Question

If $f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$ for all $x\ne 0$, $1$, then

• A. $f\left(x\right)=\frac{1}{2}(x-1+\frac{1}{x}-\frac{1}{1-x})$
• B. $f\left(x\right)=\frac{1}{2}(1+x-\frac{1}{x}+\frac{1}{1-x})$
• C. $f\left(x\right)=\frac{1}{2}(1+x-\frac{1}{x}-\frac{1}{1-x})$
• D. $f\left(x\right)=\frac{1}{2}(1-x+\frac{1}{x}-\frac{1}{1-x})$

Answer 1

The correct option is C $f\left(x\right)=\frac{1}{2}(1+x-\frac{1}{x}-\frac{1}{1-x})$

$f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$ ... (i) given

Replace '$x$' by $\left(\frac{1}{1-x}\right)$;

$\therefore f\left(\frac{1}{1-x}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$

i.e., $f\left(\frac{1}{1-x}\right)+f(1-\frac{1}{x})=\frac{1}{1-x}$ ...(ii)

In eqn (i); replace '$x$' by $(1-\frac{1}{x})$; we get

$f(1-\frac{1}{x})+f\left(\frac{1}{1-(1-\frac{1}{x})}\right)=1-\frac{1}{x}$

i.e., $f(1-\frac{1}{x})+f\left(x\right)=1-\frac{1}{x}$ ...(iii)

Now, we have these equations; and three unknowns $\left(f\right(x);f\left(\frac{1}{1-x}\right);f(1-\frac{1}{x}))$

Solving these simultaneously, to get $f\left(x\right)$; we get

$f\left(x\right)=\frac{1}{2}(1+x-\frac{1}{x}-\frac{1}{1-x})$