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

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\color{blue}\textrm{F. Fungsi Naik dan Fungsi Turun}$

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

Fungsi di atas adalah fungsi $\color{red}y=f(x)=\sin x$  untuk  $\color{red}0<x<\pi$  yang memepunya sumbu simetri di  $\color{blue}x=\displaystyle \frac{\pi }{2}=0,5\pi$. 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  

$\bullet \quad m=f'(x)>0\qquad \color{red}(\textrm{tanda positif})$

$\bullet \quad m=f'(x)=0\qquad$

$\bullet \quad m=f'(x)<0\qquad \color{red}(\textrm{tanda negatif})$

Selanjutnya perhatikan tabel berikut

$\begin{array}{|c|l|l|}\hline \color{blue}\textrm{Interval}&\: \: \: \: \color{blue}\textrm{Nilai}&\: \: \: \: \: \color{blue}\textrm{Keterangan}\\\hline x<\displaystyle \frac{\pi }{2}&f'(x)>0&\textrm{fungsi}\: \: f\: \: \textrm{naik}\\\hline \color{red}x=\displaystyle \frac{\pi }{2}&\color{red}f'(x)=0&\color{red}\textrm{tidak naik/turun}\\\hline x>\displaystyle \frac{\pi }{2}&f'(x)<0&\textrm{fungsi}\: \: f\: \: \textrm{turun}\\\hline \end{array}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah interval ketika fungsi}\\ &f(x)=\sin x+\cos x\: \: \textrm{dengan}\\ &0<x<2\pi\\ &\textrm{a}.\quad \textrm{naik}\\ &\textrm{b}.\quad \textrm{turun}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ &f(x)=\sin x+\cos x\\ &f'(x)=\cos x-\sin x\\ &\textrm{Saat}\quad \color{black}f'(x)=0,\\ &\color{black}f'(x)=\cos x-\sin x=0 \: \: \cos x=\sin x\\ &\cos x=\cos \left ( \displaystyle \frac{\pi }{2}-x \right )\\ &\: \: \: \quad x=\pm \left ( \displaystyle \frac{\pi }{2}-x \right )+k.2\pi \\ &\: \: \: \quad \begin{cases} x+x &=\displaystyle \frac{\pi }{2}+k.2\pi ,\: \: \color{red}\textrm{atau} \\ x-x &=-\displaystyle \frac{\pi }{2}+k.2\pi \end{cases}\\ &\textrm{maka}\\ &\: \: \: \quad \begin{cases} x &=\displaystyle \frac{\pi }{4}+k.\pi ,\: \: \color{red}\textrm{atau} \\ 0&=-\displaystyle \frac{\pi }{2}+k.2\pi\: \: (\color{black}\textrm{tidak memenuhi}) \end{cases}\\ &\textrm{Sehingga nilai}\: \: \color{red}x\: \: \color{blue}\textrm{yang memenuhi}:\\ &x=\displaystyle \frac{\pi }{4}\quad \textrm{dan}\quad x=\displaystyle \frac{5}{4}\pi \\ &\begin{array}{ccc|cc|ccc}\\ &&&&&&&\\\hline 0&&\displaystyle \frac{\pi }{4}&&&\displaystyle \frac{5\pi }{4}&&2\pi \end{array}\\ &\textrm{Pilih titik uji bebas, misalkan}\\ &\color{black}x=\displaystyle \frac{\pi }{6},\quad x=\frac{\pi }{3},\quad \color{blue}\textrm{dan}\quad \color{black}x=\displaystyle \frac{3\pi }{2}\\ &\textrm{untuk}\: \: \: \color{black}x=\displaystyle \frac{\pi }{6}\\ &f'(x)=\cos \left ( \displaystyle \frac{\pi }{6} \right )-\sin \left (\displaystyle \frac{\pi }{6} \right )\\ &\qquad=\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\quad \color{red}(\textrm{positif})\\ &\textrm{untuk}\: \: \: \color{black}x=\displaystyle \frac{\pi }{3}\\ &f'(x)=\cos \left ( \displaystyle \frac{\pi }{3} \right )-\sin \left (\displaystyle \frac{\pi }{3} \right )\\ &\qquad=\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}\quad \color{red}(\textrm{negatif})\\ &\textrm{dan untuk}\: \: \: \color{black}x=\displaystyle \frac{3\pi }{2}\\ &f'(x)=\cos \left ( \displaystyle \frac{3\pi }{2} \right )-\sin \left (\displaystyle \frac{3\pi }{2} \right )\\ &\qquad=0-(-1)=1\quad \color{red}(\textrm{positif})\\ &\begin{array}{ccc|cc|ccc}\\ &&&&&&&\\ &\color{red}++&&\color{black}-&\color{black}-&&\color{red}++&\\\hline 0&&\displaystyle \frac{\pi }{4}&&&\displaystyle \frac{5\pi }{4}&&2\pi \end{array}\\ &\color{black}\textrm{Berdasarkan garis bilangan di atas}\\ &\bullet \qquad f\: \: \textrm{naik pada}:\: \: \color{red}0<x<\displaystyle \frac{\pi }{4}\\ &\qquad\quad \color{black}\textrm{atau}\quad \color{red}\displaystyle \frac{5\pi }{4}<x<2\pi\\ &\bullet \qquad f\: \: \textrm{turun pada}: \color{red}\displaystyle \frac{\pi }{4}<x<\displaystyle \frac{5\pi }{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah interval ketika fungsi}\\ &f(x)=\cos^{2} x\: \: \textrm{dengan}\\ &0^{\circ}<x<360^{\circ}\\ &\textrm{a}.\quad \textrm{naik}\\ &\textrm{b}.\quad \textrm{turun}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&f(x)=\cos ^{2}x\\ &f'(x)=2\cos x\left ( -\sin x \right )=-2\sin x\cos x\\ &\, \qquad=-\sin 2x\\ &\textrm{Saat}\quad f'(x)=0\\ &-\sin 2x=0\\ &\sin 2x=0\\ &\sin 2x=\sin 0^{\circ}\\ &\color{black}\begin{array}{l|l} \begin{aligned}2x&=0^{\circ}+k.360^{\circ}\\ x&=0^{\circ}+k.180^{\circ}\\ k&=0\Rightarrow x_{1}=180^{\circ}\\ k&=1\Rightarrow x_{2}=360^{\circ} \end{aligned}&\begin{aligned}2x&=180^{\circ}+k.360^{\circ}\\ x&=90^{\circ}+k.180^{\circ}\\ k&=0\Rightarrow x_{3}=90^{\circ}\\ k&=1\Rightarrow x_{4}=270^{\circ}\\ \end{aligned}\\ \end{array}\\ &\color{black}\textrm{Lalau kita buat diagram nilai}\: \: f'(x)\: \textrm{nya}\\ &\begin{array}{cccccccccccc}\\ &&&&&&&&&&&\\ &--&&++&&&--&&&++&\\\hline 0^{\circ}&&90^{\circ}&&&180^{\circ}&&&270^{\circ}&&360^{\circ}\\ \end{array}\\ &\color{red}\textrm{Berdasar garis bilangan di atas}\\ &\color{black}(\textrm{untuk mengecek gunakan titik uji})\\ &\textrm{maka fungsi}\: \: f(x)\\ &\bullet \quad\textrm{naik}\: \: \: 90^{\circ}<x<180^{\circ}\: \: \textrm{dan}\: \: 270^{\circ}<x<360^{\circ}\\ &\bullet \quad\textrm{turun}\: \: \: 0^{\circ}<x<90^{\circ}\: \: \textrm{dan}\: \: 180^{\circ}<x<270^{\circ} \end{aligned} \end{array}$

