Vektor (Kelas X Peminatan Mat&IA K13 Revisi)

Silahkan kunjungi lembaran lama di

Tambahan Materi Vektor

A. Skalar dan Vektor

  • Besaran skalar adalah besaran yang hanya memiliki besar saja tanpa arah
  • Besaran vektor adalah besaran yang memiliki besar sekaligus arah.

Perhatikanlah tiga ilustrasi berikut ini

\begin{array}{|l|c|c|}\hline &\multicolumn{2}{c|}{\textrm{Vektor}}\\\cline{2-3} \raisebox{0.0ex}[0cm][0cm]{\quad\qquad Istilah} &\textrm{Bidang}\: \left ( \textrm{XY} \right )&\textrm{Ruang}\: \left ( \textrm{XYZ} \right )\\\cline{2-3} &\textrm{R}^{2}&\textrm{R}^{3}\\\hline \textrm{Misal titik P}&\begin{cases} \bar{p} &=(a_{1},a_{2}) \\ \bar{p} &=\begin{pmatrix} a_{1}\\ a_{2} \end{pmatrix} \end{cases}&\begin{cases} \bar{p} &=(a_{1},a_{2},a_{3}) \\ \bar{p} &= \begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix} \end{cases}\\\hline \textrm{Vektor satuan}&\bar{i},\: \bar{j}&\bar{i},\: \bar{j},\: \bar{k}\\\hline \textrm{Vektor satuan dari}\: \: \bar{p}&\multicolumn{2}{|c|}{\hat{p}=\displaystyle \frac{\bar{p}}{\left | \bar{p} \right |}}\\\hline \textrm{Vektor posisi P}&\bar{OP}=\bar{p}=a_{1}\bar{i}+a_{2}\bar{j}&\bar{OP}=\bar{p}=a_{1}\bar{i}+a_{2}\bar{j}+a_{3}\bar{k}\\\hline \textrm{Panjang vektor P}&\left |\bar{p} \right |=\sqrt{a_{1}^{2}+a_{2}^{2}}&\left |\bar{p} \right |=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\\\hline \textrm{Sifat-sifat operasi}&\multicolumn{2}{|l|}{\begin{array}{ll}\\ &\textrm{Untuk}\: \: \bar{p}=(a,b)=\begin{pmatrix} a\\ b \end{pmatrix},\: \: \bar{q}=(c,d)=\begin{pmatrix} c\\ d \end{pmatrix}\\ 1.&\bar{p}=\bar{q},\: \textrm{jika}\begin{cases} \left | \bar{p} \right |=\left | \bar{q} \right | \\ \textrm{arah}\: \bar{p}=\textrm{arah}\: \bar{q} \end{cases}\\ 2.&\textrm{Jika}\: \: \bar{w}=\bar{p}+\bar{q},\: \textrm{maka}\: \: \bar{w}=\begin{pmatrix} a+c\\ b+q \end{pmatrix}\\ 3.&\textrm{Jika}\: \: \bar{w}=k\bar{p},\: \textrm{maka}\: \: \bar{w}=\begin{pmatrix} ka\\ kb \end{pmatrix}\\ 4.&\textrm{Lawan vektor(invers)}\: \bar{p}\: \: \textrm{adalah}\: \: -\bar{p}\\ 5.&\bar{p}+\bar{q}=\bar{q}+\bar{p}\\ 6.&\bar{p}+\left ( -\bar{p} \right )=\bar{0}\\ 7.&\bar{p}+\bar{0}=\bar{0}+\bar{p}=\bar{p}\\ 8.&(k+l)\bar{p}=k\bar{p}+l\bar{p}\\ 9.&k\left (\bar{p}+\bar{q} \right )=k\bar{p}+k\bar{q} \end{array}}\\\hline \end{array}.

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Contoh Soal Fungsi Komposisi dan Invers

\begin{array}{ll}\\ \fbox{1}.&\textrm{Diketahui fungsi}\: \: f(2x)=8x-9\: \: \textrm{dan}\: \: g(3x+1)=6x+3.\\ &\textrm{Rumus untuk}\: \: \left ( f+g \right )(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&6x+8&&&\textrm{d}.&14x-6\\ \textrm{b}.&6x-8&\textrm{c}.&14x+6&\textrm{e}.&6x-6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} f(2x) &=8x-9\quad \Rightarrow \quad f(x)=f\left ( 2\left ( \displaystyle \frac{x}{2} \right ) \right )=8\left ( \displaystyle \frac{x}{2} \right )-9=4x-9 \\ g(3x+1) &=6x+3\quad \Rightarrow \quad g(x)=g\left ( 3\left ( \displaystyle \frac{x-1}{3} \right )+1 \right )=6\left ( \displaystyle \frac{x-1}{3} \right )+3\\ &\qquad\qquad\qquad\quad\qquad\: \: \: =2x+1 \end{cases}\\ \left ( f+ g \right )&=(4x-9)+(2x+1)\\ &=6x-8 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: \: f(x)=3-x,\: \: \textrm{maka}\: \: f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2x^{2}-6x+4&&&\textrm{d}.&6x+4\\ \textrm{b}.&2x^{2}+4x+6&\textrm{c}.&2x^{2}-4x-6&\textrm{e}.&-4x+6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Diketahui bahwa}\: \: f(x)=3&-x,\: \textrm{sehingga}\\ f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)&=\left ( 3-x^{2} \right )+\left ( 3-x \right )^{2}-2(3-x)\\ &=\left (3-x^{2} \right )+\left ( 9-6x+x^{2} \right )-\left ( 6-2x \right )\\ &=-x^{2}+x^{2}-6x+2x+3+9-6\\ &=-4x+6 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui beberapa fungsi memiliki sifat-sifat sebagaimana berikut ini}:\\ &(i)\quad \Phi (-x)=-\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &(ii)\quad \Phi (-x)=\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &\textrm{Jika diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{memiliki sifat}\: \: (i)\: \: \textrm{dan fungsi}\: \: h\: \: \textrm{dan}\: \: k\\ &\textrm{memiliki sifat}\: \: (ii),\: \: \textrm{maka pernyataan berikut yang salah adalah}\: ....\\ &(1)\quad (f+g)(-x)=-(f+g)(x)\\ &(2)\quad (f.k)(-x)=-(f.k)(x)\\ &(3)\quad (h-k)(-x)=(h-k)(x)\\ &(4)\quad (h-g)(-x)=(h-g)(x)\\ &\begin{array}{llllll}\\ \textrm{a}.&(1),(2)\: \textrm{dan}\: (3)&&&\textrm{d}.&(4)\: \textrm{saja}\\ \textrm{b}.&(1)\: \textrm{dan}\: (3)&\textrm{c}.&(2)\: \textrm{dan}\: (4)&\textrm{e}.&\textrm{semuanya benar} \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{SIMAK UI 2014 Mat Das})\\\\ &\textrm{Jawab}:\quad \textbf{a} \end{array}.

.\qquad \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} \Phi (-x)=-\Phi (x) & (\textrm{fungsi ganjil})\begin{cases} f & \text{ misal } f(x)=x \\ g & \text{ misal } g(x)=2x \end{cases} \\\\ \Phi (-x)=\Phi (x) & (\textrm{fungsi genap})\begin{cases} h & \text{ misal } h(x)=x^{2} \\ k & \text{ misal } k(x)=2x^{2} \end{cases} \end{cases}\\ & \end{aligned}}\\\hline \begin{aligned}(1)\quad (f+g)(-x)&=-(f+g)(x)\\ &\textrm{benar}\\ (2)\qquad (f.k)(-x)&=-(f.k)(x)\\ &\textrm{benar}\\ \end{aligned}&\begin{aligned} (3)\quad (h-k)(-x)&=(h-k)(x)\\ &\textrm{benar}\\ (4)\quad (h-g)(-x)&=(h-g)(x)\\ &\textrm{salah} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui fungsi}\: \: f:R\rightarrow R\: \: \textrm{dan}\: \: g:R\rightarrow R\: \: \textrm{dirumuskan dengan}\\ &f(x)=x-1\: \: \textrm{dan}\: \: g(x)=x^{2}+2x-3.\: \textrm{Fungsi komposisi}\: \: g\: \: \textrm{atas}\: \: f\\ &\textrm{dinotasikan dengan}\\ &\begin{array}{ll}\\ \textrm{a}.&\left ( g\circ f \right )(x)=x^{2}-4\\ \textrm{b}.&\left ( g\circ f \right )(x)=x^{2}-5\\ \textrm{c}.&\left ( g\circ f \right )(x)=x^{2}-6\\ \textrm{d}.&\left ( g\circ f \right )(x)=x^{2}-4x-4\\ \textrm{e}.&\left ( g\circ f \right )(x)=x^{2}-4x-5 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UN 2016})\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} f(x) &=x-1 \\ g(x) &=x^{2}+2x-3 \end{cases}\\ \left ( g\circ f \right )&=g\left ( f(x) \right )\\ &=\left ( f(x) \right )^{2}+2\left ( f(x) \right )-3\\ &=\left ( x-1 \right )^{2}+2\left ( x-1 \right )-3\\ &=\left (x^{2}-2x+1 \right )+\left ( 2x-2 \right )-3\\ &=x^{2}-4 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{dan}\: \: g(x)\: \: \textrm{adalah dua polinom yang berbeda dengan}\: \: f(10)=m\\ &\textrm{dan}\: \: g(10)=n.\: \textrm{Jika}\: \: f(x).h(x)=\left (\displaystyle \frac{f(x)}{g(x)}-1 \right )\left ( f(x)+g(x) \right )\: \textrm{dengan}\: \: h(10)=-\displaystyle \frac{16}{15},\\ &\textrm{maka nilai maksimum dari}\: \: \left | m+n \right |=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&8&&&\textrm{d}.&2\\ \textrm{b}.&6\qquad&\textrm{c}.&4\qquad&\textrm{e}.&0 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SIMAK UI 2014 Mat IPA})\\\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}f(x).h(x)&=\left (\displaystyle \frac{f(x)}{g(x)}-1 \right )\left ( f(x)+g(x) \right )\\ f(x).h(x)&=\left (\displaystyle \frac{f(x)-g(x)}{g(x)} \right )\left ( f(x)+g(x) \right )\\ f(x).g(x).h(x)&=f^{2}(x)-g^{2}(x)\\ f(10).g(10).h(10)&=f^{2}(10)-g^{2}(10)\\ m\times n\times -\displaystyle \frac{16}{15}&=m^{2}-n^{2}\\ 0&=15m^{2}+16mn-15n^{2}\\ 0&=\left ( 3m+5n \right )\left ( 5m-3n \right )\\ (3m+5n)&=0\Rightarrow 3m=-5n\: \: \textrm{atau}\: \: (5m-3n)=0\Rightarrow 5m=3n\\ &\qquad\qquad \displaystyle \frac{m}{n}=-\frac{5}{3}\qquad\qquad\qquad\qquad\qquad\quad \displaystyle \frac{m}{n}=\frac{3}{5}\\ \textrm{maka nilai maksi}&\textrm{mum}\: \: \left | m+n \right |=\left | 3+5 \right |=8 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: f(x)=\sqrt{2x+3}\: \: \textrm{dan}\: \: g(x)=x^{2}+1,\: \textrm{maka}\: \: \left ( f\circ g \right )(2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&2,24&&&\textrm{d}.&6\\ \textrm{b}.&3\qquad&\textrm{c}.&3,61\qquad&\textrm{e}.&6,16 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SAT Subject Test})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=f\left ( g(x) \right )\\ &=\sqrt{2g(x)+3}\\ &=\sqrt{2\left ( x^{2}+1 \right )+3}\\ \left ( f\circ g \right )(2)&=\sqrt{2\left ( 2^{2}+1 \right )+3}\\ &=\sqrt{2(5)+3}\\ &=\sqrt{13}\\ &\approx 3,61 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}}\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\: \textrm{maka}\: \: g(x+2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{x+3}&&&\textrm{d}.&x+3\\ \textrm{b}.&\displaystyle \frac{1}{x-2}\qquad&\textrm{c}.&x-2\qquad&\textrm{e}.&x+5 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UM UGM 2010 Mat Das})\\\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ f\left ( g(x) \right )&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \displaystyle \frac{1}{\sqrt{\left (g(x) \right )^{2}-2}}&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \left (g(x) \right )^{2}-2&=x^{2}+6x+7\\ \left (g(x) \right )^{2}&=x^{2}+6x+9\\ g(x)&=\sqrt{x^{2}+6x+9}=\sqrt{(x+3)^{2}}\\ g(x)&=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5 \end{aligned} \end{array}..

