Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{l}\\ 11.& (\mathbf{\text{SPMB 04}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{27}{3^{2x-1}}={81}^{-0,125}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{3}{4}}\\ \text{d}.& 1{\displaystyle \frac{1}{4}}\\ \text{e}.& 2{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{27}{3^{2x-1}}}& ={81}^{-0,125}\\ 3^{3-(2x-1)}& =3^{4(\frac{1}{8})}\\ 3-2x+1& =-{\displaystyle \frac{1}{2}}\\ -2x+4& =-{\displaystyle \frac{1}{2}}\\ -x+2& =-{\displaystyle \frac{1}{4}}\\ -x& =-2-{\displaystyle \frac{1}{4}}\\ -x& =-2{\displaystyle \frac{1}{4}}\\ x& =2{\displaystyle \frac{1}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& (\mathbf{\text{UMPTN 98}})\text{Bentuk}{\left({\displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^2}}\right)}^{-\frac{3}{4}}\\ & \text{dapat diserdernakan menjadi}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{x{y}^2}\\ \text{b}.& x\sqrt{y}\\ \text{c}.& \sqrt{x^{2}y}\\ \text{d}.& xy\sqrt{y}\\ \text{e}.& xy\sqrt{x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^2}}\right)}^{-\frac{3}{4}}& ={(x^{\frac{2}{3}-2}.y^{-\frac{3}{4}-\frac{2}{3}})}^{-\frac{3}{4}}\\ & =x^{-\frac{3}{4}(\frac{2}{3}-2)}.y^{-\frac{3}{4}(-\frac{3}{4}-\frac{2}{3})}\\ & =x^{-\frac{1}{2}+\frac{3}{2}}.y^{1+\frac{1}{2}}\\ & =x^1.y^{1\frac{1}{2}}\\ & =xy\sqrt{y}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& (\mathbf{\text{UMPTN 00}})\\ & \text{Bentuk}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ & \text{Jika}{x}_0\text{memenuhi persamaan, maka nilai}\\ & 1-{\displaystyle \frac{3}{4}{x}_0=....}\\ & \begin{array}{l}\\ \text{a}.& 1\frac{3}{16}\\ \text{b}.& 1\frac{1}{4}\\ \text{c}.& 1\frac{3}{4}\\ \text{d}.& 2\frac{1}{3}\\ \text{e}.& 2\frac{3}{4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}& ={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ 3^{-5x}& =3^{2(1-(x-2))}{.3}^{-\frac{2}{3}}\\ -5x& =2(1-(x-2))+(-{\displaystyle \frac{2}{3}}),\text{dikali}3\\ -15x& =6(3-x)+(-2)\\ -15x& =18-6x-2\\ 6x-15x& =16\\ -9x& =16\\ x& ={\displaystyle \frac{16}{-9}}\\ x_0& =-{\displaystyle \frac{16}{9},\text{selanjutnya}}\\ 1-{\displaystyle \frac{3}{4}{x}_0}& =1-{\displaystyle \frac{3}{4}\times (-\frac{16}{9})}\\ & =1+\frac{4}{3}\\ & =1+1{\displaystyle \frac{1}{3}}\\ & =2{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Diketahui}{x}^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\\ & \text{Nilai}x+x^{-1}=....\\ & \begin{array}{l}\\ \text{a}.& 7\\ \text{b}.& 8\\ \text{c}.& 8\\ \text{d}.& 10\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{x}^{\frac{1}{2}}+x^{-\frac{1}{2}}& =3\\ \text{dikadrat}& \text{kan}\\ {(x^{\frac{1}{2}}+x^{-\frac{1}{2}})}^2& =3^2\\ x+2+x^{-1}& =9\\ x+x^{-1}& =9-2\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Diketahui}{2}^{2x}+2^{-2x}=2\\ & \text{Nilai}{2}^x+2^{-x}=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& \sqrt{2}\\ \text{d}.& 3\\ \text{e}.& \sqrt{3}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{2}^{2x}+2^{-2x}& =2\\ \text{jika soal}& \text{dikuadratkan}\\ {(2^x+2^{-x})}^2& =2^{2x}+2+2^{-2x}\\ & =2^{2x}+2^{-2x}+2\\ {(2^x+2^{-x})}^2& =2+2=4\\ 2^x+2^{-x}& =\sqrt{4}\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 16.& \text{Jika}\mathbf{\text{a}}\text{dan}\mathbf{\text{b}}\text{adalah bilangan bulat positif}\\ & \text{yang memenuhi persamaan}{\mathbf{\text{a}}}^{\mathbf{\text{b}}}=2^{20}-2^{19},\\ & \text{maka nilai dari}\mathbf{\text{a}}+\mathbf{\text{b}}=....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 19\\ \text{d}.& 21\\ \text{e}.& 23\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\mathbf{\text{a}}}^{\mathbf{\text{b}}}& =2^{20}-2^{19}\\ & =2^{{}^{{}^{(19+1)}}}-2^{19}\\ & =2^{19}{.2}^1-2^{19}\\ & =2^{19}(2-1)\\ & =2^{19}\\ \text{Se}& \text{hingga nilai}\mathbf{\text{a}}+\mathbf{\text{b}}=2+19=21\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Perhatikan gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{Persamaan grafik fungsi seperti gambar di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=2^{x-2}\\ \text{b}.