SPM Practice (Long Question)

Question 6 (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