Question

Let $f:R\to R$ be such that f is injective and $f\left(x\right)f\left(y\right)=f(x+y)$ for all $x,y\in R$. If $f\left(x\right),f\left(y\right),f\left(z\right)$ are in G.P., then $x,y,z$ are in

• A. A.P. always
• B. G.P. always
• C. A.P. depending on the values of x, y, z
• D. G.P. depending on the values of x, y, z

Answer 1

Answer: (A)

A.P. always

Given $f$ is an injective function

$f\left(x\right)f\left(y\right)=f(x+y)\left(1\right)$
Also, $f\left(x\right),f\left(y\right),f\left(z\right)$ are in GP

i.e. $\left[f\right(y){]}^2=f(x\left)f\right(z)$

From $\left(1\right)$

$f\left(y\right).f\left(y\right)=f\left(x\right)f\left(z\right)f(y+y)=f(x+z)f\left(2y\right)=f(x+z)2y=x+z$

This is a condition for arithmetic progression

Hence, $x,y,z$ are in AP