Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{ll}\\ 51.&\textrm{Turunan pertama dari fungsi}\\ &g(x)=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\sin ^{2}x}\\ \textrm{c}.&\displaystyle \frac{1}{\sin^{2} x\cos ^{2}x}\\ \textrm{d}.&\displaystyle \frac{-1}{\sin ^{2}x\cos ^{2}x}\\ \textrm{e}.&\displaystyle \sin ^{2}x\cos ^{2}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(x)&=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\\ &=\frac{\sin ^{2}x+\cos ^{2}x}{\sin x\cos x}=\displaystyle \frac{1}{\sin x\cos x}\\ \color{red}\textrm{maka}&\\ g'(x)&=\displaystyle \frac{0.(\sin x\cos x)-1.\left (\cos ^{2}x -\sin ^{2}x \right )}{(\sin x\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x-\cos ^{2}x}{\sin^{2} x\cos^{2} x}\\ &=\color{red}\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Diketahui}\: \: h(x)=\cos \left ( \displaystyle \frac{3}{x} \right ), \\ &\textrm{maka}\: \: \displaystyle \frac{dh}{dx}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\sin \displaystyle \frac{3}{x}\\ \textrm{b}.&-\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{c}.&-\displaystyle \frac{3}{x}\sin \frac{3}{x}\\ \color{red}\textrm{d}.&\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{e}.&\displaystyle \frac{3}{x}\sin \frac{3}{x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\cos \displaystyle \frac{3}{x}&=-\sin \displaystyle \frac{3}{x}\left ( \displaystyle \frac{0.(x)-3.1}{x^{2}} \right )\\ &=\displaystyle \frac{-(-3)}{x^{2}}\sin \frac{3}{x}\\ &=\color{red}\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Turunan pertama dari}\: \: \tan (\cos x), \\ &\textrm{terhadap}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\sec ^{2}(\cos x)\sin x\\ \textrm{b}.&\sec ^{2}(\cos x)\sin x\\ \textrm{c}.&\sec ^{2}(\sin x)\cos x\\ \textrm{d}.&\displaystyle \sin x\\ \textrm{e}.&\displaystyle -\sin x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Misal}&\textrm{kan}\\ y&=\tan x(\cos x)\\ y'&=\sec ^{2}(\cos x)\times (-\sin x)\\ &=\color{red}-\sec ^{2}(\cos x).\sin x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&(\textbf{UN 2005})\textrm{Turunan pertama dari}\\ &f(x)=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \textrm{b}.&\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\\ \textrm{c}.&-\displaystyle \frac{2}{3}\cos^{.^{-\frac{1}{3}}} \left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \color{red}\textrm{d}.&-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ \textrm{e}.&\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Misal}&\textrm{kan}\\ f(x)&=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ f'(x)&=\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &=\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{2}}}\left ( 3x^{2}+5x \right )\times \left ( -\sin \left ( 3x^{2}+5x \right ) \right )\\ &\qquad\qquad\qquad\qquad \times (6x+5)\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &\times \cos^{-1} \left ( 3x^{2}+5x \right )\times \sin \left ( 3x^{2}+5x \right )\\ &=\color{red}-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 55.&\textrm{Persamaan garis singgung pada kurva}\\ &y=3\sin x\: \: \textrm{pada titik yang berabsis}\: \: \displaystyle \frac{\pi }{3}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )-\frac{2\sqrt{2}}{3}\\ \textrm{b}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )+\frac{2\sqrt{2}}{3}\\ \textrm{c}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{3}}{2}\\ \color{red}\textrm{d}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )+\frac{3\sqrt{3}}{2}\\ \textrm{e}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{2}}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&y=3\sin x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=3\sin \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\frac{3\sqrt{3}}{2}\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=y'=3\cos x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=3\cos \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2} \right )=\frac{3}{2}\\  &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=\color{red}\displaystyle \frac{3}{2}\left ( x-\displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{3\sqrt{3}}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 56.&\textrm{Kurva}\: \: y=\sin x+\cos x\: \: \textrm{untuk}\\ &0<x<\pi \: \: \textrm{memotong sumbu X}\\ &\textrm{di titik A. Persamaan garis}\\ &\textrm{singgung di titik A adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{b}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{2} \right )\\ \color{red}\textrm{c}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right )\\ \textrm{d}.&y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{e}.