WAEC 2017 · Paper 2 · Q13

(a)
An operation $*$ is defined by $x * y = x + y + 2xy$ , $x, y \in \mathbb R$ . (i) Calculate $(2 * 3) * 5$ . (ii) Find the truth set of $(x * 7) = (x * 5) * 2$ .
Separate values with commas, e.g. 3, −2
(b)
$PQRS$ is a trapezium in which $\overrightarrow{PQ}$ is parallel to $\overrightarrow{SR}$ and $2\overrightarrow{SR} = 3\overrightarrow{PQ}$ . If $\overrightarrow{PQ} = \begin{pmatrix} 8 \\ 6 \end{pmatrix}$ and $\overrightarrow{QR} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$ , find, in component form, the vector: (i) $\overrightarrow{RS}$ ; (ii) $\overrightarrow{PS}$ .
Show the answer
(i) $\begin{pmatrix} -12 \\ -9 \end{pmatrix}$ ; (ii) $\begin{pmatrix} -8 \\ 0 \end{pmatrix}$

Worked solution (try it first)

Write out each side with the rule.
$x * 7 = x + 7 + 14x = 15x + 7$ , and $x * 5 = x + 5 + 10x = 11x + 5$ .
For $(x * 5) * 2$ , the first number is $11x + 5$ : $(11x + 5) + 2 + 2(11x + 5)(2)$ .
This is $11x + 7 + 44x + 20$ , which is $55x + 27$ .
Solve $15x + 7 = 55x + 27$ : $-40x = 20$ , so $x = -\frac12$ .
The truth set is $\left\{-\frac12\right\}$ .

$2\overrightarrow{SR} = 3\overrightarrow{PQ}$ , so $\overrightarrow{SR} = \frac32\begin{pmatrix} 8 \\ 6 \end{pmatrix}$
$= \begin{pmatrix} 12 \\ 9 \end{pmatrix}$ .
$\overrightarrow{RS}$ goes the other way: $\overrightarrow{RS} = \begin{pmatrix} -12 \\ -9 \end{pmatrix}$ .

Go round the trapezium from $P$ : $\overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS}$
$= \begin{pmatrix} 8 - 4 - 12 \\ 6 + 3 - 9 \end{pmatrix}$
$= \begin{pmatrix} -8 \\ 0 \end{pmatrix}$ .