PAS MATEMATIKA BLOG

Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll}\\ 11.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \color{purple}\frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \color{red}\textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \color{purple}\displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=\color{red}2 \end{aligned} \end{array}$.

$\begin{aligned}&\textrm{ada cara lain yang lebih sede}\textrm{rhana, yaitu:}\\ .\qquad\: \, &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\qquad\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ &\textrm{Jika}\quad \\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga}\quad\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=\color{red}2 \end{aligned}$.

$\begin{array}{ll}\\ 13.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }=\color{red}\infty \end{array}$.

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

$\begin{array}{ll}\\ 15.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \color{red}\textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\color{red}\frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 16.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\color{blue}\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{purple}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=\color{red}-\frac{5}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \color{purple}\displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \color{red}\textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\color{red}\infty \end{aligned} \end{array}$

Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll} 1.&\textrm{Perhatikanlah pernyataan-pernyataan berikut}\\ &\textrm{a}.\quad \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{atau}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\: \: \: \: \quad \textrm{maka}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=8\\ &\textrm{b}.\quad \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{atau}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=4,\\ &\: \: \: \: \quad \textrm{maka}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=4\\ &\textrm{c}.\quad \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=4\: \: \textrm{atau}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\: \: \: \: \quad \textrm{maka}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)\: \: \textrm{tidak ada}\\ &\textrm{d}.\quad \textrm{Jika}\: \: \underset{x\rightarrow 0^{-} }{\textrm{Lim}}\: \: f(x)=3\: \: \textrm{atau}\: \: \underset{x\rightarrow 0^{+} }{\textrm{Lim}}\: \: f(x)=2,\\ &\: \: \: \: \quad \textrm{maka}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: f(x)=1\\ &\textrm{Pernyataan di atas yang tepat adalah}\: ....\\ &\textrm{a}.\quad (i)\: \: \textrm{dan}\: \: (ii)\\ &\textrm{b}.\quad (i)\: \: \textrm{dan}\: \: (iii)\\ &\textrm{c}.\quad (ii)\: \: \textrm{dan}\: \: (iii)\\ &\textrm{d}.\quad (ii)\: \: \textrm{dan}\: \: (iv)\\ &\textrm{e}.\quad (iii)\: \: \textrm{dan}\: \: (iv)\\\\ &\textbf{Jawab}:\qquad\color{red}\textrm{c}\\ &\begin{array}{|c|}\hline \begin{aligned}\textrm{Ing}&\textrm{at}\: \: \textbf{Definisi Limit}\: \: \textrm{berikut}:\\ \textrm{Mis}&\textrm{al}\: \: f\: \: \textrm{sebuah fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: c,L\in \mathbb{R}\\ 1.\: \: \: &\textrm{Limit fungsi trigonometri};\: \: \underset{x\rightarrow c }{\textrm{Lim}}\: \: f(x)=L\: \: \textbf{ada}\\ &\textrm{untuk semua nilai}\: \: x\: \: \textrm{mendekati}\: \: c\\ &\textrm{jika dan hanya jika nilai}\: \: f(x)\: \: \textrm{mendekati}\: \: L\\ 2.\: \: \: &\underset{x\rightarrow c }{\textrm{Lim}}\: \: f(x)=L\Leftrightarrow \underset{x\rightarrow c^{-} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow c^{+} }{\textrm{Lim}}\: \: f(x)=L\\ &\textrm{Limit kiri}=\textrm{limit kanan} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll} 2.&\textrm{Perhatikanlah gambar dan pernyatan-}\\ &\textrm{pernyataan berikut} \end{array}$.

$\begin{array}{ll} &\textrm{i}.\quad \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{4} }{\textrm{Lim}}\: \: f(x)=4\\ &\textrm{ii}.\: \: \: \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{3\pi }{4} }{\textrm{Lim}}\: \: f(x)=\sqrt{2}\\ &\textrm{iii}.\: \, \textrm{Nilai}\: \: \underset{x\rightarrow \pi }{\textrm{Lim}}\: \: f(x)=1\\ &\textrm{iv}.\: \, \textrm{Nilai}\: \: \underset{x\rightarrow \displaystyle \frac{5\pi }{4} }{\textrm{Lim}}\: \: f(x)=-1\\ &\textrm{Pernyataan-pernyataan yang tepat}\\ &\textrm{ditunjukkan oleh}\: ....\\\\ &\textrm{a}.\quad (i)\: \: \textrm{dan}\: \: (ii)\\ &\textrm{b}.\quad (i)\: \: \textrm{dan}\: \: (iii)\\ &\textrm{c}.\quad (ii)\: \: \textrm{dan}\: \: (iii)\\ &\textrm{d}.\quad (ii)\: \: \textrm{dan}\: \: (iv)\\ &\textrm{e}.\quad (iii)\: \: \textrm{dan}\: \: (iv)\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{a} \end{array}$.

$\begin{array}{ll} 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x),\: \: \textrm{dengan kondisi}\\ &\qquad\qquad f(x)=\begin{cases} \displaystyle \frac{x^{2}-4}{x-2} &\textrm{untuk}\: \: x\neq 2 \\ &\\ 6x & \textrm{untuk}\: \: x=2 \end{cases}\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{tidak ada}\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 4\\ &\textrm{e}.\quad 12\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sebagaimana pada soal, maka}\\ &\textrm{Harga limit kiri}:\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Dan harga limit kanan}:\\ &\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Karena limit kiri}=\textrm{limit kanan},\: \: \textrm{yaitu}\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=4,\: \: \textrm{maka}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\color{red}\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2020}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2021x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2020}{x^{2}}}{2-\displaystyle \frac{2021}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2020}{\infty ^{2}}}{2-\displaystyle \frac{2021}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=\color{red}4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\color{red}\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\times \color{purple}\displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2021}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2020x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2021}{x}}{-\sqrt{9-\displaystyle \frac{2020}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2021}{\infty }}{-\sqrt{9-\displaystyle \frac{2020}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=\color{red}-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\color{red}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=\color{red}25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \color{red}\textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e} \end{array}$

$\begin{aligned}.\qquad \: \, &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=\color{red}-\displaystyle \frac{5}{3} \end{aligned}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=\color{red}1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \color{purple}\frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }\\ &=\color{red}\infty \end{array}$.

☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝

$\color{purple}\begin{aligned}\textrm{Sebagai}&\: \: \color{black}\textbf{CATATAN}\: \textrm{di sini}\\ \textrm{Sifat-sif}&\textrm{at bilangan tak hingga}\\ (1)\: \: &\infty +\infty =\infty \\ (2)\: \: &-\infty +(-\infty )=-\infty \\ (3)\: \: &\infty \times \infty =\infty\\ (4)\: \: &-\infty +(-\infty )=\infty \\ (5)\: \: &k.\infty =\infty ,\quad k\: \: \color{blue}\textrm{positif}\\ (6)\: \: &k.(-\infty )=-\infty,\quad k\: \: \color{blue}\textrm{positif} \\ (7)\: \: &k.\infty =-\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ (8)\: \: &k.(-\infty )=\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ \textrm{yang ha}&\textrm{rus dihindari}\\ (1)\: \: &\color{red}\infty -\infty ,\quad \: \: \color{black}\textrm{bentuk tak tentu}\\ (2)\: \: &\color{red}\displaystyle \frac{\infty }{\infty },\: -\displaystyle \frac{\infty }{\infty },\: \: \color{black}\textrm{dan}\: \: \color{red}\frac{0}{0} \end{aligned}$.

