Contoh 2 Vektor

$\begin{array}{l}\\ 6.& \text{Diketahui titik}P(n,2),Q(1,-2),n>0\\ & \text{dan panjang}\overline{PQ}=5,\text{maka nilai}n\\ & \text{adalah}....\\ & \text{a}.1\\ & \text{b}.2\\ & \text{c}.3\\ & \text{d}.4\\ & \text{e}.5\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overline{PQ}& =5\\ \sqrt{(x_Q-x_P{)}^2+(y_Q-y_P{)}^2}& =5\\ (x_Q-x_P{)}^2+(y_Q-y_P{)}^2& =25\\ (1-n{)}^2+(-2-2{)}^2& =25\\ (1-n{)}^2+16& =25\\ (1-n{)}^2-3^2& =0\\ (1+3-n)(1-3-n)& =0\\ n=4\text{atau}n=-2& \end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui vektor}\overrightarrow{u}=\left(\begin{array}{c}3\\ 4\end{array}\right)\text{dan}\overrightarrow{v}=\left(\begin{array}{c}2\\ -1\end{array}\right).\\ & \text{Nilai}|\overrightarrow{u}+\overrightarrow{v}|\text{adalah}....\\ & \text{a}.\sqrt{28}\\ & \text{b}.\sqrt{30}\\ & \text{c}.\sqrt{34}\\ & \text{d}.\sqrt{44}\\ & \text{e}.\sqrt{50}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{u}+\overrightarrow{v}& =\left(\begin{array}{c}3\\ 4\end{array}\right)+\left(\begin{array}{c}2\\ -1\end{array}\right)\\ & =\left(\begin{array}{c}3+2\\ 4+(-1)\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 3\end{array}\right)\\ |\overrightarrow{u}+\overrightarrow{v}|& =\sqrt{5^2+3^2}\\ & =\sqrt{25+9}\\ & =\sqrt{34}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Vektor satuan}\overrightarrow{u}=\left(\begin{array}{c}-5\\ -12\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle -\frac{1}{13}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{b}.{\displaystyle \frac{1}{15}\left(\begin{array}{c}-5\\ -12\end{array}\right)}\\ & \text{c}.{\displaystyle -\frac{1}{17}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{d}.{\displaystyle -\frac{1}{17}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{e}.{\displaystyle \frac{1}{2}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}{\overrightarrow{e}}_{\overrightarrow{u}}& ={\displaystyle \frac{\overrightarrow{u}}{\left|\overrightarrow{u}\right|},\text{maka}}\\ {\overrightarrow{e}}_{\left(\begin{array}{c}-5\\ -12\end{array}\right)}& ={\displaystyle \frac{\left(\begin{array}{c}-5\\ -12\end{array}\right)}{\left|\left(\begin{array}{c}-5\\ -12\end{array}\right)\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-5\\ -12\end{array}\right)}{\sqrt{(-5{)}^2+(-12{)}^2}}}\\ & ={\displaystyle \frac{-\left(\begin{array}{c}5\\ 12\end{array}\right)}{\sqrt{169}}}\\ & =-{\displaystyle \frac{1}{13}\left(\begin{array}{c}5\\ 12\end{array}\right)}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika vektor}\overrightarrow{p}=\left(\begin{array}{c}8\\ 7\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}-3\\ 9\end{array}\right)\\ & \text{Hasil dari}\overrightarrow{p}+\overrightarrow{q}\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}6\\ 14\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}6\\ 13\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}6\\ 15\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}5\\ 16\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}5\\ 48\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}+\overrightarrow{q}& =\left(\begin{array}{c}8\\ 7\end{array}\right)+\left(\begin{array}{c}-3\\ 9\end{array}\right)\\ & =\left(\begin{array}{c}8-3\\ 7+9\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 16\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika vektor}\overrightarrow{p}=\left(\begin{array}{c}{}^2\log 32\\ -{}^3\log {\displaystyle \frac{1}{81}}\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}-9\\ -21\end{array}\right)\\ & \text{Hasil dari}\overrightarrow{p}+\overrightarrow{q}\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}6\\ 17\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}-6\\ -17\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}4\\ 17\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}-4\\ -17\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}5\\ -16\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}+\overrightarrow{q}& =\left(\begin{array}{c}{}^2\log 32\\ -{}^3\log {\displaystyle \frac{1}{81}}\end{array}\right)+\left(\begin{array}{c}-9\\ -21\end{array}\right)\\ & =\left(\begin{array}{c}5-9\\ 4-21\end{array}\right)\\ & =\left(\begin{array}{c}-4\\ -17\end{array}\right)\end{array}\end{array}$

Contoh 1 Vektor

$\begin{array}{l}\\ 1.& \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{maka vektor}\overrightarrow{u}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\overrightarrow{i}+5\overrightarrow{j}& & & \text{d}.& -3\overrightarrow{i}-5\overrightarrow{j}\\ \text{b}.& 5\overrightarrow{i}+3\overrightarrow{j}& \text{c}.& -3\overrightarrow{i}+5\overrightarrow{j}& \text{e}.& -5\overrightarrow{i}+3\overrightarrow{j}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Kita perhatikan lagi gambarnya}\\ & \begin{array}{rl}& \text{Vektor}\overrightarrow{u}\text{jika dinyatakan sebagai }\\ & \text{kombinasi linear adalah}\overrightarrow{u}=-3\overrightarrow{i}+5\overrightarrow{j}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Panjang vektor}\overrightarrow{p}=\left(\begin{array}{c}4\\ -8\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle \sqrt{4}}\\ & \text{b}.\sqrt{12}\\ & \text{b}.\sqrt{20}\\ & \text{d}.{\displaystyle \sqrt{80}}\\ & \text{e}.\sqrt{100}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}& =\left(\begin{array}{c}4\\ -8\end{array}\right),\text{maka besar dari vektor}\\ \overrightarrow{p}& \text{adalah}=\left|\overrightarrow{p}\right|=\sqrt{x^2+y^2}\\ & \text{Yaitu}\\ \left|\overrightarrow{p}\right|& =\sqrt{4^2+(-8{)}^2}\\ & =\sqrt{16+64}=\sqrt{80}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{l}\\ .& \text{Panjang vektor}\overrightarrow{h}\text{tersebut di atas adalah}....\\ & \text{a}.5\\ & \text{b}.7\\ & \text{c}.10\\ & \text{d}.12\\ & \text{e}.15\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Diketahui}& \overrightarrow{h}=\overline{OH}=8\overrightarrow{i}+6\overrightarrow{j}\\ \overrightarrow{h}& =\sqrt{x_H^2+y_H^2}\\ & =\sqrt{8^2+6^2}\mathbf{\text{(ingat tripel Pythagoras)}}\\ & =\sqrt{{10}^2}=10\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Vektor satuan dari}\overrightarrow{q}=3\overrightarrow{i}-4\overrightarrow{j}\text{adalah}....\\ & \text{a}.{\displaystyle \frac{4}{5}\overrightarrow{i}-\frac{3}{5}\overrightarrow{j}}\\ & \text{b}.{\displaystyle \frac{3}{5}\overrightarrow{i}-\frac{4}{5}\overrightarrow{j}}\\ & \text{c}.3\overrightarrow{i}-4\overrightarrow{j}\\ & \text{d}.4\overrightarrow{i}-3\overrightarrow{j}\\ & \text{e}.15\overrightarrow{i}-20\overrightarrow{j}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{q}& =3\overrightarrow{i}-4\overrightarrow{j}\\ & \text{Vektor satuan dari vektor}\overrightarrow{q}\text{adalah}:\\ & {\hat{e}}_{{}_{\overrightarrow{q}}}={\displaystyle \frac{1}{\left|\overrightarrow{q}\right|}.\overrightarrow{q}}\\ & \text{Sehingga}\\ {\hat{e}}_{{}_{\overrightarrow{q}}}& ={\displaystyle \frac{1}{\sqrt{3^2+(-4{)}^2}}.\left(\begin{array}{c}3\\ -4\end{array}\right)}\\ & ={\displaystyle \frac{1}{5}\left(\begin{array}{c}3\\ -4\end{array}\right)}\\ & \text{atau dalam vektor basis}\\ & ={\displaystyle \frac{3}{5}\overrightarrow{i}-\frac{4}{5}\overrightarrow{j}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Vektor berikut yang memiliki panjang}\\ & 29\text{satuan adalah}....\\ & \text{a}.{\displaystyle 18\overrightarrow{i}-19\overrightarrow{j}}\\ & \text{b}.{\displaystyle 19\overrightarrow{i}-20\overrightarrow{j}}\\ & \text{c}.20\overrightarrow{i}-21\overrightarrow{j}\\ & \text{d}.21\overrightarrow{i}-22\overrightarrow{j}\\ & \text{e}.22\overrightarrow{i}-23\overrightarrow{j}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Ingat}& \text{lah akan tigaan Pythagoras}\\ & \{\begin{array}{ll}(3,4,5)& \Rightarrow 3^2+4^2=5^2\\ (5,12,13)& \Rightarrow 5^2+{12}^2={13}^2\\ (8,15,17)& \Rightarrow \cdots \\ (20,21,29)& \Rightarrow \cdots \\ ⋮& dll\end{array}\\ \text{Sehin}& \text{gga}\\ \text{yang}& \text{paling mungkin adalah}:\\ & =\sqrt{{20}^2+{21}^2}=\sqrt{400+441}\\ & =\sqrt{841}\\ & =29\end{array}\end{array}$

Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}f(x)=x^2-2,\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2-2}& & \text{d}.{\displaystyle x}\\ \\ \text{b}.{\displaystyle x^2}& \text{c}.{\displaystyle 2x}& \text{e}.{\displaystyle 2x-2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}f(x)=x^2-2,\\ \text{maka nila}& \text{i untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{((x+h{)}^2-2)-(x^2-2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{x^2+2xh+h^2-2-x^2+2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{2xh+h^2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle (2x+h)}\\ & =2x\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui}f(x)=\sqrt{x-1},\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle 1}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.{\displaystyle 0}& \text{e}.{\displaystyle -1}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}f(x)=\sqrt{x-1},\\ \text{maka}& \text{nilai untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}\times {\displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(h+1)-1}{h\times (\sqrt{h+1}+1)}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{1}{(\sqrt{h+1}+1)}}\\ & ={\displaystyle \frac{1}{\sqrt{0+1}+1}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \mathbf{\text{(Mat Das SIMAK UI 2013)}}\\ & \text{Nilai}\underset{x\to 5}{\text{Lim}}{\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....}\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}+\sqrt{2}& & \text{d}.5\\ \text{b}.5-2\sqrt{6}& \text{c}.2\sqrt{6}& \text{e}.5+2\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 5}{\text{Lim}}& {\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times {\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}}\\ & ={\displaystyle \frac{3+2+2\sqrt{6}}{3-2}}\\ & =5+2\sqrt{6}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \mathbf{\text{(Mat IPA SBMPTN 2014)}}\\ \\ & \text{Jika}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}\\ & \text{dan}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3,}\\ \\ & \text{maka nilai}\underset{x\to a}{\text{Lim}}{\displaystyle f(x).g(x)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{14}}& & \text{d}.{\displaystyle \frac{4}{14}}\\ \\ \text{b}.{\displaystyle \frac{2}{14}}& \text{c}.{\displaystyle \frac{3}{14}}& \text{e}.{\displaystyle \frac{5}{14}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa},\\ & \begin{array}{l}\\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}& & \\ & & \\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3}& +& \\ & & \\ & & \\ 2\underset{x\to a}{\text{Lim}}{\displaystyle f(x)=1}& & \\ & & \\ \underset{x\to a}{\text{Lim}}f(x)={\displaystyle \frac{1}{2},}& & \end{array}\\ & \text{sehingga}\underset{x\to a}{\text{Lim}}f(g)={\displaystyle \frac{2}{7}}\\ & & \\ & \text{maka},\\ & \underset{x\to a}{\text{Lim}}f(x).g(x)={\displaystyle \frac{1}{2}\times \frac{2}{7}={\displaystyle \frac{2}{14}}}\end{array}\end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

$\begin{array}{l}\\ 1& \text{Nilai}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle \frac{1}{4}}\\ \\ \text{b}.-{\displaystyle \frac{1}{4}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\left({\displaystyle \frac{6-2}{2^2-4}-\frac{1}{2-2}}\right)\\ & =\left({\displaystyle \frac{4}{0}-\frac{1}{0}}\right)=\mathrm{\infty }-\mathrm{\infty }\\ & \text{hal ini tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{(x+2)}{(x-2)(x+2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{x+2}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{4-2x}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2(x-2)}{(x+2)(x-2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2}{x+2}}\right)\\ & ={\displaystyle -\frac{2}{(2+2)}}\\ & =-{\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}=....}\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle 2}& & \text{d}.{\displaystyle \frac{1}{2}}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.0& \text{e}.{\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\left({\displaystyle \frac{4-4}{2^4-4}}\right)\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}(2-\sqrt{x})}}\\ & =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{-\sqrt{x}(\sqrt{x}-2)}}\\ & =\underset{x\to 4}{\text{Lim}}-{\displaystyle \frac{(\sqrt{x}+2)}{\sqrt{x}}}\\ & =-{\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}}\\ & =-{\displaystyle \frac{2+2}{2}}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.{\displaystyle 2}\\ \\ \text{b}.{\displaystyle \frac{1}{2}}& \text{c}.1& \text{e}.{\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & ={\displaystyle \frac{(2-3+1)(1-1)}{{(1-1)}^2}}\\ & ={\displaystyle \frac{0\times 0}{0^2}}\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehi}& \text{ngga},\\ \underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & =\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{((2\sqrt{x}-1)\times (\sqrt{x}-1))(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}-1)}}\\ & =\underset{x\to 1}{\text{Lim}}(2\sqrt{x}-1)\\ & =2.1-1\\ & =2-1\\ & =1\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....}\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{1}{14}\sqrt{7}}& & \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}}\\ \\ \text{b}.-{\displaystyle \frac{1}{7}\sqrt{7}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{14}\sqrt{7}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 3}{\text{Lim}}& {\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times {\displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{(x+4)-(2x+1)}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =\underset{x\to 3}{\text{Lim}}-{\displaystyle \frac{-x+3}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{-1}{(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =-{\displaystyle \frac{1}{(\sqrt{7}+\sqrt{7})}}\\ & =-{\displaystyle \frac{1}{2\sqrt{7}}}\\ & =-{\displaystyle \frac{1}{2\sqrt{7}}\times {\displaystyle \frac{\sqrt{7}}{\sqrt{7}}}}\\ & =-{\displaystyle \frac{1}{14}\sqrt{7}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika}\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\text{maka nilai}a\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 2}& & \text{d}.{\displaystyle -2}\\ \\ \text{b}.{\displaystyle 1}& \text{c}.-1& \text{e}.{\displaystyle -3}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}& {\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6}\\ & \text{dengan bantuan limit kanan }\\ & \text{yaitu}x=2+h\Rightarrow h\to 0\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{a(2+h)-2a}{\sqrt{2(2+h)}-(2+h)}=6}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-(2+h)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah}{\sqrt{4+2h}-(2+h)}\times {\displaystyle \frac{(\sqrt{4+2h}+(2+h))}{(\sqrt{4+2h}+(2+h))}}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{4+2h-(4+4h+h^2)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{-2h-h^2}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a\times (\sqrt{4+2h}+(2+h))}{-2-h}}\\ 6& ={\displaystyle \frac{a\times (\sqrt{4+0}+(2+0))}{-2-0}}\\ 6& ={\displaystyle \frac{a(\sqrt{4}+2)}{-2}}\\ {\displaystyle \frac{a(4)}{-2}}& =6\\ a(-2)& =6\\ a& =-3\end{array}\end{array}$

