Latihan Soal 11 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{ll}\\ 96.&\textrm{Himpunan penyelesaian dari}\\ &2x-1<x+1<3-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x<1 \right \}\\ \textrm{b}.&\left \{ x|x<2 \right \}\\ \textrm{c}.&\left \{ x|1<x<2 \right \}\\ \textrm{d}.&\left \{ x|x>2 \right \}\\ \textrm{e}.&\left \{ x|x>1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x-1<x}}\, +\, \underset{\textrm{B}}{\underbrace{1<3-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x-1<x+1\\ &\qquad x<2\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+1<3-x\\ &\qquad 2x<2\\ &\qquad x<1\: ................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 97.&\textrm{Himpunan penyelesaian dari}\\ &2x+1<x<1-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-2 \right \}\\ \color{red}\textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-1<x<-2 \right \}\\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{1}{2}\right \}\\ \textrm{e}.&\left \{ x|x<1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x+1<x}} \underset{\textrm{B}}{\underbrace{\: <1-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x+1<x\\ &\qquad x<-1\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x<1-x\\ &\qquad 2x<1\\ &\qquad x<\displaystyle \frac{1}{2}\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 98.&\textrm{Himpunan penyelesaian dari}\\ &3x+14\leq x+5<3x-1\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-3 \right \}\\ \textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-3<x<-1 \right \}\\ \textrm{d}.&\left \{ x|x>3\right \}\\ \color{red}\textrm{e}.&\left \{\: \: \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{\textrm{A}}{\underbrace{4x+14\leq x}} \underset{\textrm{B}}{\underbrace{\, +\, 5 <3x-1}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 4x+14\leq x+5\\ &\qquad 3x\leq -9\\ &\qquad x\leq -3\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+5<3x-1\\ &\qquad -2x<-6\\ &\qquad x>3\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}\: \color{red}\textrm{tidak ada} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 99.&\textrm{Jika}\: \: \displaystyle \frac{1}{x}<2021\: \: \textrm{dan}\: \: \displaystyle \frac{1}{x}>2020\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2020<x<2021\\ \textrm{b}.&-2021<x<-2020\\ \textrm{c}.&\displaystyle \frac{1}{2020}<x<\displaystyle \frac{1}{2021}\\ \textrm{d}.&x<\displaystyle \frac{1}{2021}\: \: \textrm{dan}\: \: x>\displaystyle \frac{1}{2020}\\ \textrm{e}.&\textrm{semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui}:\: \color{black}\displaystyle \frac{1}{x}<2021\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}>2020\\ &\textrm{Dapat ditulis ulang dengan}\\ &\color{black}2020<\displaystyle \frac{1}{x}\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}<2021\\ &\textrm{Jika digabung menjadi}\\ &\color{black}2020<\displaystyle \frac{1}{x}<\color{black}2021 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 100.&\textrm{Jika}\: \: a>0\: \: \textrm{dan}\: \: b<0\: ,\: \textrm{maka}\\ &\textrm{pernyataan berikut yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a+b>0\\ \textrm{b}.&a-b<0\\ \textrm{c}.&a^{2}-b^{2}<0\\ \color{red}\textrm{d}.&\displaystyle \frac{a}{b}<0\\ \textrm{e}.&ab>0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{purple}\textrm{Cukup Jelas saat}\: \: \color{red}\displaystyle \frac{a}{b}=\frac{+}{-}=-<0 \end{array}$.