&y=\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\color{black}\textrm{Kurva memotong sumbu X}\\ &\textrm{di titik A, berarti}\: \: \color{red}y=0\\ &\sin x+\cos x=\color{red}0\\ &\sin x=-\cos x\\ &\displaystyle \frac{\sin x}{\cos x}=-1\Leftrightarrow \tan x=-1\\ &\tan x=\tan \left ( \displaystyle \frac{3\pi }{4} \right )\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\color{red}\textrm{Jadi, titik A-nya}:\: \: \left ( \displaystyle \frac{3\pi }{4},0 \right )\\ &\textrm{dan nilai gradien}\: \: m=y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{3\pi }{4} \right )-\sin \left ( \displaystyle \frac{3\pi }{4} \right )\\ &m=-\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}\\ &\color{black}\textrm{Persamaan garis singgung di A}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) +0\\ &\Leftrightarrow \: y=\color{red}-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec ^{2}x\: \: \textrm{pada titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \color{red}\textrm{b}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{c}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{d}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{e}.&y=4\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&y=\sec^{2} x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=\sec^{2} \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=(2)^{2}=4\\ &\color{black}\textrm{kita cari gradien}\: \: \color{red}m\: \: \color{black}\textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=2\sec^{2} x\tan x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=2\sec^{2} \left ( \color{red}\displaystyle \frac{\pi }{3} \right )\tan \left ( \color{red}\displaystyle \frac{\pi }{3} \right )\\ &\quad=2(4)\sqrt{3}=\color{red}8\sqrt{3}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=\color{red}8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Kurva berikut yang memiliki}\\ &\textrm{garis singgung dengan gradien}\\ &4\sqrt{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=2\sin x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},\sqrt{3} \right )\\ \textrm{b}.&y=\cos 2x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{12},\frac{1}{2} \right )\\ \textrm{c}.&y=\tan x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \pi ,0 \right )\\ \color{red}\textrm{d}.&y=2\sec x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},2 \right )\\ \textrm{e}.&y=\cot x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{4},1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{array}{|c|l|l|}\hline \textrm{a}&y=2\sin x&m=2\cos \displaystyle \frac{\pi }{3}\\ &y'=2\cos x&m=2.\displaystyle \frac{1}{2}=1\\\hline \textrm{b}&y=\cos 2x&m=-2\sin 2 \left (\displaystyle \frac{\pi }{12} \right )\\ &y'=-2\sin 2x&m=-2.\displaystyle \frac{1}{2}=-1\\\hline \textrm{c}&y=\tan x&m=\sec^{2} \left (\pi \right )\\ &y'=\sec^{2} x&m=(-1)^{2}=1\\\hline \color{red}\textrm{d}&y=2\sec x&\color{red}m=2\sec \left ( \displaystyle \frac{\pi }{3} \right )\tan \left (\displaystyle \frac{\pi }{3} \right )\\ &y'=2\sec x\tan x&\color{red}m=2.2.\sqrt{3}=4\sqrt{3}\\\hline \textrm{e}&y=\cot x&m=-\csc^{2} \left ( \displaystyle \frac{\pi }{4} \right )\\ &y'=-\csc^{2} x&m=-\left ( \sqrt{2} \right )^{2}=\color{red}-2\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 59.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{4}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\sqrt{3}x-\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{b}.&y=\sqrt{3}x+\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \color{red}\textrm{c}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{d}.&y=\sqrt{2}x+\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{e}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&y=\sec x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &y_{0}=\sec \left ( \color{red}\displaystyle \frac{\pi }{4} \right )=\sqrt{2}\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=y'=\sec x\tan x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &m=\sec \left ( \color{red}\displaystyle \frac{\pi }{4} \right )\tan \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}.1=\sqrt{2}\\ &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )+\sqrt{2}\\ &\Leftrightarrow \: \color{red}y=\color{red}\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sin x+\cos x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{2}\: \: \textrm{akan memotong sumbu}\\ &\textrm{Y dengan ordinatnya berupa}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{\pi }{2}+1\\ \textrm{b}.&\displaystyle \frac{\pi }{2}-1\\ \textrm{c}.&1-\displaystyle \frac{\pi }{2}\\ \textrm{d}.&2+\displaystyle \frac{\pi }{2}\\ \textrm{e}.&2-\displaystyle \frac{\pi }{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&y=\sin x+\cos x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{2}\\ &y_{0}=\sin \left ( \color{red}\displaystyle \frac{\pi }{2} \right )+\cos \left ( \displaystyle \frac{\pi }{2} \right )=1+0=1\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{\pi }{2} \right )-\sin \left ( \displaystyle \frac{\pi }{2} \right )\\ &m=0-1=-1\\ &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-1\left ( x-\displaystyle \frac{\pi }{2} \right )+1\\ &\Leftrightarrow \: \color{red}y=-x+\displaystyle \frac{\pi }{2}+1\\ &\textrm{Ordinat garis singgungnya saat}\\ &\textrm{memotong sumbu-Y adalah}:\: \: x=0,\\ &\textrm{maka}\\ &\color{red}y=-0+\displaystyle \frac{\pi }{2}+1=\color{red}\displaystyle \frac{\pi }{2}+1 \end{aligned} \end{array}$

