1.6.1 Function, SPM Practice (Simple Question Part 1)

Question 1:
Diagram below shows the relation between set M and set N in the graph form.
State
(a) the range of the relation,
(b) the type of the relation between set M and set N.

Solution: 
(a) Range of the relation = {p, r, s}.
(b) Type of the relation between set M and set N is many to one relation.

Question 2:
Diagram below shows the relation between set P and set Q.


State
(a) the object of 3,
(b) the range of the relation.

Solution: 
(a) The object of 3 is 7.
(b) The range of the relation is {–3, –1, 1, 3}.

Question 3:
Diagram below shows the function g : x → x – 2k, where k is a constant.

Find the value of k.

Solution:
$\begin{array}{l}\text{Given }g\left(6\right)=12\\ g\left(6\right)=6-2k\\ \text{ 12}=6-2k\\ \text{ 2}k=6-12\\ k=-3\end{array}$

Question 4:
Diagram below shows the relation between set M and set N in the arrow diagram.

(a) Represent the relation in the form of ordered pairs.
(b) State the domain of the relation.

Solution: 
(a) Relation in the form of ordered pairs = {(–4, 8), (3, 3), (4, 8)}.
(b) Domain of the relation = {–4, 3, 4}.

Question 5:
It is given the functions g(x) = 3x and h(x) = m – nx, where m and n are constants.
Express m in terms of n such that hg(1) = 4.

Solution:
$\begin{array}{l}hg\left(x\right)=h\left(3x\right)\\ =m-n\left(3x\right)\\ =m-3nx\\ \\ hg\left(1\right)=4\\ m-3n\left(1\right)=4\\ m-3n=4\\ m=4+3n\end{array}$

Question 6:
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 7:
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 8:
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)
$\begin{array}{l}g\left[h\left(x\right)\right]=9x^2+6x-4\\ g\left(3x+1\right)=9x^2+6x-4\\ \text{Let }y=3x+1\\ \text{thus }x=\frac{y-1}{3}\\ \text{ g}\left(y\right)=9{\left(\frac{y-1}{3}\right)}^2+6\left(\frac{y-1}{3}\right)-4\\ =\frac{\cancel{9}{\left(y-1\right)}^2}{\cancel{9}}+2\left(y-1\right)-4\\ =y^2-2y+1+2y-2-4\\ =y^2-5\\ \therefore g\left(x\right)=x^2-5\end{array}$