Q115 / 200 · WBJEE 2013 · Mathematics

Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y)=f(x+y) \quad$ for $\quad$ all $\quad x, y \in R, \quad$ if $f(x), f(y)$ and $f(z)$ are in $G P,$ then $x, y$ and $z$ are in

  • Correct answer
    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 $z$

Solution

Let the function, $f(x)=a^{kx}$ Which define in $f: R \rightarrow R$ and injective also. Now, we have $$ f(x) f(y)=f(x+y) $$ $\Rightarrow$ $a^{k x} \cdot a^{k y}=a^{k(x+y)}$ $\Rightarrow$ $a^{k(x+y)}=a^{k(x+y)}$ $\because \quad f(x), f(y)$ and $f(z)$ are in GP $\therefore$ $f(y)^{2}=f(x) \cdot f(z)$ $\Rightarrow$ $a^{2 k y}=a^{k x} \cdot a^{k z}$ $\Rightarrow$ $e^{2 k y}=e^{k(x+z)}$ On comparing, we get $2 k y=k(x+z) \Rightarrow 2 y=x+z$ $\Rightarrow x, y$ and $z$ are in $A P$.

Hence, the correct answer is option: 'A'.

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