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

 $\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Fungsi eksponensial}\\ &\qquad \textrm{dan fungsi logaritma} \end{aligned}$

• 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 

$\color{black}\begin{aligned}&\textrm{A.2 Vektor} \end{aligned}$

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Persamaan trigonometri} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{B.2 Polinom} \end{aligned}$

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Turunan fungsi trigonometri} \end{aligned}$

• 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  

$\color{black}\begin{aligned}&\textrm{C.2 Distribusi peluang binomial} \end{aligned}$

• materi distribusi binomial

MATEMATIKA WAJIB MA/SMA untuk KONDISI KHUSUS TAHUN 2020

$\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Persamaan dan pertidaksamaan}\\ &\qquad \textrm{nilai mutlak dari bentuk linear}\\ &\qquad \textrm{satu variabel} \end{aligned}$

• 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 , 

$\color{red}\begin{aligned}&\textrm{A.2 Pertidaksamaan rasional}\\ &\qquad \textrm{dan irasional satu variabel} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{A.3 Sistem persamaan linear tiga}\\ &\qquad \textrm{variabel} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{A.4 Sistem pertidaksamaan dua}\\ &\qquad \textrm{variabel-linear-linear} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{A.5 Fungsi} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.6 Fungsi Komposisi dan invers} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.7 Rasio trigonometri}\\ &\qquad \textrm{pada segitiga siku-siku} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.8 Rasio trigonometri}\\ &\qquad \textrm{sudut-sudut diberbagai}\\ &\qquad \textrm{kuadran} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.9 Aturan sinus dan cosinus} \end{aligned}$

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Program linear} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.2 Matriks} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.3 Determinan dan invers matriks}\\ &\qquad \textrm{ordo 2x2} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{B.4 Pola bilangan dan jumlah}\\ &\qquad \textrm{pada barisan aritmetika}\\ &\qquad \textrm{dan geometri} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{B.5 Limit fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.6 Turunan fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.7 Keberkaitan Turunan fungsi} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.8 Integral tak tentu}\\ &\qquad \textrm{fungsi aljabar} \end{aligned}$

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Jarak dalam ruang} \end{aligned}$

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

$\color{red}\begin{aligned}&\textrm{C.2 Statistika} \end{aligned}$

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

$\color{black}\begin{aligned}&\textrm{C.3 Aturan pencacahan} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{C.4 Peluang kejadian majumuk} \end{aligned}$