Latihan Soal 5 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{ll}\\ 41.&\textrm{Turunan pertama fungsi}\\ &h(x)=5\sin x\cos x\: \: \textrm{adalah}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\sin 2x\\ \color{red}\textrm{b}.&5\cos 2x\\ \textrm{c}.&5\sin ^{2}x\cos x\\ \textrm{d}.&5\sin ^{2}x\cos^{2} x\\ \textrm{e}.&5\sin 2x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: h(x)=5\sin x\cos x\\ h(x)&=\color{red}\displaystyle \frac{5}{2}\left ( 2\sin x\cos x \right )=\displaystyle \frac{5}{2}\sin 2x\\ h'(x)&=\color{purple}\displaystyle \frac{5}{2}\left ( \cos 2x \right ).(2)\\ &=\color{red}5\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Turunan pertama fungsi}\\ &k(x)=\cos x\tan x\: \: \textrm{adalah}\: \: k'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin x\cot x+\cos x\sec ^{2}x\\ \color{red}\textrm{b}.&-\sin x\tan x+\cos x\sec ^{2}x\\ \textrm{c}.&\sin x\tan x-\cos x\sec ^{2}x\\ \textrm{d}.&-\displaystyle \frac{1+\sin ^{2}x}{\cos x}\\ \textrm{e}.&\displaystyle \frac{1+\sin ^{2}x}{\cos x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}k(x)=\cos x\tan x\\ \textrm{guna}&\textrm{kan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'v+u.v'\\ u&=\color{black}\cos x \Rightarrow u'=-\sin x\\ v&=\color{black}\tan x \Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ k'(x)&=\left ( -\sin x \right )\tan x+\cos x.\left ( \sec ^{2}x \right )\\ &=\color{red}-\sin x\tan x+\cos x\sec ^{2}x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Jika diketahui}\: \: f(x)=\left | \tan x \right |,\: \textrm{maka}\: \: \displaystyle \frac{dy}{dx}\\ &\textrm{saat}\: \: x=k,\: \: \textrm{di mana}\: \: \displaystyle \frac{1}{2}\pi <k<\pi\\ & \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin k\\ \textrm{b}.&\cos k\\ \color{red}\textrm{c}.&-\sec ^{2}k\\ \textrm{d}.&\sec ^{2}k\\ \textrm{e}.&\cot k \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}f(x)=\left |\tan x \right |\\ \textrm{saat}&\: \: \color{red}x=k\: \: \color{blue}\textrm{dengan}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \color{black}\textrm{adal}&\color{black}\textrm{ah}:\\ f(x)&=\left | \tan x \right |,\: \: \color{black}\textrm{maka saat}\: \: \color{blue}x=k\\ f(k)&=\left | \tan k \right |=-\tan k,\: \: \color{black}\textrm{karena di}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \displaystyle \frac{dy}{dx}&=f'(k)=\color{red}-\sec ^{2}k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&\textrm{Turunan pertama}\: \: g(x)=\left | \cos x \right |\\ & \textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\left | \sin x \right |\\ \textrm{b}.&-\sin x\\ \textrm{c}.&\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \color{red}\textrm{d}.&-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{e}.&\left | \sin x \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}g(x)=\left |\cos x \right |=\sqrt{\cos ^{2}x}=\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}\\ g'(x)&=\displaystyle \frac{1}{2}\left ( \cos ^{2}x \right )^{.^{-\frac{1}{2}}}.\left ( 2\cos x \right ).\left ( -\sin x \right )\\ &=\displaystyle \frac{-2\sin x\cos x}{2\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{\sin 2x}{2\sqrt{\cos ^{2}x}}\\ &=\color{red}-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\sin x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\cos x+\sin x}{x^{2}}\\ \color{red}\textrm{b}.&\displaystyle \frac{x\cos x-\sin x}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{-x\cos x-\sin x}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{\cos x-x\sin x}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{\cos x+x\sin x}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x\Rightarrow u'=\cos x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\cos x.(x)-\sin x.1}{x^{2}}\\ &=\color{red}\displaystyle \frac{x\cos x-\sin x}{x^{2}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 46.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{1-\cos x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\sin x+\cos x+1}{x^{2}}\\ \textrm{b}.&\displaystyle \frac{x\cos x+\sin x-1}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{x\sin x-\cos x+1}{x^{2}}\\ \color{red}\textrm{d}.&\displaystyle \frac{x\sin x+\cos x-1}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{x\cos x-\sin x+1}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{1-\cos x}{x}\\ \textrm{Guna}&\textrm{kan formula}\\ &\color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=1-\cos x\Rightarrow u'=\sin x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sin x.(x)-(1-\cos x).1}{x^{2}}\\ &=\color{red}\displaystyle \frac{x\sin x+\cos x-1}{x^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\tan x}{\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1+\cos ^{2}x}{\cos ^{3}x}\\ \textrm{b}.&\displaystyle \frac{1-\cos x}{\cos ^{3}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{1+\sin x}{\cos ^{3}x}\\ \textrm{e}.&\displaystyle \frac{1-\sin ^{2}x}{\cos ^{3}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\tan x}{\cos x}\\ \textrm{Guna}&\textrm{kan formula}\\ &\color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\tan x\Rightarrow u'=\sec ^{2}x\\ v&=\cos x\Rightarrow v'=-\sin x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sec ^{2}x.(\cos x)-(\tan x).(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\sec ^{2}x.\cos x+\tan x\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\cos x+\left ( \displaystyle \frac{\sin x}{\cos x} \right )\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\cos x}+\displaystyle \frac{\sin ^{2}x}{\cos x}}{\cos ^{2}x}\\ &=\color{red}\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Turunan pertama dari}\: \: g(t)=\displaystyle \frac{\cos t+2t}{\sin t} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2\sin t+2t\cos t-1}{\sin^{2} t}\\ \textrm{b}.&\displaystyle \frac{2\sin t-2t\cos t+1}{\sin^{2} t}\\ \textrm{c}.&\displaystyle \frac{2\sin t+2t\cos t+1}{\sin^{2} t}\\ \color{red}\textrm{d}.&\displaystyle \frac{2\sin t-2t\cos t-1}{\sin^{2} t}\\ \textrm{e}.&\displaystyle \frac{-2\sin t+2t\cos t-1}{\sin^{2} t} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(t)&=\displaystyle \frac{\cos t+2t}{\sin t}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\cos t+2t\Rightarrow u'=-\sin t+2\\ v&=\sin t\Rightarrow v'=\cos t\\ \color{red}\textrm{maka}&\\ g'(t)&=\displaystyle \frac{(-\sin t+2)(\sin t)-(\cos t+2t)(\cos t)}{\sin ^{2}t}\\ &=\displaystyle \frac{-\sin ^{2}t+2\sin t-\cos ^{2}t-2t\cos t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\sin ^{2}t-\cos ^{2}t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\left (\sin ^{2}t+\cos ^{2}t \right )}{\sin ^{2}t}\\ &=\color{red}\displaystyle \frac{2\sin t-2t\cos t-1}{\sin ^{2}t} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Turunan pertama dari}\: \: h(x)=\displaystyle \frac{\sin x}{\sin x+\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x-\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\sin ^{2}x-\cos ^{2}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1}{(\sin x+\cos x)^{2}}\\ \textrm{d}.&\displaystyle \sin ^{2}x-\cos ^{2}x\\ \textrm{e}.&\displaystyle 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ h(x)&=\displaystyle \frac{\sin x}{\sin x+\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x \Rightarrow u'=\cos x\\ v&=\sin x+\cos x\Rightarrow v'=\cos x-\sin x\\ \color{red}\textrm{maka}&\\ h'(x)&=\displaystyle \frac{\cos x.(\sin x+\cos x)-\sin x.(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\cos x\sin x+\cos ^{2}x-\sin x\cos x+\sin ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x+\cos ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\color{red}\displaystyle \frac{1}{(\sin x+\cos x)^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\sin x-\cos x}{\tan x}. \: \: \textrm{Nilai}\\ &\textrm{turunan pertama fungsi}\: \: f\: \: \textrm{saat}\: \: x=45^{\circ}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle 1\\ \color{red}\textrm{d}.&\displaystyle \sqrt{2}\\ \textrm{e}.&\displaystyle \sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x-\cos x}{\tan x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x-\cos x \Rightarrow u'=\cos x+\sin x\\ v&=\tan x\Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{(\cos x+\sin x).\tan x-(\sin x-\cos x).\sec ^{2}x}{\tan ^{2}x}\\ f'\left ( 45^{\circ} \right )&=\displaystyle \frac{\left ( \displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2} \right ).1-\left ( \displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2} \right ).\left ( \sqrt{2} \right )^{2}}{1^{2}}\\ &=\displaystyle \frac{\sqrt{2}-0}{1}\\ &=\color{red}\sqrt{2} \end{aligned} \end{array}$

