Contoh 2 Vektor

$\begin{array}{ll}\\ 6.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\\ &\textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\\ & \textrm{adalah}....\\ &\textrm{a}.\quad 1\\ &\textrm{b}.\quad 2\\ &\textrm{c}.\quad 3\\ &\textrm{d}.\quad \color{red}4\\ &\textrm{e}.\quad 5\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{PQ}&=5\\ \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}&=5\\ (x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}&=25\\ (1-n)^{2}+(-2-2)^{2}&=25\\ (1-n)^{2}+16&=25\\ (1-n)^{2}-3^{2}&=0\\ (1+3-n)(1-3-n)&=0\\ \color{red}n=4\: \: \color{black}\textrm{atau}\: \: \color{red}n=-2& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui vektor}\: \: \vec{u}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{v}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\\ & \textrm{Nilai}\: \: \left | \vec{u}+\vec{v} \right |\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \sqrt{28}\\ &\textrm{b}.\quad \sqrt{30}\\ &\textrm{c}.\quad \color{red}\sqrt{34}\\ &\textrm{d}.\quad \sqrt{44}\\ &\textrm{e}.\quad \sqrt{50}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{u}+\vec{v}&=\begin{pmatrix} 3\\ 4 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &=\begin{pmatrix} 3+2\\ 4+(-1) \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ \left | \vec{u}+\vec{v} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\color{red}\sqrt{34} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Vektor satuan}\: \: \vec{u}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \color{red}\displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \frac{1}{15}\begin{pmatrix} -5\\ -12 \end{pmatrix}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{e}_{\vec{u}}&=\displaystyle \frac{\vec{u}}{\left | \vec{u} \right |},\quad \textrm{maka}\\ \vec{e}_{\begin{pmatrix} -5\\ -12 \end{pmatrix}}&=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\left | \begin{pmatrix} -5\\ -12 \end{pmatrix} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\sqrt{(-5)^{2}+(-12)^{2}}}\\ &=\displaystyle \frac{-\begin{pmatrix} 5\\ 12 \end{pmatrix}}{\sqrt{169}}\\ &=\color{red}-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} 8\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 14 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 6\\ 13 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ 48 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} 8\\ 7 \end{pmatrix}+\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ &=\begin{pmatrix} 8-3\\ 7+9 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 17 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} -6\\ -17 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 4\\ 17 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}+\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ &=\begin{pmatrix} 5-9\\ 4-21 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix} \end{aligned} \end{array}$

Contoh 1 Vektor

$\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{u}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\vec{i}+5\vec{j}&&&\textrm{d}.&-3\vec{i}-5\vec{j}\\ \textrm{b}.&5\vec{i}+3\vec{j}&\textrm{c}.&\color{red}-3\vec{i}+5\vec{j}&\textrm{e}.&-5\vec{i}+3\vec{j} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan lagi gambarnya}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{u}\: \: \textrm{jika dinyatakan sebagai }\\ &\textrm{kombinasi linear adalah}\: \: \vec{u}=\color{red}-3\vec{i}+5\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Panjang vektor}\: \: \vec{p}=\begin{pmatrix} 4\\ -8 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \sqrt{4}\\ &\textrm{b}.\quad \sqrt{12}\\ &\textrm{b}.\quad \sqrt{20}\\ &\textrm{d}.\quad \color{red}\displaystyle \sqrt{80}\\ &\textrm{e}.\quad \sqrt{100}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}&=\begin{pmatrix} 4\\ -8 \end{pmatrix},\: \: \textrm{maka besar dari vektor}\\ \vec{p}&\: \: \textrm{adalah}=\left | \vec{p} \right |=\sqrt{x^{2}+y^{2}}\\ &\textrm{Yaitu}\\ \left | \vec{p} \right |&=\sqrt{4^{2}+(-8)^{2}}\\ &=\sqrt{16+64}=\color{red}\sqrt{80} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{ll}\\ .\quad &\textrm{Panjang vektor}\: \: \vec{h}\: \: \textrm{tersebut di atas adalah}....\\ &\textrm{a}.\quad 5\\ &\textrm{b}.\quad 7\\ &\textrm{c}.\quad \color{red}10\\ &\textrm{d}.\quad 12\\ &\textrm{e}.\quad 15\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\quad \vec{h}=\overline{OH}=8\vec{i}+6\vec{j}\\ \vec{h}&=\sqrt{x_{H}^{2}+y_{H}^{2}}\\ &=\sqrt{8^{2}+6^{2}}\qquad \textbf{(ingat tripel Pythagoras)}\\ &=\sqrt{10^{2}}=\color{red}10 \end{aligned} \end{array}$

