Latihan Soal 4 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 31.& \text{Diketahui}f(x)=2\cos x-2020\\ & \text{Turunan pertama fungsi}f(x)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\sin x\\ \text{b}.& -2\sin x\\ \text{c}.& -2\sin x-2020x\\ \text{d}.& 2{\sin }^{2}x\\ \text{e}.& 2\cos x-2020x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& =2\cos x-2020\\ f^{\prime }(x)& =-2\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Jika}{f}^{\prime }(x)\text{adalah turunan pertama dari}\\ & \text{fungsi}f(x)={\sin }^{7}x,\text{maka}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 7{\cos }^{6}x\\ \text{b}.& 7{\cos }^{7}x\\ \text{c}.& 7{\sin }^{6}x\cos x\\ \text{d}.& 7{\cos }^{6}x\sin x\\ \text{e}.& 7{\cos }^{6}x{\sin }^{6}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\sin }^{7}x\\ \text{guna}& \text{kan formula}\\ & y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ f^{\prime }(x)& =7{\sin }^{6}x(\cos x)=7{\sin }^{6}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Turunan pertama fungsi}g(x)=-5{\sin }^{3}x\\ & \text{adalah}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -5{\sin }^{2}x\cos x\\ \text{b}.& -5{\sin }^2{\cos }^{2}x\\ \text{c}.& -15{\sin }^{2}x\cos x\\ \text{d}.& -15{\cos }^{3}x\\ \text{e}.& -15{\sin }^{4}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}g(x)& =-5{\sin }^{3}x\\ \text{guna}& \text{kan formula}\\ & y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ g^{\prime }(x)& =-5(3{\sin }^{2}x)(\cos x)=-15{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Jika}h(x)=4x^3+\sin x+\cos x\\ & \text{maka}{h}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 12x^2+\cos x-\sin x\\ \text{b}.& 12x^2-\cos x+\sin x\\ \text{c}.& 4x^3-\cos x-\sin x\\ \text{d}.& 4x^3-\sin x-\cos x\\ \text{e}.& 12x^3+\cos x+\sin x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}h(x)& =4x^3+\sin x+\cos x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ \text{pada}& \text{fungsi aljabarnya, yaitu}:y=4x^3\Rightarrow y^{\prime }=12x^2\\ \text{seda}& \text{ngkan fungsi transendennya mengikuti}\\ \text{turu}& \text{nan fungsi trigonometri biasa. Sehingga}\\ f^{\prime }(x)& =12x^2+\cos x-\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Jika}p(x)=-{\cos }^{4}x,\text{maka nilai}\\ & \text{maka}{p}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& \sqrt{3}\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{4}\sqrt{3}}\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}p(x)& =-{\cos }^{4}x\\ p^{\prime }(x)& =-4{\cos }^{3}x.(-\sin x)=4{\cos }^{3}x\sin x\\ p^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)& =4{\cos }^3\left({\displaystyle \frac{\pi }{3}}\right).\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & =4{\cos }^3{60}^{\circ }\times \sin {60}^{\circ }\\ & =4{\left({\displaystyle \frac{1}{2}}\right)}^3\times \left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{4}{16}\sqrt{3}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 36.& \text{Turunan pertama}q(x)={\sin }^{2}x+{\cos }^{2}x\\ & \text{adalah}{q}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\cos }^{2}x-{\sin }^{2}x\\ \text{b}.& 2{\cos }^{2}x-2{\sin }^{2}x\\ \text{c}.& \cos x-\sin x\\ \text{d}.& 2\cos x-2\sin x\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}q(x)& ={\sin }^{2}x+{\cos }^{2}x\\ \text{guna}& \text{kan formula identitas}:{\sin }^{2}x+{\cos }^{2}x=1\\ \text{Sehi}& \text{ngga soal di atas dapat dituliskan menjadi}\\ q(x)& =1,\text{maka}\\ q^{\prime }(x)& =0\\ \text{inga}& \text{t bahwa}y=a\Rightarrow {\displaystyle \frac{dy}{dx}=0}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.& -{\displaystyle \frac{1}{2}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dari}& \text{soal diketahui}:\\ f(x)& =\sin {\displaystyle \frac{\pi }{3}}\\ \text{Nila}& \text{i dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ \text{arti}& \text{nya bermakna, berapkah}{f}^{\prime }\left(x\right)?\\ \text{maka}& \\ f^{\prime }\left(x\right)& =0\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Jika}f(x)=8x-{\sin }^{3}x,\\ & \text{maka nilai}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4x^2-3{\cos }^{2}x\\ \text{b}.& 8x-3{\sin }^{2}x\cos x\\ \text{c}.& 8-3{\sin }^{2}x\cos x\\ \text{d}.& 8+{\sin }^{2}x\cos x\\ \text{e}.& 3{\sin }^{2}x\cos x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui dari soal}f(x)=8x-{\sin }^{3}x\\ \text{maka}& \text{nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=f^{\prime }(x)}\\ f^{\prime }(x)& =8-3{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Turunan pertama fungsi}f(x)=\sqrt{\sin x},\\ & \text{adalah}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2\sqrt{\sin x}}}\\ \text{b}.& {\displaystyle \frac{\cos x}{\sqrt{\sin x}}}\\ \text{c}.& {\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\\ \text{d}.& -{\displaystyle \frac{\sin x}{2\sqrt{\cos x}}}\\ \text{e}.& {\displaystyle \frac{2\cos x}{\sqrt{\sin x}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}f(x)=\sqrt{\sin x}={\sin }^{{.}^{\frac{1}{2}}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}({\sin }^{{.}^{-\frac{1}{2}}}x).(\cos x)}\\ & ={\displaystyle \frac{\cos x}{2{\sin }^{{.}^{\frac{1}{2}}}x}}\\ & ={\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Jika}{g}^{\prime }(x)\text{adalah turunan pertama}\\ & \text{fungsi}g(x)\text{dengan}g(x)=5{\tan }^{2}x,\\ & \text{maka}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 10{\cos }^{2}x\sin x\\ \text{b}.& 10{\sin }^{2}x\cos x\\ \text{c}.& {\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\\ \text{d}.& {\displaystyle \frac{10{\cos }^{3}x}{\sin x}}\\ \text{e}.& {\displaystyle \frac{10}{{\sin }^{2}x-{\cos }^{2}x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}g(x)=5{\tan }^{2}x\\ g^{\prime }(x)& =5(2\tan x).({\sec }^{2}x)\\ & =10\tan x\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & =10\left({\displaystyle \frac{\sin x}{\cos x}}\right)\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & ={\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\end{array}\end{array}$.