& f(x)=2^x-2\\ \text{c}.& f(x)=2^x-1\\ \text{d}.& f(x)={}^2\log (x-1)\\ \text{e}.& f(x)={}^2\log (x+1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{a}& \Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\ne 0(\text{salah})\\ \text{b}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-2=2^1-2=0=0(\mathbf{\text{benar}})\\ \text{c}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-1=2^1-1=1\ne 0(\text{salah})\\ \text{d}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1-1)={}^2\log 0=\\ & \mathbf{\text{tidak mungkin}}\ne 0(\text{salah})\\ \text{e}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1+1)={}^2\log 2=1\ne 0(\text{salah})\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Solusi untuk persamaan}{3}^{2x+1}={81}^{x-2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 4,5\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}{3}^{2x+1}& ={81}^{x-2}\\ \left(3^{2x}\right){.3}^1& ={\left(3^4\right)}^{x-2}\\ 3.\left(3^{2x}\right)& =3^{4x}{.3}^{-8}\\ 3.\left(3^{2x}\right)& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}}\\ 0& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}-3\left(3^{2x}\right)}\\ 0& ={\left(3^{2x}\right)}^2-3^9.\left(3^{2x}\right)\\ 0& =\left(3^{2x}\right)(\left(3^{2x}\right)-3^9)\end{array}& \begin{array}{rl}\mathbf{\text{atau}}& \\ {\left(3^{2x}\right)}^2-27.\left(3^{2x}\right)& =0\\ \left(3^{2x}\right)(\left(3^{2x}\right)-3^9)& =0\\ 3^{2x}=0\text{atau}{3}^{2x}& =3^9\\ (\text{tm})\text{atau}{3}^{2x}& =3^9\\ 2^x& =9\\ x& ={\displaystyle \frac{9}{2}}\\ \text{atau}x& =4{\displaystyle \frac{1}{2}}\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ & \text{maka nilai dari}{(a+b-c)}^{abc}=....\\ & \begin{array}{l}\\ \text{a}.& 1000\\ \text{b}.& 1\\ \text{c}.& 0\\ \text{d}.& -1\\ \text{e}.& -1000\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{r}\begin{array}{rl}{(\sqrt{2}+\sqrt{3}+\sqrt{5})}^2& =(\sqrt{2}+\sqrt{3}+\sqrt{5})\times (\sqrt{2}+\sqrt{3}+\sqrt{5})\\ & =\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ & +\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ & =2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ & =2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ & =10+\sqrt{2^2.6}+\sqrt{2^2.10}+\sqrt{2^2.15}\\ & =10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}& =\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ & \{\begin{array}{ll}a& =2\\ b& =3\\ c& =5\end{array}.\text{sesuai dengan urutannya}\\ \text{Sehingga nilai}& \\ {(a+b-c)}^{abc}& ={(2+3-5)}^{2.3.5}\\ & =0^{30}\\ & =0\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& (\mathbf{\text{UM UNDIP 2012 Math IPA}})\\ & \text{Jika}f(x)=x^3-3x^2-3x-1,\\ & \text{maka nilai dari}f(1+\sqrt[3]{2}+\sqrt[3]{4})=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& \sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& =x^3-3x^2-3x-1\\ & ={(x-1)}^3-6x\\ f(1+\sqrt[3]{2}+\sqrt[3]{4})& ={(1+\sqrt[3]{2}+\sqrt[3]{4}-1)}^3-6(1+\sqrt[3]{2}+\sqrt[3]{4})\\ & ={(\sqrt[3]{2}+\sqrt[3]{4})}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & ={\left(\sqrt[3]{2}(\sqrt[3]{2}+1)\right)}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =2(1+3\sqrt[3]{2}+3\sqrt[3]{4}+2)-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =(6+6\sqrt[3]{2}+6\sqrt[3]{4})-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =0\end{array}\end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 $\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 9\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}}& ={\displaystyle \frac{3^4{.5}^3{.7}^2}{3^3{.5}^3{.7}^2}}\\ & =3^{4-3}{.5}^{3-3}{.7}^{2-2}\\ & =3^1{.5}^0{.7}^0\\ & =3.1.1\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Bentuk sederhana dari}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle a{b}^2}\\ \\ \text{b}.& {\displaystyle a^2{b}^2}\\ \\ \text{c}.