SPM Additional Mathematics 2017, Paper 1 (Question 17 – 19)

Question 17 (3 marks):
It is given that ${\displaystyle \int \frac{5}{{\left(2x+3\right)}^n}}dx=\frac{p}{{\left(2x+3\right)}^5}+c$ , where c, n and p are constants.
Find the value of n and of p.

Solution:
$\begin{array}{l}{\displaystyle \int \frac{5}{{\left(2x+3\right)}^n}}dx={\displaystyle \int 5{\left(2x+3\right)}^{-n}}dx\\ =\frac{5{\left(2x+3\right)}^{-n+1}}{\left(-n+1\right)\times 2}+c\\ =\frac{5}{2\left(1-n\right)}\times \frac{1}{{\left(2x+3\right)}^{n-1}}+c\\ =\frac{5}{2\left(1-n\right){\left(2x+3\right)}^{n-1}}+c\\ \\ \text{Compare }\frac{5}{2\left(1-n\right){\left(2x+3\right)}^{n-1}}\\ \text{with }\frac{p}{{\left(2x+3\right)}^5}\\ \\ n-1=5\\ n=6\\ \\ \frac{5}{2\left(1-n\right)}=p\\ \frac{5}{2\left(1-6\right)}=p\\ \frac{5}{2\left(-5\right)}=p\\ p=-\frac{1}{2}\end{array}$

Question 18 (3 marks):
A straight line passes through P(3, 1) and Q(12, 7). The point R divides the line segment PQ such that 2PQ = 3RQ.
Find the coordinates of R.

Solution:



$\begin{array}{l}2PQ=3RQ\\ \frac{PQ}{RQ}=\frac{3}{2}\\ \\ \text{Point }R\\ =\left(\frac{1\left(12\right)+2\left(3\right)}{1+2},\frac{1\left(7\right)+2\left(1\right)}{1+2}\right)\\ =\left(\frac{18}{3},\frac{9}{3}\right)\\ =\left(6,3\right)\end{array}$

Question 19 (3 marks):
The variables x and y are related by the equation $y=x+\frac{r}{x^2}$ , where r is a constant. Diagram 8 shows a straight line graph obtained by plotting $\left(y-x\right)\text{ against }\frac{1}{x^2}.$

Diagram 8

Express h in terms of p and r.


Solution:

$\begin{array}{l}y=x+\frac{r}{x^2}\\ y-x=r\left(\frac{1}{x^2}\right)+0\\ Y=mX+c\\ m=r,c=0\\ \\ m=\frac{y_2-y_1}{x_2-x_1}\\ r=\frac{5p-0}{\frac{h}{2}-0}\\ \frac{hr}{2}=5p\\ hr=10p\\ h=\frac{10p}{r}\end{array}$