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Lanjutan Contoh Soal Fungsi, Persamaan, dan Pertidaksamaan Kuadrat (4)

\begin{array}{ll}\\ \fbox{17}&\textrm{Jika}\: \: x=-1\: \: \textrm{adalah akar dari persamaan kuadrat}\: x^{2}-kx-x-2k^{2}+5k-2=0,\\ &\textrm{Carilah nilai}\: \: k\: \: \textrm{dan akar persamaan kuadrat yang lain}. \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Untuk}&\: \: x=-1\: \: \textrm{maka},\\ &\qquad (-1)^{2}-k(-1)-(-1)-2k^{2}+5k-2=0\\ &\Leftrightarrow 1+k+1-2k^{2}+5k-2=0\\ &\Leftrightarrow -2k^{2}+6k=0\\ &\Leftrightarrow -2k(k-3)=0\\ &\Leftrightarrow -2k=0\quad \textrm{atau}\quad k-3=0\\ &\Leftrightarrow k=0\,\, \qquad \textrm{atau}\quad k=3\\ &\begin{array}{|l|l|}\hline k=0&k=3\\\hline \multicolumn{2}{|c|}{\textrm{selain akar} \: \: x=-1\: \: , \textrm{akar yang lain adalah}}\\\hline \begin{aligned}k&=0 \\ &\downarrow\\ &x^{2}-(0)x-x-2(0)^{2}+5(0)-2=0\\ &\Leftrightarrow x^{2}-x-2\\ &\Leftrightarrow (x+1)(x-2)\\ &\Leftrightarrow x=-1\quad \textrm{atau}\quad x=2 \end{aligned}&\begin{aligned}k&=3\\ &\downarrow\\ &x^{2}-(3)x-x-2(3)^{2}+5(3)-2=0\\ &\Leftrightarrow x^{2}-4x-5\\ &\Leftrightarrow (x+1)(x-5)\\ &\Leftrightarrow x=-1\quad \textrm{atau}\quad x=5 \end{aligned}\\\hline \end{array} \\ \textrm{Jadi},&\: \textrm{nilai}\: \: x\: \: \textrm{yang lain adalah}\: \: \begin{cases} x=2 & \text{ saat } k=0 \\ x=5 & \text{ saat } k=3 \end{cases} \end{aligned}.

\begin{array}{ll}\\ \fbox{18}&\textrm{Jika akar-akar persamaan kuadrat}\: \: px^{2}+(2p+1)x+q=0,\: \: \textrm{adalah}\\ &-2\: \: \textrm{dan}\: \: -\displaystyle \frac{1}{2}\: \: , \textrm{maka nilai dari}\: \: p+q=.... \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}\: \: x_{1}=-2\: \: \textrm{dan}\: \: x_{2}=-\displaystyle \frac{1}{2}\: \: \textrm{adalah akar-akar dari persamaan di atas}, \\ &\textrm{maka untuk}\\ &\left\{\begin{matrix} x=-2&\Rightarrow &p(-2)^{2}+(2p+1)(-2)+q=0&\Rightarrow 4p-4p-2+q=0.....\textcircled{1}\\ x=-\displaystyle \frac{1}{2}&\Rightarrow &p\left ( -\displaystyle \frac{1}{2} \right )^{2}+(2p+1)\left ( -\displaystyle \frac{1}{2} \right )+q=0&\Rightarrow \displaystyle \frac{1}{4}p-p-\frac{1}{2}+q=0.....\textcircled{2} \end{matrix}\right.\\ &\textrm{Selanjutnya dari persamaan}\: \: \textcircled{1}\: \: \textrm{diperoleh}\\ &q=2,\quad \textrm{sehingga pada persamaan}\: \: \textcircled{2}\: \: \textrm{kita memperoleh}:\: \: \displaystyle -\frac{3}{4}p-\frac{1}{2}+(2)=0 \\ &\Leftrightarrow -3p-2+4=0\Leftrightarrow p=\frac{2}{3}\\ &\textrm{Jadi},\: \: p+q=\displaystyle \frac{2}{3}+2=2\frac{2}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{19}&\textrm{Diketahui akar-akar persamaan kuadrat}\: \: x^{2}+x-3=0\: \: \textrm{adalah}\: \: \alpha \: \: \textrm{dan}\: \beta \\ &\textrm{Tentukanlah nilai dari}\: \: \alpha ^{3}-4\beta ^{2}+19\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\textrm{Diketahui}\\ &\begin{array}{|l|l|l|}\hline x^{2}+x-3=0&\begin{aligned}\alpha ^{2}+\alpha -3&=0\\ &\\ \Leftrightarrow \alpha ^{2}&=3-\alpha.....\textcircled{1} \end{aligned}&\begin{aligned}\beta ^{2}+\beta -3&=0\\ &\\ \Leftrightarrow \beta ^{2}&=3-\beta.....\textcircled{2} \end{aligned}\\\cline{2-3} \left\{\begin{matrix} \alpha +\beta =\displaystyle \frac{-b}{a}=-1\\ \alpha \beta =\displaystyle \frac{c}{a}=-3 \end{matrix}\right.&\begin{aligned}\alpha ^{3}+\alpha ^{2}-3\alpha &=0\\ &\\ \Leftrightarrow \alpha ^{3}&=3\alpha -\alpha ^{2}.....\textcircled{3} \end{aligned}&\begin{aligned}\beta ^{3}+\beta ^{2}-3\beta &=0\\ &\\ \Leftrightarrow \beta ^{3}&=3\beta -\beta ^{2} .....\textcircled{4}\end{aligned}\\\hline \end{array}\\ &\\ &\begin{aligned}\alpha ^{3}-4\beta ^{2}+19&=\left ( 3\alpha -\alpha ^{2} \right )-4\left ( 3-\beta \right )+19,\: \textnormal{perhatikan persamaan}\: \: \textcircled{3}\: \: \textrm{dan}\: \: \textcircled{2}\\ &=3\alpha -\left ( 3-\alpha \right )-12+4\beta +19\\ &=4\alpha +4\beta -3+7\\ &=4\left ( \alpha +\beta \right )+4\\ &=4(-1)+4\\ &=0 \end{aligned} \end{aligned}.

