Contoh Soal Persiapan Semester Gasal Kelas X (K13 Revisi)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi untuk}\: \: \left | 2x+5 \right |=9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&2&&&\textrm{d}.&-7\: \: \textrm{dan}\: \: 2\\ \textrm{b}.&2\: \: \textrm{dan}\: \: 7&\textrm{c}.&-7\: \: \textrm{dan}\: \: -2&\textrm{e}.&-2\: \: \textrm{dan}\: \: 7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | 2x+5 \right |&=9\\ \left ( 2x+5 \right )&=\pm 9\\ 2x&=\pm 9-5\\ x&=\displaystyle \frac{\pm 9-5}{2}\\ x&=\begin{cases} =\displaystyle \frac{+9-5}{2}=\frac{4}{2}=2 \\\\ =\displaystyle \frac{-9-5}{2}=\frac{-14}{2}=-7 \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | 3x-(4x-7) \right |=6\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -13,-1 \right \}&&&\textrm{d}.&\left \{ -13,1 \right \}\\ \textrm{b}.&\left \{ -1,13 \right \}&\textrm{c}.&\left \{ 1,13 \right \}&\textrm{e}.&\left \{ 13 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\begin{aligned}\left | 3x-(4x-7) \right |&=6\\ \left | 3x-4x+7 \right |&=6\\ \left | -x+7 \right |&=6\\ \left ( -x+7 \right )&=\pm 6\\ -x+7&=\begin{cases} 6 \Leftrightarrow -x=6-7\Leftrightarrow x=1\\\\ -6\Leftrightarrow -x=-6-7\Leftrightarrow x=13 \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | x-1 \right |=2x+1\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ -2,0 \right \}&\textrm{c}.&\left \{ -1 \right \}&\textrm{e}.&\left \{ 0 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\left | x-1 \right |&=2x+1\\ (x-1)&=\pm (2x+1)\\ (x-1)&= \begin{cases} +(2x+1) \\\\ -(2x+1) \end{cases}\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Syarat}&:\\ (x-1)&\begin{cases} x-1\geq 0 \Leftrightarrow x\geq 1\\\\ x-1<0  \Leftrightarrow x<1 \end{cases}\\ & \end{aligned}}\\\hline x\geq 1&x< 1\\\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \begin{aligned}(x-1)&=+(2x+1)\\ x-2x&=1+1\\ -x&=2\\ x&=-2 \end{aligned}&\begin{aligned}(x-1)&=-(2x+1)\\ x+2x&=-1+1\\ 3x&=0\\ x&=0 \end{aligned}\\\hline \textrm{tidak memenuhi}&\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Penyelesaian yang memenuhi untuk}\: \: \left | 3x+1 \right |=2x+9\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&\left \{ -2 \right \}&&&\textrm{d}.&\left \{ \: \: \right \}\\ \textrm{b}.&\left \{ 8 \right \}&\textrm{c}.&\left \{ -2,8 \right \}&\textrm{e}.&\textrm{setiap bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 3x+1 \right |&=2x+9\\ (3x+1)&=\pm (2x+9)\\ (3x+1)&= \begin{cases} +(2x+9) \\\\ -(2x+9) \end{cases}\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Syarat}&:\\ (3x+1)&\begin{cases} 3x+1\geq 0 \Leftrightarrow x\geq -\displaystyle \frac{1}{3}\\\\ 3x+1<0 \Leftrightarrow x<-\displaystyle \frac{1}{3} \end{cases}\\ & \end{aligned}}\\\hline x\geq -\displaystyle \frac{1}{3}&x< -\displaystyle \frac{1}{3}\\\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \begin{aligned}(3x+1)&=+(2x+9)\\ 3x-2x&=9-1\\ x&=8\\ & \end{aligned}&\begin{aligned}(3x+1)&=-(2x+9)\\ 3x+2x&=-9-1\\ 5x&=-10\\ x&=-2 \end{aligned}\\\hline \textbf{memenuhi}&\textbf{memenuhi}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jumlah akar-akar dari}\: \: x^{2}+\left | x \right |-6=0\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0&\textrm{c}.&1&\textrm{e}.&4\\\\ &&&&&(\textbf{Entrance Examination}) \end{array}\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x^{2}+\left | x \right |-6&=0\\ (\left | x \right |+3)(\left | x \right |-2)&=0\\ \left | x \right |+3=0\quad \textrm{atau}\quad \left | x \right |-2&=0\\ \left | x \right |=-3\: (\textbf{tm})\quad \textrm{atau}\quad \left | x \right |&=2\: (\textbf{mm})\\ x&=\pm 2\begin{cases} x_{1}&=2 \\ x_{2} &=-2 \end{cases}\\\\ \textrm{untuk jumlah}&\: \textrm{dari akar-akarnya adalah}:\\ x_{1}+x_{2}&=2+(-2)\\ &=0 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\: \: \left | 2x-9 \right |< 3\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&-3\leq x\leq 6&&&\textrm{d}.&3\leq x\leq 6\\ \textrm{b}.&-3<x<6&\textrm{c}.&3<x<6&\textrm{e}.&-3<x<-6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 2x-9 \right |&< 3\\ -3<2x-9&<3\\ -3+(9)<2x-9+(9)&<3+(9)&\textnormal{masing-masing ditambah 9}\\ 6<2x&<12\\ 6.\left ( \displaystyle \frac{1}{2} \right )<2x.\left ( \displaystyle \frac{1}{2} \right )&<12.\left ( \displaystyle \frac{1}{2} \right )&\textnormal{masing-masing dikali}\: \: \displaystyle \frac{1}{2}\\ 3<x&<6 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi untuk}\: \: \left | 3x+5 \right |\geq 19\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&x\leq -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x\geq 8&&&\textrm{d}.&x< -\displaystyle \frac{14}{3}\: \: \textrm{atau}\: \: x> 8\\ \textrm{b}.&x<-8\: \: \textrm{atau}\: \: x>\displaystyle \frac{14}{3}&\textrm{c}.&x\leq -8\: \: \textrm{atau}\: \: x\geq \displaystyle \frac{14}{3}&\textrm{e}.&x\leq 8\: \: \textrm{atau}\: \: x\geq -\displaystyle \frac{14}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\left | 3x+5 \right |\geq 19\\\\ (\ast )-19\geq 3x+5&\quad \textrm{atau}&(\ast \ast )\: \: 3x+5\geq 19\\ -19-5\geq 3x&\quad \textrm{atau}&3x\geq 19-5\\ -\displaystyle \frac{24}{3}\geq x&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ -8\geq x&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3}\\ x\leq -8&\quad \textrm{atau}&x\geq \displaystyle \frac{14}{3} \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{8}.&\textrm{Nilai}\: \: x\: \: \textrm{real yang memenuhi}\: \: 25-\left | 10x+5 \right |\geq \left | 40x-20 \right |\: \: \textrm{adalah}... .\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textit{NUS Entrance Examination A level})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}25&-\left | 10x+5 \right |\geq \left | 40x-20 \right |\\ 25&-5\left | 2x+1 \right |\geq 20\left | 2x-1 \right |\\ 5&-\left | 2x+1 \right |\geq 4\left | 2x-1 \right |\\ \textrm{ilustrasinya}&\quad \begin{array}{llllllllll} &&&&&&\\\hline &&-\frac{1}{2}&&&\frac{1}{2}&& \end{array}\\ \textrm{dan berikut}&\: \textrm{pembagian wilayahnya}\\ &\begin{array}{|c|c|c|}\hline -\infty < x< -\displaystyle \frac{1}{2}&-\displaystyle \frac{1}{2}\leq x< \frac{1}{2}&\displaystyle \frac{1}{2}\leq x< \infty \\\hline \begin{cases} \left | 2x+1 \right | &=-(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=-(2x-1) \end{cases}&\begin{cases} \left | 2x+1 \right | &=+(2x+1) \\ \left | 2x-1 \right | &=+(2x-1) \end{cases}\\\hline \end{array} \end{aligned} \end{array}..

