$\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | 3-\left | x \right | \right |<10\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-14\: \: \textrm{atau}\: x>12\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: x>13\\ \textrm{c}.&x<-12\: \: \textrm{atau}\: x>10\\ \textrm{d}.&0<x<10\\ \color{red}\textrm{e}.&-13<0<13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\left | 3-\left | x \right | \right |< 10\\ &-10< 3-\left | x \right |< 10\\ &-13< -\left | x \right |< 7\\ &-7< \left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ &0\leq \left | x \right |< 13\\ &\textrm{selanjutnya},\\ &\left | x \right |< 13\\ &-13< \: x< 13\\ &\textrm{HP}=\color{red}\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{l}\\ 37.&\textrm{(UM UGM 05)}\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x^{2}-3 \right |<2x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<3\\ \textrm{b}.&-3<x<1\\ \color{red}\textrm{c}.&1<x<3\\ \textrm{d}.&-3<x<-1\: \: \textrm{atau}\: \: 1<x<3\\ \textrm{e}.&x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left | x^{2}-3 \right |&<2x\\ -2x<\left ( x^{2}-3 \right )&<2x\\ \textrm{dipartisi men}&\textrm{jadi dua bagian}\\ \bullet \quad\textrm{pertama}\qquad&\\ (x^{2}-3)&>-2x\\ x^{2}+2x-3&>0\\ (x+3)(x-1)&>0\\ x<-3\: \: \textrm{atau}&\: \: x>1\\ \bullet \quad \textrm{kedua}\qquad\quad&\\ \left ( x^{2}-3 \right )&<2x\\ x^{2}-2x-3&<0\\ (x-3)(x+1)&<0\\ -1<x<3&\\ \color{black}\textrm{ambil yang}&\: \color{black}\textrm{memenuhi keduanya}\\ \textrm{berupa iris}&\textrm{an}\\ \textrm{HP}=&\color{red}\left \{ 1<x<3,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 38.&(\textrm{SPMB 05})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | x-2 \right |^{2}<4\left | x-2 \right |+12\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x\in \mathbb{R}|2\leq x\leq 8 \right \}\\ \textrm{b}.&\left \{ x\in \mathbb{R}|4<x< 8 \right \}\\ \color{red}\textrm{c}.&\left \{ x\in \mathbb{R}|-4<x< 8 \right \}\\ \textrm{d}.&\left \{ x\in \mathbb{R}|-2<x<4 \right \}\\ \textrm{e}.&\left \{ x\in \mathbb{R}|2<x<4 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{misalkan}\: \: p&=\left | x-2 \right |,\: \: \textrm{selanjutnya}\\ \left | x-2 \right |^{2}<&\, 4\left | x-2 \right |+12\\ p^{2}<&\, 4p+12\\ p^{2}-&4p-12<0\\ (p-6)&(p+2)<0\\ -2<p&<6,\: \: \color{magenta}\textrm{atau jika dikembalikan}\\ -2<&\left | x-2 \right |<6,\\ &\: \: \color{black}\textrm{ingat, nilanya tidak negatif}\\ 0\leq &\left | x-2 \right |<6\\ -6<&\: x-2<6\\ -4<&\: x<8\\ \textrm{HP}=&\color{red}\left \{ -4<x<8,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 39.&\textrm{Diketahui grafik fungsi}\: \: f(x)=mx^{2}-2mx+m\\ & \textrm{berada di atas grafik fungsi}\\ &g(x)=2x^{2}-3,\: \textrm{maka nilai}\: \: m\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&m>2&&&\textrm{d}.&-6<m<2\\ \textrm{b}.&m>6&\textrm{c}.&2<m<6&\textrm{e}.&m<-6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}f(x)&=g(x)\\ mx^{2}-2mx+m&=2x^{2}-3\\ mx^{2}-2x^{2}-2mx+m+3&=0\\ \textrm{Supaya grafik}\: \: f(x)\: \textrm{berada }&\color{blue}\textrm{di atasnya}, \\ \textrm{maka}\: \: D=B^{2}-4AC&<0\\ (m-2)x^{2}-2mx+(m+3)&=0\begin{cases} A &=m-2 \\ B &=-2m \\ C &=m+3 \end{cases}\\ B^{2}-4AC&<0\\ (-2m)^{2}-4(m-2)(m+3)&<0\\ 4m^{2}-4\left ( m^{2}+m-6 \right )&<0\\ 4m^{2}-4m^{2}-4m+24&<0\\ -4m+24&<0\\ m-6&>0\\ m&\color{red}>6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: \color{red}3<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2x-2&&&\textrm{d}.