Latihan Soal 11 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 96.&\textrm{Himpunan penyelesaian dari}\\ &2x-1<x+1<3-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x<1 \right \}\\ \textrm{b}.&\left \{ x|x<2 \right \}\\ \textrm{c}.&\left \{ x|1<x<2 \right \}\\ \textrm{d}.&\left \{ x|x>2 \right \}\\ \textrm{e}.&\left \{ x|x>1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x-1<x}}\, +\, \underset{\textrm{B}}{\underbrace{1<3-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x-1<x+1\\ &\qquad x<2\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+1<3-x\\ &\qquad 2x<2\\ &\qquad x<1\: ................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 97.&\textrm{Himpunan penyelesaian dari}\\ &2x+1<x<1-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-2 \right \}\\ \color{red}\textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-1<x<-2 \right \}\\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{1}{2}\right \}\\ \textrm{e}.&\left \{ x|x<1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x+1<x}} \underset{\textrm{B}}{\underbrace{\: <1-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x+1<x\\ &\qquad x<-1\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x<1-x\\ &\qquad 2x<1\\ &\qquad x<\displaystyle \frac{1}{2}\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 98.&\textrm{Himpunan penyelesaian dari}\\ &3x+14\leq x+5<3x-1\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-3 \right \}\\ \textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-3<x<-1 \right \}\\ \textrm{d}.&\left \{ x|x>3\right \}\\ \color{red}\textrm{e}.&\left \{\: \: \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{4x+14\leq x}} \underset{\textrm{B}}{\underbrace{\, +\, 5 <3x-1}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 4x+14\leq x+5\\ &\qquad 3x\leq -9\\ &\qquad x\leq -3\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+5<3x-1\\ &\qquad -2x<-6\\ &\qquad x>3\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}\: \color{red}\textrm{tidak ada} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 99.&\textrm{Jika}\: \: \displaystyle \frac{1}{x}<2021\: \: \textrm{dan}\: \: \displaystyle \frac{1}{x}>2020\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2020<x<2021\\ \textrm{b}.&-2021<x<-2020\\ \textrm{c}.&\displaystyle \frac{1}{2020}<x<\displaystyle \frac{1}{2021}\\ \textrm{d}.&x<\displaystyle \frac{1}{2021}\: \: \textrm{dan}\: \: x>\displaystyle \frac{1}{2020}\\ \textrm{e}.&\textrm{semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}:\: \color{black}\displaystyle \frac{1}{x}<2021\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}>2020\\ &\textrm{Dapat ditulis ulang dengan}\\ &\color{black}2020<\displaystyle \frac{1}{x}\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}<2021\\ &\textrm{Jika digabung menjadi}\\ &\color{black}2020<\displaystyle \frac{1}{x}<\color{black}2021 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 100.&\textrm{Jika}\: \: a>0\: \: \textrm{dan}\: \: b<0\: ,\: \textrm{maka}\\ &\textrm{pernyataan berikut yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a+b>0\\ \textrm{b}.&a-b<0\\ \textrm{c}.&a^{2}-b^{2}<0\\ \color{red}\textrm{d}.&\displaystyle \frac{a}{b}<0\\ \textrm{e}.&ab>0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{purple}\textrm{Cukup Jelas saat}\: \: \color{red}\displaystyle \frac{a}{b}=\frac{+}{-}=-<0 \end{array}$.

$\begin{array}{ll}\\ 101.&\textrm{Jika}\: \: 0<x+y<3\: \: \textrm{dan}\: \: 1<x-y<2\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<5\\ \textrm{b}.&\left | x \right |<1\\ \textrm{c}.&x<1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{2}<x<\frac{5}{2}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{array}{llll}\\ 0<x+y<&3&\\ 1<x-y<&2&+\\\hline \: \: 1<2x<&5&\color{black}\textrm{dibagi 2 semuanya}\\ \quad \displaystyle \frac{1}{2}<x<&\displaystyle \frac{5}{2}&\: .....\color{red}(4)\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 102.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui P, Q, dan R memancing ikan.}\\ & \textrm{Jika hasil Q lebih sedikit dari hasil R}\\ & \textrm{sedangkan jumlah hasil P dan Q lebih }\\ & \textrm{banyak dari pada dua kali hasil R,}\\ &\textrm{maka yang terbanyak mendapat ikan}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{P dan R}\\ \textrm{b}.&\textrm{P dan Q}\\ \color{red}\textrm{c}.&\textrm{P}\\ \textrm{d}.&\textrm{Q}\\ \textrm{e}.&\textrm{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\color{black}\textrm{Diketahui}:\\ &\bullet \: Q< R\: ...............\color{red}(1)\\ &\bullet \: P+Q> 2R\: ......\color{red}(2)\\ &\textrm{Sehingga untuk persamaan}\: \: \color{black}(1)\: \&\: (2)\\ &\begin{array}{llll}\\ \qquad\qquad R>&Q&\\ \qquad P+Q>&2R&+\\\hline P+Q+R>&Q+2R&\\\\ \qquad\quad\quad P>&R\: ......\color{red}(3)\\ \end{array}\\ &\textrm{dari} \: \color{red}(1)\: \color{purple}\textrm{dan}\: \color{red}(3)\: \color{purple}\textrm{diperoleh bahwa}\\ &Q<R< P\\ &\textrm{Jadi, yang terbanyak mendapat ikan}\\ &\color{red}\textrm{adalah P} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 103.&\textrm{Jika}\: \: a>0,\: b>0,\: \: \textrm{dan}\: \: a>b,\: \: \textrm{maka}\\ &\textrm{pernyataan berikut yang salah adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{a}>\frac{1}{b}\\ \textrm{b}.&a^{2}>b^{2}\\ \textrm{c}.&a^{3}>b^{3}\\ \textrm{d}.&\sqrt{a}>\sqrt{b}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{purple}\begin{aligned}&a>0,\: b>0,\: \: \textrm{dan}\: \: a>b\\ &\color{red}\textrm{Maka}\\ &\displaystyle \frac{a}{1}>\frac{b}{1},\: \: \textrm{jika dibalik}\\ &\color{red}\textrm{menjadi}\\ &\displaystyle \frac{1}{a}<\frac{1}{b} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 104.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan real, maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{2}+b^{2}\geq 2ab\\ \textrm{b}.&a^{2}+b^{2}> 2ab\\ \textrm{c}.&a^{2}+b^{2}< 2ab\\ \textrm{d}.&a^{2}+b^{2}\leq 2ab\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&a,b\in \mathbb{R}\\ &\color{red}\textrm{Maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab \end{aligned} \end{array}$

$\begin{array}{ll}\\ 105.&\textrm{Pernyataan berikut yang tepat untuk}\\ &\textrm{untuk seluruh}\: \: x\: \: \textrm{positif adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x+\displaystyle \frac{1}{x}<2\\ \textrm{b}.&x+\displaystyle \frac{1}{x}\leq 2\\ \textrm{c}.&x+\displaystyle \frac{1}{x}>2\\ \color{red}\textrm{d}.&x+\displaystyle \frac{1}{x}\geq 2\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&a,b\in \mathbb{R},\: \: a>0,\: b>0\\ &\color{red}\textrm{Mirip dengan pembahasan}\\ &\color{red}\textrm{no.19, maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab\\ &\color{black}\textrm{Saat}\: \: a=\sqrt{x},\: \: b=\displaystyle \frac{1}{\sqrt{x}}\\ &\textrm{menyebabkan}\\ &\left ( \sqrt{x} \right )^{2}+\left ( \displaystyle \frac{1}{\sqrt{x}} \right )^{2}\geq 2.\sqrt{x}.\displaystyle \frac{1}{\sqrt{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2\sqrt{x.\displaystyle \frac{1}{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2 \end{aligned} \end{array}$

DAFTAR PUSTAKA

Nugroho, P. A., Gunarto, D. 2013. BIG BANK Soal+Bahas Matematika SMA/MA Kelas 1, 2, & 3. Jakarta : Wahyumedia.
Tim BBM. 2015. Big Book Matematika SMA Kelas 1, 2, & 3. Jakarta : Cmedia

Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 86.&(\textrm{Soal SNMPTN})\\ &\textrm{Jika}\: \: x>5\: \: \textrm{dan}\: \: y<3,\: \: \textrm{maka}\\ &\textrm{nilai }\: \: x-y\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{lebih besar dari pada 1}\\ \textrm{b}.&\textrm{lebih besar dari pada 3}\\ \textrm{c}.&\textrm{lebih besar dari pada 8}\\ \textrm{d}.&\textrm{lebih besar dari pada 5}\\ \color{red}\textrm{e}.&\textrm{lebih besar dari pada 2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}\\ x&>5\: \: \color{red}\&\: \: \color{blue}y<3\\ \textrm{m}&\textrm{aka}\\ &\begin{array}{llllll}\\ x>5&\Rightarrow &x&>5\\ y<3&\Rightarrow &\color{black}-y&\color{black}>-3&\color{red}+\\\hline &&\color{red}x-y&>\color{red}2 \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 87.&\textrm{Batas pertidaksamaan}\: \: 5x-7>13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\\ \color{red}\textrm{b}.&x>4\\ \textrm{c}.&x>-4\\ \textrm{d}.&x<4\\ \textrm{e}.&-4<x<4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}5x&-7>13\\ 5x&>13+7\\ 5x&>20\\ x&\color{red}>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 88.&\textrm{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \color{red}\textrm{b}.&x<3\displaystyle \frac{1}{3}\\ \textrm{c}.&x>3\displaystyle \frac{1}{3}\\ \textrm{d}.&x>3\\ \textrm{e}.&\textrm{Semua pilihan jawaban salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}2x+3&>5x-7\\ 2x-5x&>-7-3\\ -3x&>-10\quad \color{black}\textrm{dikali (-1)}\\ 3x&<10\\ x&<\color{red}\displaystyle \frac{10}{3}=3\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 89.&(\textbf{UMPTN 01})\textrm{Jika pertidaksamaan}\\ & 2x-3a>\displaystyle \frac{3x-1}{2}+ax\: \: \textrm{mempunyai}\\ &\textrm{penyelesaian}\: \: x>5,\: \: \textrm{maka nilai}\: \: a\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{8}\\ \color{red}\textrm{c}.&\displaystyle \frac{3}{8}\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}2x-3a&>\displaystyle \frac{3x-1}{2}+ax\quad \color{black}\textrm{tiap ruas}\: (\times 2)\\ 4x-6a&>3x-1+2ax\\ 4x-3x&-2ax>-1+6a\\ x-2a&x>-1+6a\\ (1-2a)&x>-1+6a\\ x&>\displaystyle \frac{-1+6a}{1-2a}\\ \textrm{Diketa}&\textrm{hui}:\: \: x>5\: \: \color{red}\textrm{adalah penyelesaian}\\ \color{red}\textrm{maka}\: \: &\\ 5&=\displaystyle \frac{-1+6a}{1-2a}\\ 5-10a&=-1+6a\\ -6a-10&a=-1-5\\ -16a&=-6\\ a&=\displaystyle \frac{-6}{-16}\\ &=\color{red}\displaystyle \frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 90.&(\textbf{UMPTN 94})\\ &\textrm{Apabila}\: \: a<x<b\: \: \textrm{dan}\: \: a<y<b\\ & \textrm{maka berlaku}\: \: \: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a<x-y<b\\ \textrm{b}.&b-a<x-y<a-b\\ \color{red}\textrm{c}.&a-b<x-y<b-a\\ \textrm{d}.&\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)\\ \textrm{e}.&\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{array}{llllll}\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &-a>-y>-b&&\\\hline &\color{purple}\textrm{saat}&\color{black}\textrm{di susun ulang}&\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &\color{black}-b<-y<-a&\color{red}+&\\\hline &&\color{red}a-b\color{blue}<\color{red}x-y\color{blue}<&\color{red}b-a \end{array} \end{array}$.