Latihan Soal 3 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 21.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}\\ \\ \text{c}.{\displaystyle 0}\\ \\ \text{d}.{\displaystyle 3}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2}{x^2}+\frac{3x}{x^2}+\frac{4}{x^2}}}{{\displaystyle \frac{3x^2}{x^2}+\frac{2x}{x^2}+\frac{3}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{1+{\displaystyle \frac{3}{x}+\frac{4}{x^2}}}{3+{\displaystyle \frac{2}{x}+\frac{3}{x^2}}}}\\ & ={\displaystyle \frac{1+0+0}{3+0+0}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 3}\\ \\ \text{b}.{\displaystyle 1}\\ \\ \text{c}.{\displaystyle \frac{1}{3}}\\ \\ \text{d}.{\displaystyle 0}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x}{x^2}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3}{x}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & ={\displaystyle \frac{0}{9+0+0}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2}& & \\ \text{b}.{\displaystyle -1}\\ \text{c}.{\displaystyle -\frac{1}{2}}\\ \text{d}.{\displaystyle 0}\\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})\\ & =\mathrm{\infty }-\mathrm{\infty }=\mathbf{\text{tidak diperbolehkan}}\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})\\ & ={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{-2-2}{2\sqrt{1}}}\\ & =\frac{-4}{2}\\ & =-2\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -10}\\ \text{b}.{\displaystyle -5}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 5}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2}-\sqrt{x^2-10x})\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})\\ & ={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{0-(-10)}{2\sqrt{1}}}\\ & =\frac{10}{2}\\ & =5\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \mathrm{\infty }}\\ \text{b}.& {\displaystyle 5}\\ \text{c}.& {\displaystyle \sqrt{3}}\\ \text{d}.& 1\\ \text{e}.& {\displaystyle 0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}}& =\underset{u\to 0}{\text{Lim}}5\tan {\displaystyle u}\\ & =5\tan 0\\ & =5.\mathrm{\infty }\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2{\sin }^2\left({\displaystyle \frac{ab}{x}}\right)\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle ab}\\ \text{b}.& {\displaystyle a^{2}b}\\ \text{c}.& {\displaystyle a{b}^2}\\ \text{d}.& {\displaystyle (ab{)}^2}\\ \text{e}.& {\displaystyle \frac{1}{(ab{)}^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2\sin {\displaystyle \frac{ab}{x}}& =\underset{u\to 0}{\text{Lim}}{\left({\displaystyle \frac{1}{u}}\right)}^2{\sin }^2{\displaystyle abu}\\ & =\underset{u\to 0}{\text{Lim}}\left({\displaystyle \frac{{\sin }^{2}abu}{u^2}}\right)\\ & =(ab{)}^2\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}}& =100\times \underset{0}{\underbrace{\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{x}}}}\\ & =100\times 0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Nilai dari}\\ & \underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(-x)}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(x)}}\\ & =-\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (x)\cos \frac{1}{(x)}}\\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{1}{u}\cos u}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{\cos u}{u}}\\ & =-{\displaystyle \frac{1}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Asimtot tegak dari fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x=2\text{dan}x=4\\ \text{b}.& x=2\text{dan}x=3\\ \text{c}.& x=3\text{dan}x=4\\ \text{d}.& x=3\text{saja}\\ \text{e}.& x=2\text{saja}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{terjadi saat penyebut}=0.\\ & \text{Sehingga}{x}^2-5x+6=0\\ & \iff (x-2)(x-3)=0,\text{maka}\\ & x=2\text{atau}x=3\\ & \therefore \text{asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{adalah}x=2\text{dan}x=3\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Asimtot datar dari fungsi}\\ & g(x)={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& y=-3\\ \text{b}.& y=-1\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Asim}& \text{tot datar dari fungsi}\\ g(x)& ={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{untuk}}\\ g(x)& ={\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\text{terjadi saat}}\\ y& ={\displaystyle \frac{6}{-2}=-3}\\ & \mathbf{\text{atau dapat juga dicari}}\mathbf{\text{dengan}}\\ y& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{6-{\displaystyle \frac{8}{x}+\frac{2}{x^2}}}{-2+{\displaystyle \frac{5}{x}-\frac{2}{x^2}}}}\\ & ={\displaystyle \frac{6-0+0}{-2+0-0}}\\ & ={\displaystyle \frac{6}{-2}}\\ & =-3\end{array}\end{array}$

Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 11.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{{\displaystyle \frac{1}{x}(\sqrt{4x-2020}+\sqrt{8x+2021})}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{4x}{x^2}-\frac{2020}{x^2}}}+\sqrt{{\displaystyle \frac{8x}{x^2}+\frac{2021}{x^2}}})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{4}{x}-{\displaystyle \frac{2020}{x}}}}+\sqrt{{\displaystyle \frac{8}{x}+{\displaystyle \frac{2021}{x}}}})}}\\ & ={\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{-4}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle 1}\\ \text{b}.{\displaystyle 1}\\ \text{c}.{\displaystyle 2}\\ \text{d}.{\displaystyle 4}\\ \text{e}.{\displaystyle 8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\times {\displaystyle \frac{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x^2+3x-(4x^2-5x)}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x+5x}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}}}\\ & ={\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}}\\ & ={\displaystyle \frac{8}{4}}\\ & =2\end{array}\end{array}$.

$\begin{array}{rl}& \text{ada cara lain yang lebih sede}\text{rhana, yaitu:}\\ .& \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\\ & \{\begin{array}{ll}a& =4\\ b& =3\\ p& =-4\end{array}\\ & \text{Jika}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q}={\displaystyle \frac{b-p}{2\sqrt{a}}}\\ & \text{Sehingga}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}={\displaystyle \frac{3-(-5)}{2\sqrt{4}}}\\ & ={\displaystyle \frac{8}{2.2}}\\ & =2\end{array}$.

$\begin{array}{l}\\ 13.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}+\sqrt{4x^2-5x}=....\\ & \begin{array}{l}\\ \text{a}.\mathrm{\infty }\\ \text{b}.{\displaystyle 1}\\ \text{c}.{\displaystyle 2}\\ \text{d}.{\displaystyle 4}\\ \text{e}.{\displaystyle 8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}+\sqrt{4x^2-5x}\\ & =\sqrt{\mathrm{\infty }}+\sqrt{\mathrm{\infty }}=\mathrm{\infty }\end{array}$.

$\begin{array}{l}\\ 14.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\times {\displaystyle \frac{(\sqrt{3x+1}+\sqrt{3x-2})}{(\sqrt{3x+1}+\sqrt{3x-2})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}\frac{(3x+1)-(3x-2)}{(\sqrt{3x+1}+\sqrt{3x-2})}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3}{(\sqrt{3x+1}+\sqrt{3x-2})}\times \frac{{\displaystyle \frac{1}{\sqrt{x}}}}{{\displaystyle \frac{1}{\sqrt{x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3}{\sqrt{x}}}}{(\sqrt{{\displaystyle \frac{3x}{x}+\frac{1}{x}}}+\sqrt{{\displaystyle \frac{3x}{x}-\frac{2}{x}}})}}\\ & ={\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \\ \text{c}.& {\displaystyle \frac{3}{2}}\\ \\ \text{d}.& {\displaystyle \frac{7}{2}}\\ \\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\\ & \times {\displaystyle \frac{(\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7})}{(\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(4x^2+6x+8)-(4x^2-8x+7)}{\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{14x+1}{\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{14+{\displaystyle \frac{1}{x}}}{\sqrt{{\displaystyle \frac{4x^2}{x^2}+\frac{6x}{x^2}+\frac{8}{x^2}}}+\sqrt{{\displaystyle \frac{4x^2}{x^2}-\frac{8x}{x^2}+\frac{7}{x^2}}}}}\\ & ={\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}}\\ & ={\displaystyle \frac{14}{2+2}=\frac{7}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 16.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\\ & \times {\displaystyle \frac{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(2x^2+3x-1)-(x^2-5x+3)}{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+8x-4}{(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})}\times \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2}{x^2}+\frac{8x}{x^2}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{2x^2}{x^4}+\frac{3x}{x^4}-\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{x^2}{x^4}-\frac{5x}{x^4}+\frac{3}{x^4}}}}}\\ & ={\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}}\\ & ={\displaystyle \frac{1}{0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\\ & \times {\displaystyle \frac{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^2+3x+1)-(3x^2+2x+5)}{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-2x^2+x-4}{(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})}\times \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{-2x^2}{x^2}+\frac{x}{x^2}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{x^2}{x^4}+\frac{3x}{x^4}+\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{3x^2}{x^4}+\frac{2x}{x^4}+\frac{5}{x^4}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-2+{\displaystyle \frac{1}{x}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{1}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{3}{x^2}+\frac{2}{x^3}+\frac{5}{x^4}}}}}\\ & ={\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}}\\ & ={\displaystyle \frac{-2}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}((3x-2)-\sqrt{9x^2-2x+5})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \\ \text{b}.& -{\displaystyle \frac{5}{3}}\\ \\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \\ \text{d}.& 1\\ \\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}((3x-2)-\sqrt{9x^2-2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(3x-2{)}^2}-\sqrt{9x^2-2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(9x^2-12x+4}-\sqrt{9x^2-2x+5})\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(a{x}^2+bx+c}-\sqrt{p{x}^2+qx+r})\\ & \text{Jika dikerjakan dengan rumus singkat}\\ & \text{maka}\{\begin{array}{c}a=p=3\\ b=-12\\ q=-2\end{array}\\ & ={\displaystyle \frac{b-q}{2\sqrt{a}}}\\ & ={\displaystyle \frac{-12-(-2)}{2\sqrt{9}}}\\ & ={\displaystyle \frac{-10}{6}}\\ & =-\frac{5}{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{4}{9}}\\ \\ \text{c}.& {\displaystyle \frac{1}{2}}\\ \\ \text{d}.& 1\\ \\ \text{e}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\times {\displaystyle \frac{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^2}}}{\sqrt{9-\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}}\\ & ={\displaystyle \frac{2-1}{3}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{9}}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 0}& \\ \\ \text{d}.{\displaystyle 1}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x^4}{x^4}+\frac{2x^3}{x^4}-\frac{5x}{x^4}+\frac{2021}{x^4}}}{{\displaystyle \frac{2x^3}{x^4}-\frac{4x^2}{x^4}+\frac{2020}{x^4}}}}}\\ & ={\displaystyle \frac{3+{\displaystyle \frac{2}{x}-\frac{5}{x^2}+\frac{2021}{x^4}}}{{\displaystyle \frac{4}{x}-\frac{4}{x^2}+\frac{2020}{x^4}}}}\\ & ={\displaystyle \frac{3+0-0+0}{0-0+0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll}1.& \text{Perhatikanlah pernyataan-pernyataan berikut}\\ & \text{a}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=8\\ & \text{b}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=4,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=4\\ & \text{c}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)\text{tidak ada}\\ & \text{d}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=3\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=1\\ & \text{Pernyataan di atas yang tepat adalah}....\\ & \text{a}.(i)\text{dan}(ii)\\ & \text{b}.(i)\text{dan}(iii)\\ & \text{c}.(ii)\text{dan}(iii)\\ & \text{d}.(ii)\text{dan}(iv)\\ & \text{e}.(iii)\text{dan}(iv)\\ \\ & \mathbf{\text{Jawab}}:\text{c}\\ & \begin{array}{|c|}\hline \begin{array}{rl}\text{Ing}& \text{at}\mathbf{\text{Definisi Limit}}\text{berikut}:\\ \text{Mis}& \text{al}f\text{sebuah fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}c,L\in \mathbb{R}\\ 1.& \text{Limit fungsi trigonometri};\underset{x\to c}{\text{Lim}}f(x)=L\mathbf{\text{ada}}\\ & \text{untuk semua nilai}x\text{mendekati}c\\ & \text{jika dan hanya jika nilai}f(x)\text{mendekati}L\\ 2.& \underset{x\to c}{\text{Lim}}f(x)=L\iff \underset{x\to c^{-}}{\text{Lim}}f(x)=\underset{x\to c^{+}}{\text{Lim}}f(x)=L\\ & \text{Limit kiri}=\text{limit kanan}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{ll}2.& \text{Perhatikanlah gambar dan pernyatan-}\\ & \text{pernyataan berikut}\end{array}$.

