Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 61.& \text{Fungsi}f(x)=\sin x-\cos x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<{\displaystyle \frac{7\pi }{4}}}\\ \text{d}.& 0<x<{\displaystyle \frac{3\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{e}.& 0<x<{\displaystyle \frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sin x-\cos x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=\cos x+\cos x=0\\ & \sin x=-\cos x\iff {\displaystyle \frac{\sin x}{\cos x}=-1}\\ & \iff \tan x=-1\\ & \iff \tan x=\tan {\displaystyle \frac{3\pi }{4}}\\ & \iff x={\displaystyle \frac{3\pi }{4}\pm k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}}\\ & \iff k=2\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm 2\pi =\text{tm}}\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & --& & & ++& \\ 0& & {\displaystyle \frac{3\pi }{4}}& & & {\displaystyle \frac{7\pi }{4}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{2}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{1}{2}\pi +\sin {\displaystyle \frac{1}{2}\pi =0+1=1(\text{positif})}}\\ & \text{untuk}x={\displaystyle \frac{3}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{3}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{3}{2}\pi +\sin {\displaystyle \frac{3}{2}\pi =0-1=-1(\text{negatif})}}\\ & \text{untuk}x={\displaystyle \frac{11}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{11}{6}\pi }\right)}\\ & =\cos {\displaystyle \frac{11}{6}\pi +\sin {\displaystyle \frac{11}{6}\pi ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}}}\end{array}\end{array}$

$\begin{array}{l}\\ 62.& \text{Fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}<x<\pi \text{atau}{\displaystyle \frac{3\pi }{2}<x<2\pi }}\\ \text{b}.& {\displaystyle \frac{2\pi }{3}<x<\pi }\\ \text{c}.& 0<x<{\displaystyle \frac{\pi }{2}\text{atau}\pi <x<{\displaystyle \frac{3\pi }{2}}}\\ \text{d}.& {\displaystyle \frac{4\pi }{3}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{\pi }{3}<x<\pi \text{atau}{\displaystyle \frac{4\pi }{3}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)={\sin }^{2}x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x=0\\ & \iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=\pm k.2\pi \text{atau}2x=\pi \pm k.2\pi \\ & \iff x=\pm k.\pi \text{atau}x=\frac{\pi }{2}\pm k.\pi \\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{2}}\\ & \iff k=1\Rightarrow x=\pi \text{atau}x={\displaystyle \frac{\pi }{2}+\pi =\frac{3\pi }{2}}\\ & \iff k=2\Rightarrow x=2\pi \text{atau}x={\displaystyle \frac{\pi }{2}+2\pi ={\displaystyle \frac{5}{2}\pi (\text{tm})}}\\ & \begin{array}{c}\\ & & & & & & & & \\ & ++& & --& & ++& & --& \\ 0& & {\displaystyle \frac{\pi }{2}}& & \pi & & {\displaystyle \frac{3\pi }{2}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{6}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{1}{6}\pi }\right)=\sin {\displaystyle \frac{1}{3}\pi =\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{3}{4}\pi \Rightarrow f\left({\displaystyle \frac{3}{4}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{3}{4}\pi }\right)=-1(\text{negatif})\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Fungsi}f(x)={\cos }^{2}2x\text{untuk}\\ & 0^{\circ }<x<{360}^{\circ }\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {45}^{\circ }<x<{90}^{\circ }\\ \text{b}.& {135}^{\circ }<x<{180}^{\circ }\\ \text{c}.& {225}^{\circ }<x<{270}^{\circ }\\ \text{d}.& {270}^{\circ }<x<{300}^{\circ }\\ \text{e}.& {315}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ & \text{Selanjutnya}\\ & \iff -2\sin 4x=0\iff \sin 4x=0\iff \sin 4x=\sin 0^{\circ }\\ & \iff \{\begin{array}{ll}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{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}x={45}^{\circ }\\ & \iff k=1\Rightarrow x={90}^{\circ }\text{atau}x={135}^{\circ }\\ & \iff k=2\Rightarrow x={180}^{\circ }\text{atau}x={225}^{\circ }\\ & \iff k=3\Rightarrow x={270}^{\circ }\text{atau}x={315}^{\circ }\\ & \iff k=4\Rightarrow x={360}^{\circ }\text{atau}x={405}^{\circ }(\text{tm})\\ & \text{Gunakan titik uji pada}x={30}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({30}^{\circ })=-2\sin 4({30}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={60}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({60}^{\circ })=-2\sin 4({60}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{Gunakan titik uji pada}x={120}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({120}^{\circ })=-2\sin 4({120}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={150}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({150}^{\circ })=-2\sin 4({150}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{dan seterusnya}...