PAS MATEMATIKA BLOG

Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 3-\left | x \right | \right |<10\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-14\: \: \textrm{atau}\: x>12\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: x>13\\ \textrm{c}.&x<-12\: \: \textrm{atau}\: x>10\\ \textrm{d}.&0<x<10\\ \color{red}\textrm{e}.&-13<0<13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | 3-\left | x \right | \right |< 10\\ &-10< 3-\left | x \right |< 10\\ &-13< -\left | x \right |< 7\\ &-7< \left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ &0\leq \left | x \right |< 13\\ &\textrm{selanjutnya},\\ &\left | x \right |< 13\\ &-13< \: x< 13\\ &\textrm{HP}=\color{red}\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{l}\\ 37.&\textrm{(UM UGM 05)}\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x^{2}-3 \right |<2x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<3\\ \textrm{b}.&-3<x<1\\ \color{red}\textrm{c}.&1<x<3\\ \textrm{d}.&-3<x<-1\: \: \textrm{atau}\: \: 1<x<3\\ \textrm{e}.&x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | x^{2}-3 \right |&<2x\\ -2x<\left ( x^{2}-3 \right )&<2x\\ \textrm{dipartisi men}&\textrm{jadi dua bagian}\\ \bullet \quad\textrm{pertama}\qquad&\\ (x^{2}-3)&>-2x\\ x^{2}+2x-3&>0\\ (x+3)(x-1)&>0\\ x<-3\: \: \textrm{atau}&\: \: x>1\\ \bullet \quad \textrm{kedua}\qquad\quad&\\ \left ( x^{2}-3 \right )&<2x\\ x^{2}-2x-3&<0\\ (x-3)(x+1)&<0\\ -1<x<3&\\ \color{black}\textrm{ambil yang}&\: \color{black}\textrm{memenuhi keduanya}\\ \textrm{berupa iris}&\textrm{an}\\ \textrm{HP}=&\color{red}\left \{ 1<x<3,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&(\textrm{SPMB 05})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-2 \right |^{2}<4\left | x-2 \right |+12\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}|2\leq x\leq 8 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}|4<x< 8 \right \}\\ \color{red}\textrm{c}.&\left \{ x\in \mathbb{R}|-4<x< 8 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}|-2<x<4 \right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}|2<x<4 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misalkan}\: \: p&=\left | x-2 \right |,\: \: \textrm{selanjutnya}\\ \left | x-2 \right |^{2}<&\, 4\left | x-2 \right |+12\\ p^{2}<&\, 4p+12\\ p^{2}-&4p-12<0\\ (p-6)&(p+2)<0\\ -2<p&<6,\: \: \color{magenta}\textrm{atau jika dikembalikan}\\ -2<&\left | x-2 \right |<6,\\ &\: \: \color{black}\textrm{ingat, nilanya tidak negatif}\\ 0\leq &\left | x-2 \right |<6\\ -6<&\: x-2<6\\ -4<&\: x<8\\ \textrm{HP}=&\color{red}\left \{ -4<x<8,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Diketahui grafik fungsi}\: \: f(x)=mx^{2}-2mx+m\\ & \textrm{berada di atas grafik fungsi}\\ &g(x)=2x^{2}-3,\: \textrm{maka nilai}\: \: m\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&m>2&&&\textrm{d}.&-6<m<2\\ \textrm{b}.&m>6&\textrm{c}.&2<m<6&\textrm{e}.&m<-6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}f(x)&=g(x)\\ mx^{2}-2mx+m&=2x^{2}-3\\ mx^{2}-2x^{2}-2mx+m+3&=0\\ \textrm{Supaya grafik}\: \: f(x)\: \textrm{berada }&\color{blue}\textrm{di atasnya}, \\ \textrm{maka}\: \: D=B^{2}-4AC&<0\\ (m-2)x^{2}-2mx+(m+3)&=0\begin{cases} A &=m-2 \\ B &=-2m \\ C &=m+3 \end{cases}\\ B^{2}-4AC&<0\\ (-2m)^{2}-4(m-2)(m+3)&<0\\ 4m^{2}-4\left ( m^{2}+m-6 \right )&<0\\ 4m^{2}-4m^{2}-4m+24&<0\\ -4m+24&<0\\ m-6&>0\\ m&\color{red}>6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: \color{red}3<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2x-2&&&\textrm{d}.&-2\\\\ \textrm{b}.&2\quad&\textrm{c}.&8-2x\quad&\textrm{e}.&2x-8\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}\\ &=\left | x-3 \right |-\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}3<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=\color{red}2x-8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: \color{red}1<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2&&&\textrm{d}.&5\\\\ \textrm{b}.&3\quad&\textrm{c}.&4\quad&\textrm{e}.&6\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-1)^{2}}+\sqrt{(x-5)^{2}}\\ &=\left | x-1 \right |+\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}1<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-1 \right |=(x-1) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-1 \right |+\left | x-5 \right |=(x-1)+\left ( -(x-5) \right )\\ &=x-1+5-x\\ &=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 42.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\leq x\leq \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{b}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{3}{4}< x\leq 2\\ \textrm{c}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: \: x\neq \frac{3}{4}\\ \textrm{d}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \color{red}\textrm{e}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ -1\leq \displaystyle \frac{5}{4x-3}&\leq 1,\: \: \color{magenta}\textbf{jika dibalik}\\ -1\geq \displaystyle \frac{4x-3}{5}&\geq 1,\: \: \color{magenta}\textbf{bentuk ini tidak}\\ \color{magenta}\textbf{dibolehkan}&\: \color{magenta}\textbf{maka perlu diubah menjadi}\\ -1\geq \displaystyle \frac{4x-3}{5}\: \: \textrm{atau}&\: \: \displaystyle \frac{4x-3}{5}\geq 1,\: \: \color{black}\textrm{selanjutnya}\\ \bullet \quad \textrm{bagian}&\: 1\\ -1&\geq \displaystyle \frac{4x-3}{5}\Leftrightarrow \frac{4x-3}{5}\leq -1\\ 4x-3&\leq -5\\ 4x&\leq -2\\ x&\leq -\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{4x-3}{5}&\geq 1\\ 4x-3&\geq 5\\ 4x&\geq 8\\ x&\geq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&(\textrm{UMPTN 95})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2}{2x-1} \right |> 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x> 2\\ \textrm{b}.&x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{c}.&x<-1\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{d}.&-1<x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{e}.&x<-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{semua opsi bukan jawaban}\\ &\textbf{Berikut pembahasannya}\\ &\begin{aligned}\left | \displaystyle \frac{2}{2x-1} \right |&> 1\\ -1>\displaystyle \frac{2}{2x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2}{2x-1}>1,\: \color{magenta}\textbf{dibalik}\\ -1<\displaystyle \frac{2x-1}{2}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x-1}{2}<1\\ \bullet \quad \textrm{bagian}&\: 1\\ \displaystyle \frac{2x-1}{2}&>-1\\ 2x-1&>-2\\ 2x&>-1\\ x&>-\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{2x-1}{2}&<1\\ 2x-1&<2\\ 2x&<3\\ x&<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&(\textrm{UMPTN 00})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2x+7}{x-1} \right |\geq 1\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-2\leq x\leq 8\\ \textrm{b}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&-8\leq x< 1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&-2\leq x< 1\: \: \textrm{atau}\: \: 1< x\leq 8\\ \color{red}\textrm{e}.&x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{2x+7}{x-1} \right |&\geq 1\\ -1\geq \displaystyle \frac{2x+7}{x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x+7}{x-1}\geq 1\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{2x+7}{x-1}&\leq -1\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{2x+7}{x-1}&+1\leq 0\\ &\displaystyle \frac{2x+7+(x-1)}{x-1}\leq 0\\ \displaystyle \frac{3x+6}{x-1}&\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| -2\leq x< 1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{2x+7}{x-1}&\geq 1\\ \displaystyle \frac{2x+7}{x-1}&-1\geq 0\\ &\displaystyle \frac{2x+7-(x-1)}{x-1}\geq 0\\ \displaystyle \frac{x+8}{x-1}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x> 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{x-2}{x+3} \right |\leq 2\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-8\leq x< -3\\ \textrm{b}.&-8\leq x< -1\\ \textrm{c}.&-4\leq x< -3\\ \color{red}\textrm{d}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3}\\ \textrm{e}.&x\leq -4\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | \displaystyle \frac{x-2}{x+3} \right |&\leq 2\\ -2\leq \displaystyle \frac{x-2}{x+3}&\: \: \textrm{atau}\: \: \displaystyle \frac{x+2}{x+3}\leq 2\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{x-2}{x+3}&\geq -2\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{x-2}{x+3}&+2\geq 0\\ &\displaystyle \frac{x-2+2(x+3)}{x+3}\geq 0\\ \displaystyle \frac{3x+4}{x+3}&\geq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -3\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{x-2}{x+3}&\leq 2\\ \displaystyle \frac{x-2}{x+3}&-2\leq 0\\ &\displaystyle \frac{x-2-2(x+3)}{x+3}\leq 0\\ \displaystyle \frac{-x-8}{x+3}&\leq 0,\: \: \color{magenta}\textbf{koefisien \textit{x} negatif}\\ \displaystyle \frac{x+8}{x+3}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>-3,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 26.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & (x+3)(x-1)\geq (x-1)\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&1\leq x\leq 3&\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 1\\ \textrm{c}.&-3\leq x\leq -1\\ \textrm{d}.&-2\geq x\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-1\geq x\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}(x+3)(x-1)&\geq (x-1)\\ (x+3)(x-1)-(x-1)&\geq 0\\ (x-1)\left ( (x+3)-1 \right )&\geq 0\\ (x-1)(x+2)&\geq 0\\ \end{aligned}\\ &\textrm{Sehingga solusinya adalah:}\\ &\color{red}x\leq -2\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 1 \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x+3 \right |<2\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\left \{ x|x<-\displaystyle \frac{5}{3} \right \}\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \}\\ \textrm{c}.&\left \{ x|x\geq -11 \right \} \\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\ \textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x+3 \right |<2\left | x-4 \right |\\ &\left ( x+3 \right )^{2}<2^{2}\left ( x-4 \right )^{2}\\ & \textrm{dikuadratkan masing-masing ruas}\\ &x^{2}+6x+9<4\left ( x^{2}-8x+16 \right )\\ &x^{2}-4x^{2}+6x+32x+9-64<0\\ &-3x^{2}+38x-55<0\\ &3x^{2}-38x+55>0\\ &\left ( 3x-5 \right )\left (x -11 \right )>0\\\\ &\textrm{Berikut untuk}\: \textrm{garis bilangannya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Himpunan penyelesaian pertidaksamaan}\\ & \left | x^{2}+5x \right |\leq 6\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad \left \{ x|-6\leq x\leq 1 \right \}\\ &\textrm{b}.\quad \left \{ x|-3\leq x\leq -2 \right \}\\ &\textrm{c}.\quad \left \{ x|-6\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 1\right \}\\ &\textrm{d}.\quad \left \{ x|-6\leq x\leq -5 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\ &\textrm{e}.\quad \left \{ x|-5\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui bahwa}\\ &\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}\\ &\begin{array}{|c|c|}\hline \begin{aligned}&-6\leq x^{2}+5x\\ &x^{2}+5x+6\geq 0\\ &(x+3)(x+2)\geq 0 \end{aligned}&\begin{aligned}&x^{2}+5x\leq 6\\ &x^{2}+5x-6\leq 0\\ &(x+6)(x-1)\leq 0 \end{aligned}\\\hline \textbf{Lihat Gambar 1}&\textbf{Lihat Gambar 2}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ & \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-12< x< 6\\ \textrm{b}.&-30\leq x\leq 6\\ \color{red}\textrm{c}.&x\geq 6\: \: \textrm{atau}\: x\leq -30\\ \textrm{d}.&x<6\: \: \textrm{atau}\: \: x<-30\\ \textrm{e}.&x\leq 6\: \: \textrm{atau}\: \: x\geq -30\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | \displaystyle \frac{1}{2}x+6 \right |\geq 9\\ &\displaystyle \frac{1}{2}x+6\leq -9\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x+6\geq 9\\ &\displaystyle \frac{1}{2}x\leq -9-6\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 9-6\\ &\displaystyle \frac{1}{2}x\leq -15\: \: \textrm{atau}\: \: \displaystyle \frac{1}{2}x\geq 3\\ &\color{red}x\leq -30\: \: \color{black}\textrm{atau}\: \: \color{red}x\geq 6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &3\left | x+1 \right |\leq \left | x-2 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\leq x\leq -\frac{1}{4}\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{2}\leq x\leq \frac{5}{2}\\ \textrm{c}.&x\leq \displaystyle \frac{1}{4}\: \: \textrm{atau}\: x\geq \frac{5}{2}\\ \textrm{d}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq \frac{1}{4}\\ \textrm{e}.&x\leq -\displaystyle \frac{5}{2}\: \: \textrm{atau}\: x\geq -\frac{1}{4}\end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&3\left | x+1 \right |\leq \left | x-2 \right |\\ &\left (3\left | x+1 \right | \right )^{2}\leq \left (\left | x-2 \right | \right )^{2}\\ &\left ( 3x+3 \right )^{2}\leq \left (x-2 \right )^{2}\\ &(3x+3+(x-2))(3x+3-(x-2))\leq 0\\ &(4x+1)(2x+5)\leq 0\\ &\textrm{HP}=\color{red}\left \{ x|-\displaystyle \frac{5}{2}\leq x\leq -\frac{1}{4},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{l}\\ 31.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-3 \right |<3\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \textrm{b}.&-3<x<3\\ \textrm{c}.&x<-3\: \: \textrm{atau}\: x<3\\ \color{red}\textrm{d}.&x>0\: \: \textrm{atau}\: x<6\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x<6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x-3 \right |<3\\ &-3<(x-3)<3\\ &-3+3<x<3+3\\ &\color{red}0<x<6 \end{aligned} \end{array}$

