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)

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

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

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

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

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

Contoh Soal 2 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 6.& \text{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y-x>4\\ \text{b}.& y-x<4\\ \text{c}.& y+x+4>0\\ \text{d}.& y+x+4<0\\ \text{e}.& y+x<1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2y-5>2x+4y+3\\ & 2y-4y-2x-5-3>0\\ & -2y-2x-8>0\text{dibagi}(-{\displaystyle \frac{1}{2}})\\ & y+x+4<0\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jika}3x-4>5x-17\\ & \text{maka sebuah bilangan prima}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 3x-4>5x-17\\ & \iff 3x-5x>-17+4\\ & \iff -2x>-13\text{tiap ruas}(\times -1)\\ & \iff 2x<13\\ & \iff x<{\displaystyle \frac{13}{2}=6\frac{1}{2}}\\ & \text{Jadi, yang memenuhi adalah 3 dan 5}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}{\displaystyle \frac{1}{5}<\frac{1}{x}\text{dan}x<0}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{1}{5}}\\ \text{b}.& -5<x<0\\ \text{c}.& 0<x<5\\ \text{d}.& x<-5\\ \text{e}.& -{\displaystyle \frac{1}{5}<x<0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ {\displaystyle \frac{1}{5}}& <\frac{1}{x}\text{dan}x<0\\ {\displaystyle \frac{1}{5}}& <{\displaystyle \frac{1}{x}}\\ x& <5\\ x& >-5\text{karena}x<0\\ \text{Sehi}& \text{ngga}\\ -5<& x<0\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}a,b,c\text{dan}d\text{bilangan real}\\ & \text{dengan}a>b\text{dan}c>d\\ & \text{maka berlaku}\\ & (1)ac>bd\\ & (2)a+c>b+d\\ & (3)ad>bc\\ & (4)ac+bd>ad+bc\\ & \text{Pernyataan-pernyataan di atas}\\ & \text{yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& (1),(2),\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\\ \text{e}.& \text{Semua benar}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:a,b,c\text{dan}d\text{bilangan real}\\ & \text{Jelas bahwa baik bilangan positif maupun}\\ & \text{negatif termasuk semunya dibolehkan}\\ & \text{dengan}a>b\text{dan}c>d\\ & \bullet \text{Sehingga pernyataan (1)}ac>bd\\ & \text{salah saat kita coba bilangan negatif}\\ & \bullet \text{Pernyataan (2) benar karena}\\ & \begin{array}{llll}a& >& b& \\ c& >& d& +\\ a+c& >& b+d\end{array}\\ & \bullet \text{Kasusnya sama dengan poin (1)}\\ & \text{saat dicoba dengan bilangan positif}\\ & \text{tidak semuanya memenuhi}\\ & \bullet \text{Pernyataan (4) tepat juga karena}\\ & \begin{array}{l}\\ a-b>0\\ c-d>0\text{Saat dikalikan}\\ (a-b)\times (c-d)>0\\ \iff ac-ad-bc+bd>0\\ \iff ac+bd>ad+bc\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}-2<y<3\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 9<(y-2{)}^2<16\\ \text{b}.& 4<(y-2{)}^2<16\\ \text{c}.& 1<(y-2{)}^2<16\\ \text{d}.& 0\le (y-2{)}^2<16\\ \text{e}.& -1<(y-2{)}^2<16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}:-2<y<3\\ & \bullet \text{saat dikurangi}2\\ & \iff -2-2<y-2<3-2\\ & -4<y-2<1\\ & \bullet \text{Saat}-4<y-2<0\\ & (-4{)}^2<(y-2{)}^2<0^2\text{dikuadratkan}\\ & 16>(y-2{)}^2>0\\ & 0<(y-2{)}^2<16\\ & \bullet \text{Saat}0\le y-2<1\\ & 0^2\le (y-2{)}^2<1^2\\ & 0<(y-2{)}^2<1\\ & \text{Jadi},0\le (y-2)<16\end{array}\end{array}$