\\ & \begin{array}{c}\\ & & & & & & & & \\ & --& & ++& & --& & ++\\ 0& & {\displaystyle {45}^{\circ }}& & {90}^{\circ }& & {135}^{\circ }& & {180}^{\circ }\\ & --& & ++& & --& & ++\\ {180}^{\circ }& & {\displaystyle {225}^{\circ }}& & {270}^{\circ }& & {315}^{\circ }& & {360}^{\circ }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 64.& (\mathbf{\text{SBMPTN 2015}})\\ & \text{Fungsi}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{pada}0<x<\pi \text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x<\frac{11\pi }{12}}\\ \text{b}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{c}.& {\displaystyle \frac{2\pi }{3}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{3\pi }{4}<x<\pi }\\ \text{e}.& {\displaystyle \frac{3\pi }{4}<x<\frac{3\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(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\iff \sin 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{4\pi }{3}}\\ & \iff 2x={\displaystyle \frac{4\pi }{3}+k.2\pi \text{atau}2x=\pi -\frac{4\pi }{3}+k.2\pi }\\ & \iff x={\displaystyle \frac{2\pi }{3}+k.\pi \text{atau}x=-\frac{\pi }{6}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x=-{\displaystyle \frac{\pi }{6}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{5\pi }{3}\text{atau}x={\displaystyle \frac{5\pi }{6}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\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}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\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}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{2\pi }{3}}}& & {\displaystyle \frac{5\pi }{6}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 65.& \text{Fungsi}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{dengan}x>0\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x\le \frac{13\pi }{12}}\\ \text{b}.& {\displaystyle \frac{7\pi }{12}<x<\frac{11\pi }{12}}\\ \text{c}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{d}.& {\displaystyle \frac{7\pi }{6}<x\le {\displaystyle \frac{13\pi }{6}}}\\ \text{e}.& {\displaystyle \frac{7\pi }{6}<x\le \frac{11\pi }{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(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\iff \sin 2x=-{\displaystyle \frac{1}{2}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{7\pi }{6}}\\ & \iff 2x={\displaystyle \frac{7\pi }{6}+k.2\pi \text{atau}2x=\pi -\frac{7\pi }{6}+k.2\pi }\\ & \iff x={\displaystyle \frac{7\pi }{12}+k.\pi \text{atau}x=-\frac{\pi }{12}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{7\pi }{12}\text{atau}x=-{\displaystyle \frac{\pi }{12}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{19\pi }{12}\text{atau}x={\displaystyle \frac{11\pi }{12}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\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}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\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}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{7\pi }{12}}}& & {\displaystyle \frac{11\pi }{12}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 66.& \text{Titik stasioner fungsi}f(x)=\cos 3x\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1),\left({\displaystyle \frac{\pi }{4},1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{b}.& (0,1),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{6},-1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{6},1}\right),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{\pi }{2},1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},-1}\right)\\ \text{e}.& (0,1),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{2\pi }{3},1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\cos 3x\Rightarrow f^{\prime }(x)=-3\sin 3x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -\sin 3x=0\iff \sin 3x=0\iff \sin 3x=\sin 0\\ & \iff 3x=0+k.2\pi \text{atau}3x=\pi +k.2\pi \\ & \iff x=k.{\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \frac{\pi }{3}+k.\frac{2\pi }{3}}}\\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{3}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \pi }}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x=0\Rightarrow f(0)=\cos 3(0)=1\to (0,1)\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\cos 3\left({\displaystyle \frac{\pi }{3}}\right)=\cos \pi }\\ & =-1\to \left({\displaystyle \frac{\pi }{3},-1}\right)\\ & \text{dan seterusnya}\end{array}\end{array}$