Atau dapat juga dengan cara alternatif berikut

.\qquad \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&:\\ &\begin{cases} f(x) =& \displaystyle \frac{1}{\sqrt{x^{2}-2}} \\ \left (f\circ g \right )(x) & = \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \end{cases}\\ & \end{aligned}}\\\hline \begin{aligned}f(x) =y&= \displaystyle \frac{1}{\sqrt{x^{2}-2}}\\ y^{2}&=\displaystyle \frac{1}{x^{2}-2}\\ x^{2}-2&=\displaystyle \frac{1}{y^{2}}\\ x^{2}&=\displaystyle \frac{1}{y^{2}}+2\\ x&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(y)&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(x)&=\sqrt{\displaystyle \frac{1}{x^{2}}+2} \end{aligned}&\begin{aligned}g(x)&=\left (f^{-1}\circ f\circ g \right )(x)\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \right )^{2}}+2}\\ &=\sqrt{\left ( x^{2}+6x+7 \right )+2}\\ &=\sqrt{\left ( x^{2}+6x+9 \right )}\\ &=\sqrt{(x+3)^{2}}\\ &=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: g(x)=2x+4\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=4x^{2}+8x-3,\: \textrm{maka}\: \: f^{-1}(x)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&x+9&&&\textrm{d}.&\sqrt{x+1}+2\\ \textrm{b}.&\sqrt{x}+2\qquad&\textrm{c}.&x^{2}-4x-3\qquad&\textrm{e}.&\sqrt{x+7}+2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses Pencarian}}&\textrm{Hasil Invers}\\\hline \begin{aligned}g(x)=y&=2x+4\\ y-4&=2x\\ x&=\displaystyle \frac{y-4}{2}\\ f^{-1}(y)&=\displaystyle \frac{y-4}{2}\\ f^{-1}(x)&=\displaystyle \frac{x-4}{2}\\ & \end{aligned}&\begin{aligned}f(x)&=\left (f\circ g\circ g^{-1} \right )(x)\\ &=4\left ( g^{^{-1}}(x) \right )^{2}+8\left ( g^{-1}(x) \right )-3\\ &=4\left ( \displaystyle \frac{x-4}{2} \right )^{2}+8\left ( \displaystyle \frac{x-4}{2} \right )-3\\ &=\left ( \displaystyle x^{2}-8x+16 \right )+4x-16-3\\ &=x^{2}-4x-3\\ &=x^{2}-4x+4-7\\ &=(x-2)^{2}-7 \end{aligned}&\begin{aligned}f(x)=y&=(x-2)^{2}-7\\ y+7&=(x-2)^{2}\\ \sqrt{y+7}&=(x-2)\\ (x-2)&=\sqrt{y+7}\\ x&=\sqrt{y+7}+2\\ f^{-1}(y)&=\sqrt{y+7}+2\\ f^{-1}(x)&=\sqrt{x+7}+2 \end{aligned} \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: f\: \: \textrm{dan}\: \: g\: \textrm{adalah fungsi yang mempunyai invers dan memenuhi}\: \: f(2x)=g(x-3),\\ &\textrm{maka}\: \: f^{1}(x)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&g^{-1}\left ( \displaystyle \frac{x}{2}- \frac{2}{3}\right )&&&\textrm{d}.&2g^{-1}(x)-6\\ \textrm{b}.&g^{-1}\left ( \displaystyle \frac{x}{2} \right )-\displaystyle \frac{2}{3}\qquad&\textrm{c}.&g^{-1}(2x+6)\qquad&\textrm{e}.&2g^{-1}(x)+6 \end{array}\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SBMPTN 2016 Mat Das})\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&:\\ f(2x)&=g(x-3)=x,\: \textrm{sebagai misal}\\ &\begin{cases} f^{-1}(x) &=2x \\ g^{-1}(x) &=x-3 \end{cases}\\ & \end{aligned}}\\\hline \textrm{Proses}&\textrm{Hasil}\\\hline \begin{aligned}g^{-1}(x) &=x-3\\ x&=g^{-1}(x)+3\\ & \end{aligned}&\begin{aligned}f^{-1}(x) &=2x\\ &=2\left ( g^{-1}(x)+3 \right )\\ &=2g^{-1}(x)+6 \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Jika}\: \: f^{-1}(x)=\displaystyle \frac{x-1}{5}\: \: \textrm{dan}\: \: g^{-1}(x)=\displaystyle \frac{3-x}{2},\: \textrm{maka}\: \: \left (f\circ g \right )^{-1}(6)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0\qquad&\textrm{c}.&1\qquad&\textrm{e}.&3 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UMPTN 1995})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui b}&\textrm{ahwa}:\\ &\begin{cases} f^{-1}(x)&=\displaystyle \frac{x-1}{5} \\ g^{-1}(x)&=\displaystyle \frac{3-x}{2} \end{cases}\\ \left (f\circ g \right )^{-1}(x)&=\left (g^{-1}\circ f^{-1} \right )(x)\\ &=\displaystyle \frac{3-\left ( \displaystyle \frac{x-1}{5} \right )}{2}\\ \left (f\circ g \right )^{-1}(6)&=\displaystyle \frac{3-\left ( \displaystyle \frac{6-1}{5} \right )}{2}\\ &=\displaystyle \frac{3-1}{2}=\frac{2}{2}\\ &=1 \end{aligned} \end{array}.

 

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

 

Dipublikasi di Info, Matematika, Pendidikan | Meninggalkan komentar

Lanjutan Fungsi Komposisi dan Fungsi Invers (K13 Revisi)

Silahkan merujuk ke

Mengingatkan kembali (Sebagian materi berikut akan sama dengan materi tautkan di atas)

A. Operasi Aljabar Fungsi

\begin{array}{|l|l|}\hline \textrm{Aljabar Fungsi}&\textrm{Daerah Asal}\\\hline (f+g)(x)=f(x)+g(x)&D_{_{(f+g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f-g)(x)=f(x)-g(x)&D_{_{(f-g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f.g)(x)=f(x).g(x)&D_{_{(f.g)}}=D_{_{f}}\cap D_{_{g}}\\\hline \left ( \displaystyle \frac{f}{g} \right )(x)=\displaystyle \frac{f(x)}{g(x)}&D_{_{\left ( \frac{f}{g} \right )}}=D_{_{f}}\cap D_{_{g}}\: , \\ &\textrm{dengan}\: \: g(x)\neq 0 \\\hline \end{array}.

B. Fungsi Komposisi

Perhatikanlah ilustrasi berikut ini!

384

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Komposisi Fungsi}}\\\hline \textrm{Syarat}&\textrm{Sifat-sifat}\\\hline \begin{aligned}&R_{_{f}}\cap D_{_{g}}\neq \left \{ \: \right \} \end{aligned}&\begin{aligned}1.\: \: &\textrm{Tidak komutatif}\quad (f\circ g)(x)\neq (g\circ f)(x)\\ 2.\: \: &\textrm{Bersifat asosiatif}\quad f\circ (g\circ h)(x)= (f\circ g)\circ h(x)\\ 3.\: \: &\textrm{Adanya unsur dentitas}\quad (f\circ I)(x)=(I\circ f)(x)=f(x) \end{aligned}\\\hline \end{array}.

C. Fungsi Invers

Perhatikan pula islustrasi berikut

385

\begin{aligned}&\bullet \quad \textrm{Suatu fungsi}\: \: f:A\rightarrow B\: \: \textrm{memiliki fungsi invers} \: \: g:B\rightarrow A\: \: \textsl{jika dan hanya jika}\: \: f\\ &\qquad \textrm{merupakan fungsi}\: \textbf{bijektif}\\ &\bullet \quad \textrm{Jika fungsi}\: \: g\: \: \textrm{ada, maka}\: \: g\: \: \textrm{dinyatakan dengan}\: \: f^{-1}\: \: (\textrm{dibaca}:\: \: f\: \: \textrm{invers})\end{aligned}.

\begin{array}{ll}\\ \textrm{Perlu}&\textrm{diingat bahwa pada invers fungsi komposisi berlaku ketentuan sebagai berikut}\\ \blacklozenge &\left ( g\circ f \right )^{-1}(x)=\left ( f^{-1}\circ g^{-1} \right )(x)\\ \blacklozenge &\left ( f\circ g \right )^{-1}(x)=\left ( g^{-1}\circ f^{-1} \right )(x)\\ \blacklozenge &f(x)=\left ( \left (f^{-1} \right )^{-1}(x) \right )\\ \blacklozenge &x=f^{-1}\left ( f(x) \right )=\left ( f^{-1}\circ f \right )(x)=\left ( f\circ f^{-1} \right )(x)=f\left ( f^{-1}(x) \right ) \end{array}..

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}..

