Contoh Soal 9 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 41.& \text{Sebuah mesin diprogram untuk dapat}\\ & \text{begerak tiap waktu mengikuti posisi}\\ & x=2\cos 3t\text{dan}y=2\cos 2t\text{di mana}\\ & x,y\text{dalam}cm,\text{dan}t\text{dalam detik}\\ & \text{Jika kecepatakan dirumuskan dengan}\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2},\text{maka nilai}v\\ & \text{saat}t=30detik\text{adalah}...cm/detik\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{3}\\ \text{b}.& 2\sqrt{11}\\ \text{c}.& 2\sqrt{10}\\ \text{d}.& 6\\ \text{e}.& 4\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui Kecepatan gerak mesin}\\ & \{\begin{array}{ll}x=2\cos 3x& \Rightarrow {\displaystyle \frac{dx}{dt}=-6\sin 3t}\\ y=2\cos 2x& \Rightarrow {\displaystyle \frac{dy}{dt}=-4\sin 2t}\end{array}\\ & \text{Maka kecepatan mesin saat}t=30\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}\\ & v=\sqrt{{(-6\sin 3t)}^2+{(-4\sin 2t)}^2}\\ & =\sqrt{{(-6\sin 3(30))}^2+{(-4\sin 2(30))}^2}\\ & =\sqrt{{(-6(1))}^2+{(-4\left({\displaystyle \frac{1}{2}\sqrt{3}}\right))}^2}\\ & =\sqrt{36+12}=\sqrt{48}=\sqrt{16.3}\\ & =4\sqrt{3}\end{array}\end{array}$

$\begin{array}{l}\\ 42.& \text{Sebuah benda duhubungkan dengan}\\ & \text{pegas dan bergerak sepanjang sumbu}\\ & \text{X dengan formula persamaan}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dari titik}O\text{yang dapat}\\ & \text{dicapai oleh benda tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui gerak benda yang bergerak}\\ & \text{mengikuti formula}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dicapai saat}{x}^{\prime }={\displaystyle \frac{dx}{dt}=0}\\ & x^{\prime }=2\cos 2t-2\sqrt{3}\sin 2t=0\\ & \iff 2\cos 2t=2\sqrt{3}\sin 2t\\ & \iff {\displaystyle \frac{\sin 2t}{\cos 2t}={\displaystyle \frac{1}{3}\sqrt{3}}}\\ & \iff \tan 2t=\tan {30}^{\circ }\\ & \iff 2t={30}^{\circ }+k{.180}^{\circ }\\ & \iff t={15}^{\circ }+k{.90}^{\circ }\{\begin{array}{ll}k=0,& t={15}^{\circ }\\ k=1,& t={105}^{\circ }\\ k=2,& t={195}^{\circ }\\ k=3,& t={285}^{\circ }\\ k=4,& t={375}^{\circ }\\ & \text{dst}\end{array}\\ & \text{Ambil}t={15}^{\circ },\text{maka nilai}\\ & x-\text{nya adalah}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \iff x=\sin 2({15}^{\circ })+\sqrt{3}\cos 2({15}^{\circ })\\ & \iff x={\displaystyle \frac{1}{2}+\sqrt{3}\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ & \iff x={\displaystyle \frac{1}{2}+\frac{3}{2}=2}\end{array}\end{array}$

$\begin{array}{l}\\ 43.& \text{Pada kurva}y=\sin x\text{dibuat}\\ & \text{garis singgung melalui titik}\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{garis singgung tersebut memotong}\\ & \text{sumbu-X di A dan sumbu-Y di B}.\\ & \text{Luas}\mathrm{△}AOB\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{36}}\\ \text{b}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{36}}\\ \text{c}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{16}}\\ \text{d}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{18}}\\ \text{e}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{18}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikan ilustrasi berikut}\end{array}$

$.\begin{array}{rl}& \text{Misalkan koordinat titik}P\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{maka},\\ & x_p={\displaystyle \frac{2\pi }{3},y_p=k=\sin {\displaystyle \frac{2\pi }{3}={\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & \text{Persamaan garis singgung di titik P}:\\ & y=m_{x_p}(x-x_p)+y_p\\ & \{\begin{array}{ll}\left({\displaystyle \frac{2\pi }{3},k}\right)& =\left({\displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3}}\right)\\ m_{x_p}={\displaystyle \frac{dy}{dx}}& =y^{\prime }=\cos x\\ m_{x_p}& =\cos \left({\displaystyle \frac{2\pi }{3}}\right)=-\frac{1}{2}\end{array}\\ & \text{Sehingga persamaan garis singgungnya}\\ & y=(-{\displaystyle \frac{1}{2}})(x-{\displaystyle \frac{2\pi }{3}})+{\displaystyle \frac{1}{2}\sqrt{3}}\\ & \iff 2y=-x+{\displaystyle \frac{2\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{memotong sumbu-X, maka}{y}_A=0\\ & 2y_A=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 0=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & x_A={\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & \bullet \text{memotong sumbu-Y, maka}{x}_B=0\\ & 2y_B=-x_B+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 2y_B=0+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & y_B={\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{Luas}\mathrm{△}AOB=\left[AOB\right]={\displaystyle \frac{x_A.y_B}{2}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\pi }{3}+\sqrt{3}}\right).\left({\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\right)}{2}}\\ & ={\displaystyle \frac{1}{6}(2\pi +3\sqrt{3}).{\displaystyle \frac{1}{6}(2\pi +3\sqrt{3})}}\\ & ={\displaystyle \frac{1}{36}{(2\pi +3\sqrt{3})}^2}\end{array}$

$\begin{array}{l}\\ 44.& \text{Sebuah wadah penampung air hujan}\\ & \text{memiliki ukuran sisi samping 3 m dan}\\ & \text{sisi horisontal juga 3 m. Sisi samping}\\ & \text{membentuk sudut}\theta (0\le \theta \le {\displaystyle \frac{\pi }{2}})\\ & \text{dengan garis vertikal (lihat gambar)}\\ & \text{Nilai}\theta \text{supaya wadah dapat menampung}\\ & \text{air hujan maksimum adalah}....\end{array}$