Contoh Soal 1 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 1.& (\text{Soal SNMPTN})\\ & \text{Jika}x>5\text{dan}y<3,\text{maka}\\ & \text{nilai }x-y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{lebih besar dari pada 1}\\ \text{b}.& \text{lebih besar dari pada 3}\\ \text{c}.& \text{lebih besar dari pada 8}\\ \text{d}.& \text{lebih besar dari pada 5}\\ \text{e}.& \text{lebih besar dari pada 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}\\ x& >5\mathrm{\&}y<3\\ \text{m}& \text{aka}\\ & \begin{array}{l}\\ x>5& \Rightarrow & x& >5\\ y<3& \Rightarrow & -y& >-3& +\\ & & x-y& >2\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Batas pertidaksamaan}5x-7>13\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\\ \text{b}.& x>4\\ \text{c}.& x>-4\\ \text{d}.& x<4\\ \text{e}.& -4<x<4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}5x& -7>13\\ 5x& >13+7\\ 5x& >20\\ x& >4\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& x<3{\displaystyle \frac{1}{3}}\\ \text{c}.& x>3{\displaystyle \frac{1}{3}}\\ \text{d}.& x>3\\ \text{e}.& \text{Semua pilihan jawaban salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}2x+3& >5x-7\\ 2x-5x& >-7-3\\ -3x& >-10\text{dikali (-1)}\\ 3x& <10\\ x& <{\displaystyle \frac{10}{3}=3{\displaystyle \frac{1}{3}}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 01}})\text{Jika pertidaksamaan}\\ & 2x-3a>{\displaystyle \frac{3x-1}{2}+ax\text{mempunyai}}\\ & \text{penyelesaian}x>5,\text{maka nilai}a\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{8}}\\ \text{c}.& {\displaystyle \frac{3}{8}}\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}2x-3a& >{\displaystyle \frac{3x-1}{2}+ax\text{tiap ruas}(\times 2)}\\ 4x-6a& >3x-1+2ax\\ 4x-3x& -2ax>-1+6a\\ x-2a& x>-1+6a\\ (1-2a)& x>-1+6a\\ x& >{\displaystyle \frac{-1+6a}{1-2a}}\\ \text{Diketa}& \text{hui}:x>5\text{adalah penyelesaian}\\ \text{maka}& \\ 5& ={\displaystyle \frac{-1+6a}{1-2a}}\\ 5-10a& =-1+6a\\ -6a-10& a=-1-5\\ -16a& =-6\\ a& ={\displaystyle \frac{-6}{-16}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& (\mathbf{\text{UMPTN 94}})\\ & \text{Apabila}a<x<b\text{dan}a<y<b\\ & \text{maka berlaku}....\\ & \begin{array}{l}\\ \text{a}.& a<x-y<b\\ \text{b}.& b-a<x-y<a-b\\ \text{c}.& a-b<x-y<b-a\\ \text{d}.& {\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)}\\ \text{e}.& {\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{l}\\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -a>-y>-b& & \\ & \text{saat}& \text{di susun ulang}& \\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -b<-y<-a& +& \\ & & a-b<x-y<& b-a\end{array}\end{array}$

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

$\begin{array}{l}\\ & \mathbf{\text{BENTUK UMUM}}\\ & \{\begin{array}{l}ax+by<c\\ ax+by\le c\\ ax+by>c\\ ax+by\ge c\end{array}\\ \\ & \mathbf{\text{LANGKAH-LANGKAH}}\\ & \text{dalam membuat gambar grafik persamaan linear}\\ & \text{adalah sebagai berikut}:\\ & \bullet \text{membuat gambar grafik}ax+by=c\\ & \text{untuk batas wilayahnya}\\ & \bullet \text{menyelidiki wilayah yang dimaksud di sekitar}\\ & \text{garis}ax+by=c\\ & \bullet \text{ambillah sebuah titik}(x_0,y_0)\text{sembarang}\\ & \text{kemudian substitusikan ke pertidaksamaan}\\ & ax+by....c\\ & \bullet \text{jika diperoleh nilai ketaksamaan yang benar},\\ & \text{maka daerah di mana titik uji}(x_0,y_0)\\ & \text{berada merupakan wilayah penyelesaiannya}\\ & \text{demikian juga sebaliknya}\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Gambarlah himpunan penyelesaian (HP)}\\ & \text{dari pertidaksamaan linear berikut}\\ & \text{a}.3x+2y<6\\ & \text{b}.3x+2y\le 6\\ & \text{c}.3x+2y>6\\ & \text{d}.3x+2y\ge 6\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Mula}-\text{mula kita gambar garis}3x+2y=6\\ \\ & \begin{array}{|ccc|}\hline \text{Komponen}& \text{pada}& \text{pada}\\ \text{titik}& \text{sumbu}-y& \text{sumbu}-x\\ x& 0& 2\\ y& 3& 0\\ (x,y)& (0,3)& (2,0)\\ \hline\end{array}\\ \\ & \text{Selanjutnya gambar grafiknya sebagai berikut}.\end{array}\end{array}$

