Contoh Soal 4 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{l}\\ 16.& \text{Jika}f\text{dan}g\text{adalah fungsi yang }\\ & \text{mempunyai invers dan memenuhi}\\ & f(2x)=g(x-3),\text{maka}{f}^{-1}(x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& g^{-1}\left({\displaystyle \frac{x}{2}-\frac{2}{3}}\right)\\ \text{b}.& g^{-1}\left({\displaystyle \frac{x}{2}}\right)-{\displaystyle \frac{2}{3}}\\ \text{c}.& g^{-1}(2x+6)\\ \text{d}.& 2g^{-1}(x)-6\\ \text{e}.& 2g^{-1}(x)+6\end{array}\\ & (\mathbf{\text{SBMPTN 2016 Mat Das}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan bahwa}\\ & f(2x)=g(x-3)=x,\\ & \text{maka}\\ & \{\begin{array}{ll}{f}^{-1}(x)& =2x\\ g^{-1}(x)& =x-3\end{array}\end{array}\\ & \begin{array}{|cc|}\hline \text{Sintak}& \text{Hasil}\\ \begin{array}{rl}{g}^{-1}(x)& =x-3\\ x& =g^{-1}(x)+3\\ & \end{array}& \begin{array}{rl}{f}^{-1}(x)& =2x\\ & =2(g^{-1}(x)+3)\\ & =2g^{-1}(x)+6\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Jika}{f}^{-1}(x)={\displaystyle \frac{x-1}{5}\text{dan}{g}^{-1}(x)={\displaystyle \frac{3-x}{2},}}\\ & \text{maka}{(f\circ g)}^{-1}(6)=....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 2\\ \text{b}.& 0& \text{c}.& 1& \text{e}.& 3\end{array}\\ & (\mathbf{\text{UMPTN 1995}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui b}& \text{ahwa}:\\ & \{\begin{array}{ll}{f}^{-1}(x)& ={\displaystyle \frac{x-1}{5}}\\ g^{-1}(x)& ={\displaystyle \frac{3-x}{2}}\end{array}\\ {(f\circ g)}^{-1}(x)& =(g^{-1}\circ f^{-1})(x)\\ & ={\displaystyle \frac{3-\left({\displaystyle \frac{x-1}{5}}\right)}{2}}\\ {(f\circ g)}^{-1}(6)& ={\displaystyle \frac{3-\left({\displaystyle \frac{6-1}{5}}\right)}{2}}\\ & ={\displaystyle \frac{3-1}{2}=\frac{2}{2}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Invers dari}f(x)={125}^x\text{adalah}{f}^{-1}(x)\\ & \text{Nilai dari}{f}^{-1}\left(5\sqrt{5}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& {\displaystyle \frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{1}{2}}& \text{c}.& {\displaystyle \frac{1}{6}}& \text{e}.& -{\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}f(x)={125}^x,\text{maka}\\ & f(x)=y={125}^x\\ & \text{Kedua ruas dilogkan masing-masing}\\ & \log y=\log {125}^x\\ & \iff \log y=x\log 125\\ & \iff x\log 125=\log y\\ & \iff x={\displaystyle \frac{\log y}{\log 125}}\\ & \iff x={}^{125}\log y\\ & \iff f^{-1}(x)={}^{125}\log x\\ & \text{Selanjutnya}\\ & f^{-1}(x)={}^{125}\log x\\ & \iff f^{-1}(5\sqrt{5})={}^{125}\log (5\sqrt{5})\\ & \iff f^{-1}(5\sqrt{5})={}^{5^3}\log 5^{{.}^{\frac{3}{2}}}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{3}{}^5\log 5}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{1}{2}.1}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{1}{2}}\end{array}\end{array}$