\begin{array}{ll}\\ 1.&\textrm{Diketahui bahwa 2 buah fungsi}\: \: f(x)=2x+1\: \: \textrm{dan}\: \: g(x)=\sqrt{1-x}\\ &\begin{array}{ll}\\ \textrm{a}.\quad (f+g)(x)&\textrm{c}.\quad (f.g)(x)\\ \textrm{b}.\quad (f-g)(x)&\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad (f+g)(x)&=f(x)+g(x)\\ &=(2x+1)+\sqrt{1-x}\\ D_{_{(f+g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{c}.\quad (f.g)(x)&=f(x).g(x)\\ &=(2x+1)\sqrt{1-x}\\ &=\sqrt{(2x+1)^{2}(1-x)}\\ &\: \: \: \: \: \: (2x+1)^{2}(1-x)\geq 0\\ D_{_{(f.g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad (f-g)(x)&=f(x)-g(x)\\ &=(2x+1)-\sqrt{1-x}\\ D_{_{(f-g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \\ &\end{aligned}&\begin{aligned}\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)&=....................\\ &....................\\ &.................... \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{yang dinyatakan berupa pasangan berurut seperti}\\ &\textrm{berikut}:\\ &\begin{array}{ll}\\ f&=\left \{ (4,1),(0,3),(1,4),(3,6),(2,10) \right \}\\ g&=\left \{ (1,0),(3,1),(4,2),(6,3),(10,4) \right \}\\ \end{array}\\\\ &\textrm{Tentukanlah}\: \: f\circ g,\: g\circ f,\: (f\circ g)(3),\: (f\circ g)(6), (g\circ f)(1),\: \textrm{dan}\: (g\circ f)(0)\\\\ &\textrm{Jawab}:\\ &\bullet \quad f\circ g=\left \{ (...,...),(...,...),(...,...),(...,...),(...,...) \right \}\\ &\bullet \quad g\circ f=\left \{ (0,1),(1,2),(2,4),(3,3),(4,0) \right \}\\ &\bullet \quad (f\circ g)(3)=f(g(3))=....=....\\ &\bullet \quad (f\circ g)(6)=f(g(6))=....=....\\ &\bullet \quad (g\circ f)(1)=g(f(1))=g(4)=2\\ &\bullet \quad (g\circ f)(0)=g(f(0))=g(3)=1\\ &\textrm{Perhatikanlah ilustrasi berikut}\\ \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Diketahui fungsi}\: \: f,\: g,\: \textrm{dan}\: \: h\: \: \textrm{dengan}\: \: f(x)=x-4,\: g(x)=3x+2\\ &\textrm{dan}\: \: h(x)=x^{2}-1\\ &\begin{array}{ll}\\ \textrm{a}.&(f\circ g\circ h)(x)\\ \textrm{b}.&(f\circ h\circ g)(x)\\ \textrm{c}.&(h\circ g\circ f)(x)\\ \textrm{d}.&(g\circ f\circ h)(x)\\ \textrm{e}.&(g\circ h\circ f)(x)\\ \textrm{f}.&(h\circ f\circ g)(x) \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|}\hline \begin{aligned}\textrm{a}.\quad \textrm{Misal}\: \: z(x)&=(f\circ g\circ h)(x)\\ &=(f\circ g)\left (x^{2}-1 \right )\\ &=f\left ( 3\left ( x^{2}-1 \right )+2 \right )\\ &=f\left ( 3x^{2}-1 \right )\\ &=\left ( 3x^{2}-1 \right )-4\\ &=3x^{2}-5\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \textrm{Misal}\: \: y(x)&=(f\circ h\circ g)(x)\\ &=(f\circ h)\left (3x+2 \right )\\ &=f\left ( \left (3x+2 \right )^{2}-1 \right )\\ &=f\left ( 9x^{2}+12x+4-1 \right )\\ &=f\left ( 9x^{2}+12x+3 \right )\\ &=\left ( 9x^{2}+12x+3 \right )-4\\ &=9x^{2}+12x-1 \end{aligned} \\\hline \begin{aligned}\textrm{c}.\quad \textrm{Misal}\: \: w(x)&=(h\circ g\circ f)(x)\\ &=(h\circ g)\left (x-4 \right )\\ &=h\left ( 3\left ( x-4 \right )+2 \right )\\ &=h\left ( 3x-2 \right )\\ &=\left ( 3x-2 \right )^{2}-1\\ &=9x^{2}-12x+3 \end{aligned} &\begin{aligned}&\textrm{Untuk jawaban}\\ &\textrm{yang lain silahkan}\\ &\textrm{di coba sendiri}\\ &\textrm{sebagai latihan}\\ &\textrm{mandiri}\\ & \end{aligned} \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah \textbf{invers} dari fungsi-fungsi berikut ini}\\ &\begin{array}{lllllll}\\ \textrm{a}.&f(x)=3x-4&\textrm{f}.&m(x)=\sqrt[5]{4x-1},\: x\neq \displaystyle \frac{1}{4}\\ \textrm{b}.&g(x)=5-3x&\textrm{g}.&n(x)=x^{3}-4\\ \textrm{c}.&j(x)=(x-2)^{2}&\textrm{h}&o(x)=\left ( 1-x^{3} \right )^{^{\frac{1}{3}}}-3\\ \textrm{d}.&k(x)=3x^{2}-2&\textrm{i}&p(x)=\displaystyle \frac{3x-2}{x-3},\: x\neq 3\\ \textrm{e}.&l(x)=x^{2}-4x+3&\textrm{j}&q(x-2)=\displaystyle \frac{x-4}{2x-4},\: x\neq 2 \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Soal yang belum diselesaikan silahkan dicoba sendiri untuk latihan} \end{array}.

.\qquad \begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad f(x)=y&=3x-4\\ 3x-4&=y\\ 3x&=y+4\\ x&=\displaystyle \frac{y+4}{3}\\ f^{-1}(y)&=\displaystyle \frac{y+4}{3}\\ f^{-1}(x)&=\displaystyle \frac{x+4}{3} \end{aligned}&\begin{aligned}\textrm{b}.\quad g(x)=y&=5-3x\\ 5-3x&=y\\ -3x&=y-5\\ x&=\displaystyle \frac{y-5}{-3}=\frac{5-y}{3}\\ g^{-1}(y)&=\displaystyle \frac{5-y}{3}\\ g^{-1}(x)&=\displaystyle \frac{5-x}{3} \end{aligned}&\begin{aligned}\textrm{c}.\quad j(x)=y&=(x-2)^{2}\\ (x-2)^{2}&=y\\ (x-2)&=y^{^{\frac{1}{2}}}=\pm \sqrt{y}\\ x&=\pm \sqrt{y}+2\\ j^{-1}(y)&=\pm \sqrt{y}+2\\ j^{-1}(x)&=\pm \sqrt{x}+2\\ \textrm{dengan}&\: \: x\geq 0\\ & \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad k(x)=y&=3x^{2}-2\\ 3x^{2}-2&=y\\ 3x^{2}&=y+2\\ x^{2}&=\displaystyle \frac{y+2}{3}\\ x&=\pm \sqrt{\displaystyle \frac{y+2}{3}}\\ k^{-1}(y)&=\pm \sqrt{\displaystyle \frac{y+2}{3}}\\ k^{-1}(x)&=\pm \sqrt{\displaystyle \frac{x+2}{3}}\\ \textrm{dengan}&\: x\geq -2\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad m(x)=y&=\sqrt[5]{4x-1}\\ \sqrt[5]{4x-1}&=y\\ 4x-1&=y^{5}\\ 4x&=y^{5}+1\\ x&=\displaystyle \frac{y^{5}+1}{4}\\ m^{-1}(y)&=\displaystyle \frac{y^{5}+1}{4}\\ m^{-1}(x)&=\displaystyle \frac{x^{5}+1}{4}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{j}.\quad q(x-2)&=\displaystyle \frac{x-4}{2x-4}\\ q(x)&=q((x+2)-2)\\ q(x)=y&=\displaystyle \frac{(x+2)-4}{2(x+2)-4}\\ q(x)=y&=\displaystyle \frac{x-2}{2x},\: x\neq 0\\ 2xy&=x-2\\ 2xy-x&=-2\\ x(2y-1)&=-2\\ x&=\displaystyle \frac{-2}{2y-1}=\frac{2}{1-2y}\\ q^{-1}(y)&=\frac{2}{1-2y}\\ q^{-1}(x)&=\frac{2}{1-2x}\\ \textrm{dengan}&\: x\neq \displaystyle \frac{1}{2} \end{aligned}\\\hline \end{array}.

 

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

 

Dipublikasi di Matematika, Pendidikan | Meninggalkan komentar

Contoh Soal Fungsi

\begin{array}{ll}\\ \fbox{1}.&\textrm{Relasi berikut yang akan berupa fungsi adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=\sqrt{x}&&&\textrm{d}.&f(x)=\sqrt{x}-1\\\\ \textrm{b}.&f(x)=1-\sqrt{x}&\textrm{c}.&f(x)=\sqrt{x}+1&\textrm{e}.&f(x)=\left | x \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{array}{|c|l|c|l|}\hline \textrm{No}&\: \: \qquad \textrm{Fungsi}&\textrm{Grafik}&\qquad\qquad \textrm{Keterangan}\\\hline 1.\textrm{a}&f(x)=y=\sqrt{x}&y^{2}=x&\textrm{Sebuah \textbf{prapeta} akan}\\\cline{1-3} 1.\textrm{b}&f(x)=y=1-\sqrt{x}&(1-y)^{2}=x&\textrm{dapat menghasilkan}\\\cline{1-3} 1.\textrm{c}&f(x)=y=1+\sqrt{x}&(y-1)^{2}=x&\textrm{\textbf{peta} yang berbeda, sehingga}\\\cline{1-3} 1.\textrm{d}&f(x)=y=\sqrt{x}-1&(y+1)^{2}=x&\textrm{akan berupa \textbf{relasi} saja}\\\hline 1.\textrm{e}&f(x)=y=\left | x \right |&y=\begin{cases} x & \text{ jika} \: \: x\geq 0 \\ -x & \text{ jika } \: \: x<0 \end{cases}&\begin{aligned}&\textrm{Dengan \textbf{prepeta} yang berbeda}\\ &\textrm{akan menghasilkan \textbf{peta} yang}\\ &\textrm{berbeda pula (\textbf{fungsi bijektif})} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{2}.&\textrm{Fungsi dari himpunan A ke himpunan B berikut termasuk jenis fungsi}.... \end{array}..