Sumber Referensi

  1. Idris, Muhammad, Ibnu Rusdi. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: Yrama Widya.
  2. Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: Yudistira.
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Contoh Soal Fungsi, Persamaan, dan Pertidaksamaan Kuadrat (3)

Lihat contoh soal sebelumnya di sini,  di sin, dan  di sini

\begin{array}{ll}\\ \fbox{11}.&\textrm{Sketsalah grafik fungsi kuadrat} \: \: f(x)=x^{2}-6x+5 \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|c|c|c|c|}\hline \multicolumn{5}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}:\\ &f(x)=x^{2}-6x+5\left\{\begin{matrix} x_{1}\\ \\ x_{2} \end{matrix}\right.\: \: \textrm{dan}\: \: \left\{\begin{matrix} a=1\: \: \: \: \\ b=-6\\ c=5\: \: \: \end{matrix}\right.\\ & \end{aligned}}\\\hline \textrm{Istilah}&\textrm{Memotong}&\textrm{Syarat}&\textrm{Proses}&\textrm{Titik}\\\hline \textrm{Titik potonng}&\textrm{Sumbu}-X&y=0&x^{2}-6x+5=0&\begin{aligned}&(1,0)\\ &\textrm{dan}\\ &(5,0) \end{aligned}\\\cline{2-5} &\textrm{Sumbu}-Y&x=0&y=0^{2}-6(0)+5&(0,5)\\\hline \textrm{Persamaan sumbu simetri}&\multicolumn{2}{|c|}{x=\displaystyle \frac{-b}{2a}}&\multicolumn{2}{|c|}{x=\displaystyle \frac{-(-6)}{2.1}=3}\\\hline \textrm{Koordinat titik puncak}&\multicolumn{2}{|c|}{\left ( x_{p},y_{p} \right )=\left ( \frac{-b}{2a},\frac{-D}{4a} \right )}&\begin{aligned}&\left ( \frac{-(-6)}{2.1}, \frac{-\left ( (-6)^{2}-4.1.5 \right )}{4.1}\right ) \end{aligned}&(3,-4)\\\hline \end{array}.