Contoh Soal 3 Fungsi Komposisi dan Fungsi Invers

 $\begin{array}{l}\\ 11.& \text{Diketahui beberapa fungsi memiliki }\\ & \text{sifat-sifat sebagaimana berikut ini}:\\ & (i)\mathrm{\Phi }(-x)=-\mathrm{\Phi }(x)\text{untuk setiap}x\\ & (ii)\mathrm{\Phi }(-x)=\mathrm{\Phi }(x)\text{untuk setiap}x\\ & \text{Jika diketahui fungsi}f\text{dan}g\\ & \text{memiliki sifat}(i)\text{dan fungsi}h\text{dan}k\\ & \text{memiliki sifat}(ii),\text{maka pernyataan }\\ & \text{berikut yang salah adalah}....\\ & (1)(f+g)(-x)=-(f+g)(x)\\ & (2)(f.k)(-x)=-(f.k)(x)\\ & (3)(h-k)(-x)=(h-k)(x)\\ & (4)(h-g)(-x)=(h-g)(x)\\ & \begin{array}{l}\\ \text{a}.& (1),(2)\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\text{saja}\\ \text{e}.& \text{semuanya benar}\end{array}\\ & (\mathbf{\text{SIMAK UI 2014 Mat Das}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{l}\mathrm{\Phi }(-x)=-\mathrm{\Phi }(x)\\ (\text{fungsi ganjil})\{\begin{array}{ll}f& \text{ misal }f(x)=x\\ g& \text{ misal }g(x)=2x\end{array}\\ \\ \mathrm{\Phi }(-x)=\mathrm{\Phi }(x)\\ (\text{fungsi genap})\{\begin{array}{ll}h& \text{ misal }h(x)=x^2\\ k& \text{ misal }k(x)=2x^2\end{array}\end{array}\\ & \end{array}\\ & \begin{array}{|c|}\hline \begin{array}{rl}(1)(f+g)(-x)& =-(f+g)(x)\\ & \text{benar}\\ (2)(f.k)(-x)& =-(f.k)(x)\\ & \text{benar}\end{array}\\ \begin{array}{rl}(3)(h-k)(-x)& =(h-k)(x)\\ & \text{benar}\\ (4)(h-g)(-x)& =(h-g)(x)\\ & \text{salah}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Jika}f(x)={\displaystyle \frac{1}{2x-1}\text{dan}(f\circ g)(x)={\displaystyle \frac{x}{3x-2}}}\\ & \text{maka}g(x)=....\\ & (\mathbf{\text{UMPTN 1998}})\\ & \begin{array}{l}\\ \text{a}.& x+{\displaystyle \frac{1}{2}}& & & \text{d}.& 1-{\displaystyle \frac{2}{x}}\\ \\ \text{b}.& x-{\displaystyle \frac{1}{2}}& \text{c}.& 2-{\displaystyle \frac{1}{x}}& \text{e}.& 2-{\displaystyle \frac{1}{2x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 1}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{2x-1},\text{dan}}\\ & (f\circ g)(x)={\displaystyle \frac{x}{3x-2},\text{maka}}\\ & \iff (f\circ g)(x)=f(g(x))={\displaystyle \frac{x}{3x-2}}\\ & \iff {\displaystyle \frac{1}{2g(x)-1}={\displaystyle \frac{x}{3x-2}}}\\ & \iff {\displaystyle \frac{1}{2g(x)-1}={\displaystyle \frac{1}{\left({\displaystyle \frac{3x-2}{x}}\right)}}}\\ & \text{Dari bentuk di atas didapatkan}\\ & 2g(x)-1={\displaystyle \frac{3x-2}{x}}\\ & \iff 2g(x)=1+(3-{\displaystyle \frac{2}{x}})\\ & \iff 2g(x)=4-{\displaystyle \frac{2}{x}}\\ & \iff g(x)=2-{\displaystyle \frac{1}{x}}\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{2x-1},\text{dengan}}\\ & f^{-1}(x)={\displaystyle \frac{x+1}{2x}.........(\text{tunjukkan sendiri})}\\ & \text{serta}(f\circ g)(x)={\displaystyle \frac{x}{3x-2},\text{maka}}\\ & g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ & \iff g(x)={\displaystyle \frac{f(g(x))+1}{2(f(g(x)))}}\\ & \iff g(x)={\displaystyle \frac{\left({\displaystyle \frac{x}{3x-2}}\right)+1}{2\left({\displaystyle \frac{x}{3x-2}}\right)}}\\ & \iff g(x)={\displaystyle \frac{{\displaystyle \frac{4x-2}{3x-2}}}{{\displaystyle \frac{2x}{3x-2}}}}\\ & \iff g(x)={\displaystyle \frac{4x-2}{2x}}\\ & \iff g(x)=2-{\displaystyle \frac{1}{x}}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Jika}f(x)=x^2-9\text{dan}(f\circ g)(x)=x(x-6)\\ & \text{rumus fungsi}g(x)=....\\ & \begin{array}{l}\\ \text{a}.& x+3& & & \text{d}.& 3x+1\\ \text{b}.& x-3& \text{c}.& -x& \text{e}.& x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 1}}\\ & \text{Diketahui bahwa}f(x)=x^2-9,\text{dan}\\ & (f\circ g)(x)=x(x-6)=x^2-6x,\text{maka}\\ & \iff (f\circ g)(x)=f(g(x))=x^2-6x\\ & \iff {(g(x))}^2-9=x^2-6x\\ & \iff {(g(x))}^2-9=x^2-6x+9-9\\ & \iff {(g(x))}^2-9=(x-3{)}^2-9\\ & \text{Dari bentuk di atas didapatkan}\\ & g(x)=x-3\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)=x^2-9,\text{dengan}\\ & f^{-1}(x)=\sqrt{x+9}.......(\text{tunjukkan sendiri})\\ & \text{serta}(f\circ g)(x)=x^2-6x,\text{maka}\\ & g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ & \iff g(x)=\sqrt{(f(g(x)))+9}\\ & \iff g(x)=\sqrt{x^2-6x+9}\\ & \iff g(x)=\sqrt{(x-3{)}^2}\\ & \iff g(x)=x-3\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Jika}f(x)={\displaystyle \frac{1}{\sqrt{x^2-2}}\text{dan}}\\ & (f\circ g)(x)={\displaystyle \frac{1}{\sqrt{x^2+6x+7}},}\\ & \text{maka}g(x+2)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{x+3}}& & & \text{d}.& x+3\\ \text{b}.& {\displaystyle \frac{1}{x-2}}& \text{c}.& x-2& \text{e}.& x+5\end{array}\\ & (\mathbf{\text{UM UGM 2010 Mat Das}})\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}(f\circ g)(x)& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ f(g(x))& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ {\displaystyle \frac{1}{\sqrt{{(g(x))}^2-2}}}& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ {(g(x))}^2-2& =x^2+6x+7\\ {(g(x))}^2& =x^2+6x+9\\ g(x)& =\sqrt{x^2+6x+9}=\sqrt{(x+3{)}^2}\\ g(x)& =x+3\\ g(x+2)& =(x+2)+3\\ & =x+5\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{\sqrt{x^2-2}},\text{dengan}}\\ & f^{-1}(x)=\sqrt{{\displaystyle \frac{1}{x^2}+2}}.......(\text{akan ditunjukkan})\\ & \text{serta}(f\circ g)(x)={\displaystyle \frac{1}{\sqrt{x^2+6x+7}},\text{maka}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}f(x)=y& ={\displaystyle \frac{1}{\sqrt{x^2-2}}}\\ y^2& ={\displaystyle \frac{1}{x^2-2}}\\ x^2-2& ={\displaystyle \frac{1}{y^2}}\\ x^2& ={\displaystyle \frac{1}{y^2}+2}\\ x& =\sqrt{{\displaystyle \frac{1}{y^2}+2}}\\ f^{-1}(y)& =\sqrt{{\displaystyle \frac{1}{y^2}+2}}\\ f^{-1}(x)& =\sqrt{{\displaystyle \frac{1}{x^2}+2}}\end{array}& \begin{array}{rl}g(x)& =(f^{-1}\circ f\circ g)(x)\\ & =\sqrt{{\displaystyle \frac{1}{{\left({\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\right)}^2}+2}}\\ & =\sqrt{(x^2+6x+7)+2}\\ & =\sqrt{(x^2+6x+9)}\\ & =\sqrt{(x+3{)}^2}\\ & =x+3\\ g(x+2)& =(x+2)+3\\ & =x+5\\ & \end{array}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika}g(x)=2x+4\text{dan}(f\circ g)(x)=4x^2+8x-3,\\ & \text{maka}{f}^{-1}(x)=....\\ & \begin{array}{l}\\ \text{a}.& x+9\\ \text{b}.& \sqrt{x}+2\\ \text{c}.& x^2-4x-3\\ \text{d}.& \sqrt{x+1}+2\\ \text{e}.& \sqrt{x+7}+2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ccc|}\hline \text{Sintak 1}& \text{ Sintak 2}& \text{Hasil Invers}\\ \begin{array}{rl}g(x)=y& =2x+4\\ y-4& =2x\\ x& ={\displaystyle \frac{y-4}{2}}\\ f^{-1}(y)& ={\displaystyle \frac{y-4}{2}}\\ f^{-1}(x)& ={\displaystyle \frac{x-4}{2}}\\ & \end{array}& \begin{array}{rl}f(x)& =(f\circ g\circ g^{-1})(x)\\ & =4{(g^{{}^{-1}}(x))}^2+8(g^{-1}(x))-3\\ & =4{\left({\displaystyle \frac{x-4}{2}}\right)}^2+8\left({\displaystyle \frac{x-4}{2}}\right)-3\\ & =\left({\displaystyle x^2-8x+16}\right)+4x-16-3\\ & =x^2-4x-3\\ & =x^2-4x+4-7\\ & =(x-2{)}^2-7\end{array}& \begin{array}{rl}f(x)=y& =(x-2{)}^2-7\\ y+7& =(x-2{)}^2\\ \sqrt{y+7}& =(x-2)\\ (x-2)& =\sqrt{y+7}\\ x& =\sqrt{y+7}+2\\ f^{-1}(y)& =\sqrt{y+7}+2\\ f^{-1}(x)& =\sqrt{x+7}+2\end{array}\\ \hline\end{array}\end{array}$

Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{l}\\ 6.& \text{Fungsi}g:\mathbb{R}\to \mathbb{R}\text{ditentukan oleh}\\ & g(x)=x^2-x+3\text{dan}f:\mathbb{R}\to \mathbb{R}\\ & \text{sehingga}(f\circ g)(x)=3x^2-3x+4,\\ & \text{maka fungsi}f(x-2)=....\\ & \begin{array}{l}\\ \text{a}.& 2x-11& & & \text{d}.& 3x-7\\ \text{b}.& 2x-7& \text{c}.& 3x+1& \text{e}.& 3x-11\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=f(x)\\ & g(x)=x^2-x+3\end{array}\\ & (f\circ g)(x)=3x^2-3x+4\\ & \begin{array}{rl}f(g(x))& =3x^2-3x+4\\ f(x^2-& x+3)=3(x^2-x+3)-5\\ \text{Sehing}& \text{ga}\\ f(x)& =3x-5\\ f(x-2)& =3(x-2)-5\\ & =3x-11\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}g:\mathbb{R}\to \mathbb{R}\\ & \text{dengan}f(x)=x-3\text{dan}\\ & g(x)=x^2+5.\text{Jika}(f\circ g)(x)=(g\circ f)(x)\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& 4\\ \text{b}.& 2& \text{c}.& 3& \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=x-3\\ & g(x)=x^2+5\end{array}\\ & (f\circ g)(x)=(g\circ f)(x)\\ & \begin{array}{rl}f(g(x))& =g(f(x))\\ (x^2+5)-3& =(x-3{)}^2+5\\ x^2+2& =x^2-6x+14\\ 6x& =14-2\\ x& ={\displaystyle \frac{12}{6}}\\ x& =2\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}g:\mathbb{R}\to \mathbb{R}\\ & \text{dengan}f(x)=3x-10\text{dan}\\ & g(x)=4x+n.\text{Jika}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ & \text{maka nilai}n\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 15& & & \text{d}.& -10\\ \text{b}.& 10& \text{c}.& 5& \text{e}.& -15\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=3x-10\\ & g(x)=4x+n\end{array}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ & \begin{array}{rl}g(f(x))& =f(g(x))\\ 3(4x+n)-10& =4(3x-10)+n\\ 12x+3n-10& =12x-40+n\\ 2n& =-30\\ x& ={\displaystyle \frac{-30}{2}}\\ x& =-15\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}f(x)=\sqrt{2x+3}\\ & \text{dan}g(x)=x^2+1,\text{maka}(f\circ g)(2)=....\\ \\ & \begin{array}{l}\\ \text{a}.& 2,24& & & \text{d}.& 6\\ \text{b}.& 3& \text{c}.& 3,61& \text{e}.& 6,16\end{array}\\ & (\mathbf{\text{SAT Subject Test}})\\ \\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ & =\sqrt{2g(x)+3}\\ & =\sqrt{2(x^2+1)+3}\\ (f\circ g)(2)& =\sqrt{2(2^2+1)+3}\\ & =\sqrt{2(5)+3}\\ & =\sqrt{13}\\ & \approx 3,61\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Misalkan}f(x)=x^2,g(x)=2x\\ & \text{dan}h(x)=1-x.\\ & \text{Fungsi}(f\circ g\circ h)(x)=....\\ & \begin{array}{l}\\ \text{a}.& 4x^2-8x+4\\ \text{b}.& 4x^2+8x-4\\ \text{c}.& 2x^2-4x+1\\ \text{d}.& x^2-2x+1\\ \text{e}.& 4-2x+x^2\end{array}\\ \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}\\ \bullet & f(x)=x^2\\ \bullet & g(x)=2x\\ \bullet & h(x)=1-x\\ \text{mak}& \text{a}\\ & (f\circ g\circ h)=f(g(h(x)))\\ & ={(2(1-x))}^2\\ & =(2-2x{)}^2\\ & =4x^2-8x+4\end{array}\end{array}$