MATEMATIKA WAJIB MA/SMA untuk KONDISI KHUSUS TAHUN 2020-2021

 $\text{A. Kelas X (Sepuluh)}$

$\begin{array}{rl}& \text{A.1 Persamaan dan pertidaksamaan}\\ & \text{nilai mutlak dari bentuk linear}\\ & \text{satu variabel}\end{array}$

• materi persamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 
• materi pertidaksamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , 

$\begin{array}{rl}& \text{A.2 Pertidaksamaan rasional}\\ & \text{dan irasional satu variabel}\end{array}$

• materi pertidaksamaan rasional dan irasional satu variabel
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , 

$\begin{array}{rl}& \text{A.3 Sistem persamaan linear tiga}\\ & \text{variabel}\end{array}$

• materi sistem persamaan linear tiga variabel (SPLTV)
• lanjutan materi SPLTV
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.4 Sistem pertidaksamaan dua}\\ & \text{variabel-linear-linear}\end{array}$

• materi sistem pertidaksamaan dua variabel linear-linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.5 Fungsi}\end{array}$

• Materi fungsi
• Lanjutan materi fungsi
• contoh soal 1 , contoh 2 , contoh 3

$\begin{array}{rl}& \text{A.6 Fungsi Komposisi dan invers}\end{array}$

• Materi dan contoh soal
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.7 Rasio trigonometri}\\ & \text{pada segitiga siku-siku}\end{array}$

$\begin{array}{rl}& \text{A.8 Rasio trigonometri}\\ & \text{sudut-sudut diberbagai}\\ & \text{kuadran}\end{array}$

$\begin{array}{rl}& \text{A.9 Aturan sinus dan cosinus}\end{array}$

$\text{B. Kelas XI (Sebelas)}$

$\begin{array}{rl}& \text{B.1 Program linear}\end{array}$

• materi program linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{B.2 Matriks}\end{array}$

• materi matriks 
• lanjutan materi 1
• lanjutan materi 2
• contoh soal 1 , contoh 2 , contoh 3 

$\begin{array}{rl}& \text{B.3 Determinan dan invers matriks}\\ & \text{ordo 2x2}\end{array}$

• materi determinan
• materi invers matriks ordo 2x2
• contoh soal 4 , contoh 5 , contoh 6 

$\begin{array}{rl}& \text{B.4 Pola bilangan dan jumlah}\\ & \text{pada barisan aritmetika}\\ & \text{dan geometri}\end{array}$

• materi pola bilangan - induksi matematika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5

$\begin{array}{rl}& \text{B.5 Limit fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.6 Turunan fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.7 Keberkaitan Turunan fungsi}\end{array}$

$\begin{array}{rl}& \text{B.8 Integral tak tentu}\\ & \text{fungsi aljabar}\end{array}$

$\text{C. Kelas XII (Duabelas)}$

$\begin{array}{rl}& \text{C.1 Jarak dalam ruang}\end{array}$

• materi mengitung jarak di dimensi tiga
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{C.2 Statistika}\end{array}$

• materi statistika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 

$\begin{array}{rl}& \text{C.3 Aturan pencacahan}\end{array}$

• materi 
• lanjutan materi
• lanjutan materi (faktorial, permutasi, kombinasi)
• contoh soal 1, contoh 2, contoh 3
• lanjutan materi (binomial newton)
• pengayaan : fungsi pembangkit untuk kombinasi

$\begin{array}{rl}& \text{C.4 Peluang kejadian majumuk}\end{array}$

• materi
• lanjutan materi (peluang kejadian tunggal)
• lanjutan materi (peluang komplemen, peluang dua kejadian saling bebas)
• lanjutan materi (peluang kejadian saling lepas)
• contoh soal 1, contoh 2, contoh 3

MATEMATIKA PEMINATAN MA/SMA untuk KONDISI KHUSUS TAHUN 2020-2021

 $\text{A. Kelas X (Sepuluh)}$

$\begin{array}{rl}& \text{A.1 Fungsi eksponensial}\\ & \text{dan fungsi logaritma}\end{array}$

• materi eksponen 1
• contoh soal eksponen 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10 
• materi logaritma 
• materi logaritma lanjutan
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• materi logaritma lanjutan 2
• contoh soal 7 , contoh 8 , contoh 9 , contoh 10 

$\begin{array}{rl}& \text{A.2 Vektor}\end{array}$

• materi
• operasi vektor pada bidang
• lanjutan materi pada bidang
• contoh soal 
• materi perkalian skalar dua vektor pada bidang
• contoh soal 2, contoh 3
• Proyeksi ortoganl suatu vektor
• Vektor dimensi tiga
• Contoh 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7

$\text{B. Kelas XI (Sebelas)}$

$\begin{array}{rl}& \text{B.1 Persamaan trigonometri}\end{array}$

• materi persamaan trigonometri
• lanjutan materi 1 , materi 2 , materi 3
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6

$\begin{array}{rl}& \text{B.2 Polinom}\end{array}$

• materi
• lanjutan materi (operasi polinom)
• lanjutan materi (teorema sisa dan teorema faktor)
• lanjutan materi (akar-akar persamaan polinom)
• contoh soal 1, contoh 2, contoh 3, contoh 4

$\text{C. Kelas XII (Duabelas)}$

$\begin{array}{rl}& \text{C.1 Turunan fungsi trigonometri}\end{array}$

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12  

$\begin{array}{rl}& \text{C.2 Distribusi peluang binomial}\end{array}$

• materi distribusi binomial
• lanjutan materi 1 (distribusi peluang diskrit)
• Contoh soal 1 , contoh 2 (variabel acak)
• lanjutan materi 2 (distribusi peluang kontinu)
• lanjutan materi (distribusi binomial-distribusi poisson)
• contoh soal 1 (distibusi binomial)
• contoh soal 2, contoh 3

