Mathematics - Functions Question with Solution | TestHub

Put $x=y=0\Rightarrow f(0)=\frac{f(0)+f(0)}{1-[f(0){]}^2}\Rightarrow f(0)[f^2(0)+1]=0\Rightarrow f(0)=0$ (since $f^2(0)\ne -1$). Now put $y=-x$, we get

$$\begin{array}{rl}& f(0)=\frac{f(x)+f(-x)}{1-f(x)\cdot f(-x)}\\ & \Rightarrow f(x)+f(-x)=0\\ & \Rightarrow f(-x)=-f(x)\Rightarrow f(x)\text{ is an odd function. }\\ & \text{ Now }f(y-x)=\frac{f(y)+f(-x)}{1-f(-x)\cdot f(y)}\\ & \Rightarrow \frac{f(y-x)}{y-x}=\frac{f(y)-f(x)}{1+f(x)\cdot f(y)}\cdot \frac{1}{(y-x)}\end{array}$$

Taking limit as $y\to x$, we get

$$\begin{array}{rl}& \underset{y\to x}{\lim }\frac{f(y-x)-f(0)}{y-x}=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}\cdot \underset{y\to x}{\lim }\frac{1}{1+f(x)\cdot f(y)}\\ & \Rightarrow \underset{y\to x}{\lim }\frac{f(y-x)}{y-x}=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}\underset{y\to x}{\lim }\frac{1}{1+f^2(x)}\\ & \Rightarrow f^{\prime }(0)=f^{\prime }(x)\cdot \frac{1}{1+f^2(x)}\\ & \Rightarrow f^{\prime }(x)=f^{\prime }(0)\left[1+f^2(x)\right]\\ & \Rightarrow \frac{f^{\prime }(x)}{f^{\prime }(0)}=1+f^2(x).\end{array}$$