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

\begin{array}{ll}\\ \fbox{21}.&\textrm{Solusi dari persamaan eksponen}\\ & \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}\: \: \textbf{D}\\ \end{array}.

.\quad\: \, \color{blue}\begin{aligned}\left (x^{2}-5x+5 \right )^{\left ( 2x+3 \right )}&=\left (x^{2}-5x+5 \right )^{\left ( 3x-2 \right )}\\ \textbf{bentuk umum}:\: &\: h(x)^{f(x)}=h(x)^{g(x)}\\ \bullet \: \: \textrm{Pangkat sama}&, \\ 2x+3&=3x-2\\ 2x-3x&=-2-3\\ -x&=-5\Rightarrow x=5\\ \bullet \: \: \textrm{Basis}\: h(x)=1&\\ x^{2}-5x+5&=1\\ x^{2}-5x+4&=0\\ (x-1)(x-4)&=0\Rightarrow x_{1}=1\: \: \textrm{atau}\: \: x_{2}=4\\ \bullet \: \: \textrm{Basis}\: h(x)=0&\: \textrm{dengan syarat},f(x)\&g(x)> 0\\ f(1)&=2.1+3>0\\ g(1)&=3.1-2>0,\: \: dan\\ f(4)&=2.4+3>0\\ g(4)&=3.4-2>0\\ \textrm{Sehingga keduanya}&\: \textrm{memenuhi} \end{aligned}.

.\quad\: \, \color{red}\begin{aligned} \bullet \: \: \textrm{Basis}\: h(x)=0&\: \textrm{dengan syarat},f(x)\&g(x)> 0\\ x^{2}-5x+5&=0\Rightarrow x_{1,2}=\displaystyle \frac{-(-5)\pm \sqrt{5^{2}-4.1.5}}{2}\\ &\Rightarrow x_{1,2}=\displaystyle \frac{5\pm \sqrt{5}}{2}\\ f\left ( \displaystyle \frac{5+\sqrt{5}}{2} \right )&=2\left ( \displaystyle \frac{5+\sqrt{5}}{2} \right )+3>0\\ g\left ( \displaystyle \frac{5-\sqrt{5}}{2} \right )&=3\left ( \displaystyle \frac{5-\sqrt{5}}{2} \right )-2>0 \end{aligned}.

.\quad\: \, \color{magenta}\begin{aligned} \bullet \: \: \textrm{Basis}\: h(x)=-1&\: \textrm{dengan syarat},f(x)\&g(x)\\ &\textrm{keduanya ganjil atau keduanya genap}\\ x^{2}-5x+5&=-1\\ x^{2}-5x+6&=0\\ (x-2)(x-3)&=0\Leftrightarrow x_{1}=2\: \: \textrm{atau}\: \: x_{2}=3\\ \textrm{untuk}\: \: \: f(2)&=2.2+3=7\Rightarrow \textrm{ganjil}\\ g(2)&=3.2-2=6-2=4\Rightarrow \textrm{genap}\\ \textrm{sehingga}&\: \: x=2\: \: \textrm{bukan penyelesaian}\\ \textrm{untuk}\: \: \: f(3)&=2.3+3=9\Rightarrow \textrm{ganjil}\\ g(3)&=3.3-2=9-2=7\Rightarrow \textrm{ganjil}\\ maka\: \: x&=3\: \: \textrm{merupakan penyelesaian}.\\ \textrm{Jadi, solusi dari}&\: \textrm{persamaan di atas adalah}\\ &=1,3,4,5,\displaystyle \frac{5\pm \sqrt{5}}{2} \end{aligned}.

\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}.

