Lanjutan 2 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 11.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{b}.{\displaystyle y=2\sin \left({\displaystyle \frac{1}{2}\pi -x}\right)}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x+{\displaystyle \frac{1}{6}\pi })}\\ \\ & \text{d}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi +x}\right)}\\ \\ & \text{e}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi -2x}\right)}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}2\pi ,\text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (2x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kirinya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-2\sin (3x+45{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=-2\sin (3x-15{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=-2\sin (3x-45{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=2\sin (3x+15{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=2\sin (3x-45{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kanan}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{3}={120}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin 3(x-k)\\ & \text{dengan}-k\text{adalah}\\ & \mathbf{\text{besar geseran ke kanan}}{15}^{\circ }\\ & \text{Jadi},y=2\sin 3(x-15{)}^{\circ }=2\sin (3x-45{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-\cos (2x-30{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-60{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=\cos (2x+30{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=\sin (2x-80{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=\sin (2x+60{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}1\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=\cos 2(x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{15}^{\circ }\\ & \text{Jadi},y=\cos 2(x+15{)}^{\circ }=\sin (2x+30{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$