$\begin{array}{l}\\ 32.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x+4 \right |>8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x>-8\\ \textrm{b}.&x<4\: \: \textrm{atau}\: x>12\\ \textrm{c}.&x>4\: \: \textrm{atau}\: x>-12\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: x<6\\ \color{red}\textrm{e}.&x>4\: \: \textrm{atau}\: x<-12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x+4 \right |>8\\ &(x+4)<-8\: \: \textrm{atau}\: \: (x+4)>8\\ &\color{red}x<-12\: \: \color{black}\textrm{atau}\: \: \color{red}x>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{HP}=\left \{ x|-7<x<\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{b}.&\textrm{HP}=\left \{ x|x<-7\: \: \textrm{atau}\: \: x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\textrm{HP}=\left \{ x|x>-7,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\textrm{HP}=\left \{ x|-1<x<2,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\textrm{HP}=\left \{ x|x<-1\: \: \textrm{atau}\: \: x>2,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\left | \displaystyle \frac{x+1}{2} \right |>\left | \displaystyle \frac{x-2}{3} \right |\\ &\left ( \displaystyle \frac{x+1}{2} \right )^{2}>\left ( \displaystyle \frac{x-2}{3} \right )^{2}\\ &\left ( \displaystyle \frac{x+1}{2}+\frac{x-2}{3} \right )\left ( \displaystyle \frac{x+1}{2}-\displaystyle \frac{x-2}{3} \right )>0\\ &\left ( \displaystyle \frac{3(x+1)+2(x-2)}{6} \right )\left ( \displaystyle \frac{3(x+1)-2(x-2)}{6} \right )>0\\ &\left ( \displaystyle \frac{5x-1}{6} \right )\left ( \displaystyle \frac{x+7}{6} \right )>0\\ &\textrm{HP}=\color{red}\left \{ x|x<-7\: \: \color{black}\textrm{atau}\: \: \color{red}x>\displaystyle \frac{1}{5},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Penyelesaian dari pertidaksamaan}\\ &\left | \displaystyle \frac{3-2x}{-5} \right |>5\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-11\: \: \textrm{atau}\: x>14\\ \textrm{b}.&x<-14\: \: \textrm{atau}\: x>11\\ \textrm{c}.&11<x<14\\ \textrm{d}.&-14<x<-11\\ \textrm{e}.&x>14 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | \displaystyle \frac{3-2x}{-5} \right |>5\\ &\displaystyle \frac{3-2x}{-5}<-5\: \: \textrm{atau}\: \: \displaystyle \frac{3-2x}{-5}>5\\ &\displaystyle \frac{2x-3}{5}>5\: \: \textrm{atau}\: \: \displaystyle \frac{2x-3}{5}<-5\\ &2x-3>25\: \: \textrm{atau}\: \: 2x-3<-25\\ &2x>25+3\: \: \textrm{atau}\: \: 2x<-25+3\\ &x>14\: \: \textrm{atau}\: \: x<-11,\\ &\textrm{dapat juga dituliskan}\\ &\color{red}x<-11\: \: \color{black}\textrm{atau}\: \: \color{red}x>14 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 2-2\left | x+1 \right | \right |>4\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-4\: \: \textrm{atau}\: x>2\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: x>1\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: x>0\\ \textrm{d}.&x<-1\: \: \textrm{atau}\: x>3\\ \textrm{e}.&x<0\: \: \textrm{atau}\: x>4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\left | 2-2\left | x+1 \right | \right |>4\\ &2-2\left | x+1 \right |<-4\: \: \textrm{atau}\: \: 2-2\left | x+1 \right |>4\\ &-2\left | x+1 \right |<-6\: \: \textrm{atau}\: \: -2\left | x+1 \right |>2\\ &\left | x+1 \right |>3\: \: \textrm{atau}\: \: \left | x+1 \right |<-1\\ &\left\{\begin{matrix} (x+1)<-3\\ (x+1)>3 \end{matrix}\right.\: \: \textrm{atau}\: \: \left\{\begin{matrix} \left | x+1 \right |<-1\\ \color{red}\textbf{tak mungkin} \end{matrix}\right.\\ &\textrm{Selanjutnya}\: \textrm{akan didapatkan}\\ &\color{red}x<-4\: \: \color{black}\textrm{atau}\: \: \color{red}x>2 \end{aligned} \end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 16.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=\color{red}1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=\color{red}13 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | x-1 \right |=2x+1\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}&\textrm{c}.&\left \{ -1 \right \}&\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | x-1 \right |=2x+1\\ &(x-1)=\pm (2x+1)\\ &(x-1)= \begin{cases} +(2x+1) \\\\ -(2x+1) \end{cases}\\ &\begin{aligned}&\\ \color{blue}\textrm{Syarat}&:\\ (x-1)&\begin{cases} x-1\geq 0 \Leftrightarrow x\geq 1\\\\ x-1<0 \Leftrightarrow x<1 \end{cases}\\ \end{aligned}\\ &\begin{array}{|c|c|}\hline x\geq 1&x< 1\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(x-1)&=+(2x+1)\\ x-2x&=1+1\\ -x&=2\\ x&=-2 \end{aligned}&\begin{aligned}(x-1)&=-(2x+1)\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{tidak memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x+1 \right |=2x+9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ 8 \right \}&&\textrm{e}.&\textrm{setiap bilangan real}\\ \textrm{c}.&\left \{ -2,8 \right \}& \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+1 \right |=2x+9\\ &(3x+1)=\pm (2x+9)\\ &(3x+1)= \begin{cases} +(2x+9) \\\\ -(2x+9) \end{cases}\\ \end{aligned}\\ &\begin{aligned} \color{blue}\textrm{Syarat}\: \: &:\\ (3x+1)&\begin{cases} 3x+1\geq 0 \Leftrightarrow x\geq -\displaystyle \frac{1}{3}\\\\ 3x+1<0 \Leftrightarrow x<-\displaystyle \frac{1}{3} \end{cases}\\ & \end{aligned} \\ &\begin{array}{|c|c|}\hline x\geq -\displaystyle \frac{1}{3}&x< -\displaystyle \frac{1}{3}\\\hline \textrm{Proses}&\textrm{Proses}\\\hline \begin{aligned}(3x+1)&=+(2x+9)\\ 3x-2x&=9-1\\ x&=8\\ & \end{aligned}&\begin{aligned}(3x+1)&=-(2x+9)\\ 3x+2x&=-9-1\\ 5x&=-10\\ x&=-2 \end{aligned}\\\hline \color{red}\textbf{memenuhi}&\color{red}\textbf{memenuhi}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jumlah akar-akar dari}\: \: x^{2}+\left | x \right |-6=0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-1\\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&4\\\\ &&&(\textbf{Entrance Examination}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x^{2}+\left | x \right |-6&=0\\ (\left | x \right |+3)(\left | x \right |-2)&=0\\ \left | x \right |+3=0\quad \textrm{atau}\quad \left | x \right |-2&=0\\ \left | x \right |=-3\: (\textbf{tm})\quad \textrm{atau}\quad \left | x \right |&=2\: (\textbf{mm})\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned} x&=\pm 2\begin{cases} x_{1}&=2 \\ x_{2} &=-2 \end{cases}\\ &\textrm{untuk jumlah}\: \textrm{dari akar-akarnya adalah}:\\ &x_{1}+x_{2}=2+(-2)\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Penyelesaian pertidaksamaan}\: \: x^{2}+\left | x \right |-6\leq 0\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-2\leq x< 0\\ \textrm{b}.&0\leq x\leq 2\\ \textrm{c}.&-2\leq x\leq 2\\ \textrm{d}.&-3\leq x\leq 2\\ \textrm{e}.&-2\leq x\leq 3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&\textrm{Proses penyelesaian dipecah jadi 2 bagian}\\ &\textrm{yaitu}:\begin{cases} x & \geq 0 \\ x & <0 \end{cases}\\ &\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\begin{array}{|c|c|}\hline (1)&(2)\\\hline x\geq 0&x<0\\\hline \textrm{maka}\: \: \left | x \right |=x&\textrm{maka}\: \: \left | x \right |=-x\\\hline \begin{aligned}&x^{2}+(x)-6\leq 0\\ &x^{2}+x-6\leq 0\\ &(x+3)(x-2)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-3\leq x\leq 2\\ &\textrm{karena}\quad x\geq 0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}0\leq x\leq 2 \end{aligned}&\begin{aligned}&x^{2}+(-x)-6\leq 0\\ &x^{2}-x-6\leq 0\\ &(x+2)(x-3)\leq 0\\ &\color{red}\textrm{Selesaian}:\\ &-2\leq x\leq 3\\ &\textrm{karena}\quad x<0,\\ &\textrm{maka penyelesaian}\\ &\textrm{menjadi}\\ &\color{blue}-2\leq x<0 \end{aligned}\\\hline \end{array}\\ &\textrm{Gabungan dari penyelesaian (1) dan (2)}\\ &\textrm{adalah}:\quad \color{red}-2\leq x\leq 2 \end{aligned}\\\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&\textrm{Diketahui pertidaksamaan}:\: x^{2}+\left | x \right |-6\leq 0\\ &\textrm{dan perlu diingat pula bahwa}:\quad \color{red}\left | x \right |\geq 0\\ &\textrm{diubah menjadi}:\quad \left | x \right |^{2}+\left | x \right |-6\leq 0\\ &\Leftrightarrow (\left | x \right |+3)(\left | x \right |-2)\leq 0\\ &\Leftrightarrow -3\leq \left | x \right |\leq 2\\ &\textrm{karena}\quad \left | x \right |\geq 0,\: \textrm{maka}\\ &0\leq \left | x \right |\leq 2\Rightarrow \left | x \right |\leq 2\\ &\textrm{Sehingga penyelesaian menjadi}\\ &\color{red}-2\leq x\leq 2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 21.&\textrm{Seluruh bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1 }\\ &\textrm{adalah}\: ... .\\ &\begin{array}{lllllll}\\ \textrm{a}.&3<x<4&\\ \textrm{b}.&2<x<3 \\ \textrm{c}.&2<x<4 \\ \textrm{d}.&3<x<5\\ \textrm{e}.&1<x<3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Seluruh }\textrm{bilangan bilangan real}\: \: x\\ &\textrm{yang jaraknya terhadap 3 kurang dari 1, }\\ &\textrm{maksudnya adalah}:\\ &\left | x-3 \right |<1\\ &\Leftrightarrow -1<x-3<1\\ &\Leftrightarrow -1+\textbf{(3)}<x-3+\textbf{(3)}<1+\textbf{(3)}\\ &\Leftrightarrow \color{red}2<x<4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Pernyataan berikut yang tepat adalah}\: ... .\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{b}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<1\\ \textrm{c}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & -3<m<-2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{d}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\\ & 2<m<3\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{e}.&\textrm{pilihan jawaban baik a, b, c,}\\ &\textrm{maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikanlah opsi}\: \: \textbf{a}\: ,\\ \left | m \right |&<2\\ -2<m&<2\\ \textrm{sehing}&\textrm{ga untuk nilai}\: \: m\in \mathbb{R}\: \: \textrm{pada rentang}\\ -1<m&<2\: \: \: \: \textrm{akan memenuhi semua} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 2x-9 \right |< 3\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-3\leq x\leq 6&\\ \textrm{b}.&-3<x<6\\ \textrm{c}.&3<x<6\\ \textrm{d}.&3\leq x\leq 6\\ \textrm{e}.&-3<x<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 2x-9 \right |< 3\\ &\Leftrightarrow -3<2x-9<3\\ &-3+(9)<2x-9+(9)<3+(9)\\ &\textnormal{masing-masing ditambah 9}\\ &\textnormal{dan akan menjadi bentuk}\\ &6<2x<12\\ &\color{blue}6.\left ( \displaystyle \frac{1}{2} \right )<2x.\left ( \displaystyle \frac{1}{2} \right )<12.\left ( \displaystyle \frac{1}{2} \right )\\ &\textnormal{masing-masing dikali}\: \: \displaystyle \frac{1}{2}\\ &\textnormal{dan akan berubah menjadi bentuk}\\ &\color{red}3<x<6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\\ & \left | 3x+5 \right |\geq 19\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&x\leq -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x\geq 8&\\\\ \textrm{b}.&x<-8\: \: \textrm{atau}\: \: x>\displaystyle \frac{14}{3}\\\\ \textrm{c}.&x\leq -8\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{14}{3}\\\\ \textrm{d}.&x< -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x> 8\\\\ \textrm{e}.&x\leq 8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{14}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\left | 3x+5 \right |\geq 19\\\\ &(\ast )-19\geq 3x+5\quad \textrm{atau}&(\ast \ast )\: \: 3x+5\geq 19\\ &-19-5\geq 3x\quad \textrm{atau}&3x\geq 19-5\\ &-\displaystyle \frac{24}{3}\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &-8\geq x\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ &x\leq \color{red}-8\quad \color{black}\textrm{atau}&x\color{red}\geq \displaystyle \frac{14}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi}\\ & 25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\: \: \textrm{adalah}... .\\\\ &\qquad (\textbf{NUS Entrance Examination A level})\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}:\\ &\begin{aligned}&25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\\ &25-5\left | 2x+1 \right |\geq 20\left | 2x-1 \right |\\ &5-\left | 2x+1 \right |\geq 4\left | 2x-1 \right |\\ &\color{red}\textrm{ilustrasinya}\quad \color{black}\begin{array}{llllllllll} &&&&&&\\\hline &&-\frac{1}{2}&&&\frac{1}{2}&& \end{array}\\ &\textrm{dan berikut}\: \textrm{pembagian wilayahnya}\\ &\begin{array}{|c|c|c|}\hline -\infty < x< -\displaystyle \frac{1}{2}&-\displaystyle \frac{1}{2}\leq x< \frac{1}{2}&\displaystyle \frac{1}{2}\leq x< \infty \\\hline \begin{cases} \left | 2x+1 \right | &=-(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=+(2x-1) \end{cases}\\\hline \end{array}\\ &\textrm{Selanjutnya adalah} \end{aligned} \end{array}$

