$\begin{array}{ll}\\ 61.&\textrm{Fungsi}\: \: f(x)=\sin x-\cos x\: \: \textrm{dengan}\\ &0<x<2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<2\pi \\ \textrm{c}.&\displaystyle \frac{3\pi }{4}<x<\displaystyle \frac{7\pi }{4} \\ \color{red}\textrm{d}.&0<x<\displaystyle \frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{e}.&0<x<\displaystyle \frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin x-\cos x\\ &\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=\cos x+\cos x=0\\ &\sin x=-\cos x\Leftrightarrow \displaystyle \frac{\sin x}{\cos x}=-1\\ &\Leftrightarrow \tan x=-1\\ &\Leftrightarrow \tan x=\tan \displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow x=\displaystyle \frac{3\pi }{4}\pm k.\pi \\\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}\\ &\Leftrightarrow k=2\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm 2\pi =\color{red}\textrm{tm}\\ &\color{black}\begin{array}{ccccccccc}\\ &&&&&&&\\ &++&&--&&&++&\\\hline 0&&\color{red}\displaystyle \frac{3\pi }{4}&&&\color{red}\displaystyle \frac{7\pi }{4}&&2\pi \end{array}\\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{2}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{1}{2}\pi +\sin \displaystyle \frac{1}{2}\pi =0+1=1\: \: \color{red}(\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{3}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{3}{2}\pi +\sin \displaystyle \frac{3}{2}\pi =0-1=-1\: \: \color{red}(\textrm{negatif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{11}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{11}{6}\pi \right )\\ &\quad =\cos \displaystyle \frac{11}{6}\pi +\sin \displaystyle \frac{11}{6}\pi =\color{red}\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\: \: \color{red}(\textrm{positif}) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 62.&\textrm{Fungsi}\: \: f(x)=\sin^{2} x\: \: \textrm{dengan}\\ &0<x<2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{\pi }{2}<x<\pi \: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{2}<x<2\pi \\ \textrm{b}.&\displaystyle \frac{2\pi }{3}<x<\pi \\ \color{red}\textrm{c}.&0<x<\displaystyle \frac{\pi }{2}\: \: \textrm{atau}\: \: \pi <x<\displaystyle \frac{3\pi }{2} \\ \textrm{d}.&\displaystyle \frac{4\pi }{3}<x<2\pi \\ \textrm{e}.&\displaystyle \frac{\pi }{3}<x<\pi \: \: \textrm{atau}\: \: \displaystyle \frac{4\pi }{3}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin^{2} x\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=2\sin x\cos x=\sin 2x=0\\ &\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=\pm k.2\pi \: \: \textrm{atau}\: \: 2x=\pi \pm k.2\pi \\ &\Leftrightarrow x=\pm k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} \pm k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}\\ &\Leftrightarrow k=1\Rightarrow x=\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+ \pi =\frac{3\pi }{2}\\ &\Leftrightarrow k=2\Rightarrow x= 2\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+2\pi =\displaystyle \frac{5}{2}\pi \: \color{red}(\textrm{tm})\\ &\color{black}\begin{array}{ccccccccc}\\ &&&&&&&&\\ &++&&--&&++&&--&\\\hline 0&&\color{red}\displaystyle \frac{\pi }{2}&&\pi &&\color{red}\displaystyle \frac{3\pi }{2}&&2\pi \end{array} \\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{6}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{6}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{1}{6}\pi \right )=\sin \displaystyle \frac{1}{3}\pi =\frac{1}{2}\: \: \color{red}(\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{4}\pi \Rightarrow f\left ( \displaystyle \frac{3}{4}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{3}{4}\pi \right )=\color{red}-1\: \: \color{red}(\textrm{negatif}) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 63.&\textrm{Fungsi}\: \: f(x)=\cos ^{2}2x\: \: \textrm{untuk}\\ &0^{\circ}<x<360^{\circ}\: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&45^{\circ}<x<90^{\circ}\\ \textrm{b}.&135^{\circ}<x<180^{\circ}\\ \textrm{c}.&225^{\circ}<x<270^{\circ}\\ \color{red}\textrm{d}.&270^{\circ}<x<300^{\circ}\\ \textrm{e}.