Contoh 7 Vektor

$\begin{array}{l}\\ 31.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}2\\ -5\end{array}\right),\overrightarrow{q}=\left(\begin{array}{c}4\\ 3\end{array}\right),\text{maka }\\ & \text{proyeksi skalar ortogonal vektor}\overrightarrow{p}\\ & \text{pada}\overrightarrow{q}\text{adalah}....\\ & \text{a}.{\displaystyle \frac{3}{5}}\\ \\ & \text{b}.{\displaystyle \frac{7}{5}}\\ \\ & \text{c}.{\displaystyle \frac{8}{5}}\\ \\ & \text{d}.{\displaystyle \frac{9}{5}}\\ \\ & \text{e}.{\displaystyle 2}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\left|\overrightarrow{r}\right|& ={\displaystyle \frac{\overrightarrow{p}.\overrightarrow{q}}{\left|\overrightarrow{q}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -5\end{array}\right).\left(\begin{array}{c}4\\ 3\end{array}\right)}{\sqrt{4^2+3^2}}}\\ & ={\displaystyle \frac{8+(-15)}{\sqrt{25}}}\\ & =\left|{\displaystyle \frac{-7}{5}}\right|\\ & ={\displaystyle \frac{7}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Panjang Proyeksi vektor}\overrightarrow{a}=\left(\begin{array}{c}5\\ 1\end{array}\right)\\ & \text{pada}\overrightarrow{b}=\left(\begin{array}{c}0\\ 4\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle -1}\\ & \text{b}.-{\displaystyle \frac{1}{2}}\\ & \text{c}.1\\ & \text{d}.{\displaystyle 2}\\ & \text{e}.4\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}5\\ 1\end{array}\right)\bullet \left(\begin{array}{c}0\\ -4\end{array}\right)}{\left|\sqrt{0^2+(-4{)}^2}\right|}}\right|\\ & =\left|{\displaystyle \frac{0-4}{4}}\right|=|-1|=1\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Proyeksi vektor ortogonal}\overrightarrow{a}=\left(\begin{array}{c}2\\ -4\end{array}\right)\\ & \text{pada}\overrightarrow{b}=\left(\begin{array}{c}-1\\ 2\end{array}\right)\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}2\\ -1\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}2\\ -2\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}2\\ -4\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}-2\\ 4\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{c}& =\left({\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right).\overrightarrow{b}\\ & =\left({\displaystyle \frac{\left(\begin{array}{c}2\\ -4\end{array}\right)\bullet \left(\begin{array}{c}-1\\ 2\end{array}\right)}{(-1{)}^2+2^2}}\right).\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =\left({\displaystyle \frac{-2-8}{1+4}}\right).\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =-2\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ -4\end{array}\right)\end{array}\end{array}$

Contoh 6 Vektor

$\begin{array}{l}\\ 26.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}-2\\ 4\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}8\\ 4\end{array}\right),\text{maka}\\ & \text{sudut yang dibentuk vektor}\overrightarrow{p}\text{dan}\overrightarrow{q}\\ & \text{adalah}....\\ & \text{a}.0^{\circ }\\ & \text{b}.{60}^{\circ }\\ & \text{c}.{45}^{\circ }\\ & \text{d}.{60}^{\circ }\\ & \text{e}.{90}^{\circ }\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}.\overrightarrow{q}& ={\displaystyle \left|\begin{array}{c}\overrightarrow{p}\end{array}\right|.\left|\begin{array}{c}\overrightarrow{q}\end{array}\right|.\cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})}\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={\displaystyle \frac{\overrightarrow{p}.\overrightarrow{q}}{\left|\overrightarrow{p}\right|.\left|\overrightarrow{q}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-2\\ 1\end{array}\right).\left(\begin{array}{c}8\\ 4\end{array}\right)}{\sqrt{(-2{)}^2+1^2}.\sqrt{8^2+4^2}}}\\ & ={\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}}\\ & ={\displaystyle \frac{0}{40}}\\ & =0\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& =\cos {90}^{\circ }\\ \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={90}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Jika}\overline{OA}=\left(\begin{array}{c}1\\ 2\end{array}\right),\overline{OB}=\left(\begin{array}{c}4\\ 2\end{array}\right),\text{dan}\\ & \theta =\mathrm{\angle }(\overline{OA},\overline{OB}),\text{maka}\tan \theta =....\\ \\ & \text{a}.{\displaystyle \frac{3}{5}}\\ \\ & \text{b}.{\displaystyle \frac{9}{16}}\\ \\ & \text{c}.{\displaystyle \frac{3}{4}}\\ \\ & \text{d}.{\displaystyle \frac{4}{3}}\\ \\ & \text{e}.{\displaystyle \frac{16}{9}}\\ \\ & \text{Jawab}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 2\end{array}\right).\left(\begin{array}{c}4\\ 2\end{array}\right)}{\sqrt{1^2+2^2}\sqrt{4^2+2^2}}}\\ & ={\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}}\\ & ={\displaystyle \frac{8}{10}}\end{array}& \begin{array}{rl}\sin \theta & =\sqrt{1-{\cos }^2\theta }\\ & =\sqrt{1-{\left({\displaystyle \frac{8}{10}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{36}{100}}}\\ & ={\displaystyle \frac{6}{10}}\\ & \end{array}\\ \hline\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\tan \theta & ={\displaystyle \frac{\sin \theta }{\cos \theta }}\\ & ={\displaystyle \frac{{\displaystyle \frac{6}{10}}}{{\displaystyle \frac{8}{10}}}}\\ & ={\displaystyle \frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Jika}\overrightarrow{a},\overrightarrow{b}\text{dan}\overrightarrow{c}\text{adalah vektor satuan dengan}\\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\text{Nilai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\text{adalah}....\\ & \text{a}.{\displaystyle -3}\\ & \text{b}.{\displaystyle -\frac{3}{2}}\\ & \text{c}.{\displaystyle 0}\\ & \text{d}.{\displaystyle \frac{3}{2}}\\ & \text{e}.{\displaystyle 3}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Karena}& \{\begin{array}{c}\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{adalah vektor satuan, dan}\\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\end{array}\\ \text{segitig}& \text{a ABC adalah segitiga sama sisi}\\ \overrightarrow{a}.\overrightarrow{b}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{a}.\overrightarrow{c}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{b}.\overrightarrow{c}=& \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \text{Jadi, ni}& \text{lai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\\ & =(-\frac{1}{2})+(-\frac{1}{2})+(-\frac{1}{2})=-\frac{3}{2}\end{array}\end{array}$