$.\qquad\begin{array}{|c|c|c|}\hline \begin{aligned}\: -\infty < x<& -\frac{1}{2}\: ,\\ 25-\left | 10x+5 \right |&\geq \left | 40x-20 \right |\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(-(2x+1))&\geq 4(-(2x-1))\\ 5+2x+1&\geq -8x+4\\ 10x&\geq -2\\ x&\geq -\frac{2}{10}\quad (\textbf{tm})\\ & \end{aligned}&\begin{aligned} \: -\frac{1}{2} \leq x<& \frac{1}{2}\: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1)&\geq 4(-(2x-1))\\ 5-2x-1&\geq -8x+4\\ 6x&\geq 0\\ x&\geq 0\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ 0\leq x< \frac{1}{2} \right \} \end{aligned} &\begin{aligned}\: \frac{1}{2}\leq x< &\infty \: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1))&\geq 4(2x-1)\\ 5-2x-1&\geq 8x-4\\ -10x&\geq -8\\ x&\leq \frac{8}{10}\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ \frac{1}{2}\leq x\leq \frac{4}{5} \right \} \end{aligned}\\\hline \end{array}$

$.\qquad\begin{aligned}&\\ &\textrm{Sehingga yang memenuhi}\: \textrm{adalah}:\\ &=\color{red}\left \{ 0\leq x\leq \displaystyle \frac{4}{5} \right \} \end{aligned}$.

sumber soal di sini dan di sini

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 11.&\textrm{Perhatikanlah ilustrasi grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan yang memenuhi rumus tersebut }\\ &\textrm{adalah}\: ... .\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Dengan cara substitusi langsung kita}\\ & \textrm{akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4\\ &\qquad \textrm{dan sampai langkah di sini hanya ada }\\ &\qquad \textrm{1 persamaan yang memenuhi yaitu}:\\ &\qquad y=\color{red}\left | -2x+4 \right | \end{aligned} \end{array}$.