$\begin{array}{ll}.& \\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{\pi }{3}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}}\\ \text{c}.& {\displaystyle \frac{\pi }{5}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Supaya memuat dapat maksimum}\\ & \text{maka luas penampang haruslah}\\ & \text{MAKSIMUM, yaitu}\end{array}\end{array}$

gambar 1

gambar 2

$.\begin{array}{rl}& \text{Luas penampang}=\text{Luas Trapesium}\\ & \text{dengan}\{\begin{array}{ll}t& =3\sin \theta \\ n& =3\cos \theta \end{array}\\ & \text{Luas Penampang}={\displaystyle \frac{1}{2}(\sum \text{sisi sejajar})\times t}\\ & \iff L={\displaystyle \frac{1}{2}(6+2n)\times t}\\ & \iff L=(3+n)\times t\\ & \iff L=(3+3\cos \theta )\times 3\sin \theta \\ & \iff L=9\sin \theta +9\sin \theta \cos \theta \\ & \iff L=9\sin \theta +{\displaystyle \frac{9}{2}\sin 2\theta }\\ & \text{Suapa luas penampang}\text{MAKSIMUM}\\ & \text{maka}{L}^{\prime }={\displaystyle \frac{dL}{d\theta }=0}\\ & \iff L^{\prime }=9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9(2{\cos }^2\theta -1)=0\\ & \iff 2{\cos }^2\theta +\cos \theta -1=0\\ & \iff (\cos \theta +1)(2\cos \theta -1)=0\\ & \iff \cos \theta =-1\text{atau}2\cos \theta =1\\ & \iff \cos \theta =-1\text{atau}\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \theta =\cos \pi \text{atau}\cos \theta =\cos {\displaystyle \frac{\pi }{3}}\\ & \iff \theta =\pi \text{atau}\theta ={\displaystyle \frac{\pi }{3}}\end{array}$

$\begin{array}{l}\\ 45.& \text{Seseorang melempar bola dari atap}\\ & \text{sebuah rumah. Ketinggian bola saat}\\ & t(detik)\text{dinyatakan dengan persamaan}\\ & h(t)=5+{\cos }^2\pi t.\text{Kecepatan bola}\\ & \text{ditentukan dengan formula}v={\displaystyle \frac{dh}{dt}}\\ & \text{Besar kecepatan bola saat}t=0,25\\ & \text{detik adalah}....\\ & \begin{array}{ll}\text{a}.& 0\\ \text{b}.& \pi \\ \text{c}.& 2\pi \\ \text{d}.& 3\pi \\ \text{e}.& 4\pi \end{array}\\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}h(t)=5+{\cos }^2\pi t.\text{maka}\\ & v={\displaystyle \frac{dh}{dt}=2\cos \pi t(-\sin \pi t).(\pi )}\\ & \iff v=-\pi \sin 2\pi t\\ & \text{Saat}t=0,25={\displaystyle \frac{1}{4},\text{maka}}\\ & \text{besar kecepatannya adalah}:\\ & \iff v=-\pi \sin 2\pi \left({\displaystyle \frac{1}{4}}\right)\\ & \iff =-\pi \sin {\displaystyle \frac{\pi }{2}}\\ & \iff =-\pi \\ & \text{Tanda negatif menunjukkan}\\ & \text{arah kecepatan ke bawah}\\ & \text{Karena kecepatan merupakan salah}\\ & \text{satu besaran}\text{VEKTOR}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., & Akhmad G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
3. Sembiring, S., Zulkifli, M., Marsito, & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU

Contoh Soal 8 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 36.& \text{Titik stasioner fungsi}f(x)=\cos 3x\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1),\left({\displaystyle \frac{\pi }{4},1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{b}.& (0,1),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{6},-1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{6},1}\right),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{\pi }{2},1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},-1}\right)\\ \text{e}.& (0,1),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{2\pi }{3},1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\cos 3x\Rightarrow f^{\prime }(x)=-3\sin 3x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -\sin 3x=0\iff \sin 3x=0\iff \sin 3x=\sin 0\\ & \iff 3x=0+k.2\pi \text{atau}3x=\pi +k.2\pi \\ & \iff x=k.{\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \frac{\pi }{3}+k.\frac{2\pi }{3}}}\\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{3}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \pi }}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x=0\Rightarrow f(0)=\cos 3(0)=1\to (0,1)\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\cos 3\left({\displaystyle \frac{\pi }{3}}\right)=\cos \pi }\\ & =-1\to \left({\displaystyle \frac{\pi }{3},-1}\right)\\ & \text{dan seterusnya}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \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}$