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

 $\begin{array}{ll}\\ 31.&\textrm{Diketahui}\: \: f(x)=2\cos x-2020\\ &\textrm{Turunan pertama fungsi}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&2\sin x\\ \color{red}\textrm{b}.&-2\sin x\\ \textrm{c}.&-2\sin x-2020x\\ \textrm{d}.&2\sin ^{2}x\\ \textrm{e}.&2\cos x-2020x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}f(x)&=2\cos x-2020\\ f'(x)&=\color{red}-2\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: f'(x)\: \: \textrm{adalah turunan pertama dari}\\ &\textrm{fungsi}\: \: f(x)=\sin ^{7}x\: ,\: \textrm{maka}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\cos^{6} x\\ \textrm{b}.&7\cos^{7} x\\ \color{red}\textrm{c}.&7\sin^{6} x\cos x\\ \textrm{d}.&7\cos ^{6}x\sin x\\ \textrm{e}.&7\cos ^{6}x\sin ^{6}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}f(x)&=\sin ^{7}x\\ \textrm{guna}&\textrm{kan formula}\\ &\color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ f'(x)&=7\sin ^{6}x\left ( \cos x \right )=\color{red}7\sin ^{6}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Turunan pertama fungsi}\: \: g(x)=-5\sin ^{3}x\\ &\textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\sin ^{2}x\cos x\\ \textrm{b}.&-5\sin ^{2}\cos ^{2}x\\ \color{red}\textrm{c}.&-15\sin ^{2}x\cos x\\ \textrm{d}.&-15\cos ^{3}x\\ \textrm{e}.&-15\sin ^{4}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}g(x)&=-5\sin ^{3}x\\ \textrm{guna}&\textrm{kan formula}\\ &\color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ g'(x)&=-5\left ( 3\sin ^{2}x \right )(\cos x)=\color{red}-15\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: h(x)=4x^{3}+\sin x+\cos x\\ &\textrm{maka}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&12x^{2}+\cos x-\sin x\\ \textrm{b}.&12x^{2}-\cos x+\sin x\\ \textrm{c}.&4x^{3}-\cos x-\sin x\\ \textrm{d}.&4x^{3}-\sin x-\cos x\\ \textrm{e}.&12x^{3}+\cos x+\sin x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}h(x)&=4x^{3}+\sin x+\cos x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ \textrm{pada}&\: \textrm{fungsi aljabarnya, yaitu}:\color{black}y=4x^{3}\Rightarrow y'=12x^{2}\\ \textrm{seda}&\textrm{ngkan fungsi transendennya mengikuti}\\ \textrm{turu}&\textrm{nan fungsi trigonometri biasa. Sehingga}\\ f'(x)&=\color{red}12x^{2}+\cos x-\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Jika}\: \: p(x)=-\cos ^{4}x,\: \: \textrm{maka nilai}\\ &\textrm{maka}\: \: p'\left ( \displaystyle \frac{\pi }{3} \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{4}\sqrt{3}\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}p(x)&=-\cos ^{4}x\\ \color{red}p'(x)&\color{red}=-4\cos ^{3}x.(-\sin x)=4\cos ^{3}x\sin x\\ p'\left ( \displaystyle \frac{\pi }{3} \right )&=4\cos ^{3}\left ( \displaystyle \frac{\pi }{3} \right ).\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ &=4\cos ^{3}60^{\circ}\times \sin 60^{\circ}\\ &=4\left ( \displaystyle \frac{1}{2} \right )^{3}\times \left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{4}{16}\sqrt{3}\\ &=\color{red}\displaystyle \frac{1}{4}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 36.&\textrm{Turunan pertama}\: \: q(x)=\sin ^{2}x+\cos ^{2}x\\ &\textrm{adalah}\: \: q'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cos ^{2}x-\sin ^{2}x\\ \textrm{b}.&2\cos ^{2}x-2\sin ^{2}x\\ \textrm{c}.&\cos x-\sin x\\ \textrm{d}.&2\cos x-2\sin x\\ \color{red}\textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}q(x)&=\sin ^{2}x+\cos ^{2}x\\ \color{red}\textrm{guna}&\color{red}\textrm{kan formula identitas}:\: \color{black}\sin ^{2}x+\cos ^{2}x=1\\ \textrm{Sehi}&\textrm{ngga soal di atas dapat dituliskan menjadi}\\ q(x)&=1,\: \: \textrm{maka}\\ q'(x)&=0\\ \color{purple}\textrm{inga}&\color{purple}\textrm{t bahwa}\: \: \color{black}y=a\Rightarrow \displaystyle \frac{dy}{dx}=\color{red}0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Nilai dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.&-\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dari}&\: \textrm{soal diketahui}:\: \\ f(x)&=\sin \displaystyle \frac{\pi }{3}\\ \textrm{Nila}&\textrm{i dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ \textrm{arti}&\textrm{nya bermakna, berapkah}\: \: f'\left ( x \right )?\\ \color{red}\textrm{maka}&\\ f'\left ( x \right )&=\color{red}0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Jika}\: \: f(x)=8x-\sin ^{3}x,\\ &\textrm{maka nilai}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&4x^{2}-3\cos^{2}x \\ \textrm{b}.&8x-3\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&8-3\sin ^{2}x\cos x\\ \textrm{d}.&8+\sin ^{2}x\cos x\\ \textrm{e}.&3\sin ^{2}x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui dari soal}\: f(x)=8x-\sin ^{3}x\\ \color{red}\textrm{maka}&\: \textrm{nilai dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}=f'(x)\\ f'(x)&=\color{red}8-3\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Turunan pertama fungsi}\: \: f(x)=\sqrt{\sin x},\\ &\textrm{adalah}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2\sqrt{\sin x}} \\ \textrm{b}.&\displaystyle \frac{\cos x}{\sqrt{\sin x}}\\ \color{red}\textrm{c}.&\displaystyle \frac{\cos x}{2\sqrt{\sin x}}\\ \textrm{d}.&-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ \textrm{e}.&\displaystyle \frac{2\cos x}{\sqrt{\sin x}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}f(x)=\sqrt{\sin x}=\sin ^{.^{\frac{1}{2}}}x\\ f'(x)&=\displaystyle \frac{1}{2}\left ( \sin ^{.^{-\frac{1}{2}}}x \right ).(\cos x)\\ &=\displaystyle \frac{\cos x}{2\sin ^{.^{\frac{1}{2}}}x}\\ &=\color{red}\displaystyle \frac{\cos x}{2\sqrt{\sin x}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: g'(x)\: \: \textrm{adalah turunan pertama}\\ &\textrm{fungsi}\: \: g(x)\: \: \textrm{dengan}\: \: g(x)=5\tan ^{2}x,\\ &\textrm{maka}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&10\cos ^{2}x\sin x\\ \textrm{b}.&10\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&\displaystyle \frac{10\sin x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{10\cos ^{3}x}{\sin x}\\ \textrm{e}.&\displaystyle \frac{10}{\sin ^{2}x-\cos ^{2}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: g(x)=5\tan ^{2}x\\ g'(x)&=5\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )\\ &=\color{purple}10\tan x\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=10\left ( \displaystyle \frac{\sin x}{\cos x} \right )\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\color{red}\displaystyle \frac{10\sin x}{\cos ^{3}x} \end{aligned} \end{array}$.