$\begin{array}{lll}\\ 12.&\textrm{Gambarlah garfik untuk persamaan}\\ & \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\left | x \right |+\left | y \right |=4\\ &\textrm{maka untuk} \end{aligned}\\&\begin{array}{|cc|cc|}\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Perhatikanlah ilustrasinya grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}$.

$\begin{array}{lll}\\ 13.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{yang memenuhi}\: \: x+\left | x \right |+y=5\\ & \textrm{dan}\: \: x+\left | y \right |-y=10\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\\ &\begin{aligned} &\left | x \right |+x+y=5\\ &\: \: \qquad \textrm{dan}\\ &x+\left | y \right |-y=10\\ & \end{aligned} \end{aligned} \end{array}$.

$.\: \quad\begin{aligned}\begin{aligned}&\textrm{untuk}:\: \: x> 0\: ,\: y> 0\quad \textbf{(kuadran I)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5 \\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: (\textrm{ada}) \end{cases}\\ &\textrm{untuk}:\: \: x<0\: ,\: y> 0\quad \textbf{(kuadran II)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\\ &\textbf{(tidak memenuhi)} \end{cases} \end{aligned} \end{aligned}$.

$.\: \quad\begin{aligned}\begin{aligned}&\textrm{untuk}:\: \: x<0\: ,\: y<0\quad \textbf{(kuadran III)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\\ & \textbf{(tidak memenuhi)}\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\ &\textrm{untuk}:\: \: x> 0\: ,\: y<0\quad \textbf{(kuadran IV)}\\ &\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5\\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases} \end{aligned} \end{aligned}$.

$.\quad\begin{array}{ll}\\ &\textrm{Sebagai ilustrasinya, berikut grafiknya}\\ &\begin{aligned}\end{aligned}\end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Gambarlah grafik fungsi mutlak berikut}\\ &\textrm{a}.\quad y=\left | x-2 \right |\\ &\textrm{b}.\quad y=-\left | x-2 \right |\\ &\textrm{c}.\quad y=2+\left | x-2 \right |\\ &\textrm{d}.\quad y=2-\left | x-2 \right |\\ &\textrm{e}.\quad y=\left | 2+\left | x-2 \right | \right |\\ &\textrm{f}.\quad y=\left | 2-\left | x-2 \right | \right | \end{array}$.

$.\quad\begin{array}{ll}\\ &\textbf{Jawab}\quad :\\ &\textrm{Berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{a} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Dan berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{b} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Dan berikut pula ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{c} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Dan berikut pula ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{d} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Dan sebagai ilustrasinya}\\ &\textrm{lihat soal no}\: \: \textbf{c} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Dan terakhir berikut ilustrasinya}\\ &\textrm{grafik dari soal}\: \: \textbf{f} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Penyelesaian yang memenuhi untuk}\\ & \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=\color{red}1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=\color{red}13 \end{cases} \end{aligned} \end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 1.&\textrm{Nilai untuk}\: \: -\left | -2021 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2021\quad &&&\textrm{d}.&-2021^{-1}\\ \textrm{b}.&2021&\textrm{c}.&2021^{-1}\quad &\textrm{e}.&2021^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{red}\begin{aligned}-\left | -2021 \right |&=-\left ( 2021 \right )=-2021 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Nilai untuk}\: \: \left | -4 \right |-\left | -6^{2}\times 2 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-68\quad &&&\textrm{d}.&68\\ \textrm{b}.&-40&\textrm{c}.&40\quad &\textrm{e}.&76 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\left | -4 \right |-\left | -6^{2}\times 2 \right |&=\left ( 4 \right )-\left | -36\times 2 \right |\\ &=4-\left | -72 \right |\\ &=4-\left ( 72 \right )\\ &=\color{red}-68 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Nilai untuk}\: \: (-2022)-\left | -(-2021) \right |-\left | -3^{2} \right |^{2}=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-3960\quad &&&\textrm{d}.&-4068\\ \textrm{b}.&-4038&\textrm{c}.&-4050\quad &\textrm{e}.&-4124 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&(-2022)-\left | -(-2021) \right |-\left | -3^{2} \right |^{2}\\ &=-2022-2021-\left | -9 \right |^{2}\\ &=-4043-9^{2}\\ &=-4043-81\\ &=-4124 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai untuk}\: \: \left | -4 \right |^{\left | -2 \right |}-\left | -2 \right |^{\left | -4 \right |}=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-32\quad &&&\textrm{d}.&16\\ \textrm{b}.&-16&\textrm{c}.&0\quad &\textrm{e}.&32 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\left | -4 \right |^{\left | -2 \right |}-\left | -2 \right |^{\left | -4 \right |}&=4^{2}-2^{4}\\ &=16-16\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai untuk}\\ & \left | 1-2^{2} \right |+\left | 2^{2}-3^{2} \right |+\left | 3^{2}-4^{2} \right |+\cdots +\left | 2020^{2}-2021^{2} \right |...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2021^{2}\quad &&&\textrm{d}.&2021^{2}-1\\ \textrm{b}.&1-2021^{2}&\textrm{c}.&-2021\quad &\textrm{e}.&2021^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | 1-2^{2} \right |+\left | 2^{2}-3^{2} \right |+\left | 3^{2}-4^{2} \right |+\cdots +\left | 2020^{2}-2021^{2} \right |\\ &=\left ( 2^{2}-1 \right )+\left ( 3^{2}-2^{2} \right )+\left ( 4^{2}-3^{2} \right )+\cdots +\left ( 2021^{2}-2020^{2} \right )\\ &=-1+2021^{2}\\ &=\color{red}2021^{2}-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Nilai untuk}\\ & \left | \displaystyle \frac{1}{2}-1 \right |\times \left | \displaystyle \frac{1}{3}-1 \right |\times \left | \displaystyle \frac{1}{4}-1 \right |\times \left | \displaystyle \frac{1}{5}-1 \right |\times \cdots \times \left | \displaystyle \frac{1}{2021}-1 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-\displaystyle \frac{1}{2021}\quad &&&\textrm{d}.&\displaystyle \frac{2020}{2021}\\\\ \textrm{b}.&-\displaystyle \frac{2020}{2021}&\textrm{c}.&-2021\displaystyle \frac{1}{2}\quad &\textrm{e}.&\displaystyle \frac{1}{2021} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | \displaystyle \frac{1}{2}-1 \right |\times \left | \frac{1}{3}-1 \right |\times \left | \frac{1}{4}-1 \right |\times \left | \frac{1}{5}-1 \right |\times \cdots \times \left | \frac{1}{2021}-1 \right |\\ &=\left | -\displaystyle \frac{1}{2} \right |\times \left | -\frac{2}{3} \right |\times \left | -\frac{3}{4} \right |\times \left | -\frac{4}{5} \right |\times \cdots \times \left | -\frac{2020}{2021} \right |\\ &=\displaystyle \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \cdots \times \frac{2019}{2020}\times \frac{2020}{2021}\\ &=\color{red}\displaystyle \frac{1}{2021} \end{aligned} \end{array}$.

$\begin{array}{ll} 7.&\textrm{Bentuk sederhana dari}\: \: x-5y\: \: \textrm{dan}\\ &5y-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll} \textrm{a}.&\left | 5y-x \right |&\textrm{d}.&\left | y-x \right |\\ \textrm{b}.&\left | y-5x \right |&\textrm{e}.&\left |-5y-x \right |\\ \textrm{c}.&\left | x-5y \right | \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\left | x-5y \right |=m=\begin{cases} \bullet & =x-5y \\ \bullet & =-(x-5y) \end{cases} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: \left | 4m \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | 4m \right |&=16\\ (4m)&=\pm 16\\ m&=\pm \displaystyle \frac{16}{4}\\ &=\color{red}\pm 4 \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi untuk}\: \: \left | 2x+5 \right |=9\\ & \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&2&&\textrm{d}.&-7\: \: \textrm{dan}\: \: 2\\ \textrm{b}.&2\: \: \textrm{dan}\: \: 7&&\textrm{e}.&-2\: \: \textrm{dan}\: \: 7\\ \textrm{c}.&-7\: \: \textrm{dan}\: \: -2& \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | 2x+5 \right |&=9\\ \left ( 2x+5 \right )&=\pm 9\\ 2x&=\pm 9-5\\ x&=\displaystyle \frac{\pm 9-5}{2}\\ x&=\color{red}\begin{cases} =\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2 \\\\ =\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5x \right |=26\\ & \textrm{adalah}\: ... .\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}&&\textrm{d}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{b}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}&&\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \textrm{c}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\quad&\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}10-4\left | 4-5x \right |&=-26\\ -4\left | 4-5x \right |&=-36\\ \left | 4-5x \right |&=9\\ (4-5x)&=\pm 9\\ -5x&=-4\pm 9\\ x&=\displaystyle \frac{-4\pm 9}{-5}\\ x&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =\color{red}-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=\color{red}2\frac{3}{5} \end{cases} \end{aligned} \end{array}$

Problem Solving Bentuk Bilangan Riil

Seri Pemecahan Masalah

Jika pada bahasan sebelumnya kita bahas bilangan tidak nyata atau bilangan imajiner pada akar persamaan kuadrat, sekarang kita ketengahkan bahasan sebaliknya, yaitu akar nyta atau riil dari suatu persamaan kuadrat.