$\begin{array}{l}\\ 38.& \text{Nilai}x\text{pada titik stasioner}\\ & \text{fungsi}f(x)=x+\sin x\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^{\circ }\\ \text{b}.& {135}^{\circ }\\ \text{c}.& {150}^{\circ }\\ \text{d}.& {180}^{\circ }\\ \text{e}.& 360\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=x+\sin x\Rightarrow f^{\prime }(x)=1+\cos x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 1+\cos =0\iff \cos x=-1\\ & \iff \cos x=\cos {180}^{\circ }\\ & \iff x=\pm {180}^{\circ }+k{.360}^{\circ }\\ & \iff k=0\Rightarrow x=\{\begin{array}{ll}{180}^{\circ }& \text{mungkin}\\ -{180}^{\circ }& \text{tidak mungkin}\end{array}\\ & \iff k=1\Rightarrow x=\{\begin{array}{ll}{540}^{\circ }& \text{tidak mungkin}\\ {180}^{\circ }& \text{mungkin}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Nilai}y\text{pada titik stasioner}\\ & \text{fungsi}f(x)=4\cos x+\cos 2x\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5\text{dan}3\\ \text{b}.& -4\text{dan}2\\ \text{c}.& -3\text{dan}5\\ \text{d}.& -2\text{dan}4\\ \text{e}.& 3\text{dan}5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ & \Rightarrow f^{\prime }(x)=-4\sin x-2\sin 2x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -4\sin x-2\sin 2x=0\\ & \iff -4\sin x-4\sin x\cos x=0\\ & \iff -4\sin x(1+\cos x)=0\\ & \iff \sin x(1+\cos x)=0\\ & \iff \sin x=0\text{atau}1+\cos x=0\\ & \iff \sin x=0\text{atau}\cos x=-1\\ & \iff \sin x=\sin 0^{\circ }\text{atau}\cos x=\cos {180}^{\circ }\\ & \iff x=\{\begin{array}{l}{0}^{\circ }+k{.360}^{\circ }\\ {180}^{\circ }+k{.360}^{\circ }\end{array}\text{atau}x=\{\begin{array}{l}{180}^{\circ }+k{.360}^{\circ }\\ -{180}^{\circ }+k{.360}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}{180}^{\circ }\\ & \text{Nilai}y-\text{nya}\\ & x=0^{\circ }\Rightarrow f(0^{\circ })\\ & =4\cos 0^{\circ }+\cos 2(0^{\circ })=4+1=5\\ & x={180}^{\circ }\Rightarrow f({180}^{\circ })\\ & =4\cos {180}^{\circ }+\cos 2({180}^{\circ })=-4+1=-3\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \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}$

Contoh Soal 7 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 31.& \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}$

$\begin{array}{l}\\ 32.& \text{Fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}<x<\pi \text{atau}{\displaystyle \frac{3\pi }{2}<x<2\pi }}\\ \text{b}.& {\displaystyle \frac{2\pi }{3}<x<\pi }\\ \text{c}.& 0<x<{\displaystyle \frac{\pi }{2}\text{atau}\pi <x<{\displaystyle \frac{3\pi }{2}}}\\ \text{d}.& {\displaystyle \frac{4\pi }{3}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{\pi }{3}<x<\pi \text{atau}{\displaystyle \frac{4\pi }{3}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)={\sin }^{2}x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x=0\\ & \iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=\pm k.2\pi \text{atau}2x=\pi \pm k.2\pi \\ & \iff x=\pm k.\pi \text{atau}x=\frac{\pi }{2}\pm k.\pi \\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{2}}\\ & \iff k=1\Rightarrow x=\pi \text{atau}x={\displaystyle \frac{\pi }{2}+\pi =\frac{3\pi }{2}}\\ & \iff k=2\Rightarrow x=2\pi \text{atau}x={\displaystyle \frac{\pi }{2}+2\pi ={\displaystyle \frac{5}{2}\pi (\text{tm})}}\\ & \begin{array}{c}\\ & & & & & & & & \\ & ++& & --& & ++& & --& \\ 0& & {\displaystyle \frac{\pi }{2}}& & \pi & & {\displaystyle \frac{3\pi }{2}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{6}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{1}{6}\pi }\right)=\sin {\displaystyle \frac{1}{3}\pi =\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{3}{4}\pi \Rightarrow f\left({\displaystyle \frac{3}{4}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{3}{4}\pi }\right)=-1(\text{negatif})\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \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}\\ 34.& (\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}\\ 35.& \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}$