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

 $\begin{array}{ll}\\ 21.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{b}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 22.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \color{red}\textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=\color{red}0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 23.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\color{blue}\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=\color{red}-2 \end{aligned} \end{array}$.

$\begin{array}{l}\\ 24.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \color{red}\textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=\color{red}5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \color{red}\textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=\color{red}(ab)^{2} \end{aligned} \end{array}$

$\begin{array}{l}\\ 27.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \color{red}\textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=\color{red}0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\color{purple}\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 29.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \color{red}\textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: \color{red}x=2\: \: \color{black}\textrm{dan}\: \: \color{red}x=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \color{red}\textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\color{blue}\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=\color{red}-3 \end{aligned} \end{array}$

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

 $\begin{array}{ll}\\ 11.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \color{purple}\frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \color{red}\textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \color{purple}\displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=\color{red}2 \end{aligned} \end{array}$.

$\begin{aligned}&\textrm{ada cara lain yang lebih sede}\textrm{rhana, yaitu:}\\ .\qquad\: \, &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\qquad\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ &\textrm{Jika}\quad \\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga}\quad\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=\color{red}2 \end{aligned}$.

$\begin{array}{ll}\\ 13.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }=\color{red}\infty \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \color{purple}\displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=\color{red}0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \color{red}\textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\color{red}\frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 16.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\color{blue}\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{purple}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=\color{red}-\frac{5}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \color{purple}\displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \color{red}\textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\color{red}\infty \end{aligned} \end{array}$

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

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

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

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

$\begin{array}{ll} 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x),\: \: \textrm{dengan kondisi}\\ &\qquad\qquad f(x)=\begin{cases} \displaystyle \frac{x^{2}-4}{x-2} &\textrm{untuk}\: \: x\neq 2 \\ &\\ 6x & \textrm{untuk}\: \: x=2 \end{cases}\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{tidak ada}\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 4\\ &\textrm{e}.\quad 12\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sebagaimana pada soal, maka}\\ &\textrm{Harga limit kiri}:\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Dan harga limit kanan}:\\ &\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Karena limit kiri}=\textrm{limit kanan},\: \: \textrm{yaitu}\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=4,\: \: \textrm{maka}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\color{red}4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\color{red}\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2020}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2021x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2020}{x^{2}}}{2-\displaystyle \frac{2021}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2020}{\infty ^{2}}}{2-\displaystyle \frac{2021}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=\color{red}4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\color{red}\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\times \color{purple}\displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2021}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2020x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2021}{x}}{-\sqrt{9-\displaystyle \frac{2020}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2021}{\infty }}{-\sqrt{9-\displaystyle \frac{2020}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=\color{red}-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\color{red}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=\color{red}25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \color{red}\textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e} \end{array}$

$\begin{aligned}.\qquad \: \, &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=\color{red}-\displaystyle \frac{5}{3} \end{aligned}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=\color{red}1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \color{purple}\frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }\\ &=\color{red}\infty \end{array}$.

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

$\color{purple}\begin{aligned}\textrm{Sebagai}&\: \: \color{black}\textbf{CATATAN}\: \textrm{di sini}\\ \textrm{Sifat-sif}&\textrm{at bilangan tak hingga}\\ (1)\: \: &\infty +\infty =\infty \\ (2)\: \: &-\infty +(-\infty )=-\infty \\ (3)\: \: &\infty \times \infty =\infty\\ (4)\: \: &-\infty +(-\infty )=\infty \\ (5)\: \: &k.\infty =\infty ,\quad k\: \: \color{blue}\textrm{positif}\\ (6)\: \: &k.(-\infty )=-\infty,\quad k\: \: \color{blue}\textrm{positif} \\ (7)\: \: &k.\infty =-\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ (8)\: \: &k.(-\infty )=\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ \textrm{yang ha}&\textrm{rus dihindari}\\ (1)\: \: &\color{red}\infty -\infty ,\quad \: \: \color{black}\textrm{bentuk tak tentu}\\ (2)\: \: &\color{red}\displaystyle \frac{\infty }{\infty },\: -\displaystyle \frac{\infty }{\infty },\: \: \color{black}\textrm{dan}\: \: \color{red}\frac{0}{0} \end{aligned}$.

