PAS MATEMATIKA BLOG: Lanjutan 1 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

$\color{blue}\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: f(x)=x^{3}-3x^{2}-3x-1,\\ &\textrm{maka nilai dari}\: \: f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-\sqrt{2}&&&\textrm{D}.&1\\ \textrm{B}.&-1&\textrm{C}.&0&\textrm{E}.&\sqrt{2} \end{array}\\\\ &\qquad\qquad\qquad\qquad\qquad\quad(\textbf{\textit{UM UNDIP 2012 Math IPA}})\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}f(x)&=x^{3}-3x^{2}-3x-1\\ &=\left ( x- 1\right )^{3}-6x\\ f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )&=\left (1+\sqrt[3]{2}+\sqrt[3]{4}-1 \right )^{3}-6\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )\\ &=\left ( \sqrt[3]{2}+\sqrt[3]{4} \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left (\sqrt[3]{2} \left ( \sqrt[3]{2}+1 \right ) \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=2\left ( 1+3\sqrt[3]{2}+3\sqrt[3]{4}+2 \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left ( 6+6\sqrt[3]{2}+6\sqrt[3]{4} \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=0 \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: \textbf{a}\: \: \textrm{dan}\: \: \textbf{b}\: \: \textrm{adalah bilangan bulat positif yang memenuhi persamaan}\\ &\textbf{a}^{\textbf{b}}=2^{20}-2^{19},\: \textrm{maka nilai dari}\: \: \textbf{a}+\textbf{b}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&3&&&\textrm{D}.&21\\ \textrm{B}.&7&\textrm{C}.&19&\textrm{E}.&23 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\color{red}\begin{aligned}\textbf{a}^{\textbf{b}}&=2^{20}-2^{19}\\ &=2^{^{^{(19+1)}}}-2^{19}\\ &=2^{19}.2^{1}-2^{19}\\ &=2^{19}(2-1)\\ &=2^{19}\\ \textrm{Se}&\textrm{hingga nilai}\: \: \textbf{a}+\textbf{b}=2+19=21\end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{13}.&\textrm{Perhatikan gambar berikut} \end{array}$

$\color{blue}\begin{array}{ll}\\ .\quad\: \, &\textrm{Persamaan grafik fungsi seperti gambar di atas adalah}\, ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&f(x)=2^{x-2}&&&\textrm{D}.&f(x)=\, ^{2}\log (x-1)\\ \textrm{B}.&f(x)=2^{x}-2&\textrm{C}.&f(x)=2^{x}-1&\textrm{E}.&f(x)=\, ^{2}\log (x+1) \end{array}\\\\ &\color{black}\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{A}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\neq 0\: \: (\textrm{salah})\\ \textrm{B}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-2=2^{1}-2=0= 0\: \: (\color{red}\textbf{benar})\\ \textrm{C}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-1=2^{1}-1=1\neq 0\: \: (\textrm{salah})\\ \textrm{D}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1-1)=\, ^{2}\log 0=\textbf{tidak mungkin}\neq 0\: \: (\textrm{salah})\\ \textrm{E}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1+1)=\, ^{2}\log 2=1\neq 0\: \: (\textrm{salah})\\ \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{14}.&\textrm{Daerah hasil dari fungsi eksponen}\: \: y\: =x^{^{.}^{-\displaystyle \frac{2}{3}}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&y< 0&&&\textrm{D}.&y\leq 0\\ \textrm{B}.&y> 0&\textrm{C}.&y\geq 0&\textrm{E}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$

$.\qquad\: \, \color{magenta}\begin{aligned}\textrm{diketahui}&\\ y\: &=x^{^{.}^{-\displaystyle \frac{2}{3}}}\\ y^{3}\: &=x^{-2}\\ y^{3}\: &=\displaystyle \frac{1}{x^{2}},\: \textrm{atau}\\ y^{3}\times x^{2}\: &=1,\\ \textrm{sehingga}&\: \: y\: \: \textrm{tidak mungkin berharga}\: \: 0 \end{aligned}$

$\begin{array}{ll}\\ \fbox{15}.&\textrm{Solusi untuk persamaan}\: \: 3^{2x+1}=81^{x-2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&0&&&\textrm{D}.&4\displaystyle \frac{1}{2}\\\\ \textrm{B}.&2&\textrm{C}.&4&\textrm{E}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{array}{|c|c|}\hline \begin{aligned}3^{2x+1}&=81^{x-2}\\ \left (3^{2x} \right ).3^{1}&=\left (3^{4} \right )^{x-2}\\ 3.\left (3^{2x} \right )&=3^{4x}.3^{-8}\\ 3.\left (3^{2x} \right )&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}\\ 0&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}-3\left ( 3^{2x} \right )\\ 0&=\left ( 3^{2x} \right )^{2}-3^{9}.\left ( 3^{2x} \right )\\ 0&=\left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right ) \end{aligned} &\begin{aligned}\textbf{atau}\qquad\qquad\qquad&\\ \left ( 3^{2x} \right )^{2}-27.\left ( 3^{2x} \right )&=0\\ \left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right )&=0\\ 3^{2x}=0\: \: \textrm{atau}\: \: 3^{2x}&=3^{9}\\ (\textrm{tm})\quad \textrm{atau}\: \: 3^{2x}&=3^{9}\\ 2^{x}&=9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{atau}\: \: x&=4\displaystyle \frac{1}{2} \end{aligned} \\\hline \end{array} \end{array}$