$\begin{array}{ll}& \text{i}.\text{Nilai}\underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{Lim}}f(x)=4\\ & \text{ii}.\text{Nilai}\underset{x\to {\displaystyle \frac{3\pi }{4}}}{\text{Lim}}f(x)=\sqrt{2}\\ & \text{iii}.\text{Nilai}\underset{x\to \pi }{\text{Lim}}f(x)=1\\ & \text{iv}.\text{Nilai}\underset{x\to {\displaystyle \frac{5\pi }{4}}}{\text{Lim}}f(x)=-1\\ & \text{Pernyataan-pernyataan yang tepat}\\ & \text{ditunjukkan oleh}....\\ \\ & \text{a}.(i)\text{dan}(ii)\\ & \text{b}.(i)\text{dan}(iii)\\ & \text{c}.(ii)\text{dan}(iii)\\ & \text{d}.(ii)\text{dan}(iv)\\ & \text{e}.(iii)\text{dan}(iv)\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\end{array}$.

$\begin{array}{ll}3.& \text{Nilai}\underset{x\to 2}{\text{Lim}}f(x),\text{dengan kondisi}\\ & f(x)=\{\begin{array}{ll}{\displaystyle \frac{x^2-4}{x-2}}& \text{untuk}x\ne 2\\ & \\ 6x& \text{untuk}x=2\end{array}\\ & \text{adalah}....\\ & \text{a}.\text{tidak ada}\\ & \text{b}.0\\ & \text{c}.2\\ & \text{d}.4\\ & \text{e}.12\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sebagaimana pada soal, maka}\\ & \text{Harga limit kiri}:\\ & \underset{x\to 2^{-}}{\text{Lim}}f(x)=\underset{x\to 2^{-}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{-}}{\text{Lim}}(x+2)=4}\\ & \text{Dan harga limit kanan}:\\ & \underset{x\to 2^{+}}{\text{Lim}}f(x)=\underset{x\to 2^{+}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{+}}{\text{Lim}}(x+2)=4}\\ & \text{Karena limit kiri}=\text{limit kanan},\text{yaitu}\\ & \underset{x\to 2^{-}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{+}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=4,\text{maka}}}\\ & \underset{x\to 2}{\text{Lim}}f(x)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=4}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}=....}\\ \\ & \text{a}.{\displaystyle 8}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.2\\ & \text{d}.1\\ & \text{e}.{\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{8x^2}{x^2}+\frac{x}{x^2}-\frac{2020}{x^2}}}{{\displaystyle \frac{2x^2}{x^2}-\frac{2021x}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8+{\displaystyle \frac{1}{x}-\frac{2020}{x^2}}}{2-{\displaystyle \frac{2021}{x}}}}\\ & ={\displaystyle \frac{8+{\displaystyle \frac{1}{\mathrm{\infty }}-\frac{2020}{{\mathrm{\infty }}^2}}}{2-{\displaystyle \frac{2021}{\mathrm{\infty }}}}}\\ & ={\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}=....}\\ \\ & \text{a}.{\displaystyle 3}\\ & \text{b}.{\displaystyle 1}\\ & \text{c}.{\displaystyle \frac{1}{3}}\\ & \text{d}.-{\displaystyle \frac{1}{3}}\\ & \text{e}.{\displaystyle -3}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{(-\sqrt{{\displaystyle \frac{1}{x^2}}})}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{x}{x}+\frac{2021}{x}}}{-\sqrt{{\displaystyle \frac{9x^2}{x^2}-\frac{2020x}{x^2}}}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1+{\displaystyle \frac{2021}{x}}}{-\sqrt{9-{\displaystyle \frac{2020}{x}}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{2021}{\mathrm{\infty }}}}{-\sqrt{9-{\displaystyle \frac{2020}{\mathrm{\infty }}}}}={\displaystyle \frac{1}{-\sqrt{9}}}}\\ & =-{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 6.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=....}\\ \\ & \text{a}.{\displaystyle 1}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.9\\ & \text{d}.16\\ & \text{e}.25\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times {\displaystyle \frac{{\displaystyle \frac{1}{5^x}}}{{\displaystyle \frac{1}{5^x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2{\left({\displaystyle \frac{2}{5}}\right)}^x+3{\left({\displaystyle \frac{3}{5}}\right)}^x+4{\left({\displaystyle \frac{4}{5}}\right)}^x+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{2}{5}}\right)}^x+\frac{1}{3}{\left({\displaystyle \frac{3}{5}}\right)}^x+\frac{1}{4}{\left({\displaystyle \frac{4}{5}}\right)}^x+\frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{0+0+0+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{0+0+0+{\displaystyle \frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{5.1}{{\displaystyle \frac{1}{5}.1}}}\\ & =25\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \mathbf{\text{(USM UGM Mat IPA)}}\\ & \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{5}{3}}\\ \\ \text{b}.{\displaystyle \frac{2}{3}}\\ \\ \text{c}.-{\displaystyle \frac{1}{3}}\\ \\ \text{d}.-{\displaystyle \frac{2}{3}}\\ \\ \text{e}.-{\displaystyle \frac{5}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\end{array}$

$\begin{array}{rl}.& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-\sqrt[3]{{(x+1)}^3})}\\ & \text{ingat bentuk}a-b=(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})\\ & \text{dan untuk}\{\begin{array}{ll}a& =(x^3-2x^2)\\ & \\ b& ={(x+1)}^3\end{array}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-{(x+1)}^3}{\sqrt[3]{{(x^3-2x^2)}^2}+\sqrt[3]{(x^3-2x^2){(x+1)}^3}+\sqrt[3]{{\left({(x+1)}^3\right)}^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-(x^3+3x^2+3x+1)}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-5x^2+...}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & ={\displaystyle \frac{-5}{1+1+1}}\\ & =-{\displaystyle \frac{5}{3}}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 2}\\ \\ \text{d}.{\displaystyle \frac{5}{2}}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle ((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{k}-\frac{1}{k+1}))}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (1-\frac{1}{k+1})}\\ & ={\displaystyle (1-\frac{1}{\mathrm{\infty }+1})}\\ & ={\displaystyle 1-\frac{1}{\mathrm{\infty }}}\\ & =1-0\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{{\displaystyle \frac{1}{x}(\sqrt{8x-2020}+\sqrt{4x+2021})}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8x}{x^2}-\frac{2020}{x^2}}}+\sqrt{{\displaystyle \frac{4x}{x^2}+\frac{2021}{x^2}}})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8}{x}-{\displaystyle \frac{2020}{x}}}}+\sqrt{{\displaystyle \frac{4}{x}+{\displaystyle \frac{2021}{x}}}})}}\\ & ={\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{4}{0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\\ & =\sqrt{\mathrm{\infty }}+\sqrt{\mathrm{\infty }}\\ & =\mathrm{\infty }\end{array}$.

☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝

$\begin{array}{rl}\text{Sebagai}& \mathbf{\text{CATATAN}}\text{di sini}\\ \text{Sifat-sif}& \text{at bilangan tak hingga}\\ (1)& \mathrm{\infty }+\mathrm{\infty }=\mathrm{\infty }\\ (2)& -\mathrm{\infty }+(-\mathrm{\infty })=-\mathrm{\infty }\\ (3)& \mathrm{\infty }\times \mathrm{\infty }=\mathrm{\infty }\\ (4)& -\mathrm{\infty }+(-\mathrm{\infty })=\mathrm{\infty }\\ (5)& k.\mathrm{\infty }=\mathrm{\infty },k\text{positif}\\ (6)& k.(-\mathrm{\infty })=-\mathrm{\infty },k\text{positif}\\ (7)& k.\mathrm{\infty }=-\mathrm{\infty },k\text{negatif}\\ (8)& k.(-\mathrm{\infty })=\mathrm{\infty },k\text{negatif}\\ \text{yang ha}& \text{rus dihindari}\\ (1)& \mathrm{\infty }-\mathrm{\infty },\text{bentuk tak tentu}\\ (2)& {\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},-{\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},\text{dan}\frac{0}{0}}}\end{array}$.

Latihan Soal 11 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 96.& \text{Himpunan penyelesaian dari}\\ & 2x-1<x+1<3-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<1\}\\ \text{b}.& \{x|x<2\}\\ \text{c}.& \{x|1<x<2\}\\ \text{d}.& \{x|x>2\}\\ \text{e}.& \{x|x>1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x-1<x}}+\underset{\text{B}}{\underbrace{1<3-x}}\\ & \bullet \text{Bagian A}\\ & 2x-1<x+1\\ & x<2................(1)\\ & \bullet \text{Bagian B}\\ & x+1<3-x\\ & 2x<2\\ & x<1................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<1\end{array}\end{array}$

$\begin{array}{l}\\ 97.& \text{Himpunan penyelesaian dari}\\ & 2x+1<x<1-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-2\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-1<x<-2\}\\ \text{d}.& \{x|x<{\displaystyle \frac{1}{2}}\}\\ \text{e}.& \{x|x<1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x+1<x}}\underset{\text{B}}{\underbrace{<1-x}}\\ & \bullet \text{Bagian A}\\ & 2x+1<x\\ & x<-1................(1)\\ & \bullet \text{Bagian B}\\ & x<1-x\\ & 2x<1\\ & x<{\displaystyle \frac{1}{2}..................(2)}\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<-1\end{array}\end{array}$

$\begin{array}{l}\\ 98.& \text{Himpunan penyelesaian dari}\\ & 3x+14\le x+5<3x-1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-3<x<-1\}\\ \text{d}.& \{x|x>3\}\\ \text{e}.& \left\{\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{4x+14\le x}}\underset{\text{B}}{\underbrace{+5<3x-1}}\\ & \bullet \text{Bagian A}\\ & 4x+14\le x+5\\ & 3x\le -9\\ & x\le -3................(1)\\ & \bullet \text{Bagian B}\\ & x+5<3x-1\\ & -2x<-6\\ & x>3..................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}\text{tidak ada}\end{array}\end{array}$

$\begin{array}{l}\\ 99.& \text{Jika}{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2020<x<2021\\ \text{b}.& -2021<x<-2020\\ \text{c}.& {\displaystyle \frac{1}{2020}<x<{\displaystyle \frac{1}{2021}}}\\ \text{d}.& x<{\displaystyle \frac{1}{2021}\text{dan}x>{\displaystyle \frac{1}{2020}}}\\ \text{e}.& \text{semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}:{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{Dapat ditulis ulang dengan}\\ & 2020<{\displaystyle \frac{1}{x}\text{dan}{\displaystyle \frac{1}{x}<2021}}\\ & \text{Jika digabung menjadi}\\ & 2020<{\displaystyle \frac{1}{x}<2021}\end{array}\end{array}$

$\begin{array}{l}\\ 100.& \text{Jika}a>0\text{dan}b<0,\text{maka}\\ & \text{pernyataan berikut yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& a+b>0\\ \text{b}.& a-b<0\\ \text{c}.& a^2-b^2<0\\ \text{d}.& {\displaystyle \frac{a}{b}<0}\\ \text{e}.& ab>0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Cukup Jelas saat}{\displaystyle \frac{a}{b}=\frac{+}{-}=-<0}\end{array}$.

$\begin{array}{l}\\ 101.& \text{Jika}0<x+y<3\text{dan}1<x-y<2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 1<x<5\\ \text{b}.& \left|x\right|<1\\ \text{c}.& x<1\\ \text{d}.& {\displaystyle \frac{1}{2}<x<\frac{5}{2}}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{l}\\ 0<x+y<& 3& \\ 1<x-y<& 2& +\\ 1<2x<& 5& \text{dibagi 2 semuanya}\\ {\displaystyle \frac{1}{2}<x<}& {\displaystyle \frac{5}{2}}& .....(4)\end{array}\end{array}$

$\begin{array}{l}\\ 102.& \mathbf{\text{(UMPTN 1997)}}\\ & \text{Diketahui P, Q, dan R memancing ikan.}\\ & \text{Jika hasil Q lebih sedikit dari hasil R}\\ & \text{sedangkan jumlah hasil P dan Q lebih }\\ & \text{banyak dari pada dua kali hasil R,}\\ & \text{maka yang terbanyak mendapat ikan}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{P dan R}\\ \text{b}.& \text{P dan Q}\\ \text{c}.& \text{P}\\ \text{d}.& \text{Q}\\ \text{e}.& \text{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:\\ & \bullet Q<R...............(1)\\ & \bullet P+Q>2R......(2)\\ & \text{Sehingga untuk persamaan}(1)\mathrm{\&}(2)\\ & \begin{array}{l}\\ R>& Q& \\ P+Q>& 2R& +\\ P+Q+R>& Q+2R& \\ \\ P>& R......(3)\end{array}\\ & \text{dari}(1)\text{dan}(3)\text{diperoleh bahwa}\\ & Q<R<P\\ & \text{Jadi, yang terbanyak mendapat ikan}\\ & \text{adalah P}\end{array}\end{array}$

$\begin{array}{l}\\ 103.& \text{Jika}a>0,b>0,\text{dan}a>b,\text{maka}\\ & \text{pernyataan berikut yang salah adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{a}>\frac{1}{b}}\\ \text{b}.& a^2>b^2\\ \text{c}.& a^3>b^3\\ \text{d}.& \sqrt{a}>\sqrt{b}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a>0,b>0,\text{dan}a>b\\ & \text{Maka}\\ & {\displaystyle \frac{a}{1}>\frac{b}{1},\text{jika dibalik}}\\ & \text{menjadi}\\ & {\displaystyle \frac{1}{a}<\frac{1}{b}}\end{array}\end{array}$

$\begin{array}{l}\\ 104.& \text{Jika}a,b\text{bilangan real, maka}....\\ & \begin{array}{l}\\ \text{a}.& a^2+b^2\ge 2ab\\ \text{b}.& a^2+b^2>2ab\\ \text{c}.& a^2+b^2<2ab\\ \text{d}.& a^2+b^2\le 2ab\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a,b\in \mathbb{R}\\ & \text{Maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\end{array}\end{array}$

$\begin{array}{l}\\ 105.& \text{Pernyataan berikut yang tepat untuk}\\ & \text{untuk seluruh}x\text{positif adalah}....\\ & \begin{array}{l}\\ \text{a}.& x+{\displaystyle \frac{1}{x}<2}\\ \text{b}.& x+{\displaystyle \frac{1}{x}\le 2}\\ \text{c}.& x+{\displaystyle \frac{1}{x}>2}\\ \text{d}.& x+{\displaystyle \frac{1}{x}\ge 2}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& a,b\in \mathbb{R},a>0,b>0\\ & \text{Mirip dengan pembahasan}\\ & \text{no.19, maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\\ & \text{Saat}a=\sqrt{x},b={\displaystyle \frac{1}{\sqrt{x}}}\\ & \text{menyebabkan}\\ & {\left(\sqrt{x}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{x}}}\right)}^2\ge 2.\sqrt{x}.{\displaystyle \frac{1}{\sqrt{x}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2\sqrt{x.{\displaystyle \frac{1}{x}}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2}\end{array}\end{array}$