&315^{\circ}<x<360^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&f(x)=\cos ^{2}2x\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ &\textrm{Selanjutnya}\\ &\Leftrightarrow -2\sin 4x=0\Leftrightarrow \sin 4x=0\Leftrightarrow \sin 4x=\sin 0^{\circ}\\ &\Leftrightarrow \begin{cases} 4x=0^{\circ}+k.360^{\circ}&\Rightarrow x=k.90^{\circ}\\ 4x=180^{\circ}+k.360^{\circ}&\Rightarrow x=45^{\circ}+k.90^{\circ} \end{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \: \textrm{atau}\: \: x=45^{\circ}\\ &\Leftrightarrow k=1\Rightarrow x=90^{\circ}\: \: \textrm{atau}\: \: x=135^{\circ}\\ &\Leftrightarrow k=2\Rightarrow x=180^{\circ}\: \: \textrm{atau}\: \: x=225^{\circ}\\ &\Leftrightarrow k=3\Rightarrow x=270^{\circ}\: \: \textrm{atau}\: \: x=315^{\circ}\\ &\Leftrightarrow k=4\Rightarrow x=360^{\circ}\: \: \textrm{atau}\: \: x=405^{\circ}\: \: \color{red}(\textrm{tm})\\ &\textrm{Gunakan titik uji pada}\: \: x=30^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(30^{\circ})=-2\sin 4(30^{\circ})=-\sqrt{3}\: \: \color{red}(\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=60^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(60^{\circ})=-2\sin 4(60^{\circ})=\sqrt{3}\: \: \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=120^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(120^{\circ})=-2\sin 4(120^{\circ})=-\sqrt{3}\: \: \color{red}(\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=150^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(150^{\circ})=-2\sin 4(150^{\circ})=\sqrt{3}\: \: \color{red}(\textrm{positif})\\ &\color{black}\textrm{dan seterusnya}\: ...\\ &\color{black}\begin{array}{ccccccccccc}\\ &&&&&&&&\\ &--&&++&&--&&++\\\hline 0&&\color{red}\displaystyle 45^{\circ}&&90^{\circ}&&\color{red}135^{\circ}&&180^{\circ}\\ &--&&++&&--&&++\\\hline 180^{\circ}&&\color{red}\displaystyle 225^{\circ}&&270^{\circ}&&\color{red}315^{\circ}&&360^{\circ} \end{array} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 64.&(\textbf{SBMPTN 2015})\\ &\textrm{Fungsi}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ & \textrm{pada}\: \: 0<x<\pi \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}<x<\frac{11\pi }{12}\\ \textrm{b}.&\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}\\ \color{red}\textrm{c}.&\displaystyle \frac{2\pi }{3}<x<\frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{3\pi }{4}<x<\pi \\ \textrm{e}.&\displaystyle \frac{3\pi }{4}<x<\frac{3\pi }{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=\displaystyle \frac{\sin 2x+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} x+\displaystyle \frac{x\sqrt{3}}{2}}}=0\\ &\sin 2x+\displaystyle \frac{1}{2}\sqrt{3}=0\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{4\pi }{3}\\ &\Leftrightarrow 2x=\displaystyle \frac{4\pi }{3}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{4\pi }{3}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{2\pi }{3}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{6}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{6}\: \: \color{red}(\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{5\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{5\pi }{6}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=\color{black}90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{\pi }{2} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{\left ( \displaystyle \frac{\pi }{2} \right )\sqrt{3}}{2}}}=\color{red}+\\ &\qquad \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=\color{black}135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{3\pi }{4} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{1}{2}\sqrt{3}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{\left ( \displaystyle \frac{3\pi }{4} \right )\sqrt{3}}{2}}}=\color{red}-\\ &\qquad \color{red}(\textrm{negatif})\\ &\color{black}\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\color{red}\displaystyle \displaystyle \frac{2\pi }{3}&&\color{red}\displaystyle \frac{5\pi }{6} \end{array} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 65.&\textrm{Fungsi}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ & \textrm{dengan}\: \: x>0 \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}<x\leq \frac{13\pi }{12}\\ \color{red}\textrm{b}.&\displaystyle \frac{7\pi }{12}<x<\frac{11\pi }{12}\\ \textrm{c}.&\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}\\ \textrm{d}.&\displaystyle \frac{7\pi }{6}<x\leq \displaystyle \frac{13\pi }{6} \\ \textrm{e}.