Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 91.& \text{Titik A(4,-4) dicerminkan terhadap}\\ & \text{garis}y=x\tan {15}^{\circ }\text{menghasilkan}\\ & \text{bayangan}{A}^{\prime }(a,b)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}& & \text{d}.4\sqrt{3}\\ \text{b}.2\sqrt{3}& \text{c}.3\sqrt{3}& \text{e}.6\sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.15}^{\circ }& \sin {2.15}^{\circ }\\ \sin {2.15}^{\circ }& -\cos {2.15}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {30}^{\circ }& \sin {30}^{\circ }\\ \sin {30}^{\circ }& -\cos {30}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{c}2\sqrt{3}-2\\ 2+2\sqrt{3}\end{array}\right)\\ & \{\begin{array}{ll}a& =2\sqrt{3}-2\\ b& =2+2\sqrt{3}\end{array}\\ \text{mak}& \text{a nilai dari}\\ a+b& =(2\sqrt{3}-2+2+2\sqrt{3})\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 92.& \text{Lingkaran}{x}^2+y^2-5x+8y+7=0\\ & \text{ditranslasikan oleh}T=\left(\begin{array}{c}m\\ n\end{array}\right)\text{menghasilkan}\\ & \text{bayangan}{x}^2+y^2-9x+2y+6=0.\\ & \text{Nilai}m+n=...\\ & \begin{array}{l}\\ \text{a}.2& & \text{d}.5\\ \text{b}.3& \text{c}.4& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui sebuah lingkaran dengan persamaan}:\\ & x^2+y^2-5x+8y+7=0\\ \text{kar}& \text{ena akibat translasi, maka}\\ & \{\begin{array}{ll}x& =x^{\prime }-m\\ y& =y^{\prime }-n\end{array}\\ & x^2+y^2-5x+8y+7=0\\ \text{seh}& \text{ingga}\\ & \iff (x^{\prime }-m{)}^2+(y^{\prime }-n{)}^2-5(x^{\prime }-m)+8(y^{\prime }-n)+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-2m{x}^{\prime }-2n{y}^{\prime }+m^2+n^2-5x^{\prime }+5m+8y^{\prime }-8n+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-(2m+5)x^{\prime }+(8-2n)y^{\prime }+m^2+n^2+5m-8n+7=0\\ & \equiv x^{\prime 2}+y^{\prime 2}-9x^{\prime }+2y^{\prime }+6=0(\mathbf{\text{akhir bayangan}})\\ & \{\begin{array}{ll}9& =2m+5\Rightarrow m=2\\ 2& =8-2n\Rightarrow n=3\end{array}\\ \text{Jad}& \text{i , nilai}m+n=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 93.& \text{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-{\displaystyle \frac{1}{3}x\sqrt{3},\text{maka bayangan dari}}\\ & \text{titik textit{A} tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& & \\ \text{b}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& \\ \text{c}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})& \\ \text{d}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})\\ \text{e}.A^{\prime }(-1+{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ y& =-{\displaystyle \frac{1}{3}x\sqrt{3}=(-{\displaystyle \frac{1}{3}\sqrt{3}})x}\\ & =(-\tan {30}^{\circ })x=\tan ({180}^{\circ }-{30}^{\circ })x\\ & =\tan {150}^{\circ }.x\\ \text{maka}\theta & ={150}^{\circ }\Rightarrow 2\theta ={300}^{\circ }\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {300}^{\circ }& \sin {300}^{\circ }\\ \sin {300}^{\circ }& -\cos {300}^{\circ }\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ -{\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-1-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \sqrt{3}-{\displaystyle \frac{1}{2}}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 94.