\begin{array}{ll}\\ .\: \quad.&\textrm{Relasi berikut yang akan berupa fungsi adalah}....\\ &\textrm{a}.\quad \textrm{fungsi umum}\\ &\textrm{b}.\quad \textrm{fungsi satu-satu, tetapi bukan fungsi pada}\\ &\textrm{c}.\quad \textrm{fungsi pada, tetapi bukan fungsi satu-satu}\\ &\textrm{d}.\quad \textrm{fungsi pada dan satu-satu}\\ &\textrm{e}.\quad \textrm{tidak ada jawaban yang benar}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|c|l|}\hline \textrm{No}&\textrm{Keterangan}&\textrm{Alasan}\\\hline 2.\textrm{a}&\textrm{Sesuai}&\textrm{\textbf{Sesuai} definisi fungsi}\\\hline 2.\textrm{b}&\textrm{Salah}&\begin{aligned}&\textrm{Karena bukan fungsi satu-satu(fungsi injektif)}\\ &\textrm{walau benar dikatakan bukan fungsi pada (fungsi surjektif)} \end{aligned}\\\hline 2.\textrm{c}&\textrm{Salah}&\begin{aligned}&\textrm{Karena bukan fungsi pada(fungsi surjektif)}\\ &\textrm{walau benar dikatakan bukan fungsi satu-satu (fungsi injektif)} \end{aligned}\\\hline 2.\textrm{d}&\textrm{Salah}&\textrm{Jelas bukan fungsi pada dan satu-satu(fungsi bijektif)}\\\hline 2.\textrm{e}&\textrm{Salah}&\textrm{Tidak sesuai}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{3}.&\textrm{Himpunan pasangan terurut yang ditunjukkan oleh fungsi}\\ &f:x \mapsto 2-\left ( x+1 \right )^{2}\: \: \textrm{dari domain}\: \: \left \{ -1,0,1,2 \right \}\: \: \textrm{adalah}.... \\ &\begin{array}{ll}\\ \textrm{a}.&\left \{ (-1,2),(0,3),(1,5),(2,7) \right \}\\ \textrm{b}.&\left \{ (-1,2),(0,1),(1,-2),(2,-7) \right \}\\ \textrm{c}.&\left \{ (-1,1),(0,-1),(1,-4),(2,7) \right \}\\ \textrm{d}.&\left \{ (-1,0),(0,3),(1,-2),(2,7) \right \}\\ \textrm{e}.&\left \{ (-1,0),(0,-4),(1,5),(2,-7) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f:x& \mapsto 2-\left ( x+1 \right )^{2}\\ -1&\mapsto 2-\left ( -1+1 \right )^{2}=2-0=2&..........(-1,2)\\ 0&\mapsto 2-\left ( 0+1 \right )^{2}=2-1=1&...............(0,1)\\ 1&\mapsto 2-\left ( 1+1 \right )^{2}=2-4=-2&.........(1,-2)\\ 2&\mapsto 2-\left ( 2+1 \right )^{2}=2-9=-7&.........(2,-7) \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{4}.&\textrm{Dari beberapa fungsi berikut yang merupakan fungsi genap adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.&f(x)=x^{2}+\left | x \right |-1\\ \textrm{b}.&f(x)=x^{3}-\left | x \right |+x\\ \textrm{c}.&f(x)=x\left | x \right |+x\\ \textrm{d}.&f(x)=\sqrt{x-1}\\ \textrm{e}.&f(x)=4-2x \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|l|l|c|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ \textrm{Suatu fungsi}&\: \textrm{dinamakan \textbf{fungsi genap}}\\ &\textrm{jika}\: f(x)=f(-x)\\ & \end{aligned}}\\\hline \textrm{No}&f(x)&f(-x)&\textrm{Keterangan}\\\hline 4.\textrm{a}&x^{2}+\left | x \right |-1&x^{2}+\left | x \right |-1&f(x)=f(-x)\\\hline 4.\textrm{b}&x^{3}-\left | x \right |+x&-x^{3}-\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{c}&x\left | x \right |+x&-x\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{d}&\sqrt{x-1}&\sqrt{-x-1}&f(x)\neq f(-x)\\\hline 4.\textrm{e}.&4-2x&4+2x&f(x)\neq f(-x)\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui himpunan}\\ &A=\left \{ x|x\: \textrm{adalah faktor prima dari}\: 16 \right \}\\ &B=\left \{ x|x\: \textrm{adalah faktor dari}\: 16 \right \}\\ &\textrm{Banyaknya pemetaan dari}\: A\: ke\: B\: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&25\\\\ \textrm{b}.&2&\textrm{c}.&5&\textrm{e}.&32 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}A&=\left \{ x|x\: \textrm{adalah faktor prima dari}\: 16 \right \}\\ &=\left \{ 2 \right \}\Rightarrow n(A)=1\\ B&=\left \{ x|x\: \textrm{adalah faktor dari}\: 16 \right \}\\ &=\left \{ 1,2,4,8,16 \right \} \Rightarrow n(B)=5\\ \textrm{B}&\textrm{anyaknya pemetaan dari}\: A\: \textrm{ke}\: B\: \textrm{adalah}:\\ &=n(B)^{n(A)}\\ &=5^{1}\\ &=5\end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{6}.&\textrm{Diketahui bahwa}\\ &f(x)=\begin{cases} 0,&\textrm{untuk}\: x<0\\ x^{2},&\textrm{untuk}\: 0\leq x< 1\\ 2x-1,&\textrm{untuk}\: x\geq 1 \end{cases}\\ &\textrm{Nilai dari}\: f(-1)+f\left ( \displaystyle \frac{1}{2} \right )-f(3)\: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&-5\displaystyle \frac{1}{4}&&&\textrm{d}.&4\displaystyle \frac{3}{4}\\\\ \textrm{b}.&-4\displaystyle \frac{3}{4}&\textrm{c}.&4&\textrm{e}.&5\displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\\ f(x)&=\begin{cases} 0,&\textrm{untuk}\: x<0\\ x^{2},&\textrm{untuk}\: 0\leq x< 1\\ 2x-1,&\textrm{untuk}\: x\geq 1 \end{cases}\\ \textrm{maka nilai}&\: \: f(-1)+f\left ( \displaystyle \frac{1}{2} \right )-f(3)\\ &=0+\left ( \displaystyle \frac{1}{2} \right )^{2}-\left ( 2(3)-1 \right )\\ &=\displaystyle \frac{1}{4}-5\\ &=-4\displaystyle \frac{3}{4} \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika diketahui}\: \: f\left ( x+\displaystyle \frac{1}{x} \right )=x^{3}+\displaystyle \frac{1}{x^{3}}\: ,\\ &\textrm{maka nilai dari}\: \: f\left ( \displaystyle \frac{5}{2} \right )\: \: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&2\displaystyle \frac{1}{8}&&&\textrm{d}.&8\displaystyle \frac{1}{8}\\\\ \textrm{b}.&2\displaystyle \frac{1}{2}&\textrm{c}.&4\displaystyle \frac{1}{8}&\textrm{e}.&12\displaystyle \frac{1}{8} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Perhatikan}\, &\: \textrm{bahwa}:\\ (p+q)^{3}&=p^{3}+3p^{2}q+3pq^{2}+q^{3}\\ \textrm{Jika kita su}&\textrm{bstitusikan}\: \: p=x^{3}\: \: \textrm{dan}\: \: q=\displaystyle \frac{1}{x^{3}}\\ \left ( x+\displaystyle \frac{1}{x} \right )^{3}&=x^{3}+3x^{2}\left ( \displaystyle \frac{1}{x} \right )+3x\left ( \displaystyle \frac{1}{x} \right )^{2}+\left ( \displaystyle \frac{1}{x} \right )^{3}\\ &=x^{3}+\left ( \displaystyle \frac{1}{x} \right )^{3}+3x+\displaystyle \frac{3}{x}\\ &=\left ( x^{3}+\displaystyle \frac{1}{x^{3}} \right )+3\left ( x+\displaystyle \frac{1}{x} \right )\\ \textrm{sehingga}\quad&\\ f\left ( x+\displaystyle \frac{1}{x} \right )&=x^{3}+\displaystyle \frac{1}{x^{3}}\\ &=\left ( x+\displaystyle \frac{1}{x} \right )^{3}-3\left ( x+\displaystyle \frac{1}{x} \right )\\ f(u)&=u^{3}-3u,\: \: \textrm{maka}\\ f\left ( \displaystyle \frac{5}{2} \right )&=\left ( \displaystyle \frac{5}{2} \right )^{3}-3\left ( \displaystyle \frac{5}{2} \right )\\ &=\displaystyle \frac{125}{8}-\frac{15}{2}\\ &=\displaystyle \frac{65}{8}\\ &=8\displaystyle \frac{1}{8} \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{7}.&\textrm{Misal fungsi}\: \: f\: \: \textrm{terdefinisi untuk seluruh bilangan real}\: \: x.\\ &\textrm{Jika}\: \: f(p+q)=f(pq)\: \: \textrm{untuk semua}\: \: p,\: q\: \: \textrm{bilangan bulat positif dan}\: \: f(1)=2,\\ &\textrm{maka nilai}\: \: f(2018)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0&&&\textrm{d}.&3\\\\ \textrm{b}.&1&\textrm{c}.&2&\textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}\: & \textrm{bahwa}\\ f(1)&=2\: \: \textrm{dan}\\ f(p+q)&=f(pq)\\ \textrm{maka}&\\ f(2)&=f(1+1)=f(1.1)=f(1)=2\\ f(3)&=f(1+2)=f(1.2)=f(2)=f(1)=2\\ f(4)&=f(1+3)=f(1.3)=f(3)=f(2)=f(1)=2\\ f(5)&=f(1+4)=f(1.4)=f(4)=f(3)=f(2)=f(1)=2\\ \vdots &\\ f(2018)&=\cdots =\cdots =\cdots =f(2)=f(1)=2 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: a_{_{0}}=\displaystyle \frac{2}{5}\: \: \textrm{dan}\: \: a_{n+1}=2\left | a_{n} \right |-1, \textrm{maka nilai}\: \: a_{_{2018}}\: \: \textrm{adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&-0,6&&&\textrm{d}.&0,4\\\\ \textrm{b}.&-0,2&\textrm{c}.&0,2&\textrm{e}.&0,6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}\: & \textrm{bahwa}\\ a_{_{0}}&=\displaystyle \frac{2}{5}=0,4\: \: \textrm{dan}\: \: a_{n+1}=2\left | a_{n} \right |-1,\: \textrm{maka}\\ a_{_{1}}&=2\left | a_{_{0}} \right |-1=2\left | 0,4 \right |-1=0,8-1=-0,2\\ a_{_{2}}&=2\left | a_{_{1}} \right |-1=2\left | -0,2 \right |-1=2(0,2)-1=0,4-1=-0,6\\ a_{_{3}}&=2\left | a_{_{2}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2\\ a_{_{4}}&=2\left | a_{_{3}} \right |-1=2\left | 0,2 \right |-1=0,4-1=-0,6=\textbf{a}_{_{\textbf{2}}}\\ a_{_{5}}&=2\left | a_{_{4}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2=\textbf{a}_{_{\textbf{3}}}\\ a_{_{6}}&=2\left | a_{_{5}} \right |-1=2\left | 0,2 \right |-1=0,4-1=-0,6=\textbf{a}_{_{\textbf{2}}}\\ a_{_{7}}&=2\left | a_{_{6}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2=\textbf{a}_{_{\textbf{3}}}\\ \vdots &\\ a_{_{2018}}&=\cdots =\cdots =\textbf{a}_{_{\textbf{2}}}=-0,6 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Kurva}\: \: f(x)=\displaystyle \frac{10}{x^{2}-10x+25}\: \: \textrm{mempunyai asimtot vertikal pada}....\\ &\begin{array}{llllll}\\ \textrm{a}.&x=0\: \: \textrm{saja}&&&\textrm{d}.&x=0\: \: \textrm{dan}\: \: x=5\: \: \textrm{saja}\\\\ \textrm{b}.&x=5\: \: \textrm{saja}&\textrm{c}.&x=10\: \: \textrm{saja}&\textrm{e}.&x=0,\: x=5,\: \: \textrm{dan}\: \: x=10 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textbf{Asimtot vertikal}&\textrm{\textbf{(tegak)} diperoleh saat}\\ x^{2}-10x+25&=0\\ (x-5)^{2}&=0\\ x-5&=0\\ x&=5\\\\ \textrm{Ilustrasinya gamba}&\textrm{rnya adalah sebagai berikut}: \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Fungsi berikut yang tidak mempunyai asimtot vertikal adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=\displaystyle \frac{x+2}{x^{2}-3}&&&\textrm{d}.&f(x)=\displaystyle \frac{-3}{x}\\\\ \textrm{b}.&f(x)=\displaystyle \frac{x}{(x-2)^{2}}&\textrm{c}.&f(x)=\displaystyle \frac{x^{2}-9}{x+3}&\textrm{e}.&\textrm{semuanya mempunyai asimtot vertikal} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Perh}&\: \textrm{atikanlah opsi jawaban}\: \: c,\: \textrm{yaitu}:\\ f(x)&=\displaystyle \frac{x^{2}-9}{x+3}\: \\ \textrm{Jika}&\: \textrm{disederhanakan akan menjadi \textbf{fungsi linear} yaitu}:\\ f(x)&=\displaystyle \frac{x^{2}-9}{x+3}=\displaystyle \frac{(x+3)(x-3)}{x+3}=x-3\\ \textrm{sehi}&\textrm{ngga fungsi pada opsi}\: \: c\: \: \textrm{adalah berupa persamaan linear}\\ &\textrm{yang secara otomatis \textbf{tidak akan memiliki asimtot}} \end{aligned} \end{array}.

 

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

 

Dipublikasi di Info, Matematika, Pendidikan | Meninggalkan komentar

Lanjutan Materi Fungsi (K13 Revisi)

1. Fungsi Linear (Fungsi Garis Lurus)

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Fungsi}&\: \textrm{Linear adalah}:\\ &\textrm{fungsi di aman fungsi yang hanya memiliki satu}\\ &\textrm{variabel dan berpangkat satu}.\\\\ \textrm{Misal},&\: \: f:x \mapsto ax+b\\ & \end{aligned}}\\\hline \multicolumn{2}{|c|}{\textrm{Menentukan Persamaan Linear(Garis Lurus)}}\\\hline \textrm{Melalui titik}\: \: \left ( x_{1},y_{1} \right )&\textrm{Melalui titik}\: \: \left ( x_{1},y_{1} \right )\\ \textrm{dan bergradien}\: \: m&\textrm{dan}\: \: \left ( x_{2},y_{2} \right )\\\hline y=m\left ( x-x_{1} \right )+y_{1}&\displaystyle \frac{y-y_{1}}{y_{2}-y_{1}}=\displaystyle \frac{x-x_{1}}{x_{2}-x_{1}}\\ &\begin{aligned}\textrm{dengan}:&\\ m&=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \end{aligned}\\\hline \textrm{Sejajar dengan} &\textrm{Tegak lurus dengan}\\ \textrm{garis bergradien}\: \: m_{1}&\textrm{garis bergradien}\: \: m_{1}\\\hline \multicolumn{2}{|c|}{\textrm{Syarat dua garis}}\\\hline \textrm{Sejajar}\: \: m_{1}=m_{2}&\textrm{Tegak lurus}\: \: m_{1}\times m_{2}=-1\\\hline y=m_{2}\left ( x-x_{1} \right )+y_{1}&y=-\displaystyle \frac{1}{m_{2}}\left ( x-x_{1} \right )+y_{1}\\\hline \end{array}.

