Question

If $f(x+y)=f\left(x\right)\cdot f\left(y\right)$ for all $x,y\in R$ and $f\left(0\right)\ne 0$, then the function $g\left(x\right)=\frac{f\left(x\right)}{1+\left\{f\right(x){\}}^2}$ is

• A. an odd function
• B. an even function
• C. Neither an even nor an odd function
• D. an odd function for $x<0$

Answer 1

The correct option is B an even function

We know, $f(x+y)=f\left(x\right)\cdot f\left(y\right)$ is satisfied by the function $f\left(x\right)=a^{kx}$
$\Rightarrow f(-x)=a^{-kx}=\frac{1}{a^{kx}}=\frac{1}{f\left(x\right)}$
Now,
$g\left(x\right)=\frac{f\left(x\right)}{1+\left\{f\right(x){\}}^2}$
Putting $x\to -x$
$\Rightarrow g(-x)=\frac{f(-x)}{1+\left\{f\right(-x){\}}^2}\Rightarrow g(-x)=\frac{\frac{1}{f\left(x\right)}}{1+\frac{1}{\left\{f\right(x){\}}^2}}\Rightarrow g(-x)=\frac{f\left(x\right)}{1+\left\{f\right(x){\}}^2}=g\left(x\right)$
Hence, $g\left(x\right)$ is an even function.