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

Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{ll}\\ 86.&(\textrm{Soal SNMPTN})\\ &\textrm{Jika}\: \: x>5\: \: \textrm{dan}\: \: y<3,\: \: \textrm{maka}\\ &\textrm{nilai }\: \: x-y\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{lebih besar dari pada 1}\\ \textrm{b}.&\textrm{lebih besar dari pada 3}\\ \textrm{c}.&\textrm{lebih besar dari pada 8}\\ \textrm{d}.&\textrm{lebih besar dari pada 5}\\ \color{red}\textrm{e}.&\textrm{lebih besar dari pada 2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}\\ x&>5\: \: \color{red}\&\: \: \color{blue}y<3\\ \textrm{m}&\textrm{aka}\\ &\begin{array}{llllll}\\ x>5&\Rightarrow &x&>5\\ y<3&\Rightarrow &\color{black}-y&\color{black}>-3&\color{red}+\\\hline &&\color{red}x-y&>\color{red}2 \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 87.&\textrm{Batas pertidaksamaan}\: \: 5x-7>13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\\ \color{red}\textrm{b}.&x>4\\ \textrm{c}.&x>-4\\ \textrm{d}.&x<4\\ \textrm{e}.&-4<x<4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}5x&-7>13\\ 5x&>13+7\\ 5x&>20\\ x&\color{red}>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 88.&\textrm{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \color{red}\textrm{b}.&x<3\displaystyle \frac{1}{3}\\ \textrm{c}.&x>3\displaystyle \frac{1}{3}\\ \textrm{d}.&x>3\\ \textrm{e}.&\textrm{Semua pilihan jawaban salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}2x+3&>5x-7\\ 2x-5x&>-7-3\\ -3x&>-10\quad \color{black}\textrm{dikali (-1)}\\ 3x&<10\\ x&<\color{red}\displaystyle \frac{10}{3}=3\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 89.&(\textbf{UMPTN 01})\textrm{Jika pertidaksamaan}\\ & 2x-3a>\displaystyle \frac{3x-1}{2}+ax\: \: \textrm{mempunyai}\\ &\textrm{penyelesaian}\: \: x>5,\: \: \textrm{maka nilai}\: \: a\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{8}\\ \color{red}\textrm{c}.&\displaystyle \frac{3}{8}\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}2x-3a&>\displaystyle \frac{3x-1}{2}+ax\quad \color{black}\textrm{tiap ruas}\: (\times 2)\\ 4x-6a&>3x-1+2ax\\ 4x-3x&-2ax>-1+6a\\ x-2a&x>-1+6a\\ (1-2a)&x>-1+6a\\ x&>\displaystyle \frac{-1+6a}{1-2a}\\ \textrm{Diketa}&\textrm{hui}:\: \: x>5\: \: \color{red}\textrm{adalah penyelesaian}\\ \color{red}\textrm{maka}\: \: &\\ 5&=\displaystyle \frac{-1+6a}{1-2a}\\ 5-10a&=-1+6a\\ -6a-10&a=-1-5\\ -16a&=-6\\ a&=\displaystyle \frac{-6}{-16}\\ &=\color{red}\displaystyle \frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 90.&(\textbf{UMPTN 94})\\ &\textrm{Apabila}\: \: a<x<b\: \: \textrm{dan}\: \: a<y<b\\ & \textrm{maka berlaku}\: \: \: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a<x-y<b\\ \textrm{b}.&b-a<x-y<a-b\\ \color{red}\textrm{c}.&a-b<x-y<b-a\\ \textrm{d}.&\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)\\ \textrm{e}.&\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{array}{llllll}\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &-a>-y>-b&&\\\hline &\color{purple}\textrm{saat}&\color{black}\textrm{di susun ulang}&\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &\color{black}-b<-y<-a&\color{red}+&\\\hline &&\color{red}a-b\color{blue}<\color{red}x-y\color{blue}<&\color{red}b-a \end{array} \end{array}$.

$\begin{array}{ll}\\ 91.&\textrm{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y-x>4\\ \textrm{b}.&y-x<4\\ \textrm{c}.&y+x+4>0\\ \color{red}\textrm{d}.&y+x+4<0\\ \textrm{e}.&y+x<1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&2y-5>2x+4y+3\\ &2y-4y-2x-5-3>0\\ &-2y-2x-8>0\: \: \color{black}\textrm{dibagi}\: \left ( -\displaystyle \frac{1}{2} \right )\\ &\color{red}y+x+4<0 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 93.&\textrm{Jika}\: \: \displaystyle \frac{1}{5}<\frac{1}{x}\: \: \textrm{dan}\: \: x<0\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{1}{5}\\ \color{red}\textrm{b}.&-5<x<0\\ \textrm{c}.&0<x<5\\ \textrm{d}.&x<-5\\ \textrm{e}.&-\displaystyle \frac{1}{5}<x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ \displaystyle \frac{1}{5}&<\frac{1}{x}\: \: \: \textrm{dan}\: \: x<0\\ \displaystyle \frac{1}{5}&<\displaystyle \frac{1}{x}\\ x&<5 \\ x&>-5\qquad \color{black}\textrm{karena}\: \: x<0\\ \textrm{Sehi}&\textrm{ngga}\\ \color{red}-5<&\color{red}x<0 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 95.&\textrm{Jika}\: \: -2<y<3\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9<(y-2)^{2}<16\\ \textrm{b}.&4<(y-2)^{2}<16\\ \textrm{c}.&1<(y-2)^{2}<16\\ \color{red}\textrm{d}.&0\leq (y-2)^{2}<16\\ \textrm{e}.&-1<(y-2)^{2}<16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}:\: -2<y<3\\ &\color{red}\bullet \quad \textrm{saat dikurangi}\: \: 2\\ &\qquad \Leftrightarrow \: -2-2<y-2<3-2\\ &\qquad -4<y-2<1\\ &\color{red}\bullet \quad \textrm{Saat}\: \: -4<y-2<0\\ &\qquad (-4)^{2}<(y-2)^{2}<0^{2}\quad \textrm{dikuadratkan}\\ &\qquad 16>(y-2)^{2}>0\\ &\qquad 0<(y-2)^{2}<16\\ &\color{red}\bullet \quad \textrm{Saat}\: \: 0\leq y-2<1\\ &\qquad 0^{2}\leq (y-2)^{2}<1^{2}\\ &\qquad 0<(y-2)^{2}<1\\ &\textrm{Jadi}\: ,\: \: \color{red}0\leq (y-2)<16 \end{aligned} \end{array}$

Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas X

 $\begin{array}{ll}\\ 76.&\textrm{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ &\textrm{ketiga angka tersebut adalah 9. Angka kedua}\\ &\textrm{dikurangi angka pertama dan angka ketiga }\\ &\textrm{sama dengan 1. Dua kali angka pertama sama}\\ &\textrm{dengan jumlah angka kedua dan angka ketiga.}\\ &\textrm{Angka puluhan pada bilangan tersebut adalah}\\ &....\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \color{red}\textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Model matematikanya}\\ &\left\{\begin{matrix} A+B+C=9\: \: \qquad....(1)\\ 2B-A-C=1\qquad\: ....(2)\\ 2A=B+C\: \: \: \: \qquad\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle A+B+C&=9\\ \displaystyle -A+B-C&=1&+\\\hline \qquad2B&=10\\ \: \: \: \: \qquad\qquad B&=5&...(4)\\ \end{array}\\ &\textrm{Jadi},\: \textrm{bilangan kedua adalah}\: =\: B=\color{red}5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 77.&(\textbf{SIMAK UI 2010})\\ &\textrm{Jika}\: \: x+y+2z=K,\: x+2y+z=K,\\ &2x+y+z=K\: \: \textrm{dengan}\: \: K\neq 0,\: \textrm{maka}\\ &x^{2}+y^{2}+z^{2}\: \: \textrm{bila dinyatakan dalam}\: \: K\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{16}K^{2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{16}K^{2}\\ \textrm{c}.&\displaystyle \frac{4}{17}K^{2}\\ \textrm{d}.&\displaystyle \frac{3}{8}K^{2}\\ \textrm{e}.&\displaystyle \frac{2}{3}K^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+2z=K\: \: \qquad....(1)\\ x+2y+z=K\qquad\: ....(2)\\ 2x+y+z=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\color{black}\textrm{maka}\\ &\color{red}\left\{\begin{matrix} z+(x+y+z)=K\: \: \qquad....(1)\\ y+(x+y+z)=K\qquad\: ....(2)\\ x+(x+y+z)=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle x+y+2z&=K\\ \displaystyle x+2y+z&=K&\\ \displaystyle 2x+y+z&=K&+\\\hline 4x+4y+4z&=3K\\ x+y+z&=\displaystyle \frac{3}{4}K&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (4)\: \: \textrm{disubstitusikan ke}\: \: (1),(2),\: \textrm{dan}\: (3)\\ &\textrm{Jelas bahwa akan didapatkan}\\ &x=y=z=\displaystyle \frac{1}{4}K\\ &\textrm{Jadi},\: \: x^{2}+y^{2}+y^{2}=3\left ( \displaystyle \frac{1}{4}K \right )^{2}=\color{red}\displaystyle \frac{3}{16}K^{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 78.&\textrm{Diketahui}\: \: 0,15252525252...=\displaystyle \frac{p}{2q+r}\\ &\textrm{Jika jumlah}\: \: p\: \: \textrm{dan}\: \: q=\textrm{3 kali}\: \: r,\: \textrm{maka}\\ &\textrm{masing-masing harga}\: \: p,q, \: \textrm{dan}\: \: r=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 152,2819,2584\\ \textrm{b}.&\displaystyle 252,\displaystyle \frac{5638}{7},\frac{8102}{21}\\ \color{red}\textrm{c}.&\displaystyle 151,\frac{2819}{7},\frac{1292}{7}\\ \textrm{d}.&\displaystyle 151,\displaystyle \frac{2819}{7},\frac{2584}{7}\\ \textrm{e}.&\displaystyle 152,\frac{2819}{14},\frac{1292}{7} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\color{red}\begin{cases} 0,1\overline{5252}& =\displaystyle \frac{p}{2q+r}\: .....(1)\\ \quad p+q & =3r\: ............(2) \end{cases}\\ &\color{black}\textrm{dan}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \qquad \color{red}x&=0,15252525252...\\ \displaystyle 1000\color{red}x&=152,5252525252...\\ \displaystyle \quad10\color{red}x&=\: \: \: \: \: 1,5252525252...&-\\\hline \: \: 990\color{red}x&=151\\ \qquad \color{red}x&=\displaystyle \frac{151}{990},\: \: \color{black}\textrm{maka}\\ \displaystyle \frac{p}{2q+r}&=\displaystyle \frac{151}{990}\\ &\begin{cases} p &=151 \: \: .......(3)\\ 2p+r &=990 \: \: .......(4) \end{cases} \end{array}\\ &\textrm{Dari}\: \: (3)\: \textrm{diperoleh}:q=3r-p=3r-151\: ....\color{blue}(5)\\ &\textrm{Dari}\: \: (5)\: \: \textrm{disubstitusikan ke}\: \: (4)\\ &\begin{aligned}2q+r&=990\\ 2(3r-151)+r&=990\\ 6r-302+r&=990\\ 7r&=990+302=1292\\ r&=\displaystyle \frac{1292}{7}\: .....\color{blue}(6) \end{aligned}\\ &\textrm{Dari}\: \: (3)\&(6)\: \: \textrm{disubstitusikan ke}\: \: (2)\\ &\color{purple}\begin{aligned}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}\: .....\color{blue}(7) \end{aligned}\\ &\textrm{Jadi},\: \: p,q,r\: \: \textrm{adalah}\: :\: \color{red}\displaystyle 151,\frac{2819}{7},\frac{1292}{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 79.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \color{red}\textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\color{red}\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\color{black}\textrm{tidak berpenyelesaian, maka}\\ &\color{black}\textrm{ingat penyelesaian metode matrik}\\ &\color{black}\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\color{blue}\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 80.&\textrm{Diketahui}\\ &\begin{pmatrix} \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{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix}\\ &\textrm{Nilai}\: \: x,y,\: \: \textrm{dan}\: \: z\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}\\ \textrm{b}.&-1,5,1\\ \color{red}\textrm{c}.&1,5,-1\\ \textrm{d}.&-1,1,5\\ \textrm{e}.&5,1,-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{red}\left\{\begin{matrix} \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\quad \quad....(1)\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2\qquad\: ....(2)\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0\: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ \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&-\\\hline \quad\qquad \qquad \displaystyle \frac{5}{5}z&=-1&\\ \: \: \: \quad\qquad \qquad \displaystyle z&=-1&...(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lllllll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\left | \times 1 \right |&\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&\left | \times 2 \right |&-\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0&-\\\hline &&&\: \: \: \: \displaystyle \frac{5}{5}x\qquad\qquad\: =1&\\ &&&\: \: \: \quad x=-1\: ........