2. Fungsi Kuadrat

\begin{array}{|l|p{3.5cm}|l|}\hline \multicolumn{3}{|c|}{\textbf{Fungsi Kuadrat}}\\\hline \textrm{Pengertian}&\textrm{Menggambar Grafik Fungsi Kuadrat}&\textrm{Keterangan}\\\hline &\textrm{Titik potong sumbu x}&\textrm{Jika ada}\\\cline{3-3} \begin{aligned}&\textrm{Suatu fungsi yang berbentuk}\\ &f(x)=ax^{2}+bx+c\\ & a,\: b,\: c,\: \in \mathbb{R},\: a\neq 0 \end{aligned}&&\begin{aligned}&\textrm{untuk titik potong terhadap sumbu x }\\ &\textrm{Jika y = 0 maka }\: ax^{2}+bx+c=0\\ &\textrm{Selanjutnya tinggal menentukan nilai D}\\ &D=b^{2}-4ac\: \: \textrm{adalah}\\ &\: \: \: \: \: \: \: \: \: \textrm{nilai diskriminan}.\\ &\textrm{Jika} \: D>0\\ &\textrm{maka grafik memotong sumbu x}\\ &\textrm{di dua tempat berbeda}\\ &\textrm{yaitu di} \: (x_{1},0)\: \textrm{dan}\: (x_{2},0).\\ &\textrm{dan jika D = 0}\\ &\textrm{maka grafik hanya menyinggung}\\ &\textrm{sumbu x di satu titik}\\ &\textrm{yaitu di }\: (x_{1},0)\\ &\textrm{dan jika}\: D<0 \\ &\textrm{maka grafik tidak memotong}\\ &\textrm{sumbu x} \end{aligned}\\\cline{2-3} &\textrm{Titik potong sumbu y}&\begin{aligned}&\textrm{titik potong terhadap}\\ &\textrm{sumbu y, jika x = 0}\\ &y=f(x)=ax^{2}+bx+c\\ &y=f(0)=a(0)^{2}+b(0)+c\\ &y=c \end{aligned}\\\cline{2-3} &\textrm{Menentukan Sumbu Simetri (SS)}&x=\displaystyle \frac{-b}{2a}\\\cline{2-3} &\textrm{Menentukan Titik Puncak}&\left ( \displaystyle \frac{-b}{2a},\displaystyle \frac{D}{-4a} \right )\\\cline{2-3} &\textrm{Posisi grafik}&\textrm{Jika}\: a>0\: \textrm{maka grafik terbuka ke atas}\\\cline{3-3} &&\textrm{Jika}\: a<0\: \textrm{maka grafik terbuka ke bawah}\\\hline \end{array}..

Untuk menyusun grafik fungsi kuadratnya adalah sebagai berikut:

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Menyusun fungsi kuadrat}}\\\hline \textrm{Jika memotong sumbu}-\textrm{X}&\textrm{Jika menyinggung sumbu}-\textrm{X}\\ \textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan}\: \left ( x_{2},0 \right )&\textrm{di titik}\: \left ( x_{1},0 \right )\: \textrm{dan melalui}\\ \textrm{dan melalui sebuah titik lain}&\textrm{sebuah titik lain} \\\hline &\\ y=f(x)=a\left ( x-x_{1} \right )\left ( x-x_{2} \right )&y=f(x)=a\left ( x-x_{1} \right )^{2}\\ &\\\hline \multicolumn{2}{|c|}{\textrm{Jika grafik fungsi itu melalui}}\\\hline \textrm{Titik puncak}\: \: P\left ( x_{p},y_{p} \right )\: \textrm{dan}&\textrm{tiga buah titik yaitu}\: \left ( x_{1},y_{1} \right )\\ \textrm{sebuah titik lain}&\left ( x_{2},y_{2} \right )\: \: \textrm{dan}\: \: \left ( x_{3},y_{3} \right )\\\hline &\\ y=f(x)=a\left ( x-x_{p} \right )^{2}+y_{p}&y=f(x)=ax^{2}+bx+c\\ &\\\hline \end{array}.

3. Fungsi Rasional Pecahan

Fungsi rasional ada 2 macam yaitu, fungsi rasional bulat dan rasional pecahan. Selanjutnya fungsi rasional pecahan biasa disebut dengan fungsi pecahan saja.

Pada fungsi rasional biasanya akan didapatkan beberapa jenis asimtot garfik tetentu. Asimtot dari suatu grafik fungsi  y=f(x)   adalah  suatu garis lurus jenis tertentu yang tidak akan pernah dipotong oleh grafik fungsi itu sendiri, akan tetapi hanya didekati saja sampai tanpa batas.

\begin{array}{|l|c|c|c|c|c|}\hline &\multicolumn{2}{c|}{\textrm{Intersep}}&\multicolumn{3}{c|}{\textrm{Asimtot}}\\\cline{2-6} \raisebox{1.5ex}[0cm][0cm]{.\qquad\qquad \textrm{Bentuk}} &\textrm{Sumbu}-X&\textrm{Sumbu}-Y&\textrm{Tegak}&\textrm{Mendatar}&\textrm{Miring}\\\hline y=f(x)=\displaystyle \frac{ax+b}{px+q}&\left (-\displaystyle \frac{a}{b} ,0 \right )&\left ( 0,\displaystyle \frac{b}{q} \right )&x=-\displaystyle \frac{q}{p}&y=\displaystyle \frac{a}{p}&\textrm{tidak ada}\\\hline y=f(x)=\displaystyle \frac{ax+b}{px^{2}+qx+r}&\left (-\displaystyle \frac{a}{b} ,0 \right )&\left ( 0,\displaystyle \frac{b}{r} \right )&px^{2}+qx+r=0&y=0&\textrm{tidak ada}\\\hline y=f(x)=\displaystyle \frac{ax^{2}+bx+c}{px+q}&ax^{2}+bx+c=0&\left ( 0,\displaystyle \frac{c}{q} \right )&x=-\displaystyle \frac{q}{p}&ax^{2}+bx+c=0&y=\o (x)+\displaystyle \frac{m}{px+q}\\\cline{1-1}\cline{3-6} y=f(x)=\displaystyle \frac{ax^{2}+bx+c}{px^{2}+qx+r}&\textrm{lihat D}=b^{2}-4ac&\left ( 0,\displaystyle \frac{c}{r} \right )&px^{2}+qx+r=0&y=\displaystyle \frac{a}{p}&\textrm{tidak ada}\\\hline \end{array}.

Dan masih banyak fungsi-fungsi yang lain, di antaranya:

Fungsi Konstan, Fungsi Identitas, Fungsi Harga Mutlak, Fungsi Tangga, Fungsi genap ganjil, Fungsi Periodik, Fungsi Eksponen, serta Fungsi Logaritma.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: f\: \: \textrm{adalah fungsi linear dengan}\: \: f(2)-f(-2)=8,\\ & \textrm{maka nilai dari}\: \: f(4)-f(-2)\: \: \textrm{adalah}\: ....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui bahwa}:&\\ f(x)&=ax+b\\ f(2)-f(-2)&=\left (a(2)+b \right )-\left ( a(-2)+b \right )=8\\ 8&=2a+2a\\ 8&=4a\\ 2&=a\\ f(x)&=2x+b,\quad \textrm{dengan}\: \: b\: \: \textrm{konstan}\\ \textrm{Sehingga nilai}\quad&\\ f(4)-f(-2)&=\left (2(4)+b \right )-\left (2(-2)+b \right )\\ &=8+b+4-b\\ &=12 \end{aligned} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Ubahlah}\: \: 8-6x-x^{2}\: \: \textrm{ke dalam bentuk}\: \: a-(x+b)^{2},\: \textrm{selanjutnya}\\ & \textrm{tentukanlah daerah hasil dari}\: \: f(x)=8-6x-x^{2}\: \: \textrm{untuk}\: \: x\: \: \textrm{bilangan real}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(\textit{NTU Entrance Examination AO-level})\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{Diketahui}&\textrm{Mencari koordinat}\: \: \left ( x_{SS},y_{SS} \right )&\textrm{Nilai fungsi}\\\hline \begin{aligned}\textrm{Misal}\quad\qquad&\\ 8-6x-x^{2}&=f(x)\\ f(x)&=-x^{2}-6x+8\\ &=-\left ( x^{2}+6x-8 \right )\\ &=-\left ( x^{2}+6x+9-17 \right )\\ &=-\left ( (x+3)^{2}-17 \right )\\ &=-(x+3)^{2}+17\\ & \end{aligned}&\begin{aligned}f(x)&=-x^{2}-6x+8\left\{\begin{matrix} a=-1\\ b=-6\\ c=\: \: 8\: \: \end{matrix}\right.\\ \textrm{Maka}&\\ x_{SS}&=\frac{-b}{2a}=\displaystyle \frac{-(-6)}{2(-1)}\\ &=-3\\ y_{SS}&=f(-3)=-\left ( -3+3 \right )^{2}+17=17\\ \therefore &\left ( x_{SS},y_{SS} \right )=(-3,17) \end{aligned}&\begin{aligned}\textrm{Karena}&\: \: a=-1<0\\ \textrm{maka f}&\textrm{ungsi menghadap}\\ \textbf{ke ba}&\textbf{wah},\: \: \textrm{sehingga}\\ \textrm{daerah}&\: \: \textrm{hasilnya}\: \: \left (R_{f} \right )\\ \textrm{adalah}&:\\ &\left \{ -\infty <y\leq 17 \right \}\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}.

Berikut ilustrasi gambarnya

\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\: \: x^{2}+mx+m=0,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{yang menyebabkan jumlah kuadrat akar-akar mencapai}\\ &\textrm{minimum adalah}\: ....\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\: \: \: (\textit{UM UNDIP 2014 Mat Das})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}\: \: x^{2}+mx+m&=0(\textbf{persamaan kuadrat}\: \textrm{dalam}\: \: x),\: \: \textrm{maka}\\ x^{2}+mx+m&=x-(\alpha +\beta )x+(\alpha \beta )=0\begin{cases} \alpha +\beta &=-m \\ & \\ \alpha \beta &=m \end{cases}\\ \textrm{Selanjutnya}\qquad\quad\: \qquad &\\ \alpha ^{2}+\beta ^{2}&=\left ( \alpha +\beta \right )^{2}-2\alpha \beta\\ &=(-m)^{2}-2m\: \: \textrm{dan dapat kita tuliskan sebagai}\\ f(m)&=m^{2}-2m\begin{cases} a &=1 \\ b &=-2 \\ c &=0 \end{cases} \quad ,\textrm{\textbf{fungsi kuadrat} dalam}\: \: m,\\ \textrm{sehingga kita perlu men}&\textrm{cari titik}\: \: \left ( m_{SS},f\left ( m_{SS} \right ) \right ),\: \: \textrm{tetapi yang kita perlukan}\\ \textrm{cuma}\: \: m-\textrm{nya saja, yai}\, &\textrm{tu}:\: \: m=m_{SS},\: \: \textrm{dengan}\\ m_{SS}&=\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1 \end{aligned} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah asimtot mendatar, tegak, dan miring jika ada dari}\\ &\textrm{fungsi-fungsi rasional berikut}\\ &\begin{array}{llll}\\ \textrm{a}.&f(x)=\displaystyle \frac{5x^{2}+5x+1}{x^{2}-2x+1}&\textrm{c}.&f(x)=\displaystyle \frac{x-7}{3x-4}\\\\ \textrm{b}.&f(x)=\displaystyle \frac{6x}{2x^{2}-5}&\textrm{d}.&f(x)=\displaystyle \frac{x^{2}-2x-16}{2x+6} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|c|c|c|}\hline &\multicolumn{3}{c|}{\textrm{Asimtot}}\\\cline{2-4} \raisebox{1.5ex}[0cm][0cm]{\qquad\qquad\textrm{Fungsi}} & \textrm{Mendatar}&\textrm{Tegak}&\textrm{Miring}\\\hline \textcircled{1}\quad f(x)=\displaystyle \frac{5x^{2}+5x+1}{x^{2}-2x+1}&y=\displaystyle \frac{5}{1}=5&\begin{aligned}x^{2}-2x+1&=0\\ (x-1)^{2}&=0\\ x&=1 \end{aligned}&\textrm{tidak ada}\\\hline \textcircled{2}\quad f(x)=\displaystyle \frac{6x}{2x^{2}-5}&y=\displaystyle \frac{0}{2}=0&\begin{aligned}2x^{2}-5&=0\\ x^{2}&=\displaystyle \frac{5}{2}\\ x&=\pm \displaystyle \frac{1}{2}\sqrt{10} \end{aligned}&\textrm{tidak ada}\\\hline \textcircled{3}\quad f(x)=\displaystyle \frac{x-7}{3x-4}&y=\displaystyle \frac{1}{3}&\begin{aligned}3x-4&=0\\ 3x&=4\\ x&=\displaystyle \frac{4}{3} \end{aligned}&\textrm{tidak ada}\\\hline \textcircled{4}\quad f(x)=\displaystyle \frac{x^{2}-2x-16}{2x+6}&\begin{aligned}y&=\displaystyle \frac{1}{0}\: (\textbf{TD})\\ &\textrm{ini artinya}\\ &\textbf{tidak ada} \end{aligned}&\begin{aligned}2x-6&=0\\ 2x&=6\\ x&=3 \end{aligned}&\begin{aligned}\textbf{ada}\: ,\quad \textrm{yaitu}:&\\ \displaystyle \frac{x^{2}-2x-16}{2x+6}&=\left (\displaystyle \frac{1}{2}x-\frac{5}{2} \right )+\displaystyle \frac{-1}{2x+6}\\ \textrm{maka}\: \: \: \phi (x)&=\displaystyle \frac{1}{2}x-\frac{5}{2} \end{aligned}\\\hline \end{array} \end{array}..

