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 $\forall x,yϵR$. 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

Answer: (A)

A.P. always

Given : $f:R\to R$ is injevtive and $f\left(x\right)f\left(y\right)=f(x+y)$

Let $f\left(x\right)=a^x$ then $f\left(x\right)f\left(y\right)=a^x{a}^y=a^{x+y}=f(x+y)$

Also, $f\left(x\right)=f\left(y\right)⟹a^x=a^y⟹a^{x-y}=a^0⟹x-y=0⟹x=y$

$\therefore$ $a^x$ is injective

Now, $a^x,a^y,a^z$ are in G.P.

$⟹a^y=a^{x}r$, where $r$ is common ratio

Also, $a^z=a^{y}r$

$⟹\frac{a^y}{a^z}=\frac{a^x}{a^y}$

$⟹a^{2y}=a^{x+z}$

$⟹y=\frac{x+z}{2}$
$\therefore x,y,z$ are in A.P.