& {\displaystyle 5a^2{b}^5}\\ \\ \text{d}.& 5a^4{b}^5\\ \\ \text{e}.& 5a^5{a}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}}& =5a^{4-2}{b}^{2-(-3)}\\ & =5a^2{b}^5\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b}}\\ \\ \text{b}.& {\displaystyle \frac{1}{ab}}\\ \\ \text{c}.& {\displaystyle \frac{1}{a^2{b}^2}}\\ \\ \text{d}.& {\displaystyle \frac{1}{a{b}^2}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a^4{b}^4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4& ={\left(a^{1-2}{b}^{2-3}\right)}^4\\ & ={\left(a^{-1}{b}^{-1}\right)}^4\\ & ={\left({\displaystyle \frac{1}{ab}}\right)}^4\\ & ={\displaystyle \frac{1}{a^4{b}^4}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{2^{-\frac{1}{4}}\times 6^{\frac{2}{3}}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.6^{\frac{1}{4}}& & \text{d}.{\left({\displaystyle \frac{2}{3}}\right)}^{\frac{3}{4}}\\ \text{b}.6^{\frac{3}{4}}& \text{c}.6^{\frac{2}{3}}& \text{e}.{\left({\displaystyle \frac{3}{2}}\right)}^{\frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}& ={\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4{)}^{\frac{7}{12}}}{(2\times 3{)}^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times {\left(2^2\right)}^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & =2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ & =2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ & =2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ & =(2\times 3{)}^{\frac{9}{12}}\\ & =6^{\frac{9}{12}}\\ & =6^{\frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a}{b}}\\ \\ \text{b}.& {\displaystyle \frac{b}{a}}\\ \\ \text{c}.& {\displaystyle ab}\\ \\ \text{d}.& \sqrt[a]{b}\\ \\ \text{e}.& \sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}& ={\displaystyle \frac{a^{\frac{3}{4}}{b}^{\frac{-9}{2}}}{a^{-\frac{3}{2}}{b}^{-\frac{9}{4}}}}\\ & =a^{\frac{3}{4}+\frac{3}{2}}{b}^{\frac{-9}{2}+\frac{9}{4}}\\ & =a^{\frac{3+6}{4}}{b}^{\frac{-18+9}{4}}\\ & =a^{\frac{9}{4}}{b}^{-\frac{9}{4}}\\ & ={\left({\displaystyle \frac{a}{b}}\right)}^{\frac{9}{4}}\\ & =\sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}={\displaystyle \frac{1}{25}\text{adalah... .}}}\\ & \begin{array}{l}\\ \text{a}.& -5\\ \text{b}.& -4\\ \text{c}.& -3\\ \text{d}.& 1\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}}& ={\displaystyle \frac{1}{25}}\\ 25.{\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}}& =1\\ 5^2{.5}^2{.5}^{-\frac{4x+12}{6}}& =5^0\\ 5^{2+2-\frac{2}{3}x-2}& =5^0\\ {\displaystyle 2+2-\frac{2}{3}x-2}& =0\\ 2-{\displaystyle \frac{2}{3}x}& =0\\ -{\displaystyle \frac{2}{3}x}& =-2\\ x& =3\end{array}\end{array}$

$\begin{array}{l}\\ 7.& (\mathbf{\text{UM-UGM 03}})\text{Nilai}x\text{yang memenuhi}\\ & {\left({\displaystyle \frac{1}{25}}\right)}^{x-\frac{5}{2}}=\sqrt{{\displaystyle \frac{625}{5^{2-x}}}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{8}{5}}\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{25}}\right)}^{x-\frac{5}{2}}& =\sqrt{{\displaystyle \frac{625}{5^{2-x}}}}\\ 5^{-2x+5}& =5^{\frac{1}{2}(4-(2-x))}\\ -2x+5& ={\displaystyle \frac{1}{2}(4-2+x)}\\ -4x+10& =2+x\\ -5x& =2-10\\ x& ={\displaystyle \frac{-8}{-5}}\\ & =\frac{8}{5}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi}\\ & 2^{\frac{x}{3}-1}=16\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 10\\ \text{c}.& 15\\ \text{d}.& 20\\ \text{e}.& 25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{2}^{\frac{x}{3}-1}& =16\\ 2^{\frac{x}{3}-1}& =2^4\\ {\displaystyle \frac{x}{3}-1}& =4\\ {\displaystyle \frac{x}{3}}& =5\\ x& =15\end{array}\end{array}$

$\begin{array}{l}\\ 9.& (\mathbf{\text{SPMB 05}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{\sqrt[3]{(0,08{)}^{7-2x}}}{(0,2{)}^{-4x+5}}=1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -2\\ \text{c}.& -1\\ \text{d}.& 0\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt[3]{(0,08{)}^{7-2x}}}{(0,2{)}^{-4x+5}}}& =1\\ \sqrt[3]{(0,08{)}^{7-2x}}& =(0,2{)}^{-4x+5}\\ (0,2{)}^{\frac{3(7-2x)}{3}}& =(0,2{)}^{-4x+5}\\ {\displaystyle 7-2x}& =-4x+5\\ 4x-2x& =5-7\\ 2x& =-2\\ x& =-1\end{array}\end{array}$

