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}$

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{l}\\ 36.& \text{Persamaan}{}^x\log 2+{}^x\log (3x-4)=2\\ & \text{mempunyai akar}{x}_1\text{dan}{x}_2,\text{maka}\\ & \text{nilai}{x}_1+x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 6\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & {}^x\log 2+{}^x\log (3x-4)=2\\ & \iff {}^x\log 2(3x-4)=2\\ & \iff {}^x\log 6x-8=2\\ & \iff 6x-8=x^2\\ & \iff x^2-6x+8=0,\text{dengan}\{\begin{array}{ll}a& =1\\ b& =-6\\ c& =8\end{array}\\ & \iff x_1+x_2=-{\displaystyle \frac{b}{a}}\\ & \iff x_1+x_2=-{\displaystyle \frac{-6}{1}=6}\\ & \text{Alternatif 2}\\ & \iff x^2-6x+8=0\\ & \iff (x-2)(x-4)\\ & \iff x_1=2\text{atau}{x}_2=4\\ & \iff x_1+x_2=2+4=6\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Jika}{x}_1\text{dan}{x}_2\text{memenuhi}\\ & (\log x)(2\log x-3)=\log 100\\ & \text{maka}{x}_1\times x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 100\\ \text{b}.& 10\sqrt{10}\\ \text{c}.& \sqrt{10}\\ \text{d}.& -\sqrt{10}\\ \text{e}.& -10\sqrt{10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& (\log x)(2\log x-3)=\log 100\\ & \iff (\log x)(2\log x-3)=2\\ & \iff 2{\log }^{2}x-3\log x-2=0\{\begin{array}{ll}a& =2\\ b& =-3\\ c& =-2\end{array}\\ & \iff \log x_1+\log x_2=-{\displaystyle \frac{-3}{2}=\frac{3}{2}}\\ & \iff \log (x_1\times x_2)=1{\displaystyle \frac{1}{2}}\\ & \iff (x_1\times x_2)={10}^{1\frac{1}{2}}=10\sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Persamaan}\\ & {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{mempunyai dua akar yaitu}{x}_1\text{dan}{x}_2\\ & \text{Nilai}{x}_1\times x_2=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -5\\ \text{c}.& 2\\ \text{d}.& 5\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & {10}^{2^{{}^2}\log x}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{adalah persamaan kuadrat dalam}{10}^{{}^{{}^2}\log x}\\ & \text{Misalkan}p={10}^{{}^{{}^2}\log x},\text{maka persamaan}\\ & \text{menjadi}{p}^2-7p+10=0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =10\end{array}\\ & \text{Karena nilai}{p}_1\times p_2={\displaystyle \frac{c}{a}\text{maka}}\\ & {10}^{{}^{{}^2}\log x_1}\times {10}^{{}^{{}^2}\log x_2}={\displaystyle \frac{10}{1}=10}\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}=10\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}={10}^1\\ & \iff {}^2\log x_1+{}^2\log x_2=1\\ & \iff {}^2\log x_1\times x_2=1\\ & \iff x_1\times x_2=2^1=2\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Nilai}x\text{yang memenuhi}\\ & {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & {}^x\log (x+12)-{}^x\log 4^3=-1\\ & {}^x\log {\displaystyle \frac{x+12}{64}=-1}\\ & {\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}}\\ & x+12={\displaystyle \frac{64}{x}}\\ & x^2+12x-64=0\\ & (x+16)(x-4)=0\\ & x=-16\text{atau}x=4\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\text{dan}6\\ \text{b}.& -2\text{dan}6\\ \text{c}.& -1\\ \text{d}.& -2\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & {}^x\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & {}^x\log (2x-3)(x+2)=1+\log (x+6)\\ & {}^x\log {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=1}\\ & {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=x^1}\\ & (2x^2+x-6)=x^2+6x\\ & x^2-5x-6=0\\ & (x+1)(x-6)=0\\ & x=-1\text{atau}x=6\end{array}\end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