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

Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 86.& (\text{Soal SNMPTN})\\ & \text{Jika}x>5\text{dan}y<3,\text{maka}\\ & \text{nilai }x-y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{lebih besar dari pada 1}\\ \text{b}.& \text{lebih besar dari pada 3}\\ \text{c}.& \text{lebih besar dari pada 8}\\ \text{d}.& \text{lebih besar dari pada 5}\\ \text{e}.& \text{lebih besar dari pada 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}\\ x& >5\mathrm{\&}y<3\\ \text{m}& \text{aka}\\ & \begin{array}{l}\\ x>5& \Rightarrow & x& >5\\ y<3& \Rightarrow & -y& >-3& +\\ & & x-y& >2\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 87.& \text{Batas pertidaksamaan}5x-7>13\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\\ \text{b}.& x>4\\ \text{c}.& x>-4\\ \text{d}.& x<4\\ \text{e}.& -4<x<4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}5x& -7>13\\ 5x& >13+7\\ 5x& >20\\ x& >4\end{array}\end{array}$

$\begin{array}{l}\\ 88.& \text{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& x<3{\displaystyle \frac{1}{3}}\\ \text{c}.& x>3{\displaystyle \frac{1}{3}}\\ \text{d}.& x>3\\ \text{e}.& \text{Semua pilihan jawaban salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}2x+3& >5x-7\\ 2x-5x& >-7-3\\ -3x& >-10\text{dikali (-1)}\\ 3x& <10\\ x& <{\displaystyle \frac{10}{3}=3{\displaystyle \frac{1}{3}}}\end{array}\end{array}$

$\begin{array}{l}\\ 89.& (\mathbf{\text{UMPTN 01}})\text{Jika pertidaksamaan}\\ & 2x-3a>{\displaystyle \frac{3x-1}{2}+ax\text{mempunyai}}\\ & \text{penyelesaian}x>5,\text{maka nilai}a\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{8}}\\ \text{c}.& {\displaystyle \frac{3}{8}}\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}2x-3a& >{\displaystyle \frac{3x-1}{2}+ax\text{tiap ruas}(\times 2)}\\ 4x-6a& >3x-1+2ax\\ 4x-3x& -2ax>-1+6a\\ x-2a& x>-1+6a\\ (1-2a)& x>-1+6a\\ x& >{\displaystyle \frac{-1+6a}{1-2a}}\\ \text{Diketa}& \text{hui}:x>5\text{adalah penyelesaian}\\ \text{maka}& \\ 5& ={\displaystyle \frac{-1+6a}{1-2a}}\\ 5-10a& =-1+6a\\ -6a-10& a=-1-5\\ -16a& =-6\\ a& ={\displaystyle \frac{-6}{-16}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 90.& (\mathbf{\text{UMPTN 94}})\\ & \text{Apabila}a<x<b\text{dan}a<y<b\\ & \text{maka berlaku}....\\ & \begin{array}{l}\\ \text{a}.& a<x-y<b\\ \text{b}.& b-a<x-y<a-b\\ \text{c}.& a-b<x-y<b-a\\ \text{d}.& {\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)}\\ \text{e}.& {\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{l}\\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -a>-y>-b& & \\ & \text{saat}& \text{di susun ulang}& \\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -b<-y<-a& +& \\ & & a-b<x-y<& b-a\end{array}\end{array}$.

$\begin{array}{l}\\ 91.& \text{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y-x>4\\ \text{b}.& y-x<4\\ \text{c}.& y+x+4>0\\ \text{d}.& y+x+4<0\\ \text{e}.& y+x<1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2y-5>2x+4y+3\\ & 2y-4y-2x-5-3>0\\ & -2y-2x-8>0\text{dibagi}(-{\displaystyle \frac{1}{2}})\\ & y+x+4<0\end{array}\end{array}$

$\begin{array}{l}\\ 92.& \text{Jika}3x-4>5x-17\\ & \text{maka sebuah bilangan prima}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 3x-4>5x-17\\ & \iff 3x-5x>-17+4\\ & \iff -2x>-13\text{tiap ruas}(\times -1)\\ & \iff 2x<13\\ & \iff x<{\displaystyle \frac{13}{2}=6\frac{1}{2}}\\ & \text{Jadi, yang memenuhi adalah 3 dan 5}\end{array}\end{array}$

$\begin{array}{l}\\ 93.& \text{Jika}{\displaystyle \frac{1}{5}<\frac{1}{x}\text{dan}x<0}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{1}{5}}\\ \text{b}.& -5<x<0\\ \text{c}.& 0<x<5\\ \text{d}.& x<-5\\ \text{e}.& -{\displaystyle \frac{1}{5}<x<0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ {\displaystyle \frac{1}{5}}& <\frac{1}{x}\text{dan}x<0\\ {\displaystyle \frac{1}{5}}& <{\displaystyle \frac{1}{x}}\\ x& <5\\ x& >-5\text{karena}x<0\\ \text{Sehi}& \text{ngga}\\ -5<& x<0\end{array}\end{array}$

$\begin{array}{l}\\ 94.& \text{Jika}a,b,c\text{dan}d\text{bilangan real}\\ & \text{dengan}a>b\text{dan}c>d\\ & \text{maka berlaku}\\ & (1)ac>bd\\ & (2)a+c>b+d\\ & (3)ad>bc\\ & (4)ac+bd>ad+bc\\ & \text{Pernyataan-pernyataan di atas}\\ & \text{yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& (1),(2),\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\\ \text{e}.& \text{Semua benar}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:a,b,c\text{dan}d\text{bilangan real}\\ & \text{Jelas bahwa baik bilangan positif maupun}\\ & \text{negatif termasuk semunya dibolehkan}\\ & \text{dengan}a>b\text{dan}c>d\\ & \bullet \text{Sehingga pernyataan (1)}ac>bd\\ & \text{salah saat kita coba bilangan negatif}\\ & \bullet \text{Pernyataan (2) benar karena}\\ & \begin{array}{llll}a& >& b& \\ c& >& d& +\\ a+c& >& b+d\end{array}\\ & \bullet \text{Kasusnya sama dengan poin (1)}\\ & \text{saat dicoba dengan bilangan positif}\\ & \text{tidak semuanya memenuhi}\\ & \bullet \text{Pernyataan (4) tepat juga karena}\\ & \begin{array}{l}\\ a-b>0\\ c-d>0\text{Saat dikalikan}\\ (a-b)\times (c-d)>0\\ \iff ac-ad-bc+bd>0\\ \iff ac+bd>ad+bc\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 95.& \text{Jika}-2<y<3\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 9<(y-2{)}^2<16\\ \text{b}.& 4<(y-2{)}^2<16\\ \text{c}.& 1<(y-2{)}^2<16\\ \text{d}.& 0\le (y-2{)}^2<16\\ \text{e}.& -1<(y-2{)}^2<16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}:-2<y<3\\ & \bullet \text{saat dikurangi}2\\ & \iff -2-2<y-2<3-2\\ & -4<y-2<1\\ & \bullet \text{Saat}-4<y-2<0\\ & (-4{)}^2<(y-2{)}^2<0^2\text{dikuadratkan}\\ & 16>(y-2{)}^2>0\\ & 0<(y-2{)}^2<16\\ & \bullet \text{Saat}0\le y-2<1\\ & 0^2\le (y-2{)}^2<1^2\\ & 0<(y-2{)}^2<1\\ & \text{Jadi},0\le (y-2)<16\end{array}\end{array}$

Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{l}\\ 76.& \text{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ & \text{ketiga angka tersebut adalah 9. Angka kedua}\\ & \text{dikurangi angka pertama dan angka ketiga }\\ & \text{sama dengan 1. Dua kali angka pertama sama}\\ & \text{dengan jumlah angka kedua dan angka ketiga.}\\ & \text{Angka puluhan pada bilangan tersebut adalah}\\ & ....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 6\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Model matematikanya}\\ & \{\begin{array}{c}A+B+C=9....(1)\\ 2B-A-C=1....(2)\\ 2A=B+C....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle A+B+C}& =9\\ {\displaystyle -A+B-C}& =1& +\\ 2B& =10\\ B& =5& ...(4)\end{array}\\ & \text{Jadi},\text{bilangan kedua adalah}=B=5\end{array}\end{array}$

