1.7.2 Function, SPM Practice (Long Question)

Question 3:
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

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

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