Berikut permasalahannya

(sumber soal dari blog saya sendiri di wordpress)

$\color{blue}\begin{aligned}&\textrm{Akar riil terbesar untuk persamaan}\\ &\color{black}\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^{2}-11x-4\\ &\textrm{adalah}\: \: p+\sqrt{q+\sqrt{r}}\: \: \textrm{dengan}\: p,\: q,\: \textrm{dan}\: r\: \textrm{adalah}\\ &\textrm{bilangan asli}.\: \: \textrm{Tentukanlah nilai}\: \: p+q+r\\\\ &\color{black}\textbf{Solusi}:\\ \end{aligned}$.

$\begin{aligned}&\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^{2}-11x-4\\ &\frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1=x^{2}-11x\\ &\frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19}=x^{2}-11x\\ &\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^{2}-11x\\ &\frac{x(x-19)+x(x-3)}{(x-3)(x-19)}+\frac{x(x-17)+x(x-5)}{(x-5)(x-17)}=x^{2}-11x\\ &\frac{2x^{2}-22x}{x^{2}-22x+57}+\frac{2x^{2}-22}{x^{2}-22x+85}=x^{2}-11x\\ &\left ( x^{2}-11x \right )\left ( \frac{2}{x^{2}-22x+57}+\frac{2}{x^{2}-22x+85} \right )\\ &\qquad\qquad\qquad\qquad =x^{2}-11x,\quad \color{red}\textrm{misal}\: \: t=x^{2}-22x\\ &\left ( \frac{2}{t+57}+\frac{2}{t+85} \right )=\frac{x^{2}-11x}{x^{2}-11x}=1\\ &2\left ( t+85 \right )+2\left ( t+57 \right )=(t+57)(t+85)\\ &2t+170+2t+114=t^{2}+142t+4845\\ &0=t^{2}+138t+4731\\ &\color{red}t^{2}+138t+4731=0\: \: \left\{\begin{matrix} a=1\\ b=138\\ c=4731 \end{matrix}\right.\\ &t_{1,2}=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ &t_{1,2}=\displaystyle \frac{-138\pm \sqrt{138^{2}-4.1.4731}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{19044-18924}}{2}\\ &=\displaystyle \frac{-138\pm \sqrt{120}}{2}\\ &=\displaystyle \frac{-138\pm 2\sqrt{30}}{2}\\ &=-69\pm \sqrt{30} \end{aligned}$.

$\color{red}\begin{aligned}&\color{black}\textrm{Selanjutnya}\\ &t_{1,2}=-69\pm \sqrt{30}\\ &x^{2}-22x=-69\pm \sqrt{30}\\ &x^{2}-22x+69\pm \sqrt{30}=0\\ &x^{2}-22x+69+\sqrt{30}=0\\ &\textrm{atau}\quad x^{2}-22x+69-\sqrt{30}\\ &\\ &\color{black}\textrm{dengan cara yang} \: \: \color{black}\textrm{semisal diatas}\\ &\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69+\sqrt{30} \right )}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{22^{2}-4\left ( 69-\sqrt{30} \right )}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{484-276+4\sqrt{30}}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm \sqrt{208-4\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm \sqrt{208+4\sqrt{30}}}{2}\\ &x_{1,2}=\displaystyle \frac{22\pm 2\sqrt{52-\sqrt{30}}}{2}\\ &\textrm{atau}\qquad x_{3,4}=\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}\\ &x_{1,2}=11\pm \sqrt{52-\sqrt{30}}\\ &\textrm{atau}\qquad x_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ &\\ &\color{black}\textrm{Maka}, \\ &\left\{\begin{matrix} x_{1}=11+\sqrt{52-\sqrt{30}}\\ \\ x_{2}=11-\sqrt{52-\sqrt{30}} \end{matrix}\right.\\ &\textrm{atau}\qquad \left\{\begin{matrix} x_{3}=11+\sqrt{52+\sqrt{30}}\\ \\ x_{4}=11-\sqrt{52+\sqrt{30}} \end{matrix}\right. \end{aligned}$.

$\begin{aligned}\textrm{Selanjutnya nilai}&\: \textrm{yang paling pas sesuai soal adalah}:\\ &\color{red}x_{3}=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \textrm{Sehingga nilai}\: \: \: \: \, \, &p+q+r=11+52+30=93 \end{aligned}$.

Problem Solving Bentuk Bilangan Imajiner (Bilangan Tidak Nyata)

Seri Pemecahan Masalah

Suatu ketika saya sharing-sharing mengenai soal bentuk perpangkatan dari salah seorang teman yang kebetulan memang soalnya membuat penasayaran untuk ditemukan jawabannya.

Berikut soalnya

Saat saya melihat soalnya dengan pangkat berupa angka yang seolah berpola tapi agak susah dicari hungan antara keduanya. Yang satu bilangan utuh yang satu lagi bentuk pecahan (bilangan pada soal, bukan pada yang diketahui). Tapi ada sedikit petunjuk yang mensiratkan soal di atas akan segera dapat dipecahkankan, yaitu posisi yang diketahui $x+\displaystyle \frac{1}{x}=-1$ adalah salah satu bentuk persamaan kuadrat dengan akar kemungkinan rasional atau imajiner/khayal/tidak nyata dan pangkat pada soal yang semuanya menunjukkan kelipatan 3, yaitu 1234567891011 dan yang satunya posisi penyebut dengan pangkat 1110987654321 dengan basis/bilangan pokok perpangkatannya sama dengan yang diketahui dari soal yaitu $a$.

Sebelumnya saya pernah menyinggung mengenai istilah definit positif dan definit negatif (silahkan klik di sini) yang kurang lebih istilah tersebut sangat berkaitan dengan akar persamaan kuadrat yang berbentuk imajiner.

Ok, kita kembali ke arah penyelesaian soal di atas, yaitu:

$\begin{aligned}&a+\displaystyle \frac{1}{a}=-1\: \Leftrightarrow\: a^{2}+1=-a\\ &\Leftrightarrow a^{2}+a+1=0\\ &\Leftrightarrow a_{1,2}=\color{red}\displaystyle \frac{-1\pm \sqrt{-3}}{2}=\displaystyle \frac{-1\pm \sqrt{3.(-1)}}{2}\\ &\: \quad\qquad =\displaystyle \frac{-1\pm \sqrt{3}\sqrt{-1}}{2}=\frac{-1\pm \sqrt{3}i}{2}\\ &\: \quad\qquad \: \textrm{dengan}\: \: i=\sqrt{-1} \end{aligned}$.

$\begin{aligned}&\textrm{Misalkan kita pilih}\: \: a=\color{red}\displaystyle \frac{-1+ \sqrt{3}i}{2}\\ &\textrm{maka nilai dari}\\ &\displaystyle \frac{1}{a}=\displaystyle \frac{1}{\displaystyle \frac{-1+ \sqrt{3}i}{2}}=\displaystyle \frac{2}{-1+ \sqrt{3}i}=\displaystyle \frac{2}{ \sqrt{3}i-1}\\ &\: \quad =\displaystyle \frac{2}{ \sqrt{3}i-1}\times \displaystyle \frac{\sqrt{3}i+1}{\sqrt{3}i+1}=\displaystyle \frac{2(\sqrt{3}i+1)}{-3-1}\\ &\: \quad= -\displaystyle \frac{2(\sqrt{3}i+1)}{-4}=\displaystyle \frac{\sqrt{3}i+1}{-2}\quad \textrm{atau}\\ &\displaystyle \frac{1}{a}=\color{blue}\displaystyle \frac{-1-\sqrt{3}i}{2} \end{aligned}$.

Penjabaran bentuk pangkat dari salah satu akar ternyata membentuk pola yang unik sebagaimana bentuk berikut:

$\begin{aligned}&\begin{cases} a &=\displaystyle \frac{-1+ \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a} & =\displaystyle \frac{-1-\sqrt{3}i}{2} \end{cases},\quad \begin{cases} a^{2} &=\displaystyle \frac{-1- \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{2}} & =\displaystyle \frac{-1+\sqrt{3}i}{2} \end{cases}\\ &\qquad\qquad\begin{cases} a^{3} &=1 \\ \displaystyle \frac{1}{a^{3}} & =1 \end{cases}\\ &\begin{cases} a^{4} &=\displaystyle \frac{-1+ \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{4}} & =\displaystyle \frac{-1-\sqrt{3}i}{2} \end{cases}\quad \begin{cases} a^{5} &=\displaystyle \frac{-1- \sqrt{3}i}{2} \\ \displaystyle \frac{1}{a^{5}} & =\displaystyle \frac{-1+\sqrt{3}i}{2} \end{cases}\\ &\qquad\qquad\begin{cases} a^{6} &=1 \\ \displaystyle \frac{1}{a^{6}} & =1 \end{cases}\\ &\: \quad\vdots \\ &\cdots \quad \cdots \quad \begin{cases} a^{9} &=1 \\ \displaystyle \frac{1}{a^{9}} & =1 \end{cases}\\ &\cdots \quad \cdots \quad \begin{cases} a^{12} &=1 \\ \displaystyle \frac{1}{a^{12}} & =1 \end{cases}\\ &\cdots \quad \cdots \quad \begin{cases} a^{15} &=1 \\ \displaystyle \frac{1}{a^{15}} & =1 \end{cases}\\ &\textrm{dan seterusnya}\\ & \end{aligned}$.

Jadi, setiap pangkat kelipatan 3 ternyata sama dengan 1, sehingga ini mengakibatkan soal di atas dapat dituliskan lagi dengan

$\begin{aligned}&\color{red}\textrm{Perhatikan lagi bentuk soal}\\ &a^{1234567891011}+\displaystyle \frac{1}{a^{11100987654321}}\\ &=a^{3m}+\displaystyle \frac{1}{a^{3n}}=1+\displaystyle \frac{1}{1}=1+1=\color{red}2 \end{aligned}$.

