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.
$$f'\left( x \right)$$ does not exist
B.
$$f'\left( x \right) = 0$$ for all $$x$$
C.
$$f'\left( 0 \right) < f'\left( 1 \right)$$
D.
none of these
Answer :
$$f'\left( x \right) = 0$$ for all $$x$$
Solution :
$$\eqalign{
& {\text{Putting }}x = 0,\,y = 0,\,\,{\text{we get }}2f\left( 0 \right){\text{ + }}{\left\{ {f\left( 0 \right)} \right\}^2}{\text{ = 1}} \cr
& \Rightarrow f\left( 0 \right) = \sqrt 2 - 1\,\,\,\,\left\{ {\because \,f\left( 0 \right) > 0} \right\} \cr
& {\text{Putting }}y = x,\,\,\,2f\left( x \right) + {\left\{ {f\left( x \right)} \right\}^2} = 1 \cr
& {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,{\text{ }} \cr
& {\text{2}}f'\left( x \right) + 2f\left( x \right).f'\left( x \right) = 0\,\,\,\,\,{\text{or}},\,\,f'\left( x \right)\left\{ {1 + f\left( x \right)} \right\} = 0 \cr
& \Rightarrow f'\left( x \right) = 0,\,\,{\text{because }}f\left( x \right) > 0 \cr} $$