Berikut ilustrasi untuk jawaban a)

Berikut untuk ilustrasi jawaban b)

Berikut untuk ilustrasi jawaban c)

Dan berikut pula ilustrasi untuk jawaban d)

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.
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Fungsi (K13 Revisi)

Materi terkait sebelumnya terdapat di

A. Notasi, Domain, dan Range Suatu Fungsi

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\\textrm{\textbf{Fungsi}}\: &\textrm{atau \textbf{pemetaan} dari A ke B adalah}\\ &\textrm{suatu relasi khusus yang memasangkan setiap}\: \: x\in A\\ &\textrm{ke tepat satu}\: \: y\in B.\\ & \end{aligned}}\\\hline \textrm{Notasi}&f:x \rightarrow y\: \: \: \textrm{atau}\: \: \: f:x \rightarrow f(x) \\\hline \textrm{Dibaca}&\textrm{fungsi}\: \: f\: \: \textrm{memetakan}\: \: x\in A\: \: \textrm{ke}\: \: y\in B\\\hline \: \: \: \: A&\textrm{Domain atau daerah asal fungsi atau}\quad D_{f}\\\hline \: \: \: \: \, x&\textrm{prapeta(sebelum dipetakan)}\\\hline \: \: \: \: B&\textrm{Kodomain atau daerah kawan fungsi atau}\quad K_{f}\\\hline \: \: \: \: \, y&\textrm{peta(bayangan dari prapeta) adalah Range atau}\quad R_{f}\\\hline \end{array}.

Perhatikanlah ilustrasi berikut!

\begin{array}{l|clllllllllll} \multicolumn{3}{c}{\begin{matrix} \: \: \textbf{Input}\\ \: (x) \end{matrix}}&&&&&&&&\\ \multicolumn{1}{l}{\textrm{\: }}&\Downarrow&&&&&&&&&\\ \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\\cline{1-1}\cline{3-3} &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\\cline{1-1}\cline{3-3} \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\ \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\\cline{4-8} \multicolumn{5}{|l}{\textrm{\: }}&&\multicolumn{2}{l|}{\textrm{\: }}&&\textbf{Output}\\\cline{9-9} \multicolumn{5}{|c}{\quad \LARGE\textbf{Proses}}&&\multicolumn{2}{l}{\textrm{\: }}&\Rightarrow &\: \: \: f(x)\\\cline{9-9} \multicolumn{5}{|l}{\textrm{.}}&&\multicolumn{2}{l|}{\textrm{\: }}& \\\cline{1-3}\cline{4-8} \end{array}.

Sebagai misal,

\begin{array}{l|clllllllllll} \multicolumn{3}{c}{\: \, \,\textbf{x}}&&&&&&&&\\ \multicolumn{1}{l}{\textrm{\: }}&\Downarrow&&&&&&&&&\\ \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\\cline{1-1}\cline{3-3} &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\ &&\multicolumn{2}{|l}{\textrm{\: }}&&&&&&\\\cline{1-1}\cline{3-3} \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\ \multicolumn{3}{|l|}{\textrm{\: }}&&&&&&&&\\\cline{4-8} \multicolumn{5}{|l}{\textrm{\: }}&&\multicolumn{2}{l|}{\textrm{\: }}&&\\\cline{9-9} \multicolumn{5}{|c}{\quad \LARGE\textbf{Proses}}&&\multicolumn{2}{l}{\textrm{\: }}&\Rightarrow &\: \: \: \textbf{3x+2}\\\cline{9-9} \multicolumn{5}{|l}{\textrm{.}}&&\multicolumn{2}{l|}{\textrm{\: }}& \\\cline{1-3}\cline{4-8} \end{array}.

Selanjutnya ilustrasi di atas dapat notasikan sebagai:

\begin{aligned}\LARGE\boxed{f:x\rightarrow f(x)=3x+2}&\\ &\textit{\underline{\textbf{dibaca}}}:\\ &"\textrm{Fungsi \textbf{\textit{f}} memetakan \textbf{\textit{x}} ke \textbf{3\textit{x}+2}}" \end{aligned}.

B. Sifat-Sifat Fungsi

\begin{array}{|c|c|l|}\hline \multicolumn{3}{|c|}{\textrm{Sifat-sifat fungsi}\: \: f : A\rightarrow B}\\\hline \textrm{Injektif(satu-satu)}&\textrm{Surjektif(pada)}&\textrm{Bijektif(korespondensi satu-satu)}\\\hline \begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan A memiliki}\\ &\textrm{bayangan berbeda di}\\ &\textrm{himpunan B} \end{aligned}&\begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan di B}\\ &\textrm{mempunyai prapeta}\\ &\textrm{di himpunan A} \end{aligned}&\begin{aligned}&\textrm{Jika fungsi yang injektif dan}\\ &\textrm{sekaligus juga surjektif}\\ &\\ & \end{aligned}\\\hline \end{array}..

C. Operasi Aljabar Fungsi

\begin{array}{|l|l|}\hline \textrm{Aljabar Fungsi}&\textrm{Daerah Asal}\\\hline (f+g)(x)=f(x)+g(x)&D_{_{(f+g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f-g)(x)=f(x)-g(x)&D_{_{(f-g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f.g)(x)=f(x).g(x)&D_{_{(f.g)}}=D_{_{f}}\cap D_{_{g}}\\\hline \left ( \displaystyle \frac{f}{g} \right )(x)=\displaystyle \frac{f(x)}{g(x)}&D_{_{\left ( \frac{f}{g} \right )}}=D_{_{f}}\cap D_{_{g}}\: , \\ &\textrm{dengan}\: \: g(x)\neq 0 \\\hline \end{array}..

D. Macam-Macam Fungsi

\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textrm{Macam-Macam Fungsi}}\\\hline &\textrm{Fungsi Konstan}&f(x)=c\\\cline{2-3} \raisebox{0.25ex}[0cm][0cm]{Fungsi Linear} &\textrm{Fungsi Identitas}&f(x)=x\\\cline{2-3} &\textrm{Fungsi linear/garis lurus}&f(x)=ax+b\\\hline \textrm{Fungsi Kuadrat}&\textrm{Fungsi Kuadrat/parabola}&f(x)=ax^{2}+bx+c,\quad a\neq 0\\\hline \textrm{Fungsi Rasional}&\textrm{Fungsi Pecahan}&f(x)=\displaystyle \frac{p(x)}{q(x)}\\\hline &\textrm{Fungsi Modulus(nilai mutlak)}&f(x)=\left | x \right |\\\cline{2-3} \raisebox{0.00ex}[0cm][0cm]{Fungsi Khusus} &\textrm{Fungsi tangga}&f(x)=\left \lfloor x \right \rfloor\\\cline{2-3} &\textrm{Fungsi genap dan ganjil}&\begin{cases} \textrm{Fungsi ganjil} & f(-x)=-f(x) \\ \textrm{Fungsi genap} & f(-x)=f(x) \end{cases}\\\hline \end{array}..

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Diketahui 2 humpuan sebagai berikut}:\\ &\begin{cases} \textrm{P} & =\left \{ -2,-1,0,1,2 \right \} \\ \textrm{Q} & =\left \{ 0,1,2,5,7 \right \} \end{cases}\\ &\textrm{Di antara relasi dari P ke Q berikut manakah yang merupakan fungsi}\\ &\textrm{a}.\quad \textrm{A}=\left \{ (-2,0),(-1,0),(0,0),(1,0),(2,0) \right \}\\ &\textrm{b}.\quad \textrm{B}=\left \{ (-2,1),(-1,2),(0,5),(1,7),(-2,2) \right \}\\ &\textrm{c}.\quad \textrm{C}=\left \{ (-2,0),(-1,1),(0,2),(1,5),(2,7) \right \}\\\\ &\textrm{Jawab}:\\ &\textrm{Semuanya Fungsi kecuali}\textbf{ poin b)} \end{array}..

\begin{array}{ll}\\ 2.&\textrm{Relasi berikut yang merupakan fungsi adalah}.... \end{array}.

