1.6.2 Function, SPM Practice (Paper 1 Question 11 – 20)

Question 11 (SPM 2017 – 3 marks):
Diagram 5 shows the graph of the function f : x → |1 – 2x| for the domain –2 ≤ x ≤ 4.

Diagram 5

State
(a) the object of 7,
(b) the image of 3,
(c) the domain of 0 ≤ f(x) ≤ 5.


Solution:
(a)
The object of 7 is 4.

(b)
f (x) = |1 – 2x|
f (3) = |1 – 2(3)|
= |1 – 6|
= |–5|
= 5

The image of 3 is 5.

(c)
|1 – 2x| = 5
1 – 2x = ±5
Given when f(x) = 5, x = –2.

When f(x) = –5
1 – 2x = –5
2x = 6
x = 3

Domain: –2 ≤ x ≤ 3.

Question 12 (SPM 2017 – 4 marks):
Given the function g : x → 2x – 8, find
$\begin{array}{l}\left(\text{a}\right)g^{-1}\left(x\right),\\ \left(\text{b}\right)\text{ the value of }p\text{ such that }{g}^2\left(\frac{3p}{2}\right)=30.\end{array}$

Solution:
(a)
$\begin{array}{l}\text{Let }y=g\left(x\right)\\ =2x-8\\ 2x-8=y\\ 2x=y+8\\ x=\frac{y+8}{2}\\ \text{Thus, }{g}^{-1}\left(x\right)=\frac{x+8}{2}\end{array}$

(b)
$\begin{array}{l}g\left(x\right)=2x-8\\ g^2\left(x\right)=g\left[g\left(x\right)\right]\\ =g\left(2x-8\right)\\ =2\left(2x-8\right)-8\\ =4x-16-8\\ =4x-24\\ g^2\left(\frac{3p}{2}\right)=30\\ 4\left(\frac{3p}{2}\right)-24=30\\ 6p=54\\ p=9\end{array}$

Question 13 (SPM 2018 – 4 marks):
Diagram 9 shows the relation between set A, set B and set C.

Diagram 9

It is given that set A maps to set B by the function $\frac{x+1}{2}$ and maps to set C by fg : x → x² + 2x + 4.
(a) Write the function which maps set A to set B by using the function notation.
(b) Find the function which maps set B to set C.


Solution:

(a)
$g:x\to \frac{x+1}{2}$

(b)

$\begin{array}{l}g\left(x\right)=\frac{x+1}{2}\\ fg\left(x\right)=x^2+2x+4\\ f\left[g\left(x\right)\right]=x^2+2x+4\\ f\left(\frac{x+1}{2}\right)=x^2+2x+4\\ \\ \text{Let }\frac{x+1}{2}=y\\ x+1=2y\\ x=2y-1\\ \\ \therefore f\left(y\right)={\left(2y-1\right)}^2+2\left(2y-1\right)+4\\ f\left(y\right)=4y^2-4y+1+4y-2+4\\ f\left(y\right)=4y^2+3\\ f\left(x\right)=4x^2+3\\ \\ \text{Thus, function which maps set }B\text{ to set }C\\ \text{is }f\left(x\right)=4x^2+3\end{array}$

Question 14:
$\begin{array}{l}\text{The function }f\text{ is denoted by }f:x\to \frac{1+x}{1-x},x\ne 1.\\ \text{Find }{f}^2,f^3,f^4\text{ and hence write down the functions }{f}^{51}\text{ and }{f}^{52}\text{.}\end{array}$

Solution:
$\begin{array}{l}f\left(x\right)=\frac{1+x}{1-x},x\ne 1\\ f^2\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{1+x}{1-x}\right)\\ =\frac{1+\left(\frac{1+x}{1-x}\right)}{1-\left(\frac{1+x}{1-x}\right)}=\frac{\frac{1-x+1+x}{\cancel{1-x}}}{\frac{1-x-1-x}{\cancel{1-x}}}\\ =\frac{2}{-2x}=-\frac{1}{x}\\ \\ f^3\left(x\right)=f\left[f^2\left(x\right)\right]=f\left(-\frac{1}{x}\right)\\ =\frac{1+\left(-\frac{1}{x}\right)}{1-\left(-\frac{1}{x}\right)}=\frac{\frac{x-1}{\cancel{x}}}{\frac{x+1}{\cancel{x}}}\\ =\frac{x-1}{x+1}\\ \\ f^4\left(x\right)=f\left[f^3\left(x\right)\right]=f\left(\frac{x-1}{x+1}\right)\\ =\frac{1+\left(\frac{x-1}{x+1}\right)}{1-\left(\frac{x-1}{x+1}\right)}=\frac{\frac{x+1+x-1}{\cancel{x+1}}}{\frac{x+1-x+1}{\cancel{x+1}}}\\ =\frac{2x}{2}=x\\ \\ f^5\left(x\right)=f\left[f^4\left(x\right)\right]=f\left(x\right)=\frac{1+x}{1-x}\left(\text{recurring}\right)\\ \\ \therefore f^{51}\left(x\right)=f^3\left[f^{48}\left(x\right)\right]=f^3\left(x\right)\\ =\frac{x-1}{x+1}\\ f^{52}\left(x\right)=f^4\left[f^{48}\left(x\right)\right]=f^4\left(x\right)=x\end{array}$