Contoh Soal 13 Statistika

$\begin{array}{ll} 56.&\textrm{Simpangan baku dari data berikut}:\\ &6,7,4,5,3\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{2}\quad &\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\quad&\textrm{e}.&\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,4,5,3\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+4+5+3}{5}=\displaystyle \frac{25}{5}=\color{red}5 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((6-5)^{2}+(7-5)^{2}+(4-5)^{2}+(5-5)^{2}+(3-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(1^{2}+2^{2}+1^{2}+0+2^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 1+4+1+4 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(10)}=\sqrt{\displaystyle \frac{10}{5}}=\color{red}\sqrt{2}\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 57.&\textbf{UN 2010}\\ &\textrm{Simpangan baku dari data berikut}:\\ &2,3,4,5,6\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\sqrt{15}&&&\textrm{d}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \sqrt{10}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&\sqrt{2} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,3,4,5,6\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+3+4+5+6}{5}=\displaystyle \frac{20}{5}=\color{red}4 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((2-5)^{2}+(3-5)^{2}+(4-5)^{2}+(5-5)^{2}+(6-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(3^{2}+2^{2}+1^{2}+0+1^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 9+4+1+1 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(15)}=\sqrt{\displaystyle \frac{15}{5}}=\color{red}\sqrt{3}\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 58.&\textrm{Simpangan baku dari data berikut}:\\ &7,9,11,13,15\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&2,4&&&\textrm{d}.&2,8\\ \textrm{b}.&\displaystyle 2,5\quad &\textrm{c}.&\displaystyle 2,7\quad&\textrm{e}.&2,9 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &7,9,11,13,15\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{7+9+11+13+15}{5}=\displaystyle \frac{55}{5}=\color{red}11 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((7-11)^{2}+(9-11)^{2}+(11-11)^{2}+(13-11)^{2}+(15-11)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(4^{2}+2^{2}+0+2^{2}+4^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 16+4+4+16 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(40)}=\sqrt{\displaystyle \frac{40}{5}}=\color{red}\sqrt{8}=2,82..\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 59.&\textrm{Simpangan baku dari data berikut}:\\ &2,4,4,5,6,6,7,8,9,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&4\sqrt{3}&&&\textrm{d}.&\displaystyle \frac{2}{5}\sqrt{30}\\ \textrm{b}.&2\displaystyle \frac{2}{5}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,4,4,5,6,6,7,8,9,9\quad \\ &\textrm{Simpangan bakunya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+4+4+5+6+6+7+8+9+9}{10}=\displaystyle \frac{60}{10}=\color{red}6 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left((2-6)^{2}+2(4-6)^{2}+(5-6)^{2}+2(6-6)^{2}+(7-6)^{2}+(8-6)^{2}+2(9-6)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left(4^{2}+2.2^{2}+1^{2}+0+1^{2}+2^{2}+2.3^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left( 16+8+1+1+4+18 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}(48)}=\sqrt{\displaystyle \frac{48}{10}}=\sqrt{\displaystyle \frac{120}{25}}=\color{red}\displaystyle \frac{2}{5}\sqrt{30}\\ \end{aligned} \end{array}$.

Contoh Soal 12 Statistika

$\begin{array}{ll} 51.&\textrm{Simpangan kuartil dari data}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \: \textrm{adalah}\: \: 1\displaystyle \frac{1}{2}.\: \textrm{Jika median datanya}\\ & \textrm{adalah}\: \: 5\displaystyle \frac{1}{2},\: \textrm{maka rata-rata data}\: \: \textrm{tersebut adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{4}&&&\textrm{d}.&\textrm{5}\displaystyle \frac{1}{2}\\ \textrm{b}.&\textrm{4}\displaystyle \frac{1}{2}\quad &\textrm{c}.&\textrm{5}\quad&\textrm{e}.&\textrm{6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \Rightarrow \: n=6\\&\color{red}\textrm{karena mediannya}=M_{e}=Q_{2}=5\frac{1}{2}=\frac{11}{2}\\&\color{blue}\textrm{ data menjadi}\: :\: \color{black}a\: \: 3\: \: 5\: \: 6\: \: 7\: \: 8\: \: \color{blue}\textrm{atau}\color{black}\: \: 3\: \: a\: \: 5\: \: 6\: \: 7\: \: 8\\ &\textrm{Simpangan kuartilnya adalah}\: 1\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2}\\ &Q_{d}=\displaystyle \frac{1}{2}\left (Q_{3}-Q_{1} \right ) =\displaystyle \frac{3}{2}\: \Rightarrow\: Q_{3}-Q_{1}=3\\&\textrm{maka}\\ &3=Q_{3}-Q_{1}=x_{._{\frac{3}{4}n+\frac{1}{2}}}-x_{._{\frac{1}{4}n+\frac{1}{2}}}=x_{._{\frac{3}{4}6+\frac{1}{2}}}-x_{._{\frac{1}{4}6+\frac{1}{2}}} \\ &\quad\: =\left (x_{._{5}}-x_{._{2}} \right )=7-x_{._{2}}=\color{red}3\: \color{black}\Rightarrow\: x_{._{2}}=4\\ &\textrm{Jadi, rata-ratanya adalah}:\\ &\overline{x}=\frac{3+4+5+6+7+8}{6}=\color{red}5,5 \end{aligned} \end{array}$.

$\begin{array}{ll} 52.&\textrm{Simpangan kuartil dari data}\\ &\textrm{berikut}\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{10}&&&\textrm{d}.&\textrm{6}\\ \textrm{b}.&\textrm{9}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{-}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\color{blue}\textrm{data durutkan}\\ &50,50,50,53,53,53,53\\ &61,61,61,61,61,70,70,70\\ &\textrm{Simpangan kuartil adalah}\: Q_{d},\\ &Q_{d}=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( 61-53 \right )=\displaystyle \frac{1}{2}\times 8=4\\ &\textrm{Jadi},\: Q_{d}=4 \end{aligned} \end{array}$.

$\begin{array}{ll} 53.&\textrm{Simpangan rata-rata dari data berikut}:\\ &6\quad 4\quad 2\quad 8\quad 10\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{2}&&&\textrm{d}.&\textrm{3},0\\ \textrm{b}.&\textrm{2},4\quad &\textrm{c}.&\textrm{2},5\quad&\textrm{e}.&\textrm{3},5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6\quad 4\quad 2\quad 8\quad 10\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+4+2+8+10}{5}=\displaystyle \frac{30}{5}=\color{red}6 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 6-6 \right|+\left| 4-6 \right|+\left| 2-6 \right|+\left| 8-6 \right|+\left| 10-6 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 0 \right|+\left| -2 \right|+\left| -4 \right|+\left| 2 \right|+\left| 4 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( 0+2+4+2+4 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}(12)=\displaystyle \frac{12}{5}=\color{red}2,4\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 54.&\textrm{Simpangan rata-rata dari data berikut}:\\ &10,8,7,10,7,5,8,6,10,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1,0}&&&\textrm{d}.&\textrm{8,0}\\ \textrm{b}.&\textrm{1,4}\quad &\textrm{c}.&\textrm{6,0}\quad&\textrm{e}.&\textrm{14,0} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &10,8,7,10,7,5,8,6,10,9\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{10+8+7+10+7+5+8+6+10+9}{10}\\ &\: \: \: \, =\displaystyle \frac{80}{10}=\color{red}8 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| 5-8 \right|+\left| 6-8 \right|+2\left| 7-8 \right|+2\left| 8-8 \right|+\left| 9-8 \right|+3\left| 10-8 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| -3 \right|+\left| -2 \right|+2\left| -1 \right|+2\left| 0 \right|+\left| 1 \right|+3\left| 2 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( 3+2+2+0+1+6 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}(14)=\displaystyle \frac{14}{10}=\color{red}1,4\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 55.&\textrm{Nilai variansi dari data}\\ &6,7,7,8,8,8,8,12\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1}&&&\textrm{d}.&\textrm{8}\\ \textrm{b}.&\displaystyle \frac{26}{8}\quad &\textrm{c}.&\displaystyle \frac{11}{4}\quad&\textrm{e}.&\textrm{22} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{c}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,7,8,8,8,8,12\quad \\ &\textrm{Variannya adalah}: \\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+7+8+8+8+8+12}{8}=\displaystyle \frac{64}{8}=\color{red}8 \\ &\textrm{maka nilai}\\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( (6-8)^{2}+2(7-6)^{2}+4(8-8)^{2}+(12-8)^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 2^{2}+2.1+4.0+4^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 4+2+0+16 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}(22)=\color{red}\displaystyle \frac{11}{4}=\color{black}2,75\\ \end{aligned} \end{array}$.

Contoh Soal 11 Statistika

$\begin{array}{ll} 46.&\textrm{Rata-rata dari data yang disajikan }\\ &\textrm{dengan hitogram berikut adalah}\: ....\\ \end{array}$.

$.\qquad\begin{array}{ll} &\begin{array}{lllllll}\\ \textrm{a}.&41,372&&&\textrm{d}.&43,135\\ \textrm{b}.&42,150\quad&\textrm{c}.&43,125\quad&\textrm{e}.&44,250 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned} &\begin{array}{|c|c|c|c|}\hline \begin{aligned}&\textrm{Berat}\\ &\textrm{Badan} \end{aligned}&\color{red}\textrm{f}_{1}&\begin{aligned}&\textrm{Nilai Tengah}\\ &\qquad(\textrm{x}_{1}) \end{aligned}&\color{red}\textrm{f}_{1}\textrm{x}_{1}\\\hline 30-34&5&32&160\\ 35-39&7&37&259\\ 40-44&12&42&504\\ 45-49&9&47&423\\ 50-54&4&52&208\\ 55-59&3&57&171\\\hline &\sum \textrm{f}_{1}=\color{red}40&&\sum \textrm{f}_{1}\textrm{x}_{1}=\color{red}1725\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: \overline{\textrm{x}}\: \: \textrm{adalah}:\\ &\overline{\textrm{x}}=\displaystyle \frac{\sum \textrm{f}_{1}\textrm{x}_{1}}{\sum \textrm{f}_{1}}=\displaystyle \frac{1725}{40}=\color{red}43,125 \end{aligned} \end{array}$.