Contoh Soal 6 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 25.& \text{Persamaan garis singgung pada kurva}\\ & y=3\sin x\text{pada titik yang berabsis}{\displaystyle \frac{\pi }{3}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})-\frac{2\sqrt{2}}{3}}\\ \text{b}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})+\frac{2\sqrt{2}}{3}}\\ \text{c}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{3}}{2}}\\ \text{d}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})+\frac{3\sqrt{3}}{2}}\\ \text{e}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{2}}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& y=3\sin x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0=3\sin \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)=\frac{3\sqrt{3}}{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=3\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=3\cos \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}}\right)=\frac{3}{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y={\displaystyle \frac{3}{2}(x-{\displaystyle \frac{\pi }{3}})+{\displaystyle \frac{3\sqrt{3}}{2}}}\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Kurva}y=\sin x+\cos x\text{untuk}\\ & 0<x<\pi \text{memotong sumbu X}\\ & \text{di titik A. Persamaan garis}\\ & \text{singgung di titik A adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{b}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{2}})\\ \text{c}.& y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\\ \text{d}.& y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{e}.& y=\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Kurva memotong sumbu X}\\ & \text{di titik A, berarti}y=0\\ & \sin x+\cos x=0\\ & \sin x=-\cos x\\ & {\displaystyle \frac{\sin x}{\cos x}=-1\iff \tan x=-1}\\ & \tan x=\tan \left({\displaystyle \frac{3\pi }{4}}\right)\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \text{Jadi, titik A-nya}:\left({\displaystyle \frac{3\pi }{4},0}\right)\\ & \text{dan nilai gradien}m=y^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{3\pi }{4}}\right)-\sin \left({\displaystyle \frac{3\pi }{4}}\right)\\ & m=-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Persamaan garis singgung di A}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})+0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y={\sec }^{2}x\text{pada titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{b}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{c}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{d}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{e}.& y=4\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& y={\sec }^{2}x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0={\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)=(2{)}^2=4\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=2{\sec }^{2}x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=2{\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)\tan \left({\displaystyle \frac{\pi }{3}}\right)\\ & =2(4)\sqrt{3}=8\sqrt{3}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Kurva berikut yang memiliki}\\ & \text{garis singgung dengan gradien}\\ & 4\sqrt{3}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=2\sin x\text{pada titik}\left({\displaystyle \frac{\pi }{3},\sqrt{3}}\right)\\ \text{b}.& y=\cos 2x\text{pada titik}\left({\displaystyle \frac{\pi }{12},\frac{1}{2}}\right)\\ \text{c}.& y=\tan x\text{pada titik}\left({\displaystyle \pi ,0}\right)\\ \text{d}.& y=2\sec x\text{pada titik}\left({\displaystyle \frac{\pi }{3},2}\right)\\ \text{e}.& y=\cot x\text{pada titik}\left({\displaystyle \frac{\pi }{4},1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{|cll|}\hline \text{a}& y=2\sin x& m=2\cos {\displaystyle \frac{\pi }{3}}\\ & y^{\prime }=2\cos x& m=2.{\displaystyle \frac{1}{2}=1}\\ \text{b}& y=\cos 2x& m=-2\sin 2\left({\displaystyle \frac{\pi }{12}}\right)\\ & y^{\prime }=-2\sin 2x& m=-2.{\displaystyle \frac{1}{2}=-1}\\ \text{c}& y=\tan x& m={\sec }^2\left(\pi \right)\\ & y^{\prime }={\sec }^{2}x& m=(-1{)}^2=1\\ \text{d}& y=2\sec x& m=2\sec \left({\displaystyle \frac{\pi }{3}}\right)\tan \left({\displaystyle \frac{\pi }{3}}\right)\\ & y^{\prime }=2\sec x\tan x& m=2.2.\sqrt{3}=4\sqrt{3}\\ \text{e}& y=\cot x& m=-{\csc }^2\left({\displaystyle \frac{\pi }{4}}\right)\\ & y^{\prime }=-{\csc }^{2}x& m=-{\left(\sqrt{2}\right)}^2=-2\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sec x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=\sqrt{3}x-{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{b}.& y=\sqrt{3}x+{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{c}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{d}.& y=\sqrt{2}x+{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{e}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& y=\sec x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & y_0=\sec \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=\sec x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & m=\sec \left({\displaystyle \frac{\pi }{4}}\right)\tan \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}.1=\sqrt{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})+\sqrt{2}\\ & \iff y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sin x+\cos x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{2}\text{akan memotong sumbu}}\\ & \text{Y dengan ordinatnya berupa}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}+1}\\ \text{b}.& {\displaystyle \frac{\pi }{2}-1}\\ \text{c}.& 1-{\displaystyle \frac{\pi }{2}}\\ \text{d}.& 2+{\displaystyle \frac{\pi }{2}}\\ \text{e}.& 2-{\displaystyle \frac{\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& y=\sin x+\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{2}}\\ & y_0=\sin \left({\displaystyle \frac{\pi }{2}}\right)+\cos \left({\displaystyle \frac{\pi }{2}}\right)=1+0=1\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{\pi }{2}}\right)-\sin \left({\displaystyle \frac{\pi }{2}}\right)\\ & m=0-1=-1\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-1(x-{\displaystyle \frac{\pi }{2}})+1\\ & \iff y=-x+{\displaystyle \frac{\pi }{2}+1}\\ & \text{Ordinat garis singgungnya saat}\\ & \text{memotong sumbu-Y adalah}:x=0,\\ & \text{maka}\\ & y=-0+{\displaystyle \frac{\pi }{2}+1={\displaystyle \frac{\pi }{2}+1}}\end{array}\end{array}$

Lanjutan Materi (9) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MENYELESAIKAN MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

(APLIKASI TITIK STASIONER)

$\begin{array}{l}\\ 1.& \text{Masalah Maksimum minimum}\\ 2.& \text{Kecepatan dan percepatan}\end{array}$

Aplikasi dari titik stasioner ini yang sering muncul dalam kasus maksimum-manimum khususnya berkaitan dengan fungsi trigonometri di samping juga masalah kecepatan dan percepatan. Berikut ilustrasi contoh-contohnya

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Perhatikanlah gambar berikut}\\ & \text{Gambar di bawah menunjukkan}\\ & \text{trapesium PQRS dengan}\\ & PS=RS=QR=4cm\text{dan}\\ & \mathrm{\angle }SPQ=\mathrm{\angle }RQB=2\theta \text{radian}\\ & \text{dengan}\theta \text{sudut lancip}\end{array}$

$\begin{array}{l}\\ .& \text{a}.\text{Nyatakanlah luas trapesium}\\ & \text{dalam fungsi}\theta \\ & \text{b}.\text{Tentukanlah besar}\theta \text{agar}\\ & \text{luas trapesium maksimum}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \text{Luas Trapesium}\\ & ={\displaystyle \frac{1}{2}\times \text{jmlh sisi sjjr}\times \text{tinggi}}\\ & {\text{L}}_{Trapesium}=\left({\displaystyle \frac{PQ+SR}{2}}\right)\times SA\\ & {\text{L}}_T=\left({\displaystyle \frac{PA+4+BQ+4}{2}}\right)\times 4\sin 2\theta \\ & =(4\cos 2\theta +4+4\cos 2\theta +4)\times 2\sin 2\theta \\ & =(8+8\cos 2\theta )\times 4\sin \theta \cos \theta \\ & =\left(8(1+\cos 2\theta )\right)\times 4\sin \theta \cos \theta \\ & =\left(8(2{\cos }^2\theta )\right)\times 4\sin \theta \cos \theta \\ & =64\sin \theta {\cos }^3\theta \\ \text{b}& \text{Supaya luas maksimum adalah}\\ & \text{nilai stasioner fungsi luas}=0\text{yaitu}:\\ & {\text{L}}_T^{{}^{\prime }}=0\\ & \begin{array}{rl}& {\text{L}}_T=U.V\{\begin{array}{ll}U& =64\sin \theta \\ V& ={\cos }^3\theta \end{array}\\ & {\text{L}}_T^{{}^{\prime }}=U^{\prime }V+U{V}^{\prime }\\ & =64\cos \theta .{\cos }^3\theta +64\sin \theta (-3{\cos }^2\theta \sin \theta )\\ & =64{\cos }^2\theta ({\cos }^2\theta -3{\sin }^2\theta )\end{array}\\ & \text{Karena syarat luas maksimum}\\ & {\text{L}}_T^{{}^{\prime }}=0,\text{maka}\\ & 64{\cos }^2\theta ({\cos }^2\theta -3{\sin }^2\theta )=0\\ & \begin{array}{r}\\ 64{\cos }^2\theta =0& \text{V}& ({\cos }^2\theta -3{\sin }^2\theta )=0\\ \cos \theta =0& & {\cos }^2\theta =3{\sin }^2\theta \\ \theta ={\displaystyle \frac{\pi }{2}}& & {\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }={\displaystyle \frac{1}{3}}}\\ & & {\tan }^2\theta ={\displaystyle \frac{1}{3}}\\ & & \tan \theta =\sqrt{{\displaystyle \frac{1}{3}}}\\ & & \tan \theta ={\displaystyle \frac{1}{3}\sqrt{3}}\\ & & \theta ={\displaystyle \frac{\pi }{6}={30}^{\circ }}\end{array}\\ & \text{Jadi},\theta ={\displaystyle \frac{\pi }{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 2& \text{Sebuah partikel bergerak mengikuti}\\ & \text{sebuah lintasan yang dinyatakan dalam}\\ & s=6\cos 3t+{\sin }^{2}t+t^2+5\text{dalam meter}\\ & \text{Jika waktu yang ditempuh dalam}t\text{detik}\\ & \text{tentukanlah kecepatan saat}t={\displaystyle \frac{\pi }{2}\text{detik}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}v={\displaystyle \frac{ds}{dt}}\\ & \text{maka},\\ & v=-18\sin 3t+2\sin t\cos t+2t\\ & \text{Kecepatan saat}t={\displaystyle \frac{\pi }{2}\text{detik}}\\ & v=-18\sin 3t+\sin 2t+2t\\ & v=-18\sin 3\left(\frac{\pi }{2}\right)+\sin 2\left(\frac{\pi }{2}\right)+2\left(\frac{\pi }{2}\right)\\ & =-18(-1)+0+\pi \\ & =18+\pi \end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., & Akhmad G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