$\begin{array}{l}\\ 77.& (\mathbf{\text{SIMAK UI 2010}})\\ & \text{Jika}x+y+2z=K,x+2y+z=K,\\ & 2x+y+z=K\text{dengan}K\ne 0,\text{maka}\\ & x^2+y^2+z^2\text{bila dinyatakan dalam}K\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{16}{K}^2}\\ \text{b}.& {\displaystyle \frac{3}{16}{K}^2}\\ \text{c}.& {\displaystyle \frac{4}{17}{K}^2}\\ \text{d}.& {\displaystyle \frac{3}{8}{K}^2}\\ \text{e}.& {\displaystyle \frac{2}{3}{K}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+2z=K....(1)\\ x+2y+z=K....(2)\\ 2x+y+z=K....(3)\end{array}\\ & \text{maka}\\ & \{\begin{array}{c}z+(x+y+z)=K....(1)\\ y+(x+y+z)=K....(2)\\ x+(x+y+z)=K....(3)\end{array}\\ & \text{Saat}(1)+(2)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle x+y+2z}& =K\\ {\displaystyle x+2y+z}& =K& \\ {\displaystyle 2x+y+z}& =K& +\\ 4x+4y+4z& =3K\\ x+y+z& ={\displaystyle \frac{3}{4}K}& ...(4)\end{array}\\ & \text{Saat}(4)\text{disubstitusikan ke}(1),(2),\text{dan}(3)\\ & \text{Jelas bahwa akan didapatkan}\\ & x=y=z={\displaystyle \frac{1}{4}K}\\ & \text{Jadi},x^2+y^2+y^2=3{\left({\displaystyle \frac{1}{4}K}\right)}^2={\displaystyle \frac{3}{16}{K}^2}\end{array}\end{array}$

$\begin{array}{l}\\ 78.& \text{Diketahui}0,15252525252...={\displaystyle \frac{p}{2q+r}}\\ & \text{Jika jumlah}p\text{dan}q=\text{3 kali}r,\text{maka}\\ & \text{masing-masing harga}p,q,\text{dan}r=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 152,2819,2584}\\ \text{b}.& {\displaystyle 252,{\displaystyle \frac{5638}{7},\frac{8102}{21}}}\\ \text{c}.& {\displaystyle 151,\frac{2819}{7},\frac{1292}{7}}\\ \text{d}.& {\displaystyle 151,{\displaystyle \frac{2819}{7},\frac{2584}{7}}}\\ \text{e}.& {\displaystyle 152,\frac{2819}{14},\frac{1292}{7}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Dari soal diketahui}\\ & \{\begin{array}{ll}0,1\overline{5252}& ={\displaystyle \frac{p}{2q+r}.....(1)}\\ p+q& =3r............(2)\end{array}\\ & \text{dan}\\ & \begin{array}{l}\\ {\displaystyle x}& =0,15252525252...\\ {\displaystyle 1000x}& =152,5252525252...\\ {\displaystyle 10x}& =1,5252525252...& -\\ 990x& =151\\ x& ={\displaystyle \frac{151}{990},\text{maka}}\\ {\displaystyle \frac{p}{2q+r}}& ={\displaystyle \frac{151}{990}}\\ & \{\begin{array}{ll}p& =151.......(3)\\ 2p+r& =990.......(4)\end{array}\end{array}\\ & \text{Dari}(3)\text{diperoleh}:q=3r-p=3r-151....(5)\\ & \text{Dari}(5)\text{disubstitusikan ke}(4)\\ & \begin{array}{rl}2q+r& =990\\ 2(3r-151)+r& =990\\ 6r-302+r& =990\\ 7r& =990+302=1292\\ r& ={\displaystyle \frac{1292}{7}.....(6)}\end{array}\\ & \text{Dari}(3)\mathrm{\&}(6)\text{disubstitusikan ke}(2)\\ & \begin{array}{rl}p+q& =3r\\ 151+q& =3\left({\displaystyle \frac{1292}{7}}\right)\\ q& ={\displaystyle \frac{3876}{7}-151}\\ & ={\displaystyle \frac{3876-1057}{7}}\\ & ={\displaystyle \frac{2819}{7}.....(7)}\end{array}\\ & \text{Jadi},p,q,r\text{adalah}:{\displaystyle 151,\frac{2819}{7},\frac{1292}{7}}\end{array}\end{array}$

$\begin{array}{l}\\ 79.& \text{Perhatikanlah sistem persamaan berikut}\\ & \{\begin{array}{ll}3x+2y-5z& =3\\ 2x-6y+kz& =9\\ 5x-4y-z& =5\end{array}\\ & \text{agar sistem persamaan ini tidak}\\ & \text{memiliki penyelesaian, maka nilai}k=....\\ & \begin{array}{l}\\ \text{a}.& -4\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Agar sistem persamaan}\\ & \{\begin{array}{ll}3x+2y-5z& =3\\ 2x-6y+kz& =9\\ 5x-4y-z& =5\end{array}\\ & \text{tidak berpenyelesaian, maka}\\ & \text{ingat penyelesaian metode matrik}\\ & \text{buatlah penyebutnya}=0,\text{yaitu}:\\ & \left|\begin{array}{ccc}3& 2& -5\\ 2& -6& k\\ 5& -4& -1\end{array}\right|=0\\ & \text{Selanjutnya}\\ & 3\left|\begin{array}{cc}-6& k\\ -4& -1\end{array}\right|-2\left|\begin{array}{cc}2& k\\ 5& -1\end{array}\right|-5\left|\begin{array}{cc}2& -6\\ 5& -4\end{array}\right|=0\\ & 3(6+4k)-2(-2-5k)-5(-8+30)=0\\ & 18+12k+4+10k+40-150=0\\ & 22x=88\\ & x=4\end{array}\end{array}$

$\begin{array}{l}\\ 80.& \text{Diketahui}\\ & \left(\begin{array}{ccc}{\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}\\ {\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}& -{\displaystyle \frac{4}{5}}\\ -{\displaystyle \frac{2}{5}}& {\displaystyle \frac{1}{10}}& {\displaystyle \frac{1}{10}}\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)\\ & \text{Nilai}x,y,\text{dan}z\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}}\\ \text{b}.& -1,5,1\\ \text{c}.& 1,5,-1\\ \text{d}.& -1,1,5\\ \text{e}.& 5,1,-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}{\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1....(1)}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2....(2)}\\ -{\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0....(3)}\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z}& =1& \\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z}& =2& -\\ {\displaystyle \frac{5}{5}z}& =-1& \\ {\displaystyle z}& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z}& =1& |\times 1|& {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1}\\ -{\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z}& =0& |\times 2|& -{\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0}& -\\ & & & {\displaystyle \frac{5}{5}x=1}& \\ & & & x=-1........(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{akan didapatkan}\\ & y=5\\ & \text{Jadi},(x,y,z)=(1,5,-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 81.& \text{Diketahui suatu fungsi kuadrat}\\ & f(x)=a{x}^2+bx+c.\text{Jika fungsi}\\ & (-1,0),(1,4),\text{dan}(2,9),\text{maka}\\ & \text{fungsi yang dimaksud adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle f(x)=x^2-2x+3}\\ \text{b}.& f(x)=x^2+2x+3\\ \text{c}.& f(x)=x^2+2x-3\\ \text{d}.& f(x)=x^2-2x-3\\ \text{e}.& f(x)=x^2+2x+1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}(-1,0)\Rightarrow f(-1)=a-b+c=0....(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4....(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9....(3)\end{array}\\ & \text{Saat}(1)\mathrm{\&}(2),\text{didapatkan}\\ & b=2...............(4)\\ & \text{Saat}(1)\mathrm{\&}(3),\text{didapatkan}\\ & \begin{array}{l}\\ 4a+2b+c& =9& \\ a-b+c& =0& -\\ 3a+3b& =9& \\ a+b& =3& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \{\begin{array}{ll}a& =1\\ c& =1\end{array}\\ & \text{Jadi},f(x)=a{x}^2+bx+c=x^2+2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 82.& \text{Diketahui persamaan}\{\begin{array}{ll}x-y& =2\\ kx+y& =3\end{array}\\ & \text{memiliki solusi}(x,y)\text{di kuadran I}\\ & \text{Jika dan hanya jika nilai}k\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle k=-1}\\ \text{b}.& k>-1\\ \text{c}.& k<{\displaystyle \frac{3}{2}}\\ \text{d}.& 0<k<{\displaystyle \frac{3}{2}}\\ \text{e}.& -1<k<{\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x-y=2....(1)\\ kx+y=3....(2)\end{array}\\ & \text{Dengan metode matriks didapatkan}\\ & x={\displaystyle \frac{\left|\begin{array}{cc}2& -1\\ 3& 1\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}}}\\ & \text{Dengan cara yang sama pula}\\ & y={\displaystyle \frac{\left|\begin{array}{cc}1& 2\\ k& 3\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{3-2k}{k+1}}}\\ & \text{Supaya memiliki solusi di kwadran I},\\ & \text{maka baik}x\text{maupun}y\\ & \text{haruslah positif, akibatnya}:\\ & k+1>0\Rightarrow k>-1\\ & \text{Sebagai akibat yang lain adalah}:\\ & 3-2k>0\Rightarrow k<{\displaystyle \frac{3}{2}}\\ & \text{Jadi},-1<k<{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 83.& \text{Diketahui sistem persamaan}\\ & y+{\displaystyle \frac{2}{x+z}=4}\\ & 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ & {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\\ & \text{Nilai}y+\sqrt{x^2-2xz+y^2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 3}\\ \text{b}.& 5\\ \text{c}.& 7\\ \text{d}.& 9\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}y+{\displaystyle \frac{2}{x+z}=4}\\ 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\end{array}\\ & \text{Jika disederhanakan beberapa bagian}\\ & \{\begin{array}{ll}y+2A& =4....(1)\\ 5y+18B& =18....(2)\\ 8A-6B& =3....(3)\end{array}\\ & \text{Saat}(1)+(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ y+2A& =4& |\times 5|& 5y+10A=20\\ 5y+3(8A-3)& =18& |\times 1|& 5y+24A=27& -\\ & & & -14A=-7\\ & & & A={\displaystyle \frac{1}{2}...(4)}\\ \text{maka}B={\displaystyle \frac{1}{6}\mathrm{\&}}& y=3& & \\ \text{akibatnya}\\ \{\begin{array}{ll}x& =1\\ z& =1\end{array}\end{array}\\ & \text{Jadi},y+\sqrt{x^2-2xz+z^2}=3+0=3\end{array}\end{array}$

