Question

If for all $x,y$ the function f is defined by; $f\left(x\right)+f\left(y\right)+f\left(x\right)\cdot f\left(y\right)=1$ and $f\left(x\right)>0$.When $f\left(x\right)$ is differentiable $f^{\prime }\left(x\right)=$,

• A. $-1$
• B. $1$
• C. $0$
• D. cannot be determined

Answer 1

The correct option is C $0$

Here, $f\left(x\right)+f\left(y\right)+f\left(x\right)\cdot f\left(y\right)=1$ .....(i)
Substitute $x=y=0$, we get
$2f\left(0\right)+{\left\{f\right(0\left)\right\}}^2=1\Rightarrow {\left\{f\right(0\left)\right\}}^2+2f\left(0\right)-1=0$
$f\left(0\right)=\frac{-2\pm \sqrt{4+4}}{2}=-1-\sqrt{2}$ and $-1+\sqrt{2}$
As $f\left(0\right)>0\Rightarrow f\left(0\right)=\sqrt{2}-1$ [neglecting $f\left(0\right)=-1-\sqrt{2}$ as f(0) is positive]
Again, putting $y=x$ is Eq, (i), $2f\left(x\right)+{\left\{f\right(x\left)\right\}}^2=1$
On differentiating w.r.t. x, $2f^{\prime }\left(x\right)+2f\left(x\right)f^{\prime }\left(x\right)=0$
$2f^{\prime }\left(x\right)\{1+f(x\left)\right\}=0\Rightarrow f^{\prime }\left(x\right)=0$ because $f\left(x\right)>0$
Thus, $f^{\prime }\left(x\right)=0$ when $f\left(x\right)>0$