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

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

Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &=\color{red}2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad \color{red}5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &=\color{red}5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \color{red}\displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}=\color{red}\displaystyle \frac{2}{14} \end{aligned} \end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 1&\textrm{Nilai}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle \frac{1}{4}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{4}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&= \left ( \displaystyle \frac{6-2}{2^{2}-4}-\frac{1}{2-2} \right )\\ &=\left ( \displaystyle \frac{4}{0}-\frac{1}{0} \right )=\color{blue}\infty -\infty \\ &\: \: \textrm{hal ini tidak diperkenankan}\\ \textrm{Sehingga},\, \qquad\qquad\qquad &\\ \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{(x+2)}{(x-2)(x+2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{x+2}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{4-2x}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2(x-2)}{(x+2)(x-2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2}{x+2} \right )\\ &=\displaystyle -\frac{2}{(2+2)}\\ &=\color{red}-\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle 2&&\textrm{d}.\quad \displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle 2\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&= \left ( \displaystyle \frac{4-4}{2^{4}-4} \right )\\ &=\color{blue}\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\: \: \: \: \quad&\\ \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{\sqrt{x}\left ( 2-\sqrt{x} \right )} \\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{-\sqrt{x}\left ( \sqrt{x}-2 \right )}\\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: -\displaystyle \frac{\left ( \sqrt{x}+2 \right )}{\sqrt{x}}\\ &=-\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}\\ &=-\displaystyle \frac{2+2}{2}\\ &=\color{red}-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \color{red}1\quad &\textrm{e}.\quad \displaystyle 4\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 1}{\textrm{Lim}}\:& \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &= \displaystyle \frac{\left ( 2-3+1 \right )\left ( 1-1 \right )}{\left ( 1-1 \right )^{2}} \\ &=\displaystyle \frac{0\times 0}{0^{2}}\\ &=\color{blue}\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehi}&\textrm{ngga},\\ \underset{x\rightarrow 1}{\textrm{Lim}}\: &\: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \left ( 2\sqrt{x}-1 \right )\times \left ( \sqrt{x}-1 \right ) \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-1 \right )}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \left ( 2\sqrt{x}-1 \right )\\ &=2.1-1\\ &=2-1\\ &=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle \frac{1}{14}\sqrt{7}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{7}\sqrt{7}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{14}\sqrt{7}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 3}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times \displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x+4 \right )-\left ( 2x+1 \right )}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: -\displaystyle \frac{-x+3}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{-1}{\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=-\displaystyle \frac{1}{\left ( \sqrt{7}+\sqrt{7} \right )}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\times \displaystyle \frac{\sqrt{7}}{\sqrt{7}}\\ &=\color{red}-\displaystyle \frac{1}{14}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\: \: \textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad \displaystyle -2\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad -1\quad &\textrm{e}.\quad \displaystyle \color{red}-3\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: &\: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6\\ &\textrm{dengan bantuan limit kanan }\\ &\textrm{yaitu}\: \: x=2+h\: \Rightarrow \: h\rightarrow 0\\ \underset{h\rightarrow 0}{\textrm{Lim}}\: &\: \displaystyle \frac{a\left ( 2+h \right )-2a}{\sqrt{2\left ( 2+h \right )}-\left ( 2+h \right )}=6\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-\left ( 2+h \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah}{\sqrt{4+2h}-\left ( 2+h \right )}\times \displaystyle \frac{\left ( \sqrt{4+2h}+\left ( 2+h \right ) \right )}{\left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{4+2h-\left ( 4+4h+h^{2} \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2h-h^{2}}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2-h}\\ 6&=\displaystyle \frac{a\times \left (\sqrt{4+0} +\left ( 2+0 \right ) \right )}{-2-0}\\ 6&=\displaystyle \frac{a\left ( \sqrt{4}+2 \right )}{-2}\\ \displaystyle \frac{a(4)}{-2}&=6\\ a(-2)&=6\\ a&=\color{red}-3 \end{aligned} \end{array}$

