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

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

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

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

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

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

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

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

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

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

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