WAEC 2023 · Paper 2 · Q9

Two functions $g$ and $h$ are defined on the set of real numbers, $\mathbb R$ , by $g: x \to \dfrac{2 + x}{x^2 + x - 20}$ and $h: x \to x + 3$ .

(a)(i)
Find $g \circ h(x)$ ;
(a)(ii)
the values of $x$ for which $g \circ h(x)$ is undefined.
Separate values with commas, e.g. 3, −2
(b)
Express $g \circ h(x)$ in partial fractions.

Worked solution (try it first)

(a)(i)

$g \circ h(x) = g(x + 3)$ : replace every $x$ in $g$ by $x + 3$ .
The top: $2 + (x + 3) = x + 5$ .
The bottom: $(x + 3)^2 + (x + 3) - 20 = x^2 + 6x + 9 + x + 3 - 20$
$= x^2 + 7x - 8$ .
So $g \circ h(x) = \dfrac{x + 5}{x^2 + 7x - 8}$ .

(ii)

Factorise the bottom: $x^2 + 7x - 8 = (x + 8)(x - 1)$ .
It is zero, so $g \circ h$ is undefined, at $x = 1$ and $x = -8$ .

(b)

Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 8}$ and multiply through: $x + 5 = A(x + 8) + B(x - 1)$ .
Put $x = 1$ : $6 = 9A$ , so $A = \frac23$ .
Put $x = -8$ : $-3 = -9B$ , so $B = \frac13$ .
So $g \circ h(x) = \dfrac{2}{3(x - 1)} + \dfrac{1}{3(x + 8)}$ .

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