$\begin{array}{ll}\\ 91.&\textrm{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y-x>4\\ \textrm{b}.&y-x<4\\ \textrm{c}.&y+x+4>0\\ \color{red}\textrm{d}.&y+x+4<0\\ \textrm{e}.&y+x<1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&2y-5>2x+4y+3\\ &2y-4y-2x-5-3>0\\ &-2y-2x-8>0\: \: \color{black}\textrm{dibagi}\: \left ( -\displaystyle \frac{1}{2} \right )\\ &\color{red}y+x+4<0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 92.&\textrm{Jika}\: \: 3x-4>5x-17\\ &\textrm{maka sebuah bilangan prima}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&11\\ \textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&3x-4>5x-17\\ &\Leftrightarrow 3x-5x>-17+4\\ &\Leftrightarrow -2x>-13\quad \color{black}\textrm{tiap ruas}\: (\times -1)\\ &\Leftrightarrow 2x<13\\ &\Leftrightarrow \color{red}x<\displaystyle \frac{13}{2}=6\frac{1}{2}\\ &\color{black}\textrm{Jadi, yang memenuhi adalah 3 dan 5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 93.&\textrm{Jika}\: \: \displaystyle \frac{1}{5}<\frac{1}{x}\: \: \textrm{dan}\: \: x<0\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{1}{5}\\ \color{red}\textrm{b}.&-5<x<0\\ \textrm{c}.&0<x<5\\ \textrm{d}.&x<-5\\ \textrm{e}.&-\displaystyle \frac{1}{5}<x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ \displaystyle \frac{1}{5}&<\frac{1}{x}\: \: \: \textrm{dan}\: \: x<0\\ \displaystyle \frac{1}{5}&<\displaystyle \frac{1}{x}\\ x&<5 \\ x&>-5\qquad \color{black}\textrm{karena}\: \: x<0\\ \textrm{Sehi}&\textrm{ngga}\\ \color{red}-5<&\color{red}x<0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 94.&\textrm{Jika}\: \: a,b,c\: \: \textrm{dan}\: \: d\: \: \textrm{bilangan real}\\ &\textrm{dengan}\: \: a>b\: \: \textrm{dan}\: \: c>d\\ &\textrm{maka berlaku}\\ &(1)\quad ac>bd\\ &(2)\quad a+c>b+d\\ &(3)\quad ad>bc\\ &(4)\quad ac+bd>ad+bc\\ &\textrm{Pernyataan-pernyataan di atas}\\ & \textrm{yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(1),(2),\: \: \textrm{dan}\: \: (3)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \color{red}\textrm{c}.&(2)\: \: \textrm{dan}\: \: (4)\\ \textrm{d}.&(4)\\ \textrm{e}.&\textrm{Semua benar} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}:\: \color{black}a,b,c\: \: \textrm{dan}\: \: d\: \: \color{blue}\textrm{bilangan real}\\ &\color{red}\textrm{Jelas bahwa baik bilangan positif maupun} \\ &\color{red}\textrm{negatif termasuk semunya dibolehkan}\\ &\textrm{dengan}\: \: \color{black}a>b\: \: \textrm{dan}\: \: c>d\\ &\bullet \quad\textrm{Sehingga pernyataan (1)}\quad ac>bd\\ &\qquad\textrm{salah saat kita coba bilangan negatif}\\ &\bullet \quad \textrm{Pernyataan (2) benar karena}\\ &\qquad \color{blue}\begin{array}{llll} \color{black}a&>&\color{black}b&\\ \color{black}c&>&\color{black}d&\color{red}+\\\hline \color{red}a+c&>&\color{red}b+d\\ \end{array}\\ &\bullet \quad \textrm{Kasusnya sama dengan poin (1)}\\ &\qquad \textrm{saat dicoba dengan bilangan positif}\\ &\qquad \color{red}\textrm{tidak semuanya memenuhi}\\ &\bullet \quad \textrm{Pernyataan (4) tepat juga karena}\\ &\qquad \color{blue}\begin{array}{ll}\\ a-b>0\\ c-d>0\qquad \color{black}\textrm{Saat dikalikan}\\\hline \color{red}(a-b)\times \color{red}(c-d)>0\\ \Leftrightarrow \color{red}ac\color{black}-ad-bc\color{red}+bd>0\\ \Leftrightarrow \color{red}ac+bd>\color{black}ad+bc \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 95.&\textrm{Jika}\: \: -2<y<3\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9<(y-2)^{2}<16\\ \textrm{b}.&4<(y-2)^{2}<16\\ \textrm{c}.&1<(y-2)^{2}<16\\ \color{red}\textrm{d}.&0\leq (y-2)^{2}<16\\ \textrm{e}.&-1<(y-2)^{2}<16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}:\: -2<y<3\\ &\color{red}\bullet \quad \textrm{saat dikurangi}\: \: 2\\ &\qquad \Leftrightarrow \: -2-2<y-2<3-2\\ &\qquad -4<y-2<1\\ &\color{red}\bullet \quad \textrm{Saat}\: \: -4<y-2<0\\ &\qquad (-4)^{2}<(y-2)^{2}<0^{2}\quad \textrm{dikuadratkan}\\ &\qquad 16>(y-2)^{2}>0\\ &\qquad 0<(y-2)^{2}<16\\ &\color{red}\bullet \quad \textrm{Saat}\: \: 0\leq y-2<1\\ &\qquad 0^{2}\leq (y-2)^{2}<1^{2}\\ &\qquad 0<(y-2)^{2}<1\\ &\textrm{Jadi}\: ,\: \: \color{red}0\leq (y-2)<16 \end{aligned} \end{array}$

Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 76.&\textrm{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ &\textrm{ketiga angka tersebut adalah 9. Angka kedua}\\ &\textrm{dikurangi angka pertama dan angka ketiga }\\ &\textrm{sama dengan 1. Dua kali angka pertama sama}\\ &\textrm{dengan jumlah angka kedua dan angka ketiga.}\\ &\textrm{Angka puluhan pada bilangan tersebut adalah}\\ &....\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \color{red}\textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Model matematikanya}\\ &\left\{\begin{matrix} A+B+C=9\: \: \qquad....(1)\\ 2B-A-C=1\qquad\: ....(2)\\ 2A=B+C\: \: \: \: \qquad\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle A+B+C&=9\\ \displaystyle -A+B-C&=1&+\\\hline \qquad2B&=10\\ \: \: \: \: \qquad\qquad B&=5&...(4)\\ \end{array}\\ &\textrm{Jadi},\: \textrm{bilangan kedua adalah}\: =\: B=\color{red}5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 77.&(\textbf{SIMAK UI 2010})\\ &\textrm{Jika}\: \: x+y+2z=K,\: x+2y+z=K,\\ &2x+y+z=K\: \: \textrm{dengan}\: \: K\neq 0,\: \textrm{maka}\\ &x^{2}+y^{2}+z^{2}\: \: \textrm{bila dinyatakan dalam}\: \: K\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{16}K^{2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{16}K^{2}\\ \textrm{c}.&\displaystyle \frac{4}{17}K^{2}\\ \textrm{d}.&\displaystyle \frac{3}{8}K^{2}\\ \textrm{e}.&\displaystyle \frac{2}{3}K^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+2z=K\: \: \qquad....(1)\\ x+2y+z=K\qquad\: ....(2)\\ 2x+y+z=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\color{black}\textrm{maka}\\ &\color{red}\left\{\begin{matrix} z+(x+y+z)=K\: \: \qquad....(1)\\ y+(x+y+z)=K\qquad\: ....(2)\\ x+(x+y+z)=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle x+y+2z&=K\\ \displaystyle x+2y+z&=K&\\ \displaystyle 2x+y+z&=K&+\\\hline 4x+4y+4z&=3K\\ x+y+z&=\displaystyle \frac{3}{4}K&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (4)\: \: \textrm{disubstitusikan ke}\: \: (1),(2),\: \textrm{dan}\: (3)\\ &\textrm{Jelas bahwa akan didapatkan}\\ &x=y=z=\displaystyle \frac{1}{4}K\\ &\textrm{Jadi},\: \: x^{2}+y^{2}+y^{2}=3\left ( \displaystyle \frac{1}{4}K \right )^{2}=\color{red}\displaystyle \frac{3}{16}K^{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 78.&\textrm{Diketahui}\: \: 0,15252525252...=\displaystyle \frac{p}{2q+r}\\ &\textrm{Jika jumlah}\: \: p\: \: \textrm{dan}\: \: q=\textrm{3 kali}\: \: r,\: \textrm{maka}\\ &\textrm{masing-masing harga}\: \: p,q, \: \textrm{dan}\: \: r=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 152,2819,2584\\ \textrm{b}.&\displaystyle 252,\displaystyle \frac{5638}{7},\frac{8102}{21}\\ \color{red}\textrm{c}.&\displaystyle 151,\frac{2819}{7},\frac{1292}{7}\\ \textrm{d}.&\displaystyle 151,\displaystyle \frac{2819}{7},\frac{2584}{7}\\ \textrm{e}.&\displaystyle 152,\frac{2819}{14},\frac{1292}{7} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\color{red}\begin{cases} 0,1\overline{5252}& =\displaystyle \frac{p}{2q+r}\: .....(1)\\ \quad p+q & =3r\: ............(2) \end{cases}\\ &\color{black}\textrm{dan}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \qquad \color{red}x&=0,15252525252...\\ \displaystyle 1000\color{red}x&=152,5252525252...\\ \displaystyle \quad10\color{red}x&=\: \: \: \: \: 1,5252525252...&-\\\hline \: \: 990\color{red}x&=151\\ \qquad \color{red}x&=\displaystyle \frac{151}{990},\: \: \color{black}\textrm{maka}\\ \displaystyle \frac{p}{2q+r}&=\displaystyle \frac{151}{990}\\ &\begin{cases} p &=151 \: \: .......(3)\\ 2p+r &=990 \: \: .......(4) \end{cases} \end{array}\\ &\textrm{Dari}\: \: (3)\: \textrm{diperoleh}:q=3r-p=3r-151\: ....\color{blue}(5)\\ &\textrm{Dari}\: \: (5)\: \: \textrm{disubstitusikan ke}\: \: (4)\\ &\begin{aligned}2q+r&=990\\ 2(3r-151)+r&=990\\ 6r-302+r&=990\\ 7r&=990+302=1292\\ r&=\displaystyle \frac{1292}{7}\: .....\color{blue}(6) \end{aligned}\\ &\textrm{Dari}\: \: (3)\&(6)\: \: \textrm{disubstitusikan ke}\: \: (2)\\ &\color{purple}\begin{aligned}p+q&=3r\\ 151+q&=3\left ( \displaystyle \frac{1292}{7} \right )\\ q&=\displaystyle \frac{3876}{7}-151\\ &=\displaystyle \frac{3876-1057}{7}\\ &=\displaystyle \frac{2819}{7}\: .....\color{blue}(7) \end{aligned}\\ &\textrm{Jadi},\: \: p,q,r\: \: \textrm{adalah}\: :\: \color{red}\displaystyle 151,\frac{2819}{7},\frac{1292}{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 79.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \color{red}\textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\color{red}\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\color{black}\textrm{tidak berpenyelesaian, maka}\\ &\color{black}\textrm{ingat penyelesaian metode matrik}\\ &\color{black}\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\color{blue}\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 80.&\textrm{Diketahui}\\ &\begin{pmatrix} \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5}\\ \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & -\displaystyle \frac{4}{5}\\ -\displaystyle \frac{2}{5} & \displaystyle \frac{1}{10} & \displaystyle \frac{1}{10} \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix}\\ &\textrm{Nilai}\: \: x,y,\: \: \textrm{dan}\: \: z\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}\\ \textrm{b}.&-1,5,1\\ \color{red}\textrm{c}.&1,5,-1\\ \textrm{d}.&-1,1,5\\ \textrm{e}.&5,1,-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{red}\left\{\begin{matrix} \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\quad \quad....(1)\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2\qquad\: ....(2)\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0\: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z&=2&-\\\hline \quad\qquad \qquad \displaystyle \frac{5}{5}z&=-1&\\ \: \: \: \quad\qquad \qquad \displaystyle z&=-1&...(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lllllll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\left | \times 1 \right |&\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z&=0&\left | \times 2 \right |&-\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0&-\\\hline &&&\: \: \: \: \displaystyle \frac{5}{5}x\qquad\qquad\: =1&\\ &&&\: \: \: \quad x=-1\: ........(5) \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{akan didapatkan}\\ &y=5\\ &\textrm{Jadi},\: \: (x,y,z)=\color{red}(1,5,-1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 81.&\textrm{Diketahui suatu fungsi kuadrat}\\ &f(x)=ax^{2}+bx+c.\: \: \textrm{Jika fungsi}\\ &(-1,0),(1,4),\: \textrm{dan}\: \: (2,9),\: \: \textrm{maka}\\ &\textrm{fungsi yang dimaksud adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle f(x)=x^{2}-2x+3\\ \textrm{b}.&f(x)=x^{2}+2x+3\\ \textrm{c}.&f(x)=x^{2}+2x-3\\ \textrm{d}.&f(x)=x^{2}-2x-3\\ \color{red}\textrm{e}.&f(x)=x^{2}+2x+1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} (-1,0)\Rightarrow f(-1)=a-b+c=0\: ....\color{red}(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4\: ....\color{red}(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9\: ....\color{red}(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)\&(2),\: \textrm{didapatkan}\\ &b=2\: \: ...............\color{blue}(4)\\ &\textrm{Saat}\: \: (1)\&(3),\: \textrm{didapatkan}\\ &\color{blue}\begin{array}{llll}\\ 4a+2b+c&=9&\\ \: \: \: \: a-b+c&=0&-\\\hline \quad\qquad \qquad 3a+3b&=9&\\ \: \: \: \quad\qquad \qquad a+b&=3&...(5) \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{cases} a &=1 \\ c & =1 \end{cases}\\ &\textrm{Jadi},\: \: f(x)=ax^{2}+bx+c=\color{red}x^{2}+2x+1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 82.&\textrm{Diketahui persamaan}\begin{cases} x-y & =2 \\ kx+y & =3 \end{cases}\\ &\textrm{memiliki solusi}\: \: (x,y)\: \: \textrm{di kuadran I}\\ &\textrm{Jika dan hanya jika nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle k=-1\\ \textrm{b}.&k>-1\\ \textrm{c}.&k<\displaystyle \frac{3}{2}\\ \textrm{d}.&0<k<\displaystyle \frac{3}{2}\\ \color{red}\textrm{e}.&-1<k<\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x-y=2\: \: \: \quad....(1)\\ kx+y=3\quad\: ....(2)\end{matrix}\right.\\ &\textrm{Dengan metode matriks didapatkan}\\ &\color{blue}x=\displaystyle \frac{\begin{vmatrix} 2 & -1\\ 3& 1 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}\\ &\textrm{Dengan cara yang sama pula}\\ &\color{blue}y=\displaystyle \frac{\begin{vmatrix} 1 & 2\\ k & 3 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{3-2k}{k+1}\\ &\textrm{Supaya memiliki solusi di kwadran I},\\ &\textrm{maka baik}\: \: x\: \: \textrm{maupun}\: \: y\\ &\textrm{haruslah positif, akibatnya}:\\ &\color{red} k+1>0\Rightarrow k>-1\\ &\textrm{Sebagai akibat yang lain adalah}:\\ &3-2k>0\Rightarrow k<\displaystyle \frac{3}{2}\\ &\textrm{Jadi},\: \: \color{red}-1<k<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 83.&\textrm{Diketahui sistem persamaan}\\ &y+\displaystyle \frac{2}{x+z}=4\\ &5y+\displaystyle \frac{18}{2x+y+z}=18\\ &\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\\ &\textrm{Nilai}\: \: y+\sqrt{x^{2}-2xz+y^{2}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 3\\ \textrm{b}.&5\\ \textrm{c}.&7\\ \textrm{d}.&9\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} y+\displaystyle \frac{2}{x+z}=4\qquad\quad\\ 5y+\displaystyle \frac{18}{2x+y+z}=18\\ \displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\end{matrix}\right.\\ &\textrm{Jika disederhanakan beberapa bagian}\\ &\begin{cases} y+2A & =4\: ....(1) \\ 5y+18B & =18\: ....(2) \\ 8A-6B & =3\: ....(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2)\&(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ y+2A&=4&\left | \times 5 \right |&5y+10A=20\\ 5y+3(8A-3)&=18&\left | \times 1 \right |&5y+24A=27&-\\\hline &&&\: \: \quad-14A=-7\\ &&&\: \: \: \: \: \: \: \qquad A=\displaystyle \frac{1}{2}...(4)\\ \textrm{maka}\: B=\displaystyle \frac{1}{6}\: \& &y=3&&\\ \textrm{akibatnya}\\ \begin{cases} x &=1 \\ z &=1 \end{cases} \end{array} \\ &\textrm{Jadi},\: \: y+\sqrt{x^{2}-2xz+z^{2}}=3+0=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 84.&\textrm{Diberikan}\: \: a,b,\: \textrm{dan}\: \: c \: \: \textrm{adalah angka-angka}\\ &\textrm{dari bilangan 3 digit yang memenuhi}\\ &49a+7b+c=286.\: \: \textrm{Nilai dari}\: \: a+b+c\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&16\\ \textrm{b}.&17\\ \textrm{c}.&18\\ \textrm{d}.&19\\ \textrm{e}.&20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{blue}49a+7b+c=286\\ &\textrm{Nilai maksimum}\: \: a\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}49\times 5=245,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}245+7b+c=286\Rightarrow 7b+c=286-245=41\\ &\textrm{Nilai maksimum}\: \: b\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}7\times 5=35,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}35+c=41\Rightarrow c=41-35=6\\ &\color{black}\textrm{Sehingga}\: \: \color{blue}a,b,\: \: \color{black}\textrm{dan}\: \: \color{blue}c\: \: \color{black}\textrm{adalah}\: \: \color{blue}5,5,\: \: \color{black}\textrm{dan}\: \: \color{blue}6\\ &\textrm{Jadi},\: \textrm{nilai}\: \: \color{red}a+b+c=5+5+6=16 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 85.&\textrm{Diketahui sistem persamaan}\\ &(2x+3y)^{.^{\log (x-y+2z)}}=1\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &5x+3y+8z=2\\ &\textrm{Himpunan penyelesaian yang}\\ &\textrm{memenuhi adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ \displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \}\\ \textrm{b}.&\left \{ -\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6} \right \}\\ \textrm{c}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6} \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6} \right \}\\ \color{red}\textrm{e}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\color{blue}\textrm{Untuk persamaan}\: \: (1)\\ &(2x+3y)^{.^{\log (x-y+2z)}}=(2x+3y)^{0}\\ &\Leftrightarrow (x-y+2z)=10^{0}=1\\ &\color{blue}\textrm{Untuk persamaan}\: \: (2)\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &\Leftrightarrow 3^{2x+y+z+3(3z+2y+x)}=3^{4}\\ &\Leftrightarrow 5x+7y+10z=4\\ &\color{blue}\textrm{Sehingga sistem persamaan akan terlihat}\\ &\left\{\begin{matrix} x-y+2z=1\: \: \qquad....(1)\\ 5x+7y+10z=4\quad\: ....(2)\\ 5x+3y+8z=2\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (2)\&(3),\: \textrm{maka}\\ &\begin{array}{llll}\\ 5x+7y+10z&=4&\\ 5x+3y+8z&=2&-\\\hline \qquad 4y\quad+2z&=2\\ \qquad 2y\quad+z&=1\: ...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llll}\\ 5x-5y+10z&=5&\\ 5x+7y+10z&=4&-\\\hline \quad -12y\quad&=1\\ \: \: \: \: \qquad y\quad&=-\displaystyle \frac{1}{12}\: ...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (5)\: \: \textrm{disubstistusikan ke}\: \: (4)\\ &\color{blue}\begin{aligned}2y+z&=1\\ 2\left ( -\displaystyle \frac{1}{12} \right )+z&=1\\ z&=1+\displaystyle \frac{1}{6}\\ z&=\displaystyle \frac{7}{6} \end{aligned}\\ &\textrm{Cukup jelas juga}\: \: x=....\\ &\textrm{Jadi},\: \textrm{pilihannya adalah}\: \: \color{red}e \end{aligned} \end{array}$

