PAS MATEMATIKA BLOG: Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

$\color{blue}\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: k-l^{k-l}\: \: \textrm{untuk}\: \:k=2\: \: \textrm{dan}\: \: l=-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&-18&&&\textrm{D}.&18\\ \textrm{B}.&-14&\textrm{C}.&14&\textrm{E}.&256 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}k-l^{k-l}&=(2-(-2))^{2-(-2)}\\ &=4^{4}\\ &=256 \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{2}.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{t+8}\left (7^{3t+1}\times 3^{4t-1} \right )^{2}}{(7^{2t}\times 3^{3t+2})^{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&49&&&\textrm{D}.&7^{2t+2}\\ \textrm{B}.&9&\textrm{C}.&7&\textrm{E}.&3^{2t-1} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{3^{t+8}\left (7^{3t+1}\times 3^{4t-1} \right )^{2}}{(7^{2t}\times 3^{3t+2})^{3}}&=\displaystyle \frac{3^{t+8+2(4t-1)}\times 7^{2(3t+1)}}{7^{3.2t}\times 3^{3(3t+2)}}\\ &=\displaystyle \frac{3^{t+8t+8-2}\times 7^{6t+2}}{3^{9t+6}\times 7^{6t}}\\ &=3^{0}\times 7^{2}\\ &=1\times 49\\ &=49 \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika}\: \: \displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}=\frac{2^{m}3^{n}}{5^{k}}\:, \: \textrm{nilai}\: \: m+n+k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&7&&&\textrm{D}.&10\\ \textrm{B}.&8&\textrm{C}.&9&\textrm{E}.&11 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{(0,24)^{3}(0,243)^{5}}{(3,6)^{7}}&=\displaystyle \frac{\left (\frac{24}{100} \right )^{3}\times \left ( \frac{243}{1000} \right )^{5}}{\left (\frac{36}{10} \right )^{7}}\\ &=\displaystyle \frac{(8\times 3)^{3}\times \left (3^{5} \right )^{5}}{(2^{2}\times 3^{2})^{7}}\times \displaystyle \frac{10^{7}}{100^{3}\times 1000^{5}}\\ &=\displaystyle \frac{(2^{3}\times 3)^{3}\times 3^{25}}{(2^{14}\times 3^{14})}\times \displaystyle \frac{10^{7}}{\left ( 10^{2} \right )^{3}\times \left ( 10^{3} \right )^{5}}\\ &=2^{9-14}.3^{3+25-14}.10^{7-6-15}\\ &=2^{-5}.3^{14}.10^{-14}=2^{-5}.3^{14}.(2.5)^{-14}\\ &=2^{-5-14}.3^{21}.5^{-14}\\ &=\displaystyle \frac{2^{-19}.3^{14}}{5^{14}}\\ \textrm{Sehingga}\: \: &m+n+k=-19+14+14=9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{4}.&\textrm{Perhatikanlah grafik berikut} \end{array}$