\color{magenta}\begin{array}{ll}\\ \fbox{23}.&\textrm{(\textbf{Prestasi Maksima 2007}) Carilah semua bilangan real}\: \: x \\ &\textrm{yang memenuhi persamaan eksponensial}\: \: \: \: \displaystyle \frac{8^{x}+27^{x}}{12^{x}+18^{x}}=\frac{7}{6}\\\\ &\textrm{Solusi}:\\ &\color{black}\begin{aligned}\textrm{Untuk jawaban No}.23\quad\qquad\quad & \\\\ \displaystyle \frac{8^{x}+27^{x}}{12^{x}+18^{x}}&=\frac{7}{6}\\ \displaystyle \frac{2^{3x}+3^{3x}}{(2^{2}.3)^{x}+(2.3^{2})^{x}}&=\frac{7}{6}\Leftrightarrow \displaystyle \frac{2^{3x}+3^{3x}}{2^{2x}.3^{x}+2^{x}.3^{2x}}=\frac{7}{6}\\ \displaystyle \frac{\left (2^{x} \right )^{3}+\left (3^{x} \right )^{3}}{\left (2^{x} \right )^{2}.3^{x}+2^{x}.\left (3^{x} \right )^{2}}&=\frac{7}{6}\\ \begin{cases} 2^{x} &=p \\ 3^{x} &=q \end{cases}\\ \displaystyle \frac{p^{3}+q^{3}}{p^{2}q+pq^{2}}&=\frac{7}{6}\Leftrightarrow \displaystyle \frac{(p+q)\left ( p^{2}-pq+q^{2} \right )}{pq(p+q)}=\frac{7}{6}\\ \displaystyle \frac{p^{2}-pq+q^{2}}{pq}&=\frac{7}{6}\\ 6p^{2}-6pq+6q^{2}&=7pq\\ 6p^{2}-13pq+6q^{2}&=0\\ (2p-3q)(3p-2q)&=0\\ 2p-3q=0\quad \textrm{V}\quad 3p-2q&=0\\ 2p=3q\quad \textrm{V}\quad 3p&=2q\\ 2.2^{x}=3.3^{x}\quad \textrm{V}\quad 3.2^{x}&=2.3^{x}\\ x=-1\qquad \textrm{V}\qquad x&=1\\ \textrm{Jadi},\: \: \textrm{HP}=\left \{ x|x=-1\: \: \textrm{atau}\: \: x=1 \right \}& \end{aligned} \end{array}.

\color{magenta}\begin{array}{ll}\\ \fbox{24}.&\textrm{Jika}\: \: f(x)=b^{x},\: \: \textrm{di mana konstan positif},\\\\ &\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}= ....\\ &\begin{array}{lllllllll}\\ \textrm{A}.&f\left ( x^{2} \right )&&&\textrm{D}.&f(x+1)-f(x-1)\\ \textrm{B}.&f(x+1)f(x-1)&\textrm{C}.&f(x+1)+f(x-1)&\textrm{E}.&f\left ( x^{2}-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{E}\\ &\color{black}\begin{aligned}\displaystyle \frac{f\left ( x^{2}+x \right )}{f(x+1)}&=\frac{b^{x^{2}+x}}{b^{x+1}}\\ &=b^{x^{2}+x-(x+1)}\\ &=b^{x^{2}-1}\\ &=f\left ( x^{2}-1 \right ) \end{aligned} \end{array}..

\color{magenta}\begin{array}{ll}\\ \fbox{25}.&\textrm{Misal}\: \: n\: \: \textrm{suatu bilangan asli dan}\: \: x\: \: \textrm{adalah bilangan riil positif}\\ &\textrm{Jika}\: \: \: \: 2x^{n}+\displaystyle \frac{3}{x^{.^{-\displaystyle \frac{n}{2}}}}-2=0,\: \textrm{maka nilai}\: \: \: \displaystyle \frac{2}{x^{n}+\displaystyle \frac{1}{4}}=....\\ &\textrm{Solusi}:\\ &\color{black}\begin{aligned}2x^{n}+\displaystyle \frac{3}{x^{.^{-\displaystyle \frac{n}{2}}}}-2&=0\\ 2\left ( x^{.^{\displaystyle \frac{n}{2}}} \right )^{2}+3.\left ( x^{.^{\displaystyle \frac{n}{2}}} \right )-2&=0\\ \textrm{Misalkan}\: \: \: \: p=x^{.^{\displaystyle \frac{n}{2}}}&\: \: \textrm{maka}\\ 2p^{2}+3p-2&=0\\ (p+2)(2p-1)&=0\Leftrightarrow p_{1}=-2\: \: \textbf{atau}\: \: p_{2}=\frac{1}{2}\\ \textrm{Ambil nilai}\: \: p\: \: \textrm{yang}&>0,\: \: \textrm{yaitu}\: \: p=\frac{1}{2}\\ \textrm{maka nilai}\: \: \displaystyle \frac{2}{x^{n}+\displaystyle \frac{1}{4}}&=\displaystyle \frac{2}{p^{2}+\displaystyle \frac{1}{4}}=\displaystyle \frac{2}{\displaystyle \frac{1}{4}+\frac{1}{4}}=\displaystyle \frac{2}{\displaystyle \frac{1}{2}}=4 \end{aligned} \end{array}..

\color{blue}\begin{array}{ll}\\ \fbox{26}.&\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}.

Sumber Referensi

  1. Kanginan, Marthen, Ghany Akhmad, Hadi Nurdiansyah. 2014. Matematika untuk SMA/MA Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Yrama Widya.
  2. Soetiyono, dkk. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program IPA. Jakarta: Yudistira.
  3. Sukino. 2013. Matematika Kelompok Peminatan Matematika dan Ilmu Alam untuk SMA/MA Kelas x. Jakarta: Erlangga.
  4. Susianto, Bambang. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.
  5. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

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