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

 $\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Persamaan dan pertidaksamaan}\\ &\qquad \textrm{nilai mutlak dari bentuk linear}\\ &\qquad \textrm{satu variabel} \end{aligned}$

• 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 , 

$\color{red}\begin{aligned}&\textrm{A.2 Pertidaksamaan rasional}\\ &\qquad \textrm{dan irasional satu variabel} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{A.3 Sistem persamaan linear tiga}\\ &\qquad \textrm{variabel} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{A.4 Sistem pertidaksamaan dua}\\ &\qquad \textrm{variabel-linear-linear} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{A.5 Fungsi} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{A.6 Fungsi Komposisi dan invers} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{A.7 Rasio trigonometri}\\ &\qquad \textrm{pada segitiga siku-siku} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.8 Rasio trigonometri}\\ &\qquad \textrm{sudut-sudut diberbagai}\\ &\qquad \textrm{kuadran} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.9 Aturan sinus dan cosinus} \end{aligned}$

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Program linear} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.2 Matriks} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.3 Determinan dan invers matriks}\\ &\qquad \textrm{ordo 2x2} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.4 Pola bilangan dan jumlah}\\ &\qquad \textrm{pada barisan aritmetika}\\ &\qquad \textrm{dan geometri} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{B.5 Limit fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.6 Turunan fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.7 Keberkaitan Turunan fungsi} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.8 Integral tak tentu}\\ &\qquad \textrm{fungsi aljabar} \end{aligned}$

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Jarak dalam ruang} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{C.2 Statistika} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{C.3 Aturan pencacahan} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{C.4 Peluang kejadian majumuk} \end{aligned}$

• 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

 $\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Fungsi eksponensial}\\ &\qquad \textrm{dan fungsi logaritma} \end{aligned}$

• 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 

$\color{black}\begin{aligned}&\textrm{A.2 Vektor} \end{aligned}$

• 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

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Persamaan trigonometri} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{B.2 Polinom} \end{aligned}$

• 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

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Turunan fungsi trigonometri} \end{aligned}$

• 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  

$\color{black}\begin{aligned}&\textrm{C.2 Distribusi peluang binomial} \end{aligned}$

• 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

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

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

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

$\begin{array}{ll}\\ 14.&\textrm{Peluang seseorang sembih dari }\\ &\textrm{penyakit jantung adalah 0,6}\\ &\textrm{Jika 7 orang penderita ini menjalani }\\ &\textrm{operasi, maka peluang 3 sampai}\\ &\textrm{6 orang sembuh adalah}... .\\ &\textrm{a}.\quad \displaystyle 0,0629\qquad\qquad\quad\qquad \quad\textrm{d}.\quad \color{red}0,6822\\ &\textrm{b}.\quad \displaystyle 0,2613\quad \textrm{c}.\quad 0,2898\quad\quad \: \textrm{e}.\quad 0,9720\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang sembuh}=0,6,\qquad \textrm{maka}\: \: \\ &q=\textbf{Peluang tidak sembuh}=1-0,6=0,4\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &P\left ( 3\leq X\leq 6 \right )=P\left ( X\leq 6 \right )-P\left ( X\leq 3 \right )\\ &=C_{4}^{7}(0,6)^{4}(0,4)^{3}+C_{5}^{7}(0,6)^{5}(0,4)^{2}+C_{6}^{7}(0,6)^{6}(0,4)^{1}\\ &=35\times 0,0082944+21\times 0,0124416+7\times 0,0186624\\ &=0,290304+0,2612736+0,1306368\\ &=\color{red}0,6822144 \end{aligned} \end{array}$