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

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: 0<x+y<3\: \: \textrm{dan}\: \: 1<x-y<2\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<5\\ \textrm{b}.&\left | x \right |<1\\ \textrm{c}.&x<1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{2}<x<\frac{5}{2}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{array}{llll}\\ 0<x+y<&3&\\ 1<x-y<&2&+\\\hline \: \: 1<2x<&5&\color{black}\textrm{dibagi 2 semuanya}\\ \quad \displaystyle \frac{1}{2}<x<&\displaystyle \frac{5}{2}&\: .....\color{red}(4)\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 17.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui P, Q, dan R memancing ikan.}\\ & \textrm{Jika hasil Q lebih sedikit dari hasil R}\\ & \textrm{sedangkan jumlah hasil P dan Q lebih }\\ & \textrm{banyak dari pada dua kali hasil R,}\\ &\textrm{maka yang terbanyak mendapat ikan}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{P dan R}\\ \textrm{b}.&\textrm{P dan Q}\\ \color{red}\textrm{c}.&\textrm{P}\\ \textrm{d}.&\textrm{Q}\\ \textrm{e}.&\textrm{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{Diketahui}:\\ &\bullet \: Q< R\: ...............\color{red}(1)\\ &\bullet \: P+Q> 2R\: ......\color{red}(2)\\ &\textrm{Sehingga untuk persamaan}\: \: \color{black}(1)\: \&\: (2)\\ &\begin{array}{llll}\\ \qquad\qquad R>&Q&\\ \qquad P+Q>&2R&+\\\hline P+Q+R>&Q+2R&\\\\ \qquad\quad\quad P>&R\: ......\color{red}(3)\\ \end{array}\\ &\textrm{dari} \: \color{red}(1)\: \color{blue}\textrm{dan}\: \color{red}(3)\: \color{blue}\textrm{diperoleh bahwa}\\ &Q<R< P\\ &\textrm{Jadi, yang terbanyak mendapat ikan}\\ &\color{red}\textrm{adalah P} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika}\: \: a>0,\: b>0,\: \: \textrm{dan}\: \: a>b,\: \: \textrm{maka}\\ &\textrm{pernyataan berikut yang salah adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{a}>\frac{1}{b}\\ \textrm{b}.&a^{2}>b^{2}\\ \textrm{c}.&a^{3}>b^{3}\\ \textrm{d}.&\sqrt{a}>\sqrt{b}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&a>0,\: b>0,\: \: \textrm{dan}\: \: a>b\\ &\color{red}\textrm{Maka}\\ &\displaystyle \frac{a}{1}>\frac{b}{1},\: \: \textrm{jika dibalik}\\ &\color{red}\textrm{menjadi}\\ &\displaystyle \frac{1}{a}<\frac{1}{b} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan real, maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{2}+b^{2}\geq 2ab\\ \textrm{b}.&a^{2}+b^{2}> 2ab\\ \textrm{c}.&a^{2}+b^{2}< 2ab\\ \textrm{d}.&a^{2}+b^{2}\leq 2ab\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&a,b\in \mathbb{R}\\ &\color{red}\textrm{Maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Pernyataan berikut yang tepat untuk}\\ &\textrm{untuk seluruh}\: \: x\: \: \textrm{positif adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x+\displaystyle \frac{1}{x}<2\\ \textrm{b}.&x+\displaystyle \frac{1}{x}\leq 2\\ \textrm{c}.&x+\displaystyle \frac{1}{x}>2\\ \color{red}\textrm{d}.&x+\displaystyle \frac{1}{x}\geq 2\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&a,b\in \mathbb{R},\: \: a>0,\: b>0\\ &\color{red}\textrm{Mirip dengan pembahasan}\\ &\color{red}\textrm{no.19, maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab\\ &\color{black}\textrm{Saat}\: \: a=\sqrt{x},\: \: b=\displaystyle \frac{1}{\sqrt{x}}\\ &\textrm{menyebabkan}\\ &\left ( \sqrt{x} \right )^{2}+\left ( \displaystyle \frac{1}{\sqrt{x}} \right )^{2}\geq 2.\sqrt{x}.\displaystyle \frac{1}{\sqrt{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2\sqrt{x.\displaystyle \frac{1}{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2 \end{aligned} \end{array}$

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)

$\begin{array}{ll}\\ 11.&\textrm{Himpunan penyelesaian dari}\\ &2x-1<x+1<3-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x<1 \right \}\\ \textrm{b}.&\left \{ x|x<2 \right \}\\ \textrm{c}.&\left \{ x|1<x<2 \right \}\\ \textrm{d}.&\left \{ x|x>2 \right \}\\ \textrm{e}.&\left \{ x|x>1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x-1<x}}\, +\, \underset{\textrm{B}}{\underbrace{1<3-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x-1<x+1\\ &\qquad x<2\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+1<3-x\\ &\qquad 2x<2\\ &\qquad x<1\: ................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Himpunan penyelesaian dari}\\ &2x+1<x<1-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-2 \right \}\\ \color{red}\textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-1<x<-2 \right \}\\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{1}{2}\right \}\\ \textrm{e}.&\left \{ x|x<1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x+1<x}} \underset{\textrm{B}}{\underbrace{\: <1-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x+1<x\\ &\qquad x<-1\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x<1-x\\ &\qquad 2x<1\\ &\qquad x<\displaystyle \frac{1}{2}\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Himpunan penyelesaian dari}\\ &3x+14\leq x+5<3x-1\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-3 \right \}\\ \textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-3<x<-1 \right \}\\ \textrm{d}.&\left \{ x|x>3\right \}\\ \color{red}\textrm{e}.&\left \{\: \: \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{4x+14\leq x}} \underset{\textrm{B}}{\underbrace{\, +\, 5 <3x-1}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 4x+14\leq x+5\\ &\qquad 3x\leq -9\\ &\qquad x\leq -3\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+5<3x-1\\ &\qquad -2x<-6\\ &\qquad x>3\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}\: \color{red}\textrm{tidak ada} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: \displaystyle \frac{1}{x}<2021\: \: \textrm{dan}\: \: \displaystyle \frac{1}{x}>2020\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2020<x<2021\\ \textrm{b}.&-2021<x<-2020\\ \textrm{c}.&\displaystyle \frac{1}{2020}<x<\displaystyle \frac{1}{2021}\\ \textrm{d}.&x<\displaystyle \frac{1}{2021}\: \: \textrm{dan}\: \: x>\displaystyle \frac{1}{2020}\\ \textrm{e}.&\textrm{semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}:\: \color{black}\displaystyle \frac{1}{x}<2021\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}>2020\\ &\textrm{Dapat ditulis ulang dengan}\\ &\color{black}2020<\displaystyle \frac{1}{x}\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}<2021\\ &\textrm{Jika digabung menjadi}\\ &\color{black}2020<\displaystyle \frac{1}{x}<\color{black}2021 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: a>0\: \: \textrm{dan}\: \: b<0\: ,\: \textrm{maka}\\ &\textrm{pernyataan berikut yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a+b>0\\ \textrm{b}.&a-b<0\\ \textrm{c}.&a^{2}-b^{2}<0\\ \color{red}\textrm{d}.&\displaystyle \frac{a}{b}<0\\ \textrm{e}.&ab>0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\textrm{Cukup Jelas saat}\: \: \color{red}\displaystyle \frac{a}{b}=\frac{+}{-}=-<0 \end{array}$

