Question

Let $\alpha, \beta$ and $\gamma$ be three positive real numbers Let $f ( x )=\alpha x ^5+\beta x ^3+\gamma x , x \in R$ and $g: R \rightarrow R$ be such that $g(f(x))=x$ for all $x \in R$ If $a_1, a_2, a_3, \ldots, a_n$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\frac{1}{n} \displaystyle\sum_{i=1}^n f\left(a_i\right)\right)\right)$is equal to :

• A. 0
• B. 3
• C. 9
• D. 27

Answer 1

Answer: (A)

0

Answer 2

The correct option is (A) : 0
Consider a case when α = β = 0 then
$f(x) = yx$
$g(x)=\frac{x}{y}$
$\frac{1}{n}\sum{^{n}_{i=1}}f(a_i)⇒\frac{y}{n}(a_1+a_2+....+a_n)$
$=0$
$⇒f(g(0))⇒f(0)$
$⇒0$