Contoh Soal 3 Distribusi Binomial

$\begin{array}{l}\\ 11.& \text{Suatu tes dengan pilihan jawaban }\\ & \text{benar-salah berjumlah 8 soal}\\ & \text{Supaya lulus tes, peserta diharuskan }\\ & \text{menjawab benar minimal 50}\mathrm{\%}\\ & \text{Peluang seseorang dianggap lulus tes }\\ & \text{adalah}....\\ & \text{a}.{\displaystyle 0,2188\text{d}.0,6367}\\ & \text{b}.{\displaystyle 0,2734\text{c}.0,3633\text{e}.0,7266}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang benar}}={\displaystyle \frac{1}{2},\text{dan}}\\ & q=\mathbf{\text{Peluang Salah}}=1-{\displaystyle \frac{1}{2}=\frac{1}{2}}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & \text{maka}\\ & P(X=50\mathrm{\%}(8)=4)=\left(\begin{array}{c}8\\ 4\end{array}\right)\times {\left({\displaystyle \frac{1}{2}}\right)}^4\times {\left(\frac{1}{2}\right)}^{8-4}\\ & ={\displaystyle \frac{8!}{4!\times 4!}{\left({\displaystyle \frac{1}{2}}\right)}^{4+4}}\\ & =70\times {\displaystyle \frac{1}{256}}\\ & =0,2734\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Sebuah kotak berisi 20 bola dengan }\\ & \text{rincian 12 boal berwarna kuning dan }\\ & \text{sisanya berwarna hijau. Dari kotak}\\ & \text{diambil 6 bola secara acak. Peluang}\\ & \text{terambil 4 bola hijau adalah}....\\ & \text{a}.{\displaystyle 0,1238\text{d}.0,8132}\\ & \text{b}.{\displaystyle 0,1382\text{c}.0,3110\text{e}.0,9590}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang bola kuning}}\\ & ={\displaystyle \frac{C_1^{12}}{C_1^{20}}={\displaystyle \frac{12}{20}=\frac{3}{5},}}\\ & q=\mathbf{\text{Peluang bola hijau}}=1-{\displaystyle \frac{3}{5}=\frac{2}{5}}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & \text{maka}\\ & f(4)=\left(\begin{array}{c}6\\ 4\end{array}\right)\times {\left({\displaystyle \frac{2}{5}}\right)}^4\times {\left(\frac{3}{5}\right)}^{6-4}\\ & ={\displaystyle \frac{6!}{2!\times 4!}\left({\displaystyle \frac{16}{625}}\right)\times \left({\displaystyle \frac{9}{25}}\right)}\\ & =15\times {\displaystyle \frac{144}{15625}=\frac{2160}{15625}}\\ & =0,1382\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Dua dadu dilambungkan 5 kali}\\ & \text{Peluang muncul pasangan mata dadu}\\ & \text{berjumlah 4 sampai dengan 7 }\\ & \text{sebanyak 4 kali adalah}....\\ & \text{a}.{\displaystyle 0,1503\text{d}.0,1583}\\ & \text{b}.{\displaystyle 0,1553\text{c}.0,1563\text{e}.0,1593}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang mata dadu berjumlah 4 sampai 7}}\\ & ={\displaystyle \frac{18}{36}=\frac{1}{2},\text{dan}}\\ & q=\mathbf{\text{Peluang bola hijau}}=1-{\displaystyle \frac{1}{2}=\frac{1}{2}}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & f(4)=P(X=4)=\left(\begin{array}{c}5\\ 4\end{array}\right)\times {\left({\displaystyle \frac{1}{2}}\right)}^4\times {\left(\frac{1}{2}\right)}^{5-4}\\ & ={\displaystyle \frac{5!}{1!\times 4!}\left({\displaystyle \frac{1}{16}}\right)\times \left(\frac{1}{2}\right)}\\ & =5\times {\displaystyle \frac{1}{32}=\frac{5}{32}}\\ & =0,1563\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Peluang seseorang sembih dari }\\ & \text{penyakit jantung adalah 0,6}\\ & \text{Jika 7 orang penderita ini menjalani }\\ & \text{operasi, maka peluang 3 sampai}\\ & \text{6 orang sembuh adalah}....\\ & \text{a}.{\displaystyle 0,0629\text{d}.0,6822}\\ & \text{b}.{\displaystyle 0,2613\text{c}.0,2898\text{e}.0,9720}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang sembuh}}=0,6,\text{maka}\\ & q=\mathbf{\text{Peluang tidak sembuh}}=1-0,6=0,4\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & \text{maka}\\ & P(3\le X\le 6)=P(X\le 6)-P(X\le 3)\\ & =C_4^7(0,6{)}^4(0,4{)}^3+C_5^7(0,6{)}^5(0,4{)}^2+C_6^7(0,6{)}^6(0,4{)}^1\\ & =35\times 0,0082944+21\times 0,0124416+7\times 0,0186624\\ & =0,290304+0,2612736+0,1306368\\ & =0,6822144\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Peluang seseorang mendapatkan reaksi }\\ & \text{buruk setelah disuntik adalah 0,0005}\\ & \text{Dari 4000 orang yang disuntik, maka }\\ & \text{peluang seseorang mendapatkan reaksi}\\ & \text{ada 2 orang adalah}.....\\ & \text{a}.{\displaystyle \frac{1}{2}{e}^{-2}}\\ & \text{b}.e^{-2}\\ & \text{c}.2e^{-2}\\ & \text{d}.{\displaystyle \frac{1}{2}{e}^2}\\ & \text{e}.2e^2\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Di atas adalah contoh kasus }\\ & \text{permasalahan}\mathbf{\text{Distribusi Poisson}}\\ & P(X=x)=f(x)=\{\begin{array}{c}{\displaystyle \frac{e^{-\lambda }.{\lambda }^x}{x!},x=0,1,2,3,\cdots }\\ 0,\text{untuk}x\text{yang lain}\end{array}\\ & P(X=2)={\displaystyle \frac{e^{-np}.(np{)}^2}{2!}}\\ & ={\displaystyle \frac{e^{-(4000.0,0005)}.(4000.0,0005{)}^2}{2!}}\\ & ={\displaystyle \frac{e^{-2}{.2}^2}{2}}\\ & =2e^{-2}\end{array}\end{array}$

