Matematik Tambahan SPM 2017, Kertas 1 (Soalan 4 – 7)

Soalan 4 (3 markah):
Rajah 4 menunjukkan trapezium ABCD.

Rajah 4

$\begin{array}{l}\text{Diberi }\underset{˜}{p}=\left(\begin{array}{l}3\\ 4\end{array}\right)\text{ dan }\underset{˜}{q}=\left(\begin{array}{l}k-1\\ 2\end{array}\right)\text{, dengan}\\ \text{keadaan }k\text{ ialah pemalar, cari nilai }k\text{.}\end{array}$


Penyelesaian:

$\begin{array}{l}\underset{˜}{p}=m\underset{˜}{q}\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=m\left(\begin{array}{l}k-1\\ 2\end{array}\right)\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=\left(\begin{array}{l}mk-m\\ 2m\end{array}\right)\\ \\ mk-m=3\mathrm{\dots \dots \dots .}\left(1\right)\\ 2m=4\mathrm{\dots \dots \dots \dots \dots .}\left(2\right)\\ \\ \text{Dari}\left(2\right):\\ 2m=4\\ m=2\\ \\ \text{Gantikan }m=2\text{ ke dalam }\left(1\right):\\ 2k-2=3\\ 2k=3+2\\ 2k=5\\ k=\frac{5}{2}\end{array}$

Soalan 5 (3 markah):
$\begin{array}{l}\text{Diberi }\frac{{25}^{h+3}}{{125}^{p-1}}\text{=1, ungkapkan }p\text{ dalam}\\ \text{sebutan }h\text{.}\end{array}$

Penyelesaian:
$\begin{array}{l}\frac{{25}^{h+3}}{{125}^{p-1}}=1\\ {25}^{h+3}={125}^{p-1}\\ {\left(5^2\right)}^{h+3}={\left(5^3\right)}^{p-1}\\ 5^{2h+6}=5^{3p-3}\\ 2h+6=3p-3\\ 3p=2h+9\\ p=\frac{2h+9}{3}\end{array}$

Soalan 6 (3 markah):
$\begin{array}{l}\text{Selesaikan persamaan:}\\ {\log }_{m}324-{\log }_{\sqrt{m}}2m=2\end{array}$

Penyelesaian:
$\begin{array}{l}{\log }_{m}324-{\log }_{\sqrt{m}}2m=2\\ {\log }_{m}324-\frac{{\log }_{m}2m}{{\log }_m{m}^{\frac{1}{2}}}=2\\ {\log }_{m}324-2\left(\frac{{\log }_{m}2m}{{\log }_{m}m}\right)=2\\ {\log }_{m}324-2{\log }_{m}2m=2\\ {\log }_{m}324-{\log }_m{\left(2m\right)}^2=lo{g}_m{m}^2\\ {\log }_m\left(\frac{324}{4m^2}\right)=lo{g}_m{m}^2\\ \frac{324}{4m^2}=m^2\\ 4m^4=324\\ m^4=81\\ m=\pm 3\left(-3\text{ ditolak}\right)\end{array}$

Soalan 7 (2 markah):
Diberi bahawa sebutan ke-n bagi suatu janjang geometri ialah $T_n=\frac{3r^{n-1}}{2},r\ne k.$  
Nyatakan
(a) nilai k,
(b) sebutan pertama bagi janjang itu.

Penyelesaian:
(a)
k = 0, k = 1 atau k = -1 (salah satu daripada jawapan ini).

(b)
$\begin{array}{l}{T}_n=\frac{3}{2}{r}^{n-1}\\ T_1=\frac{3}{2}{r}^{1-1}\\ =\frac{3}{2}{r}^0\\ =\frac{3}{2}\left(1\right)\\ =\frac{3}{2}\end{array}$