&-2\\\\ \textrm{b}.&2\quad&\textrm{c}.&8-2x\quad&\textrm{e}.&2x-8\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}\\ &=\left | x-3 \right |-\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}3<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=\color{red}2x-8 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: \color{red}1<x<5\: \: \color{black}\textrm{maka penyelesaian untuk}\\ &\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{lllllll}\\ \textrm{a}.&2&&&\textrm{d}.&5\\\\ \textrm{b}.&3\quad&\textrm{c}.&4\quad&\textrm{e}.&6\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\sqrt{x^{2}-2x+1}+\sqrt{x^{2}-10x+25}\\ &=\sqrt{(x-1)^{2}}+\sqrt{(x-5)^{2}}\\ &=\left | x-1 \right |+\left | x-5 \right |\\ &\textrm{ingat bahwa saat}\: \: \color{red}1<x<5\\ &\textrm{maka}\: \begin{cases} \left | x-1 \right |=(x-1) \\ \left | x-5 \right |=-(x-5) \end{cases},\\ &\color{blue}\textrm{sehingga}\\ &=\left | x-1 \right |+\left | x-5 \right |=(x-1)+\left ( -(x-5) \right )\\ &=x-1+5-x\\ &=\color{red}4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 42.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{5}{4x-3} \right |\leq 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\leq x\leq \frac{3}{4}\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{b}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: \displaystyle \frac{3}{4}< x\leq 2\\ \textrm{c}.&-\displaystyle \frac{1}{2}\leq x\leq 2,\: \: x\neq \frac{3}{4}\\ \textrm{d}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x>\frac{3}{4}\\ \color{red}\textrm{e}.&x\leq -\displaystyle \frac{1}{2}\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{5}{4x-3} \right |&\leq 1\\ -1\leq \displaystyle \frac{5}{4x-3}&\leq 1,\: \: \color{magenta}\textbf{jika dibalik}\\ -1\geq \displaystyle \frac{4x-3}{5}&\geq 1,\: \: \color{magenta}\textbf{bentuk ini tidak}\\ \color{magenta}\textbf{dibolehkan}&\: \color{magenta}\textbf{maka perlu diubah menjadi}\\ -1\geq \displaystyle \frac{4x-3}{5}\: \: \textrm{atau}&\: \: \displaystyle \frac{4x-3}{5}\geq 1,\: \: \color{black}\textrm{selanjutnya}\\ \bullet \quad \textrm{bagian}&\: 1\\ -1&\geq \displaystyle \frac{4x-3}{5}\Leftrightarrow \frac{4x-3}{5}\leq -1\\ 4x-3&\leq -5\\ 4x&\leq -2\\ x&\leq -\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{4x-3}{5}&\geq 1\\ 4x-3&\geq 5\\ 4x&\geq 8\\ x&\geq 2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 43.&(\textrm{UMPTN 95})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2}{2x-1} \right |> 1\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x> 2\\ \textrm{b}.&x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{c}.&x<-1\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{d}.&-1<x<2\: \: \textrm{dan}\: \: x\neq \displaystyle \frac{1}{2}\\ \textrm{e}.