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

Contoh Soal 2 Distribusi Binomial

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

$\begin{array}{ll}\\ 7.&\textrm{Pada pengundian terhadap mata uang identik},\\ &\textrm{sebanyak 10 kali, peluang distribusi binomial} \\ &\textrm{untuk mendapatkan 7 muka gambar adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.\quad \displaystyle 0,2653&&\textrm{d}.\quad \displaystyle 0,7522\\ \textrm{b}.\quad \displaystyle \color{red}0,1172&\textrm{c}.\quad \displaystyle 0,2653&\textrm{e}.\quad 0,2422 \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Uraian berikut sekaligus tambahan}\\ &\textrm{penjelasan pada uraian jawaban}\\ &\color{blue}\textrm{soal no. 6 di atas}\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ &\textrm{Ingat sebuah koin ada 2 muka}\\ &\textrm{yaitu muka gambar (G) dan angka (A)}\\ &\color{red}\textrm{misalkan}\\ &A=\textrm{kejadian muncul muka gambar}\\ &\textrm{maka peluangnya adalah}\: \: \displaystyle \frac{1}{2}\\ &\textrm{Selanjutnya di sini disimbolkan dengan}\: \: \: \color{blue}p=\displaystyle \frac{1}{2}\\ &\color{red}\textrm{Demikian juga misalkan}\\ &B=\textrm{kejadian muncul muka angka}\\ &\textrm{maka peluang juga}\: \displaystyle \frac{1}{2}\\ &\textrm{Di sini dituliskan dengan}\: \: \: \color{blue}q=\displaystyle \frac{1}{2}\\ f(7)&=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{10-7}\\ &=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{3}\\ &=\displaystyle \frac{10!}{7!\times (10-7)!}\left ( \displaystyle \frac{1}{2} \right )^{7+3}\\ &=\displaystyle \frac{10.9.8.\not{7!}}{\not{7!}.3.2.1}\left ( \displaystyle \frac{1}{1024} \right ) \\ &=\color{red}0,1172 \end{aligned} \end{array}$