$.\text{berikut ilustrasinya}$

$\begin{array}{l}\\ 29.& \text{Jika}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ },\left|\overrightarrow{a}\right|=4\text{dan}\\ & \left|\overrightarrow{b}\right|=3,\text{maka}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})\text{adalah}....\\ & \text{a}.2\\ & \text{b}.4\\ & \text{c}.6\\ & \text{d}.8\\ & \text{e}.10\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})& =\overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}\\ & =\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|\cos 0^{\circ }-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {60}^{\circ }\\ & ={\left|\overrightarrow{a}\right|}^2-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|.{\displaystyle \frac{1}{2}}\\ & =4^2-4.3.{\displaystyle \frac{1}{2}}\\ & =16-6\\ & =10\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Diketahui}\left|\overrightarrow{a}\right|=\sqrt{3},\left|\overrightarrow{b}\right|=1,\text{dan}|\overrightarrow{a}-\overrightarrow{b}|=1\\ & \text{maka panjang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}& & & \text{d}.& 2\sqrt{2}\\ \text{b}.& \sqrt{5}& \text{c}.& \sqrt{7}& \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Diketahui}& \text{sebagaimana pada soal}\\ {|\overrightarrow{a}-\overrightarrow{b}|}^2& ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ 1^2& ={\left(\sqrt{3}\right)}^2+1^2-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta & =3\\ \text{maka pan}& \text{jang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}\\ |\overrightarrow{a}+\overrightarrow{b}|& =\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }\\ & =\sqrt{{\left(\sqrt{3}\right)}^2+1^2+3}\\ & =\sqrt{3+1+3}\\ & =\sqrt{7}\end{array}\end{array}$

Contoh 5 Vektor

 $\begin{array}{l}\\ 21.& \text{Jika}\overrightarrow{g}=\left(\begin{array}{c}{3}^{x+y}\\ 5\end{array}\right)\text{dan}\overrightarrow{h}=\left(\begin{array}{c}81\\ {\displaystyle \frac{y+7}{2}}\end{array}\right)\\ & \text{sehingga}\overrightarrow{g}=\overrightarrow{h}\text{nilai dari}4x-3y=....\\ & \text{a}.-5\\ & \text{b}.-1\\ & \text{c}.0\\ & \text{d}.5\\ & \text{e}.10\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\overrightarrow{g}=\overrightarrow{h}\\ \left(\begin{array}{c}{3}^{x+y}\\ 5\end{array}\right)& =\left(\begin{array}{c}81\\ {\displaystyle \frac{y+7}{2}}\end{array}\right)\\ 3^{x+y}& =81=3^4\iff x+y=4\\ {\displaystyle \frac{y+7}{2}}& =5\iff y=10-7=3,\text{sehingga}\\ x+y& =4\iff x+3=4\iff x=4-3=1,\\ & \text{maka}\\ 4x-3y& =4(1)-3(3)\\ & =4-9=-5\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Vektor}\overrightarrow{m}=\left(\begin{array}{c}-2\\ 5\end{array}\right)\text{searah }\\ & \text{dengan vektor}....\\ & \text{a}.{\displaystyle \left(\begin{array}{c}2\\ -5\end{array}\right)}\\ & \text{b}.{\displaystyle \left(\begin{array}{c}2\\ 5\end{array}\right)}\\ & \text{c}.{\displaystyle \left(\begin{array}{c}-6\\ 15\end{array}\right)}\\ & \text{d}.{\displaystyle \left(\begin{array}{c}-4\\ 5\end{array}\right)}\\ & \text{e}.{\displaystyle \left(\begin{array}{c}-3\\ 10\end{array}\right)}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}& \text{Vektor}\overrightarrow{m}\text{searah dengan vektor}k.\overrightarrow{m}\\ & k.\overrightarrow{m}=k\left(\begin{array}{c}-2\\ 5\end{array}\right),\text{dengan}k\text{skalar positif}\\ & \begin{array}{|cc|}\hline \text{a}& \text{b}\\ \left(\begin{array}{c}2\\ -5\end{array}\right)=-\left(\begin{array}{c}-2\\ 5\end{array}\right)& \left(\begin{array}{c}2\\ 5\end{array}\right)=...\\ \text{c}& \text{d}\\ \left(\begin{array}{c}-6\\ 15\end{array}\right)=3\left(\begin{array}{c}-2\\ 5\end{array}\right)& \left(\begin{array}{c}-4\\ 5\end{array}\right)=...\\ \text{e}& \\ \left(\begin{array}{c}-3\\ 10\end{array}\right)=...& \\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Jika vektor}\overline{AC}=\overrightarrow{p},\overline{BC}=\overrightarrow{q}\text{dan}\\ & \overline{AD}:\overline{DC}=1:2,\text{maka vektor}\\ & \overline{BD}\text{bila dinyatakan}\\ & \text{dalam}\overrightarrow{p}\text{dan}\overrightarrow{q}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}(3\overrightarrow{p}-2\overrightarrow{q})}\\ \text{b}.{\displaystyle \frac{1}{3}(3\overrightarrow{q}-2\overrightarrow{p})}\\ \text{c}.{\displaystyle \frac{1}{3}(\overrightarrow{p}-2\overrightarrow{q})}\\ \text{d}.{\displaystyle (\overrightarrow{p}-{\displaystyle \frac{1}{3}\overrightarrow{q}})}\\ \text{e}.{\displaystyle \frac{1}{3}(\overrightarrow{p}-\overrightarrow{q})}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overline{AC}:\overline{CD}& =3:-2\\ \overline{CD}& =-{\displaystyle \frac{2}{3}\overline{AC}}\\ & ={\displaystyle -\frac{2}{3}\overrightarrow{p}}\\ \text{maka},& \\ \overline{BD}& =\overline{BC}+\overline{CD}\\ & =\overrightarrow{q}+(-{\displaystyle \frac{2}{3}\overrightarrow{p}})\\ & ={\displaystyle \frac{1}{3}(3\overrightarrow{q}-2\overrightarrow{p})}\end{array}\end{array}$

$.\text{Gambar berikut untuk soal 24}$

$\begin{array}{l}\\ 24.& \text{Jika vektor}\overline{AC}=\overrightarrow{p},\overline{BC}=\overrightarrow{q}\text{dan}\\ & \overline{AD}:\overline{DC}=1:2,\text{maka vektor}\overline{BD}\\ & \text{bila dinyatakan dalam}\overrightarrow{p}\text{dan}\overrightarrow{q}\text{adalah}....\\ & \text{a}.{\displaystyle \frac{1}{3}(3\overrightarrow{p}-2\overrightarrow{q})}\\ & \text{b}.{\displaystyle \frac{1}{3}(3\overrightarrow{q}-2\overrightarrow{p})}\\ & \text{c}.{\displaystyle (\overrightarrow{p}-{\displaystyle \frac{1}{3}\overrightarrow{q}})}\\ & \text{d}.{\displaystyle \frac{1}{3}(\overrightarrow{p}-2\overrightarrow{q})}\\ & \text{e}.{\displaystyle \frac{1}{3}(\overrightarrow{p}-\overrightarrow{q})}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overline{AC}:\overline{CD}& =3:-2\\ \overline{CD}& =-{\displaystyle \frac{2}{3}\overline{AC}}\\ & ={\displaystyle -\frac{2}{3}\overrightarrow{p}}\\ \text{maka},& \\ \overline{BD}& =\overline{BC}+\overline{CD}\\ & =\overrightarrow{q}+(-{\displaystyle \frac{2}{3}\overrightarrow{p}})\\ & ={\displaystyle \frac{1}{3}(3\overrightarrow{q}-2\overrightarrow{p})}\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Jika titik}A(2,6)\text{dan}B(5,3)\text{demikian }\\ & \text{juga titik}P\text{terletak pada}\overline{AB}\\ & \text{dengan}\overline{AP}:\overline{PB}=2:1,\text{maka vektor }\\ & \text{posisi}\overrightarrow{p}\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}4\\ 4\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}4\\ 5\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}-4\\ 4\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}4\\ 2\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}-4\\ 6\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overline{AP}:\overline{PB}& =2:1\\ \overline{AP}& =2\overline{PB}\\ \overrightarrow{p}-\overrightarrow{a}& =2(\overrightarrow{b}-\overrightarrow{p})\\ \overrightarrow{p}+2\overrightarrow{p}& =\overrightarrow{a}+2\overrightarrow{b}\\ 3\overrightarrow{p}& =\overrightarrow{a}+2\overrightarrow{b}\\ \overrightarrow{p}& ={\displaystyle \frac{1}{3}(\overrightarrow{a}+2\overrightarrow{b})}\\ & ={\displaystyle \frac{1}{3}\left(\begin{array}{c}2+2.5\\ 6+2.3\end{array}\right)}\\ & ={\displaystyle \frac{1}{3}\left(\begin{array}{c}12\\ 12\end{array}\right)}\\ & =\left(\begin{array}{c}4\\ 4\end{array}\right)\end{array}\end{array}$

