$\begin{array}{ll}\\ 51.&\textrm{Diketahui}\: \: f(x)=\cos ^{2}2x\: .\: \textrm{Jika}\\ &f''(x)=a\sin ^{2}bx+c\cos ^{2}dx,\: \textrm{nilai untuk}\\ &\displaystyle \frac{a-b}{c-d}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5}{3}\\ \textrm{b}.&\displaystyle \frac{2}{3}\\ \color{red}\textrm{c}.&-\displaystyle \frac{3}{5}\\ \textrm{d}.&-\displaystyle \frac{6}{5}\\ \textrm{e}.&-\displaystyle \frac{9}{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{purple}\begin{aligned}&f(x)=\cos ^{2}2x\\ &f'(x)=2\cos 2x(-\sin 2x)(2)\\ &\: \qquad =-4\sin 2x\cos 2x\\ &\color{blue}f''(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ &\: \: \quad\quad=8\sin ^{2}2x-8\cos ^{2}2x\\ &\textrm{Bandingkan dengan}\\ &\color{red}f''(x)=a\sin ^{2}bx+c\cos ^{2}dx\\ &\textrm{maka},\: \: a=8,\: b=2,\: c=-8,\: d=2\\ &\textrm{Jadi},\: \displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 52.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\: .\: \textrm{Jika}\\ &f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\: ,\: \textrm{nilai dari}\\ &m.n=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\\ &f'(x)=\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &\, \qquad =\displaystyle \frac{-\sin ^{2}x-\cos ^{2}x+0}{\sin ^{2}+2\sin x\cos x+\cos ^{2}x}\\ &\, \qquad=\displaystyle \frac{-1}{1+\sin 2x}\\ &f''(x)=\displaystyle \frac{0-((-1).2\cos 2x)}{\left ( \sin 2x+1 \right )^{2}}=\frac{2\cos 2x}{\left ( \sin 2x+1 \right )^{2}}\\ &\color{red}\textrm{Bandingkan dengan yang diketahui}\\ &\color{black}f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\\ &\begin{cases} m &=2 \\ n &=2 \end{cases}\\ &\textrm{Jadi},\: \: m.n=2.1=2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 53.&\textrm{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\: \: \textrm{dengan}\: \: 0\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{4},0 \right )\\ \color{red}\textrm{b}.&\left ( \displaystyle \frac{\pi }{2},0 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{2},1 \right )\\ \textrm{e}.&\left ( \pi ,1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin 2x\\ &f'(x)=2\cos 2x\Rightarrow f''(x)=-4\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=0+k.2\pi \: \: \textrm{atau}\: \: 2x=\pi +k.2\pi \\ &\Leftrightarrow x=0+k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} +k.\pi \\ &\Leftrightarrow \color{black}x=0,\: \color{red}x=\displaystyle \frac{\pi }{2},\: x=\pi \: ,\: x=\displaystyle \frac{3\pi }{2}\: \: \textrm{atau}\: \: \color{black}x=2\pi\\ &\bullet f\left ( \displaystyle \frac{\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{2},0 \right )\\ &\bullet f\left ( \displaystyle \pi \right )=\sin 2\left ( \displaystyle \pi \right )=0\Rightarrow \color{red}\left ( \displaystyle \pi ,0 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{3\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{2},0 \right ) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 54.&\textrm{Diketahui fungsi}\: \: f(x)=-3\cos 2x+1\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},2 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{2\pi }{3},\frac{5}{2} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{3\pi}{2} ,4 \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{5\pi }{3},\frac{5}{2} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=-3\cos 2x+1\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=6\sin 2x\Rightarrow f''(x)=12\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &12\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{atau}\: \: x=\frac{7\pi }{4}\\ &\bullet f\left ( \displaystyle \frac{\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{3\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{5\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{5\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{7\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{7\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{7\pi }{4},1 \right ) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 55.&\textrm{Diketahui fungsi}\: \: f(x)=\sin^{2} x+2\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left ( \displaystyle \frac{\pi }{4},\frac{5}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{11}{4} \right )\\ \textrm{c}.&\left ( \displaystyle \pi ,2 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{4\pi }{3},\frac{11}{4} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{11\pi }{6},\frac{9}{4} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\sin^{2} x+2\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=2\sin x\cos x\Rightarrow f'(x)=\sin 2x\\ &f''(x)=2\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{atau}\: \: x=\frac{7\pi }{4}\\ &\bullet f\left ( \displaystyle \frac{\pi }{4} \right )=\sin^{2} \left ( \displaystyle \frac{\pi }{4} \right )+2=\frac{5}{2}\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{4},\frac{5}{2} \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{4} \right )=\sin^{2} \left ( \displaystyle \frac{3\pi }{4} \right )+2=\frac{5}{2}\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{4},\frac{5}{2} \right )\\ &\bullet f\left ( \displaystyle \frac{5\pi }{4} \right )=\sin^{2} \left ( \displaystyle \frac{5\pi }{4} \right )+2=\frac{5}{2}\Rightarrow \color{red}\left ( \displaystyle \frac{5\pi }{4},\frac{5}{2} \right )\\ &\bullet f\left ( \displaystyle \frac{7\pi }{4} \right )=\sin^{2} \left ( \displaystyle \frac{7\pi }{4} \right )+2=\frac{5}{2}\Rightarrow \color{red}\left ( \displaystyle \frac{7\pi }{4},\frac{5}{2} \right ) \end{aligned} \end{array}$
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