Mathematics - Functions Question with Solution | TestHub

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

Put $x=\frac{1}{1-t}$ $1-x=1-\frac{1}{1-t}=-\frac{t}{1-t}$.

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

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

Put $\frac{x-1}{x}=t$.

$x-1=xt$ $x=\frac{1}{1-t}$.

$1-x=1-\frac{1}{1-t}=\frac{-t}{1-t}$,

$\frac{1}{1-x}=\frac{t-1}{t}$.

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

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

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

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

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