Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketiga

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

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\cos x}\\ \\ & \text{b}.{\displaystyle y=\cos 2x}\\ \\ & \text{c}.{\displaystyle y=\cos {\displaystyle \frac{1}{2}x}}\\ \\ & \text{d}.{\displaystyle y=2\cos 2x}\\ \\ & \text{e}.{\displaystyle y=2\cos {\displaystyle \frac{1}{2}x}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\\ & \mathbf{\text{atas dan ke bawah}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ }=\pi ,}\\ & \text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\cos 2x\end{array}\end{array}$.

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

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (2x+\pi )}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{d}.{\displaystyle y=\sin (2x+{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{e}.{\displaystyle y=2\sin {\displaystyle (x+{\displaystyle \frac{1}{2}\pi })}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{1}={360}^{\circ }=2\pi ,}\\ & \text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (x+{\displaystyle kx})\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{\displaystyle \frac{1}{2}\pi \text{atau}{90}^{\circ }}\\ & \text{Jadi},y=2\sin (x+{\displaystyle \frac{1}{2}\pi {)}^{\circ }}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos 2x-2\cos x=-1\text{untuk}0<x<2\pi \\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{3}{2}\pi ,2\pi }}\}}\\ \\ & \text{b}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi ,2\pi }}\}}\\ \\ & \text{c}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi ,{\displaystyle \frac{3}{2}\pi }}\}}\\ \\ & \text{d}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi }}\}}\\ \\ & \text{e}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos 2x-2\cos x=-1\\ & \iff \cos 2x-2\cos x+1=0\\ & \iff (2{\cos }^{2}x-1)-2\cos x+1=0\\ & \iff 2\cos x(\cos x-1)=0\\ & \iff \cos x=0\text{atau}\cos x=1\\ & \iff \cos x=\cos {\displaystyle \frac{1}{2}\pi \text{atau}\cos x=\cos 0}\\ & \iff x_{1,2}=\pm {\displaystyle \frac{1}{2}\pi +k.2\pi \text{atau}{x}_3=k.2\pi }\\ & \text{maka}\\ & k=0\Rightarrow x_1=-{\displaystyle \frac{1}{2}\pi (\text{tm})\text{atau}}\\ & x_2={\displaystyle \frac{1}{2}\pi (\text{mm})}\\ & x_3={\displaystyle 0(\text{mm})}\\ & k=1\Rightarrow x_1={\displaystyle \frac{3}{2}\pi (\text{mm})}\\ & x_2={\displaystyle \frac{5}{2}\pi (\text{tm})}\\ & x_3={\displaystyle 2\pi (\text{mm})}\end{array}\end{array}\end{array}$