$\begin{array}{ll} 47.&(\textbf{UN Mat IPA 2006})\\ &\textrm{Perhatikan gambar berikut}\end{array}$.

$.\qquad\begin{array}{ll} &\textrm{Berat badan pada suatu kelas tersaji dengan}\\ &\textrm{bentuk histogram seperti pada gambar di atas}\\ &\textrm{Rata-rata berat badan tersebut adalah}\: ....\: \textrm{Kg}\\ &\begin{array}{lllllll}\\ \textrm{a}.&64,5&&&\textrm{d}.&66\\ \textrm{b}.&65\quad&\textrm{c}.&65,5\quad&\textrm{e}.&66,5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned} &\begin{array}{|c|c|c|c|}\hline \begin{aligned}&\textrm{Berat}\\ &\textrm{Badan} \end{aligned}&\color{red}\textrm{f}_{1}&\begin{aligned}&\textrm{Nilai Tengah}\\ &\qquad(\textrm{x}_{1}) \end{aligned}&\color{red}\textrm{f}_{1}\textrm{x}_{1}\\\hline 50-54&4&52&208\\ 55-59&6&57&342\\ 60-64&8&62&496\\ 65-69&10&67&670\\ 70-74&8&72&576\\ 75-79&4&77&308\\\hline &\sum \textrm{f}_{1}=\color{red}40&&\sum \textrm{f}_{1}\textrm{x}_{1}=\color{red}2600\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: \overline{\textrm{x}}\: \: \textrm{adalah}:\\ &\overline{\textrm{x}}=\displaystyle \frac{\sum \textrm{f}_{1}\textrm{x}_{1}}{\sum \textrm{f}_{1}}=\displaystyle \frac{2600}{40}=\color{red}65 \end{aligned} \end{array}$.

$\begin{array}{ll} 48.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&3\\ 150-154&5\\ 155-159&17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{Kuartil bawah dapat dinyatakan dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&149,5+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{b}.&150+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{c}.&155+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{d}.&154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{e}.&155,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{Ditanyakan kuartil bawah, maka hal ini}\\ &=Q_{1}\: \Rightarrow \: x_{._{\frac{1}{4}n}}=x_{._{\frac{1}{4}50}}=x_{._{12,5}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&\color{red}3\\ 150-154&\color{red}5\\ \color{blue}155-159&\color{blue}17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{1}\: \: \textrm{adalahh}:\\ &Q_{1}=L+\left( \displaystyle \frac{\frac{1}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 \end{aligned} \end{array}$.

$\begin{array}{ll} 49.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&2\\ 10-14&8\\ 15-19&10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{Median dari tabel di atas adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&15,0&&&\textrm{d}.&16,5\\ \textrm{b}.&15,5\quad&\textrm{c}.&16,0\quad&\textrm{e}.&17,0 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{e}\\ &\begin{aligned} &\textrm{Diketahui}\: n=\color{blue}30\\ &\textrm{Ditanyakan median, maka formulanya}\\ &=Q_{2}\: \Rightarrow \: x_{._{\frac{2}{4}n}}=x_{._{\frac{1}{2}.30}}=x_{._{15}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&\color{red}2\\ 10-14&\color{red}8\\ \color{blue}15-19&\color{blue}10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{2}\: \: \textrm{adalah}:\\ &Q_{2}=L+\left( \displaystyle \frac{\frac{2}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}14,5+\left( \displaystyle \frac{15-10}{10} \right).5\\ &\quad\: \: =14,5+\frac{25}{10}=14,5+2,5=\color{red}17,0 \end{aligned} \end{array}$.

$\begin{array}{ll} 50.&\textrm{Jangkauan antarkuartil dari data}\\ &\textrm{berikut}:\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{20}&&&\textrm{d}.&\textrm{5}\\ \textrm{b}.&\textrm{10}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{a}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\color{blue}\textrm{data durutkan}\\ &25\: \: 28\: \: 35\: \: 36\: \: 38\: \: 40\: \: 42\: \: 43\\ &45\: \: 54\: \: 55\: \: 56\: \: 64\: \: 70\: \: 82\\ &\textrm{Jangkauan antarkuartil adalah}\: H,\\ &H= Q_{3}-Q_{1} \\ &\quad\: =x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}}\\ &\quad\: =x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}}\\ &\quad\: =\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =56-36\\ &\textrm{Jadi},\: H=Q_{3}-Q_{1}=\color{red}20 \end{aligned} \end{array}$.

Contoh Soal 10 Statistika

Soal sebelumnya (yaitu Contoh Soal 9 Statistika) klik di sini

$\begin{array}{ll} 41.&\textrm{Median dan modus dari data berikut}\\ &\color{red}3,6,7,8,4,5,9,6\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7 dan 5}&&&\textrm{d}.&\textrm{5 dan 6}\frac{1}{2}\\ \textrm{b}.&\textrm{6 dan 6} &\textrm{c}.&\textrm{6 dan 7}&\textrm{e}.&\textrm{5 dan 6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\textrm{Diketahui}\: \: \color{blue} n=8\\ &\textrm{Datum diurutkan dari kecil ke besar}\\ &:3,4,5,\color{red}6,6\color{black},7,8,9\\ &\bullet \: \: \textbf{modus} =M_{o}=6,\\ &\bullet \: \: \textbf{mean} =\overline{x}=6 \end{aligned} \end{array}$.

$\begin{array}{ll} 42.&\textrm{Hasil tes matematika di suatu kelas yang}\\ &\textrm{diikuti tes 49 siswa menghasilkan nilai}\\ &\textrm{rata-rata 7}.\: \textrm{Jika Andi ikut tes susulan}\\ &7,04.\: \textrm{Nilai Andi adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7,5}&&&\textrm{d}.&\textrm{9}\\ \textrm{b}.&\textrm{8} &\textrm{c}.&\textrm{8,5}&\textrm{e}.&\textrm{9,5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}\textrm{Misalkan}&\: \: \color{red}y\color{black}=\textrm{besar nilai Andi}\\ \overline{x}_{gabungan}&=\displaystyle \frac{n.\overline{x}+\color{red}y}{n+1}\\ 7,04&=\displaystyle \frac{49\times 7+\color{red}y}{49+1}\\ 7,04&=\displaystyle \frac{343+\color{red}y}{50}\\ 343+\color{red}y&=50\times \left ( 7,04 \right )\\ 343+\color{red}y&=352\\ y&=352-343\\ &=\color{red}9 \end{aligned} \end{array}$.

$\begin{array}{ll} 43.&\textrm{Mean dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&10\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{60,5}&&&\textrm{d}.&\textrm{58,5}\\ \textrm{b}.&\textrm{60} &\textrm{c}.&\textrm{59,5}&\textrm{e}.&\textrm{57} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}&\begin{array}{|c|c|c|c|c|}\hline \textrm{Ukuran}&x_{i}&d_{i}=x_{i}-x_{s}&f_{i}&f_{i}\times d_{i}\\\hline 50-54&52&-10&4&-40\\\hline 55-59&57&-5&6&-30\\\hline 60-64&62&0&10&0\\\hline \textrm{Jumlah}&&&20&-70\\\hline \end{array}\\ &\overline{x}=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}\times d_{1}}{\displaystyle \sum_{i=0}^{n}f_{i}}=62+\displaystyle \frac{-70}{20}\\ &\: \: =62-3,5=\color{red}58,5 \end{aligned} \end{array}$.

$\begin{array}{ll} 44.&\textrm{Median dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&1\\ 50-52&6\\ 53-55&6\\ 56-58&7\\ 59-61&4\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{55,5}&&&\textrm{d}.&\textrm{53,5}\\ \textrm{b}.&\textrm{55} &\textrm{c}.&\textrm{54,5}&\textrm{e}.&\textrm{53} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&\color{red}1\\ 50-52&\color{red}6\\ \color{blue}53-55&\color{blue}6\\ 56-58&7\\ 59-61&4\\\hline &24\\\hline \end{array}\\ &\textrm{Median posisi datumnya}:\\ &\textrm{datum ke}-\left ( \displaystyle \frac{24}{2} \right )=x_{._{12}}\\ &\textrm{dan terletak di interval}\\ &53-55,\: \: \textrm{dengan}\: \: f_{k}=1+6=7\\ &\textrm{serta}\: \: c=3\\ &M_{e}=Q_{2}=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\\ &M_{e}=Q_{2}=52,2+3\left ( \displaystyle \frac{\displaystyle \frac{24}{2}-7}{6} \right )\\ &\: \quad=52,5+3\left ( \displaystyle \frac{12-7}{6} \right )\\ &\: \quad=52,5+\left ( \displaystyle \frac{5}{2} \right )\\ &\: \quad=52,5+2,5\\ &\: \quad=\color{red}55 \end{aligned} \end{array}$.

