Question

If for all $x,y$ the function $f$ is defined by $f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right)=1$ and $f\left(x\right)>0$, then

• A. $f^{\prime }\left(x\right)$ does not exist
• B. $f^{\prime }\left(x\right)=0$ for all $x$
• C. $f^{\prime }\left(0\right)<f^{\prime }\left(1\right)$
• D. None of these

Answer 1

The correct option is B $f^{\prime }\left(x\right)=0$ for all $x$

Putting $x=0,y=0,$ we get

$2f\left(0\right)+{\left\{f\left(0\right)\right\}}^2=1\Rightarrow f\left(0\right)=\sqrt{2}-1(\because f\left(x\right)>0)$

Putting $y=x,2f\left(x\right)+{\left\{f\left(x\right)\right\}}^2=1$

Differentiating w.r.t. $x,$ we get

$2f^{\prime }\left(x\right)+2f\left(x\right).f^{\prime }\left(x\right)=0f^{\prime }\left(x\right)\{1+f\left(x\right)\}=0$

$\Rightarrow f^{\prime }\left(x\right)=0,$ because $f\left(x\right)>0$