Contoh 4 Vektor

 $\begin{array}{l}\\ 16.& \text{Diketahui titik A(-1,1,0) dan titik B(1,-2,2)}\\ & \text{maka panjang vektor}\overrightarrow{BA}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2}& & & \text{d}.& \sqrt{17}\\ \text{b}.& \sqrt{5}& \text{c}.& \sqrt{9}& \text{e}.& \sqrt{21}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui}\text{sebagaimana pada soal}\\ & \text{maka pan}\text{jang vektor}\overrightarrow{BA}\text{adalah}\\ & \left|\overrightarrow{BA}\right|=\sqrt{(1-(-1){)}^2+(-2-1{)}^2+(2-0{)}^2}\\ & =\sqrt{2^2+(-3{)}^2+2^2}\\ & =\sqrt{4+9+4}\\ & =\sqrt{17}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Vektor satuan untuk vektor}\overrightarrow{a}=\left(\begin{array}{ccc}2,& 1,& -2\end{array}\right)=....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{ccc}-{\displaystyle \frac{2}{3},}& -{\displaystyle \frac{1}{3},}& {\displaystyle \frac{2}{3}}\end{array}\right)& \\ \\ \text{b}.& \left(\begin{array}{ccc}{\displaystyle \frac{2}{3},}& {\displaystyle \frac{1}{3},}& -{\displaystyle \frac{2}{3}}\end{array}\right)& \\ \\ \text{c}.& \left(\begin{array}{ccc}{\displaystyle \frac{2}{4},}& {\displaystyle \frac{1}{4},}& -{\displaystyle \frac{2}{4}}\end{array}\right)& \\ \\ \text{d}.& \left(\begin{array}{ccc}-{\displaystyle \frac{2}{4},}& -{\displaystyle \frac{1}{4},}& {\displaystyle \frac{2}{4}}\end{array}\right)\\ \\ \text{e}.& \left(\begin{array}{ccc}-{\displaystyle \frac{2}{9},}& -{\displaystyle \frac{1}{9},}& {\displaystyle \frac{2}{9}}\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Vektor }& \text{satuan dari vektor}\overrightarrow{a}\text{adalah}:\\ \hat{a}& ={\displaystyle \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{ccc}2,& 1,& -2\end{array}\right)}{\sqrt{2^2+1^2+(-2{)}^2}}}\\ & ={\displaystyle \frac{\left(\begin{array}{ccc}2,& 1,& -2\end{array}\right)}{\sqrt{9}}}\\ & ={\displaystyle \frac{\left(\begin{array}{ccc}2,& 1,& -2\end{array}\right)}{3}}\\ & =\left(\begin{array}{ccc}{\displaystyle \frac{2}{3},}& {\displaystyle \frac{1}{3},}& -{\displaystyle \frac{2}{3}}\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Jika titik A(-2,3,5) dan B(4,1,-3)},\\ & \text{maka vektor posisi AB adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{ccc}-6,& 2,& 8\end{array}\right)& \\ \text{b}.& \left(\begin{array}{ccc}8,& 2,& -6\end{array}\right)& \\ \text{c}.& \left(\begin{array}{ccc}6,& -2,& -8\end{array}\right)& \\ \text{d}.& \left(\begin{array}{ccc}-8& -2,& 6\end{array}\right)\\ \text{e}.& \left(\begin{array}{ccc}2,& 4,& 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Vektor posisi}& \text{dari}\overrightarrow{AB}\text{adalah}:\\ \overrightarrow{AB}& =\overrightarrow{OB}-\overrightarrow{OA}\\ & =\left(\begin{array}{c}4\\ 1\\ -3\end{array}\right)-\left(\begin{array}{c}-2\\ 3\\ 5\end{array}\right)\\ & =\left(\begin{array}{c}4+2\\ 1-3\\ -3-5\end{array}\right)\\ & =\left(\begin{array}{c}6\\ -2\\ -8\end{array}\right)\mathbf{\text{atau}}\\ & =\left(\begin{array}{ccc}6,& -2,& -8\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}{}^2\log 8x\\ {({}^2\log x)}^y\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}5\\ 8\end{array}\right)\\ & \text{sehingga}\overrightarrow{p}=\overrightarrow{q}\text{nilai dari}x.y=....\\ & \text{a}.6\\ & \text{b}.12\\ & \text{c}.18\\ & \text{d}.24\\ & \text{e}.30\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\overrightarrow{p}=\overrightarrow{q}\\ \left(\begin{array}{c}{}^2\log 8x\\ {({}^2\log x)}^y\end{array}\right)& =\left(\begin{array}{c}5\\ 8\end{array}\right),\text{maka}\\ 8x& =2^5=32\\ \iff x& ={\displaystyle \frac{32}{8}=4}\\ {({}^2\log 4)}^y& =8\\ \iff 2^y& =8=2^3\\ \iff y& =3\\ \text{Sehingga}& \\ x.y& =4\times 3=12\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}3x\\ 4x+y\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}{\displaystyle \frac{2x-4}{2}}\\ 6\end{array}\right)\\ & \text{sehingga}\overrightarrow{p}=\overrightarrow{q}\text{nilai dari}2x+y=....\\ & \text{a}.-12\\ & \text{b}.0\\ & \text{c}.8\\ & \text{d}.9\\ & \text{e}.19\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\overrightarrow{p}=\overrightarrow{q}\\ \left(\begin{array}{c}3x\\ 4x+y\end{array}\right)& =\left(\begin{array}{c}{\displaystyle \frac{2x-4}{2}}\\ 6\end{array}\right)\\ 3x& ={\displaystyle \frac{2x-4}{2}\iff 6x=2x-4}\\ \iff x& =-1\\ 4(-1)+y& =6\iff -4+y=6\\ \iff y& =6+4\\ y& =10\\ x+y& =(-1)+10=9\end{array}\end{array}$

Contoh 3 Vektor

$\begin{array}{l}\\ 11.& \text{Jika vektor}\overrightarrow{a}=\left(\begin{array}{c}6\\ -4\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}3\\ 2\end{array}\right),\\ & \text{maka}3\overrightarrow{a}-2\overrightarrow{b}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{c}12\\ -16\end{array}\right)& & \text{d}.\left(\begin{array}{c}24\\ 16\end{array}\right)\\ \text{b}.\left(\begin{array}{c}24\\ -16\end{array}\right)& \text{c}.\left(\begin{array}{c}12\\ 16\end{array}\right)& \text{e}.\left(\begin{array}{c}-12\\ -16\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}3\overrightarrow{a}-2\overrightarrow{b}& =3\left(\begin{array}{c}6\\ -4\end{array}\right)-2\left(\begin{array}{c}3\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}18-6\\ -12-4\end{array}\right)\\ & =\left(\begin{array}{c}12\\ -16\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Diketahui jajar genjang ABCD }\\ & \text{dengan titik E adalah perpotongan }\\ & \text{diagonal jajar genjang}.\end{array}$