$\begin{array}{l}\\ 84.& \text{Diberikan}a,b,\text{dan}c\text{adalah angka-angka}\\ & \text{dari bilangan 3 digit yang memenuhi}\\ & 49a+7b+c=286.\text{Nilai dari}a+b+c\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 16\\ \text{b}.& 17\\ \text{c}.& 18\\ \text{d}.& 19\\ \text{e}.& 20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & 49a+7b+c=286\\ & \text{Nilai maksimum}a\text{adalah}5\\ & 49\times 5=245,\text{akibatnya}:\\ & 245+7b+c=286\Rightarrow 7b+c=286-245=41\\ & \text{Nilai maksimum}b\text{adalah}5\\ & 7\times 5=35,\text{akibatnya}:\\ & 35+c=41\Rightarrow c=41-35=6\\ & \text{Sehingga}a,b,\text{dan}c\text{adalah}5,5,\text{dan}6\\ & \text{Jadi},\text{nilai}a+b+c=5+5+6=16\end{array}\end{array}$

$\begin{array}{l}\\ 85.& \text{Diketahui sistem persamaan}\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=1\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & 5x+3y+8z=2\\ & \text{Himpunan penyelesaian yang}\\ & \text{memenuhi adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\right\}\\ \text{b}.& \{-{\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6}}\}\\ \text{c}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6}}\}\\ \text{d}.& \left\{{\displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6}}\right\}\\ \text{e}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Untuk persamaan}(1)\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=(2x+3y{)}^0\\ & \iff (x-y+2z)={10}^0=1\\ & \text{Untuk persamaan}(2)\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & \iff 3^{2x+y+z+3(3z+2y+x)}=3^4\\ & \iff 5x+7y+10z=4\\ & \text{Sehingga sistem persamaan akan terlihat}\\ & \{\begin{array}{c}x-y+2z=1....(1)\\ 5x+7y+10z=4....(2)\\ 5x+3y+8z=2....(3)\end{array}\\ & \text{Saat}(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ 5x+7y+10z& =4& \\ 5x+3y+8z& =2& -\\ 4y+2z& =2\\ 2y+z& =1...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ 5x-5y+10z& =5& \\ 5x+7y+10z& =4& -\\ -12y& =1\\ y& =-{\displaystyle \frac{1}{12}...(5)}\end{array}\\ & \text{Dari persamaan}(5)\text{disubstistusikan ke}(4)\\ & \begin{array}{rl}2y+z& =1\\ 2(-{\displaystyle \frac{1}{12}})+z& =1\\ z& =1+{\displaystyle \frac{1}{6}}\\ z& ={\displaystyle \frac{7}{6}}\end{array}\\ & \text{Cukup jelas juga}x=....\\ & \text{Jadi},\text{pilihannya adalah}e\end{array}\end{array}$

DAFTAR PUSTAKA

1. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade MAtematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
2. Kanginan, M. 2016. Matematika untuk SMA-MA/SMK-MAK Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA
3. Kurnianingsih, S. 2008. SPM Matematika SMA dan MA Program IPS Siap Tuntas Menghadapi Ujian. Jakarta: ESIS
4. Susianto, B. 2011. Soal dan Pembahasan Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO
5. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI

Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 66.& \text{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ & \text{B, dan C bersama-sama dalam 2 jam saja.}\\ & \text{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ & \text{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ & \text{B dan C bersama-sama dalam waktu 3 jam,}\\ & \text{maka sistem persamaan berikut yang memenuhi}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{\begin{array}{ll}A+B+C& =2\\ A+B& ={\displaystyle \frac{12}{5}}\\ B+C& =3\end{array}\\ \text{b}.& \{\begin{array}{ll}A+B+C& ={\displaystyle \frac{1}{2}}\\ A+B& ={\displaystyle \frac{5}{12}}\\ B+C& ={\displaystyle \frac{1}{3}}\end{array}\\ \text{c}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =2\\ {\displaystyle \frac{1}{A}+\frac{1}{B}}& ={\displaystyle \frac{12}{5}}\\ {\displaystyle \frac{1}{B}+\frac{1}{C}}& =3\end{array}\\ \text{d}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{A}+\frac{1}{B}}& ={\displaystyle \frac{5}{12}}\\ {\displaystyle \frac{1}{B}+\frac{1}{C}}& ={\displaystyle \frac{1}{3}}\end{array}\\ \text{e}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =2\\ {\displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}}& ={\displaystyle \frac{12}{5}}\\ {\displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =3\end{array}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Per}& \text{hatikan bahwa}:\text{Waktu penyelesaian}\\ \text{sua}& \text{tu pekerjaan adalah termasuk}\\ \text{per}& \text{bandingan berbalik nilai},\text{maka}\\ \bullet & A,B,\text{dan}C\text{dalam 2 jam, artinya}:\\ & {\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\text{demikian juga}}\\ \bullet & A\text{dan}B\text{bersama-sama selesai dalam}\\ & \text{2 jam 24 menit atau}{\displaystyle \frac{12}{5}\text{jam}:}\\ & {\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}}\\ \bullet & B\text{dan}C\text{selesai dalam 3 jam}:\\ & {\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 67.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{c}x+y+4z=15\\ x-y+z=2\\ x+2y-3z=-4\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{(-1,1,3)\}\\ \text{b}.& \{(1,2,3)\}\\ \text{c}.& \{(-2,1,1)\}\\ \text{d}.& \{(3,2,-1)\}\\ \text{e}.& \{(1,-2,3)\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Semunya dikerjakan dengan metode}\\ & \text{matriks}(\mathbf{\text{Cara Cramer}})\\ & \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{ccc}15& 1& 4\\ 2& -1& 1\\ -4& 2& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{15\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|+4\left|\begin{array}{cc}2& -1\\ -4& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1}\\ y& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 15& 4\\ 1& 2& 1\\ 1& -4& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|-15\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2}\\ z& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 1& 15\\ 1& -1& 2\\ 1& 2& -4\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right|-1\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|+15\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3}\end{array}\end{array}$