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$3x+2y<6$

Kita dapat menggunakan titik uji untuk memastikan kondisi gambar di atas, yaitu di antaranya

$\begin{array}{|ccc|}\hline \text{Titik}& \text{Pengujian}& \text{Keterangan}\\ & \text{Uji}& 3x+2y<6\\ (0,0)& 3(0)+2(0)=0<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,1)& 3(0)+2(1)=2<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,0)& 3(1)+2(0)=3<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,1)& 3(1)+2(1)=5<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,2)& 3(0)+2(2)=4<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (2,0)& 3(2)+2(0)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (2,2)& 3(2)+2(2)=10>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (0,3)& 3(0)+2(3)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,0)& 3(3)+2(0)=9>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,3)& 3(3)+2(3)=15>\mathbf{\text{6}}& \text{Di luar wilayah}\\ ⋮& ⋮& ⋮\\ \hline\end{array}$

Dan berikut untuk wilayah yang memenuhi  $"3x+2y\le 6$

$\begin{array}{l}\\ 2.& \text{Selesaikanlah pertidaksamaan berikut}\\ & \text{a}.12x+2>4x+6\\ & \text{b}.2-3x<6-x\\ & \text{c}.6x+1\ge 2\\ & \text{d}.{\displaystyle \frac{2-3x}{2}<\frac{3-x}{3}}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{a}.12x& +2>4x+6\\ 12x& -4x>6-2\\ 8x& >4\\ x& >{\displaystyle \frac{1}{2}}\end{array}\\ & \begin{array}{rl}\text{b}.& 2-3x<6-x\\ & -3x+x<6-2\\ & -2x<4\text{dikali}(-1)\\ & 2x>-4(\text{tanda berubah})\\ & x>-2\end{array}\\ & \begin{array}{rl}\text{c}.6x& +1\ge 2\\ 6x& \ge 2-1\\ x& \ge {\displaystyle \frac{1}{6}}\end{array}\\ & \begin{array}{rl}\text{d}.{\displaystyle \frac{2-3x}{2}}& <\frac{3-x}{3}\\ 3(2-3x)& <2(3-x)\\ 6-9x& <6-2x\\ -9x+2x& <6-6\\ -7x& <0\text{di kali}(-1)\\ 7x& >0(\text{tanda berubah})\\ x& >{\displaystyle \frac{0}{7}}\\ x& >0\end{array}\end{array}$

DAFTAR PUSTAKA

1. Heryadi, D. 2007. Modul Matematikauntuk SMK Kelas X. Bogor: YUDHISTIRA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo. PT. TIGA SERANGKAI PUSTAKA MANDIRI.

Contoh Soal 9 Statistika

$\begin{array}{l}\\ 37.& (\mathbf{\text{SMBTN 2013}})\\ & \text{Median dan rata-rata dari data yang}\\ & \text{terdiri dari empat bilangan asli yang}\\ & \text{telah diurutkan mulai dari yang terkecil}\\ & \text{adalah 7. Jika data tersebut tidak}\\ & \text{memiliki modus dan selisih antara data}\\ & \text{dan data terkecil adalah 8, maka hasil}\\ & \text{kali terbesar dari datum kedua dan}\\ & \text{keempat adalah}....\\ & \begin{array}{l}\\ \text{a}.& 39\\ \text{b}.& 44\\ \text{c}.& 48\\ \text{d}.& 55\\ \text{e}.& 66\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui data}:x_1,x_2,x_3,\text{dan}{x}_4\\ & \bullet \text{Modus}:\text{tidak ada}\\ & \text{berarti}\text{semua datumnya berbeda}\\ & \bullet \text{Median}:7\\ & \text{maka}{\displaystyle \frac{x_2+x_3}{2}=7\Rightarrow x_2+x_3=14...(1)}\\ & \bullet \text{Rata-rata (Mean)}:7\\ & \text{berarti}{\displaystyle \frac{x_1+x_2+x_3+x_4}{4}=7}\\ & \Rightarrow x_1+x_2+x_3+x_4=28..........(2)\\ & \bullet \text{Jangkauan}:x_4-x_1=8.....(3)\\ & \text{SUBSTITUSI}\\ & \text{Dari persamaan (1) ke (2) diperoleh}:\\ & x_1+x_4=14...........(5)\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (3) dan (4) diperoleh}:\\ & x_1=3\text{dan}{x}_4=11\\ & \text{KEMUNGKINAN}\\ & \text{nilai}{x}_2\text{dan}{x}_3\text{adalah}:3<x_2;x_3<11\\ & ⧫x_2=4\to x_3=10<11✓\\ & ⧫x_2=5\to x_3=9<11✓\\ & ⧫x_2=6\to x_3=8<11✓\\ & ⧫x_2=7\to x_3=7<11\times \\ & \text{KESIMPULAN}\\ & \text{Hasil kali terbesar}{x}_2\times x_3=6\times 11=66\end{array}\end{array}$