Contoh Soal 2 Distribusi Binomial

 $\begin{array}{l}\\ 6.& \text{Pengundian terhadap mata uang }\\ & \text{yang homogen sebanyak 10 kali}\\ & \text{Peluang untuk mendapatkan 6 }\\ & \text{muka angka adalah}....\\ & \text{a}.0,1172\\ & \text{b}.0,2051\\ & \text{c}.0,2461\\ & \text{d}.0,2651\\ & \text{e}.0,2852\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang Angka}}={\displaystyle \frac{1}{2},\text{dan}}\\ & q=\mathbf{\text{Bukan Angka}}\\ & =\mathbf{\text{Peluang Gambar}}=1-{\displaystyle \frac{1}{2}=\frac{1}{2}}\\ & f(x)=P(x;n;p)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ & \text{maka}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & f(6)=P(X=6)=\left(\begin{array}{c}10\\ 6\end{array}\right)\times {\left({\displaystyle \frac{1}{2}}\right)}^6\times {\left(\frac{1}{2}\right)}^{10-6}\\ & ={\displaystyle \frac{10!}{6!\times 4!}{\left({\displaystyle \frac{1}{2}}\right)}^{6+4}}\\ & =210\times {\displaystyle \frac{1}{1024}}\\ & =0,2051\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Pada pengundian terhadap mata uang identik},\\ & \text{sebanyak 10 kali, peluang distribusi binomial}\\ & \text{untuk mendapatkan 7 muka gambar adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 0,2653}& & \text{d}.{\displaystyle 0,7522}\\ \text{b}.{\displaystyle 0,1172}& \text{c}.{\displaystyle 0,2653}& \text{e}.0,2422\end{array}\\ \\ & \text{Jawab}:\\ & \text{Uraian berikut sekaligus tambahan}\\ & \text{penjelasan pada uraian jawaban}\\ & \text{soal no. 6 di atas}\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ & \text{Ingat sebuah koin ada 2 muka}\\ & \text{yaitu muka gambar (G) dan angka (A)}\\ & \text{misalkan}\\ & A=\text{kejadian muncul muka gambar}\\ & \text{maka peluangnya adalah}{\displaystyle \frac{1}{2}}\\ & \text{Selanjutnya di sini disimbolkan dengan}p={\displaystyle \frac{1}{2}}\\ & \text{Demikian juga misalkan}\\ & B=\text{kejadian muncul muka angka}\\ & \text{maka peluang juga}{\displaystyle \frac{1}{2}}\\ & \text{Di sini dituliskan dengan}q={\displaystyle \frac{1}{2}}\\ f(7)& =\left(\begin{array}{c}10\\ 7\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^7{\left({\displaystyle \frac{1}{2}}\right)}^{10-7}\\ & =\left(\begin{array}{c}10\\ 7\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^7{\left({\displaystyle \frac{1}{2}}\right)}^3\\ & ={\displaystyle \frac{10!}{7!\times (10-7)!}{\left({\displaystyle \frac{1}{2}}\right)}^{7+3}}\\ & ={\displaystyle \frac{10.9.8.\text{⧸}7!}{\text{⧸}7!.3.2.1}\left({\displaystyle \frac{1}{1024}}\right)}\\ & =0,1172\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Sebuah uang logam dilempar sebanyak 8}\\ & \text{kali. Peluang muncul gambar sebanyak}\\ & \text{5 kali adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{32}}& & & \text{d}.& {\displaystyle \frac{7}{32}}\\ \\ \text{b}.& {\displaystyle \frac{4}{32}}& \text{c}.& {\displaystyle \frac{5}{32}}& \text{e}.& {\displaystyle \frac{9}{32}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(5)& =\left(\begin{array}{c}8\\ 5\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^5{\left({\displaystyle \frac{1}{2}}\right)}^{8-5}\\ & =\left(\begin{array}{c}8\\ 5\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^5{\left({\displaystyle \frac{1}{2}}\right)}^3\\ & ={\displaystyle \frac{8!}{5!\times (8-5)!}{\left({\displaystyle \frac{1}{2}}\right)}^{5+3}}\\ & ={\displaystyle \frac{8.7.6.5!}{5!.3.2.1}\left({\displaystyle \frac{1}{256}}\right)}\\ & ={\displaystyle \frac{8.7}{256}}\\ & ={\displaystyle \frac{7}{32}}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Pada pelemparan sebuah koin sebanyak 4 kali}\\ & \text{Peluang didapatkannya dua angka pada}\\ & \text{pelemparan tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 0,123}& & \text{d}.{\displaystyle 0,232}\\ \text{b}.{\displaystyle 0,135}& \text{c}.{\displaystyle 0,154}& \text{e}.0,375\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(2)& =\left(\begin{array}{c}4\\ 2\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^2{\left({\displaystyle \frac{1}{2}}\right)}^{4-2}\\ & =\left(\begin{array}{c}4\\ 2\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^2{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{4!}{2!\times (4-2)!}{\left({\displaystyle \frac{1}{2}}\right)}^{2+2}}\\ & ={\displaystyle \frac{4.3.2!}{2!.2.1}\left({\displaystyle \frac{1}{16}}\right)}\\ & =0,375\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Dari data survei didapatkan bahwa}\\ & \text{satu dari lima orang telah berkunjung}\\ & \text{ke dokter dalam sembarang bulan yang}\\ & \text{ditanyakan. Jika 10 orang dipilih secara}\\ & \text{acak, peluang 3 orang telah berkunjung}\\ & \text{ke dokter bulan lalu adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,125}& & & \text{d}.& {\displaystyle 0,201}\\ \\ \text{b}.& {\displaystyle 0,174}& \text{c}.& {\displaystyle 0,182}& \text{e}.& {\displaystyle 0,423}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(3)& =\left(\begin{array}{c}10\\ 3\end{array}\right){\left({\displaystyle \frac{1}{5}}\right)}^3{\left({\displaystyle \frac{4}{5}}\right)}^{10-3}\\ & =\left(\begin{array}{c}10\\ 3\end{array}\right){\left({\displaystyle \frac{1}{5}}\right)}^3{\left({\displaystyle \frac{4}{5}}\right)}^7\\ & ={\displaystyle \frac{10!}{3!\times 7!}\left({\displaystyle \frac{1}{125}}\right)\left({\displaystyle \frac{4^7}{5^7}}\right)}\\ & =\cdots \\ & ={\displaystyle 0,201}\end{array}\end{array}$