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

$\begin{array}{ll}\\ 6.&\textrm{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y-x>4\\ \textrm{b}.&y-x<4\\ \textrm{c}.&y+x+4>0\\ \color{red}\textrm{d}.&y+x+4<0\\ \textrm{e}.&y+x<1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&2y-5>2x+4y+3\\ &2y-4y-2x-5-3>0\\ &-2y-2x-8>0\: \: \color{black}\textrm{dibagi}\: \left ( -\displaystyle \frac{1}{2} \right )\\ &y+x+4<0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Jika}\: \: 3x-4>5x-17\\ &\textrm{maka sebuah bilangan prima}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&11\\ \textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&3x-4>5x-17\\ &\Leftrightarrow 3x-5x>-17+4\\ &\Leftrightarrow -2x>-13\quad \color{black}\textrm{tiap ruas}\: (\times -1)\\ &\Leftrightarrow 2x<13\\ &\Leftrightarrow x<\displaystyle \frac{13}{2}=6\frac{1}{2}\\ &\color{black}\textrm{Jadi, yang memenuhi adalah 3 dan 5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \displaystyle \frac{1}{5}<\frac{1}{x}\: \: \textrm{dan}\: \: x<0\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{1}{5}\\ \color{red}\textrm{b}.&-5<x<0\\ \textrm{c}.&0<x<5\\ \textrm{d}.&x<-5\\ \textrm{e}.&-\displaystyle \frac{1}{5}<x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ \displaystyle \frac{1}{5}&<\frac{1}{x}\: \: \: \textrm{dan}\: \: x<0\\ \displaystyle \frac{1}{5}&<\displaystyle \frac{1}{x}\\ x&<5 \\ x&>-5\qquad \color{black}\textrm{karena}\: \: x<0\\ \textrm{Sehi}&\textrm{ngga}\\ \color{red}-5<&\color{red}x<0 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: -2<y<3\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9<(y-2)^{2}<16\\ \textrm{b}.&4<(y-2)^{2}<16\\ \textrm{c}.&1<(y-2)^{2}<16\\ \color{red}\textrm{d}.&0\leq (y-2)^{2}<16\\ \textrm{e}.&-1<(y-2)^{2}<16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}:\: -2<y<3\\ &\color{red}\bullet \quad \textrm{saat dikurangi}\: \: 2\\ &\qquad \Leftrightarrow \: -2-2<y-2<3-2\\ &\qquad -4<y-2<1\\ &\color{red}\bullet \quad \textrm{Saat}\: \: -4<y-2<0\\ &\qquad (-4)^{2}<(y-2)^{2}<0^{2}\quad \textrm{dikuadratkan}\\ &\qquad 16>(y-2)^{2}>0\\ &\qquad 0<(y-2)^{2}<16\\ &\color{red}\bullet \quad \textrm{Saat}\: \: 0\leq y-2<1\\ &\qquad 0^{2}\leq (y-2)^{2}<1^{2}\\ &\qquad 0<(y-2)^{2}<1\\ &\textrm{Jadi}\: ,\: \: \color{red}0\leq (y-2)<16 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 1.&(\textrm{Soal SNMPTN})\\ &\textrm{Jika}\: \: x>5\: \: \textrm{dan}\: \: y<3,\: \: \textrm{maka}\\ &\textrm{nilai }\: \: x-y\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{lebih besar dari pada 1}\\ \textrm{b}.&\textrm{lebih besar dari pada 3}\\ \textrm{c}.&\textrm{lebih besar dari pada 8}\\ \textrm{d}.&\textrm{lebih besar dari pada 5}\\ \color{red}\textrm{e}.&\textrm{lebih besar dari pada 2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}\\ x&>5\: \: \color{red}\&\: \: \color{blue}y<3\\ \textrm{m}&\textrm{aka}\\ &\begin{array}{llllll}\\ x>5&\Rightarrow &x&>5\\ y<3&\Rightarrow &\color{black}-y&\color{black}>-3&\color{red}+\\\hline &&\color{red}x-y&>\color{red}2 \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 2.&\textrm{Batas pertidaksamaan}\: \: 5x-7>13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\\ \color{red}\textrm{b}.&x>4\\ \textrm{c}.&x>-4\\ \textrm{d}.&x<4\\ \textrm{e}.&-4<x<4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}5x&-7>13\\ 5x&>13+7\\ 5x&>20\\ x&\color{red}>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \color{red}\textrm{b}.&x<3\displaystyle \frac{1}{3}\\ \textrm{c}.&x>3\displaystyle \frac{1}{3}\\ \textrm{d}.&x>3\\ \textrm{e}.&\textrm{Semua pilihan jawaban salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}2x+3&>5x-7\\ 2x-5x&>-7-3\\ -3x&>-10\quad \color{black}\textrm{dikali (-1)}\\ 3x&<10\\ x&<\color{red}\displaystyle \frac{10}{3}=3\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 01})\textrm{Jika pertidaksamaan}\\ & 2x-3a>\displaystyle \frac{3x-1}{2}+ax\: \: \textrm{mempunyai}\\ &\textrm{penyelesaian}\: \: x>5,\: \: \textrm{maka nilai}\: \: a\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{8}\\ \color{red}\textrm{c}.&\displaystyle \frac{3}{8}\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}2x-3a&>\displaystyle \frac{3x-1}{2}+ax\quad \color{black}\textrm{tiap ruas}\: (\times 2)\\ 4x-6a&>3x-1+2ax\\ 4x-3x&-2ax>-1+6a\\ x-2a&x>-1+6a\\ (1-2a)&x>-1+6a\\ x&>\displaystyle \frac{-1+6a}{1-2a}\\ \textrm{Diketa}&\textrm{hui}:\: \: x>5\: \: \color{red}\textrm{adalah penyelesaian}\\ \color{red}\textrm{maka}\: \: &\\ 5&=\displaystyle \frac{-1+6a}{1-2a}\\ 5-10a&=-1+6a\\ -6a-10&a=-1-5\\ -16a&=-6\\ a&=\displaystyle \frac{-6}{-16}\\ &=\color{red}\displaystyle \frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&(\textbf{UMPTN 94})\\ &\textrm{Apabila}\: \: a<x<b\: \: \textrm{dan}\: \: a<y<b\\ & \textrm{maka berlaku}\: \: \: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a<x-y<b\\ \textrm{b}.&b-a<x-y<a-b\\ \color{red}\textrm{c}.&a-b<x-y<b-a\\ \textrm{d}.&\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)\\ \textrm{e}.&\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{array}{llllll}\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &-a>-y>-b&&\\\hline &\color{purple}\textrm{saat}&\color{black}\textrm{di susun ulang}&\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &\color{black}-b<-y<-a&\color{red}+&\\\hline &&\color{red}a-b\color{blue}<\color{red}x-y\color{blue}<&\color{red}b-a \end{array} \end{array}$