(5) \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{akan didapatkan}\\ &y=5\\ &\textrm{Jadi},\: \: (x,y,z)=\color{red}(1,5,-1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 81.&\textrm{Diketahui suatu fungsi kuadrat}\\ &f(x)=ax^{2}+bx+c.\: \: \textrm{Jika fungsi}\\ &(-1,0),(1,4),\: \textrm{dan}\: \: (2,9),\: \: \textrm{maka}\\ &\textrm{fungsi yang dimaksud adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle f(x)=x^{2}-2x+3\\ \textrm{b}.&f(x)=x^{2}+2x+3\\ \textrm{c}.&f(x)=x^{2}+2x-3\\ \textrm{d}.&f(x)=x^{2}-2x-3\\ \color{red}\textrm{e}.&f(x)=x^{2}+2x+1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} (-1,0)\Rightarrow f(-1)=a-b+c=0\: ....\color{red}(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4\: ....\color{red}(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9\: ....\color{red}(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)\&(2),\: \textrm{didapatkan}\\ &b=2\: \: ...............\color{blue}(4)\\ &\textrm{Saat}\: \: (1)\&(3),\: \textrm{didapatkan}\\ &\color{blue}\begin{array}{llll}\\ 4a+2b+c&=9&\\ \: \: \: \: a-b+c&=0&-\\\hline \quad\qquad \qquad 3a+3b&=9&\\ \: \: \: \quad\qquad \qquad a+b&=3&...(5) \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{cases} a &=1 \\ c & =1 \end{cases}\\ &\textrm{Jadi},\: \: f(x)=ax^{2}+bx+c=\color{red}x^{2}+2x+1 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 83.&\textrm{Diketahui sistem persamaan}\\ &y+\displaystyle \frac{2}{x+z}=4\\ &5y+\displaystyle \frac{18}{2x+y+z}=18\\ &\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\\ &\textrm{Nilai}\: \: y+\sqrt{x^{2}-2xz+y^{2}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 3\\ \textrm{b}.&5\\ \textrm{c}.&7\\ \textrm{d}.&9\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} y+\displaystyle \frac{2}{x+z}=4\qquad\quad\\ 5y+\displaystyle \frac{18}{2x+y+z}=18\\ \displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\end{matrix}\right.\\ &\textrm{Jika disederhanakan beberapa bagian}\\ &\begin{cases} y+2A & =4\: ....(1) \\ 5y+18B & =18\: ....(2) \\ 8A-6B & =3\: ....(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2)\&(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ y+2A&=4&\left | \times 5 \right |&5y+10A=20\\ 5y+3(8A-3)&=18&\left | \times 1 \right |&5y+24A=27&-\\\hline &&&\: \: \quad-14A=-7\\ &&&\: \: \: \: \: \: \: \qquad A=\displaystyle \frac{1}{2}...(4)\\ \textrm{maka}\: B=\displaystyle \frac{1}{6}\: \& &y=3&&\\ \textrm{akibatnya}\\ \begin{cases} x &=1 \\ z &=1 \end{cases} \end{array} \\ &\textrm{Jadi},\: \: y+\sqrt{x^{2}-2xz+z^{2}}=3+0=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 84.&\textrm{Diberikan}\: \: a,b,\: \textrm{dan}\: \: c \: \: \textrm{adalah angka-angka}\\ &\textrm{dari bilangan 3 digit yang memenuhi}\\ &49a+7b+c=286.\: \: \textrm{Nilai dari}\: \: a+b+c\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&16\\ \textrm{b}.&17\\ \textrm{c}.&18\\ \textrm{d}.&19\\ \textrm{e}.&20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{blue}49a+7b+c=286\\ &\textrm{Nilai maksimum}\: \: a\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}49\times 5=245,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}245+7b+c=286\Rightarrow 7b+c=286-245=41\\ &\textrm{Nilai maksimum}\: \: b\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}7\times 5=35,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}35+c=41\Rightarrow c=41-35=6\\ &\color{black}\textrm{Sehingga}\: \: \color{blue}a,b,\: \: \color{black}\textrm{dan}\: \: \color{blue}c\: \: \color{black}\textrm{adalah}\: \: \color{blue}5,5,\: \: \color{black}\textrm{dan}\: \: \color{blue}6\\ &\textrm{Jadi},\: \textrm{nilai}\: \: \color{red}a+b+c=5+5+6=16 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 85.&\textrm{Diketahui sistem persamaan}\\ &(2x+3y)^{.^{\log (x-y+2z)}}=1\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &5x+3y+8z=2\\ &\textrm{Himpunan penyelesaian yang}\\ &\textrm{memenuhi adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ \displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \}\\ \textrm{b}.&\left \{ -\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6} \right \}\\ \textrm{c}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6} \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6} \right \}\\ \color{red}\textrm{e}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\color{blue}\textrm{Untuk persamaan}\: \: (1)\\ &(2x+3y)^{.^{\log (x-y+2z)}}=(2x+3y)^{0}\\ &\Leftrightarrow (x-y+2z)=10^{0}=1\\ &\color{blue}\textrm{Untuk persamaan}\: \: (2)\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &\Leftrightarrow 3^{2x+y+z+3(3z+2y+x)}=3^{4}\\ &\Leftrightarrow 5x+7y+10z=4\\ &\color{blue}\textrm{Sehingga sistem persamaan akan terlihat}\\ &\left\{\begin{matrix} x-y+2z=1\: \: \qquad....(1)\\ 5x+7y+10z=4\quad\: ....(2)\\ 5x+3y+8z=2\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (2)\&(3),\: \textrm{maka}\\ &\begin{array}{llll}\\ 5x+7y+10z&=4&\\ 5x+3y+8z&=2&-\\\hline \qquad 4y\quad+2z&=2\\ \qquad 2y\quad+z&=1\: ...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llll}\\ 5x-5y+10z&=5&\\ 5x+7y+10z&=4&-\\\hline \quad -12y\quad&=1\\ \: \: \: \: \qquad y\quad&=-\displaystyle \frac{1}{12}\: ...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (5)\: \: \textrm{disubstistusikan ke}\: \: (4)\\ &\color{blue}\begin{aligned}2y+z&=1\\ 2\left ( -\displaystyle \frac{1}{12} \right )+z&=1\\ z&=1+\displaystyle \frac{1}{6}\\ z&=\displaystyle \frac{7}{6} \end{aligned}\\ &\textrm{Cukup jelas juga}\: \: x=....\\ &\textrm{Jadi},\: \textrm{pilihannya adalah}\: \: \color{red}e \end{aligned} \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