WAEC 2008 · Paper 2 · Q1

Functions

Two functions $f$ and $g$ are defined on the set $\mathbb{R}$ of real numbers by $f : x \to \dfrac{x + 1}{x - 2}$ , $x \neq 2$ , and $g : x \to \dfrac{kx + 3}{x}$ , $x \neq 0$ .

(a)
Find $f^{-1}$ , the inverse of $f$ .
(b)
Given that $g \circ f^{-1}(4) = 6$ , find the value of $k$ .

Worked solution (try it first)

(a)

Let $y = \dfrac{x + 1}{x - 2}$ and make $x$ the subject.
Multiply up: $y(x - 2) = x + 1$ .
Expand: $xy - 2y = x + 1$ .
Collect the $x$ terms: $xy - x = 2y + 1$ , so $x(y - 1) = 2y + 1$ .
Divide: $x = \dfrac{2y + 1}{y - 1}$ .
So $f^{-1}(x) = \dfrac{2x + 1}{x - 1}$ , $x \neq 1$ .

(b)

First find $f^{-1}(4)$ : put $x = 4$ into $\dfrac{2x + 1}{x - 1}$ to get $\dfrac93$ , which is 3.
Then $g(3) = \dfrac{3k + 3}{3} = k + 1$ .
Set $k + 1 = 6$ , so $k = 5$ .

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