$\color{red}\begin{array}{ll}\\ \fbox{16}.&\textrm{Jika diketahui}\: \: x^{x^{x^{x^{\cdots }}}}=2019,\\ &\textrm{maka nilai dari}\: \: x=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&\sqrt{2019}&&&\textrm{D}.&\sqrt{2019}^{\sqrt{2019}}\\ \textrm{B}.&\sqrt[2019]{2019}&\textrm{C}.&2019^{\sqrt{2019}}&\textrm{E}.&\sqrt{2019\sqrt{2019}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\\x^{x^{x^{x^{\cdots }}}}&=2019\\ \textrm{maka},&\\ x^{2019}&=2019\\ x&=\sqrt[2019]{2019} \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika}\: \: \sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ &\textrm{maka nilai dari}\: \: \left ( a+b-c \right )^{abc}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&1000&&&\textrm{D}.&-1\\ \textrm{B}.&1&\textrm{C}.&0&\textrm{E}.&-1000 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\begin{aligned}\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )^{2}&=\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\times \left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\\ &=\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ &\qquad\qquad+\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ &=2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ &=2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ &=10+\sqrt{2^{2}.6}+\sqrt{2^{2}.10}+\sqrt{2^{2}.15}\\ &=10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}&=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ &\begin{cases} a &=2 \\ b & =3 \\ c &=5 \end{cases}.\qquad \textrm{sesuai dengan urutannya}\\ \textrm{Sehingga nilai}\qquad &\\ \left ( a+b-c \right )^{abc}&=\left ( 2+3-5 \right )^{2.3.5}\\ &=0^{30}\\ &=0 \end{aligned} \end{aligned} \end{array}$

$\color{red}\begin{array}{ll}\\ \fbox{18}.&\textrm{Nilai}\: \: \sqrt[3]{\displaystyle \frac{16}{\displaystyle \sqrt[3]{\frac{16}{\displaystyle \sqrt[3]{\displaystyle \frac{16}{...}}}}}}=....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&8&&&\textrm{D}.&\displaystyle \frac{1}{4}\\ \textrm{B}.&4&\textrm{C}.&2&\textrm{E}.&\displaystyle \frac{1}{16} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}&=....\\ \textrm{misalkan}&\\ x&=\displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}\\ x&=\sqrt[3]{\displaystyle \frac{16}{x}},\qquad \textrm{masing-masing ruas dipangkatkan 3, sehingga}\\ x^{3}&=\displaystyle \frac{16}{x}\\ x^{3}.x&=16\\ x^{4}&=16\\ x&=\sqrt[4]{16}=\sqrt[4]{2^{4}}\\ &=2\\ \therefore \displaystyle \sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{\sqrt[3]{\frac{16}{...}}}}}}&=2 \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{19}.&\textrm{Jumlah semua nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: 2019^{x^{2}-7x+7}=2020^{x^{2}-7x+7}\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-7&&&\textrm{D}.&5\\ \textrm{B}.&-5&\textrm{C}.&-3&\textrm{E}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}\begin{aligned}2019^{x^{2}-7x+7}&=2020^{x^{2}-7x+7}\\ \textrm{karena bas}&\textrm{is tidak sama}\\ \textrm{maka pang}&\textrm{kat haruslah}=0,\: \textrm{yaitu}:\\ x^{2}-7x+7&=0,\: \: \textrm{jumlah semua nilai yang memenuhi adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}\\ \textrm{karena}\qquad&x^{2}-7x+7=0\begin{cases} \textrm{akar-akarnya yaitu}\begin{cases} x_{1} \\ x_{2} \end{cases} \\ \textrm{koefisien} /\textrm{konstan}\begin{cases} a=1 \\ b=-7 \\ c=7 \end{cases} \end{cases}\\ \therefore \: \: x_{1}+x_{2}&=-\displaystyle \frac{-7}{1}\\ &=7 \end{aligned} \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{20}.&\textrm{Nilai eksak dari}\\ &\displaystyle \frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+\frac{1}{10^{-2018}+1}+...+\displaystyle \frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1} \\ &\begin{array}{lllllllll}\\ \textrm{A}.&2020&&&\textrm{D}.&2021,5\\ \textrm{B}.&2020,5&\textrm{C}.&2021&\textrm{E}.&2022 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\textrm{Misal},\: \: x&=\frac{1}{10^{-2020}+1}+\frac{1}{10^{-2019}+1}+\frac{1}{10^{-2018}+1}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ \textrm{maka},\, \: x&=\frac{1}{\frac{1}{10^{2020}}+1}+\frac{1}{\frac{1}{10^{2019}}+1}+\frac{1}{\frac{1}{10^{2018}}+1}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{10^{2019}}{1+10^{2019}}+\frac{10^{2018}}{1+10^{2018}}+...+\frac{1}{10^{2018}+1}+\frac{1}{10^{2019}+1}+\frac{1}{10^{2020}+1}\\ &=\frac{10^{2020}}{1+10^{2020}}+\frac{1}{10^{2020}+1}+\frac{10^{2019}}{1+10^{2019}}+\frac{1}{10^{2019}+1}+\frac{10^{2018}}{1+10^{2018}}++\frac{1}{10^{2018}+1}...+\frac{1}{1^{0}+1}\\ &=\frac{10^{2020}+1}{10^{2020}+1}+\frac{10^{2019}+1}{10^{2019}+1}+\frac{10^{2018}+1}{10^{2018}+1}+...+\frac{10^{1}+1}{10^{1}+1}+\frac{1}{1^{0}+1}\\ &=\underset{\textrm{sebanyak}\: 2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ &=2020,5 \end{aligned} \end{array}$

PAS MATEMATIKA BLOG

Lanjutan 1 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

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