$\begin{array}{ll}\\ 31.&\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}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin x-\cos x\\ &\color{black}\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 =\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\: \: \color{red}(\textrm{positif}) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 32.&\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}\\ &\color{blue}\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 )=-1\: \: \color{red}(\textrm{negatif}) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 33.&\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}\\ &\color{blue}\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}\\ 34.&(\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}\\ &\color{blue}\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}\\ 35.&\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}\\ &\color{blue}\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}$
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