JAMB 1983 · UME · Q40

If $f(x) = \dfrac{1}{x - 1} + \dfrac{x - 1}{x^2 - 1}$ , find $f(1 - x)$ .

Worked solution (try it first)

Factorise $x^2 - 1 = (x - 1)(x + 1)$ and cancel $x - 1$ : the second fraction is $\frac{1}{x + 1}$ , so $f(x) = \frac{1}{x - 1} + \frac{1}{x + 1}$ .
Replace every $x$ by $1 - x$ : $f(1 - x) = \frac{1}{(1 - x) - 1} + \frac{1}{(1 - x) + 1}$ , which is $\frac{1}{-x} + \frac{1}{2 - x}$ .
Take the minus sign out of each bottom: $\frac{1}{2 - x} = -\frac{1}{x - 2}$ .
So $f(1 - x) = -\dfrac1x - \dfrac{1}{x - 2}$ , option C.