&\displaystyle \frac{7\pi }{6}<x\leq \frac{11\pi }{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=\displaystyle \frac{\sin 2x+\displaystyle \frac{1}{2}}{2\sqrt{\sin^{2} x+\displaystyle \frac{x}{2}}}=0\\ &\sin 2x+\displaystyle \frac{1}{2}=0\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{7\pi }{6}\\ &\Leftrightarrow 2x=\displaystyle \frac{7\pi }{6}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{7\pi }{6}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{7\pi }{12}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{12}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{7\pi }{12}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{12}\: \: \color{red}(\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{19\pi }{12}\: \: \textrm{atau}\: \: x=\displaystyle \frac{11\pi }{12}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=\color{black}90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{\pi }{2} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{1}{2}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{\pi }{2} \right )+\displaystyle \frac{\left ( \displaystyle \frac{\pi }{2} \right )}{2}}}=\color{red}+\\ &\qquad \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=\color{black}135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\left ( \displaystyle \frac{3\pi }{4} \right )=\displaystyle \frac{\sin 2\left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{1}{2}}{\sqrt{\sin^{2} \left ( \displaystyle \frac{3\pi }{4} \right )+\displaystyle \frac{\left ( \displaystyle \frac{3\pi }{4} \right )}{2}}}=\color{red}-\\ &\qquad \color{red}(\textrm{negatif})\\ &\color{black}\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\color{red}\displaystyle \displaystyle \frac{7\pi }{12}&&\color{red}\displaystyle \frac{11\pi }{12} \end{array} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 66.&\textrm{Titik stasioner fungsi}\: \: f(x)=\cos 3x\\ & \textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1),\left ( \displaystyle \frac{\pi }{4},1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \textrm{b}.&(0,1),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{6},-1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{6},1 \right ),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{\pi }{2},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-1 \right ) \\ \color{red}\textrm{e}.&(0,1),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{2\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\cos 3x\Rightarrow \color{red}f'(x)=-3\sin 3x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &-\sin 3x=0\Leftrightarrow \sin 3x=0\Leftrightarrow \sin 3x=\sin 0\\ &\Leftrightarrow 3x=0+k.2\pi \: \: \textrm{atau}\: \: 3x=\pi +k.2\pi \\ &\Leftrightarrow x=k.\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3} +k.\frac{2\pi}{3} \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \pi \\ &\color{red}\textrm{Sekarang kita tentukan nilai dan titiknya}\\ &x=0\Rightarrow f(0)=\cos 3(0)=1\rightarrow (0,1)\\ &x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )=\cos 3\left ( \displaystyle \frac{\pi }{3} \right )=\cos \pi \\ &\qquad=-1\rightarrow \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ &\color{red}\textrm{dan seterusnya} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 67.&\textrm{Titik stasioner fungsi}\: \: f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{6},-1 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{6},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{\pi }{3},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{6},-1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{2},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\Rightarrow \color{red}f'(x)=2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\Leftrightarrow \cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\\ &\color{black}\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\pm \displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \begin{cases} x & =\displaystyle \frac{\pi }{3}+k.\pi \\ x & =-\displaystyle \frac{\pi }{6}+k.\pi \end{cases}\\ &\Leftrightarrow k=0\Rightarrow \begin{cases} x & =\displaystyle \frac{\pi }{3} \\ x & =-\displaystyle \frac{\pi }{6}\: \: \color{red}(\textrm{tm}) \end{cases}\\ &\Leftrightarrow k=1\Rightarrow \begin{cases} x & =\displaystyle \frac{4\pi }{3}\: \: \color{red}\textrm{tm} \\ x & =\displaystyle \frac{5\pi }{6} \end{cases}\\ &\color{red}\textrm{Sekarang kita tentukan nilai dan titiknya}\\ &x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )=\sin \left (2.\displaystyle \frac{\pi }{3}- \displaystyle \frac{\pi }{6} \right )=\sin \frac{\pi}{2}=1 \\ &\qquad=1\rightarrow \left ( \displaystyle \frac{\pi }{3},1 \right )\\ &x=\displaystyle \frac{5\pi }{6}\Rightarrow f\left ( \displaystyle \frac{5\pi }{6} \right )=\sin \left (2.\displaystyle \frac{5\pi }{6}- \displaystyle \frac{\pi }{6} \right )\\ &\qquad=\sin \frac{3\pi}{2}=\color{red}-1\rightarrow \left ( \displaystyle \frac{5\pi }{6},-1 \right ) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 68.