$\begin{array}{l}\\ 38.& (\mathbf{\text{SIMAK UI 2012}})\\ & \text{Diketahui bahwa jika Deni mendapatkan}\\ & \text{nilai 75 pada ulangan yang akan datang}\\ & \text{maka rata-rata nilai ulangannya adalah 82.}\\ & \text{Jika Deni mendapatkan nilai 93, maka}\\ & \text{rata-rata nilai ulangannya adalah 85.}\\ & \text{Banyak ulangan yang telah diikuti Deni}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 6\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{MISAL}\\ & n=\text{banyak ulangan yang dijalani oleh Deni}\\ & x=\text{Total nilai ulangannya Deni}\\ & \text{MODEL MATEMATIKA}\\ & \bullet {\displaystyle \frac{x+75}{n+1}=82}\\ & \iff x+75=82(n+1)\iff x=82n+82-75\\ & \iff x=82n+7..................(1)\\ & \bullet {\displaystyle \frac{x+93}{n+1}=85}\\ & \iff x+93=85(n+1)\iff x=85n+85-93\\ & \iff x=85n-8..................(2)\\ & \text{KESAMAAN}\\ & \text{Dari persamaan (1) dan (2), maka}\\ & x=x\\ & 85n-8=82n+7\\ & 85n-82n=7+8\iff 3n=15\iff n=5\end{array}\end{array}$

$\begin{array}{l}\\ 39.& (\mathbf{\text{SIMAK UI}})\\ & \text{Jika rata-rata 20 bilangan bulat nonnegatif}\\ & \text{berbeda adalah 20, maka bilangan terbesar}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 210\\ \text{b}.& 229\\ \text{c}.& 230\\ \text{d}.& 239\\ \text{e}.& 240\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{x_1+x_2+x_3+...+x_{18}+x_{19}+x_{20}}{20}=20}\\ & x_1+x_2+x_3+...+x_{18}+x_{19}+x_{20}=400\\ & \text{Yang mungkin bilangan bulat nonnegatif}\\ & \text{dan berbeda adalah}:0,1,2,3,....dst\\ & 0+1+2+..+18+x_{20}=400\\ & {\displaystyle \frac{18\times 19}{2}+x_{20}=400}\\ & 171+x_{20}=400\\ & x_{20}=400-171=229\end{array}\end{array}$.

$\begin{array}{ll}40.& \mathbf{\text{SPMB 2006}}\\ & \text{Jika jangkauan dari data terurut}:x-1,2x-1,3x,\\ & 5x-3,4x+3,6x+2\text{adalah}18,\text{maka mediannya}\\ & \text{ adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{9}& & & \text{d}.& \text{21}\\ \text{b}.& \text{10},5& \text{c}.& \text{12}& \text{e}.& \text{24},8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & x-1,2x-1,3x,5x-3,4x+3,6x+2\\ & \text{Dan diketahui pula nilai jangkauannya}=18\\ & J=x_{{}_{max}}-x_{{}_{min}}=18\\ & \iff (6x+2)-(x-1)=18\\ & \iff 5x+3=18\\ & \iff 5x=15\\ & \iff x=3\\ & \text{Sehingga datanya}:2,3,9,12,15,20\\ & \text{Selanjutnya ditentukan mediannya}=M_e=Q_{{}_2}\\ & \text{karena}n=6,\text{datum ke}-{\displaystyle \frac{2}{4}(6+1)=3,5}\\ & M_e=Q_2=x_{{.}_3}+0,5(x_{{.}_4}-x_{{.}_3})=9+0,5(12-9)\\ & =9+0,5(3)=9+1,5=10,5\\ & \text{Jadi},J=10,5\end{array}\end{array}$.

