\

SPM Additional Mathematics 2023, Paper 2 (Question 1 & 2)


Question 1:
Solve the following simultaneous equations: $$ \begin{gathered} 2 p-q-r=6 \\ p+2 q-4 r=8 \\ 3 p+q-r=2 \end{gathered} $$ [5 marks]
Solution:
$$ \begin{aligned} &\begin{aligned} & 2 p-q-r=6 \ldots(1)\\ & p+2 q-4 r=8 \ldots(2) \\ & 3 p+q-r=2 \ldots .(3) \end{aligned}\\ &\text { From (2), } p=8-2 q+4 r \ldots \text { (4) } \end{aligned} $$
$$ \begin{aligned} &\text { Substitute (4) into (1), }\\ &\begin{aligned} 2(8-2 q+4 r)-q-r & =6 \\ 16-4 q+8 r-q-r & =6 \\ -5 q+7 r & =-10\ldots(5) \end{aligned} \end{aligned} $$

$$ \begin{aligned} &\text { Substitute (4) into (3), }\\ &\begin{aligned} 3(8-2 q+4 r)+q-r & =2 \\ 24-6 q+12 r+q-r & =2 \\ -5 q+11 r & =-22 \ldots(6) \end{aligned} \end{aligned} $$
$$ \begin{aligned} & \text { Equation }(6)-(5) \\ & -5 q+11 r=-22 \ldots .(6) \\ & -5 q+7 r=-10 \ldots(5) \\ & 4 r=-12 \\ & r=-3 \end{aligned} $$

$$ \begin{aligned} & \text { Substitute } r=-3 \text { into }(5), \\ & \begin{aligned} -5 q+7(-3) & =-10 \\ -5 q-21 & =-10 \\ -5 q & =11 \\ q & =-\frac{11}{5} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { From (4), }\\ &\begin{aligned} & p=8-2\left(-\frac{11}{5}\right)+4(-3) \\ & p=\frac{2}{5} \\ & \therefore p=\frac{2}{5}, q=-\frac{11}{5}, r=-3 \end{aligned} \end{aligned} $$

Question 2:
Diagram 1 shows quadrilateral BCDE and right angled triangle ABC.


It is given that
$$ \overrightarrow{A B}=2 \underline{x}, \overrightarrow{A C}=3 \underline{y}, \overrightarrow{A B}=\frac{2}{3} \overrightarrow{A E} \text { and } \overrightarrow{C D}=\frac{1}{3} \underline{x}+\underline{y} . $$
(a) Express in terms of x and y of
(i) BC,
(ii) ED.
[3 marks]

(b) Show that ED and BC are parallel. [2 marks]

Solution:


(a)(i)
$$ \begin{aligned} & \overrightarrow{B C}=\overrightarrow{B A}+\overrightarrow{A C} \\ & \overrightarrow{B C}=-\overrightarrow{A B}+\overrightarrow{A C} \\ & \overrightarrow{B C}=-2 \underline{x}+3 \underline{y} \end{aligned} $$

(a)(ii)

$$ \begin{aligned} & \overrightarrow{E D}=\overrightarrow{E A}+\overrightarrow{A C}+\overrightarrow{C D} \\ & \overrightarrow{E D}=-\overrightarrow{A E}+\overrightarrow{A C}+\overrightarrow{C D} \\ & \overrightarrow{E D}=-\frac{3}{2} \overrightarrow{A B}+3 \underline{y}+\frac{1}{3} \underline{x}+\underline{y} \\ & \overrightarrow{E D}=-\frac{3}{2}(2 \underline{x})+4 \underline{y}+\frac{1}{3} \underline{x} \\ & \overrightarrow{E D}=-3 \underline{x}+4 \underline{y}+\frac{1}{3} \underline{x} \\ & \overrightarrow{E D}=4 \underline{y}-\frac{8}{3} \underline{x} \end{aligned} $$

(b)
$$ \begin{aligned} & \overrightarrow{E D}=-\frac{8}{3} \underline{x}+4 \underline{y} \\ \overrightarrow{E D} & =\frac{4}{3}(-2 \underline{x}+3 \underline{y}) \\ \overrightarrow{E D} & =\frac{4}{3}(\overrightarrow{B C}) \\ \therefore \overrightarrow{E D} & =\frac{4}{3} \overrightarrow{B C} \\ \therefore & \overrightarrow{E D} / / \overrightarrow{B C} \end{aligned} $$

Leave a Comment