WAEC 2014 · Paper 2 · Q13

(a)
Two functions $f$ and $g$ are defined by $f : x \to 2x^2 - 1$ and $g : x \to 3x + 2$ , where $x$ is a real number. (i) If $f(x - 1) - 7 = 0$ , find the values of $x$ . (ii) Evaluate $\dfrac{f\left(-\frac12\right) \cdot g(3)}{f(4) - g(5)}$ .
Separate values with commas, e.g. 3, −2
(b)
An operation $(*)$ is defined on the set $\mathbb R$ of real numbers by $m * n = \dfrac{-n}{m^2 + 1}$ , where $m, n \in \mathbb R$ . If $m = -3$ and $n = -10$ , show whether or not $(*)$ is commutative.
Show the answer
$m * n = 1$ but $n * m = \frac{3}{101}$ , so $*$ is not commutative

Worked solution (try it first)

Replace every $x$ in $f(x) = 2x^2 - 1$ by $(x - 1)$ : $f(x - 1) = 2(x - 1)^2 - 1$ .
So $2(x - 1)^2 - 1 - 7 = 0$ , which gives $(x - 1)^2 = 4$ .
Then $x - 1 = 2$ or $x - 1 = -2$ .
So $x = 3$ or $x = -1$ .

$f\left(-\frac12\right) = 2 \times \frac14 - 1$
$= -\frac12$ .
$g(3) = 9 + 2 = 11$ .
$f(4) = 32 - 1 = 31$ .
$g(5) = 15 + 2 = 17$ .
So the value is $\frac{-\frac12 \times 11}{31 - 17} = \frac{-\frac{11}{2}}{14}$
$= -\frac{11}{28}$ .