Untuk gambarnya perhatikanlah ilustrasi gambar di bawah ini

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\begin{array}{ll}\\ \fbox{12}.&\textrm{Diketahu}i\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\\ &x^{2}-x-2=0,\: \: \textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \alpha +\beta \: \: \textrm{dan}\: \: \alpha \beta&\textrm{e}.\quad \alpha ^{2}+\beta ^{2}\\ \textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&\textrm{f}.\quad \alpha ^{2}-\beta ^{2}\\ \textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}\\ \textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}\\ \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}-x-2=0\left\{\begin{matrix} \alpha \\ \\ \beta \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=-1\\ c=-2 \end{matrix}\right.\\ & \end{aligned}}\\\hline \begin{aligned}\textrm{a}.\quad &\alpha +\beta =-\frac{b}{a}=-\frac{(-1)}{1}=1\\ &\alpha \beta =\frac{c}{a}=\frac{(-2)}{1}=-2 \end{aligned}&\begin{aligned}\textrm{e}.\quad \alpha ^{2}+\beta ^{2}&=\left ( \alpha +\beta \right )^{2}-2\alpha \beta \\ &=1^{2}-2(-2)\\ &=1+4=5 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&=\frac{D}{a^{2}}\\ &=\frac{b^{2}-4ac}{a^{2}}\\ &=\frac{(-1)^{2}-4.(1).(-2)}{(1)^{2}}\\ &=1+8=9 \end{aligned}&\begin{aligned}\textrm{f}.\quad \alpha ^{2}-\beta ^{2}&=\left ( \alpha +\beta \right )\left ( \alpha -\beta \right )\\ &=(1).(9)=9\\ &\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&=\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }\\ &=\displaystyle \frac{5}{-2}\\ &=-\frac{5}{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}&=\frac{(\alpha -2)+(\beta -2)}{(\alpha -2).(\beta -2)}\\ &=\displaystyle \frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4}\\ &=\displaystyle \frac{(-1)-4}{(-2)-2(-1)+4}\\ &=\displaystyle \frac{-5}{-2+2+4}\\ &=-\frac{5}{4} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&=\frac{\alpha +\beta }{\alpha \beta }\\ &=\displaystyle \frac{(-1)}{(-2)}\\ &=\frac{1}{2} \end{aligned}&\begin{aligned}\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}&=\frac{\alpha ^{3}+\beta ^{3}}{(\alpha \beta )^{2}}\\ &=\displaystyle \frac{(\alpha +\beta )^{3}-3\alpha \beta (\alpha +\beta )}{(\alpha \beta )^{2}}\\ &=.... \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Diketahu}i\: \: p \: \: \textrm{dan}\: \: q \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\\ &x^{2}+2x-5=0,\: \: \textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad p^{2}+q^{2}&\textrm{e}.\quad (p-3)^{2}+(q-3)^{2}\\ \textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&\textrm{f}.\quad p^{2}q+pq^{2}\\ \textrm{c}.\quad p^{3}+q^{3}&\textrm{g}.\quad (p+q)^{2}-(p-q)^{2}\\ \textrm{d}.\quad p^{3}-q^{3}&\textrm{h}.\quad (p^{3}+q^{3})-(p^{3}-q^{3})\\ \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}+2x-5=0\left\{\begin{matrix} p \\ \\ q \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=2\\ c=-5 \end{matrix}\right.\\ & \end{aligned}}\\\hline \begin{aligned}\textrm{a}.\quad p^{2}+q^{2}&=(p+q)^{2}-2pq\\ &=(-\frac{b}{a})^{2}-2\left ( \frac{c}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{2}-2\left ( \frac{(-5)}{1} \right )\\ &=4+10\\ &=14 \end{aligned}&\begin{aligned}\textrm{e}.\quad &(p-3)^{2}+(q-3)^{2}\\ &=p^{2}-6p+9+q^{2}-6q+9\\ &=p^{2}+q^{2}-6(p+q)+18\\ &=14-6(-2)+18\\ &=14+12+18\\ &=44 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&=\frac{p^{2}+q^{2}}{pq}\\ &=\displaystyle \frac{14}{-5}\\ &=-\frac{14}{5} \end{aligned}&\begin{aligned}\textrm{e}.\quad p^{2}q+pq^{2}&=pq(p+q)\\ &=(-5)(-2)\\ &=10\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad p^{3}+q^{3}&=(p+q)^{3}-3pq(p+q)\\ &=\left ( -\frac{b}{a} \right )^{3}-3\left ( \frac{c}{a} \right )\left ( -\frac{b}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{3}-3\left ( \frac{(-5)}{1} \right )\left ( -\frac{2}{1} \right )\\ &=-8-30\\ &=-38\\ & \end{aligned}&\begin{aligned}\textrm{d}.\quad p^{3}&-q^{3}\\ &=(p-q)^{3}+3pq(p-q)\\ &=\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )^{^{3}}+3\left ( \frac{c}{a} \right )\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )\\ &=\left ( \displaystyle \frac{\sqrt{2^{2}-4.1.(-5)}}{1} \right )^{3}+3\left ( \frac{-5}{1} \right )....\\ &=....\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Untuk jawaban soal yang lain silahkan coba sendiri sebagai latihan}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{14}&\textrm{Tentukanlah akar-akar persamaan kuadrat dengan rumus kuadrat}\\ &\begin{array}{ll}\\ \textrm{a}.\quad x^{2}-2=0&\textrm{f}.\quad 2p^{2}-5p-12=0\\ \textrm{b}.\quad x^{2}+3x-1=0&\textrm{g}.\quad 3q^{2}-11q+10=0\\ \textrm{c}.\quad x^{2}+2x-3=0&\textrm{h}.\quad 4x^{2}+11x+6=0\\ \textrm{d}.\quad x^{2}+5x-6=0&\textrm{i}.\quad 5z^{2}-z-4=0\\ \textrm{e}.\quad x^{2}-7x-8=0&\textrm{j}.\quad 6x^{2}+17x+7=0 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad x^{2}&-2=0\\ &\begin{cases} a & =1 \\ b & =0 \\ c & =-2 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x_{1,2}&=\displaystyle \frac{0\pm \sqrt{0^{2}-4.1.(-2)}}{2.1}\\ &=\displaystyle \frac{\pm \sqrt{8}}{2}=\frac{\pm \sqrt{4.2}}{2}\\ &=\displaystyle \frac{\pm 2\sqrt{2}}{2}\\ &=\pm \sqrt{2}\\ x_{1}&=\sqrt{2}\quad \textrm{atau}\quad x_{2}=-\sqrt{2}\end{aligned}&\begin{aligned}\textrm{i}.\quad 5z^{2}&-z-4=0\\ &\begin{cases} a & =5 \\ b & =-1 \\ c & =-4 \end{cases}\\ z_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ z_{1,2}&=\displaystyle \frac{-(-1)\pm \sqrt{(-1)^{2}-4.5.(-4)}}{2.5}\\ &=\displaystyle \frac{1\pm \sqrt{1+80}}{10}=\frac{1\pm \sqrt{81}}{10}\\ &=\displaystyle \frac{1\pm 9}{10}\\ z_{1}&=\displaystyle \frac{1+9}{10}=1\quad \textrm{atau}\quad z_{2}=\frac{1-9}{10}=\frac{-8}{10}=-\frac{4}{5}\\ &\end{aligned}\\\hline \end{array}.