Contoh Soal 3 Polinom

$\begin{array}{l}\\ 11.& \text{Jika polinom}f(x)\text{dibagi oleh}\\ & (x-a)(x-b)\text{dan}a\ne b,\text{maka}\\ & \text{sisa pembagiannya adalah}....\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{b}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{c}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{d}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{e}.{\displaystyle {\displaystyle \frac{x-a}{b-a}f(b)+\frac{x-a}{b-a}f(a)}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misal sisa pembagiannya}:s(x)=px+q\\ & \text{Saat}f(x)\text{dibagi}(x-a)(x-b)\text{berarti}\\ & \bullet x=a\Rightarrow s(a)=f(a)=ap+q....(1)\\ & \bullet x=b\Rightarrow s(b)=f(b)=bp+q......(2)\\ & \text{Persamaan}(1)\text{dan}(2)\text{dieliminasi}\\ & \begin{array}{l}\\ ap& +& q& =& f(a)\\ bp& +& q& =& f(b)& -\\ ap& -& bp& =& f(a)-f(b)\\ & & p& =& {\displaystyle \frac{f(a)-f(b)}{a-b}}& \end{array}\\ & \text{Dari persamaan}(1),\\ & f(a)=ap+q\\ & f(a)=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+q\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\left({\displaystyle \frac{a-b}{a-b}}\right)\\ & q={\displaystyle \frac{-bf(a)-af(b)}{a-b}}\\ & \text{Sehingga}\\ & s(x)=px+q\\ & =\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)x+\left({\displaystyle \frac{-bf(a)-af(b)}{a-b}}\right)\\ & ={\displaystyle \frac{f(a)x-f(b)x-bf(a)+af(b)}{a-b}}\\ & ={\displaystyle \frac{(x-b)f(a)+(a-x)f(b)}{a-b}}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Diketahui}f(x)\text{dibagi oleh}x-2\text{bersisa 5},\\ & \text{dan dibagi}x-3\text{bersisa 7. Jia}f(x)\\ & \text{dibagi oleh}{x}^2-5x+6\text{akan memiliki sisa}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x-2}& & \text{d}.{\displaystyle 2x+1}\\ \text{b}.{\displaystyle 2x-4}& \text{c}.{\displaystyle x+2}& \text{e}.2x+3\end{array}\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}f(x)& =(x-2).h(x)+5\\ f(x)& =(x-3).h(x)+7\\ f(x)& =(x^2-5x+6).H(x)+s(x)\\ f(x)& =(x-2)(x-3).H(x)+px+q\\ f(2)& =(2-2)(2-3).H(x)+2p+q=5\\ & \Rightarrow 0+2p+q=5.................(1)\\ f(3)& =(3-2)(3-3).H(x)+3p+q=7\\ & \Rightarrow 0+3p+q=7.................(2)\\ \text{Dari}& \text{persamaan}(1)\text{dan}(2)\\ \text{saat}& \text{persamaan (1) dikurangi persamaan (2)}\\ & -p=-2\\ & p=2\\ & \text{maka},q=1\\ & \text{Sehingga},\\ & s(x)=px+q=2x+1\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& f(x)\text{dibagi}(x-2)\text{sisa}5\Rightarrow f(2)=5\\ & f(x)\text{dibagi}(x-3)\text{sisa}7\Rightarrow f(3)=7\\ & \text{maka},\\ & s(x)={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\\ & ={\displaystyle \frac{x-3}{2-3}(5)+\frac{x-2}{3-2}(7)}\\ & ={\displaystyle \frac{5x-15}{-1}+\frac{7x-14}{1}}\\ & =15-5x+7x-14\\ & =2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Polinom}f(x)\text{dibagi oleh}(2x-4)\text{bersisa 6},\\ & \text{dibagi oleh}(x+4)\text{bersisa 24}.\\ & \text{Dan polinom}g(x)\text{dibagi oleh}(2x-4)\text{bersisa 5},\\ & \text{dibagi oleh}(x+4)\text{bersisa 2}.\\ & \text{Jika}h(x)=f(x).g(x),\text{maka}h(x)\\ & \text{dibagi}(2x^2+4x-16)\text{akan sisa}....\\ & \begin{array}{l}\\ \text{a}.-3x+24& & \text{d}.-6x+36\\ \text{b}.-3x+36& \text{c}.6x+24& \text{e}.12x+3\end{array}\\ \\ & \text{Jawab}:\\ & \text{Langkah pertama}\\ & \begin{array}{rl}f(x)& =(2x-4).h(x{)}_1+6\\ f(x)& =(x+4).h(x{)}_2+24\\ f(x)& =(2x-4)(x+4).H_1(x)+p_{1}x+q_1\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{1}x+q_1=-3x+12\end{array}\\ & \text{Langkah kedua}\\ & \begin{array}{rl}g(x)& =(2x-4).h(x{)}_3+5\\ g(x)& =(x+4).h(x{)}_4+2\\ g(x)& =(2x-4)(x+4).H_2(x)+p_{2}x+q_2\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{2}x+q_2={\displaystyle \frac{1}{2}x+4}\end{array}\\ & \text{Langkah ketiga}\\ & \begin{array}{rl}& h(x)=f(x)\times g(x)\\ & =((2x-4)(x+4)H_1(x)+(-3x+12))\\ & \times ((2x-4)(x+4)H_2(x)+{\displaystyle \frac{1}{2}x+4})\\ & \text{maka}\\ & \bullet h(2)=(0+(-3.2+12))(0+{\displaystyle \frac{1}{2}.2+4})=6.5=30\\ & \bullet h(-4)=(0+(-3.-4+12))(0+{\displaystyle \frac{1}{2}.-4+4})=24.2=48\\ & \text{Dengan pembagi}2x^2+x-16,\text{maka sisanya}:s_3(x)=p_{3}x+q_3\\ & \text{saat}x=2\Rightarrow 2p+q=30\\ & \text{saat}x=-4\Rightarrow -4p+q=48\\ & \text{selanjutnya dengan eliminasi-substitusi diperoleh}p=-3,q=36\\ & \text{sehingga}s(x)=px+q=-3x+36\end{array}\end{array}$

$\begin{array}{l}\\ 14.& (\text{KSM 2015})\text{Diketahui}f(x)\text{adalah polinom}\\ & (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ & \text{dengan}{x}_1,x_2,x_3,x_4,\text{dan}{x}_5\text{adalah}\\ & \text{bilangan bulat berbeda}.\text{Jika}f(104)=2012,\\ & \text{maka nilai}{x}_1+x_2+x_3+x_4+x_5\text{sama dengan}....\\ & \text{a}.13\\ & \text{b}.14\\ & \text{c}.16\\ & \text{d}.17\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}:\\ f(x)& =(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ f(104)& =(104-x_1)(104-x_2)(104-x_3)(104-x_4)(104-x_5)=2012\\ & =2012=1\times 2\times 503\\ & =(-1)\times (1)\times (-2)\times (2)\times (503)\\ \text{maka}& \{\begin{array}{ll}(104-x_1)& =-2\Rightarrow x_1=106\\ (104-x_2)& =-1\Rightarrow x_2=105\\ (104-x_3)& =1\Rightarrow x_3=103\\ (104-x_4)& =2\Rightarrow x_4=102\\ (104-x_5)& =503\Rightarrow x_5=-399\end{array}\\ \text{sehin}& \text{gga},\\ & x_1+x_2+x_3+x_4+x_5=106+105+103+102+(-399)=17\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Tentukanlah suku banyak}f(x)\text{sedemikian}\\ & \text{sehingga}f(x)\text{terbagi oleh}{x}^2+1,\\ & \text{sedangkan}f(x)+1\text{terbagi oleh}{x}^3+x^2+1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =(x^2+1).h_{1}x\\ f(x)+1& =(x^2+1).h_{1}x+1\\ \text{supaya}& f(x)+1\text{terbagi habis oleh}{x}^3+x^2+1,\\ & \text{maka akan ada bilangan bulat}k,(k\ne 0)\\ k& ={\displaystyle \frac{f(x)+1}{x^3+x^2+1}}\\ & ={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}}\\ k=1\Rightarrow & 1={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}\text{maka}{h}_{1}x=x}\\ \text{sehingga}& f(x)=x^3+x^2\\ \text{untuk ni}& \text{lai}k\text{yang lain, tak ditemukan}\end{array}\end{array}$

