$\begin{array}{ll}\\ 11.&\textrm{Diketahui beberapa fungsi memiliki }\\ &\textrm{sifat-sifat sebagaimana berikut ini}:\\ &(i)\quad \Phi (-x)=-\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &(ii)\quad \Phi (-x)=\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &\textrm{Jika diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\\ & \textrm{memiliki sifat}\: \: (i)\: \: \textrm{dan fungsi}\: \: h\: \: \textrm{dan}\: \: k\\ &\textrm{memiliki sifat}\: \: (ii),\: \: \textrm{maka pernyataan }\\ &\textrm{berikut yang salah adalah}\: ....\\ &(1)\quad (f+g)(-x)=-(f+g)(x)\\ &(2)\quad (f.k)(-x)=-(f.k)(x)\\ &(3)\quad (h-k)(-x)=(h-k)(x)\\ &(4)\quad (h-g)(-x)=(h-g)(x)\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&(1),(2)\: \textrm{dan}\: (3)\\ \textrm{b}.&(1)\: \textrm{dan}\: (3)\\ \textrm{c}.&(2)\: \textrm{dan}\: (4)\\ \textrm{d}.&(4)\: \textrm{saja}\\ \textrm{e}.&\textrm{semuanya benar} \end{array}\\ & (\textbf{SIMAK UI 2014 Mat Das})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} \Phi (-x)=-\Phi (x) \\ (\textrm{fungsi ganjil})\begin{cases} f & \text{ misal } f(x)=x \\ g & \text{ misal } g(x)=2x \end{cases} \\\\ \Phi (-x)=\Phi (x)\\ (\textrm{fungsi genap})\begin{cases} h & \text{ misal } h(x)=x^{2} \\ k & \text{ misal } k(x)=2x^{2} \end{cases} \end{cases}\\ & \end{aligned}\\ &\begin{array}{|c|}\hline \begin{aligned}(1)\quad (f+g)(-x)&=-(f+g)(x)\\ &\textrm{benar}\\ (2)\qquad (f.k)(-x)&=-(f.k)(x)\\ &\textrm{benar}\\ \end{aligned}\\\hline \begin{aligned} (3)\quad (h-k)(-x)&=(h-k)(x)\\ &\textrm{benar}\\ (4)\quad (h-g)(-x)&=(h-g)(x)\\ &\color{blue}\textrm{salah} \end{aligned}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{2x-1}\: \: \textrm{dan}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2}\\ &\textrm{maka}\: \: g(x)=\: ....\\ &(\textbf{UMPTN 1998})\\ &\begin{array}{llllll}\\ \textrm{a}.&x+\displaystyle \frac{1}{2}&&&\textrm{d}.&1-\displaystyle \frac{2}{x}\\\\ \textrm{b}.&x-\displaystyle \frac{1}{2}&\color{red}\textrm{c}.&2-\displaystyle \frac{1}{x}&\textrm{e}.&2-\displaystyle \frac{1}{2x} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dan}\\ &(f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{1}{\left (\displaystyle \frac{3x-2}{x} \right )}\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\color{magenta}2g(x)-1=\displaystyle \frac{3x-2}{x}\\ &\Leftrightarrow 2g(x)=1+\left ( 3-\displaystyle \frac{2}{x} \right )\\ &\Leftrightarrow 2g(x)=4-\displaystyle \frac{2}{x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dengan}\\ &f^{-1}(x)=\displaystyle \frac{x+1}{2x}.........(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\displaystyle \frac{f(g(x))+1}{2\left ( f(g(x)) \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\left ( \displaystyle \frac{x}{3x-2} \right )+1}{2\left ( \displaystyle \frac{x}{3x-2} \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\displaystyle \frac{4x-2}{3x-2}}{\displaystyle \frac{2x}{3x-2}}\\ &\Leftrightarrow g(x)=\displaystyle \frac{4x-2}{2x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: f(x)=x^{2}-9\: \: \textrm{dan}\: \: (f\circ g)(x)=x(x-6)\\ &\textrm{rumus fungsi}\: \: g(x)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+3&&&\textrm{d}.&3x+1\\ \textrm{b}.&x-3&\color{red}\textrm{c}.&-x&\textrm{e}.&x \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dan}\\ &(f\circ g)(x)=x(x-6)=x^{2}-6x,\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x+9-9\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=(x-3)^{2}-9\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\qquad g(x)=x-3 \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{x+9}\: .......(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=x^{2}-6x,\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\sqrt{\left ( f(g(x)) \right )+9}\\ &\Leftrightarrow g(x)=\sqrt{x^{2}-6x+9}\\ &\Leftrightarrow g(x)=\sqrt{(x-3)^{2}}\\ &\Leftrightarrow g(x)=x-3 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\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&\color{red}\textrm{e}.&x+5 \end{array}\\ & (\textbf{UM UGM 2010 Mat Das})\\\\ &\textrm{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\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}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}},\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{\displaystyle \frac{1}{x^{2}}+2}\: .......(\textrm{akan ditunjukkan})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\: \: \textrm{maka}\\ &\begin{array}{|c|c|}\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} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: g(x)=2x+4\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=4x^{2}+8x-3,\\ & \textrm{maka}\: \: f^{-1}(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+9\\ \textrm{b}.&\sqrt{x}+2\\ \textrm{c}.&x^{2}-4x-3\\ \textrm{d}.&\sqrt{x+1}+2\\ \color{red}\textrm{e}.&\sqrt{x+7}+2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{Sintak 1}&\textrm{ Sintak 2}&\textrm{Hasil Invers}\\\hline \begin{aligned}g(x)=y&=2x+4\\ y-4&=2x\\ x&=\displaystyle \frac{y-4}{2}\\ f^{-1}(y)&=\displaystyle \frac{y-4}{2}\\ f^{-1}(x)&=\displaystyle \frac{x-4}{2}\\ & \end{aligned}&\begin{aligned}f(x)&=\left (f\circ g\circ g^{-1} \right )(x)\\ &=4\left ( g^{^{-1}}(x) \right )^{2}+8\left ( g^{-1}(x) \right )-3\\ &=4\left ( \displaystyle \frac{x-4}{2} \right )^{2}+8\left ( \displaystyle \frac{x-4}{2} \right )-3\\ &=\left ( \displaystyle x^{2}-8x+16 \right )+4x-16-3\\ &=x^{2}-4x-3\\ &=x^{2}-4x+4-7\\ &=(x-2)^{2}-7 \end{aligned}&\begin{aligned}f(x)=y&=(x-2)^{2}-7\\ y+7&=(x-2)^{2}\\ \sqrt{y+7}&=(x-2)\\ (x-2)&=\sqrt{y+7}\\ x&=\sqrt{y+7}+2\\ f^{-1}(y)&=\sqrt{y+7}+2\\ f^{-1}(x)&=\sqrt{x+7}+2 \end{aligned} \\\hline \end{array} \end{array}$
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