Lanjutan 1 Contoh Soal Persiapan Semester Gasal 2019/2020 Kelas X Wajib

\color{magenta}\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai dari}\: \: \left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}=...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3\pi -10}{5}\\ \textrm{b}.&\displaystyle \frac{3\pi +5}{2}\\ \textrm{c}.&\displaystyle \frac{-3\pi +5}{2}\\ \textrm{d}.&\displaystyle \frac{-3\pi +10}{2}\\ \textrm{e}.&\displaystyle \frac{3\pi -10}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\color{black}\begin{aligned}\left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}&=\left (2\pi -5 \right )-\displaystyle \frac{\pi }{2},\qquad \textrm{sebagai catatan bahwa}\: \: \: \pi =3,14...\\ &=\displaystyle \frac{2(2\pi )-2(5)-\pi }{2}\\ &=\displaystyle \frac{3\pi -10}{2} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Nilai untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: \left | 2x-5 \right |=11\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{hanya}\: \: 3\\ \textrm{b}.&\textrm{hanya}\: \: 8\\ \textrm{c}.&-3\: \: \textrm{atau}\: \: 8\\ \textrm{d}.&3\: \: \textrm{atau}\: \: -8\\ \textrm{e}.&3\: \: \textrm{atau}\: \: 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 2x-5 \right |&=11\\ (2x-5)&=\pm 11\\ 2x&=5\pm 11\\ 2x&=\begin{cases} 5+11 & =16 \\ \textrm{atau}&\\ 5-11 &=-6 \end{cases}\\ x&=\begin{cases} 8 & \\ &\textrm{atau} \\ -3 & \end{cases} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Himpunan penyelesaian untuk}\: \: -2\left | x-5 \right |=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0 \right \}\\ \textrm{b}.&\left \{ \: \: \right \}\\ \textrm{c}.&\left \{ 4,5 \right \}\\ \textrm{d}.&\left \{ 1,9 \right \}\\ \textrm{e}.&\left \{ -1,9 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}-2\left | x-5 \right |&=8\\ \left | x-5 \right |&=-4\: ,\quad \textrm{karena bernilai negatif maka tidak ada harga}\: \: x\: \: \textrm{yang memenuhi}\\\\ \therefore \: \: \: \textbf{HP}&=\left \{ \: \: \right \} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi}\: \: \left | 4k \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | 4k \right |&=16\\ (4k)&=\pm 16\\ k&=\pm \displaystyle \frac{16}{4}\\ &=\pm 4 \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai}\: \: p\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5p \right |=26\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}&&&\textrm{d}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{b}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\quad&\textrm{c}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\quad&\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}10-4\left | 4-5p \right |&=-26\\ -4\left | 4-5p \right |&=-36\\ \left | 4-5p \right |&=9\\ (4-5p)&=\pm 9\\ -5p&=-4\pm 9\\ p&=\displaystyle \frac{-4\pm 9}{-5}\\ p&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=2\frac{3}{5} \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ .\: \: \: \: \: \: &\textrm{Persamaan yang memenuhi rumus tersebut adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Dengan cara substitusi langsung kita akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4\: \: \textrm{dan}\\ &\qquad \textrm{sampai langkah di sini hanya ada 1 persamaan yang memenuhi yaitu}:\: \: y=\left | -2x+4 \right | \end{aligned}\end{array}.

\begin{array}{lll}\\ \fbox{13}.&\textrm{Gambarlah garfik untuk persamaan}\: \: \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+\left | y \right |=4\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}

Berikut sebagai ilustrasinya

\begin{array}{lll}\\ \fbox{14}.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\: \: x+\left | x \right |+y=5\: \: \textrm{dan}\: \: x+\left | y \right |-y=10\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+x+y=5\\ &\: \: \qquad \textrm{dan}\\ &x+\left | y \right |-y=10\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0\quad \textbf{(kuadran I)}&&&x> 0\: ,\: y<0\quad \textbf{(kuadran IV)}\\\hline \begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5 \\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: (\textrm{ada}) \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5\\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0\quad \textbf{(kuadran II)}&&&x<0\: ,\: y<0\quad \textbf{(kuadran III)}\\\hline \begin{cases} \left | x+x+y=5 \right | & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: \textbf{(tidak memenuhi)} \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: \textbf{(tidak memenuhi)}\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\\hline \end{array} \end{array}

. Beriut ilustrasi grafiknya

Jadi, nilai x dan y yang memenuhi untuk persamaan di atas adalah  x = 4  dan  y = -3

\color{magenta}\begin{array}{ll}\\ \fbox{15}.&\textrm{Gambarlah grafik fungsi mutlak dari}\\ &\textrm{a}.\quad y=\left | x-2 \right |\\ &\textrm{b}.\quad y=-\left | x-2 \right |\\ &\textrm{c}.\quad y=2+\left | x-2 \right |\\ &\textrm{d}.\quad y=2-\left | x-2 \right |\\ &\textrm{e}.\quad y=\left | 2+\left | x-2 \right | \right |\\ &\textrm{f}.\quad y=\left | 2-\left | x-2 \right | \right | \end{array}.

Berikut grafik dari soal di atas

untuk Soal poin a)
untuk jawaban poin b)
untuk jawaban poin c)
untuk jawaban poin d)
untuk jawaban poin f)
sSedangkan poin e) silahkan tentukan atau buat sendiri grafiknya

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