SPM Additional Mathematics 2018, Paper 1 (Question 20 – 22)

Question 20 (3 marks):
It is given that the curve y = (p – 2)x² – x + 7, where p is a constant, intersects with the straight line y = 3x + 5 at two points.
Find the range of values of p.

Solution:
$\begin{array}{l}y=\left(p-2\right)x^2-x+7\mathrm{\dots \dots \dots }\left(1\right)\\ y=3x+5\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots }\left(2\right)\\ \\ \text{Substitute }\left(1\right)\text{ into }\left(2\right):\\ \left(p-2\right)x^2-x+7=3x+5\\ \left(p-2\right)x^2-4x+2=0\\ a=\left(p-2\right),b=-4,c=2\\ b^2-4ac>0\\ {\left(-4\right)}^2-4\left(p-2\right)\left(2\right)>0\\ 16-8p+16>0\\ -8p>-32\\ 8p<32\\ p<4\end{array}$

Question 21 (3 marks):
It is given that the quadratic equation hx² – 3x + k = 0, where h and k are constants has roots β and 2β.
Express h in terms of k.

Solution:
$\begin{array}{l}h{x}^2-3x+k=0\\ a=h,b=-3,c=k\\ \\ \text{SOR}=-\frac{b}{a}=-\frac{\left(-3\right)}{h}=\frac{3}{h}\\ \\ \text{POR}=\frac{c}{a}=\frac{k}{h}\\ \\ \text{Given roots}=\beta \text{ and 2}\beta .\\ \text{SOR}=\beta +\text{2}\beta =3\beta ;\\ \text{POR}=\beta \left(\text{2}\beta \right)=2{\beta }^2\\ \frac{3}{h}=\text{3}\beta \\ \beta =\frac{1}{h}\mathrm{\dots \dots \dots \dots ..}\left(1\right)\\ \frac{k}{h}=2{\beta }^2\mathrm{\dots \dots \dots ..}\left(2\right)\\ \\ \text{Substitute }\left(1\right)\text{ into }\left(2\right):\\ \frac{k}{h}=2{\left(\frac{1}{h}\right)}^2\\ \frac{k}{h}=\frac{2}{h^2}\\ \frac{h^2}{h}=\frac{2}{k}\\ h=\frac{2}{k}\end{array}$

Question 22 (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)

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