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 $x$

$f\left(x\right)+f\left(y\right)+f\left(x\right).f\left(y\right)=1$
put $y=0$
$\Rightarrow f\left(x\right)+f\left(0\right)+f\left(x\right).f\left(0\right)=1$
$\Rightarrow f\left(x\right)=\frac{1-f\left(0\right)}{1+f\left(0\right)}=C$ (constant)
$\therefore f^{\prime }\left(x\right)=0\forall x$