$\begin{array}{ll} 45.&\textrm{Modus dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&8\\ 65-69&16\\ 70-74&10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\textrm{67,32}&&&\textrm{d}.&\textrm{70,12}\\ \textrm{b}.&\textrm{67,36} &\textrm{c}.&\textrm{67,56}&\textrm{e}.&\textrm{70,36} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&\color{red}8\\ \color{blue}65-69&\color{blue}16\\ 70-74&\color{red}10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{modus terletak pada kelas}\\ &\textrm{interval dengan frekuensi}\\ &\textrm{terbesar, yaitu}:16.\: \textrm{Kelas intervalnya}\\ &65-69,\: \: \textrm{dengan}\: \: c=5\\ &\textrm{serta}\: \: \begin{cases} \triangle _{1} & =f-f_{1}=16-8=8 \\ \triangle _{2} & =f-f_{2}=16-10=6 \end{cases}\\ &M_{o}=L+c\left ( \displaystyle \frac{\triangle _{1}}{\triangle _{1}+\triangle _{2}} \right )\\ &\quad\: \: =64,5+5\left ( \displaystyle \frac{8}{8+6} \right )\\ &\quad\: \: =64,5+\displaystyle \frac{40}{14}\\ &\quad\: \: =64,5+2,857\\ &\quad\: \: =\color{red}67,36 \end{aligned} \end{array}$.

Koefisien Keragaman (Koefisien Variansi)

A. Pengertian

Pada bahasan ini untuk membandingkan dua atau lebih distribusi data yang sejenis dapat digunakan koefisien keragaman. Koefisien variansi adalah nilai dari standar deviasi suatu data dibagi dengan rata-ratanya.

B. Formula koefisien Variansi

Jika diketahui $S$ adalah simpangan baku dan $\overline{x}$ adalah rataan hitung suatu data, maka koefidien variansinya (V) dirumuskan dengan:

$V=\displaystyle \frac{S}{\overline{x}}\times 100%$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

Contoh 1

Coba perhatikan lagi data pada halaman ini di sini, dengan datanya adalah:

$\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&47-49&50-52&53-55&56-58&59-61\\\hline \textrm{Frek}&2&4&6&5&3\\\hline \end{array}$.

Dari perhitungan untuk data tersebut didapatkan besar rataan hitungnya adalah 54,45 dan simpangan bakunya adalah 3,58, maka koefisien dari variansi dari data tersebut adalah:

$\begin{aligned}V&=\displaystyle \frac{S}{\overline{x}}\times 100\%\\ &=\displaystyle \frac{3,58}{54,45}\times 100\%\\ &=\color{red}6,57\% \end{aligned}$.

Contoh 2

Diketahui nilai ulangan matematika suatu kelas di suatu waktu memiliki rataan 78 dengan simpangan bakunya adalah 7, sedangkan untuk nilai ulangan kimia dari kelas tersebut mendapatkan rataan 62 dan simpangan bakunya adalah 6. Tentukanlah mata pelajaran mana dari keduanya yang telah diuhikan itu yang memiliki penyebaran data yang lebih kecil

Jawab:

Dari data di atas, jika kita hanya berpatokan pada hasil simpangan baku kedua mapel yang telah diujikan tersebut tentunya mapel kimia akan memiliki persebaran yang lebih kecil dari pada mapel matematika. Akan tetapi adalah perhitungan yang lebih baik tentang permasalahan di atas, yaitu dengan menggunkan rumus koefisien variansi sebagaimana perhitungan berikut ini:

$\begin{array}{|c|c|}\hline \textrm{Mapel Matematika}&\textrm{Mapel Kimia}\\\hline \begin{aligned}V&=\displaystyle \frac{7}{78}\times 100\%\\ &=8,97\% \end{aligned}&\begin{aligned}V&=\displaystyle \frac{6}{62}\times 100\%\\ &=9,68\% \end{aligned}\\\hline \end{array}$

Tampak dari perhitungan koefisien variansi di atas bahwa nilai ulangan mapel matematika memiliki sebaran relatif lebih kecil dari pada hasil ulangan mapel kimia.

C. Angka Baku

Misalkan ada suatu permasalahan seorang siswa saat ulangan matematika mendapatkan nilai 8 di mana rataan kelasnya adalah 6,5 dan simpangan bakunya adalah 2. Sedangkan untuk hasil ulangan kimia ia berhasil mendapatkan nilai 9 yang rataan kelasnya 7,5 dan simpangan bakunya 3. Pertanyaannnya adalah hasil yang didapatkan anak tersebut kedudukannya mana yang lebih baik?

Untuk menjawab pertanyaan di atas kita dapat menggunkan angka baku, yaitu $z=\displaystyle \frac{x-\overline{x}}{S}$.

Berdasarkan nilai kita bisa tentukan angka baku nilai siswa tersebut, yaitu:

$\begin{aligned}\textrm{matematika}\: :\: z&=\displaystyle \frac{8-6,5}{2}=\color{blue}0,75\\ \textrm{fisika}\qquad\quad\: :\: z&=\displaystyle \frac{9-7,2}{3}=\color{red}0,60 \end{aligned}$.

Dari perhitungan angka bakunya, tampak bahwa nilai ulangan matematika siswa tersebut lebih besar dari angka baku fisikanya. Hal ini menunjukkan nilai matematika siswa tersebut adalah yang lebih baik.

Ukuran Penyebaran Data Berkelompok (Materi Kelas XII Matematika Wajib) (Bagian 2)

B. 2 Data Berkelompok

$\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Data Dispersi}&\textrm{Keterangan}\\\hline 1.&\textrm{Jangkauan}&\begin{aligned}\textrm{a}.\: \: &\textrm{selisih titik tengah}\\ &\textrm{kelas tertinggi dengan}\\ &\textrm{titik tengah kelas}\\ &\textrm{terendah}\\ \textrm{b}.\: \: &\textrm{selisih tepi atas kelas}\\ &\textrm{kelas tertinggi dengan}\\ &\textrm{tepi bawah kelas}\\ &\textrm{terendah} \end{aligned}\\\hline 2.&H&Q_{3}-Q_{1}\\\hline 3.&Q_{d}&\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 4.&SR&\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}} \\\hline 5.&S^{2}&\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i} \left (x_{i}-\overline{x} \right )^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}\\\hline 6.&S&\sqrt{\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i} \left (x_{i}-\overline{x} \right )^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukanlah nilai simpangan rata-ratanya}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&\colorbox{white}{47-49}&\colorbox{white}{50-52}&\colorbox{white}{53-55}&\colorbox{white}{56-58}&\colorbox{white}{59-61}\\\hline \textrm{Frek}&2&4&6&5&3\\\hline \end{array}\\\\ &\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{array}{ll} &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&f_{i}.x_{i}&\left | x_{i}-\overline{x} \right |&f_{i}.\left | x_{i}-\overline{x} \right |\\\hline 47-49&48&2&96&6,45&12,49\\\hline 50-52&51&4&204&3,45&13,8\\\hline 53-55&\colorbox{yellow}{54}&6&\colorbox{yellow}{324}&\colorbox{yellow}{0,45}&\colorbox{yellow}{2,7}\\\hline 56-58&57&5&285&2,55&12,75\\\hline 59-61&60&3&180&5,55&16,65\\\hline \textrm{Jumlah}&&20&1089&&58,8\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\begin{aligned}\overline{x}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.x_{i}}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=54+\displaystyle \frac{1089}{20}=54+0,45=\color{red}54,45 \end{aligned}\\ &\begin{aligned}SR&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{58,8}{20}\\ &=\color{red}2,94 \end{aligned}\\ &\textrm{Jadi, simpangan rata-ratanya adalah}\: SR=2,94 \end{array} \\\\ &\textbf{Alternatif 2}\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&d_{i}&f_{i}.d_{i}&\left | x_{i}-\overline{x} \right |&f_{i}.\left | x_{i}-\overline{x} \right |\\\hline \colorbox{white}{47-49}&48&2&-6&-12&6,45&12,49\\\hline \colorbox{white}{50-52}&51&4&-3&-12&3,45&13,8\\\hline \colorbox{yellow}{53-55}&\colorbox{yellow}{54}&6&\colorbox{yellow}0&\colorbox{yellow}0&\colorbox{yellow}{0,45}&\colorbox{yellow}{2,7}\\\hline \colorbox{white}{56-58}&57&5&3&15&2,55&12,75\\\hline \colorbox{white}{59-61}&60&3&6&18&5,55&16,65\\\hline \textrm{Jumlah}&&20&&9&&58,8\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\begin{aligned}\overline{x}&=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=54+\displaystyle \frac{9}{20}=54+0,45=\color{red}54,45 \end{aligned}\\ &\begin{aligned}SR&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{58,8}{20}\\ &=\color{red}2,94 \end{aligned}\\ &\textrm{Jadi, simpangan rata-ratanya adalah}\: SR=2,94 \end{array}$.

$\begin{array}{ll} 2.&\textrm{Tentukanlah nilai varian/ragamnya}\\ &\textrm{dari data soal no.1 di atas}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&\left | x_{i}-\overline{x} \right |& (x_{i}-\overline{x})^{2}&f_{i}. (x_{i}-\overline{x})^{2} \\\hline 47-49&48&2&6,45&41,6025&83,205\\\hline 50-52&51&4&3,45&11,9025&47,61\\\hline 53-55&54&6&324&0,2025&1,215\\\hline 56-58&57&5&285&6,5025&32,5125\\\hline 59-61&60&3&180&30,8025&92,4075\\\hline \textrm{Jumlah}&&20&&&256,95\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\textrm{dan}\: \: \overline{x}=\color{red}54,45\: (\textrm{lihat soal no.1})\\ &\textrm{maka}\\ &\begin{aligned}S^{2}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}. (x_{i}-\overline{x})^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{256,95}{20}\\ &=\color{red}12,8475 \end{aligned}\\ &\textrm{Jadi, varian/ragamnya adalah}\\ & S^{2}=12,8475 \end{array}$.

$\begin{array}{ll} 3.&\textrm{Tentukanlah nilai simpangan baku dari}\\ &\textrm{dari data soal no.1 di atas}\\\\ &\textbf{Jawab}:\\ &S=\sqrt{S^{2}}=\sqrt{12,8475}\approx \color{red}3,58 \end{array}$.