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

$\begin{array}{ll}\\ &\textbf{BENTUK UMUM}\\ &\color{blue}\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ &\textbf{LANGKAH-LANGKAH}\\ &\color{purple}\textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: \: \color{red}ax+by=c\\ &\quad \: \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar}\\ &\quad \: \: \textrm{garis} \: \: \color{red}ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \color{red}\left ( x_{0},y_{0} \right )\: \color{black}\textrm{sembarang}\\ &\: \: \quad \textrm{kemudian substitusikan ke pertidaksamaan}\\ &\quad \: \: \color{red}ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar},\\ &\: \: \quad \textrm{maka daerah di mana titik uji}\: \color{red}\left ( x_{0},y_{0} \right )\\ &\: \: \quad \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

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

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$\color{red}3x+2y<6$

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

$\begin{array}{|c|c|c|}\hline \color{black}\color{purple}\textrm{Titik}&\color{black}\textrm{Pengujian}&\color{black}\textrm{Keterangan}\\ &\color{blue}\textrm{Uji}&\color{red}3x+2y<6\\\hline (0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Dalam wilayah}\\\hline (2,0)&3(2)+2(0)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (2,2)&3(2)+2(2)=10>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (0,3)&3(0)+2(3)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,0)&3(3)+2(0)=9>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,3)&3(3)+2(3)=15>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline \vdots &\vdots&\vdots \\\hline \end{array}$

Dan berikut untuk wilayah yang memenuhi  $"\color{red}3x+2y\leq 6$

$\begin{array}{ll}\\ 2.&\textrm{Selesaikanlah pertidaksamaan berikut}\\ &\textrm{a}.\quad 12x+2>4x+6\\ &\textrm{b}.\quad 2-3x<6-x\\ &\textrm{c}.\quad 6x+1\geq 2\\ &\textrm{d}.\quad \displaystyle \frac{2-3x}{2}<\frac{3-x}{3}\\\\ &\textrm{Jawab}\\ &\color{purple}\begin{aligned}\color{black}\textrm{a}.\: \: 12x&+2>4x+6\\ 12x&-4x>6-2\\ 8x&>4\\ x&>\displaystyle \frac{1}{2} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{b}.\: \: &2-3x<6-x\\ &-3x+x<6-2\\ &-2x<4\: \: \color{red}\textrm{dikali}\: \: (-1)\\ &2x>-4\: \: (\color{black}\textrm{tanda berubah})\\ &x>-2 \end{aligned}\\ &\color{purple}\begin{aligned}\color{black}\textrm{c}.\: \: 6x&+1\geq 2\\ 6x&\geq 2-1\\ x&\geq \displaystyle \frac{1}{6} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{d}.\: \: \: \: \color{blue}\displaystyle \frac{2-3x}{2}&<\frac{3-x}{3}\\ 3(2-3x)&<2(3-x)\\ 6-9x&<6-2x\\ -9x+2x&<6-6\\ -7x&<0\: \: \color{red}\textrm{di kali}\: \: (-1)\\ 7x&>0\: \: (\color{black}\textrm{tanda berubah})\\ x&>\displaystyle \frac{0}{7}\\ x&>0 \end{aligned} \end{array}$

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}{ll}\\ 37.&(\textbf{SMBTN 2013})\\ &\textrm{Median dan rata-rata dari data yang}\\ &\textrm{terdiri dari empat bilangan asli yang}\\ &\textrm{telah diurutkan mulai dari yang terkecil}\\ &\textrm{adalah 7. Jika data tersebut tidak}\\ &\textrm{memiliki modus dan selisih antara data}\\ &\textrm{dan data terkecil adalah 8, maka hasil}\\ &\textrm{kali terbesar dari datum kedua dan}\\ &\textrm{keempat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&39\\ \textrm{b}.&44\\ \textrm{c}.&48\\ \textrm{d}.&55\\ \color{red}\textrm{e}.&66 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui data}\: :\: \color{black}x_{1},x_{2},x_{3},\: \: \textrm{dan}\: \: x_{4}\\ &\bullet \quad\textrm{Modus}\: :\: \textrm{tidak ada}\\ &\textrm{berarti}\: \color{red}\textrm{semua datumnya berbeda}\\ &\bullet \quad \textrm{Median} \: :\: 7\\ &\textrm{maka}\: \: \displaystyle \frac{x_{2}+x_{3}}{2}=7\color{black}\Rightarrow x_{2}+x_{3}=14\: ...\color{red}(1)\\ &\bullet \quad\textrm{Rata-rata (Mean)}\: :\: 7\\ &\textrm{berarti}\: \: \displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=7\\ &\color{black}\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}=28\: ..........\color{red}(2)\\ &\bullet \quad\textrm{Jangkauan}\: :\: x_{4}-x_{1}=8\: .....\color{red}(3)\\ &\color{black}\textrm{SUBSTITUSI}\\ &\textrm{Dari persamaan (1) ke (2) diperoleh}:\\ &x_{1}+x_{4}=14\: ...........\color{red}(5)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (3) dan (4) diperoleh}:\\ &x_{1}=3\: \: \textrm{dan}\: \: x_{4}=11\\ &\color{black}\textrm{KEMUNGKINAN}\\ &\textrm{nilai}\: x_{2}\: \textrm{dan}\: x_{3}\: \textrm{adalah}:\: 3< x_{2}\: ;\: x_{3}<11\\ &\blacklozenge \: x_{2}=4\rightarrow x_{3}=10<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=5\rightarrow x_{3}=9<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=6\rightarrow x_{3}=8<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=7\rightarrow x_{3}=7<11\: \: \color{red}\times \\ &\color{black}\textrm{KESIMPULAN}\\ &\textrm{Hasil kali terbesar}\: \: \color{red}x_{2}\times x_{3}=6\times 11=66 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 39.&(\textbf{SIMAK UI})\\ &\textrm{Jika rata-rata 20 bilangan bulat nonnegatif}\\ &\textrm{berbeda adalah 20, maka bilangan terbesar}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&210\\ \textrm{b}.&229\\ \textrm{c}.&230\\ \textrm{d}.&239\\ \textrm{e}.&240 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\color{black}\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}}{20}=\color{red}20\\ &x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}=\color{red}400\\ &\color{black}\textrm{Yang mungkin bilangan bulat nonnegatif}\\ &\textrm{dan berbeda adalah}\: :\: \color{red}0, 1,2,3,....dst\\ &0+1+2+..+18+x_{20}=\color{red}400\\ &\displaystyle \frac{18\times 19}{2}+x_{20}=\color{red}400\\ &171+x_{20}=\color{red}400\\ &x_{20}=\color{red}400-171=229 \end{aligned} \end{array}$.