Contoh Soal 4 Polinom

$\begin{array}{l}\\ 16.& \text{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ & \text{adalah}{x}_1,x_2,x_3,...,x_{2017}\\ & \text{Tentukan nilai dari}\\ & {\displaystyle \frac{1}{1-x_1}+\frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{x^{2018}-1}{x-1}}& =x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ \text{perlu dii}& \text{ngat bahwa kondisi ini mensyaratkan}x\ne 1,\\ & \text{sehingga}\\ \\ x^{2018}-1& =0\\ x^{2018}& =1\\ x& =\pm 1,\text{pilih}x=-1\\ \text{maka}& \text{nilai dari}\\ {\displaystyle \frac{1}{1-x_1}+}& \frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+{\displaystyle \frac{1}{2}}}}}\\ & ={\displaystyle \frac{2017}{2}}\end{array}\end{array}$

Contoh Soal 2 Polinom

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}\\ & {\displaystyle \frac{f(x)}{x-2}=h(x)+{\displaystyle \frac{3}{x-2}}}\\ & \text{dan}{\displaystyle \frac{f(x)}{x-1}=h(x)+{\displaystyle \frac{2}{x-1},}}\\ & \text{jika}{\displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+{\displaystyle \frac{s(x)}{(x-2)(x-1)},}}\\ & \text{maka}s(x)=....\\ & \begin{array}{l}\\ \text{a}.x+1& & \text{d}.2x-1\\ \text{b}.x+2& \text{c}.2x+1& \text{e}.x-2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{f(x)}{x-2}=h(x)+{\displaystyle \frac{3}{x-2}}}\\ & \Rightarrow f(x)=(x-2).h(x)+3\Rightarrow f(2)=3\\ & {\displaystyle \frac{f(x)}{x-1}=h(x)+{\displaystyle \frac{2}{x-1}}}\\ & \Rightarrow f(x)=(x-1).h(x)+2\Rightarrow f(1)=2\\ & {\displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+{\displaystyle \frac{s(x)}{(x-2)(x-1)}}}\\ & \text{maka}f(x)=(x-2)(x-1).h(x)+s(x)\\ & f(x)=(x-2)(x-1).h(x)+px+q\\ & f(2)=2p+q=3\\ & f(1)=p+q=2,\\ & \text{sehingga dengan }\text{eliminasi akan diperoleh}\\ p& =1\text{dan}\\ & q=1\\ & \text{Jadi},px+q=x+1\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jika}{x}^4+2mx-n\text{dibagi}{x}^2-1\\ & \text{bersisa}2x-1,\text{maka nilai}m\\ & \text{dan}n\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.m=-1\text{dan}n=2\\ \text{b}.m=1\text{dan}n=-2\\ \text{c}.m=1\text{dan}n=2\\ \text{d}.m=-1\text{dan}n=-2\\ \text{e}.m=-2\text{dan}n=1\end{array}\\ \\ & \text{Jawab}:\\ & \text{dengan Horner-Kino didapatkan}\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=x^4+2mx-n\\ \text{Pembagai}:& p(x)=(x-1)(x+1)=x^2-1\\ & :1\text{dari}-\frac{-1}{1},\text{sedang}0=-\left(\frac{0}{1}\right)\\ \text{Hasil bagi}:& h(x)=x^2+1\\ \text{Sisa bagi}:& s(x)=2mx+(1-n)=2x-1\end{array}$

$.\begin{array}{rl}& \text{Sehingga},\\ & \bullet 2m=2\Rightarrow m=1\\ & \bullet 1-n=-1\Rightarrow n=2\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}f(x)=x^4-k{x}^2+5\text{habis dibagi}\\ & (x-1)\text{maka}f(x)\text{juga habis dibagi oleh}....\\ & \begin{array}{l}\\ \text{a}.x+1& & \text{d}.x+5\\ \text{b}.2x+1& \text{c}.3x+1& \text{e}.2x+5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =x^4-k{x}^2+5\\ f(1)& =(1{)}^4-k(1{)}^2+5\\ 0& =1-k+5\\ k& =6\\ f(x)& =x^4-6x^2+5\\ & =(x^2-1)(x^2-5)\\ & =(x-1)(x+1)(x^2-5)\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}(m-2)\text{adalah faktor dari}2m^3+3tm+4,\\ & \text{maka nilai}t\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{10}{3}}& & \text{d}.-{\displaystyle \frac{3}{10}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}& \text{c}.{\displaystyle \frac{3}{10}}& \text{e}.-{\displaystyle \frac{10}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(m)& =2m^3+3tm+4\\ f(2)& =2(2{)}^3+3t(2)+4\\ 0& =16+6t+4\\ -6t& =20\\ t& =-{\displaystyle \frac{10}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{(KSM MA Kab/Kota 2015)Nilai terkecil}n\\ & \text{yang mengkin sehingga}n.(n+1).(n+2)\\ & \text{habis dibagi 24 adalah}....\\ & \begin{array}{c}\\ \text{a}.1\\ \text{b}.2\\ \text{c}.3\\ \text{d}.4\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}k& ={\displaystyle \frac{n.(n+1).(n+2)}{24}}\\ & ={\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}}\\ & \text{maka}n=2\end{array}\end{array}$

Contoh Soal 1 Polinom

$\begin{array}{l}\\ 1.& \text{Jika}g(x)=2x^3+x^2-x+1,\\ & \text{maka}g(1)=....\\ & \begin{array}{l}\\ \text{a}.-2& & \text{d}.2\\ \text{b}.-1& \text{c}.1& \text{e}.3\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}g(x)& =2x^3+x^2-x+1\\ g(1)& =2(1{)}^3+(1{)}^2-(1)+1\\ & =2+1-1+1\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}p(y)=5y^4+2r^2{y}^3+y^2+1\text{dan}\\ & q(y)=4y^5+3r{y}^2-3y-1\\ & \text{serta}p(-1)=q(-1),\text{maka nilai}r\\ & \text{sama dengan}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{2}\text{dan}3}& & \text{d}.-{\displaystyle \frac{3}{2}}\\ \text{b}.{\displaystyle -\frac{3}{2}\text{dan}3}& \text{c}.{\displaystyle \frac{3}{2}\text{dan}-3}& \text{e}.3\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}p(-1)& =q(-1)\\ 5(-1{)}^4+2r^2(-1{)}^3+(-1{)}^2+1& =4(-1{)}^5+3r(-1{)}^2-3(-1)-1\\ 5-2r^2+1+1& =-4+3r+3-1\\ 9-3r-2r^2& =0\\ {\displaystyle \frac{(-6-2r)(-3+2r)}{2}}& =0,\text{ingat pemfaktoran}\\ (-3-r)(-3+2r)& =0\\ r=-3\vee r& ={\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui}f(x)\text{berderajat}n.\\ & \text{Jika pembaginya berbentuk}(a{x}^2+bx+c),\\ & \text{dengan}a\ne 0,\text{maka hasil baginya berderajat}....\\ & \begin{array}{l}\\ \text{a}.n-1& & \text{d}.3\\ \text{b}.n-2& \text{c}.n-3& \text{e}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Suku banyak (polinom)}\\ & =\text{pembagi}\times \text{hasil bagi}+\text{sisa}\\ & x^n+...=(a{x}^2+bx+c)\times (x^{n-2}+...)+(mx+n)\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Hasil bagi dan sisanya jika}(6x^4-3x^2+x-1)\\ & \text{dibagi oleh}(2x-1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+\frac{1}{8}}}& \text{dan}& -{\displaystyle \frac{7}{8}}\\ \text{b}.3x^3+3x^2-{\displaystyle \frac{3}{4}x+1}& \text{dan}& -7\\ \text{c}.x^3+{\displaystyle \frac{3}{2}{x}^2-3x+\frac{1}{8}}& \text{dan}& {\displaystyle \frac{7}{8}}\\ \text{d}.x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+1}}& \text{dan}& {\displaystyle \frac{1}{8}}\\ \text{e}.3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x-\frac{1}{8}}}& \text{dan}& -{\displaystyle \frac{7}{8}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \begin{array}{rrrrrrr}x=\frac{1}{2}& & 6& 0& -3& 1& -1\\ & & & 3& \frac{3}{2}& -\frac{3}{4}& \frac{1}{8}& +& \\ & & 6& 3& -\frac{3}{2}& \frac{1}{4}& \boxed{{\displaystyle -\frac{7}{8}}}\end{array}\\ & \text{Selanjutnya}\\ & \{\begin{array}{ll}\text{Hasil bagi}:& {\displaystyle \frac{6x^3+3x^2-\frac{3}{2}x+\frac{1}{4}}{2}}\\ & =3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+{\displaystyle \frac{1}{8}}}}\\ & \\ \text{Sisa bagi}:& -{\displaystyle \frac{7}{8}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Hasil bagi dan sisanya jika}(x^4-x^3-x^2+x-1)\\ & \text{dibagi oleh}(x-2)(x+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.x^2+1& \text{dan}& 2x+1\\ \text{b}.x^2+1& \text{dan}& 2x-1\\ \text{c}.x^2-1& \text{dan}& 2x+1\\ \text{d}.x^2-1& \text{dan}& 2x-1\\ \text{e}.2x^2-1& \text{dan}& x+1\end{array}\\ \\ & \text{Jawab}:\\ & \text{Dengan cara}\mathbf{\text{Horner-Kino}}\text{diperoleh}\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=x^4-x^3-x^2+x-1\\ \text{Pembagai}:& p(x)=(x-2)(x+1)=x^2-x-2\\ & :2\text{dari}-\frac{-2}{1},\text{sedang}1=-\left(\frac{-1}{1}\right)\\ \text{Hasil bagi}:& h(x)=x^2+1\\ \text{Sisa bagi}:& s(x)=2x+1\end{array}$

