Contoh Soal 3 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 11.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{5}{4x-3}}\right|\le 1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\le x\le \frac{3}{4}\text{atau}x\ge 2}\\ \text{b}.& x\le -{\displaystyle \frac{1}{2}\text{atau}{\displaystyle \frac{3}{4}<x\le 2}}\\ \text{c}.& -{\displaystyle \frac{1}{2}\le x\le 2,x\ne \frac{3}{4}}\\ \text{d}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x>\frac{3}{4}}\\ \text{e}.& x\le -{\displaystyle \frac{1}{2}\text{atau}x\ge 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{5}{4x-3}}\right|& \le 1\\ -1\le {\displaystyle \frac{5}{4x-3}}& \le 1,\mathbf{\text{jika dibalik}}\\ -1\ge {\displaystyle \frac{4x-3}{5}}& \ge 1,\mathbf{\text{bentuk ini tidak}}\\ \mathbf{\text{dibolehkan}}& \mathbf{\text{maka perlu diubah menjadi}}\\ -1\ge {\displaystyle \frac{4x-3}{5}\text{atau}}& {\displaystyle \frac{4x-3}{5}\ge 1,\text{selanjutnya}}\\ \bullet \text{bagian}& 1\\ -1& \ge {\displaystyle \frac{4x-3}{5}\iff \frac{4x-3}{5}\le -1}\\ 4x-3& \le -5\\ 4x& \le -2\\ x& \le -{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{4x-3}{5}}& \ge 1\\ 4x-3& \ge 5\\ 4x& \ge 8\\ x& \ge 2\end{array}\end{array}$

$\begin{array}{l}\\ 12.& (\text{UMPTN 95})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2}{2x-1}}\right|>1\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{c}.& x<-1\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{d}.& -1<x<2\text{dan}x\ne {\displaystyle \frac{1}{2}}\\ \text{e}.& x<-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{semua opsi bukan jawaban}}\\ & \mathbf{\text{Berikut pembahasannya}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2}{2x-1}}\right|& >1\\ -1>{\displaystyle \frac{2}{2x-1}}& \text{atau}{\displaystyle \frac{2}{2x-1}>1,\mathbf{\text{dibalik}}}\\ -1<{\displaystyle \frac{2x-1}{2}}& \text{atau}{\displaystyle \frac{2x-1}{2}<1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x-1}{2}}& >-1\\ 2x-1& >-2\\ 2x& >-1\\ x& >-{\displaystyle \frac{1}{2}}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x-1}{2}}& <1\\ 2x-1& <2\\ 2x& <3\\ x& <{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& (\text{UMPTN 00})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{2x+7}{x-1}}\right|\ge 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\le x\le 8\\ \text{b}.& x\le -8\text{atau}x\ge -2\\ \text{c}.& -8\le x<1\text{atau}x>1\\ \text{d}.& -2\le x<1\text{atau}1<x\le 8\\ \text{e}.& x\le -8\text{atau}-2\le x<1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{2x+7}{x-1}}\right|& \ge 1\\ -1\ge {\displaystyle \frac{2x+7}{x-1}}& \text{atau}{\displaystyle \frac{2x+7}{x-1}\ge 1}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{2x+7}{x-1}}& \le -1(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{2x+7}{x-1}}& +1\le 0\\ & {\displaystyle \frac{2x+7+(x-1)}{x-1}\le 0}\\ {\displaystyle \frac{3x+6}{x-1}}& \le 0\\ {\text{HP}}_1=& \{x|-2\le x<1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{2x+7}{x-1}}& \ge 1\\ {\displaystyle \frac{2x+7}{x-1}}& -1\ge 0\\ & {\displaystyle \frac{2x+7-(x-1)}{x-1}\ge 0}\\ {\displaystyle \frac{x+8}{x-1}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}-2\le x<1\text{atau}x>1,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \left|{\displaystyle \frac{x-2}{x+3}}\right|\le 2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\le x<-3\\ \text{b}.& -8\le x<-1\\ \text{c}.& -4\le x<-3\\ \text{d}.& x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3}}\\ \text{e}.& x\le -4\text{atau}x\ge 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|{\displaystyle \frac{x-2}{x+3}}\right|& \le 2\\ -2\le {\displaystyle \frac{x-2}{x+3}}& \text{atau}{\displaystyle \frac{x+2}{x+3}\le 2}\\ \bullet \text{bagian}& 1\\ {\displaystyle \frac{x-2}{x+3}}& \ge -2(\mathbf{\text{tidak boleh kali silang}})\\ {\displaystyle \frac{x-2}{x+3}}& +2\ge 0\\ & {\displaystyle \frac{x-2+2(x+3)}{x+3}\ge 0}\\ {\displaystyle \frac{3x+4}{x+3}}& \ge 0\\ {\text{HP}}_1=& \{x|x<-3\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\\ \bullet \text{bagian}& 2\\ {\displaystyle \frac{x-2}{x+3}}& \le 2\\ {\displaystyle \frac{x-2}{x+3}}& -2\le 0\\ & {\displaystyle \frac{x-2-2(x+3)}{x+3}\le 0}\\ {\displaystyle \frac{-x-8}{x+3}}& \le 0,\mathbf{\text{koefisien textit{x} negatif}}\\ {\displaystyle \frac{x+8}{x+3}}& \ge 0\\ {\text{HP}}_2=& \{x|x\le -8\text{atau}x>-3,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x\le -8\text{atau}x\ge -{\displaystyle \frac{4}{3},x\in \mathbb{R}}\}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\text{UMPTN 01})\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2}{x+1}\le \left|x\right|\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \{x|x\le -2\text{atau}x\ge 1\}\\ \text{b}.& \{x|x\le -2\text{atau}0\le x\le 1\}\\ \text{c}.& \{x|x\ge 1\}\\ \text{d}.& \{x|x<-1\text{atau}x\ge 1\}\\ \text{e}.& \{x|-1<x\le 1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left|x\right|& \ge {\displaystyle \frac{2}{x+1}\text{berakibat}}\\ {\displaystyle \frac{-2}{x+1}\ge x}& \text{atau}x\ge {\displaystyle \frac{2}{x+1}}\\ \bullet \text{bagian}& 1\\ x\le {\displaystyle \frac{-2}{x+1}}& (\mathbf{\text{tidak boleh kali silang}})\\ x+{\displaystyle \frac{2}{x+1}}& \le 0\\ & {\displaystyle \frac{x(x+1)+2}{x+1}\le 0}\\ {\displaystyle \frac{x^2+x+2}{x+1}}& \le 0\iff {\displaystyle \frac{\text{Definit positif}}{x+1}\le 0}\\ {\text{HP}}_1=& \{x|x<-1,x\in \mathbb{R}\}\\ \bullet \text{bagian}& 2\\ x& \ge {\displaystyle \frac{2}{x+3}}\\ x-& {\displaystyle \frac{2}{x+1}\ge 0}\\ & {\displaystyle \frac{x(x+1)-2}{x+1}\ge 0}\\ & {\displaystyle \frac{x^2+x-2}{x+1}\ge 0}\\ & {\displaystyle \frac{(x+2)(x-1)}{x+1}\ge 0}\\ {\text{HP}}_2=& \{x|-2\le x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\\ \text{HP}={\text{HP}}_1+{\text{HP}}_2& =\{x|x<-1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\end{array}$