Soal lanjutannya (yaitu Contoh Soal 10 Statistika) silahkan klik di sini

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2013. Matematika Wajib untuk SMA/MA Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Contoh Soal 9 Fungsi Logaritma (Pemecahan Masalah Olimpiade)

$\begin{array}{l}\\ 41.& \text{Jika}x={}^{15}\log 75\text{dan}y={}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125},}\\ & \text{maka nilai}5x+3y-2xy\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 1\\ \text{c}.& 3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 5x+3y-2xy\\ & =5({}^{15}\log 75)+3\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & -2({}^{15}\log 75)\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & =5\left({\displaystyle \frac{\log 75}{\log 15}}\right)+3\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & -2\left({\displaystyle \frac{\log 75}{\log 15}}\right)\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & =5\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)+3\left({\displaystyle \frac{\log -\log 125}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)\left({\displaystyle \frac{\log 9-\log 125}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3+\log 5}}\right)\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+2\log 5}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+2\log 5}{\log 3+\log 5}}\right)\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ \\ & \text{Misalkan}\log 3=A,\log 5=B\end{array}\end{array}$

$.\begin{array}{rl}& \text{Selanjutnya}\\ & =5\left({\displaystyle \frac{A+2B}{A+B}}\right)+3\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & -2\left({\displaystyle \frac{A+2B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & =\left({\displaystyle \frac{5A+10B}{A+B}}\right)+\left({\displaystyle \frac{6A-9B}{A-B}}\right)\\ & -\left({\displaystyle \frac{2A+4B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & ={\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{5A^2-5AB+10AB-10B^2}{A^2-B^2}}\\ & +{\displaystyle \frac{6A^2+6AB-9AB-9B^2}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{7A^2-7B^2}{A^2-B^2}}\\ & ={\displaystyle \frac{7(A^2-B^2)}{A^2-B^2}}\\ & =7\end{array}$

$\begin{array}{l}\\ 42.& \text{Diberikan}A={}^6\log 16\text{dan}B={}^{12}\log 27\\ & \text{Terdapat bilangan-bilangan bulat positif}\\ & a,b,\text{dan}c\text{sehingga}(A+a)(B+b)=c\\ & \text{Nilai dari}a+b+c\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& 23\\ \text{b}.& 24\\ \text{c}.& 27\\ \text{d}.& 30\\ \text{e}.& 34\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{....}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & A={}^6\log 16={\displaystyle \frac{\log 16}{\log 6}={\displaystyle \frac{\log 2^4}{\log 2.3}=\frac{4\log 2}{\log 2+\log 3}}}\\ & \iff \log 2+\log 3={\displaystyle \frac{4\log 2}{A}...........(1)}\\ & B={}^{12}\log 27={\displaystyle \frac{\log 27}{\log 12}=\frac{\log 3^3}{\log 2^2.3}=\frac{3\log 3}{2\log 2+\log 3}}\\ & \iff 2\log 2+\log 3={\displaystyle \frac{3\log 3}{B}.........(2)}\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (1) dan (2) diperoleh}:\\ & \bullet \log 2={\displaystyle \frac{3\log 3}{B}-{\displaystyle \frac{4\log 2}{A}}}\\ & \iff \log 2={\displaystyle \frac{3A\log 3-4B\log 2}{AB}}\\ & \iff AB\log 2=3A\log 3-4B\log 2\\ & \iff AB\log 2+4B\log 2=3A\log 3\\ & \iff (AB+4B)\log 2=3A\log 3\\ & \iff {\displaystyle \frac{\log 2}{\log 3}={\displaystyle \frac{3A}{AB+4B}..........(3)}}\\ & \bullet \log 3={\displaystyle \frac{8\log 2}{A}-{\displaystyle \frac{3\log 3}{B}}}\\ & \iff \log 3={\displaystyle \frac{8B\log 2-3A\log 3}{AB}}\\ & \iff AB\log 3=8B\log 2-3A\log 3\\ & \iff AB\log 3+3A\log 3=8B\log 2\\ & \iff (AB+3A)\log 3=8B\log 2\\ & \iff \frac{\log 2}{\log 3}={\displaystyle \frac{AB+3A}{8B}...........(4)}\\ & \text{KESAMAAN}\\ & \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ & {\displaystyle \frac{AB+3A}{8B}={\displaystyle \frac{3A}{AB+4B}}}\\ & \iff (AB+3A)(AB+4B)=(8B).(3A)\\ & \iff (B+3)(A+4)=24\\ & \iff (A+4)(B+3)=24\\ & \text{KESIMPULAN}\\ & a=4,b=3,\text{dan}c=24,\\ & \text{maka}a+b+c=4+3+24=31\end{array}\end{array}$