&x<-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{semua opsi bukan jawaban}\\ &\textbf{Berikut pembahasannya}\\ &\begin{aligned}\left | \displaystyle \frac{2}{2x-1} \right |&> 1\\ -1>\displaystyle \frac{2}{2x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2}{2x-1}>1,\: \color{magenta}\textbf{dibalik}\\ -1<\displaystyle \frac{2x-1}{2}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x-1}{2}<1\\ \bullet \quad \textrm{bagian}&\: 1\\ \displaystyle \frac{2x-1}{2}&>-1\\ 2x-1&>-2\\ 2x&>-1\\ x&>-\displaystyle \frac{1}{2}\\ \bullet \quad \textrm{bagian}&\: 2\\ \displaystyle \frac{2x-1}{2}&<1\\ 2x-1&<2\\ 2x&<3\\ x&<\displaystyle \frac{3}{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 44.&(\textrm{UMPTN 00})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{2x+7}{x-1} \right |\geq 1\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-2\leq x\leq 8\\ \textrm{b}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&-8\leq x< 1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&-2\leq x< 1\: \: \textrm{atau}\: \: 1< x\leq 8\\ \color{red}\textrm{e}.&x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\left | \displaystyle \frac{2x+7}{x-1} \right |&\geq 1\\ -1\geq \displaystyle \frac{2x+7}{x-1}&\: \: \textrm{atau}\: \: \displaystyle \frac{2x+7}{x-1}\geq 1\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{2x+7}{x-1}&\leq -1\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{2x+7}{x-1}&+1\leq 0\\ &\displaystyle \frac{2x+7+(x-1)}{x-1}\leq 0\\ \displaystyle \frac{3x+6}{x-1}&\leq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| -2\leq x< 1,\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{2x+7}{x-1}&\geq 1\\ \displaystyle \frac{2x+7}{x-1}&-1\geq 0\\ &\displaystyle \frac{2x+7-(x-1)}{x-1}\geq 0\\ \displaystyle \frac{x+8}{x-1}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>1,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: -2\leq x< 1\: \: \textrm{atau}\: \: x> 1,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 45.&(\textrm{UMPTN 01})\\ &\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\left | \displaystyle \frac{x-2}{x+3} \right |\leq 2\: \: \textrm{adalah}...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&-8\leq x< -3\\ \textrm{b}.&-8\leq x< -1\\ \textrm{c}.&-4\leq x< -3\\ \color{red}\textrm{d}.&x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3}\\ \textrm{e}.&x\leq -4\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left | \displaystyle \frac{x-2}{x+3} \right |&\leq 2\\ -2\leq \displaystyle \frac{x-2}{x+3}&\: \: \textrm{atau}\: \: \displaystyle \frac{x+2}{x+3}\leq 2\\ \bullet \qquad \textrm{bagian}&\: \: 1\\ \displaystyle \frac{x-2}{x+3}&\geq -2\: \: \color{magenta}(\textbf{tidak boleh kali silang})\\ \displaystyle \frac{x-2}{x+3}&+2\geq 0\\ &\displaystyle \frac{x-2+2(x+3)}{x+3}\geq 0\\ \displaystyle \frac{3x+4}{x+3}&\geq 0\\ \textrm{HP}_{1}=&\color{black}\left \{x| x< -3\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \}\\ \bullet \qquad \textrm{bagian}&\: \: 2\\ \displaystyle \frac{x-2}{x+3}&\leq 2\\ \displaystyle \frac{x-2}{x+3}&-2\leq 0\\ &\displaystyle \frac{x-2-2(x+3)}{x+3}\leq 0\\ \displaystyle \frac{-x-8}{x+3}&\leq 0,\: \: \color{magenta}\textbf{koefisien \textit{x} negatif}\\ \displaystyle \frac{x+8}{x+3}&\geq 0\\ \textrm{HP}_{2}=&\color{black}\left \{x|x\leq -8\: \: \textrm{atau}\: \: x>-3,\: x\in \mathbb{R} \right \}\\ \textrm{HP}=\textrm{HP}_{1}+\textrm{HP}_{2}&=\color{red}\left \{ x|x\leq -8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{4}{3},\: x\in \mathbb{R} \right \} \end{aligned} \end{array}$
Tidak ada komentar:
Posting Komentar
Informasi