Question

Let $f:R\to R$ be such that $f$ is injective and $f\left(x\right)f\left(y\right)=f(x+y)$ $\forall x,y\in R$. 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 value of $x,y,z$
• D. G.P. depending on the value of $x,y,z$

Answer 1

The correct option is A A.P. always

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

Since, $f(y{)}^2=f(x\left)f\right(z)$ as $f\left(x\right),f\left(y\right),f\left(z\right)$ are in G.P.,
$\Rightarrow f\left(2y\right)=f(x+z)$
It is given that $f$ is injective.
$\Rightarrow 2y=x+z$
$\Rightarrow x,y,z$ are in A.P.