$\begin{array}{ll} 40.&\textbf{SPMB 2006}\\ &\textrm{Jika jangkauan dari data terurut}:x-1,\: 2x-1,\: 3x,\\ &5x-3,\: 4x+3,\: 6x+2\: \: \textrm{adalah}\: \: 18,\: \textrm{maka mediannya}\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{9}&&&\textrm{d}.&\textrm{21}\\ \textrm{b}.&\textrm{10},5\quad &\textrm{c}.&\textrm{12}\quad&\textrm{e}.&\textrm{24},8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &x-1,\: 2x-1,\: 3x,\: 5x-3,\: 4x+3,\: 6x+2\\ &\textrm{Dan diketahui pula nilai jangkauannya}\: =18\\ &J=x_{_{max}}-x_{_{min}}=18\\ &\Leftrightarrow (6x+2)-(x-1)=18\\ &\Leftrightarrow 5x+3=18\\ &\Leftrightarrow 5x=15\\ &\Leftrightarrow x=3\\ &\textrm{Sehingga datanya}:\: \color{red}2,3,9,12,15,20\\ &\textrm{Selanjutnya ditentukan mediannya}=M_{e}=Q_{_{2}}\\ &\textrm{karena}\: \: n=6,\: \textrm{datum ke}-\displaystyle \frac{2}{4}(6+1)=3,5\\ & M_{e}=Q_{2}=x_{._{3}}+0,5(x_{._{4}}-x_{._{3}})=9+0,5(12-9)\\ &\: \: \: \quad =9+0,5(3)=9+1,5=10,5\\ &\textrm{Jadi},\: J=\color{red}10,5 \end{aligned} \end{array}$.

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}{ll}\\ 41.&\textrm{Jika}\: \: x=\: ^{15}\log 75\: \: \textrm{dan}\: \: y=\: ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125},\\ &\textrm{maka nilai}\: \: 5x+3y-2xy\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&1\\ \textrm{c}.&3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\color{black}5x+3y-2xy\\ &=5\left ( ^{15}\log 75 \right )+3\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &\qquad -2\left ( ^{15}\log 75 \right )\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 )\\ &\qquad -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 )\\ &\qquad -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 )\\ &\qquad -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 )\\ &\qquad -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 )\\\\ &\color{red}\textrm{Misalkan}\: \: \color{black}\log 3=A,\: \: \log 5=B \end{aligned} \end{array}$

$.\qquad\color{purple}\begin{aligned} &\color{red}\textrm{Selanjutnya}\\ &=5\left ( \displaystyle \frac{A+2B}{A+B} \right )+3\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &\qquad -2\left ( \displaystyle \frac{A+2B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\left ( \displaystyle \frac{5A+10B}{A+B} \right )+\left ( \displaystyle \frac{6A-9B}{A-B} \right )\\ &\qquad -\left ( \displaystyle \frac{2A+4B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{5A^{2}-5AB+10AB-10B^{2}}{A^{2}-B^{2}}\\ &\quad +\displaystyle \frac{6A^{2}+6AB-9AB-9B^{2}}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{7A^{2}-7B^{2}}{A^{2}-B^{2}}\\ &=\displaystyle \frac{7\left ( A^{2}-B^{2} \right )}{A^{2}-B^{2}}\\ &=7 \end{aligned}$