Soal yang belum dibahas silahkan dikerjakan sendiri sebagai latihan.

\begin{array}{ll}\\ \fbox{15}&\textrm{Carilah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\end{array}\\ \textrm{Jawab}:\\ \begin{aligned}\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}&=\frac{2}{x^{2}-10x-69}\\ \displaystyle \frac{1}{\left (x^{2}-10x-37 \right )+8}+\frac{1}{\left (x^{2}-10x-37 \right )-8}&=\frac{2}{\left (x^{2}-10x-37 \right )-32}\\ \textrm{Misalkan}\: \: x^{2}-10x-37=p, \: \: \textrm{maka}\qquad&\\ \displaystyle \frac{1}{p+8}+\frac{1}{p-8}&=\frac{2}{p-32}\\ \displaystyle \frac{p-8+p+8}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{2p}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{p}{p^{2}-64}&=\frac{1}{p-32}\\ p^{2}-32p&=p^{2}-64\\ p&=\displaystyle \frac{-64}{-32}\\ p&=2,\quad \textnormal{kita kembali ke bentuk semula}\\ x^{2}-10x-37&=2\\ x^{2}-10x-39&=0\\ (x-13)(x+3)&=0\\ x=13\: \: \textrm{atau}\: \: x=-3& \end{aligned}\\ \textrm{Jadi},\: \: x=13.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Tunjukkan bahwa untuk}\: \: m\in \textrm{rasional},\: \textrm{maka kedua akar persamaan}\\ &\textrm{a}.\quad x^{2}+(m+2)x+2m=0,\: \textrm{adalah rasional juga}\\ &\textrm{b}.\quad 2x^{2}+(m+4)x+(m-1)=0,\: \textrm{selalu memiliki dua akar real yang berlainan}\\ &\textrm{c}.\quad x^{2}+(m+4)x-2m^{2}-m+3=0,\: \textrm{selalu memiliki dua akar real dan rasional}\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textrm{Persamaan kuadrat}}\\\hline x^{2}+(m+2)x+2m=0&2x^{2}+(m+4)x+(m-1)=0&x^{2}+(m+4)x-2m^{2}-m+3=0\\\hline \begin{aligned}a&=1,\: b=(m+2),\: c=2m \end{aligned}&\begin{aligned}a&=2,\: b=m+4,\: c=m-1 \end{aligned}&\begin{aligned}a&=1,\: b=m+4,\: c=-2m^{2}-m+3 \end{aligned}\\\hline \multicolumn{3}{|c|}{\textrm{Jenis-jenis akar berdasarkan nilai}\: D,\: \textrm{di mana}\: D=b^{2}-4ac}\\\hline \begin{aligned}D&=(m+2)^{2}-4.1.(2m)\\ &=m^{2}+4m+4-8m\\ &=m^{2}-4m+4\\ &=(m-2)^{2} \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.2.(m-1)\\ &=m^{2}+8m+16-8m+8\\ &=m^{2}+24\\ & \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.1.(-2m^{2}-m+3)\\ &=m^{2}+8m+16+8m^{2}+4m-12\\ &=9m^{2}+12m+4\\ &=(3m+2)^{2} \end{aligned}\\\hline \multicolumn{3}{|c|}{\textrm{Kesimpulan}}\\\hline \begin{aligned}&\textrm{2 akar rasional} \end{aligned}&\begin{aligned}&\textrm{2 akar real dan berbeda} \end{aligned}&\begin{aligned}&\textrm{2 akar rasional} \end{aligned}\\\hline \end{array}.

 

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Tambahan Contoh Soal 6 (Bentuk Pangkat, Akar, dan Logaritma)