$.\text{Cara di atas}$  full matriks-Cramer

$\begin{array}{l}\\ 68.& \text{Hasil dari}xyz\text{yang memenuhi}\\ & \{\begin{array}{c}x+y+z=2\\ x-y+z=-2\\ x-y-z=2\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\\ \text{b}.& -4\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=2.....(1)\\ x-y+z=-2.....(2)\\ x-y-z=2.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y+z& =-2& -\\ 2y& =4& \\ y& =2& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y-z& =2& +\\ 2x& =4& \\ x& =2& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (2)+(2)+z& =2\\ z& =-2\end{array}\\ & \text{Jadi},xyz=(2).(2).(-2)=-8\end{array}\end{array}$

$.\text{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{l}\\ 69.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=-6\\ x-2y+z=3\\ -2x+y+z=9\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -30\\ \text{b}.& -15\\ \text{c}.& 5\\ \text{d}.& 30\\ \text{e}.& 35\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=-6....(1)\\ x-2y+z=3....(2)\\ -2x+y+z=9....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ x-2y+z& =3& -\\ 3y& =-9& \\ y& =-3& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ -2x+y+z& =9& -\\ 3x& =-15& \\ x& =-5& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (-5)+(-3)+z& =-6\\ z& =2\end{array}\\ & \text{Jadi},xyz=(-5).(-3).(2)=30\end{array}\end{array}$

$\begin{array}{l}\\ 70.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+2y+z=4\\ 3x+y+2z=-5\\ x-2y+2z=-6\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -96\\ \text{b}.& -24\\ \text{c}.& 24\\ \text{d}.& 32\\ \text{e}.& 96\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+2y+z=4.......(1)\\ 3x+y+2z=-5......(2)\\ x-2y+2z=-6.......(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& |\times 1|& x+2y+z& =4\\ 3x+y+2z& =-5& |\times 2|& 6x+2y+4z& =-10& -\\ & & & -5x-3z& =14& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& \\ x-2y+2z& =-6& +\\ 2x+3z& =-2& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ -5x-3z& =14& \\ 2x+3z& =-2& +\\ -3x& =12& \\ x& =-4& .....(6)\\ \text{didapat pula}& z& =2......(7)\end{array}\\ & \text{Dari persamaan}(6)\mathrm{\&}(7)\text{didapatkan}\\ & \begin{array}{rl}x+2y+z& =4\\ (-4)+2y+2& =4\\ y& =3\end{array}\\ & \text{Jadi},xyz=(-4).(3).(2)=-24\end{array}\end{array}$.

$\begin{array}{l}\\ 71.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y-z=1\\ 2x-y+2z=9\\ x+3y-z=7\end{array}\\ & \text{Nilai}{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{13}{12}}\\ \text{d}.& {\displaystyle \frac{5}{4}}\\ \text{e}.& {\displaystyle \frac{7}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y-z=1....(1)\\ 2x-y+2z=9....(2)\\ x+3y-z=7....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& \\ 2x-y+2z& =9& +\\ 3x+z& =10& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& |\times 3|& 3x+3y-3z& =3& \\ x+3y-z& =7& |\times 1|& x+3y-z& =7& -\\ & & & 2x-2z& =-4& \\ & & & x-z& =2& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ 3x+z& =10& \\ x-z& =-2& +\\ 4x& =8& \\ x& =2& .....(6)\\ \text{didapat pula}& z& =4......(7)\end{array}\\ & \text{Dari persamaan}(1)\mathrm{\&}(3)\text{didapatkan juga}\\ & \begin{array}{l}\\ x+y-z& =1& \\ x+3y-z& =-7& -\\ -2y& =-6& \\ y& =3& ....(8)\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}}}\end{array}\end{array}$

$\begin{array}{l}\\ 72.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=5\\ x+y-4z=10\\ -2x+y+z=0\end{array}\\ & \text{Nilai dari}{\displaystyle \frac{xz}{y}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{6}{13}}\\ \text{b}.& {\displaystyle -\frac{5}{13}}\\ \text{c}.& {\displaystyle -\frac{1}{13}}\\ \text{d}.& {\displaystyle \frac{1}{13}}\\ \text{e}.& {\displaystyle \frac{7}{13}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=5.....(1)\\ x+y-4z=10.....(2)\\ -2x+y+z=0.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ x+y-4z& =10& -\\ 5z& =-5& \\ z& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ -2x+y+z& =0& -\\ 3x& =5& \\ x& ={\displaystyle \frac{5}{3}}& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}x+y+z& =5\\ {\displaystyle \frac{5}{3}+y-1}& =5\\ y& =5+1-{\displaystyle \frac{5}{3}=\frac{13}{3}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{xz}{y}={\displaystyle \frac{\left({\displaystyle \frac{5}{3}}\right).(-1)}{{\displaystyle \frac{13}{3}}}=-{\displaystyle \frac{5}{13}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4\end{array}\\ & \text{adalah}\{(x,y,z)\},\text{maka}x+3z=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8....(1)\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10.....(2)\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4...........(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10& -\\ -{\displaystyle \frac{1}{x}-\frac{1}{z}}& =-2\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =8& |\times 2|& {\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}}& =16\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& |\times 1|& {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& -\\ & & & {\displaystyle \frac{2}{x}+\frac{6}{z}}& =12\\ & & \iff & {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2-\\ {\displaystyle \frac{2}{z}}& =4& \\ z& ={\displaystyle \frac{1}{2}......(6)}\\ x& =2-{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\end{array}\\ & \text{Jadi},x+3z={\displaystyle \frac{3}{2}+3.\frac{1}{2}=3}\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \text{Diketahui tiga buah bilangan berturut-turut}\\ & a,b,\text{dan}c.\text{Rata-rata dari ke tiga bilangan}\\ & \text{itu adalah 12. Bilangan kedua sama dengan}\\ & \text{jumlah bilangan yang lain dikurangi 12}.\\ & \text{Jika bilangan ke tiga sama dengan jumlah}\\ & \text{bilangan yang lain, maka nilai}2a+b-c=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle 42}\\ \text{b}.& -{\displaystyle 36}\\ \text{c}.& -{\displaystyle 18}\\ \text{d}.& -{\displaystyle 12}\\ \text{e}.& -{\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Model matematika dari persamaan di atas}\\ & \{\begin{array}{c}a+b+c=36....(1)\\ -a+b-x=12....(2)\\ a+b-c=0....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ -a+b-c& =12& +\\ 2b& =48& \\ b& =24& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ a+b-c& =0& -\\ 2c& =36& \\ c& =18& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}a+b+c& =36\\ a+24+18& =36\\ a& =36-42\\ & =-6\end{array}\\ & \text{Jadi},2a+b-c=2(-6)+24-18=-6\end{array}\end{array}$

$\begin{array}{l}\\ 75.& \text{Jumlah uang terdiri atas koin pecahan}Rp500,00\\ & Rp200,00danRp100,00\text{dengan nilai total}\\ & Rp100.000,00.\text{Jika nilai uang pecahan 500-an}\\ & \text{setengah dari nilai uang pecahan 200-an, tetapi}\\ & \text{tiga kali uang pecahan 100-an, maka banyak koin}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 460\\ \text{b}.& 440\\ \text{c}.& 420\\ \text{d}.& 380\\ \text{e}.& 350\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Model matematika dari kasus di atas}\\ & \{\begin{array}{c}A(500)+B(200)+C(100)=100.000....(1)\\ A(500)={\displaystyle \frac{1}{2}B(200)....(2)}\\ A(500)=3C(100)....(3)\end{array}\\ & \text{Dari persamaan}(2)\text{didapatkan}\\ & 2A(500)=B(200)\\ & \text{Dari persamaan}(3)\text{akan didapatkan}\\ & {\displaystyle \frac{1}{3}A(500)=C(100)}\\ & \text{Dari persamaan}(1)\text{maka},\\ & A(500)+B(200)+C(100)=100.000\\ & A(500)+2A(500)+{\displaystyle \frac{1}{3}A(500)=100.000}\\ & {\displaystyle \frac{10}{3}A(500)=100.000\iff A(500)=30.000}\\ & \text{maka akan didapatkan}\\ & B(200)=2(30.000)=60.000\\ & C(100)={\displaystyle \frac{1}{3}(30.000)=10.000}\\ & \{\begin{array}{ll}A(500)& =30.000\Rightarrow A={\displaystyle \frac{30.000}{500}=60}\\ B(200)& =60.000\Rightarrow B={\displaystyle \frac{60.000}{200}=300}\\ C(100)& =10.000\Rightarrow C={\displaystyle \frac{10.000}{100}=100}\end{array}\\ & \text{Jadi},A+B+C=60+300+100=460\end{array}\end{array}$.