$\begin{array}{l}\\ .& \text{Jika}\overline{AB}=\overrightarrow{b}\text{dan}\overline{AD}=\overrightarrow{a},\text{maka}\overline{CE}\\ & \text{bila dinyatakan dalam}\overrightarrow{a}\text{dan}\overrightarrow{b}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})}& \\ \text{b}.{\displaystyle \frac{1}{2}(\overrightarrow{a}-\overrightarrow{b})}& \\ \text{c}.{\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})}& \\ \text{d}.{\displaystyle -\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})}\\ \text{e}.-{\displaystyle \frac{1}{2}(2\overrightarrow{a}+\overrightarrow{b})}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overline{AC}& =\overline{AD}+\overline{DC}\\ \overline{CA}& =\overline{CD}+\overline{DA}\\ \overline{CE}& ={\displaystyle \frac{1}{2}\overline{CA}}\\ & ={\displaystyle \frac{1}{2}(-\overrightarrow{b}-\overrightarrow{a})}\\ & =-{\displaystyle \frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Pada segi enam beraturan ABCDEF},\\ & \text{jika}\overrightarrow{AB}=\overrightarrow{u}\text{dan}\overrightarrow{AF}=\overrightarrow{v}\text{maka vektor}\\ & \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ & \begin{array}{l}\\ \text{a}.& 2\overrightarrow{u}+2\overrightarrow{v}& & & \text{d}.& 6\overrightarrow{u}+6\overrightarrow{v}\\ \text{b}.& 4\overrightarrow{u}+4\overrightarrow{v}& \text{c}.& 5\overrightarrow{u}+5\overrightarrow{v}& \text{e}.& 8\overrightarrow{u}+8\overrightarrow{v}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$

$\begin{array}{rl}.& \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ & =\overrightarrow{AB}+(\overrightarrow{AO}+\overrightarrow{OC})+2\overrightarrow{AO}+(\overrightarrow{AO}+\overrightarrow{OE})+\overrightarrow{AF}\\ & =\overrightarrow{u}+(2\overrightarrow{u}+\overrightarrow{v})+2(\overrightarrow{v}+\overrightarrow{u})+(2\overrightarrow{v}+\overrightarrow{u})+\overrightarrow{v}\\ & =6\overrightarrow{u}+6\overrightarrow{v}\end{array}$

$\begin{array}{l}\\ 14.& \text{Perhatikanlah juga ilustrasi gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{maka vektor}\overrightarrow{w}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8\overrightarrow{i}-6\overrightarrow{j}-13\overrightarrow{k}& \\ \text{b}.& 8\overrightarrow{i}-13\overrightarrow{j}-6\overrightarrow{k}\\ \text{c}.& 6\overrightarrow{i}-8\overrightarrow{j}-13\overrightarrow{k}& \\ \text{d}.& -6\overrightarrow{i}+8\overrightarrow{j}-13\overrightarrow{k}\\ \text{e}.& -6\overrightarrow{i}-13\overrightarrow{j}+8\overrightarrow{k}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Kita perhatikan juga ilustrasi }\\ & \text{gambarnya semisal dengan soal No.1}\\ & \text{Misalkan titiknya adalah titik }\\ & \text{W dengan koordinat (8,-6,-13)},\\ & \text{maka vektor posisi titik }\\ & \text{W tersebut adalah}\overrightarrow{OW}=\overrightarrow{w}\\ & \text{di mana}\\ & \begin{array}{rl}& \text{Vektor}\overrightarrow{w}\text{jika dinyatakan }\\ & \text{dalam kombinasi linear adalah}\\ & \overrightarrow{w}=8\overrightarrow{i}-6\overrightarrow{j}-13\overrightarrow{k}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika titik Z(4,-5,2)},\text{maka panjang }\\ & \text{vektor posisi titik Z adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& 5\sqrt{2}\\ \text{b}.& 2\sqrt{5}& \text{c}.& 3\sqrt{5}& \text{e}.& 5\sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Vektor posisi}& \text{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\overrightarrow{z}& =\left(\begin{array}{ccc}4,& -5,& 2\end{array}\right),\\ \text{Dan panjang}& \text{vektor}\overrightarrow{z}\text{ini adalah}\\ \left|\overrightarrow{z}\right|& =\sqrt{4^2+(-5{)}^2+2^2}\\ & =\sqrt{16+25+4}\\ & =\sqrt{45}=\sqrt{9\times 5}\\ & =3\sqrt{5}\end{array}\end{array}$.

Contoh 2 Vektor

$\begin{array}{l}\\ 6.& \text{Diketahui titik}P(n,2),Q(1,-2),n>0\\ & \text{dan panjang}\overline{PQ}=5,\text{maka nilai}n\\ & \text{adalah}....\\ & \text{a}.1\\ & \text{b}.2\\ & \text{c}.3\\ & \text{d}.4\\ & \text{e}.5\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overline{PQ}& =5\\ \sqrt{(x_Q-x_P{)}^2+(y_Q-y_P{)}^2}& =5\\ (x_Q-x_P{)}^2+(y_Q-y_P{)}^2& =25\\ (1-n{)}^2+(-2-2{)}^2& =25\\ (1-n{)}^2+16& =25\\ (1-n{)}^2-3^2& =0\\ (1+3-n)(1-3-n)& =0\\ n=4\text{atau}n=-2& \end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui vektor}\overrightarrow{u}=\left(\begin{array}{c}3\\ 4\end{array}\right)\text{dan}\overrightarrow{v}=\left(\begin{array}{c}2\\ -1\end{array}\right).\\ & \text{Nilai}|\overrightarrow{u}+\overrightarrow{v}|\text{adalah}....\\ & \text{a}.\sqrt{28}\\ & \text{b}.\sqrt{30}\\ & \text{c}.\sqrt{34}\\ & \text{d}.\sqrt{44}\\ & \text{e}.\sqrt{50}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{u}+\overrightarrow{v}& =\left(\begin{array}{c}3\\ 4\end{array}\right)+\left(\begin{array}{c}2\\ -1\end{array}\right)\\ & =\left(\begin{array}{c}3+2\\ 4+(-1)\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 3\end{array}\right)\\ |\overrightarrow{u}+\overrightarrow{v}|& =\sqrt{5^2+3^2}\\ & =\sqrt{25+9}\\ & =\sqrt{34}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Vektor satuan}\overrightarrow{u}=\left(\begin{array}{c}-5\\ -12\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle -\frac{1}{13}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{b}.{\displaystyle \frac{1}{15}\left(\begin{array}{c}-5\\ -12\end{array}\right)}\\ & \text{c}.{\displaystyle -\frac{1}{17}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{d}.{\displaystyle -\frac{1}{17}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ & \text{e}.{\displaystyle \frac{1}{2}\left(\begin{array}{c}5\\ 12\end{array}\right)}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}{\overrightarrow{e}}_{\overrightarrow{u}}& ={\displaystyle \frac{\overrightarrow{u}}{\left|\overrightarrow{u}\right|},\text{maka}}\\ {\overrightarrow{e}}_{\left(\begin{array}{c}-5\\ -12\end{array}\right)}& ={\displaystyle \frac{\left(\begin{array}{c}-5\\ -12\end{array}\right)}{\left|\left(\begin{array}{c}-5\\ -12\end{array}\right)\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-5\\ -12\end{array}\right)}{\sqrt{(-5{)}^2+(-12{)}^2}}}\\ & ={\displaystyle \frac{-\left(\begin{array}{c}5\\ 12\end{array}\right)}{\sqrt{169}}}\\ & =-{\displaystyle \frac{1}{13}\left(\begin{array}{c}5\\ 12\end{array}\right)}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika vektor}\overrightarrow{p}=\left(\begin{array}{c}8\\ 7\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}-3\\ 9\end{array}\right)\\ & \text{Hasil dari}\overrightarrow{p}+\overrightarrow{q}\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}6\\ 14\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}6\\ 13\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}6\\ 15\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}5\\ 16\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}5\\ 48\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}+\overrightarrow{q}& =\left(\begin{array}{c}8\\ 7\end{array}\right)+\left(\begin{array}{c}-3\\ 9\end{array}\right)\\ & =\left(\begin{array}{c}8-3\\ 7+9\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 16\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika vektor}\overrightarrow{p}=\left(\begin{array}{c}{}^2\log 32\\ -{}^3\log {\displaystyle \frac{1}{81}}\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}-9\\ -21\end{array}\right)\\ & \text{Hasil dari}\overrightarrow{p}+\overrightarrow{q}\text{adalah}....\\ & \text{a}.\left(\begin{array}{c}6\\ 17\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}-6\\ -17\end{array}\right)\\ & \text{c}.\left(\begin{array}{c}4\\ 17\end{array}\right)\\ & \text{d}.\left(\begin{array}{c}-4\\ -17\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}5\\ -16\end{array}\right)\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}+\overrightarrow{q}& =\left(\begin{array}{c}{}^2\log 32\\ -{}^3\log {\displaystyle \frac{1}{81}}\end{array}\right)+\left(\begin{array}{c}-9\\ -21\end{array}\right)\\ & =\left(\begin{array}{c}5-9\\ 4-21\end{array}\right)\\ & =\left(\begin{array}{c}-4\\ -17\end{array}\right)\end{array}\end{array}$

