Question 1:
The following information refers to the equation of the straight lines AB and CD.
Find the value of p, if both of the straight lines are parallel. [2 marks]
Answer:
$$ \begin{aligned} A B: 2 y-8 x+1 & =0 \\ 2 y & =8 x-1 \\ y & =\frac{8}{2} x-\frac{1}{2} \\ y & =4 x-\frac{1}{2} \end{aligned} $$
$$ \begin{aligned} &C D: y=6 p x+3\\ &\begin{aligned} A B \| C D \rightarrow m_{C D} & =m_{A B} \\ 6 p & =4 \\ p & =\frac{4}{6} \\ p & =\frac{2}{3} \end{aligned} \end{aligned} $$
The following information refers to the equation of the straight lines AB and CD.
Find the value of p, if both of the straight lines are parallel. [2 marks]
Answer:
$$ \begin{aligned} A B: 2 y-8 x+1 & =0 \\ 2 y & =8 x-1 \\ y & =\frac{8}{2} x-\frac{1}{2} \\ y & =4 x-\frac{1}{2} \end{aligned} $$
$$ \begin{aligned} &C D: y=6 p x+3\\ &\begin{aligned} A B \| C D \rightarrow m_{C D} & =m_{A B} \\ 6 p & =4 \\ p & =\frac{4}{6} \\ p & =\frac{2}{3} \end{aligned} \end{aligned} $$
Question 2:
Diagram 2 shows the graph of the straight line Y against X.
(a) Determine the gradient of the straight line. [1 mark]
(b) $$ \text { Hence, express } y \text { in terms of } x \text { such that } X=x^2 y \text { and } Y=\frac{y}{x} \text {. } $$
[2 marks]
Answer:
(a) $$ \begin{aligned} & m=\frac{3-0}{0-(-6)} \\ & m=\frac{1}{2} \end{aligned} $$
(b) $$ \begin{aligned} Y & =m X+C \\ \frac{y}{x} & =\frac{1}{2}\left(x^2 y\right)+3 \\ y & =\frac{1}{2} x^3 y+3 x \\ 2 y & =x^3 y+6 x \\ 2 y-x^3 y & =6 x \\ y\left(2-x^3\right) & =6 x \\ y & =\frac{6 x}{2-x^3} \end{aligned} $$
Diagram 2 shows the graph of the straight line Y against X.
(a) Determine the gradient of the straight line. [1 mark](b) $$ \text { Hence, express } y \text { in terms of } x \text { such that } X=x^2 y \text { and } Y=\frac{y}{x} \text {. } $$
[2 marks]
Answer:
(a) $$ \begin{aligned} & m=\frac{3-0}{0-(-6)} \\ & m=\frac{1}{2} \end{aligned} $$
(b) $$ \begin{aligned} Y & =m X+C \\ \frac{y}{x} & =\frac{1}{2}\left(x^2 y\right)+3 \\ y & =\frac{1}{2} x^3 y+3 x \\ 2 y & =x^3 y+6 x \\ 2 y-x^3 y & =6 x \\ y\left(2-x^3\right) & =6 x \\ y & =\frac{6 x}{2-x^3} \end{aligned} $$
Question 3:
Listing out all terms of the sequence is not accepted for this question.
Given a sequence of number 9, 3, -3, … , -105.
(a) Find the number of terms of the sequence. [2 marks]
(b) Hence, find the sum of all the terms. [2 marks]
Answer:
(a) $$ \begin{aligned} &9,3,-3, \ldots,-105\\ &\begin{aligned} d & =3-9 \quad, \quad a=9 \\ & =-6 \end{aligned} \end{aligned} $$
$$ \begin{aligned} T_n & =a+(n-1) d \\ -105 & =9+(n-1)(-6) \\ -105 & =9-6 n+6 \\ 6 n & =15+105 \\ 6 n & =120 \\ n & =\frac{120}{6} \\ n & =20 \end{aligned} $$
(b) $$ \begin{aligned} & S_n=\frac{n}{2}(2 a+(n-1) d) \\ & S_{20}=\frac{20}{2}[2(9)+(20-1)(-6)] \\ & S_{20}=10(-96) \\ & S_{20}=-960 \end{aligned} $$
Listing out all terms of the sequence is not accepted for this question.
Given a sequence of number 9, 3, -3, … , -105.
(a) Find the number of terms of the sequence. [2 marks]
(b) Hence, find the sum of all the terms. [2 marks]
Answer:
(a) $$ \begin{aligned} &9,3,-3, \ldots,-105\\ &\begin{aligned} d & =3-9 \quad, \quad a=9 \\ & =-6 \end{aligned} \end{aligned} $$
$$ \begin{aligned} T_n & =a+(n-1) d \\ -105 & =9+(n-1)(-6) \\ -105 & =9-6 n+6 \\ 6 n & =15+105 \\ 6 n & =120 \\ n & =\frac{120}{6} \\ n & =20 \end{aligned} $$
(b) $$ \begin{aligned} & S_n=\frac{n}{2}(2 a+(n-1) d) \\ & S_{20}=\frac{20}{2}[2(9)+(20-1)(-6)] \\ & S_{20}=10(-96) \\ & S_{20}=-960 \end{aligned} $$