\begin{array}{ll}\\ \fbox{46}&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\\ &\textrm{b}.\quad ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &\textrm{c}.\quad \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &\textrm{d}.\quad ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &\textrm{e}.\quad^{3}\log 54+\: ^{3}\log 6-2.\, ^{3}\log 2\\ &\textrm{f}.\quad 2.\, ^{5}\log 15+\: ^{5}\log 4-2.\, ^{5}\log 6\end{array}\\\\\\ \textrm{Jawab}:\\\\.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}&=\: ^{6}\log 8\times \frac{9}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=2\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}&=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \displaystyle \frac{1}{6}.\, ^{2}\log &25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=-\frac{1}{3} \end{aligned}&\begin{aligned}\textrm{d}.\quad ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )&=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=-2\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{47}&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=3\sqrt{5} \end{aligned}&\begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=1\\ &\\ &\\ &\\ &\end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ \fbox{48}.&\textrm{Diketahui bahwa}\: \: ^{^{3}}\log 7=m,\: \textrm{maka nilai untuk}\\ &\textrm{a}.\quad ^{^{27}}\log \sqrt[3]{\displaystyle \frac{1}{49}}\\ &\textrm{b}.\quad \displaystyle \frac{^{^{5}}\log \sqrt{27}}{^{^{5}}\log 7\sqrt{7}}\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{l|l}\\ \begin{aligned}\textrm{a}.\quad ^{^{27}}\log \sqrt[3]{\displaystyle \frac{1}{49}}&=\: ^{^{3^{3}}}\log 7^{^{-\frac{2}{3}}}\\ &=\displaystyle \frac{\left ( -\frac{2}{9} \right )}{3}.\: ^{^{3}}\log 7\\ &=-\frac{2}{9}m\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \displaystyle \frac{^{^{5}}\log \sqrt{27}}{^{^{5}}\log 7\sqrt{7}}&=\: ^{^{7\sqrt{7}}}\log \sqrt{27}\\ &=\: ^{^{7^{\frac{3}{2}}}}\log 3^{^{\frac{3}{2}}}\\ &=\displaystyle \frac{\frac{3}{2}}{\frac{3}{2}}\: ^{7}\log 3\\ &=\displaystyle \frac{1}{^{^{3}}\log 7}\\ &=\displaystyle \frac{1}{m} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{49}.&\textrm{Diketahui}\: \: ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\: \: \textrm{maka}\: \: ^{^{44}}\log 66=....\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}&\textnormal{titik-titik dipilih dengan memperhatikan}\: \: numerus\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}&\textnormal{dipilih}\: \: numerus\: \: 11\: \: \textrm{yang memiliki bilangan basis 3}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\frac{1+p+pq}{2+pq} \end{aligned}.

\begin{array}{ll}\\ \fbox{50}.&\textrm{(AIME 1984)Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy \end{array}\\\\\\ \textrm{Jawab}:\\\\.

\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5&\Rightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7&\Rightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &&-----------\: \: \\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|&\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|&==\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\\ &&-------------\: \:\ominus \\ &&-\frac{8}{9}.\: ^{^{^{2}}}\log x \: \: \: \: \: \: \: \: =\frac{5}{3}-7=-\frac{16}{3} \end{aligned}.

\begin{aligned}^{^{^{2}}}\log x&=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ ^{^{^{2}}}\log x&=6\\ x&=2^{6}=64........(3)\\ \textrm{Sehingga}&,\: \textrm{dengan menggunakan persamaan 1 kita akan mendapatkan}\\ \frac{1}{3}.\: ^{^{^{2}}}\log x&+\: ^{^{^{2}}}\log y=5\: \: \Rightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ \frac{1}{3}.6&+\: ^{^{^{2}}}\log y=5\\ &\: \: \: \: \: \: \: ^{^{^{2}}}\log y=5-2=3\\ &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y=2^{3}=8\\ \textrm{Jadi}, \: \: \textrm{nilai dari}&\: \: xy=64\times 8=512\end{aligned}.

 

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Dirgahayu Indonesia ang ke-71

Bersukurlah kita sebagai warga negara Indonesia yang telah menghirup udara kemerdekaan yang telah lebih dari 7 dasawarsa tepatnya ke-71. Mari kita isi kemerdekaan ini yang sudah 71 tahun dengan berbagai aktivitas yang poitif yang kita tujukan demi kecintaan kita pada bumi pertiwi ini. Ke depan semoga bangsa ini menjadi bangsa yang tambah makmur dan sejahtera, amin.

Salam merdeka dan sukses untuk kita semua.

Berikut

Youtube yel-yel paskibra MA Futuhiyah Jeketro 2016

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Lanjutan Kumpulan Contoh Soal Integral (3)

Lihat Contoh Soal sebelumnya di sini

\begin{array}{ll}\\ \fbox{26}.&\textrm{Tentukanah integral berikut}?\\ &\textrm{a}.\quad \int dt\\ &\textrm{b}.\quad \int 4\: dw\\ &\textrm{c}.\quad \int \left ( x^{3}+5 \right )\: dx\\ &\textrm{d}.\quad \int \left ( x^{\frac{3}{2}}-2\sqrt{x}+1 \right )dx\\ &\textrm{e}.\quad \int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx\\ \end{array}\\\\\\ \textrm{Jawab}:\\ \begin{array}{l}\\ \begin{aligned}\textrm{a}.\quad \int dt=t+c \end{aligned}\\ \begin{aligned}\textrm{b}.\quad \int 4\: dw=4w+C \end{aligned}\\ \begin{aligned}\textrm{c}.\quad \int (x^{3}+5)\: dx&=\displaystyle \frac{1}{3+1}x^{3+1}+5x+C\\ &=\displaystyle \frac{1}{4}x^{4}+5x+C \end{aligned}\\ \begin{aligned}\textrm{d}.\quad \int (x^{\frac{3}{2}}+2\sqrt{x}+1)\: dx&=\int \left (x^{\frac{3}{2}}+2x^{\frac{1}{2}}+1 \right )\: dx\\ &=\displaystyle \frac{1}{\frac{3}{2}+1}x^{\frac{3}{2}+1}+\frac{2}{\frac{1}{2}+1}x^{\frac{1}{2}+1} +x+C\\ &=\displaystyle \frac{1}{\frac{5}{2}}x^{\frac{5}{2}}+\frac{2}{\frac{3}{2}}x^{\frac{3}{2}}+x+C\\ &=\frac{2}{5}x^{\frac{5}{2}}+\frac{4}{3}x^{\frac{3}{2}}+x+C\\ &\quad \textrm{atau dapat juga kita menyatakan dengan}\\ &=\frac{2}{5}x^{2\frac{1}{2}}+\frac{4}{3}x^{1\frac{1}{2}}+x+C\\ &=\frac{2}{5}x^{2}\sqrt{x}+\frac{4}{3}x\sqrt{x}+x+C \end{aligned}\\ \begin{aligned}\textrm{e}.\quad \int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx&=.....\\ &\quad \textrm{silahkan dicoba sendiri sebagai latihan}\end{aligned} \end{array}.