$\begin{array}{l}\\ 67.& \text{Titik stasioner fungsi}f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1)\text{dan}\left({\displaystyle \frac{\pi }{6},-1}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{6},1}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},-1}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},1}\right)\text{dan}\left({\displaystyle \frac{5\pi }{6},-1}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{2},-1}\right)\text{dan}\left({\displaystyle \pi ,1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\Rightarrow f^{\prime }(x)=2\cos (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 2\cos (2x-{\displaystyle \frac{\pi }{6}})=0\iff \cos (2x-{\displaystyle \frac{\pi }{6}})=0\\ & \cos (2x-{\displaystyle \frac{\pi }{6}})=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff (2x-{\displaystyle \frac{\pi }{6}})=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & \iff x={\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}+k.\pi }\\ x& =-{\displaystyle \frac{\pi }{6}+k.\pi }\end{array}}\\ & \iff k=0\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}}\\ x& =-{\displaystyle \frac{\pi }{6}(\text{tm})}\end{array}\\ & \iff k=1\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{4\pi }{3}\text{tm}}\\ x& ={\displaystyle \frac{5\pi }{6}}\end{array}\\ & \text{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}\\ & =1\to \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)}\\ & =\sin \frac{3\pi }{2}=-1\to \left({\displaystyle \frac{5\pi }{6},-1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 68.& \text{Nilai}x\text{pada titik stasioner}\\ & \text{fungsi}f(x)=x+\sin x\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^{\circ }\\ \text{b}.& {135}^{\circ }\\ \text{c}.& {150}^{\circ }\\ \text{d}.& {180}^{\circ }\\ \text{e}.& 360\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=x+\sin x\Rightarrow f^{\prime }(x)=1+\cos x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 1+\cos =0\iff \cos x=-1\\ & \iff \cos x=\cos {180}^{\circ }\\ & \iff x=\pm {180}^{\circ }+k{.360}^{\circ }\\ & \iff k=0\Rightarrow x=\{\begin{array}{ll}{180}^{\circ }& \text{mungkin}\\ -{180}^{\circ }& \text{tidak mungkin}\end{array}\\ & \iff k=1\Rightarrow x=\{\begin{array}{ll}{540}^{\circ }& \text{tidak mungkin}\\ {180}^{\circ }& \text{mungkin}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 69.& \text{Nilai}y\text{pada titik stasioner}\\ & \text{fungsi}f(x)=4\cos x+\cos 2x\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5\text{dan}3\\ \text{b}.& -4\text{dan}2\\ \text{c}.& -3\text{dan}5\\ \text{d}.& -2\text{dan}4\\ \text{e}.& 3\text{dan}5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ & \Rightarrow f^{\prime }(x)=-4\sin x-2\sin 2x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -4\sin x-2\sin 2x=0\\ & \iff -4\sin x-4\sin x\cos x=0\\ & \iff -4\sin x(1+\cos x)=0\\ & \iff \sin x(1+\cos x)=0\\ & \iff \sin x=0\text{atau}1+\cos x=0\\ & \iff \sin x=0\text{atau}\cos x=-1\\ & \iff \sin x=\sin 0^{\circ }\text{atau}\cos x=\cos {180}^{\circ }\\ & \iff x=\{\begin{array}{l}{0}^{\circ }+k{.360}^{\circ }\\ {180}^{\circ }+k{.360}^{\circ }\end{array}\text{atau}x=\{\begin{array}{l}{180}^{\circ }+k{.360}^{\circ }\\ -{180}^{\circ }+k{.360}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}{180}^{\circ }\\ & \text{Nilai}y-\text{nya}\\ & x=0^{\circ }\Rightarrow f(0^{\circ })\\ & =4\cos 0^{\circ }+\cos 2(0^{\circ })=4+1=5\\ & x={180}^{\circ }\Rightarrow f({180}^{\circ })\\ & =4\cos {180}^{\circ }+\cos 2({180}^{\circ })=-4+1=-3\end{array}\end{array}$

$\begin{array}{l}\\ 70.& \text{Nilai stasioner fungsi}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}}\\ & \text{untuk}0\le x\le 2\pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},\frac{1}{2}}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-\frac{1}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3}}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow f^{\prime }(x)={\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}}}\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & {\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}=0\iff 2\cos x-1=0}\\ & \iff \cos x={\displaystyle \frac{1}{2}\iff \cos x=\cos {\displaystyle \frac{\pi }{3}}}\\ & \iff x=\pm {\displaystyle \frac{\pi }{3}+k.2\pi }\\ & \iff k=0\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}\iff x=\{\begin{array}{ll}{\displaystyle \frac{\pi }{3}}& \text{memenuhi}\\ -{\displaystyle \frac{\pi }{3}}& \text{tidak memenuhi}\end{array}}\\ & \iff k=1\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}+2\pi \iff x=\{\begin{array}{ll}{\displaystyle \frac{7\pi }{3}}& \text{tidak memenuhi}\\ {\displaystyle \frac{5\pi }{3}}& \text{memenuhi}\end{array}}\\ & \text{Titiknya adalah}\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{3}}}{2-\cos {\displaystyle \frac{\pi }{3}}}={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{\pi }{3},{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\\ & x={\displaystyle \frac{5\pi }{3}\Rightarrow f\left({\displaystyle \frac{5\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{5\pi }{3}}}{2-\cos {\displaystyle \frac{5\pi }{3}}}={\displaystyle \frac{-{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{5\pi }{3},-{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\end{array}\end{array}$.