 

Lanjutan Materi (8) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

 MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{G. Nilai Stasioner}$

Jika fungsi  $y=f(x)$  kontinu dan diferensiabel di  $x=f^{\prime }(a)=0$ , maka fungsi tersebut mempunyai nilai stasioner di  $x=a$.

$\begin{array}{l}\\ \text{a}.& \text{Suatu fungsi memiliki nilai stasioner}\\ & \text{adalah}{f}^{\prime }(x)=0\text{untuk suatu nilai}x\\ \text{b}& \text{Jika fungsi}f(x)\text{mempunyai nilai}f(a)\\ & \text{di}x=a,\text{maka titik}(a,f(a))\text{adalah}\\ & \mathbf{\text{titik stasioner}}\end{array}$

Selanjutnya titik stasioner disebut juga dengan titik kritis atau titik ekstrim dan titik stasioner ini terbagi dalam 3 macam

• titik maksimum
• titik minimum, dan 
• titik belok

Sebagai ilustrasi pada fungsi trigonometri, perhatikanlah ilustrasi fungsi sinus berikut

$\begin{array}{l}\\ \underset{\begin{array}{c}\Downarrow \\ \overbrace{\begin{array}{ccc}\begin{array}{rl}& \text{Maksimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{maksimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)<0\\ & \text{Pada contoh di atas}\\ & \text{Titik A,C,E}\\ & \end{array}& \begin{array}{rl}& \text{Minimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{minimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)>0\\ & \text{Pada contoh di atas}\\ & \text{Titik B,D}\\ & \end{array}& \begin{array}{rl}& \text{Belok}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{belok}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)=0\\ & \text{Pada contoh di atas}\\ & \text{Titik}(-\pi ,0)\\ & (0^{\circ },0),(\pi ,0),(2\pi ,0)\end{array}\end{array}}\end{array}}{\begin{array}{c}\text{Stasioner}\\ f^{\prime }(x)=0\text{saat}x=a\end{array}}\end{array}$

Sebagai catatan bahwa, nilai maksimum dan minimum yang telah di dapatkan sampai dengan memasukkan titik ujinya adalah sebenarnya titik maksimum atau minimum LOKAL dalam selang yang diberikan. Supaya menjadi nilai maksimum atau minimum mutlak, maka nilai-nilai dari nilai stasioner ini harus dibandingkan dengan nilai-nilai FUNGSI pada titik-titik ujung intervalnya yang diberikan tersebut.

$\boxed{\text{CONTOH SOAL}}$

Pada contoh soal LANJUTAN MATERI (7) lihat di sini tentang fungsi naik fungsi turun

$\begin{array}{l}\\ 1.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga ada dua absis yang memenuhi}\\ & \text{sebagai titik STASIONER},\text{yaitu}\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x=\frac{5\pi }{4}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{4}}\\ & f\left({\displaystyle \frac{\pi }{4}}\right)=\sin \left({\displaystyle \frac{\pi }{4}}\right)-\cos \left({\displaystyle \frac{\pi }{4}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}=\sqrt{2}}\\ & \text{untuk}x={\displaystyle \frac{5\pi }{4}}\\ & f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin \left({\displaystyle \frac{5\pi }{4}}\right)+\cos \left({\displaystyle \frac{5\pi }{4}}\right)\\ & =-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Jadi titik stasionernya}:\left({\displaystyle \frac{\pi }{4},2}\right)\mathrm{\&}\left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)\\ & \text{Langkah berikutnya gunakanlah titik}\\ & \text{uji di sekitar nilai stasioner yaitu}:\\ & \begin{array}{ccccccccc}& & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & \pi & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Selanjutnya}\\ & \text{Untuk}{f}^{\prime }(x)=\cos x-\sin x\\ & x=0\Rightarrow f^{\prime }(0)=\cos 0-\sin 0\\ & =1+0=1>0(\text{positif})\\ & x=\pi \Rightarrow f^{\prime }(\pi )=\cos \pi -\sin \pi \\ & =-1+0=-1<0(\text{negatif})\\ & x=0\Rightarrow f^{\prime }(2\pi )=\cos 2\pi -\sin 2\pi \\ & =1+0=1>0(\text{positif})\\ & \begin{array}{|cccccl|}\hline x& 0& {\displaystyle \frac{\pi }{4}}& \pi & {\displaystyle \frac{5\pi }{4}}& 2\pi \\ f^{\prime }(x)& +& 0& -& 0& +\\ & & --& & & \\ \text{Garfik}& /& & \mathrm{\setminus }& & /\\ & & & & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}& \\ \hline\end{array}\\ & \text{Dari tabel di atas didapatkan}\\ & \{\begin{array}{ll}\left({\displaystyle \frac{\pi }{4},\sqrt{2}}\right)& \text{titik balik maksimum}\\ \left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)& \text{titik balik minimum}\end{array}\end{array}\end{array}$