Untuk No. 26. e  pembahasannya di sini

\begin{array}{ll}\\ \fbox{27}.&\textrm{Tentukanlah integral dari}\: \: \int (x+2)\: dx\: \: \textrm{dengan cara}\\ &\textrm{a})\quad \textrm{substitusi}\\ &\textrm{b})\quad \textrm{parsial}\end{array}\\\\\\ \textrm{Jawab}:\\\\.

\begin{aligned}\textbf{Cara substitusi}&\\ &\\ \int (x+2)\: dx&=........?\\ &\textrm{Misalkan}\: \: m=x+2\\ &\qquad\qquad \, dm=dx\\ \textrm{maka},&\\ \int (x+2)\: dx&=\int \underset{\begin{matrix} |\\ \textbf{m} \end{matrix}}{\underbrace{(x+2)}}.\underset{\begin{matrix} |\\ \textbf{dm} \end{matrix}}{\underbrace{dx}}\\ &=\frac{1}{2}m^{2}+C=\frac{1}{2}(x+2)^{2}+C\\ &=\frac{1}{2}\left ( x^{2}+4x+4 \right )+C=\frac{1}{2}x^{2}+2x+\underset{\begin{matrix} |\\ \textbf{C} \end{matrix}}{\underbrace{2+C}}\\ &=\frac{1}{2}x^{2}+2x+C \end{aligned}.

\begin{aligned}\textbf{Cara parsial}&\\ &\\ \int (x+2)\: dx&=\int \underset{\begin{matrix} |\\ \textbf{u} \end{matrix}}{\underbrace{1}}.\underset{\begin{matrix} |\\ \textbf{dv} \end{matrix}}{\underbrace{(x+2)\: dx}}\left\{\begin{matrix} u=1\quad \rightarrow \quad du=0\qquad \\ \\ v=\underset{\begin{matrix} |\\ =\frac{1}{2}x^{2}+2x+C \end{matrix}}{\int dv}\quad \leftarrow \quad dv=(x+2)dx \end{matrix}\right.\\ &=\textbf{u.v}-\int \textbf{v.du}\\ &=1.\left (\frac{1}{2}x^{2}+2x+C \right )-\int \left (\frac{1}{2}x^{2}+2x+C \right ).0=\frac{1}{2}x^{2}+2x+C \end{aligned}.

atau

dapat kita bandingkan sebagai berikut:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textbf{Cara biasa}&\\ &\\\int (x+2)\: dx&=\frac{1}{2}x^{2}+2x+C \end{aligned}}\\\hline \begin{aligned}\textbf{Cara substitusi}&\\ &\\ \int (x+2)\: dx&=........?\\ &\textrm{Misalkan}\: \: m=x+2\\ &\qquad\qquad \, dm=dx\\ \textrm{maka},&\\ \int (x+2)\: dx&=\int \underset{\begin{matrix} |\\ \textbf{m} \end{matrix}}{\underbrace{(x+2)}}.\underset{\begin{matrix} |\\ \textbf{dm} \end{matrix}}{\underbrace{dx}}\\ &=\frac{1}{2}m^{2}+C\\ &=\frac{1}{2}(x+2)^{2}+C\\ &=\frac{1}{2}\left ( x^{2}+4x+4 \right )+C\\ &=\frac{1}{2}x^{2}+2x+\underset{\begin{matrix} |\\ \textbf{C} \end{matrix}}{\underbrace{2+C}}\\ &=\frac{1}{2}x^{2}+2x+C \end{aligned} &\begin{aligned}\textbf{Cara parsial}&\\ &\\ \int (x+2)\: dx&=\int \underset{\begin{matrix} |\\ \textbf{u} \end{matrix}}{\underbrace{1}}.\underset{\begin{matrix} |\\ \textbf{dv} \end{matrix}}{\underbrace{(x+2)\: dx}}\left\{\begin{matrix} u=1\quad \rightarrow \quad du=0\qquad \\ \\ v=\underset{\begin{matrix} |\\ =\frac{1}{2}x^{2}+2x+C \end{matrix}}{\int dv}\quad \leftarrow \quad dv=(x+2)dx \end{matrix}\right.\\ &=\textbf{u.v}-\int \textbf{v.du}\\ &=1.\left (\frac{1}{2}x^{2}+2x+C \right )-\int \left (\frac{1}{2}x^{2}+2x+C \right ).0\\ &=\frac{1}{2}x^{2}+2x+C \end{aligned} \\\hline \end{array}.