DAFTAR PUSTAKA

Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade MAtematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
Kanginan, M. 2016. Matematika untuk SMA-MA/SMK-MAK Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA
Kurnianingsih, S. 2008. SPM Matematika SMA dan MA Program IPS Siap Tuntas Menghadapi Ujian. Jakarta: ESIS
Susianto, B. 2011. Soal dan Pembahasan Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO
Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI

Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 66.&\textrm{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ &\textrm{B, dan C bersama-sama dalam 2 jam saja.}\\ &\textrm{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ &\textrm{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ &\textrm{B dan C bersama-sama dalam waktu 3 jam,}\\ &\textrm{maka sistem persamaan berikut yang memenuhi}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\begin{cases} A+B+C&=2 \\ A+B & =\displaystyle \frac{12}{5} \\ B+C &=3 \end{cases}\\ \textrm{b}.&\begin{cases} A+B+C&=\displaystyle \frac{1}{2} \\ A+B & =\displaystyle \frac{5}{12} \\ B+C &=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{c}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{12}{5} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=3 \end{cases}\\ \color{red}\textrm{d}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{2} \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{5}{12} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{e}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}& =\displaystyle \frac{12}{5} \\ \displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=3 \end{cases} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Per}&\textrm{hatikan bahwa}:\color{red}\textrm{Waktu penyelesaian}\\ \color{red}\textrm{sua}&\color{red}\textrm{tu pekerjaan adalah termasuk}\\ \color{red}\textrm{per}&\color{red}\textrm{bandingan berbalik nilai},\: \color{blue}\textrm{maka}\\ \bullet \: \: \: &A,B,\: \textrm{dan}\: C \: \textrm{dalam 2 jam, artinya}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\: \color{blue}\textrm{demikian juga}\\ \bullet \: \: \: &A\: \textrm{dan}\: B\: \textrm{bersama-sama selesai dalam}\\ &\textrm{2 jam 24 menit atau}\: \displaystyle \frac{12}{5}\: \textrm{jam}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}\\ \bullet \: \: \: &B\: \textrm{dan}\: C\: \textrm{selesai dalam 3 jam}:\\ &\color{black}\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 67.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \color{red}\textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\color{blue}\textrm{matriks}\: (\color{black}\textbf{Cara Cramer})\\ &\begin{aligned} \color{blue}x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ \color{blue}y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ \color{blue}z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \color{blue}\textrm{Cara di atas}$ full matriks-Cramer

$\begin{array}{ll}\\ 68.&\textrm{Hasil dari}\: \: xyz\: \: \textrm{yang memenuhi}\\ &\left\{\begin{matrix} x+y+z=2\quad\\ x-y+z=-2\: \\ x-y-z=2\quad \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-8\\ \textrm{b}.&-4\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=2\quad.....(1)\\ x-y+z=-2\: .....(2)\\ x-y-z=2\quad .....(3) \end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=2&\\ x-y+z&=-2&-\\\hline \: \, \quad2y&=4&\\ \qquad\quad y&=2&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lcc}\\ x+y+z&=2&\\ x-y-z&=2&+\\\hline 2x&=4&\\ \qquad\quad x&=2&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (2)+(2)+z&=2\\ z&=-2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(2).(2).(-2)=\color{red}-8 \end{aligned} \end{array}$

