Contoh Soal 11 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 51.& \text{Diketahui}f(x)={\cos }^{2}2x.\text{Jika}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx,\text{nilai untuk}\\ & {\displaystyle \frac{a-b}{c-d}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{3}}\\ \text{b}.& {\displaystyle \frac{2}{3}}\\ \text{c}.& -{\displaystyle \frac{3}{5}}\\ \text{d}.& -{\displaystyle \frac{6}{5}}\\ \text{e}.& -{\displaystyle \frac{9}{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)\\ & =-4\sin 2x\cos 2x\\ & f^{″}(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ & =8{\sin }^{2}2x-8{\cos }^{2}2x\\ & \text{Bandingkan dengan}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx\\ & \text{maka},a=8,b=2,c=-8,d=2\\ & \text{Jadi},{\displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 52.& \text{Diketahui}f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}.\text{Jika}}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2},\text{nilai dari}}\\ & m.n=....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 8\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}}\\ & f^{\prime }(x)={\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{-{\sin }^{2}x-{\cos }^{2}x+0}{{\sin }^2+2\sin x\cos x+{\cos }^{2}x}}\\ & ={\displaystyle \frac{-1}{1+\sin 2x}}\\ & f^{″}(x)={\displaystyle \frac{0-((-1).2\cos 2x)}{{(\sin 2x+1)}^2}=\frac{2\cos 2x}{{(\sin 2x+1)}^2}}\\ & \text{Bandingkan dengan yang diketahui}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2}}\\ & \{\begin{array}{ll}m& =2\\ n& =2\end{array}\\ & \text{Jadi},m.n=2.1=2\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\text{dengan}0\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},0}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{2},0}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{2},1}\right)\\ \text{e}.& (\pi ,1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\Rightarrow f^{″}(x)=-4\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=0+k.2\pi \text{atau}2x=\pi +k.2\pi \\ & \iff x=0+k.\pi \text{atau}x=\frac{\pi }{2}+k.\pi \\ & \iff x=0,x={\displaystyle \frac{\pi }{2},x=\pi ,x={\displaystyle \frac{3\pi }{2}\text{atau}x=2\pi }}\\ & \bullet f\left({\displaystyle \frac{\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{\pi }{2},0}\right)\\ & \bullet f\left({\displaystyle \pi }\right)=\sin 2\left({\displaystyle \pi }\right)=0\Rightarrow \left({\displaystyle \pi ,0}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{3\pi }{2},0}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 54.& \text{Diketahui fungsi}f(x)=-3\cos 2x+1\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},2}\right)\\ \text{b}.& \left({\displaystyle \frac{2\pi }{3},\frac{5}{2}}\right)\\ \text{c}.& \left({\displaystyle \frac{3\pi }{2},4}\right)\\ \text{d}.& \left({\displaystyle \frac{5\pi }{4},1}\right)\\ \text{e}.& \left({\displaystyle \frac{5\pi }{3},\frac{5}{2}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=-3\cos 2x+1\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=6\sin 2x\Rightarrow f^{″}(x)=12\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 12\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{3\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{3\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{5\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{5\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{7\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{7\pi }{4},1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Diketahui fungsi}f(x)={\sin }^{2}x+2\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{11}{4}}\right)\\ \text{c}.& \left({\displaystyle \pi ,2}\right)\\ \text{d}.& \left({\displaystyle \frac{4\pi }{3},\frac{11}{4}}\right)\\ \text{e}.& \left({\displaystyle \frac{11\pi }{6},\frac{9}{4}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x+2\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=2\sin x\cos x\Rightarrow f^{\prime }(x)=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{3\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{5\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{5\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{7\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{7\pi }{4},\frac{5}{2}}\right)\end{array}\end{array}$