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

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

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

Contoh Soal 3 Polinom

$\begin{array}{ll}\\ 11.&\textrm{Jika polinom}\: \: f(x)\: \: \textrm{dibagi oleh}\\ &(x-a)(x-b)\: \: \textrm{dan}\: \: a\neq b\: ,\: \textrm{maka}\\ &\textrm{sisa pembagiannya adalah}\: ....\\ &\begin{array}{lllllll}\\ &\textrm{a}.\quad \displaystyle \displaystyle \frac{x-a}{a-b}f(a)+\frac{x-a}{b-a}f(b)\\\\ &\textrm{b}.\quad \displaystyle \displaystyle \frac{x-a}{a-b}f(b)+\frac{x-a}{b-a}f(a)\\\\ &\textrm{c}.\quad \displaystyle \displaystyle \color{red}\frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)\\\\ &\textrm{d}.\quad \displaystyle \displaystyle \frac{x-b}{a-b}f(b)+\frac{x-a}{b-a}f(a)\\\\ &\textrm{e}.\quad \displaystyle \displaystyle \frac{x-a}{b-a}f(b)+\frac{x-a}{b-a}f(a)\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Misal sisa pembagiannya}:\: \color{red}s(x)=px+q\\ &\textrm{Saat}\: \: f(x)\: \: \textrm{dibagi}\: \: (x-a)(x-b)\: \: \textrm{berarti}\\ &\bullet \quad x=a\Rightarrow s(a)=f(a)=ap+q\: ....(1)\\ &\bullet \quad x=b\Rightarrow s(b)=f(b)=bp+q\: ......(2)\\ &\textrm{Persamaan}\: \: (1)\: \: \textrm{dan}\: \: (2)\: \: \textrm{dieliminasi}\\ &\color{blue}\begin{array}{llllllll}\\ ap&+&q&=&f(a)\\ bp&+&q&=&f(b)&-\\\hline ap&-&bp&=&f(a)-f(b)\\ &&p&=&\color{purple}\displaystyle \frac{f(a)-f(b)}{a-b}& \end{array}\\ &\textrm{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}\\ &\textrm{Sehingga}\\ &s(x)=px+q\\ &\qquad =\left ( \displaystyle \frac{f(a)-f(b)}{a-b} \right )x+\left ( \displaystyle \frac{-bf(a)-af(b)}{a-b} \right )\\ &\qquad =\displaystyle \frac{f(a)x-f(b)x-bf(a)+af(b)}{a-b}\\ &\qquad =\displaystyle \frac{(x-b)f(a)+(a-x)f(b)}{a-b}\\ &\qquad =\displaystyle \frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)\\ &\qquad =\color{red}\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 13.&\textrm{Polinom}\: \: f(x)\: \: \textrm{dibagi oleh}\: \: (2x-4)\: \: \textrm{bersisa 6},\\ &\textrm{dibagi oleh}\: \: (x+4)\: \: \textrm{bersisa 24}.\\ &\textrm{Dan polinom}\: \: g(x)\: \: \textrm{dibagi oleh}\: \: (2x-4)\: \: \textrm{bersisa 5},\\ & \textrm{dibagi oleh}\: \: (x+4)\: \: \textrm{bersisa 2}.\\ &\textrm{Jika}\: \: h(x)=f(x).g(x),\: \: \textrm{maka}\: \: h(x)\\ &\textrm{dibagi}\: \: (2x^{2}+4x-16)\: \: \textrm{akan sisa}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3x+24&&\textrm{d}.\quad -6x+36\\ \textrm{b}.\quad \color{red}-3x+36&\textrm{c}.\quad 6x+24&\textrm{e}.\quad 12x+3 \end{array}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Langkah pertama}\\ &\begin{aligned}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}\\ &\textrm{Gunakanlah cara sebagai mana}\\ &\textrm{contoh soal No. 12 di atas yang}\\ \color{magenta}\textrm{Alte}&\color{magenta}\textrm{natif 2}\\ \textrm{mak}&\textrm{a}\quad p_{1}x+q_{1}=-3x+12 \end{aligned} \\ &\color{blue}\textrm{Langkah kedua}\\ &\begin{aligned}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}\\ &\textrm{Gunakanlah cara sebagai mana}\\ &\textrm{contoh soal No. 12 di atas yang}\\ \color{magenta}\textrm{Alte}&\color{magenta}\textrm{natif 2}\\ \textrm{mak}&\textrm{a}\quad p_{2}x+q_{2}=\displaystyle \frac{1}{2}x+4 \end{aligned} \\ &\color{blue}\textrm{Langkah ketiga}\\ &\begin{aligned}&h(x)=\color{red}f(x)\times g(x)\\ &=\left ( (2x-4)(x+4)H_{1}(x)+(-3x+12) \right )\\ &\qquad\qquad\qquad \times \left ( (2x-4)(x+4)H_{2}(x)+\displaystyle \frac{1}{2}x+4 \right )\\ &\textrm{maka}\\ &\bullet \quad h(2)=\left ( 0+(-3.2+12) \right )\left ( 0+\displaystyle \frac{1}{2}.2+4 \right )=6.5=30\\ &\bullet \quad h(-4)=\left ( 0+(-3.-4+12) \right )\left ( 0+\displaystyle \frac{1}{2}.-4+4 \right )=24.2=48\\ &\textrm{Dengan pembagi}\: \: 2x^{2}+x-16,\: \textrm{maka sisanya}:\: s_{3}(x)=p_{3}x+q_{3}\\ &\textrm{saat}\: \: x=2\qquad \Rightarrow 2p+q=30\\ &\textrm{saat}\: \: x=-4\: \: \Rightarrow -4p+q=48\\ &\textrm{selanjutnya dengan eliminasi-substitusi diperoleh}\: \: p=-3,\: q=36\\ &\textrm{sehingga}\: \: s(x)=px+q=\color{red}-3x+36 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\color{purple}(\textrm{KSM 2015})\color{black}\textrm{Diketahui}\: \: f(x)\: \: \textrm{adalah polinom}\\ & (x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ &\textrm{dengan}\: \: \: x_{1},\: x_{2},\: x_{3},\: x_{4},\: \: \textrm{dan}\: \: x_{5}\: \: \textrm{adalah}\\ &\textrm{bilangan bulat berbeda}.\: \textrm{Jika}\: \: f(104)=2012,\\ &\textrm{maka nilai} \: \: \: x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}\: \: \textrm{sama dengan}....\\ &\textrm{a}.\quad 13\\ &\textrm{b}.\quad 14\\ &\textrm{c}.\quad 16\\ &\textrm{d}.\quad \color{red}17\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}:\\ f(x)&=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ f(104)&=(104-x_{1})(104-x_{2})(104-x_{3})(104-x_{4})(104-x_{5})=2012\\ &=2012=1\times 2\times 503\\ &=(-1)\times (1)\times (-2)\times (2)\times (503)\\ \textrm{maka}&\: \: \begin{cases} (104-x_{1}) &=-2\Rightarrow x_{1}=106 \\ (104-x_{2}) &=-1\Rightarrow x_{2}=105 \\ (104-x_{3}) &=1\Rightarrow x_{3}=103 \\ (104-x_{4}) &=2\Rightarrow x_{4}=102 \\ (104-x_{5}) &=503\Rightarrow x_{5}=-399 \\ \end{cases}\\ \textrm{sehin}&\textrm{gga},\\ &x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}=106+105+103+102+(-399)=\color{red}17 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Tentukanlah suku banyak}\: \: f(x)\: \: \textrm{sedemikian}\\ &\textrm{sehingga}\: \: f(x)\: \: \textrm{terbagi oleh}\: \: x^{2}+1,\\ &\textrm{sedangkan}\: \: f(x)+1\: \: \textrm{terbagi oleh}\: \: x^{3}+x^{2}+1\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\left ( x^{2}+1 \right ).h_{1}x\\ f(x)+1&=\left ( x^{2}+1 \right ).h_{1}x+1\\ \textrm{supaya}\: \: \: &f(x)+1\: \: \textrm{terbagi habis oleh}\: \: x^{3}+x^{2}+1 ,\\ & \textrm{maka akan ada bilangan bulat}\: \: k,\: \: \left ( k\neq 0 \right )\\ k&=\displaystyle \frac{f(x)+1}{x^{3}+x^{2}+1}\\ &=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\\ k=1\Rightarrow &1=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\: \: \textrm{maka}\: \: h_{1}x=x\\ \textrm{sehingga}&\: \: f(x)=\color{red}x^{3}+x^{2}\\ \textrm{untuk ni}&\textrm{lai}\: \: k\: \: \textrm{yang lain, tak ditemukan} \end{aligned} \end{array}$