$\begin{array}{ll}\\ .\quad\: \, &\textrm{Rumus fungsi untuk grafik tersebut di ata adalah}\, ....\\ &\begin{array}{llllllll}\\ \textrm{A}.&f(x)=1+3^{2x-1}&&&\textrm{D}.&f(x)=-1+3^{2x-1}\\ \textrm{B}.&f(x)=1-3^{2x-1}&\textrm{C}.&f(x)=1-\left ( \displaystyle \frac{1}{3} \right )^{2x-1}&\textrm{E}.&f(x)=-1+\left ( \displaystyle \frac{1}{3} \right )^{2x-1} \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{kita uji titik-titiknya, misal di titik}\: \: (1,2)\: \: \textrm{dan}\: \: \left (\displaystyle \frac{1}{2},0 \right )\\ &\begin{aligned}\textrm{A}&\Rightarrow (1,2)\Rightarrow f(1)=1+3^{2.1-1}=1+3=4\neq 2\: \: (\textrm{salah})\\ \textrm{B}&\Rightarrow (1,2)\Rightarrow f(1)=1-3^{2.1-1}=1-3=-2\neq 2\: \: (\textrm{salah})\\ \textrm{C}&\Rightarrow (1,2)\Rightarrow f(1)=1-\left ( \displaystyle \frac{1}{3} \right )^{2.1-1}=1-\frac{1}{3}=\frac{2}{3}\neq 2\: \: (\textrm{salah})\\ \textrm{D}&\Rightarrow (1,2)\Rightarrow f(1)=-1+3^{2.1-1}=-1+3=2= 2\: \: (\textbf{benar})\\ \textrm{E}&\Rightarrow (1,2)\Rightarrow f(1)=-1+\left ( \displaystyle \frac{1}{3} \right )^{2.1-1}=-1+\frac{1}{3}=-\frac{2}{3}\neq 2\: \: (\textrm{salah})\\ \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{5}.&\textrm{Nilai}\: \: \displaystyle \frac{10^{2020}-10^{2019}}{9}\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle \frac{1}{9}&&&\textrm{D}.&\displaystyle \frac{10^{2019}}{9}\\\\ \textrm{B}.&\displaystyle \frac{10}{9}&\textrm{C}.&10^{2014}&\textrm{E}.&10^{2019} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\begin{aligned}\displaystyle \frac{10^{2020}-10^{2019}}{9}&=\displaystyle \frac{10^{2019}\left ( 10^{1}-1 \right )}{9}\\ &=10^{2019}.\displaystyle \frac{9}{9}\\ &=10^{2019} \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle 1,1&&&\textrm{D}.&\displaystyle 1,4\\\\ \textrm{B}.&\displaystyle 1,2&\textrm{C}.&1,3&\textrm{E}.&1,5\\\\ &&&&&(\textbf{\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai}\: \: \left ( -\displaystyle \frac{1}{16} \right )^{^{^{\frac{2}{3}}}}=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle -0,25&&&\textrm{D}.&\displaystyle 0,35\\\\ \textrm{B}.&\displaystyle -0,16&\textrm{C}.&0,16&\textrm{E}.&\textrm{nilainya tidak real}\\\\ &&&&&(\textbf{\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\left ( -\displaystyle \frac{1}{16} \right )^{^{^{\frac{2}{3}}}}&=\sqrt[3]{\left ( -\displaystyle \frac{1}{16} \right )^{2}}\\ &=\sqrt[3]{\displaystyle \frac{1}{256}}\\ &=\sqrt[3]{\displaystyle \frac{1}{64\times 4}}\\ &=\sqrt[3]{\displaystyle \frac{1}{64}}\times \sqrt[3]{\displaystyle \frac{1}{4}}\\ &=\displaystyle \frac{1}{4}\times \sqrt[3]{\displaystyle \frac{1}{4}\times \frac{2}{2}}\\ &=\displaystyle \frac{1}{4}\times \sqrt[3]{\displaystyle \frac{1}{8}}\times \sqrt[3]{2}\\ &=\displaystyle \frac{1}{4}\times \frac{1}{2}\times \sqrt[3]{2}\\ &=\displaystyle \frac{1}{8}\times \sqrt[3]{2}\\ &=(0,125)\times (1,...)\\ &\approx 0,16 \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\: \: \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{A}.&\displaystyle \frac{a^{2}}{a-b}&&&\textrm{D}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{B}.&\displaystyle \frac{a^{2}}{a-1}&\textrm{C}.&\displaystyle \frac{b-a}{ab}&\textrm{E}.&\displaystyle \frac{1}{a-b}\\\\ &&&&&(\textbf{\color{black}\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{\textit{UM UGM 2016 Mat Das}})\\ &\begin{array}{llllllll}\\ \textrm{A}.&0&&&\textrm{D}.&3\\ \textrm{B}.&1&\textrm{C}.&2&\textrm{E}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{10}.&\textrm{Bentuk sederhana dari}\: \: \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{\textit{SIMAK UI 2012 Mat IPA}})\\ &\begin{array}{llllllll}\\ \textrm{A}.&2-\sqrt{2}&&&\textrm{D}.&2+5\sqrt{2}\\ \textrm{B}.&8-\sqrt{2}&\textrm{C}.&-2+\sqrt{2}&\textrm{E}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\textrm{misalkan},\\ &\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=8-\sqrt{2} \end{aligned} \end{array}$

PAS MATEMATIKA BLOG

Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

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