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Statistika (KTSP MA/SMA Kelas XI)

A. Pengertian Statistika

Statistika adalah suatu cabang ilmu dari matematika yang mempelajari cara pengumpulan data, penyusunan data, penyajian data serta pengelolaan data untuk dianalisa.

Beberapa istilah penting dalam statistika di antaranya sebagai berikut:

\begin{tabular}{|l|p{11.5cm}|}\hline Istilah&Pengertian dan atau Penjelasan\\\hline Statistika&Lihat pengertian di atas\\ Statistik&Hasil pengolahan data.\\ Statistika deskriptif&Statistika baik yang berkenaan dengan kegiatan pengumpulan, penyajian, penyederhanaan atau penganalisaan, serta penentuan khusus dari suatu data tanpa penarikan suatu kesimpulan.\\ populasi&Keseluruhan objek yang akan diteliti.\\ Sampel (Contoh)&Bagian dari populasi yang diamati.\\ Data&Kumpulan dari datum.\\ Datum&Informasi atau catatan keterangan dari penelitian.\\ Data kualitatif&Data yang menunjukkan sifat atau kondisi objek.\\ Data kuantitatif&Data yang menunjukkan jumlah objek.\\ Data ukuran (Data kontinu)&Data yang diperoleh dengan cara mengukur besaran objek.\\ Data cacahan (Data diskrit)&Data yang diperoleh dengan cara mencacah, membilang atau menghitung banyak objek.\\\hline \end{tabular}.

B. Statistika Deskriptif (Data Tunggal)

\begin{array}{|ll|ll|}\hline \multicolumn{2}{|l|}{\textrm{Ukuran Pemusatan(Tendensi Sentral)}}&\multicolumn{2}{|l|}{\textrm{Ukuran Penyebaran(Dispersi)}}\\\hline \multicolumn{2}{|l|}{\textrm{Mean}\left ( \bar{x} \right )/\textrm{Rataan hitung}}&\multicolumn{2}{|l|}{\textrm{Jangkauan}\left ( J \right )/\textrm{Rentang}\left ( R \right )}\\ &\bar{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}&&J=R=x_{maks}-x_{min}\\\hline \multicolumn{2}{|l|}{\textrm{Median}\left ( M_{e} \right )/\textrm{Nilai datum tengah}}&\multicolumn{2}{|l|}{\textrm{Hamparan/Jangkauan antarkuartil}}\\ &\textrm{Data Ganjil}:\quad M_{e}=x_{\frac{n+1}{2}}&&H=Q_{3}-Q_{1}\\ &\textrm{Data Genap}:\quad M_{e}=\displaystyle \frac{1}{2}\left ( x_{\frac{n}{2}}+x_{\frac{n}{2}+1} \right )&&\\\hline \multicolumn{2}{|l|}{\textrm{Modus}\left (M_{o} \right )}&\multicolumn{2}{|l|}{\textrm{Simpangan kuartil}\left ( Q_{d} \right )}\\ &M_{o}:\: \textrm{Nilai datum dengan frekuensi terbesar} &&Q_{q}=\displaystyle \frac{1}{2}H\\\hline \multicolumn{2}{|l|}{\textrm{Kuartil}\left ( Q\right )}&\multicolumn{2}{|l|}{\textrm{Langkah}\left ( L \right )}\\ &\textrm{Data Ganjil}:\quad \begin{cases} Q_{1}= & x_{\frac{1}{4}(n+1)}\\ Q_{2}= & x_{\frac{2}{4}(n+1)}\\ Q_{3}= & x_{\frac{3}{4}(n+1)} \end{cases}&&L=\displaystyle \frac{3}{2}H\\ &\textrm{Data Genap}:\quad \begin{cases} Q_{1}= & x_{\frac{1}{4}n+\frac{1}{2}}\\ Q_{2}= & x_{\frac{2}{4}n+\frac{1}{2}}\\ Q_{3}= & x_{\frac{3}{4}n+\frac{1}{2}} \end{cases}&&\\\hline &&\multicolumn{2}{|l|}{\textrm{Pagar dalam dan Pagar luar}}\\\hline \end{array}.

\begin{array}{|ll|ll|}\hline \multicolumn{2}{|l|}{\textrm{Ukuran Pemusatan(Tendensi Sentral)}}&\multicolumn{2}{|l|}{\textrm{Ukuran Penyebaran(Dispersi)}}\\\hline &&\multicolumn{2}{|l|}{\textrm{Pagar dalam dan Pagar luar}}\\ &&&\bullet \: \textrm{Pagar dalam}:Q_{1}-L\\ &&&\bullet \: \textrm{Pagar luar}:Q_{3}+L\\ &&&\begin{cases} \textrm{Data} & \textrm{normal } \\ &:Q_{1}-L\leq x_{i}\leq Q_{3}+L\\ \textrm{Data} & \textrm{tak normal(pencilan)}\\ &:\begin{cases} x_{i}<Q_{1}-L \\ x_{i}>Q_{3}+L \end{cases} \end{cases}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Simpangan rata}\left ( SR \right )}\\ &&&SR=\displaystyle \frac{\sum \left | x_{i}-\bar{x} \right |}{n}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Ragam/varians}\left ( s^{2} \right )}\\ &&&s^{2}=\displaystyle \frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Simpangan baku}\left ( s=\sqrt{s^{2}} \right )}\\\hline \end{array}.

