Find the function $f(x)$ when $f(f(x))=1-x$, for $x\in [0,1]$ 
The function f(x) is continuous and $f(f(x))=1-x$, for $x\in [0,1]$ then,


(A) $f(\frac{1}{8})+f(\frac{7}{8})=3$


(B) $f(\frac{2}{3})+f(\frac{1}{3})=2$


(C) $f(\frac{5}{6})+f(\frac{1}{6})=1$


(D) None of These

My approach is as follow $f\left( {f\left( x \right)} \right) = 1 - x \Rightarrow f'\left( {f\left( x \right)} \right) \times f'\left( x \right) =  - 1$
$f'\left( {f\left( 0 \right)} \right) \times f'\left( 0 \right) =  - 1\& f'\left( {f\left( 1 \right)} \right) \times f'\left( 1 \right) =  - 1$
$f\left( {f\left( 0 \right)} \right) = 1;f\left( {f\left( 1 \right)} \right) = 0$
Not able to procced from here

From $f(f(x)) =1-x$ follows
$$1-f(x)=f(f(f(x)))=f(1-x) .$$
Hence,
$$f(x) +f(1-x) =1$$
for all $x\in[0, 1]$.