Contoh Soal 4 Polinom

$\begin{array}{ll}\\ 16.&\textrm{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ & \textrm{adalah}\: \: x_{1},\: x_{2},\: x_{3},...,x_{2017}\\ &\textrm{Tentukan nilai dari}\\ & \displaystyle \frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{x^{2018}-1}{x-1}&=x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ \textrm{perlu dii}&\textrm{ngat bahwa kondisi ini mensyaratkan}\: \: x\neq 1,\\ & \textrm{sehingga}\\\\ x^{2018}-1&=0\\ x^{2018}&=1\\ x&=\pm 1,\: \: \: \: \textrm{pilih}\: \: x=-1\\ \textrm{maka}\: \quad &\textrm{nilai dari}\\ \displaystyle \frac{1}{1-x_{1}}+&\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+\displaystyle \frac{1}{2}}} \\ &=\color{red}\displaystyle \frac{2017}{2} \end{aligned} \end{array}$

Contoh Soal 2 Polinom

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

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

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

$.\qquad \begin{aligned}&\textrm{Sehingga},\\ &\bullet \quad 2m=2\Rightarrow m=\color{red}1\\ &\bullet \quad 1-n=-1\Rightarrow n=\color{red}2 \end{aligned}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: f(x)=x^{4}-kx^{2}+5\: \: \textrm{habis dibagi}\\ &(x-1)\: \: \textrm{maka}\: \: f(x)\: \: \textrm{juga habis dibagi oleh}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x+1&&\textrm{d}.\quad x+5\\ \textrm{b}.\quad 2x+1&\textrm{c}.\quad 3x+1&\textrm{e}.\quad 2x+5 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=x^{4}-kx^{2}+5\\ f(1)&=(1)^{4}-k(1)^{2}+5\\ 0&=1-k+5\\ k&=6\\ f(x)&=x^{4}-6x^{2}+5\\ &=(x^{2}-1)(x^{2}-5)\\ &=(x-1)\color{red}(x+1)\color{black}(x^{2}-5) \end{aligned} \end{array}$

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

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

Contoh Soal 1 Polinom

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: g(x)=2x^{3}+x^{2}-x+1,\\ &\textrm{maka}\: \: g(1)=....\: \: \\ &\begin{array}{lll}\\ \textrm{a}.\quad -2&&\textrm{d}.\quad 2\\ \textrm{b}.\quad -1&\textrm{c}.\quad 1&\textrm{e}.\quad \color{red}3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}g(x)&=2x^{3}+x^{2}-x+1\\ g(1)&=2(1)^{3}+(1)^{2}-(1)+1\\ &=2+1-1+1\\ &=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: p(y)=5y^{4}+2r^{2}y^{3}+y^{2}+1\: \: \textrm{dan}\\ & q(y)=4y^{5}+3ry^{2}-3y-1\: \: \\ &\textrm{serta}\: \: p(-1)=q(-1),\: \: \textrm{maka nilai}\: \: r\\ & \textrm{sama dengan}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: 3&&\textrm{d}.\quad -\displaystyle \frac{3}{2}\\ \textrm{b}.\quad \displaystyle -\frac{3}{2}\: \: \textrm{dan}\: \: 3&\textrm{c}.\quad \color{red}\displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: -3&\textrm{e}.\quad 3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}p(-1)&=q(-1)\\ 5(-1)^{4}+2r^{2}(-1)^{3}+(-1)^{2}+1&=4(-1)^{5}+3r(-1)^{2}-3(-1)-1\\ 5-2r^{2}+1+1&=-4+3r+3-1\\ 9-3r-2r^{2}&=0\\ \displaystyle \frac{(-6-2r)(-3+2r)}{2}&=0,\qquad \textrm{ingat pemfaktoran}\\ (-3-r)(-3+2r)&=0\\ r=-3\quad \vee \quad r&=\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{berderajat}\: \: n.\\ &\textrm{Jika pembaginya berbentuk}\: \: \left ( ax^{2}+bx+c \right ),\\ &\textrm{dengan}\: \: a\neq 0,\: \: \textrm{maka hasil baginya berderajat}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad n-1&&\textrm{d}.\quad 3\\ \textrm{b}.\quad \color{red}n-2&\textrm{c}.\quad n-3&\textrm{e}.\quad 2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Suku banyak (polinom)}\\ &=\textrm{pembagi}\times \textrm{hasil bagi}+\textrm{sisa}\\ &x^{n}+...=\left ( ax^{2}+bx+c \right )\times \color{red}\left ( x^{n-2}+... \right )\color{black}+\left (mx+n \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (6x^{4}-3x^{2}+x-1 \right )\\ & \textrm{dibagi oleh}\: \: \left ( 2x-1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\frac{1}{8}&\textrm{dan}&\color{red}-\displaystyle \frac{7}{8}\\ \textrm{b}.\quad 3x^{3}+3x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&-7\\ \textrm{c}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-3x+\frac{1}{8}&\textrm{dan}&\displaystyle \frac{7}{8}\\ \textrm{d}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&\displaystyle \frac{1}{8}\\ \textrm{e}.\quad 3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x-\frac{1}{8}&\textrm{dan}&-\displaystyle \frac{7}{8} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\begin{array}{rr|rrrrrrr} \color{blue}x=\frac{1}{2}&&6&0&-3&1&-1\\ &&&3&\frac{3}{2}&-\frac{3}{4}&\frac{1}{8}&+&\\\hline &&6&3&-\frac{3}{2}&\frac{1}{4}&\color{red}\boxed{-\frac{7}{8}} \end{array}\\ &\textrm{Selanjutnya}\\ &\begin{cases} \textrm{Hasil bagi}: & \displaystyle \frac{6x^{3}+3x^{2}-\frac{3}{2}x+\frac{1}{4}}{2}\\ &=\color{red}3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\displaystyle \frac{1}{8} \\ & \\ \textrm{Sisa bagi}: & -\displaystyle \frac{7}{8} \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (x^{4}-x^{3}-x^{2}+x-1 \right )\\ &\textrm{dibagi oleh}\: \: \left ( x-2 \right )\left ( x+1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x^{2}+1&\textrm{dan}&\color{red}2x+1\\ \textrm{b}.\quad x^{2}+1&\textrm{dan}&2x-1\\ \textrm{c}.\quad x^{2}-1&\textrm{dan}&2x+1\\ \textrm{d}.\quad x^{2}-1&\textrm{dan}&2x-1\\ \textrm{e}.\quad 2x^{2}-1&\textrm{dan}&x+1\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Dengan cara}\: \: \textbf{Horner-Kino}\: \: \textrm{diperoleh} \end{array}$