$\begin{array}{rl}& \text{Sebagai CATATAN bahwa}:\\ & \text{Nilai ujung intervalnya adalah}:\\ & \{\begin{array}{ll}x=0& \Rightarrow f(0)=\sin 0+\cos 0=0+1=1\\ & \text{titiknya}(0,1)\\ x=2\pi & \Rightarrow f(2\pi )=\sin 2\pi +\cos 2\pi =0+1=1\\ & \text{dan titiknya}(2\pi ,1)\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin 2x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\\ & \text{Stasioner saat}{f}^{\prime }(x)=0\\ & 2\cos 2x=0\\ & \cos 2x=0\\ & \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & x=\pm {\displaystyle \frac{\pi }{4}+k.\pi }\\ & \text{saat}k=0,\Rightarrow x={\displaystyle \frac{\pi }{4}}\\ & \text{saat}k=1,\Rightarrow x={\displaystyle \frac{5\pi }{4}\text{dan}x={\displaystyle \frac{3\pi }{4}}}\\ & \text{saat}k=2,\Rightarrow x={\displaystyle \frac{7\pi }{4}}\\ & \text{Nilai stasionernya dari absis di atas}:\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)=-1\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{5\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{7\pi }{4}}\right)=-1\\ & \\ & \text{SILAHKAN LANJUTKAN SENDIRI}\end{array}\end{array}$

Lanjutan Materi (7) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{F. Fungsi Naik dan Fungsi Turun}$

Dalam menentukan interval-interval di mana fungsi naik atau turun perhatikan dulu ilustrasi berikut ini

Fungsi di atas adalah fungsi $y=f(x)=\sin x$  untuk  $0<x<\pi$  yang memepunya sumbu simetri di  $x={\displaystyle \frac{\pi }{2}=0,5\pi }$. Semua garis singgung yang berada di sebelah kiri sumbu simetri akan mempunyai nilai positif dan semunya garis singgung yang berada di sebelah kanan sumbu simetri bernilai negatif tetapi garis singgung yang tepat pada sumbu simetri memiliki nilai nol.

Pada bahasan sebelumnya-lihat di sini-telah dijelaskan bahwa gradien suatu garis singgung seperti disinggung di atas merupakan nilai dari turunan fungsi pada titik singgung tersebut.

Perhatikanlah gambar ilustrasi berikut

Untuk  

$\bullet m=f^{\prime }(x)>0(\text{tanda positif})$

$\bullet m=f^{\prime }(x)=0$

$\bullet m=f^{\prime }(x)<0(\text{tanda negatif})$

Selanjutnya perhatikan tabel berikut

$\begin{array}{|cll|}\hline \text{Interval}& \text{Nilai}& \text{Keterangan}\\ x<{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)>0& \text{fungsi}f\text{naik}\\ x={\displaystyle \frac{\pi }{2}}& f^{\prime }(x)=0& \text{tidak naik/turun}\\ x>{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)<0& \text{fungsi}f\text{turun}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0<x<2\pi \\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga nilai}x\text{yang memenuhi}:\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x={\displaystyle \frac{5}{4}\pi }}\\ & \begin{array}{c}\\ & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Pilih titik uji bebas, misalkan}\\ & x={\displaystyle \frac{\pi }{6},x=\frac{\pi }{3},\text{dan}x={\displaystyle \frac{3\pi }{2}}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{6}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{6}}\right)-\sin \left({\displaystyle \frac{\pi }{6}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{\pi }{3}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{3}}\right)-\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & ={\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}(\text{negatif})}\\ & \text{dan untuk}x={\displaystyle \frac{3\pi }{2}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{3\pi }{2}}\right)-\sin \left({\displaystyle \frac{3\pi }{2}}\right)\\ & =0-(-1)=1(\text{positif})\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & -& -& & ++& \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Berdasarkan garis bilangan di atas}\\ & \bullet f\text{naik pada}:0<x<{\displaystyle \frac{\pi }{4}}\\ & \text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\\ & \bullet f\text{turun pada}:{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)={\cos }^{2}x\text{dengan}\\ & 0^{\circ }<x<{360}^{\circ }\\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& f(x)={\cos }^{2}x\\ & f^{\prime }(x)=2\cos x(-\sin x)=-2\sin x\cos x\\ & =-\sin 2x\\ & \text{Saat}{f}^{\prime }(x)=0\\ & -\sin 2x=0\\ & \sin 2x=0\\ & \sin 2x=\sin 0^{\circ }\\ & \begin{array}{ll}\begin{array}{rl}2x& =0^{\circ }+k{.360}^{\circ }\\ x& =0^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_1={180}^{\circ }\\ k& =1\Rightarrow x_2={360}^{\circ }\end{array}& \begin{array}{rl}2x& ={180}^{\circ }+k{.360}^{\circ }\\ x& ={90}^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_3={90}^{\circ }\\ k& =1\Rightarrow x_4={270}^{\circ }\end{array}\end{array}\\ & \text{Lalau kita buat diagram nilai}{f}^{\prime }(x)\text{nya}\\ & \begin{array}{c}\\ & & & & & & & & & & & \\ & --& & ++& & & --& & & ++& \\ 0^{\circ }& & {90}^{\circ }& & & {180}^{\circ }& & & {270}^{\circ }& & {360}^{\circ }\end{array}\\ & \text{Berdasar garis bilangan di atas}\\ & (\text{untuk mengecek gunakan titik uji})\\ & \text{maka fungsi}f(x)\\ & \bullet \text{naik}{90}^{\circ }<x<{180}^{\circ }\text{dan}{270}^{\circ }<x<{360}^{\circ }\\ & \bullet \text{turun}{0}^{\circ }<x<{90}^{\circ }\text{dan}{180}^{\circ }<x<{270}^{\circ }\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