$\begin{array}{ll}\\ 101.&\textrm{Jika}\: \: 0<x+y<3\: \: \textrm{dan}\: \: 1<x-y<2\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<5\\ \textrm{b}.&\left | x \right |<1\\ \textrm{c}.&x<1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{2}<x<\frac{5}{2}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{array}{llll}\\ 0<x+y<&3&\\ 1<x-y<&2&+\\\hline \: \: 1<2x<&5&\color{black}\textrm{dibagi 2 semuanya}\\ \quad \displaystyle \frac{1}{2}<x<&\displaystyle \frac{5}{2}&\: .....\color{red}(4)\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 102.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui P, Q, dan R memancing ikan.}\\ & \textrm{Jika hasil Q lebih sedikit dari hasil R}\\ & \textrm{sedangkan jumlah hasil P dan Q lebih }\\ & \textrm{banyak dari pada dua kali hasil R,}\\ &\textrm{maka yang terbanyak mendapat ikan}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{P dan R}\\ \textrm{b}.&\textrm{P dan Q}\\ \color{red}\textrm{c}.&\textrm{P}\\ \textrm{d}.&\textrm{Q}\\ \textrm{e}.&\textrm{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\color{black}\textrm{Diketahui}:\\ &\bullet \: Q< R\: ...............\color{red}(1)\\ &\bullet \: P+Q> 2R\: ......\color{red}(2)\\ &\textrm{Sehingga untuk persamaan}\: \: \color{black}(1)\: \&\: (2)\\ &\begin{array}{llll}\\ \qquad\qquad R>&Q&\\ \qquad P+Q>&2R&+\\\hline P+Q+R>&Q+2R&\\\\ \qquad\quad\quad P>&R\: ......\color{red}(3)\\ \end{array}\\ &\textrm{dari} \: \color{red}(1)\: \color{purple}\textrm{dan}\: \color{red}(3)\: \color{purple}\textrm{diperoleh bahwa}\\ &Q<R< P\\ &\textrm{Jadi, yang terbanyak mendapat ikan}\\ &\color{red}\textrm{adalah P} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 103.&\textrm{Jika}\: \: a>0,\: b>0,\: \: \textrm{dan}\: \: a>b,\: \: \textrm{maka}\\ &\textrm{pernyataan berikut yang salah adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{a}>\frac{1}{b}\\ \textrm{b}.&a^{2}>b^{2}\\ \textrm{c}.&a^{3}>b^{3}\\ \textrm{d}.&\sqrt{a}>\sqrt{b}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{purple}\begin{aligned}&a>0,\: b>0,\: \: \textrm{dan}\: \: a>b\\ &\color{red}\textrm{Maka}\\ &\displaystyle \frac{a}{1}>\frac{b}{1},\: \: \textrm{jika dibalik}\\ &\color{red}\textrm{menjadi}\\ &\displaystyle \frac{1}{a}<\frac{1}{b} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 104.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan real, maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{2}+b^{2}\geq 2ab\\ \textrm{b}.&a^{2}+b^{2}> 2ab\\ \textrm{c}.&a^{2}+b^{2}< 2ab\\ \textrm{d}.&a^{2}+b^{2}\leq 2ab\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&a,b\in \mathbb{R}\\ &\color{red}\textrm{Maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab \end{aligned} \end{array}$

$\begin{array}{ll}\\ 105.&\textrm{Pernyataan berikut yang tepat untuk}\\ &\textrm{untuk seluruh}\: \: x\: \: \textrm{positif adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x+\displaystyle \frac{1}{x}<2\\ \textrm{b}.&x+\displaystyle \frac{1}{x}\leq 2\\ \textrm{c}.&x+\displaystyle \frac{1}{x}>2\\ \color{red}\textrm{d}.&x+\displaystyle \frac{1}{x}\geq 2\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&a,b\in \mathbb{R},\: \: a>0,\: b>0\\ &\color{red}\textrm{Mirip dengan pembahasan}\\ &\color{red}\textrm{no.19, maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab\\ &\color{black}\textrm{Saat}\: \: a=\sqrt{x},\: \: b=\displaystyle \frac{1}{\sqrt{x}}\\ &\textrm{menyebabkan}\\ &\left ( \sqrt{x} \right )^{2}+\left ( \displaystyle \frac{1}{\sqrt{x}} \right )^{2}\geq 2.\sqrt{x}.\displaystyle \frac{1}{\sqrt{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2\sqrt{x.\displaystyle \frac{1}{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2 \end{aligned} \end{array}$

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