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 $f\left(x\right),f\left(y\right)$ and $f\left(z\right)$ are in GP, then $x,y$ and $z$ are in

• A. AP always
• B. GP always
• C. AP depending on the values of $x,y$ and $z$
• D. GP depending on the values of $x,y$ and $z$

Answer 1

The correct option is A AP always

Let the funtion $f\left(x\right)=a^{kx}$
which define in $f:R\to R$ and injective also.

Now, we have
$f\left(x\right)f\left(y\right)=f(x+y)$

$\Rightarrow$ $a^{kx}.a^{ky}=a^{k(x+y)}$

$\Rightarrow$ $a^{k(x+y)}=a^{k(x+y)}$

$\because$ $f\left(x\right),f\left(y\right)$ and $f\left(z\right)$ are in GP

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

$\Rightarrow$ $a^{2ky}=a^{kx}.a^{kz}$

$\Rightarrow$ $e^{2ky}=e^{k(x+z)}$

On comparing, we get
$2ky=k(x+z)$ $\Rightarrow$ $2y=x+z$

$\Rightarrow$ $x,y$ and $z$ are in AP