$\begin{array}{rl}& \text{Sehingga},\\ & x^4-x^3-x^2+x-1\\ & =(x^2-x-2)(x^2+1)+2x+1\end{array}$

Contoh Soal 1 Distribusi Binomial

$\begin{array}{l}\\ 1.& \text{Manakah yang merupakan data diskrit dari pernyataan berikut}\\ & \text{a}.\text{Suhu Badan Anton ketika sakit mencapai}{40}^{\circ }C\\ & \text{b}.\text{Kecepatan mobil yang sedang melaju adalah}100km/jam\\ & \text{c}.\text{Tinggi tiang bendaera di madrasah Budi adalah 4 m}\\ & \text{d}.\text{Jumlah guru yang mengajar di MA Futuhiyah }\\ & \text{sebanyak 30 orang}\\ & \text{e}.\text{Berat bayi yang baru lahir adalah 3.500 gram}\\ \\ & \text{Jawab}:\\ & \text{Alasannya dikarena hasil mencacah}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika Anda mengumpulkan nilai raport}\\ & \text{teman-teman sekelas Anda untuk pelajaran}\\ & \text{ matematika, maka data yang Anda peroleh }\\ & \text{adalah}....\\ & \text{a}.\text{data diskrit}\\ & \text{b}.\text{data kontinu}\\ & \text{c}.\text{data kualitatif}\\ & \text{d}.\text{Populasi}\\ & \text{e}.\text{Sampel}\\ \\ & \text{Jawab}:\\ & \text{Dengan catatan nilainya cacah}\end{array}$

$\begin{array}{l}\\ 3.& \text{Ukuran yang dihitung dari seluruh data }\\ & \text{dalam populasi adalah}....\\ & \text{a}.\text{data kuantitatif}\\ & \text{b}.\text{data kualitatif}\\ & \text{c}.\text{Statistik}\\ & \text{d}.\text{Statistika}\\ & \text{e}.\text{Parameter}\\ \\ & \text{Jawab}:\\ & \text{Parameter adalah ukuran dari }\\ & \text{seluruh data atau populasi}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui distribusi peluang suatu }\\ & \text{variabel acak diskrit sebagai berikut}\\ & \begin{array}{|ccccc|}\hline x& 0& 1& 2& 3\\ f(x)& m& 0,26& 3m& 0,42\\ \hline\end{array}\\ & \text{Peluang nilai X minimal berharga 2 adalah}\\ & \text{a}.0,24\\ & \text{b}.0,34\\ & \text{c}.0,42\\ & \text{d}.0,58\\ & \text{e}.0,66\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}X\text{adalah variabel }\\ & \text{acak diskrit, maka}\sum f(x)=1\\ & F(c)=P(X\le c)={\displaystyle \sum _{x=0}^{x=c}f(x)}\\ & =f(0)+f(1)+f(2)+f(3)+\cdots +f(c)=1\\ & \text{dalam hal soal}\text{di atas, maka kita tentukan}\\ & \text{nilai}m\text{dulu}\\ & F(3)=P(X\le 3)={\displaystyle \sum _{x=0}^{x=3}f(x)}\\ & =f(0)+f(1)+f(2)+f(3)=1\\ & 1=m+0,26+3m+0,42=4m+0,68\\ & 4m=1-0.68=0,32\\ & m=0.08,\text{sehingga}\\ & P(2\le X\le 3)=f(2)+f(3)=3m+0,42\\ & =3(0,08)+0,42=0,24+0,42=0,66\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui fungsi peluang suatu }\\ & \text{variabel acak kontinu adalah}\\ & f(y)=\{\begin{array}{c}0,\text{untuk textit{y} yang lain}\\ \\ {\displaystyle \frac{2y+k}{50},\text{untuk}0\le y\le 5}\end{array}\\ & \text{Nilai}P(|Y-1|\le 2)\text{adalah}....\\ & \text{a}.{\displaystyle \frac{7}{25}\text{d}.\frac{14}{25}}\\ \\ & \text{b}.{\displaystyle \frac{9}{25}\text{c}.\frac{12}{25}\text{e}.\frac{18}{25}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\int }_0^5{\displaystyle \frac{2y+k}{50}dy=1}\\ & 1={\int }_0^5{\displaystyle \frac{2y+k}{50}dy}\\ & 50={\int }_0^5(2y+k)dy\\ & 50=y^2+ky{|}_0^5=5^2+5k=25+5k\\ & k=5\\ & P(|Y-1|\le 2)=P(-2\le Y-1\le 2)\\ & =P(-1\le Y\le 3)\\ & =f(-1)+f(0)+f(1)+f(2)+f(3)\\ & ={\int }_0^3\left({\displaystyle \frac{2y+5}{50}}\right)dy\\ & ={\displaystyle \frac{1}{50}(y^2+5y){|}_0^3}\\ & ={\displaystyle \frac{1}{50}(9+15)={\displaystyle \frac{24}{50}=\frac{12}{25}}}\end{array}\end{array}$