Contoh 1 Vektor

$\begin{array}{l}\\ 1.& \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{maka vektor}\overrightarrow{u}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\overrightarrow{i}+5\overrightarrow{j}& & & \text{d}.& -3\overrightarrow{i}-5\overrightarrow{j}\\ \text{b}.& 5\overrightarrow{i}+3\overrightarrow{j}& \text{c}.& -3\overrightarrow{i}+5\overrightarrow{j}& \text{e}.& -5\overrightarrow{i}+3\overrightarrow{j}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Kita perhatikan lagi gambarnya}\\ & \begin{array}{rl}& \text{Vektor}\overrightarrow{u}\text{jika dinyatakan sebagai }\\ & \text{kombinasi linear adalah}\overrightarrow{u}=-3\overrightarrow{i}+5\overrightarrow{j}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Panjang vektor}\overrightarrow{p}=\left(\begin{array}{c}4\\ -8\end{array}\right)\text{adalah}....\\ & \text{a}.{\displaystyle \sqrt{4}}\\ & \text{b}.\sqrt{12}\\ & \text{b}.\sqrt{20}\\ & \text{d}.{\displaystyle \sqrt{80}}\\ & \text{e}.\sqrt{100}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}& =\left(\begin{array}{c}4\\ -8\end{array}\right),\text{maka besar dari vektor}\\ \overrightarrow{p}& \text{adalah}=\left|\overrightarrow{p}\right|=\sqrt{x^2+y^2}\\ & \text{Yaitu}\\ \left|\overrightarrow{p}\right|& =\sqrt{4^2+(-8{)}^2}\\ & =\sqrt{16+64}=\sqrt{80}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{l}\\ .& \text{Panjang vektor}\overrightarrow{h}\text{tersebut di atas adalah}....\\ & \text{a}.5\\ & \text{b}.7\\ & \text{c}.10\\ & \text{d}.12\\ & \text{e}.15\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Diketahui}& \overrightarrow{h}=\overline{OH}=8\overrightarrow{i}+6\overrightarrow{j}\\ \overrightarrow{h}& =\sqrt{x_H^2+y_H^2}\\ & =\sqrt{8^2+6^2}\mathbf{\text{(ingat tripel Pythagoras)}}\\ & =\sqrt{{10}^2}=10\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Vektor satuan dari}\overrightarrow{q}=3\overrightarrow{i}-4\overrightarrow{j}\text{adalah}....\\ & \text{a}.{\displaystyle \frac{4}{5}\overrightarrow{i}-\frac{3}{5}\overrightarrow{j}}\\ & \text{b}.{\displaystyle \frac{3}{5}\overrightarrow{i}-\frac{4}{5}\overrightarrow{j}}\\ & \text{c}.3\overrightarrow{i}-4\overrightarrow{j}\\ & \text{d}.4\overrightarrow{i}-3\overrightarrow{j}\\ & \text{e}.15\overrightarrow{i}-20\overrightarrow{j}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{q}& =3\overrightarrow{i}-4\overrightarrow{j}\\ & \text{Vektor satuan dari vektor}\overrightarrow{q}\text{adalah}:\\ & {\hat{e}}_{{}_{\overrightarrow{q}}}={\displaystyle \frac{1}{\left|\overrightarrow{q}\right|}.\overrightarrow{q}}\\ & \text{Sehingga}\\ {\hat{e}}_{{}_{\overrightarrow{q}}}& ={\displaystyle \frac{1}{\sqrt{3^2+(-4{)}^2}}.\left(\begin{array}{c}3\\ -4\end{array}\right)}\\ & ={\displaystyle \frac{1}{5}\left(\begin{array}{c}3\\ -4\end{array}\right)}\\ & \text{atau dalam vektor basis}\\ & ={\displaystyle \frac{3}{5}\overrightarrow{i}-\frac{4}{5}\overrightarrow{j}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Vektor berikut yang memiliki panjang}\\ & 29\text{satuan adalah}....\\ & \text{a}.{\displaystyle 18\overrightarrow{i}-19\overrightarrow{j}}\\ & \text{b}.{\displaystyle 19\overrightarrow{i}-20\overrightarrow{j}}\\ & \text{c}.20\overrightarrow{i}-21\overrightarrow{j}\\ & \text{d}.21\overrightarrow{i}-22\overrightarrow{j}\\ & \text{e}.22\overrightarrow{i}-23\overrightarrow{j}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Ingat}& \text{lah akan tigaan Pythagoras}\\ & \{\begin{array}{ll}(3,4,5)& \Rightarrow 3^2+4^2=5^2\\ (5,12,13)& \Rightarrow 5^2+{12}^2={13}^2\\ (8,15,17)& \Rightarrow \cdots \\ (20,21,29)& \Rightarrow \cdots \\ ⋮& dll\end{array}\\ \text{Sehin}& \text{gga}\\ \text{yang}& \text{paling mungkin adalah}:\\ & =\sqrt{{20}^2+{21}^2}=\sqrt{400+441}\\ & =\sqrt{841}\\ & =29\end{array}\end{array}$

Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}f(x)=x^2-2,\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2-2}& & \text{d}.{\displaystyle x}\\ \\ \text{b}.{\displaystyle x^2}& \text{c}.{\displaystyle 2x}& \text{e}.{\displaystyle 2x-2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}f(x)=x^2-2,\\ \text{maka nila}& \text{i untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{((x+h{)}^2-2)-(x^2-2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{x^2+2xh+h^2-2-x^2+2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{2xh+h^2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle (2x+h)}\\ & =2x\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui}f(x)=\sqrt{x-1},\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle 1}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.{\displaystyle 0}& \text{e}.{\displaystyle -1}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}f(x)=\sqrt{x-1},\\ \text{maka}& \text{nilai untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}\times {\displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(h+1)-1}{h\times (\sqrt{h+1}+1)}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{1}{(\sqrt{h+1}+1)}}\\ & ={\displaystyle \frac{1}{\sqrt{0+1}+1}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \mathbf{\text{(Mat Das SIMAK UI 2013)}}\\ & \text{Nilai}\underset{x\to 5}{\text{Lim}}{\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....}\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}+\sqrt{2}& & \text{d}.5\\ \text{b}.5-2\sqrt{6}& \text{c}.2\sqrt{6}& \text{e}.5+2\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 5}{\text{Lim}}& {\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times {\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}}\\ & ={\displaystyle \frac{3+2+2\sqrt{6}}{3-2}}\\ & =5+2\sqrt{6}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \mathbf{\text{(Mat IPA SBMPTN 2014)}}\\ \\ & \text{Jika}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}\\ & \text{dan}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3,}\\ \\ & \text{maka nilai}\underset{x\to a}{\text{Lim}}{\displaystyle f(x).g(x)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{14}}& & \text{d}.{\displaystyle \frac{4}{14}}\\ \\ \text{b}.{\displaystyle \frac{2}{14}}& \text{c}.{\displaystyle \frac{3}{14}}& \text{e}.{\displaystyle \frac{5}{14}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa},\\ & \begin{array}{l}\\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}& & \\ & & \\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3}& +& \\ & & \\ & & \\ 2\underset{x\to a}{\text{Lim}}{\displaystyle f(x)=1}& & \\ & & \\ \underset{x\to a}{\text{Lim}}f(x)={\displaystyle \frac{1}{2},}& & \end{array}\\ & \text{sehingga}\underset{x\to a}{\text{Lim}}f(g)={\displaystyle \frac{2}{7}}\\ & & \\ & \text{maka},\\ & \underset{x\to a}{\text{Lim}}f(x).g(x)={\displaystyle \frac{1}{2}\times \frac{2}{7}={\displaystyle \frac{2}{14}}}\end{array}\end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

