SPM Additional Mathematics 2018, Paper 2 (Question 1 & 2)

Question 1 (6 marks):
The sum of the first n terms of an arithmetic progression, S_n is given by $S_n=\frac{3n\left(n-33\right)}{2}.$  
Find
(a) the sum of the first 10 terms,
(b) the first term and the common difference,
(c) the value of q, given that q^th term is the first positive term of the progression.


Solution:
(a)
$\begin{array}{l}{S}_n=\frac{3n\left(n-33\right)}{2}\\ S_{10}=\frac{3\left(10\right)\left(10-33\right)}{2}\\ S_{10}=-345\end{array}$

(b)
$\begin{array}{l}{S}_n=\frac{3n\left(n-33\right)}{2}\\ S_1=\frac{3\left(1\right)\left(1-33\right)}{2}\\ S_1=-48\\ T_1=S_1=-48\\ \\ \text{First term, }a=T_1=-48\\ T_n=S_n-S_{n-1}\\ T_2=S_2-S_1\\ T_2=\frac{3\left(2\right)\left(2-33\right)}{2}-\left(-48\right)\\ T_2=-45\\ \\ \text{Common difference, }d\\ =T_2-T_1\\ =-45-\left(-48\right)\\ =3\end{array}$


(c)
$\begin{array}{l}\text{First positive term, }{T}_q>0\\ T_q>0\\ a+\left(q-1\right)d>0\\ -48+\left(q-1\right)3>0\\ -48+3q-3>0\\ 3q>51\\ q>17\\ \text{Thus, }q=18.\end{array}$

Question 2 (8 marks):
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if $g\left(k+2\right)=\frac{1}{7}h\left(5\right),$
(iii) hg(x).

(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.


Solution:
(a)(i)
$\begin{array}{l}h\left(x\right)=1-3x\\ h\left(5\right)=1-3\left(5\right)\\ =-14\end{array}$

(a)(ii)
$\begin{array}{l}g\left(x\right)=2x-3\\ g\left(k+2\right)=\frac{1}{7}h\left(5\right)\\ 2\left(k+2\right)-3=\frac{1}{7}\left(-14\right)\\ 2k+4-3=-2\\ 2k=-3\\ k=-\frac{3}{2}\end{array}$

(a)(iii)
$\begin{array}{l}g\left(x\right)=2x-3,h\left(x\right)=1-3x\\ hg\left(x\right)=h\left(2x-3\right)\\ =1-3\left(2x-3\right)\\ =1-6x+9\\ =10-6x\end{array}$


(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16