$\begin{array}{ll}\\ 42.&\textrm{Diberikan}\: \: A=\: ^{6}\log 16\: \: \textrm{dan}\: \: B=\: ^{12}\log 27\\ &\textrm{Terdapat bilangan-bilangan bulat positif}\\ &a,\: b,\: \: \textrm{dan}\: \: c\: \: \textrm{sehingga}\: \: (A+a)(B+b)=c\\ &\textrm{Nilai dari}\: \: a+b+c\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&23\\ \textrm{b}.&24\\ \textrm{c}.&27\\ \textrm{d}.&30\\ \textrm{e}.&34 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{....}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{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}\\ &\Leftrightarrow \color{black}\log 2+\log 3=\displaystyle \frac{4\log 2}{A}\: ...........\color{red}(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}\\ &\Leftrightarrow \color{black}2\log 2+\log 3=\displaystyle \frac{3\log 3}{B}\: .........\color{red}(2)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (1) dan (2) diperoleh}:\\ &\bullet \quad \log 2=\displaystyle \frac{3\log 3}{B}-\displaystyle \frac{4\log 2}{A}\\ &\qquad \Leftrightarrow \log 2=\displaystyle \frac{3A\log 3-4B\log 2}{AB}\\ &\qquad \Leftrightarrow AB\log 2=3A\log 3-4B\log 2\\ &\qquad \Leftrightarrow AB\log 2+4B\log 2=3A\log 3\\ &\qquad \Leftrightarrow (AB+4B)\log 2=3A\log 3\\ &\qquad \Leftrightarrow \displaystyle \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{3A}{AB+4B}\: ..........\color{red}(3)\\ &\bullet \quad \log 3=\displaystyle \frac{8\log 2}{A}-\displaystyle \frac{3\log 3}{B}\\ &\qquad \Leftrightarrow \log 3=\displaystyle \frac{8B\log 2-3A\log 3}{AB}\\ &\qquad \Leftrightarrow AB\log 3=8B\log 2-3A\log 3\\ &\qquad \Leftrightarrow AB\log 3+3A\log 3=8B\log 2\\ &\qquad \Leftrightarrow (AB+3A)\log 3=8B\log 2\\ &\qquad \Leftrightarrow \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{AB+3A}{8B}...........\color{red}(4)\\ &\color{black}\textrm{KESAMAAN}\\ &\qquad\quad \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ &\color{purple}\displaystyle \frac{AB+3A}{8B}=\color{purple}\displaystyle \frac{3A}{AB+4B}\\ &\qquad \Leftrightarrow (AB+3A)(AB+4B)=(8B).(3A)\\ &\qquad \Leftrightarrow (B+3)(A+4)=24\\ &\qquad \Leftrightarrow (A+4)(B+3)=24\\ &\color{black}\textrm{KESIMPULAN}\\ &a=4,\: b=3,\: \: \textrm{dan}\: \: c=24,\\ &\color{purple}\textrm{maka}\: \: \color{black}a+b+c=4+3+24=\color{red}31 \end{aligned} \end{array}$

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{ll}\\ 36.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

 $\begin{array}{ll}\\ 32.&\textrm{Agar}\: \: \log \left ( x^{2}-1 \right )<0\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<1\\ \textrm{b}.&-\sqrt{2}<x<\sqrt{2}\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&x<-\sqrt{2}\: \: \textrm{atau}\: \: x>\sqrt{2}\\ \color{red}\textrm{e}.&-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\log \left ( x^{2}-1 \right )<0\\ &\textrm{Diketahui}\: \: \color{red}\log f(x)<0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-1>0\\ &\Leftrightarrow x<-1\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: \log \left ( x^{2}-1 \right )<0\\ &\log \left ( x^{2}-1 \right )<\log 1\\ &\Leftrightarrow x^{2}-1<1\\ &\Leftrightarrow x^{2}-2<0\\ &\Leftrightarrow x^{2}-\left ( \sqrt{2} \right )^{2}<0\\ &\Leftrightarrow -\sqrt{2}<x<\sqrt{2}\\ &\textrm{Jadi},\: \: \color{red}-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari}\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{b}.&\left \{ x|-\sqrt{3}<x<-1\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{c}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{d}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{e}.&\left \{ x|\sqrt{3}<x<2 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &\textrm{Diketahui}\: \: \color{red}^{.^{\frac{1}{2}}}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-3>0\\ &\Leftrightarrow x<-\sqrt{3}\: \: \textrm{atau}\: \: x>\sqrt{3}\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>\: ^{.^{\frac{1}{2}}}\log 1\\ &\Leftrightarrow x^{2}-3<1\quad \left (\color{black}\textrm{karena basisnya}\: \: \displaystyle \frac{1}{2}<1 \right )\\ &\Leftrightarrow x^{2}-4<0\\ &\Leftrightarrow x^{2}-2^{2}>0\\ &\Leftrightarrow -2<x<2\\ &\textrm{Jadi},\: \: \color{red}-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &^{2}\log \left ( x^{2}-x \right )\leq 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-1\leq x\leq 2,\: x\neq 1\: \: \textrm{atau}\: \: x\neq 0\\ \color{red}\textrm{c}.&-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2\\ \textrm{d}.&-1< x\leq 0\: \: \textrm{atau}\: \: 1\leq x< 2\\ \textrm{e}.&-1\leq x\leq 0\: \: \textrm{atau}\: \: 1\leq x\leq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{2}\log \left ( x^{2}-x \right )\leq 1\\ &\textrm{Diketahui}\: \: \color{red}^{2}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-x>0\Leftrightarrow x(x-1)>0\\ &\Leftrightarrow x<0\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{2}\log \left ( x^{2}-x \right )\leq 1\\ &^{2}\log \left ( x^{2}-x \right )\leq \: ^{2}\log 2\\ &\Leftrightarrow x^{2}-x\leq 2\\ &\Leftrightarrow x^{2}-x-2\leq 0\\ &\Leftrightarrow (x+1)(x-2)\leq 0\\ &\Leftrightarrow -1\leq x\leq 2\\ &\textrm{Jadi},\: \: \color{red}-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left | \log (x+1) \right |> 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{b}.&x<-9\: \: \textrm{atau}\: \: x>9\\ \color{red}\textrm{c}.&-1<x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{d}.&-9< x<0,9\\ \textrm{e}.&-0,9<x<9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Ingat bahwa}\\ &\left | x \right |>A\Leftrightarrow \color{black}x<-A\: \: \textrm{atau}\: \: x>A,\: \: \color{red}A>0\\ &\Leftrightarrow \log (x+1)<-1\: \: \textrm{atau}\: \: \log (x+1)>1\\ &\color{red}\textrm{Syarat (1) buat keduanya},\: \: \color{red}f(x)>0\\ &(x+1)>0\Leftrightarrow x>-1\\ &\color{red}\textrm{Syarat (2)},\: \: \log \left ( x+1 \right )<-1\\ &\log (x+1)<\log 10^{-1}\\ &x+1<\displaystyle \frac{1}{10}\Leftrightarrow x<-\frac{9}{10}\\ &\color{red}\textrm{Syarat (3)},\: \: \log \left ( x+1 \right )> 1\\ &\log (x+1)>\log 10^{1}\\ &(x+1)>10\Leftrightarrow x>9\\ &\textrm{Jadi},\: \: \color{red}-1<x<-0,9\: \: \textrm{atau}\: \: x>9 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

