1.7.1 Function, SPM Practice (Paper 2 Questions 1 – 5)

Question 1:
The function f and g is defined by
$\begin{array}{l}f:x\mapsto 2x-3\\ g:x\mapsto \frac{2}{x};x\ne 0\end{array}$
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f^-1 ,
Calculate the value of x such that ff(x) = gf(x).

Solution:
(a)
$\begin{array}{l}ff\left(x\right)=f\left[f\left(x\right)\right]\\ =f\left(2x-3\right)\\ =2\left(2x-3\right)-3\\ =4x-9\\ ff:x\mapsto 4x-9\end{array}$

(b)
$\begin{array}{l}gf\left(x\right)=g\left[f\left(x\right)\right]\\ =g\left(2x-3\right)\\ =\frac{2}{2x-3}\\ gf:x\mapsto \frac{2}{2x-3}\end{array}$

(c)
$\begin{array}{l}\text{Let }{f}^{-1}\left(x\right)=y,\\ \text{thus }f\left(y\right)=x\\ 2y-3=x\\ y=\frac{x+3}{2}\\ \therefore f^{-1}\left(x\right)=\frac{x+3}{2}\\ f^{-1}:x\mapsto \frac{x+3}{2}\\ \\ \text{When }ff\left(x\right)=gf\left(x\right),\\ 4x-9=\frac{2}{2x-3}\\ \left(4x-9\right)\left(2x-3\right)=2\\ 8x^2-30x+27=2\\ 8x^2-30x+25=0\\ \left(4x-5\right)\left(2x-5\right)=0\\ 4x-5=0\text{ or }2x-5=0\\ x=\frac{5}{4}\text{ or }x=\frac{5}{2}\end{array}$

Question 2:
The function f and g is defined by
$\begin{array}{l}f\left(x\right)=3x-2\\ g\left(x\right)=\frac{3}{x},x\ne 0\\ \text{Find}\\ \text{(a) }{f}^{-1}\left(2\right),\\ \text{(b) g}f\left(-3\right),\\ \text{(c) function }h\text{ if }hf\left(x\right)=3x+2,\\ \text{(d) function }k\text{ if }fk\left(x\right)=4x-7.\end{array}$

Solution:
(a)
$\begin{array}{l}\text{Let }{f}^{-1}\left(2\right)=x,\\ \text{thus }f\left(x\right)=2\\ \text{ 3}x-2=2\\ 3x=4\\ x=\frac{4}{3}\\ f^{-1}\left(2\right)=\frac{4}{3}\end{array}$

(b)
$\begin{array}{l}gf\left(-3\right)=g\left[3\left(-3\right)-2\right]\\ =g\left(-11\right)\\ =-\frac{3}{11}\end{array}$

(c)
$\begin{array}{l}h\left[f\left(x\right)\right]=3x+2\\ h\left(3x-2\right)=3x+2\\ \text{Let }y=3x-2\\ \text{thus }x=\frac{y+2}{3}\\ h\left(y\right)=3\left(\frac{y+2}{3}\right)+2\\ =y+2+2\\ =y+4\\ \therefore h\left(x\right)=x+4\end{array}$

(d)
$\begin{array}{l}f\left[k\left(x\right)\right]=4x-7\\ 3k\left(x\right)-2=4x-7\\ 3k\left(x\right)=4x-5\\ k\left(x\right)=\frac{4x-5}{3}\end{array}$

Question 3:
In diagram below, the function g maps set P to set Q and the function h maps set Q to set R.



Find
(a) in terms of x, the function
(i) which maps set Q to set P,
(ii) h(x).
(b) the value of x such that gh(x) = 8x + 1.



Solution:
(a)(i)
$\begin{array}{l}g\left(x\right)=3x+2\\ \text{Let }{g}^{-1}\left(x\right)=y\\ g\left(y\right)=x\\ 3y+2=x\\ y=\frac{x-2}{3}\\ g^{-1}\left(x\right)=\frac{x-2}{3}\end{array}$

(a)(ii)
$\begin{array}{l}hg\left(x\right)=12x+5\\ h\left(3x+2\right)=12x+5\to \boxed{g\left(x\right)=3x+2}\\ \text{Let }u=3x+2\\ x=\frac{u-2}{3}\\ h\left(u\right)=12\left(\frac{u-2}{3}\right)+5\\ =4u-8+5\\ =4u-3\\ h\left(x\right)=4x-3\end{array}$

(b)
$\begin{array}{l}gh\left(x\right)=g\left(4x-3\right)\\ =3\left(4x-3\right)+2\\ =12x-9+2\\ =12x-7\\ 12x-7=8x+1\\ 4x=8\\ x=2\end{array}$

Question 4:
(c) function h if hg (x) = 12x + 5 [3 marks]

Question 5:
Given that f : x → hx + k and f² : x → 4x + 15.  
(a) Find the value of h and of k.
(b) Take the value of h > 0, find the values of x for which f (x² ) = 7x

Solution:
(a)
Given f (x) = hx + k
f² (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h² x + hk + k

f² (x) = 4x + 15
h² x + hk + k = 4x + 15
h² = 4
h = ± 2
when, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5 f (x² ) = 7x
2 (x² ) + 5 = 7x
2x² – 7x + 5 = 0
(2x – 5)(x –1) = 0
2x – 5 = 0 or x –1= 0
x = 5/2 x = 1