PAS MATEMATIKA BLOG: 2019

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

Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

$\color{magenta}\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai untuk}\: \: -\left | -2019 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2019\quad &&&\textrm{d}.&-2019^{-1}\\ \textrm{b}.&2019&\textrm{c}.&2019^{-1}\quad &\textrm{e}.&2019^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\color{blue}\begin{aligned}-\left | -2019 \right |&=-\left ( 2019 \right )=-2019 \end{aligned} \end{array}$

$\color{magenta}\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai untuk}\: \: \left | -4 \right |-\left | -6^{2}\times 2 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-68\quad &&&\textrm{d}.&68\\ \textrm{b}.&-40&\textrm{c}.&40\quad &\textrm{e}.&76 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\color{black}\begin{aligned}\left | -4 \right |-\left | -6^{2}\times 2 \right |&=\left ( 4 \right )-\left | -36\times 2 \right |\\ &=4-\left | -72 \right |\\ &=4-\left ( 72 \right )\\ &=-68 \end{aligned} \end{array}$

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

$\color{blue}\begin{array}{ll}\\ \fbox{21}.&\textrm{Solusi dari}\\ &\displaystyle \left ( x^{2}-5x+5 \right )^{2x+3}=\left ( x^{2}-5x+5 \right )^{3x-2} \: \: \textrm{adalah}\: ....\\\\ &\begin{matrix} \textrm{A}. & 1,2,4,5, \displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2}\\ & & \\ \textrm{B}. & 1,3,4,5, \displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2} \\ & & \\ \textrm{C}. & 1,2,4,5, \displaystyle \frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2} \\ & & \\ \textrm{D}. & 1,3,4,5, \displaystyle \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \\ & & \\ \textrm{E}. & 1,2,3,6, \displaystyle \frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2} \\ & & \end{matrix} \\\\ &\textrm{Jawab}:\quad \textbf{dicoba buat latihan}\\ &\begin{aligned} \end{aligned} \end{array}$

$\color{blue}\begin{array}{ll}\\ \fbox{22}.&\textrm{Jumlah akar-akar persamaan}\: \: 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{A}.&-2&&&\textrm{D}.&1\\ \textrm{B}.&-1&\textrm{C}.&0&\textrm{E}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{D}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ \fbox{23}.&\textrm{Diketahui persamaan}\\\\ &\displaystyle \frac{x^{2}}{10000}=\displaystyle \frac{1000}{x^{2\left ( {{^{10}\log x}} \right )-8}} \\\\ &\textrm{Hasil kali dari nilai-nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&10^{2}&&&\textrm{D}.&10^{7}\\ \textrm{B}.&10^{3}&\textrm{C}.&10^{4}&\textrm{E}.&10^{8} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{B}\\ &\begin{aligned}\displaystyle \frac{x^{2}}{10000}&=\displaystyle \frac{1000}{x^{2\left ( ^{10}\log x \right )-8}}\\ x^{2}.x^{2\left ( ^{10}\log x \right )-8}&=10^{3}.10^{4}\\ x^{2\left ( ^{10}\log x \right )-6}&=10^{7}\\ x^{2\log x-6}&=10^{7},\quad \textrm{di}-log-\textrm{kan masing-masing ruas}\\ \log x^{\log x^{2}-6}&= \log 10^{7}\\ \left ( 2\log x-6 \right )\log x&=7\\ 2\left ( \log x \right )^{2}-6\left ( \log x \right )-7&=0\\ \textrm{persamaan kuadrat yan}&\textrm{g variabelnya berupa}-log.\\ \textrm{Sehingga hasil kali dari}\, &\: \textrm{nilai-nilai}\: \: x-\textrm{nya adalah}:\\ \alpha +\beta &=-\displaystyle \frac{b}{a}\\ \log x_{1}+\log x_{2}&=-\displaystyle \frac{b}{a},\qquad \textrm{ingat bahwa}\: \: \log x=\, ^{10}\log x\\ \log x_{1}.x_{2}&=-\displaystyle \frac{-6}{2}\\ ^{10}\log x_{1}.x_{2}&=3\\ x_{1}.x_{2}&=10^{3} \end{aligned} \end{array}$

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

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

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

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

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

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

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

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

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

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

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

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

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

Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Peminatan

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

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

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

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

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

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

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

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

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

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

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

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Turunan (Kelas XII MIPA)

A. Pendahuluan

Mengenal laju perubahan untuk nilai suatu fungsi

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Laju Perubahan}}\\\hline \textrm{Laju erubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}$

Perhatikanlah ilustrasi gambar berikut

Misalkan diketahui fungsi $y=f(x)$ terdefinisi untuk semua harga x di sekitar $x=k$ . Jika $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ ada, maka bentuk $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ disebut turunan dari fungsi $f(x)$ saat $x=k$ .

B. Notasi

Notasi turunan fungsi dilambangkan dengan ${f}'(k)$ dengan ${f}'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ .
Lambang ${f}'(k)$ dibaca f aksen k desebut turunan atau derivatif untuk fungsi $f(x)$ terhadap $x$ saat $x=k$ .
Jika limitnya ada, dapat dikatakan fungsi $f(x)$ diferensiabel(dapat didiferebsialkan) saat $x=k$ dan bentuk limitnya selanjutna dilambangkan dengan ${f}'(k)$ .
Misalkan fungsi $f(x)$ memiliki turunan ${f}'(x)$ . Jika ${f}'(k)$ tidak terdefinisi maka $f(x)$ tidak diferensiabel di $x=k$ .

Bentuk umum turunan fungsi ang selanjutnya disebut juga turunan pertama fungsi $y$ terhadap $x$ dapat dinotasikan dengan berbagai bentuk berikut yaitu:

$\color{blue}{y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}$

$\colorbox{yellow}{\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}}$

$\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: g(x)=3x-5, \textrm{hitunglah laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{g(x)=3x-5}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned} g(x)&=3x-5\\ g(2)&=3.2-5=1\\ {g}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{g(x)-g(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=3 \end{aligned}&\begin{aligned}g(2)\: \,&=1\\ g(2+&h)=3(2+h)-5=3h+1\\ g(2+&h)-g(2)=3h\\ {g}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{g(2+h)-g(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=3\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Jadi, laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\: \: \textrm{adalah 3}}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2017x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \begin{aligned}f(x)&=2017x^{2}\\ f(x+h)&=2017(x+h)^{2}\\ &=2017\left ( x^{2}+2xh+h^{2} \right )\\ &=2017x^{2}+4034xh+2017h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2017x^{2}+4034xh+2017h^{2} \right )-\left (2017x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4034xh+2017h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4034x+2017h\\ &=4034x \end{aligned}&\begin{aligned}{f}'(x)&=4034x\\ \textrm{maka},&\\ {f}'(1)&=4034.1\\ &=4034\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\: \: \textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}\textrm{Untuk}&\: \textrm{jawaban poin a dan b adalah sebagai berikut}\\ {f}'(x)&=-\displaystyle \frac{2}{x^{3}}\\ {f}'(a)&=-\displaystyle \frac{2}{a^{3}}\\ \textrm{maka},&\\ {f}'(0)&=-\displaystyle \frac{2}{0^{3}}\\ &=-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa bilangan}\: \: 0\\ &\textrm{maka}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$