& \text{Bayangan titik A(2,4) dicerminkan }\\ & \text{terhadap garis}y-x=0\text{dilanjutkan}\\ & \text{ke garis}x\sqrt{3}-3y=0\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(2+\sqrt{3},-1+2\sqrt{3})& & \\ \text{b}.A^{\prime }(2+\sqrt{3},1-2\sqrt{3})& \\ \text{c}.A^{\prime }(1-\sqrt{3},-2+\sqrt{3})& \\ \text{d}.A^{\prime }(-2+\sqrt{3},1+2\sqrt{3})\\ \text{e}.A^{\prime }(2-\sqrt{3},1-2\sqrt{3})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ & \{\begin{array}{ll}x\sqrt{3}-3y=0& \iff y={\displaystyle \frac{1}{3}\sqrt{3}x}\\ & \iff y=\tan {30}^{\circ }.x\\ \\ x-y=0& \iff y=x\end{array}\\ \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.30}^{\circ }& \sin {2.30}^{\circ }\\ \sin {2.30}^{\circ }& -\cos {2.30}^{\circ }\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}\\ {\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}\sqrt{3}+2\\ -1+2\sqrt{3}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 95.& \text{Jika}{T}_1=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\text{dan}{T}_2=\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\\ & \text{maka bayangan garis}x+y+1=0\\ & \text{oleh}{T}_2\circ T_1\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.x-2y-1=0& & \\ \text{b}.x+2y-1=0& \\ \text{c}.x+2y+1=0& \\ \text{d}.x-2y+1=0\\ \text{e}.x+y-1=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T_2\circ T_1\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2+5& -4+5\\ -1+3& -2+3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}3x+y\\ 2x+y\end{array}\right)\\ \text{Dipe}& \text{roleh}\\ & \begin{array}{l}\\ x^{\prime }& =3x+y\\ y^{\prime }& =2x+y-\\ x^{\prime }-y^{\prime }& =x\\ \iff x& =x^{\prime }-y^{\prime }....(1)\\ \text{maka}\\ y& =x^{\prime }-3x\\ & =x^{\prime }-3(x^{\prime }-y^{\prime })\\ & =3y^{\prime }-2x^{\prime }....(2)\end{array}\\ \text{Sehin}& \text{gga}\\ x+y& +1=0\\ x^{\prime }-& y^{\prime }+3y^{\prime }-2x^{\prime }+1=0\\ -x^{\prime }+& 2y^{\prime }+1=0\\ x^{\prime }-& 2y^{\prime }-1=0\\ \text{maka}& \text{bayangan garisnya}\\ x-2& y-1=0\end{array}\end{array}$.

$\begin{array}{l}\\ 96.& \text{Garis}2x+y+4=0\text{ditranslasikan}\\ & \text{oleh}\left(\begin{array}{c}-2\\ 5\end{array}\right)\text{dilanjutkan transformasi}\\ & \text{oleh}\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\text{persamaan bayangannya}\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.2x+y+3=0& & \\ \text{b}.2x-3y+3=0& \\ \text{c}.2x+3y+3=0& \\ \text{d}.3x+2y+3=0\\ \text{e}.3x-2y+3=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-2\\ 5\end{array}\right)=\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x-2+2y+10\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x+2y+8\\ y+5\end{array}\right)\\ \text{Diper}& \text{oleh}\\ & \begin{array}{l}\\ x^{″}& =x+2y+8\\ 2y^{″}& =2y+10-\\ x^{″}-2y^{″}& =x-2\\ \iff x& =x^{″}-2y^{″}+2....(1)\\ \text{maka}\\ y& =y^{″}-5....(2)\end{array}\\ \text{sehin}& \text{gga}\\ 2x+& y+4=0\\ 2(x^{″}& -2y^{″}+2)+(y^{″}-5)+4=0\\ 2x^{″}-& 3y^{″}+3=0\\ \text{maka}& \text{bayangan garisnya}\\ 2x-& 3y+3=0\end{array}\end{array}$.