 $\begin{array}{l}\\ 32.& \text{Agar}\log (x^2-1)<0\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& -1<x<1\\ \text{b}.& -\sqrt{2}<x<\sqrt{2}\\ \text{c}.& x<-1\text{atau}x>1\\ \text{d}.& x<-\sqrt{2}\text{atau}x>\sqrt{2}\\ \text{e}.& -\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \log (x^2-1)<0\\ & \text{Diketahui}\log f(x)<0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-1>0\\ & \iff x<-1\text{atau}x>1\\ & \text{Syarat (2)},\log (x^2-1)<0\\ & \log (x^2-1)<\log 1\\ & \iff x^2-1<1\\ & \iff x^2-2<0\\ & \iff x^2-{\left(\sqrt{2}\right)}^2<0\\ & \iff -\sqrt{2}<x<\sqrt{2}\\ & \text{Jadi},-\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Himpunan penyelesaian dari}\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\}\\ \text{b}.& \{x|-\sqrt{3}<x<-1\text{atau}\sqrt{3}<x<2\}\\ \text{c}.& \{x|-2<x<-\sqrt{3}\}\\ \text{d}.& \{x|-2<x<-\sqrt{3}\}\\ \text{e}.& \{x|\sqrt{3}<x<2\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & \text{Diketahui}{}^{{.}^{\frac{1}{2}}}\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-3>0\\ & \iff x<-\sqrt{3}\text{atau}x>\sqrt{3}\\ & \text{Syarat (2)},{}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>{}^{{.}^{\frac{1}{2}}}\log 1\\ & \iff x^2-3<1\left(\text{karena basisnya}{\displaystyle \frac{1}{2}<1}\right)\\ & \iff x^2-4<0\\ & \iff x^2-2^2>0\\ & \iff -2<x<2\\ & \text{Jadi},-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai}x\text{yang memenuhi}\\ & {}^2\log (x^2-x)\le 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<0\text{atau}x>1\\ \text{b}.& -1\le x\le 2,x\ne 1\text{atau}x\ne 0\\ \text{c}.& -1\le x<0\text{atau}1<x\le 2\\ \text{d}.& -1<x\le 0\text{atau}1\le x<2\\ \text{e}.& -1\le x\le 0\text{atau}1\le x\le 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^2\log (x^2-x)\le 1\\ & \text{Diketahui}{}^2\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-x>0\iff x(x-1)>0\\ & \iff x<0\text{atau}x>1\\ & \text{Syarat (2)},{}^2\log (x^2-x)\le 1\\ & {}^2\log (x^2-x)\le {}^2\log 2\\ & \iff x^2-x\le 2\\ & \iff x^2-x-2\le 0\\ & \iff (x+1)(x-2)\le 0\\ & \iff -1\le x\le 2\\ & \text{Jadi},-1\le x<0\text{atau}1<x\le 2\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai}x\text{yang memenuhi}\\ & |\log (x+1)|>1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<-0,9\text{atau}x>9\\ \text{b}.& x<-9\text{atau}x>9\\ \text{c}.& -1<x<-0,9\text{atau}x>9\\ \text{d}.& -9<x<0,9\\ \text{e}.& -0,9<x<9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Ingat bahwa}\\ & \left|x\right|>A\iff x<-A\text{atau}x>A,A>0\\ & \iff \log (x+1)<-1\text{atau}\log (x+1)>1\\ & \text{Syarat (1) buat keduanya},f(x)>0\\ & (x+1)>0\iff x>-1\\ & \text{Syarat (2)},\log (x+1)<-1\\ & \log (x+1)<\log {10}^{-1}\\ & x+1<{\displaystyle \frac{1}{10}\iff x<-\frac{9}{10}}\\ & \text{Syarat (3)},\log (x+1)>1\\ & \log (x+1)>\log {10}^1\\ & (x+1)>10\iff x>9\\ & \text{Jadi},-1<x<-0,9\text{atau}x>9\end{array}\end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

 MENGINGAT KEMBALI

$\text{E. Sifat-Sifat Eksponen dan Logaritma}$

$\text{E.1 Persamaan Eksponen dan Logaritma}$

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

Selanjutnya

$\begin{array}{|lll|}\hline \text{No}& \text{Bentuk}& \text{Syarat}\\ 1.& a^{f(x)}=1& a\ne 0,\text{maka}f(x)=0\\ 2.& a^{f(x)}=a^p& a>0,a\ne 1,\text{maka}f(x)=p\\ 3.& a^{f(x)}=a^{g(x)}& a>0,a\ne 1,\text{maka}f(x)=g(x)\\ 4.& a^{f(x)}=b^{f(x)}& a\ne 0,b\ne 0,\text{maka}f(x)=0\\ 5.& f(x{)}^{g(x)}=1& \{\begin{array}{ll}f(x)=1& \\ g(x)=0,& \text{jika}f(x)\ne 0\\ f(x)=-1,& \text{jika}g(x)=\text{genap}\end{array}\\ 6.& f(x{)}^{g(x)}=f(x{)}^{h(x)}& \{\begin{array}{ll}(i).g(x)=h(x)& \\ (ii).f(x)=1& \\ (iii).f(x)=0,& g(x)>0,h(x)>0\\ (iv).f(x)=-1,& g(x)\text{dan}h(x)\\ & \text{keduanya ganjil}\\ & \text{atau genap}\end{array}\\ 7.& g(x{)}^{f(x)}=h(x{)}^{f(x)}& \{\begin{array}{ll}(i).g(x)=h(x)& \\ (ii).f(x)=0,& g(x)\ne 0,h(x)\ne 0\end{array}\\ 8.& \begin{array}{rl}& A{\left(a^{f(x)}\right)}^2\\ & +B\left(a^{f(x)}\right)\\ & +C=0\end{array}& a>0,a\ne 1\\ \hline\end{array}$

