Lanjutan 2 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

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\begin{array}{ll}\\ \fbox{17}.&\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, maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \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}\\\\ \textbf{silahk}&\textbf{an cek sendiri untuk opsi jawaban yang lain} \end{aligned} \end{array}
\begin{array}{ll}\\ \fbox{18}.&\textrm{Himpunan penyelesaian dari 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{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \} \quad&\textrm{c}.&\left \{ x|x\geq -11 \right \} \quad&\textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \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}\quad \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}\\ &\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&&&&&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{r}{\begin{matrix} \displaystyle \frac{3}{5}\\ \end{matrix}}&&&\multicolumn{2}{l}{11}&\\ \end{array} \end{aligned} \end{array}
\begin{array}{ll}\\ \fbox{19}.&\textrm{Himpunan penyelesaian dari 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 \textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}}\\\hline -6\leq x^{2}+5x&x^{2}+5x\leq 6\\\hline \begin{aligned}x^{2}+5x+6&\geq 0\\ (x+3)(x+2)&\geq 0 \end{aligned}&\begin{aligned}x^{2}+5x-6&\leq 0\\ (x+6)(x-1)&\leq 0 \end{aligned}\\\hline &\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-3}&&&\multicolumn{2}{r}{-2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{c}{1}&\\ \end{array} \\ &\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Kesimpulan}:&\\ &\begin{array}{ccc|cccc|ccc|ccc|cccccccc}\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{2}{c}{.}&&\multicolumn{2}{l}{.}&&\multicolumn{2}{r}{.}\\\cline{4-7}\cline{11-13} &&&&&&&&&&&&&&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{r}{-3}&&\multicolumn{2}{l}{-2}&&\multicolumn{2}{l}{1}&\\ \end{array} \\ & \end{aligned}}\\\hline \end{array} \end{array}
\begin{array}{ll}\\ \fbox{20}.&\textbf{(USM UGM Mat IPA)}\textrm{Semua nilai \textit{x} yang memenuhi}\: \: x\left | x-2 \right |<x-2\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad x<-1\: \: \textrm{atau}\: \: 1<x<2\\ &\textrm{b}.\quad x<-2\\ &\textrm{c}.\quad -2<x<-1\\ &\textrm{d}.\quad x<-1\\ &\textrm{e}.\quad -2<x<1\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ &x\left | x-2 \right |<x-2\\ & \end{aligned}}\\\hline x<2&x\geq 2\\\hline \begin{aligned}x\left | x-2 \right |&<x-2\\ x(2-x)&<x-2\\ 2x-x^{2}&<x-2\\ -x^{2}+x+2&<0\\ x^{2}-x-2&>0\\ (x-2)(x+1)&>0\end{aligned}&\begin{aligned}x\left | x-2 \right |&<x-2\\ x(x-2)&<x-2\\ x^{2}-2x&<x-2\\ x^{2}-3x+2&<0\\ (x-1)(x-2)&<0\\ & \end{aligned}\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-1}&&&\multicolumn{2}{c}{2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{c}{1}&&&\multicolumn{2}{c}{2}&\\ \end{array} \\\hline \textbf{ada yang memenuhi}&\textbf{tidak ada yang memenuhi}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{yang memenuhi}:&\\ &x<-1\\ & \end{aligned}}\\\hline \end{array} \end{array}

Sumber Referensi

  1. Budhi, W. S. 2014. Bupena Matematika SMA/MA Kelas X Kelompok Wajib. Jakarta: Erlangga.
  2. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.
  3. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

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