&\textrm{Nilai}\: \: x\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=x+\sin x\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&90^{\circ}\\ \textrm{b}.&135^{\circ}\\ \textrm{c}.&150^{\circ}\\ \color{red}\textrm{d}.&180^{\circ}\\ \textrm{e}.&360 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=x+\sin x\Rightarrow \color{red}f'(x)=1+\cos x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &1+\cos =0\Leftrightarrow \cos x=-1\\ &\Leftrightarrow \cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\pm 180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow k=0\Rightarrow x=\begin{cases} 180^{\circ} & \color{blue}\textrm{mungkin} \\ -180^{\circ} & \color{red}\textrm{tidak mungkin} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\begin{cases} 540^{\circ} & \color{red}\textrm{tidak mungkin} \\ 180^{\circ} & \color{blue}\textrm{mungkin} \end{cases} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 69.&\textrm{Nilai}\: \: y\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=4\cos x+\cos 2x\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\: \: \textrm{dan}\: \: 3\\ \textrm{b}.&-4\: \: \textrm{dan}\: \: 2\\ \color{red}\textrm{c}.&-3\: \: \textrm{dan}\: \: 5\\ \textrm{d}.&-2\: \: \textrm{dan}\: \: 4\\ \textrm{e}.&3\: \: \textrm{dan}\: \: 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ &\Rightarrow \color{red}f'(x)=-4\sin x-2\sin 2x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &-4\sin x-2\sin 2x=0\\ &\Leftrightarrow -4\sin x-4\sin x\cos x=0\\ &\Leftrightarrow -4\sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \color{black}\sin x=0\: \: \color{red}\textrm{atau}\: \: \color{black}1+\cos x=0\\ &\Leftrightarrow \color{black}\sin x=0\: \: \color{red}\textrm{atau}\: \: \color{black}\cos x=-1\\ &\Leftrightarrow \color{black}\sin x=\sin 0^{\circ}\: \: \color{red}\textrm{atau}\: \: \color{black}\cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\begin{cases} 0^{\circ} +k.360^{\circ} \\ 180^{\circ} +k.360^{\circ} \end{cases}\: \textrm{atau}\: \: x=\begin{cases} 180^{\circ} +k.360^{\circ} \\ -180^{\circ} +k.360^{\circ} \end{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \textrm{atau}\: 180^{\circ}\\ &\color{red}\textrm{Nilai}\: \: y-\textrm{nya}\\ &\color{black}x=0^{\circ}\Rightarrow f(0^{\circ})\\ &\qquad=4\cos 0^{\circ}+\cos 2(0^{\circ})=\color{red}4+1=5\\ &x=180^{\circ}\Rightarrow f(180^{\circ})\\ &\qquad=\color{black}4\cos 180^{\circ}+\cos 2(180^{\circ})=-4+1=\color{red}-3 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 70.&\textrm{Nilai stasioner fungsi}\\ &\quad\quad\quad f(x)=\displaystyle \frac{\sin x}{2-\cos x}\\ &\textrm{untuk}\: \: 0\leq x\leq 2\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},\frac{1}{2} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-\frac{1}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow \color{red}f'(x)=\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}=0\Leftrightarrow 2\cos x-1=0\\ &\Leftrightarrow \cos x=\displaystyle \frac{1}{2}\Leftrightarrow \cos x=\cos \displaystyle \frac{\pi }{3}\\ &\Leftrightarrow x=\pm \displaystyle \frac{\pi }{3}+k.2\pi \\ &\Leftrightarrow k=0\Rightarrow x=\pm \displaystyle \frac{\pi }{3}\Leftrightarrow x=\begin{cases} \displaystyle \frac{\pi }{3} & \color{black}\textrm{memenuhi} \\ -\displaystyle \frac{\pi }{3} & \color{red}\textrm{tidak memenuhi} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\pm \displaystyle \frac{\pi }{3}+2\pi \Leftrightarrow x=\begin{cases} \displaystyle \frac{7\pi }{3} & \color{red}\textrm{tidak memenuhi} \\ \displaystyle \frac{5\pi }{3} & \color{black}\textrm{memenuhi} \end{cases}\\ &\color{black}\textrm{Titiknya adalah}\\ &\color{black}x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )\\ &\qquad=\color{black}\displaystyle \frac{\sin \displaystyle \frac{\pi }{3}}{2-\cos \displaystyle \frac{\pi }{3}}=\color{red}\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \left ( \displaystyle \frac{\pi }{3},\displaystyle \frac{1}{3}\sqrt{3} \right )\\ &\color{black}x=\displaystyle \frac{5\pi }{3}\Rightarrow f\left ( \displaystyle \frac{5\pi }{3} \right )\\ &\qquad=\color{black}\displaystyle \frac{\sin \displaystyle \frac{5\pi }{3}}{2-\cos \displaystyle \frac{5\pi }{3}}=\color{red}\displaystyle \frac{-\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=-\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \color{red}\left ( \displaystyle \frac{5\pi }{3},-\displaystyle \frac{1}{3}\sqrt{3} \right ) \end{aligned} \end{array}$.
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