Question

Let $f(x+y)=f(x.f(y\left)\right)\forall x,y\in R$ and $y\in R$ and $f\left(0\right)\ne 0$. Then the function $\varphi$ defined by $\varphi \left(X\right)=\frac{f\left(x\right)}{1+\left(f\right(x){)}^2}$ is

• A. Neither odd nor even function
• B. An odd function
• C. Constant function
• D. An even function

Answer 1

The correct option is D An even function

$f(x+y)=f\left(x\right)f\left(y\right),\forall x,y\in R$
$f\left(0\right)\ne 0.f(0+0)=f\left(0\right)f\left(0\right)$
$\Rightarrow f\left(0\right)=1\text{as}f\left(0\right)\ne 0$
Let us put $y=–x$ we get
$f\left(0\right)=f\left(x\right)f(–x)=1$
$\Rightarrow f(-x)=\frac{1}{f\left(x\right)}$
Now, $\varphi (-x)=\frac{f(-x)}{1+\left(f\right(-x){)}^2}=\frac{\frac{1}{f\left(x\right)}}{1+{\left(\frac{1}{f\left(x\right)}\right)}^2}=\frac{f\left(x\right)}{\left(f\right(x){)}^2+1}$
$=\varphi \left(x\right)$
$\Rightarrow \varphi \left(x\right)$ is an even function.