$\begin{array}{l}\\ 97.& \text{Diketahui}M\text{adalah pencerminan terhadap}\\ & \text{garis}y=-x\text{dan}T\text{adalah transformasi}\\ & \text{yang dinyatakan oleh matriks}\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\\ & \text{Koordinat bayangan titik}A(2,-8)\text{oleh}\\ & \text{transformasi}M\text{dilanjutkan oleh}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-10,2)& & \\ \text{b}.(-2,-10)& \\ \text{c}.(10,2)& \\ \text{d}.(-10,-2)\\ \text{e}.(2,10)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T\circ M\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}0-3& -2+0\\ 0+1& 0+0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}-3& -2\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{c}-6+16\\ 2+0\end{array}\right)\\ & =\left(\begin{array}{c}10\\ 2\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 98.& \text{Jika}W\text{adalah transformasi oleh}\\ & \text{matriks}\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{maka titik mula}\\ & \text{dari}{W}^{\prime }(-2,5)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-11,-2)& & \\ \text{b}.(11,-2)& \\ \text{c}.(-2,11)& \\ \text{d}.(2,11)\\ \text{e}.(12,11)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dimi}& \text{salkan}:\\ A& =\left(\begin{array}{c}-2\\ 5\end{array}\right),\text{dan}\\ W& =\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{serta}X=\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{mak}& \text{a}\\ & \begin{array}{|c|}\hline \begin{array}{rl}A& =BX\\ B^{-1}A& =B^{-1}BX\\ B^{-1}A& =I.X\\ B^{-1}A& =X\\ X& =B^{-1}A\end{array}\\ \hline\end{array}\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|}\left(\begin{array}{cc}1& 0\\ -3& 1\end{array}\right)\left(\begin{array}{c}-2\\ 5\end{array}\right)}\\ & =1.\left(\begin{array}{c}-2+0\\ 6+5\end{array}\right)\\ & =\left(\begin{array}{c}-2\\ 11\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 99.& \text{Jika setiap titik pada grafik dengan}\\ & \text{dengan persamaan}y=\sqrt{x}\text{dicerminkan}\\ & \text{terhadap garis}y=x,\text{maka persamaan}\\ & \text{grafik yang dihasilkan adalah}...\\ & \begin{array}{l}\\ \text{a}.y=x^2,x\ge 0& & \\ \text{b}.y=-\sqrt{x},x\ge 0& \\ \text{c}.y=-x^2,x\le 0& \\ \text{d}.y=\sqrt{-x},x\le 0\\ \text{e}.y=-\sqrt{-x},x\le 0\end{array}\\ \\ & \mathbf{\text{UMB Tahun 2011 Kode 152}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ y& =\sqrt{x},\text{atau}{y}^2=x\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 1}}\\ \text{mak}& \text{a}\text{saat dicerminkan terhadap}\\ \text{gari}& \text{s}y=x,\text{adalah}{x}^2=y\\ \text{atau}& y=x^2.\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 2}}\\ \text{Jika}& \text{ingin dikerjakan dengan rumus}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =M_{x=y}\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)\\ \text{Sela}& \text{njutnya hasilnya disubstitusikan}\\ \text{ke p}& \text{ersamaan}y=\sqrt{x}\Rightarrow x^{\prime }=\sqrt{y^{\prime }}\\ \sqrt{y^{\prime }}& =x^{\prime }\text{maka}\\ y^{\prime }& ={\left(x^{\prime }\right)}^2\text{selanjutnya}\\ y& =x^2\end{array}\end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{l}\\ 100.& \text{Transformasi}T\text{adalah pencerminan}\\ & \text{terhadap garis}y={\displaystyle \frac{x}{3}\text{dilanjutkan oleh}}\\ & \text{pencerminan terhadap garis}y=-3x.\\ & \text{Matriks yang bersesuian dengan}\\ & \text{transformasi}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)& \\ \text{c}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& \\ \text{d}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ \text{e}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{SBMPTN Tahun 2013 Kode 433}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \text{sebu}& \text{ah persamaan garis lurus}\\ \text{dapa}& \text{t dituliskan dengan}:y=mx\\ \text{Dike}& \text{tahui pula bahwa ada 2 garis}:\\ y_1& ={\displaystyle \frac{1}{3}x\text{dan}{y}_2=-3x}\\ \text{seba}& \text{gai representasi transformasi}T.\\ \text{Kare}& \text{na}{m}_1\times m_2=\left({\displaystyle \frac{1}{3}}\right)(-3)=-1\\ \text{bera}& \text{rti 2 garis di atas saling tegak}\\ \text{luru}& \text{s dan hal ini seperti rotasi 2}\\ \text{kali}& {90}^{\circ }\text{atau}{180}^{\circ }\\ \text{Jadi},& T=\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ \iff & T=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.