$.\quad\: \: \color{black}\textrm{Cara di atas}$ full eliminasi-substitusi

$\begin{array}{ll}\\ 69.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=-6\quad\\ x-2y+z=3\quad\: \\ -2x+y+z=9\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-30\\ \textrm{b}.&-15\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&30\\ \textrm{e}.&35 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=-6\quad ....(1)\\ x-2y+z=3\quad\: ....(2)\\ -2x+y+z=9\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ x-2y+z&=3&-\\\hline \: \: \: \quad 3y&=-9&\\ \qquad\quad y&=-3&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ -2x+y+z&=9&-\\\hline 3x&=-15&\\ \qquad\quad x&=-5&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (-5)+(-3)+z&=-6\\ z&=2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-5).(-3).(2)=\color{red}30 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 70.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+2y+z=4\: \: \qquad\\ 3x+y+2z=-5\quad\: \\ x-2y+2z=-6\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-96\\ \color{red}\textrm{b}.&-24\\ \textrm{c}.&24\\ \textrm{d}.&32\\ \textrm{e}.&96 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+2y+z=4\: \qquad.......(1)\\ 3x+y+2z=-5\quad\: ......(2)\\ x-2y+2z=-6\quad .......(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llclll}\\ x+2y+z&=4&\left | \times 1 \right |&\: \: x+2y+z&=4\\ 3x+y+2z&=-5&\left | \times 2 \right |&6x+2y+4z&=-10&-\\\hline &&&-5x\: \: \quad-3z&=14&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lll}\\ x+2y+z&=4&\\ x-2y+2z&=-6&+\\\hline 2x\: \: \: \, \quad +3z&=-2&...(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ -5x-3z&=14&\\ 2x+3z&=-2&+\\\hline -3x&=12&\\ \qquad\quad x&=-4&.....(6)\\ \color{red}\textrm{didapat pula}&z&=2......(7) \end{array}\\ &\textrm{Dari persamaan}\: \: (6)\&(7)\: \: \textrm{didapatkan}\\ &\color{red}\begin{aligned}x+2y+z&=4\\ (-4)+2y+2&=4\\ y&=3 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-4).(3).(2)=\color{red}-24 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 71.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad\\ 2x-y+2z=9\quad\: \\ x+3y-z=7\: \: \: \: \quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{3}{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{13}{12}\\ \textrm{d}.&\displaystyle \frac{5}{4}\\ \textrm{e}.&\displaystyle \frac{7}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad....(1)\\ 2x-y+2z=9\quad\: ....(2)\\ x+3y-z=7\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y-z&=1&\\ 2x-y+2z&=9&+\\\hline 3x\: \: \qquad+z&=10&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y-z&=1&\left | \times 3 \right |&3x+3y-3z&=3&\\ x+3y-z&=7&\left | \times 1 \right |&\quad x+3y-z&=7&-\\\hline &&&2x\quad \: \: \quad-2z&=-4&\\ &&&\: \: x\quad \: \: \: \: \quad-z&=2&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ 3x+z&=10&\\ x-z&=-2&+\\\hline 4x&=8&\\ \qquad\quad x&=2&.....(6)\\ \color{red}\textrm{didapat pula}&z&=4......(7) \end{array} \\ &\textrm{Dari persamaan}\: \: (1)\&(3)\: \: \textrm{didapatkan juga}\\ &\begin{array}{lll}\\ x+y-z&=1&\\ x+3y-z&=-7&-\\\hline \quad -2y&=-6&\\ \qquad\qquad y&=3&....(8) \end{array}\\ &\textrm{Jadi},\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\color{red}\frac{13}{12} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 72.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad\\ x+y-4z=10\quad \\ -2x+y+z=0 \quad \end{matrix}\right.\\ &\textrm{Nilai dari}\: \: \displaystyle \frac{xz}{y}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\frac{6}{13}\\ \color{red}\textrm{b}.&\displaystyle -\frac{5}{13}\\ \textrm{c}.&\displaystyle -\frac{1}{13}\\ \textrm{d}.&\displaystyle \frac{1}{13}\\ \textrm{e}.&\displaystyle \frac{7}{13} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad.....(1)\\ x+y-4z=10\quad .....(2)\\ -2x+y+z=0 \quad .....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y+z&=5&\\ x+y-4z&=10&-\\\hline \: \: \qquad \: \: \: \: \: 5z&=-5&\\ \: \: \qquad\quad \: \: \: z&=-1&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y+z&=5&\\ -2x+y+z&=0&-\\\hline 3x\quad \: \quad&=5&\\ \: \: \quad \: \: \: \: \quad x&=\displaystyle \frac{5}{3}&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}x+y+z&=5\\ \displaystyle \frac{5}{3}+y-1&=5\\ y&=5+1-\displaystyle \frac{5}{3}=\frac{13}{3} \end{aligned}\\ &\textrm{Jadi},\: \: \displaystyle \frac{xz}{y}=\displaystyle \frac{\left ( \displaystyle \frac{5}{3} \right ).(-1)}{\displaystyle \frac{13}{3}}=\color{red}-\displaystyle \frac{5}{13} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Himpunan penyelesaian dari}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8 \\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10 \\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4 \end{cases}\\ &\textrm{adalah}\: \: \left \{ (x,y,z) \right \},\: \textrm{maka}\: \: x+3z=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8\: ....(1)\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10\: .....(2)\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4\: ...........(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}&=8\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=10&-\\\hline -\displaystyle \frac{1}{x}\: \: \: \: \: -\frac{1}{z}&=-2\\ \displaystyle \frac{1}{x}\: \: \: \: \: \: \: \: +\frac{1}{z}&=2&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lllllll}\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=8&\left | \times 2 \right |&\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}&=16\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&\left | \times 1 \right |&\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&-\\\hline &&&\displaystyle \frac{2}{x}\: \: \: \: \: +\frac{6}{z}&=12\\ &&\Leftrightarrow &\displaystyle \frac{1}{x}\: \: \: \: \: +\frac{3}{z}&=6&...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{3}{z}&=6\\ \displaystyle \frac{1}{x}+\frac{1}{z}&=2\: \: \: -\\\hline \qquad\displaystyle \frac{2}{z}&=4&\\ \qquad z&=\displaystyle \frac{1}{2}\: \: ......(6)\\ \qquad x&=2-\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2} \end{array}\\ &\textrm{Jadi},\: \: x+3z=\displaystyle \frac{3}{2}+3.\frac{1}{2}=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 74.&\textrm{Diketahui tiga buah bilangan berturut-turut}\\ &a,\: b,\: \textrm{dan}\: c.\: \textrm{Rata-rata dari ke tiga bilangan}\\ &\textrm{itu adalah 12. Bilangan kedua sama dengan}\\ &\textrm{jumlah bilangan yang lain dikurangi 12}.\\ &\textrm{Jika bilangan ke tiga sama dengan jumlah}\\ &\textrm{bilangan yang lain, maka nilai}\: \: 2a+b-c=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle 42\\ \textrm{b}.&-\displaystyle 36\\ \textrm{c}.&-\displaystyle 18\\ \textrm{d}.&-\displaystyle 12\\ \color{red}\textrm{e}.&-\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Model matematika dari persamaan di atas}\\ &\left\{\begin{matrix} a+b+c=36\: \: \qquad....(1)\\ -a+b-x=12\quad\: ....(2)\\ a+b-c=0\: \: \: \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ a+b+c&=36&\\ -a+b-c&=12&+\\\hline \: \: \: \: \: \: \: \: \: \: \: 2b&=48&\\ \: \: \qquad\quad \: \: \: b&=24&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ a+b+c&=36&\\ a+b-c&=0&-\\\hline \quad\qquad2c &=36&\\ \: \: \quad \: \: \: \: \quad c&=18&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}a+b+c&=36\\ a+24+18&=36\\ a&=36-42\\ &=-6 \end{aligned} \\ &\textrm{Jadi},\: \: 2a+b-c=2(-6)+24-18=\color{red}-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 75.&\textrm{Jumlah uang terdiri atas koin pecahan}\: \: Rp500,00\\ &Rp200,00\: \: dan\: \: Rp100,00\: \: \textrm{dengan nilai total}\\ &Rp100.000,00.\: \textrm{Jika nilai uang pecahan 500-an}\\ &\textrm{setengah dari nilai uang pecahan 200-an, tetapi}\\ &\textrm{tiga kali uang pecahan 100-an, maka banyak koin}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&460\\ \textrm{b}.&440\\ \textrm{c}.&420\\ \textrm{d}.&380\\ \textrm{e}.&350 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Model matematika dari kasus di atas}\\ &\left\{\begin{matrix} A(500)+B(200)+C(100)=100.000\: ....(1)\\ A(500)=\displaystyle \frac{1}{2}B(200)\qquad\qquad\qquad\qquad\: ....(2)\\ A(500)=3C(100)\qquad\qquad\qquad\: \: \, \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Dari persamaan}\: \: (2)\: \textrm{didapatkan}\\ &2A(500)=B(200)\\ &\textrm{Dari persamaan}\: \: (3)\: \textrm{akan didapatkan}\\ &\displaystyle \frac{1}{3}A(500)=C(100)\\ &\textrm{Dari persamaan}\: \: (1)\: \: \textrm{maka},\\ &A(500)+B(200)+C(100)=100.000\\ &A(500)+2A(500)+\displaystyle \frac{1}{3}A(500)=100.000\\ &\displaystyle \frac{10}{3}A(500)=100.000\Leftrightarrow A(500)=30.000\\ &\textrm{maka akan didapatkan}\\ &B(200)=2(30.000)=60.000\\ &C(100)=\displaystyle \frac{1}{3}(30.000)=10.000\\ &\color{red}\begin{cases} A(500) &=30.000\Rightarrow \color{black}A=\displaystyle \frac{30.000}{500}=60 \\ B(200) &=60.000\Rightarrow \color{black}B=\displaystyle \frac{60.000}{200}=300 \\ C(100) &=10.000\Rightarrow \color{black}C=\displaystyle \frac{10.000}{100}=100 \end{cases}\\ &\textrm{Jadi},\: \: A+B+C=60+300+100=\color{red}460 \end{aligned} \end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 56.&\textrm{Nilai}\: \: x\: \: \textrm{berikut yang tidak memenuhi}\\ &\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \color{red}\textrm{b}.&-1\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\displaystyle \frac{x-3}{x^{2}+2x+1}\leq 0\\ &\displaystyle \frac{(x-3)}{(x+1)^{2}}\leq 0\\ &\textrm{Pembuat nol}\\ &\begin{cases} x =3 ,\: \textrm{boleh digunakan}\\ x =-1,\: \textrm{tetapi}\: \: x\neq -1, \end{cases}\\ &\textrm{sehingga}\: \: -1\: \: \textrm{tidak digunakan}\\ &\color{red}\begin{array}{ccc|cccc|ccccc}\\ &&&&&&&&&&\\ &-&-&-&-&-&&+&+&&\\\hline &&-1&&&&3&&&&\\ &&&&&&&&&&\\ \end{array}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 57.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-5}\leq x \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<0\\\\ \textrm{b}.&x<1\: \: \textrm{atau}\: \: x\geq 5\\\\ \color{red}\textrm{c}.&\displaystyle \frac{5}{6}\leq x\leq 1\: \: \textrm{atau}\: \: x\geq 5\\\\ \textrm{d}.&\displaystyle \frac{5}{6}\leq x< 1\: \: \textrm{atau}\: \: 5<x<6\\\\ \textrm{e}.&x\geq 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sqrt{6x-5}&\leq x\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-5&\leq x^{2}\\ -x^{2}+6x-5&\leq 0\\ x^{2}-6x+5&\geq 0\\ (x-1)(x-5)&\geq 0\\ x\leq 1\: \: \textrm{atau}&\: \: x\geq 5\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-5&\geq 0\\ 6x&\geq 5\\ x&\geq \displaystyle \frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x+6}>6 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>7\\ \textrm{b}.&x\geq 7\\ \textrm{c}.&x<7\\ \textrm{d}.&x>1\\ \textrm{e}.&x\geq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\sqrt{6x+6}&>6\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x+6&>36\\ x+1&>6\\ x&>7\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x+6&\geq 0\\ 6x&\geq 6\\ x&\geq \displaystyle \frac{6}{6}\\ x&\geq 1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 59.&\textrm{Penyelesaian pertidaksamaan}\\ &x+2>\displaystyle \sqrt{10-x^{2}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x\leq \sqrt{10}\\ \color{red}\textrm{b}.&1<x\leq \sqrt{10}\\ \textrm{c}.&-3<x\leq \sqrt{10}\\ \textrm{d}.&-\sqrt{10}\leq x\leq \sqrt{10}\\ \textrm{e}.&x< -3\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}x+2&>\displaystyle \sqrt{10-x^{2}} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x^{2}+4x+4&>10-x^{2}\\ 2x^{2}+4x&+4-10>0\\ 2x^{2}+4x-&6>0\\ x^{2}+2x-&3>0\\ (x+3)&(x-1)>0\\ x<-3&\: \: \textrm{atau}\: \: x>1\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 10-x^{2}&\geq 0\\ x^{2}-10&\leq 0\\ (x-\sqrt{10})&(x+\sqrt{10})\leq 0\\ -\sqrt{10}\leq x&\leq \sqrt{10} \end{aligned} \end{array}$