Selanjutnya,

\begin{array}{|c|c|c|}\hline \begin{aligned}\textrm{\underline{Untuk}}:\: \: -\infty < x<& -\frac{1}{2}\: ,\\ 25-\left | 10x+5 \right |&\geq \left | 40x-20 \right |\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(-(2x+1))&\geq 4(-(2x-1))\\ 5+2x+1&\geq -8x+4\\ 10x&\geq -2\\ x&\geq -\frac{2}{10}\quad (\textbf{tm})\\ & \end{aligned}&\begin{aligned}\textrm{\underline{Untuk}}:\: \: -\frac{1}{2} \leq x<& \frac{1}{2}\: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1)&\geq 4(-(2x-1))\\ 5-2x-1&\geq -8x+4\\ 6x&\geq 0\\ x&\geq 0\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ 0\leq x< \frac{1}{2} \right \} \end{aligned} &\begin{aligned}\textrm{\underline{Untuk}}:\: \: \frac{1}{2}\leq x< &\infty \: ,\\ 25-5\left | 2x+1 \right |&\geq 20\left | 2x-1 \right |\\ 5-(2x+1))&\geq 4(2x-1)\\ 5-2x-1&\geq 8x-4\\ -10x&\geq -8\\ x&\leq \frac{8}{10}\quad (\textbf{mm})\\ \textrm{yang memenuhi}&=\left \{ \frac{1}{2}\leq x\leq \frac{4}{5} \right \} \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Sehingga yang memenuhi}&\: \textrm{adalah}:\\ &=\left \{ 0\leq x\leq \displaystyle \frac{4}{5} \right \} &\\ & \end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: (x+3)(x-1)\geq (x-1)\: \: \textrm{adalah}... .\\ &\begin{array}{llllll}\\ \textrm{a}.&1\leq x\leq 3&&&\textrm{d}.&-2\geq x\: \: \textrm{atau}\: \: x\geq 3\\ \textrm{b}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 1&\textrm{c}.&-3\leq x\leq -1&\textrm{e}.&-1\geq x\: \: \textrm{atau}\: \: x\geq 3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}(x+3)(x-1)&\geq (x-1)\\ (x+3)(x-1)-(x-1)&\geq 0\\ (x-1)\left ( (x+3)-1 \right )&\geq 0\\ (x-1)(x+2)&\geq 0\\ \textrm{Berikut garis bilangan}&\textrm{nya}\\ &\begin{array}{lll|llll|llll} &\multicolumn{3}{c}{.}&&&\multicolumn{3}{c}{.}&\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{3}{l}{.}&&\\\cline{1-3}\cline{8-9} +&+&&-&-&-&&&+&+\\\hline &&\multicolumn{2}{l}{\textcircled{-2}}&&&\multicolumn{2}{l}{\textcircled{1}}&&\\ \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\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}\\\\ &\textrm{Jawab}:\quad \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 di atasnya, 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&>6 \end{aligned} \end{array}.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
Pos ini dipublikasikan di Info, Matematika, Pendidikan. Tandai permalink.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout /  Ubah )

Foto Google

You are commenting using your Google account. Logout /  Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout /  Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout /  Ubah )

Connecting to %s