**Soalan 14 (4 markah):**

Rajah menunjukkan lengkung

*y*=

*g*(

*x*). Garis lurus ialah tangen kepada lengkung itu.

**Rajah**

Diberi

*g*’(

*x*) = –4

*x*+ 8, cari persamaan lengkung itu.

*Penyelesaian*:$\begin{array}{l}\text{Diberi}g\text{'}\left(x\right)=-4x+8\\ \text{Titikmaximumapabila}g\text{'}\left(x\right)=0\\ -4x+8=0\\ 4x=8\\ x=2\\ \text{Maka,titikmaximumialah}\left(2,11\right).\\ \\ g\text{'}\left(x\right)=-4x+8\\ {\displaystyle \int g\text{'}\left(x\right)}={\displaystyle \int \left(-4x+8\right)}dx\\ g\left(x\right)=\frac{-4{x}^{2}}{2}+8x+c\\ g\left(x\right)=-2{x}^{2}+8x+c\\ \\ \text{Gantikan}\left(2,11\right)\text{kedalam}g\left(x\right):\\ 11=-2{\left(2\right)}^{2}+8\left(2\right)+c\\ c=3\\ \\ \text{Maka,persamaanlengkungialah}\\ g\left(x\right)=-2{x}^{2}+8x+3\end{array}$

**Soalan 15 (3 markah):**

Diberi bahawa $\int \frac{5}{{\left(2x+3\right)}^{n}}}dx=\frac{p}{{\left(2x+3\right)}^{5}}+c$ , dengan keadaan

*c*,

*n*dan

*p*ialah pemalar.

Cari nilai

*n*dan nilai

*p*.

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