$\begin{array}{l}\\ 60.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{3x+7}\geq x-1 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<6\\ \textrm{b}.&-1\leq x<6\\ \textrm{c}.&x\geq -\displaystyle \frac{7}{3}\\ \color{red}\textrm{d}.&-\displaystyle \frac{7}{3}\leq x\leq 6\\ \textrm{e}.&-\displaystyle \frac{7}{3}\leq x\leq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sqrt{3x+7}&\geq x-1 \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 3x+7&\geq x^{2}-2x+1\\ -x^{2}+3x&+2x+7-1\geq 0\\ -x^{2}+5x+&6\geq 0\\ x^{2}-5x-&6\leq 0\\ (x+1)&(x-6)\leq 0\\ -1\leq x&\leq 6\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 3x+7&\geq 0\\ 3x&\geq -7\\ x&\geq -\displaystyle \frac{7}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 61.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x\geq -5\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: \: x\geq -4\\ \textrm{c}.&x<13\\ \color{red}\textrm{d}.&4\leq x< 13\\ \textrm{e}.&-5\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\sqrt{2x-8}&<\sqrt{x+5}\\ (1)\: \: \, \textrm{kuadratkan}&\\ 2x-8&<x+5\\ x&<13\\ (2)\quad 2x-8\geq 0&\\ x&\geq 4\\ (3)\: \: \quad x+5\geq 0&\\ x&\geq -5\\ \textrm{perhatikan}&\textrm{lah garis bilangannya berikut}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 62.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-4}< \sqrt{2x+8} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-4<x\leq \displaystyle \frac{2}{3}\\ \textrm{b}.&-4<x<3\\ \color{red}\textrm{c}.&\displaystyle \frac{2}{3}\leq x< 3\\ \textrm{d}.&2<x\leq 4\\ \textrm{e}.&-4\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle \sqrt{6x-4}&< \sqrt{2x+8} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-4&<2x+8\\ 6x-2x&<8+4\\ 4x&<12\\ x&<3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-4&\geq 0\\ 6x&\geq 4\\ x&\geq \displaystyle \frac{2}{3}\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 2x+8&\geq 0\\ 2x&\geq -8\\ x&\geq -4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 63.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{x+3}> \sqrt{12-2x} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3<x\leq 6\\ \color{red}\textrm{b}.&-3<x\leq 6\\ \textrm{c}.&-6<x\leq 3\\ \textrm{d}.&-6<x\leq -3\\ \textrm{e}.&x<3\: \: \textrm{atau}\: \: x>6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \displaystyle \sqrt{x+3}&> \sqrt{12-2x} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x+3&>12-2x\\ x+2x&>12-3\\ 3x&>9\\ x&>3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ x+3&\geq 0\\ x&\geq -3\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 12-2x&\geq 0\\ 2x-12&\leq 0\\ 2x&\leq 12\\ x&\leq 6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 64.&(\textbf{SBMPTN 2013 Mat Das})\\ &\textrm{Jika}\: \: 1<m<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x>-3\\ \textrm{b}.&x<-4\\ \color{red}\textrm{c}.&-4<x<0\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: \: x>0\\ \textrm{e}.&x<-3\: \: \textrm{atau}\: \: x>-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\\ &\color{black}\textrm{dengan kondisi}\: \: 1<m<2\\ &\textrm{Perhatikanlah penyebutnya yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi penyebutnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena penyebutnya}\: :\: -x^{2}+3x-3m,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=3,\: \&\: c=-3m,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}3^{2}-4(-1)(-3m)=\color{black}9-12m\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}12<12m<24\\ &\Leftrightarrow \: \color{red}-12>-12m>-24\\ &\Leftrightarrow \: \color{red}9-12>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-3>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-13<\color{black}9-12m\color{red}<-3\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat penyebut berupa}\: \: -x^{2}+3x-3m\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{x(x+4)}{\underset{\color{red}definit\: negatif}{\underbrace{-x^{2}+3x-3m}}}>0\color{black}\Leftrightarrow \frac{x(x+4)}{-}>0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+4)<0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+4)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-4\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}-4<x<0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 65.&\textrm{Jika}\: \: 1<a<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-3\: \: \textrm{atau}\: \: x>0\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{d}.&-3<x<0\\ \textrm{e}.&-2\leq x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\\ &\textrm{untuk membedakan}\: \: a\: \: \textrm{pada persamaan}\\ &\textrm{kuadrat dengan}\: \: a\: \: \textrm{di atas, selanjutnya}\\ &\textrm{kita menuliskan}\: a\: \textrm{di atas dengan}:\: \color{red}m\\ &\color{black}\textrm{karena}\: \: 1<a<2\: \: \textrm{diubah}:1<m<2\\ &\textrm{Perhatikanlah pembilang yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi pembilangnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena pebilangnya}\: :\: \color{red}-x^{2}+2mx-6,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=2m,\: \&\: c=-6,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}(2m)^{2}-4(-1)(-6)=\color{black}4m^{2}-24\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}1^{2}<m^{2}<2^{2}\color{blue}\Leftrightarrow \color{red}1<m^{2}<4\\ &\Leftrightarrow \: \color{red}4<4m^{2}<16\\ &\Leftrightarrow \: \color{red}4-24<\color{black}4m^{2}-24\color{red}<16-24\\ &\Leftrightarrow \: \color{red}-20<\color{black}4m^{2}-24\color{red}<-8\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat pembilangnya berupa}\: \: -x^{2}+2mx-6\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{\overset{\color{red}definit\: negatif}{\overbrace{\color{blue}-x^{2}+2mx-6}}}{x(x+3)}\leq 0\color{black}\Leftrightarrow \frac{-}{x(x+3)}\leq 0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+3)> 0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+3)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-3\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}x<-3\: \: \textrm{atau}\: \: x>0 \end{aligned} \end{array}$.