$\text{E.2 Pertidaksamaan Eksponen dan Logaritma}$

Berikut sifat pertidaksamaan Eksponen

$\begin{array}{|ll|}\hline a>1& 0<a<1\\ a^{f(x)}\le a^{g(x)}\Rightarrow f(x)\le g(x)& a^{f(x)}\le a^{g(x)}\Rightarrow f(x)\ge g(x)\\ a^{f(x)}<a^{g(x)}\Rightarrow f(x)<g(x)& a^{f(x)}<a^{g(x)}\Rightarrow f(x)>g(x)\\ a^{f(x)}\ge a^{g(x)}\Rightarrow f(x)\ge g(x)& a^{f(x)}\ge a^{g(x)}\Rightarrow f(x)\le g(x)\\ a^{f(x)}>a^{g(x)}\Rightarrow f(x)>g(x)& a^{f(x)}>a^{g(x)}\Rightarrow f(x)<g(x)\\ \hline\end{array}$

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

$\begin{array}{|ll|}\hline a>1& 0<a<1\\ \begin{array}{rl}& {}^a\log f(x)\le {}^a\log g(x)\\ & \Rightarrow f(x)\le g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)\le {}^a\log g(x)\\ & \Rightarrow f(x)\ge g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)<{}^a\log g(x)\\ & \Rightarrow f(x)<g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)<{}^a\log g(x)\\ & \Rightarrow f(x)>g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)\ge {}^a\log g(x)\\ & \Rightarrow f(x)\ge g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)\ge {}^a\log g(x)\\ & \Rightarrow f(x)\le g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)>{}^a\log g(x)\\ & \Rightarrow f(x)>g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)>{}^a\log g(x)\\ & \Rightarrow f(x)<g(x)\end{array}\\ \hline\end{array}$

Contoh Soal 6 Matriks

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 2004}})\\ & \text{Nilai-nilai}x\text{agar matriks}\\ & \left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4\text{atau}5\\ \text{b}.& -2\text{atau}2\\ \text{c}.& -4\text{atau}5\\ \text{d}.& -6\text{atau}4\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{supaya matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers},\text{maka}\\ & \text{determinan matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)=0\\ & \text{Sehingga}\\ & \left|\begin{array}{cc}5x& 5\\ 4& x\end{array}\right|=0\\ & \iff 5x^2-20=0\\ & \iff x^2=4\\ & \iff x=\pm 2\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{UM UGM 2005}})\\ & \text{Matriks}\left(\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right)\\ & \text{tidak memiliki invers untuk}\\ & \text{nilai}x=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-2\\ \text{b}.& -1\text{atau}0\\ \text{c}.& -1\text{atau}1\\ \text{d}.& -1\text{atau}2\\ \text{e}.& 1\text{atau}2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Mirip dengan pembahasan no. 26}\\ & \begin{array}{rl}& \text{Nilai}\left|\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right|=0\\ & \iff x-x^2-(-2)=0\\ & \iff 2+x-x^2=0\\ & \iff x^2-x-2=0\\ & \iff (x-2)(x+1)=0\\ & \iff x=2\text{atau}x=-1\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{Mat Das SIMAK UI 2014}})\\ & \text{Jika matriks}\text{A}\text{adalah invers}\\ & \text{dari matriks}{\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\text{dan}}\\ & \text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\text{maka nilai}2x+y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{10}{3}}\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& {\displaystyle \frac{9}{7}}\\ \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Misalkan diketahui matriks}\\ & \text{B}={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right),}\\ & \text{maka}\text{A}={\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\\ & \text{selanjutnya}\text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)=A^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right),\\ & \text{ingat bahwa}{\left({\mathbf{\text{A}}}^{-1}\right)}^{-1}=\mathbf{\text{A}}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\left({\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\right)}^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}\left(\begin{array}{c}-1-9\\ 4+15\end{array}\right)=\left(\begin{array}{c}{\displaystyle -\frac{10}{3}}\\ {\displaystyle \frac{19}{3}}\end{array}\right)}\\ & 2x+y=2(-{\displaystyle \frac{10}{3}})+\frac{19}{3}\\ & ={\displaystyle \frac{-20+19}{3}=-{\displaystyle \frac{1}{3}}}\end{array}\end{array}$