 \sum n f (a_{i}) \Rightarrow \frac{y}{n} (a_{1} + a_{2} + \dots .. + a_{n})$
$= 0$
$\Rightarrow f (g (0)) \Rightarrow f (0)$
$\Rightarrow 0$