C. Penyajian Data

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Penyajian Data}}\\\hline \textrm{Bentuk Diagram}&\textrm{BentukDaftar Distribusi Frekuensi}\\\hline \begin{aligned}\blacklozenge &\: \textrm{Diagram garis}\\ \blacklozenge &\: \textrm{Diagram batang daun}\\ \blacklozenge &\: \textrm{Diagram kotak garis}\\ &\\ & \end{aligned}&\begin{aligned}\blacklozenge &\: \textrm{Daftar distribusi data tunggal}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi data berkelompok}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi relatif}\\ \blacklozenge &\: \textrm{Daftar distribusi kumulatif}\\ \blacklozenge &\: \textrm{Histogram, poligon frekuensi, dan ogif }\end{aligned}\\\hline \end{array}.

D. Statistika Deskriptif (Data Berkelompok)

\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textbf{Data Deskriptif Berkelompok}}\\\hline \textrm{Rataan Hitung}\left ( \bar{\textrm{x}} \right )&\textrm{Modus}\left ( \textrm{M}_{o} \right )&\textrm{Kuartil}\left ( \textrm{Q} \right )\\\hline \bar{\textrm{x}}=\bar{x}_{s}+\displaystyle \frac{\sum f_{i}.d_{i}}{\sum f_{i}}&\textrm{M}_{o}=t_{p}+\displaystyle p\left ( \frac{d_{1}}{d_{1}+d_{2}} \right )&\textrm{Q}_{i}=t_{p}+\displaystyle p\left ( \frac{\displaystyle \frac{i.n}{4}-\sum f_{i}}{f_{b}} \right )\\\hline \multicolumn{3}{|c|}{\textbf{Keterangan}}\\\hline \begin{aligned}\bar{x}_{s}=&\textrm{rataan sementara}\\ x_{i}=&\textrm{titik tengahinterval}\\ &\textrm{kelas ke}-i\\ d_{i}=&x_{i}-\bar{x}_{s}\\ f_{i}=&\textrm{frekuensi kelas ke}-i\\ &\\ &\\ &\\ &\\ &\end{aligned}&\begin{aligned}t_{p}=&\textrm{tepi bawah kelas modus}\\ p=&\textrm{panjang interval kelas}\\ d_{1}=&f_{0}-f_{-1}\\ d_{2}=&f_{0}-f_{+1}\\ f_{0}=&\textrm{frekuensi kelas modus}\\ f_{-1}=&\textrm{frekuensi sebelum kelas modus}\\ f_{+1}=&\textrm{frekuensi setelah kelas modus}\\ &\\ &\\ & \end{aligned}&\begin{aligned}tp=&\textrm{tepi bawah kuartil ke}-i\\ p=&\textrm{panjang interval kelas}\\ n=&\textrm{banyaknya data}\\ \sum f_{i}=&\textrm{jumlah semua frekuensi}\\ &\textrm{sebelum kelas kurtil ke}-i\\ f_{q}=&\textrm{frekuensi kelas kuartil ke}-i\\ Q_{i}=&\textrm{kuartil ke}-i\\ Q_{1}=&\textrm{kuartil bawah}\\ Q_{2}=&\textrm{kuatil tengah/median}\\ Q_{3}=&\textrm{kuatil atas} \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\fbox{{CONTOH SOAL}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Berikut yang}\: \: bukan\: \: \textrm{merupakan ukuran penyebaran data adalah} ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \textrm{rentang}&&\textrm{d}.\quad\textrm{kuartil}\\ \textrm{b}.\quad\textrm{varians}&\textrm{c}.\quad\textrm{jarak antarkuartil}&\textrm{e}.\quad\textrm{simpangan baku}\\ &&\\ &&(\textbf{SMBB TELKOM 2006}) \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textrm{d}.\quad \textrm{karena kuartil merupakan ukuran pemusatan data}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Simpangan kuartil dari data :}\: 71, 70,68,40,45,48,52,53,53,67, 62\: \: \textrm{adalah} ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 8&&\textrm{d}.\quad 18\\ \textrm{b}.\quad 10\qquad&\textrm{c}.\quad 15\qquad &\textrm{e}.\quad 20 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\

Sumber Referensi

  1. Johanes, Kastolan, dan Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Sosial Kurikulum Berbasis Kompetensi 2004. Jakarta: Yudistira.
  2. Kanginan, Marthen, Yuza Terzalgi. 2014. Matematikauntuk SMA-SMK/SMK Kelas XI. Bandung: SEWU.
  3. Sobirin. 2006. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 2 IPA. Jakarta: Kawan Pustaka.
  4. Tampomas, Husein. 1999. Seribu Pena  Matematika SMU Jilid 2 kelas 2. Jakarta: Erlangga.
  5. Wirodikromo, Sartono. 2007. Matematika untuk SMA Kelas XI. Jakarta: Erlangga.

 

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