Short Question 14 & 15

Question 14 (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 15 (4 marks):
Diagram shows the curve y = g(x). The straight line is a tangent to the curve.
Diagram

Given g’(x) = –4x + 8, find the equation of the curve.


Solution:
$\begin{array}{l}\text{Given }g\text{'}\left(x\right)=-4x+8\\ \text{Maximum point when }g\text{'}\left(x\right)=0\\ -4x+8=0\\ 4x=8\\ x=2\\ \text{Thus, maximum point is }\left(2,11\right).\\ \\ g\text{'}\left(x\right)=-4x+8\\ {\displaystyle \int g\text{'}\left(x\right)}={\displaystyle \int \left(-4x+8\right)}dx\\ g\left(x\right)=\frac{-4x^2}{2}+8x+c\\ g\left(x\right)=-2x^2+8x+c\\ \\ \text{Substitute }\left(2,11\right)\text{ into }g\left(x\right):\\ 11=-2{\left(2\right)}^2+8\left(2\right)+c\\ c=3\\ \\ \text{Thus, the equation of curve is}\\ g\left(x\right)=-2x^2+8x+3\end{array}$