MATEMATIKA PEMINATAN MA/SMA untuk KONDISI KHUSUS TAHUN 2020

 $\text{A. Kelas X (Sepuluh)}$

$\begin{array}{rl}& \text{A.1 Fungsi eksponensial}\\ & \text{dan fungsi logaritma}\end{array}$

• materi eksponen 1
• contoh soal eksponen 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10 
• materi logaritma 
• materi logaritma lanjutan
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• materi logaritma lanjutan 2
• contoh soal 7 , contoh 8 , contoh 9 , contoh 10 

$\begin{array}{rl}& \text{A.2 Vektor}\end{array}$

$\text{B. Kelas XI (Sebelas)}$

$\begin{array}{rl}& \text{B.1 Persamaan trigonometri}\end{array}$

• materi persamaan trigonometri
• lanjutan materi 1 , materi 2 , materi 3
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6

$\begin{array}{rl}& \text{B.2 Polinom}\end{array}$

$\text{C. Kelas XII (Duabelas)}$

$\begin{array}{rl}& \text{C.1 Turunan fungsi trigonometri}\end{array}$

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12  

$\begin{array}{rl}& \text{C.2 Distribusi peluang binomial}\end{array}$

• materi distribusi binomial

MATEMATIKA WAJIB MA/SMA untuk KONDISI KHUSUS TAHUN 2020

$\text{A. Kelas X (Sepuluh)}$

$\begin{array}{rl}& \text{A.1 Persamaan dan pertidaksamaan}\\ & \text{nilai mutlak dari bentuk linear}\\ & \text{satu variabel}\end{array}$

• materi persamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 
• materi pertidaksamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , 

$\begin{array}{rl}& \text{A.2 Pertidaksamaan rasional}\\ & \text{dan irasional satu variabel}\end{array}$

• materi pertidaksamaan rasional dan irasional satu variabel
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , 

$\begin{array}{rl}& \text{A.3 Sistem persamaan linear tiga}\\ & \text{variabel}\end{array}$

• materi sistem persamaan linear tiga variabel (SPLTV)
• lanjutan materi SPLTV
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.4 Sistem pertidaksamaan dua}\\ & \text{variabel-linear-linear}\end{array}$

• materi sistem pertidaksamaan dua variabel linear-linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.5 Fungsi}\end{array}$

$\begin{array}{rl}& \text{A.6 Fungsi Komposisi dan invers}\end{array}$

$\begin{array}{rl}& \text{A.7 Rasio trigonometri}\\ & \text{pada segitiga siku-siku}\end{array}$

$\begin{array}{rl}& \text{A.8 Rasio trigonometri}\\ & \text{sudut-sudut diberbagai}\\ & \text{kuadran}\end{array}$

$\begin{array}{rl}& \text{A.9 Aturan sinus dan cosinus}\end{array}$

$\text{B. Kelas XI (Sebelas)}$

$\begin{array}{rl}& \text{B.1 Program linear}\end{array}$

• materi program linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{B.2 Matriks}\end{array}$

• materi matriks 
• lanjutan materi 1
• lanjutan materi 2
• contoh soal 1 , contoh 2 , contoh 3 

$\begin{array}{rl}& \text{B.3 Determinan dan invers matriks}\\ & \text{ordo 2x2}\end{array}$

• materi determinan
• materi invers matriks ordo 2x2
• contoh soal 4 , contoh 5 , contoh 6 

$\begin{array}{rl}& \text{B.4 Pola bilangan dan jumlah}\\ & \text{pada barisan aritmetika}\\ & \text{dan geometri}\end{array}$

• materi pola bilangan - induksi matematika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5

$\begin{array}{rl}& \text{B.5 Limit fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.6 Turunan fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.7 Keberkaitan Turunan fungsi}\end{array}$

$\begin{array}{rl}& \text{B.8 Integral tak tentu}\\ & \text{fungsi aljabar}\end{array}$

$\text{C. Kelas XII (Duabelas)}$

$\begin{array}{rl}& \text{C.1 Jarak dalam ruang}\end{array}$

• materi mengitung jarak di dimensi tiga
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{C.2 Statistika}\end{array}$