.\qquad \begin{array}{|c|l|l|l|}\hline \textrm{Poin}&\textrm{Jenis}&\textrm{Keterangan}&\textrm{Keterangan lebih lanjut}\\\hline \textrm{a}&\textrm{Fungsi}&\begin{aligned}&\textrm{Sesuai definisi}\\ \end{aligned}&\\ &&\begin{aligned}&\textrm{\textbf{yaitu}}:\\ &\textrm{Setiap prepeta(anggota himpunan A)}\\ &\textrm{memiliki peta di himpunan B}\\ &\textrm{tepat satu} \end{aligned}&\begin{aligned}&\textrm{bukan fungsi \textbf{injektif}}\\ &\textrm{dan bukan pula fungsi \textbf{surjektif}} \end{aligned}\\\cline{1-3} \textrm{b}&\textrm{Fungsi}&\textrm{Sama di atas}&\\\hline \textrm{c}&\textrm{Bukan Fungsi}&\textrm{Tidak sesuai definisi}&\textrm{hanya berupa relasi saja}\\\hline \textrm{d}&\textrm{Fungsi}&\textrm{Sesuai definisi}&\textrm{Fungsi \textbf{bijektif}}\\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah daerah asal dari fungsi beberapa berikut}:\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=x-3&\textrm{g}.&f(x)=\displaystyle \frac{\left | x \right |}{x}\\ \textrm{b}.&f(x)=\displaystyle \frac{6}{x^{2}-2x-8}&\textrm{h}.&f(x)=\left \lfloor x \right \rfloor\qquad \textrm{catatan}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah bulat terbesar atau sama dengan}\: \: x\\ \textrm{c}.&f(x)=\displaystyle \frac{x^{2}-3x}{x^{2}-2x-15}&\textrm{i}.&f(x)=\left | x \right |+\left \lfloor x \right \rfloor\\ \textrm{d}.&y+2=x^{2}-5x+5&\textrm{j}.&f(x)=\sqrt{x^{2}-16}\\ \textrm{e}.&f(x)=\left | x-3 \right |&\textrm{k}.&f(x)=\sqrt{2x^{2}-50}\\ \textrm{f}.&f(x)=3-\left | 2x-1 \right |&\textrm{l}.&f(x)=\displaystyle \frac{2\sqrt{x}}{x-3} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline (\textrm{a})&(\textrm{b})&(\textrm{c})\\\hline \begin{aligned}f(x)&=x-3\\ \textrm{selu}&\textrm{ruh bilangan real}\\ &x\: \: \textrm{akan terdefinisi}\\ &\textrm{atau tetap bernilai}\\ &\textrm{real}\\ \textrm{sehi}&\textrm{ngga},\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned}f(x)&=\displaystyle \frac{6}{x^{2}-2x-8}\\ \textrm{terd}&\textrm{efinisi ketika}\\ &\textrm{penyebut \textit{tidak} sama}\\ &\textrm{dengan}\: \: 0, \: \: \textrm{yaitu}:\\ &\begin{aligned}x^{2}-2x-8&\neq 0\\ (x-4)(x+2)&\neq 0\\ x\neq 4\: \: \textrm{dan}\: \: x&\neq -2 \end{aligned}\\ D_{f}&=\left \{ x|x\in \mathbb{R},\: x\neq 4\: \: \textrm{dan}\: \: x\neq -2 \right \} \end{aligned}&\begin{aligned}f(x)&=\left | x-3 \right |\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ \textrm{teta}&\textrm{pi pada \textit{range} fungsinya}\\ &\textrm{hanya akan berupa}\\ &\textrm{bilangan positif saja}.\\ \textrm{yait}&\textrm{u}:\\ R_{f}&=\left \{ y|y\in \mathbb{R},\: y\geq 0 \right \}\\ & \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ 4.&\textrm{Jika}\: \left | x \right |\: \textrm{menyatakan nilai mutlak}\\ &\textrm{dan}\: \left \lfloor x \right \rfloor\: \textrm{menyatakan bilangan bulat terbesarnatau sama dengan}\: x\\ &\textrm{misalkan}\: \: \left \lfloor 1,6 \right \rfloor=1,\: \left \lfloor \pi \right \rfloor=3\\ &\textrm{Jika diberikan}\: \: f(x)=\left | x \right |+\left \lfloor x \right \rfloor,\: \textrm{maka tentukanlah nilai untuk}\\ &\textrm{a}.\quad f\left ( -3,5 \right )+f\left ( 2,5 \right )\\ &\textrm{b}.\quad f\left ( -1,5 \right )+f\left ( 3,5 \right )\\\\ &\textrm{Jawab}:\\ &\begin{array}{l}\\ \begin{aligned}\textrm{a}.f\left ( -3,5 \right )+f\left ( 2,5 \right )&=\left | -3,5 \right |+\left \lfloor -3,5 \right \rfloor+\left | 2,5 \right |+\left \lfloor 2,5 \right \rfloor\\ &=3,5+\left ( -4 \right )+2,5+2\\ &=4 \end{aligned}\\ \begin{aligned}\textrm{b}.f\left ( -1,5 \right )+f\left ( 3,5 \right )&=\left | -1,5 \right |+\left \lfloor -1,5 \right \rfloor+\left | 3,5 \right |+\left \lfloor 3,5 \right \rfloor\\ &=1,5+(-2)+3,5+3\\ &=6 \end{aligned} \end{array} \end{array}.

\begin{array}{ll}\\ 5.&\textrm{Jika diketahui relasi}\: \: f\: \: \textrm{dengan kondisi}\\ &\begin{array}{l} (\textrm{a}).\quad f(1)=1\\ (\textrm{b}).\quad f(2x)=4f(x)+6\\ (\textrm{c}).\quad f(x+2)=f(x)+12x+12 \end{array}\\ &\textrm{maka nilai}\: \: f(14)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(1)&=1\\ f(2.1)=f(2)&=4f(1)+6=4.1+6=10\\ f(1+2)=f(3)&=f(1)+12.1+12\\ f(3)&=1+12+12=25\\ f(3+2)=f(5)&=f(3)+12.3+12\\ f(5)&=25+36+12=73\\ f(5+2)=f(7)&= f(5)+12.5+12\\ f(7)&=73+60+12=145\\ f(7.2)=f(14)&=4.f(7)+6\\ f(14)&=4.145+6=580+6\\ &=586 \end{aligned} \end{array}.

\begin{array}{ll}\\ 6.&\textbf{(OSK 2013)}\textrm{Fungsi}\: \: f\: \: \textrm{didefinisikan oleh}\: \: f(x)=\displaystyle \frac{kx}{2x+3},\quad x=-\displaystyle \frac{3}{2}.\\ & \textrm{Tentukanlah nilai}\: \: k\: \: \textrm{agar}\: \: f(f(x))=x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{kx}{2x+3},\quad x\neq -\frac{2}{3}\\ f\left ( f(x) \right )&=x\\ x&=f\left ( f(x) \right )\\ x&=\displaystyle \frac{k\left ( \displaystyle \frac{kx}{2x+3} \right )}{2\left ( \displaystyle \frac{kx}{2x+3} \right )+3}\\ x&=\displaystyle \frac{\displaystyle \frac{k^{2}x}{2x+3}}{\displaystyle \frac{2kx+3(2x+3)}{2x+3}}\\ x&=\displaystyle \frac{k^{2}x}{2kx+6x+9}\\ 2kx+6x+9&=k^{2}\\ 0&=k^{2}-2xk-6x-9\\ 0&=(k+3)(k-2x-3)\\ &\quad k=-3\: \: \textrm{atau}\: \: k=2x+3 \end{aligned} \end{array}.

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Contoh Soal Persiapan Semester Gasal Peminatan Kelas X (K13 Revisi)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: \displaystyle \frac{10^{2018}-10^{2017}}{9}\: \: \textrm{sama dengan}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{9}&&&\textrm{d}.&\displaystyle \frac{10^{2017}}{9}\\\\ \textrm{b}.&\displaystyle \frac{10}{9}&\textrm{c}.&10^{2012}&\textrm{e}.&10^{2017} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{10^{2018}-10^{2017}}{9}&=\displaystyle \frac{10^{2017}\left ( 10^{1}-1 \right )}{9}\\ &=10^{2017}.\displaystyle \frac{9}{9}\\ &=10^{2017} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\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{3}.&\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}.

\begin{array}{ll}\\ \fbox{4}.&\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{\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{5}.&\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{6}.&\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}.

\begin{array}{ll}\\ \fbox{7}.&\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\qquad\qquad\qquad (\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}.

\begin{array}{ll}\\ \fbox{8}.&\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}\\ &\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{9}.&\sqrt{\displaystyle \frac{ \sqrt{41}+4}{\sqrt{41}-4}}-\sqrt{\displaystyle \frac{ \sqrt{41}-4}{\sqrt{41}+4}}=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-\displaystyle \frac{8}{5}&&&\textrm{d}.&\displaystyle \frac{8}{5}\\ \textrm{b}.&0&\textrm{c}.&\displaystyle \frac{16}{5}&\textrm{e}.&5\sqrt{41} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Misalkan saja}\: \: &\: \begin{cases} \sqrt{\displaystyle \frac{ \sqrt{41}+4}{\sqrt{41}-4}} & =\sqrt{\displaystyle \frac{a}{b}}=x \\\\ \sqrt{\displaystyle \frac{ \sqrt{41}-4}{\sqrt{41}+4}} & =\sqrt{\displaystyle \frac{b}{a}}=\displaystyle \frac{1}{x} \end{cases} \\ \sqrt{\displaystyle \frac{ \sqrt{41}+4}{\sqrt{41}-4}}-\sqrt{\displaystyle \frac{ \sqrt{41}-4}{\sqrt{41}+4}}&=x-\displaystyle \frac{1}{x}\\ &=\displaystyle \frac{x^{2}-1}{x}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{a}{b} \right )-1}{\sqrt{\displaystyle \frac{a}{b}}}=\displaystyle \frac{a-b}{\sqrt{ab}}\\ &=\displaystyle \frac{(\sqrt{41}+4)-(\sqrt{41}-4)}{\sqrt{\left ( \sqrt{41} \right )^{2}-4^{2}}}\\ &=\displaystyle \frac{8}{5} \end{aligned} \end{array}.

