Soalan 1:
Selesaikan persamaan serentak berikut:
$$ 4 x+y+2 z=0 \quad, \quad-2 x+3 y+z=8 \quad, \quad x-y-z=-\frac{1}{2} $$
[5 markah]
Penyelesaian:
$$ \begin{aligned} 4 x+y+2 z & =0 … (1) \\ -2 x+3 y+z & =8 … (2) \\ x-y-z & =-\frac{1}{2} … (3)\end{aligned} $$
$$ \text { Dari (1) : } y=-4 x-2 z … (4) $$
$$ \begin{aligned} &\text { Ganti (4) ke dalam (2) }\\ &\begin{aligned} -2 x+3(-4 x-2 z)+z & =8 \\ -2 x-12 x-6 z+z & =8 \\ -14 x-5 z & =8 \\ x & =\frac{-5 z-8}{14} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Ganti (4) dan (5) ke dalam (3) }\\ &\begin{aligned} \left(\frac{-5 z-8}{14}\right)-\left[-4\left(\frac{-5 z-8}{14}\right)-2 z\right]-z & =-\frac{1}{2} \\ -5 z-8-20 z-32+28 z-14 z & =-7 \\ -11 z & =33 \\ z & =-3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & x=\frac{-5(-3)-8}{14}=\frac{1}{2} \\ & y=-4\left(\frac{1}{2}\right)-2(-3)=4 \\ & x=\frac{1}{2} ; y=4 ; z=-3 \end{aligned} $$
Selesaikan persamaan serentak berikut:
$$ 4 x+y+2 z=0 \quad, \quad-2 x+3 y+z=8 \quad, \quad x-y-z=-\frac{1}{2} $$
[5 markah]
Penyelesaian:
$$ \begin{aligned} 4 x+y+2 z & =0 … (1) \\ -2 x+3 y+z & =8 … (2) \\ x-y-z & =-\frac{1}{2} … (3)\end{aligned} $$
$$ \text { Dari (1) : } y=-4 x-2 z … (4) $$
$$ \begin{aligned} &\text { Ganti (4) ke dalam (2) }\\ &\begin{aligned} -2 x+3(-4 x-2 z)+z & =8 \\ -2 x-12 x-6 z+z & =8 \\ -14 x-5 z & =8 \\ x & =\frac{-5 z-8}{14} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Ganti (4) dan (5) ke dalam (3) }\\ &\begin{aligned} \left(\frac{-5 z-8}{14}\right)-\left[-4\left(\frac{-5 z-8}{14}\right)-2 z\right]-z & =-\frac{1}{2} \\ -5 z-8-20 z-32+28 z-14 z & =-7 \\ -11 z & =33 \\ z & =-3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & x=\frac{-5(-3)-8}{14}=\frac{1}{2} \\ & y=-4\left(\frac{1}{2}\right)-2(-3)=4 \\ & x=\frac{1}{2} ; y=4 ; z=-3 \end{aligned} $$
Soalan 2:
(a) $$ \text { Adakah } \sqrt{0.25} \text { suatu surd? Justifikasikan jawapan anda } $$
[1 markah]
(b)(i) Buktikan
$$ \log _a b=\frac{\log _c b}{\log _c a} $$
(ii)
$$ \text { Seterusnya, ungkapkan } \log _3 h+\log _9 k \text { sebagai satu logaritma tunggal dalam asas } 3 \text {. } $$
[6 markah]
Penyelesaian:
(a) $$ \sqrt{0.25} \text { bukan suatu surd kerana ia merupakan nombor nisbah. } $$
(b)(i) $$ \begin{aligned} \log _{\mathrm{a}} b & =x \\ b & =a^x \\ \log _{\mathrm{c}} b & =\log _{\mathrm{c}} a^x \\ \log _{\mathrm{c}} b & =x \log _{\mathrm{c}} a \\ x & =\frac{\log _{\mathrm{c}} b}{\log _{\mathrm{c}} a} \\ \log _{\mathrm{a}} b & =\frac{\log _{\mathrm{c}} b}{\log _{\mathrm{c}} a} \end{aligned} $$
(b)(ii) $$ \begin{aligned} \log _3 h+\log _9 k & =\log _3 h+\frac{\log _3 k}{\log _3 9} \\ & =\log _3 h+\frac{\log _3 k}{\log _3 3^2} \\ & =\log _3 h+\frac{\log _3 k}{2} \\ & =\log _3 h+\frac{1}{2} \log _3 k \\ & =\log _3 h+\log _3 k^{\frac{1}{2}} \\ & =\log _3 h k^{\frac{1}{2}} \end{aligned} $$
(a) $$ \text { Adakah } \sqrt{0.25} \text { suatu surd? Justifikasikan jawapan anda } $$
[1 markah]
(b)(i) Buktikan
$$ \log _a b=\frac{\log _c b}{\log _c a} $$
(ii)
$$ \text { Seterusnya, ungkapkan } \log _3 h+\log _9 k \text { sebagai satu logaritma tunggal dalam asas } 3 \text {. } $$
[6 markah]
Penyelesaian:
(a) $$ \sqrt{0.25} \text { bukan suatu surd kerana ia merupakan nombor nisbah. } $$
(b)(i) $$ \begin{aligned} \log _{\mathrm{a}} b & =x \\ b & =a^x \\ \log _{\mathrm{c}} b & =\log _{\mathrm{c}} a^x \\ \log _{\mathrm{c}} b & =x \log _{\mathrm{c}} a \\ x & =\frac{\log _{\mathrm{c}} b}{\log _{\mathrm{c}} a} \\ \log _{\mathrm{a}} b & =\frac{\log _{\mathrm{c}} b}{\log _{\mathrm{c}} a} \end{aligned} $$
(b)(ii) $$ \begin{aligned} \log _3 h+\log _9 k & =\log _3 h+\frac{\log _3 k}{\log _3 9} \\ & =\log _3 h+\frac{\log _3 k}{\log _3 3^2} \\ & =\log _3 h+\frac{\log _3 k}{2} \\ & =\log _3 h+\frac{1}{2} \log _3 k \\ & =\log _3 h+\log _3 k^{\frac{1}{2}} \\ & =\log _3 h k^{\frac{1}{2}} \end{aligned} $$