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

$\qquad\begin{aligned}&\textrm{Sehingga},\\ &x^{4}-x^{3}-x^{2}+x-1\\ &\qquad =\color{red}\left ( x^{2}-x-2 \right )\left ( x^{2}+1 \right )+2x+1 \end{aligned}$

Contoh Soal 1 Distribusi Binomial

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

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

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

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

$\begin{array}{ll}\\ 5.&\textrm{Diketahui fungsi peluang suatu }\\ &\textrm{variabel acak kontinu adalah}\\ &f(y)=\left\{\begin{matrix} 0,\quad \textrm{untuk \textit{y} yang lain}\\\\ \displaystyle \frac{2y+k}{50},\: \: \textrm{untuk}\: \: 0\leq y\leq 5 \end{matrix}\right.\\ &\textrm{Nilai}\: \: P\left ( \left | Y-1 \right |\leq 2 \right )\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{7}{25}\qquad\qquad\qquad\qquad \textrm{d}.\quad \frac{14}{25}\\\\ &\textrm{b}.\quad \displaystyle \frac{9}{25}\qquad \textrm{c}.\quad \color{red}\frac{12}{25}\qquad\quad \color{black}\textrm{e}.\quad \frac{18}{25}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\int_{0}^{5}\displaystyle \frac{2y+k}{50}\: dy=1\\ &1=\int_{0}^{5}\displaystyle \frac{2y+k}{50}\: dy\\ &50=\int_{0}^{5}(2y+k)dy\\ &50=y^{2}+ky|_{0}^{5}=5^{2}+5k=25+5k\\ &k=5\\ &\color{blue}P(\left | Y-1 \right |\leq 2)=P\left ( -2\leq Y-1\leq 2 \right )\\ &=P\left ( -1\leq Y\leq 3 \right )\\ &=f(-1)+f(0)+f(1)+f(2)+f(3)\\ &=\int_{0}^{3}\left ( \displaystyle \frac{2y+5}{50} \right )dy\\ &=\displaystyle \frac{1}{50}\left ( y^{2}+5y \right )|_{0}^{3}\\ &=\displaystyle \frac{1}{50}\left ( 9+15 \right )=\displaystyle \frac{24}{50}=\color{red}\frac{12}{25} \end{aligned} \end{array}$