 MENGINGAT KEMBALI

$\color{blue}\textrm{E. Sifat-Sifat Eksponen dan Logaritma}$

$\color{blue}\textrm{E.1  Persamaan Eksponen dan Logaritma}$

$\color{blue}\begin{array}{|l|l|}\hline  \color{red}\textrm{Eksponens}&\color{black}\textrm{Logaritma}\\\hline \displaystyle a^{n}=\underset{n\: \: faktor}{\underbrace{a\times a\times a\times \cdots \times a}}&\color{black}^{a}\log b=c\: \Rightarrow \: a^{c}=b\\\hline \bullet \quad a^{p}\times a^{q}=a^{p+q}&\bullet \quad \color{black}^{a}\log x+\: ^{a}\log y=\: ^{a}\log xy\\\hline \bullet \quad a^{p}: a^{q}=a^{p-q}&\bullet \quad \color{black}^{a}\log x-\: ^{a}\log y=\: ^{a}\log \displaystyle \frac{x}{y}\\\hline \bullet \quad \left ( a^{p} \right )^{q}=a^{p.q}&\bullet \quad ^{a}\log x=\: \displaystyle \frac{^{m}\log x}{^{m}\log a}\\\hline \bullet \quad \displaystyle \sqrt[q]{a^{p}}=\displaystyle a^{ \left (\frac{p}{q} \right )}&\bullet \quad ^{a}\log b\: \times \: ^{b}\log c=\: ^{a}\log c\\\hline \bullet \quad \left ( a\times b \right )^{p}=a^{p}\: \times \: b^{p}&\bullet \quad ^{a^{m}}\log b^{n}=\displaystyle \frac{n}{m}\times \: ^{a}\log b\\\hline \bullet \quad \left ( \displaystyle \frac{a}{b} \right )^{p}=\displaystyle \frac{\displaystyle a^{p}}{\displaystyle b^{p}}&\bullet \quad \displaystyle a^{\: {^{a}}\log b}=b\\\hline \bullet \quad a^{-p}=\displaystyle \frac{1}{\displaystyle a^{p}}&\bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{b}\log a}\\\hline \bullet \quad a^{0}=1,\: \: \: \: \: a\neq 0&\bullet \quad ^{a}\log 1=0\\\hline \bullet \quad a^{1}=1&\bullet \quad \color{black}^a\log a=1\\\hline \begin{cases} a,b\: \in \mathbb{R} \\ p,q\: \in \mathbb{Q} \end{cases}&\begin{cases} a\neq 0 &\\ a>0&(\textrm{bilangan pokok}) \\ x,y>0 & (\textrm{numerus}) \end{cases}\\\hline \end{array}$

Selanjutnya

$\color{blue}\begin{array}{|l|l|l|}\hline \textrm{No}&\qquad\textrm{Bentuk}&\qquad\qquad\color{red}\textrm{Syarat}\\\hline 1.&a^{f(x)}=1&a\neq 0,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 2.&a^{f(x)}=a^{p}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=p\\\hline 3.&a^{f(x)}=a^{g(x)}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=g(x)\\\hline 4.&a^{f(x)}=b^{f(x)}&a\neq 0,\: b\neq 0\: ,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 5.&f(x)^{g(x)}=1&\begin{cases} f(x)=1 & \\ g(x)=0, & \textrm{jika}\: \: f(x)\neq 0 \\ f(x)=-1, & \textrm{jika}\: \: g(x)=\: \textrm{genap} \end{cases}\\\hline 6.&\color{black}f(x)^{g(x)}=f(x)^{h(x)}&\color{red}\begin{cases} (i).\quad g(x)=h(x)& \\ (ii).\quad f(x)=1& \\ (iii).\quad f(x)=0,&g(x)>0,\: \: h(x)>0 \\ (iv).\quad f(x)=-1,&g(x)\: \textrm{dan}\: h(x)\: \: \\ &\color{black}\textrm{keduanya ganjil}\\ &\color{black}\textrm{atau genap} \end{cases}\\\hline 7.&g(x)^{f(x)}=h(x)^{f(x)}&\begin{cases} (i).\quad g(x) =h(x)& \\ (ii).\quad f(x)=0, & g(x)\neq 0,\: h(x)\neq 0 \end{cases}\\\hline 8.&\begin{aligned}&\color{black}A\left ( a^{f(x)} \right )^{2}\\ &\quad \color{black}+B\left ( a^{f(x)} \right )\\ &\qquad \color{black}+C=0 \end{aligned}&\color{red}a>0,\: \: a\neq 1\\\hline \end{array}$

$\color{blue}\textrm{E.2  Pertidaksamaan Eksponen dan Logaritma}$

Berikut sifat pertidaksamaan Eksponen

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad\qquad \color{black}0<a<1\\\hline a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\leq g(x)&a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\geq g(x)\\\hline a^{f(x)}< a^{g(x)}\Rightarrow f(x)< g(x)&a^{f(x)}< a^{g(x)}\Rightarrow f(x)> g(x)\\\hline a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\geq g(x)&a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\leq g(x)\\\hline a^{f(x)}> a^{g(x)}\Rightarrow f(x)> g(x)&a^{f(x)}> a^{g(x)}\Rightarrow f(x)< g(x)\\\hline \end{array}$

Untuk pertidaksamaan logaritma (dengan syarat  $\left (f(x)>0\: \: \textrm{dan}\: \: g(x)>0 \right )$ ) adalah sebagai berikut:

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad \color{black}0<a<1\\\hline \begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}&\begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}&\begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}\\\hline \end{array}$