• materi statistika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 

$\begin{array}{rl}& \text{C.3 Aturan pencacahan}\end{array}$

$\begin{array}{rl}& \text{C.4 Peluang kejadian majumuk}\end{array}$

Contoh Soal 4 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 16.& \text{Jika}0<x+y<3\text{dan}1<x-y<2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 1<x<5\\ \text{b}.& \left|x\right|<1\\ \text{c}.& x<1\\ \text{d}.& {\displaystyle \frac{1}{2}<x<\frac{5}{2}}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{l}\\ 0<x+y<& 3& \\ 1<x-y<& 2& +\\ 1<2x<& 5& \text{dibagi 2 semuanya}\\ {\displaystyle \frac{1}{2}<x<}& {\displaystyle \frac{5}{2}}& .....(4)\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \mathbf{\text{(UMPTN 1997)}}\\ & \text{Diketahui P, Q, dan R memancing ikan.}\\ & \text{Jika hasil Q lebih sedikit dari hasil R}\\ & \text{sedangkan jumlah hasil P dan Q lebih }\\ & \text{banyak dari pada dua kali hasil R,}\\ & \text{maka yang terbanyak mendapat ikan}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{P dan R}\\ \text{b}.& \text{P dan Q}\\ \text{c}.& \text{P}\\ \text{d}.& \text{Q}\\ \text{e}.& \text{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:\\ & \bullet Q<R...............(1)\\ & \bullet P+Q>2R......(2)\\ & \text{Sehingga untuk persamaan}(1)\mathrm{\&}(2)\\ & \begin{array}{l}\\ R>& Q& \\ P+Q>& 2R& +\\ P+Q+R>& Q+2R& \\ \\ P>& R......(3)\end{array}\\ & \text{dari}(1)\text{dan}(3)\text{diperoleh bahwa}\\ & Q<R<P\\ & \text{Jadi, yang terbanyak mendapat ikan}\\ & \text{adalah P}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Jika}a>0,b>0,\text{dan}a>b,\text{maka}\\ & \text{pernyataan berikut yang salah adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{a}>\frac{1}{b}}\\ \text{b}.& a^2>b^2\\ \text{c}.& a^3>b^3\\ \text{d}.& \sqrt{a}>\sqrt{b}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a>0,b>0,\text{dan}a>b\\ & \text{Maka}\\ & {\displaystyle \frac{a}{1}>\frac{b}{1},\text{jika dibalik}}\\ & \text{menjadi}\\ & {\displaystyle \frac{1}{a}<\frac{1}{b}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}a,b\text{bilangan real, maka}....\\ & \begin{array}{l}\\ \text{a}.& a^2+b^2\ge 2ab\\ \text{b}.& a^2+b^2>2ab\\ \text{c}.& a^2+b^2<2ab\\ \text{d}.& a^2+b^2\le 2ab\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a,b\in \mathbb{R}\\ & \text{Maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Pernyataan berikut yang tepat untuk}\\ & \text{untuk seluruh}x\text{positif adalah}....\\ & \begin{array}{l}\\ \text{a}.& x+{\displaystyle \frac{1}{x}<2}\\ \text{b}.& x+{\displaystyle \frac{1}{x}\le 2}\\ \text{c}.& x+{\displaystyle \frac{1}{x}>2}\\ \text{d}.& x+{\displaystyle \frac{1}{x}\ge 2}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& a,b\in \mathbb{R},a>0,b>0\\ & \text{Mirip dengan pembahasan}\\ & \text{no.19, maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\\ & \text{Saat}a=\sqrt{x},b={\displaystyle \frac{1}{\sqrt{x}}}\\ & \text{menyebabkan}\\ & {\left(\sqrt{x}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{x}}}\right)}^2\ge 2.\sqrt{x}.{\displaystyle \frac{1}{\sqrt{x}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2\sqrt{x.{\displaystyle \frac{1}{x}}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Nugroho, P. A., Gunarto, D. 2013. BIG BANK Soal+Bahas Matematika SMA/MA Kelas 1, 2, & 3. Jakarta : Wahyumedia.
2. Tim BBM. 2015. Big Book Matematika SMA Kelas 1, 2, & 3. Jakarta : Cmedia

Contoh Soal 3 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 11.& \text{Himpunan penyelesaian dari}\\ & 2x-1<x+1<3-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<1\}\\ \text{b}.& \{x|x<2\}\\ \text{c}.& \{x|1<x<2\}\\ \text{d}.& \{x|x>2\}\\ \text{e}.& \{x|x>1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x-1<x}}+\underset{\text{B}}{\underbrace{1<3-x}}\\ & \bullet \text{Bagian A}\\ & 2x-1<x+1\\ & x<2................(1)\\ & \bullet \text{Bagian B}\\ & x+1<3-x\\ & 2x<2\\ & x<1................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<1\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Himpunan penyelesaian dari}\\ & 2x+1<x<1-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-2\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-1<x<-2\}\\ \text{d}.& \{x|x<{\displaystyle \frac{1}{2}}\}\\ \text{e}.& \{x|x<1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x+1<x}}\underset{\text{B}}{\underbrace{<1-x}}\\ & \bullet \text{Bagian A}\\ & 2x+1<x\\ & x<-1................(1)\\ & \bullet \text{Bagian B}\\ & x<1-x\\ & 2x<1\\ & x<{\displaystyle \frac{1}{2}..................(2)}\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<-1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Himpunan penyelesaian dari}\\ & 3x+14\le x+5<3x-1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-3<x<-1\}\\ \text{d}.& \{x|x>3\}\\ \text{e}.& \left\{\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{4x+14\le x}}\underset{\text{B}}{\underbrace{+5<3x-1}}\\ & \bullet \text{Bagian A}\\ & 4x+14\le x+5\\ & 3x\le -9\\ & x\le -3................(1)\\ & \bullet \text{Bagian B}\\ & x+5<3x-1\\ & -2x<-6\\ & x>3..................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}\text{tidak ada}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Jika}{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2020<x<2021\\ \text{b}.& -2021<x<-2020\\ \text{c}.& {\displaystyle \frac{1}{2020}<x<{\displaystyle \frac{1}{2021}}}\\ \text{d}.& x<{\displaystyle \frac{1}{2021}\text{dan}x>{\displaystyle \frac{1}{2020}}}\\ \text{e}.& \text{semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}:{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{Dapat ditulis ulang dengan}\\ & 2020<{\displaystyle \frac{1}{x}\text{dan}{\displaystyle \frac{1}{x}<2021}}\\ & \text{Jika digabung menjadi}\\ & 2020<{\displaystyle \frac{1}{x}<2021}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Jika}a>0\text{dan}b<0,\text{maka}\\ & \text{pernyataan berikut yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& a+b>0\\ \text{b}.& a-b<0\\ \text{c}.& a^2-b^2<0\\ \text{d}.& {\displaystyle \frac{a}{b}<0}\\ \text{e}.& ab>0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Cukup Jelas saat}{\displaystyle \frac{a}{b}=\frac{+}{-}=-<0}\end{array}$