Berikut ilustrasi grafik untuk pertanyaan soal no. 10

\begin{array}{ll}\\ \fbox{10}.&\textrm{Fungsi yang bersesuaian dengan grafik di atas adalah}....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=2^{x}&&&\textrm{d}.&f(x)=3^{x+1}\\ \textrm{b}.&f(x)=2^{x+1}&\textrm{c}.&f(x)=3^{2x-2}&\textrm{e}.&f(x)=3^{x-2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Dari soal terdapat}\: &\: \textit{titik uji}\: \: \textrm{yaitu}\: \: (2,1)\: \: \textrm{dan}\: \: (3,3)\: \: \textrm{yang secara langsung dapat kita substitusikan ke fungsi}\\ f(2)&=\begin{cases} \textrm{a} & =2^{2}=4 \\ \textrm{b} & =2^{2+1}=2^{3}=8\\ \textrm{c} & =3^{2.2-2}=3^{2}=9 \\ \textrm{d} & =3^{2+1}=3^{3}=27 \\ \textrm{e} & =3^{2-2}=3^{0}=1\: \: .............\: \: (\textrm{cukup jelas}) \end{cases}\\\\ \textrm{Sehingga untuk}\: \: \: \: \: \, &f(3)=3^{3-2}=3^{1}=3\: \: .............\: \: (\textrm{juga cukup jelas}) \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: a^{3}-a-1=0\: ,\: \textrm{maka nilai untuk}\: \: a^{4}+a^{3}-a^{2}-2a+9=...\\\\ &\textrm{Jawab}:\\ &\begin{aligned}a^{4}+a^{3}-a^{2}-2a+9&=a^{4}-a^{2}-a+a^{3}-a-1+10\\ &=a\left ( a^{3}-a-1 \right )+\left ( a^{3}-a-1 \right )+10\\ &=a(0)+0+10\\ &=10 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{6}{2^{^{^{\frac{4}{3}}}}+2^{^{^{\frac{2}{3}}}}+1}=....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{6}{2^{^{^{\frac{4}{3}}}}+2^{^{^{\frac{2}{3}}}}+1}&=\displaystyle \frac{6}{2^{^{^{\frac{4}{3}}}}+2^{^{^{\frac{2}{3}}}}+1}\times \displaystyle \frac{2^{^{^{\frac{2}{3}}}}-1}{2^{^{^{\frac{2}{3}}}}-1}\\ &=\displaystyle \frac{6\left ( 2^{^{^{\frac{2}{3}}}}-1 \right )}{2^{2}-1}\\ &=2\left ( 2^{^{^{\frac{2}{3}}}}-1 \right )\: \: \: atau\\ &=2\left ( \sqrt[3]{4}-1 \right ) \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Jika}\: \: x>0\: ,\: x\neq 1\: \textrm{dan}\: y>0,\: \: \textrm{maka persamaan}\: \: x^{n}=y\: \textrm{dapat dituliskan dengan}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle ^{^{x}}\log y=n&&&\textrm{d}.&\displaystyle ^{^{x}}\log n=y\\\\ \textrm{b}.&\displaystyle ^{^{y}}\log x=n&\textrm{c}.&^{^{n}}\log x=y&\textrm{e}.&^{^{n}}\log y=x \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ & \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Jika}\: \: x\neq 1\: , \: \textrm{maka}\: \: \displaystyle \frac{1}{^{^{a}}\log x}+\displaystyle \frac{1}{\: ^{^{b}}\log x}\: =\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle ^{^{ab}}\log x&&&\textrm{d}.&\displaystyle ^{^{x^{^{2}}}}\log (a+b)\\\\ \textrm{b}.&\displaystyle ^{^{(a+b)}}\log x&\textrm{c}.&^{^{(a+b)}}\log 2x&\textrm{e}.&^{^{x}}\log ab \end{array}\\\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\displaystyle \frac{1}{^{^{a}}\log x}+\displaystyle \frac{1}{\: ^{^{b}}\log x}&=\, ^{^{x}}\log a\, + \, ^{^{x}}\log b \\ &=\, ^{^{x}}\log ab \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika}\: \: m=0,\overline{333}\: \: \textrm{dan}\: \: n=\sqrt{90-\sqrt{90-\sqrt{90-\sqrt{...}}}}\: , \: \textrm{maka}\: \: \displaystyle ^{^{m}}\log n\: =\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle -3&&&\textrm{d}.&\displaystyle 1\\\\ \textrm{b}.&\displaystyle -2&\textrm{c}.&-1&\textrm{e}.&2 \end{array}\\\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}^{^{m}}\log n&=\, ^{^{0,\overline{333}}}\log \sqrt{90-\sqrt{90-\sqrt{90-\sqrt{...}}}}\\ &=\, ^{^{0,\overline{333}}}\log \sqrt{10.9-\sqrt{10.9-\sqrt{10.9-\sqrt{...}}}}\\ &=\, ^{^{^{\frac{1}{3}}}}\log 9\\ &=\, ^{^{^{\frac{1}{3}}}}\log \left ( \displaystyle \frac{1}{3} \right )^{-2}\\ &=-2 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Nilai dari}\: \: \displaystyle \frac{1}{1+\, ^{^{a}}\log bc}+\, \displaystyle \frac{1}{1+\, ^{^{b}}\log ac}+\, \displaystyle \frac{1}{1+\, ^{^{c}}\log ab}\: =\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{2}{3}&&&\textrm{d}.&\displaystyle 2\\\\ \textrm{b}.&\displaystyle 1&\textrm{c}.&\displaystyle \frac{3}{2}&\textrm{e}.&\displaystyle \frac{5}{2} \end{array}\\\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\displaystyle \frac{1}{1+\, ^{^{a}}\log bc}+\, \displaystyle \frac{1}{1+\, ^{^{b}}\log ac}+\, \displaystyle \frac{1}{1+\, ^{^{c}}\log ab}&=\displaystyle \frac{1}{^{^{a}}\log a+\, ^{^{a}}\log bc}+\, \displaystyle \frac{1}{^{^{b}}\log b+\, ^{^{b}}\log ac}+\, \displaystyle \frac{1}{^{^{c}}\log c+\, ^{^{c}}\log ab}\\ &=\displaystyle \frac{1}{^{^{a}}\log abc}+\, \displaystyle \frac{1}{^{^{b}}\log abc}+\, \displaystyle \frac{1}{^{^{c}}\log abc}\\ &=\, ^{^{abc}}\log a+\, ^{^{abc}}\log b+\, ^{^{abc}}\log c\\ &=\, ^{^{abc}}\log abc\\ &=1 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Diketahui}\: \: abc=2^{6}\: \: \textrm{dan}\: \: \displaystyle \left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log bc \right )+\left ( ^{^{2}}\log b \right )\left ( ^{^{2}}\log c \right )=10.\\ &\textrm{Jika}\: \: a\: ,\: b\: ,\: c\: >\: 0,\: \textrm{tunjukkan bahwa}\\\\ &\displaystyle \sqrt{^{^{2}}\log \, ^{2}\, a+^{^{2}}\log \, ^{2}\, b+^{^{2}}\log \, ^{2}\, c}=4\\\\ &\qquad\qquad\qquad\qquad\qquad (\textbf{\textit{SIMAK UI 2012 Mat Das}})\\\\ &\textbf{Bukti}:\\ &\begin{aligned}\displaystyle \left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log bc \right )+\left ( ^{^{2}}\log b \right )\left ( ^{^{2}}\log c \right )&=\displaystyle \left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log b+\, ^{^{2}}\log c \right )+\left ( ^{^{2}}\log b \right )\left ( ^{^{2}}\log c \right )\\ &=\left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log b \right )+\left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log c \right )+\left ( ^{^{2}}\log b \right )\left ( ^{^{2}}\log c \right )\\ &=10\\ \textrm{Masih ingat bentuk}\qquad x^{2}+y^{2}+z^{2}&=(x+y+z)^{2}-2(xy+yz+xz)\\ ^{^{2}}\log \, ^{2}\, a+^{^{2}}\log \, ^{2}\, b+^{^{2}}\log \, ^{2}\, c&=\left ( ^{^{2}}\log a+\, ^{^{2}}\log b+\, ^{^{2}}\log c \right )^{2}\\ &\qquad -2\left ( \left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log b \right )+\left ( ^{^{2}}\log a \right )\left ( ^{^{2}}\log c \right )+\left ( ^{^{2}}\log b \right )\left ( ^{^{2}}\log c \right ) \right )\\ &=\left ( ^{^{2}}\log abc \right )^{2}-2(10)\\ &=\left ( ^{^{2}}\log 2^{6} \right )^{2}-20\\ &=(6)^{2}-20\\ &=36-20\\ &=16\\ \sqrt{^{^{2}}\log \, ^{2}\, a+^{^{2}}\log \, ^{2}\, b+^{^{2}}\log \, ^{2}\, c}&=\sqrt{16}\\ &=4\qquad\quad \blacksquare \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Jika}\: \: ^{^{^{pq}}}\log p=4\: ,\: \textrm{maka}\: \: \displaystyle ^{^{^{pq}}}\log \displaystyle \frac{\sqrt[3]{p}}{\sqrt{q}}=\, ....\\ &\begin{array}{llllllllll}\\ \textrm{a}.&-3&&&&\textrm{d}.&\displaystyle \frac{29}{42}\\\\ \textrm{b}.&-\displaystyle \frac{3}{4}&\textrm{c}.&-\displaystyle \frac{1}{6}&&\textrm{e}.&\displaystyle \frac{17}{6} \end{array}\\\\ &\qquad\qquad\qquad\qquad\qquad (\textbf{\textit{SIMAK UI 2014 Mat Das}})\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Dari soal diketahui bahwa}\quad ^{^{^{pq}}}\log p&=4\\ \displaystyle \frac{\log p}{\log pq}&=4\\ \log p&=4\left ( \log p+\log q \right )\\ -3\log p&=4\log q\\ \log p&=-\displaystyle \frac{4}{3}\log q \\ \textrm{selanjutnya maka}\qquad \displaystyle ^{^{^{pq}}}\log \displaystyle \frac{\sqrt[3]{p}}{\sqrt{q}}&=\displaystyle \frac{\log \displaystyle \frac{\sqrt[3]{p}}{\sqrt{q}}}{\log pq}\\ &=\displaystyle \frac{\log p^{^{^{\frac{1}{3}}}}-\log q^{^{^{\frac{1}{2}}}}}{\log p+\, \log q}\\ &=\displaystyle \frac{\displaystyle \frac{1}{3}\log p-\frac{1}{2}\log q}{\log p+\, \log q}\\ &=\displaystyle \frac{\displaystyle \frac{1}{3}\left ( -\frac{4}{3}\log b \right )-\displaystyle \frac{1}{2}\log b}{-\displaystyle \frac{4}{3}\log b+\, \log b}\\ &=\displaystyle \frac{\left ( -\displaystyle \frac{4}{9}-\frac{1}{2} \right )\log b}{\left ( -\displaystyle \frac{4}{3}+1 \right )\log b}\\ &=\displaystyle \frac{-\displaystyle \frac{17}{18}}{-\displaystyle \frac{1}{3}}\\ &=\displaystyle \frac{17}{6} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{19}.&\textrm{Diketahui sistem persamaan}\\ &\begin{array}{rr}\\ ^{^{10}}\log (2000ab)-\left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log b \right )=4&,\\ ^{^{10}}\log (2bc)-\left ( ^{^{10}}\log b \right )\left ( ^{^{10}}\log c \right )=1&,\\ ^{^{10}}\log (ac)-\left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log c \right )=0&, \end{array}\\\\ &\textrm{akan memiliki solusi}\: \: \left ( a_{1},b_{1},c_{1} \right )\: \: \textrm{dan}\: \: \left ( a_{2},b_{2},c_{2} \right ).\\ &\textrm{Nilai untuk}\: \: b_{1}+b_{2}=\, ....\\\\ &\textrm{Jawab}:25\\ \end{array}.

\begin{array}{ll}\\ .\qquad&\begin{aligned}^{^{10}}\log (ac)-\left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log c \right )&=0\\ ^{^{10}}\log (a)+\, ^{^{10}}\log (c)-\left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log c \right )&=0\\ \left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log c \right )-\, ^{^{10}}\log (a)-\, ^{^{10}}\log (c)&=0\\ \left ( ^{^{10}}\log a \right )\left ( ^{^{10}}\log c \right )-\, ^{^{10}}\log (a)-\, ^{^{10}}\log (c)+1&=1\\ \left ( ^{^{10}}\log a-1 \right )\left ( ^{^{10}}\log c-1\right )&=1\\ \left ( ^{^{10}}\log a-1 \right )&=\displaystyle \frac{1}{\left ( ^{^{10}}\log b-1 \right )}\\ \left ( ^{^{10}}\log a-\, ^{^{10}}\log 10 \right )&=\displaystyle \frac{1}{\left ( ^{^{10}}\log c-\, ^{^{10}}\log 10 \right )}\\ \left ( ^{^{10}}\log \displaystyle \frac{a}{10} \right )&=\displaystyle \frac{1}{\left ( ^{^{10}}\log \displaystyle \frac{c}{10} \right )}\\ \left ( ^{^{10}}\log \displaystyle \frac{a}{10} \right )&=\, ^{^{^{\displaystyle \frac{c}{10}}}}\log 10\\ \textrm{akan dipenuhi saat}\quad &a=c=1\: \: \textrm{atau}\: \: a=c=100\\\\ \textrm{untuk}\quad \: a=c=1,\qquad \textrm{maka} \qquad\qquad &\\ ^{^{10}}\log (2b(1))-\left ( ^{^{10}}\log b \right )\left ( ^{^{10}}\log (1) \right )&=1\\ ^{^{10}}\log 2b-0&=1\\ 2b&=10^{1}\\ b&=5\\ \textrm{untuk}\quad a=c=100,\quad \textrm{maka} \qquad\qquad &\\ ^{^{10}}\log (2b(100))-\left ( ^{^{10}}\log b \right )\left ( ^{^{10}}\log (100) \right )&=1\\ ^{^{10}}\log 100+\, ^{^{10}}\log 2b-\, \left ( ^{^{10}}\log b \right ).2&=1\\ 2+\, ^{^{10}}\log 2b-\, ^{^{10}}\log b^{2}&=1\\ ^{^{10}}\log \displaystyle \frac{2b}{b^{2}}&=-1\\ \displaystyle \frac{2b}{b^{2}}&=10^{-1}=\displaystyle \frac{1}{10}\\ 20b&=b^{2}\\ 20&=b \end{aligned} \end{array}.

Sumber Referensi

  1. Idris, M., Ibnu Rusdi. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: Yrama Widya.
  2. Kanginan, M., Hadi Nurdiansyah, Ghany Ahmad. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Imu-Ilmu Alam (Cetakan VI). Bandung: Yrama Widya.
  3. Sembiring, S., Mahmun Zulkifli, Marsito, dan Ibnu Rusdi. 2016. Matematika untuk Siswa Kelas X SMA/MA Kelompok Peminatan Matematika dan Imu-Ilmu Alam (Cetakan II). Bandung: Srikandi Empat Widya Utama.
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