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 46.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2}{x+1}\leq \left | x \right |\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{b}.&\left \{ x|x\leq -2\: \: \textrm{atau}\: \: 0\leq x\leq 1 \right \}\\ \textrm{c}.&\left \{ x|x\geq 1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1 \right \}\\ \textrm{e}.&\left \{ x|-1< x\leq 1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\left | x \right |\geq \displaystyle \frac{2}{x+1}\quad\quad\quad \color{black}\textrm{berakibat}\\ &\displaystyle \frac{-2}{x+1}\geq x\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{2}{x+1}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 1\\ &x\leq \displaystyle \frac{-2}{x+1}\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ &x+\displaystyle \frac{2}{x+1}\leq 0\\ &\displaystyle \frac{x(x+1)+2}{x+1}\leq 0\\ &\displaystyle \frac{x^{2}+x+2}{x+1}\leq 0\Leftrightarrow \displaystyle \frac{\textrm{Definit positif}}{x+1}\leq 0\\ &\begin{aligned}&\textrm{HP}_{1}=\color{black}\left \{x| x< -1,\: x\in \mathbb{R} \right \}\\ &\bullet \qquad \color{red}\textrm{bagian}\: \: 2\\ &x\geq \displaystyle \frac{2}{x+3}\\ &x-\displaystyle \frac{2}{x+1}\geq 0\\ &\displaystyle \frac{x(x+1)-2}{x+1}\geq 0\\ &\displaystyle \frac{x^{2}+x-2}{x+1}\geq 0\\ &\displaystyle \frac{(x+2)(x-1)}{x+1}\geq 0\\ &\textrm{HP}_{2}=\color{black}\left \{x|-2\leq x< -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ &\textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}=\color{red}\left \{ x|x<-1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 47.&\textrm{Diketahui pertidaksamaan}\: \: \displaystyle \frac{x+10}{x-9}\leq 0\\ &\textrm{dan diberikan beberapa nilai berikut}\\ &(\textrm{i})\quad x=-6\: \, \qquad\qquad (\textrm{iii})\quad x=-14\\ &(\textrm{ii})\, \, \, \: x=-10\qquad\quad\quad (\textrm{iv})\quad x=-18\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\textrm{di atas adalah ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(\textrm{i})\: \: \textrm{dan} \: \: (\textrm{ii})\\ \textrm{b}.&(\textrm{i})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{c}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iii})\\ \textrm{d}.&(\textrm{ii})\: \: \textrm{dan}\: \: (\textrm{iv})\\ \textrm{e}.&(\textrm{iii})\: \: \textrm{dan}\: \: \: (\textrm{iv}) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{x+10}{x-9}&\leq 0\\ \textrm{HP}=&\color{red}\left \{ x|-10\leq x< 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 48.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x< 6\\ \textrm{b}.&2\leq x< 3\\ \color{red}\textrm{c}.&2< x< 3\: \: \textrm{atau}\: \: x>6\\ \textrm{d}.&x<3\: \: \textrm{atau}\: \: 3<x<6\\ \textrm{e}.&x<2\: \: \textrm{atau}\: \: x>3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\\ &\Leftrightarrow \displaystyle \frac{6}{x-3}-\frac{8}{x-2}<0\\ &\Leftrightarrow \displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}<0\\ &\Leftrightarrow \displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{-2x+12}{(x-2)(x-3)}<0\\ &\Leftrightarrow \displaystyle \frac{2x-12}{(x-2)(x-3)}>0\\ &\Leftrightarrow \displaystyle \frac{2(x-6)}{(x-2)(x-3)}>0\\ &\textrm{HP}=\color{red}\left \{ x|2<x<3\: \: \textrm{atau}\: \: x>6,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 49.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\: \:\textrm{ adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x\leq -9\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{b}.&-9\leq x< 0\: \: \textrm{atau}\: \: x\geq 9\\ \textrm{c}.&-9\leq x< 0\: \: \textrm{atau}\: \: 0<x\leq 9\\ \textrm{d}.&-9< x\leq 9\\ \textrm{e}.&x\in \mathbb{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-81}{x^{2}}\geq 0\\ &\displaystyle \frac{(x+9)(x-9)}{x^{2}}\geq 0\\ &\textrm{HP}=\color{red}\left \{ x|x\leq -9\: \: \textrm{atau}\: \: x\geq 9,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}-4}{x+2}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>2\\ \textrm{b}.&-2\leq x< 2\\ \textrm{c}.&x<-2\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<-2\: \: \textrm{atau}\: \: -2< x< 2\\ \textrm{e}.&x\geq -2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{x^{2}-4}{x+2}>0\\ &\displaystyle \frac{(x+2)(x-2)}{(x+2)}>0\\ &(x-2)>0\\ &\color{red}x>2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 51.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}<0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|-9< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|-6< x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \}\\ \color{red}\textrm{d}.&\left \{ x|-9< x< -6\: \: \textrm{atau}\: \: \displaystyle \frac{5}{2}<x<5,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x< -9\: \: \textrm{atau}\: \: -6< x< \displaystyle \frac{5}{2}\: \: \textrm{atau}\: \: x<5,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \frac{x^{2}+x-30}{2x^{2}+13x-45}&<0\\ \displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}&<0\\ \color{red}\textrm{Cukup jelas}& \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 52.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{2x+6}{x-4}\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-10<x<4\\ \color{red}\textrm{b}.&-10\leq x<4\\ \textrm{c}.&-4<x\leq 10\\ \textrm{d}.&x\leq -10\: \: \textrm{atau}\: \: x\geq 4\\ \textrm{e}.&x<-10\: \: \textrm{atau}\: \: x\geq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{2x+6}{x-4}&\leq 1\\ \displaystyle \frac{2x+6}{x-4}-1&\leq 0\\ \displaystyle \frac{2x+6-(x-4)}{x-4}&\leq 0\\ \displaystyle \frac{x+10}{x-4}&\leq 0\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \displaystyle \frac{x^{2}-x}{x+3}\geq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-3\: \: \textrm{atau}\: \: -1\leq x\leq 3\\ \color{red}\textrm{b}.&-3< x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{c}.&-3\leq x\leq 3\\ \textrm{d}.&-3\leq x\leq -1\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{e}.&-3\leq x\leq -1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{x^{2}-x}{x+3}&\geq 1\\ \displaystyle \frac{x^{2}-x}{x+3}&-1\geq 0\\ \displaystyle \frac{x^{2}-x-(x+3)}{x+3}&\geq 0\\ \displaystyle \frac{x^{2}-2x-3}{x+3}&\geq 0\\ \displaystyle \frac{(x-3)(x+1)}{x+3}&\geq 0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+2+\displaystyle \frac{1}{x+4}>0\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\: \: \textrm{atau}\: \: x\geq -3\\ \textrm{b}.&x<-4\: \: \textrm{atau}\: \: x>-3\\ \textrm{c}.&-4\leq x\leq -3\\ \color{red}\textrm{d}.&x>-4\\ \textrm{e}.&-4\leq x\leq -3\: \: \textrm{atau}\: \: x>-3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}x+2+\displaystyle \frac{1}{x+4}&>0\\ \displaystyle \frac{(x+2)(x+4)+1}{(x+4)}&>0\\ \displaystyle \frac{x^{2}+6x+8+1}{x+4}&>0\\ \displaystyle \frac{x^{2}+6x+9}{x+4}&>0\\ \displaystyle \frac{(x+3)^{2}}{(x+4)}&>0\\ x&>-4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 55.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x+3<\displaystyle \frac{x^{2}+6x+11}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x< -3\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{b}.&\left \{ x|0\leq x\leq 11,\: x\in \mathbb{R} \right \}\\ \textrm{c}.&\left \{ x|x<-11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \}\\ \textrm{d}.&\left \{ x|x<0\: \: \textrm{atau}\: \: x>11,\: x\in \mathbb{R} \right \}\\ \textrm{e}.&\left \{ x|x\leq 11\: \: \textrm{atau}\: \: x>0,\: x\in \mathbb{R} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}x+3<\displaystyle \frac{x^{2}+6x+11}{x}&\\ x+3-\displaystyle \frac{x^{2}+6x+11}{x}&<0\\ \displaystyle \frac{x(x+3)-\left (x^{2}+6x+11 \right )}{x}&<0\\ \displaystyle \frac{x^{2}+3x-x^{2}-6x-11}{x}&<0\\ \displaystyle \frac{-3x-11}{x}&<0\\ \displaystyle \frac{3x+11}{x}&>0 \end{aligned} \end{array}$.