If `f(x+y)=f(x)dotf(y)` for all real `x , ya n df(0)!=0,` then prove that the function `g(x)=(f(x))/(1+{f(x)}^2)` is an even function.

Explanation

Given `f(x+y)=f(x)*f(y). " (1) " `
Put `x=y=0. ` Then `f(0)=1.`
Putting `y= -x` in (1), we get
`f(0)=f(x)f(-x)`
or ` f(-x)=`
Now, `g(x)=`
` :. g(-x)=(f(-x))/(1+{f(-x)}^(2))=((1)/(f(x)))/(1+(1)/({f(x)}^(2)))=(f(x))/(1+{f(x)}^(2))=g(x)`