SPM Practice (Short Question)

Question 1:
Given the function g : x → 3x – 2, find  
(a) the value of x when g(x) maps onto itself,
(b) the value of k such that g(2 – k) = 4k.

Solution:
(a)
$\begin{array}{l}g\left(x\right)=x\\ 3x-2=x\\ 3x-x=2\\ 2x=2\\ x=1\end{array}$

(b)
$\begin{array}{l}g\left(x\right)=3x-2\\ g\left(2-k\right)=4k\\ 3\left(2-k\right)-2=4k\\ 6-3k-2=4k\\ -7k=-4\\ k=\frac{4}{7}\end{array}$

Question 2:
Given the functions f : x → px + 1, g : x → 3x – 5 and fg(x) = 3px + q.  
Express p in terms of q.

Solution:
$\begin{array}{l}f\left(x\right)=px+1,g\left(x\right)=3x-5\\ fg\left(x\right)=p\left(3x-5\right)+1\\ =3px-5p+1\\ \\ \text{Given }fg\left(x\right)=3px+q\\ 3px-5p+1=3px+q\\ -5p+1=q\\ -5p=q-1\\ 5p=1-q\\ p=\frac{1-q}{5}\end{array}$

Question 3:
Given the functions h : x → 3x + 1, and gh : x → 9x² + 6x – 4, find  
(a) h^-1 (x),
(b) g(x).

Solution:
(a)
$\begin{array}{l}\text{Let }{h}^{-1}\left(x\right)=y,\\ \text{thus }h\left(y\right)=x\\ 3y+1=x\\ \text{3}y=x-1\\ y=\frac{x-1}{3}\\ \therefore h^{-1}\left(x\right)=\frac{x-1}{3}\\ h^{-1}:x\mapsto \frac{x-1}{3}\end{array}$

(b)
$\begin{array}{l}g\left[h\left(x\right)\right]=9x^2+6x-4\\ g\left(3x+1\right)=9x^2+6x-4\\ \text{Let }y=3x+1\\ \text{thus }x=\frac{y-1}{3}\\ \text{ g}\left(y\right)=9{\left(\frac{y-1}{3}\right)}^2+6\left(\frac{y-1}{3}\right)-4\\ =\frac{\cancel{9}{\left(y-1\right)}^2}{\cancel{9}}+2\left(y-1\right)-4\\ =y^2-2y+1+2y-2-4\\ =y^2-5\\ \therefore g\left(x\right)=x^2-5\end{array}$