$\begin{array}{l}\\ 1& \text{Nilai}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle \frac{1}{4}}\\ \\ \text{b}.-{\displaystyle \frac{1}{4}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\left({\displaystyle \frac{6-2}{2^2-4}-\frac{1}{2-2}}\right)\\ & =\left({\displaystyle \frac{4}{0}-\frac{1}{0}}\right)=\mathrm{\infty }-\mathrm{\infty }\\ & \text{hal ini tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{(x+2)}{(x-2)(x+2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{x+2}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{4-2x}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2(x-2)}{(x+2)(x-2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2}{x+2}}\right)\\ & ={\displaystyle -\frac{2}{(2+2)}}\\ & =-{\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}=....}\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle 2}& & \text{d}.{\displaystyle \frac{1}{2}}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.0& \text{e}.{\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\left({\displaystyle \frac{4-4}{2^4-4}}\right)\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}(2-\sqrt{x})}}\\ & =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{-\sqrt{x}(\sqrt{x}-2)}}\\ & =\underset{x\to 4}{\text{Lim}}-{\displaystyle \frac{(\sqrt{x}+2)}{\sqrt{x}}}\\ & =-{\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}}\\ & =-{\displaystyle \frac{2+2}{2}}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.{\displaystyle 2}\\ \\ \text{b}.{\displaystyle \frac{1}{2}}& \text{c}.1& \text{e}.{\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & ={\displaystyle \frac{(2-3+1)(1-1)}{{(1-1)}^2}}\\ & ={\displaystyle \frac{0\times 0}{0^2}}\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehi}& \text{ngga},\\ \underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & =\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{((2\sqrt{x}-1)\times (\sqrt{x}-1))(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}-1)}}\\ & =\underset{x\to 1}{\text{Lim}}(2\sqrt{x}-1)\\ & =2.1-1\\ & =2-1\\ & =1\end{array}\end{array}$

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

$\begin{array}{l}\\ 5.& \text{Jika}\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\text{maka nilai}a\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 2}& & \text{d}.{\displaystyle -2}\\ \\ \text{b}.{\displaystyle 1}& \text{c}.-1& \text{e}.{\displaystyle -3}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}& {\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6}\\ & \text{dengan bantuan limit kanan }\\ & \text{yaitu}x=2+h\Rightarrow h\to 0\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{a(2+h)-2a}{\sqrt{2(2+h)}-(2+h)}=6}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-(2+h)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah}{\sqrt{4+2h}-(2+h)}\times {\displaystyle \frac{(\sqrt{4+2h}+(2+h))}{(\sqrt{4+2h}+(2+h))}}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{4+2h-(4+4h+h^2)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{-2h-h^2}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a\times (\sqrt{4+2h}+(2+h))}{-2-h}}\\ 6& ={\displaystyle \frac{a\times (\sqrt{4+0}+(2+0))}{-2-0}}\\ 6& ={\displaystyle \frac{a(\sqrt{4}+2)}{-2}}\\ {\displaystyle \frac{a(4)}{-2}}& =6\\ a(-2)& =6\\ a& =-3\end{array}\end{array}$

MATEMATIKA WAJIB MA/SMA untuk KONDISI KHUSUS TAHUN 2020-2021

 $\text{A. Kelas X (Sepuluh)}$

$\begin{array}{rl}& \text{A.1 Persamaan dan pertidaksamaan}\\ & \text{nilai mutlak dari bentuk linear}\\ & \text{satu variabel}\end{array}$

• materi persamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 
• materi pertidaksamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , 

$\begin{array}{rl}& \text{A.2 Pertidaksamaan rasional}\\ & \text{dan irasional satu variabel}\end{array}$

• materi pertidaksamaan rasional dan irasional satu variabel
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , 

$\begin{array}{rl}& \text{A.3 Sistem persamaan linear tiga}\\ & \text{variabel}\end{array}$

• materi sistem persamaan linear tiga variabel (SPLTV)
• lanjutan materi SPLTV
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.4 Sistem pertidaksamaan dua}\\ & \text{variabel-linear-linear}\end{array}$

• materi sistem pertidaksamaan dua variabel linear-linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.5 Fungsi}\end{array}$

• Materi fungsi
• Lanjutan materi fungsi
• contoh soal 1 , contoh 2 , contoh 3

$\begin{array}{rl}& \text{A.6 Fungsi Komposisi dan invers}\end{array}$

• Materi dan contoh soal
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\begin{array}{rl}& \text{A.7 Rasio trigonometri}\\ & \text{pada segitiga siku-siku}\end{array}$

$\begin{array}{rl}& \text{A.8 Rasio trigonometri}\\ & \text{sudut-sudut diberbagai}\\ & \text{kuadran}\end{array}$

$\begin{array}{rl}& \text{A.9 Aturan sinus dan cosinus}\end{array}$

$\text{B. Kelas XI (Sebelas)}$

$\begin{array}{rl}& \text{B.1 Program linear}\end{array}$

• materi program linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{B.2 Matriks}\end{array}$

• materi matriks 
• lanjutan materi 1
• lanjutan materi 2
• contoh soal 1 , contoh 2 , contoh 3 

$\begin{array}{rl}& \text{B.3 Determinan dan invers matriks}\\ & \text{ordo 2x2}\end{array}$

• materi determinan
• materi invers matriks ordo 2x2
• contoh soal 4 , contoh 5 , contoh 6 

$\begin{array}{rl}& \text{B.4 Pola bilangan dan jumlah}\\ & \text{pada barisan aritmetika}\\ & \text{dan geometri}\end{array}$

• materi pola bilangan - induksi matematika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5

$\begin{array}{rl}& \text{B.5 Limit fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.6 Turunan fungsi aljabar}\end{array}$

$\begin{array}{rl}& \text{B.7 Keberkaitan Turunan fungsi}\end{array}$

$\begin{array}{rl}& \text{B.8 Integral tak tentu}\\ & \text{fungsi aljabar}\end{array}$

$\text{C. Kelas XII (Duabelas)}$

$\begin{array}{rl}& \text{C.1 Jarak dalam ruang}\end{array}$

• materi mengitung jarak di dimensi tiga
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\begin{array}{rl}& \text{C.2 Statistika}\end{array}$

• materi statistika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 

$\begin{array}{rl}& \text{C.3 Aturan pencacahan}\end{array}$

• materi 
• lanjutan materi
• lanjutan materi (faktorial, permutasi, kombinasi)
• contoh soal 1, contoh 2, contoh 3
• lanjutan materi (binomial newton)
• pengayaan : fungsi pembangkit untuk kombinasi

$\begin{array}{rl}& \text{C.4 Peluang kejadian majumuk}\end{array}$

• materi
• lanjutan materi (peluang kejadian tunggal)
• lanjutan materi (peluang komplemen, peluang dua kejadian saling bebas)
• lanjutan materi (peluang kejadian saling lepas)
• contoh soal 1, contoh 2, contoh 3