$\begin{array}{l}\\ 10.& (\mathbf{\text{SPMB 05}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{\sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}}{27}=3^{x+1}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -16\\ \text{b}.& -7\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}}{27}}& =3^{x+1}\\ \sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}& =27\times 3^{x+1}\\ \sqrt[3]{3^{-2(2-x)}}& =3^3{.3}^{x+1}\\ 3^{\frac{-4+2x}{3}}& =3^{3+(x+1)}\\ {\displaystyle \frac{-4+2x}{3}}& =4+x\\ -4+2x& =12+3x\\ 2x-3x& =12+4\\ -x& =16\\ x& =-16\end{array}\end{array}$

Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{l}\\ 91.& \text{(UM UNBRAW)}\\ & \text{Nilai maksimum dari fungsi}\\ & f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6\\ \text{b}.& 24\\ \text{c}.& 26\\ \text{d}.& 32\\ \text{e}.& 92\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & f(x)=4{\cos }^{2}x+4{\sin }^{2}x+10{\sin }^{2}x+12\sin 2x+10\\ & f(x)=4+5(1-\cos 2x)+12\sin 2x+10\\ & f(x)=19-5\cos 2x+12\sin 2x\\ & f(x)=19+12\sin 2x-5\cos 2x\\ & f(x)=19+\sqrt{{12}^2+(-5{)}^2}\cos (2x-\theta )\\ & f(x)=19+13\cos (2x-\theta )\\ & \text{Karena nilai}\cos (2x-\theta )=\pm 1,\text{maka}\\ & f(x{)}_{maks}=19+13=32\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., & Akhmad G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
3. Sembiring, S., Zulkifli, M., Marsito, & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
4. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.
5. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 3 untuk Kelas XII SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Latihan Soal 9 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 81.& \text{Diketahui}f(x)={\cos }^{2}2x.\text{Jika}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx,\text{nilai untuk}\\ & {\displaystyle \frac{a-b}{c-d}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{3}}\\ \text{b}.& {\displaystyle \frac{2}{3}}\\ \text{c}.& -{\displaystyle \frac{3}{5}}\\ \text{d}.& -{\displaystyle \frac{6}{5}}\\ \text{e}.& -{\displaystyle \frac{9}{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)\\ & =-4\sin 2x\cos 2x\\ & f^{″}(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ & =8{\sin }^{2}2x-8{\cos }^{2}2x\\ & \text{Bandingkan dengan}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx\\ & \text{maka},a=8,b=2,c=-8,d=2\\ & \text{Jadi},{\displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 82.& \text{Diketahui}f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}.\text{Jika}}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2},\text{nilai dari}}\\ & m.n=....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 8\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}}\\ & f^{\prime }(x)={\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{-{\sin }^{2}x-{\cos }^{2}x+0}{{\sin }^2+2\sin x\cos x+{\cos }^{2}x}}\\ & ={\displaystyle \frac{-1}{1+\sin 2x}}\\ & f^{″}(x)={\displaystyle \frac{0-((-1).2\cos 2x)}{{(\sin 2x+1)}^2}=\frac{2\cos 2x}{{(\sin 2x+1)}^2}}\\ & \text{Bandingkan dengan yang diketahui}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2}}\\ & \{\begin{array}{ll}m& =2\\ n& =2\end{array}\\ & \text{Jadi},m.n=2.1=2\end{array}\end{array}$

$\begin{array}{l}\\ 83.& \text{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\text{dengan}0\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},0}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{2},0}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{2},1}\right)\\ \text{e}.& (\pi ,1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\Rightarrow f^{″}(x)=-4\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=0+k.2\pi \text{atau}2x=\pi +k.2\pi \\ & \iff x=0+k.\pi \text{atau}x=\frac{\pi }{2}+k.\pi \\ & \iff x=0,x={\displaystyle \frac{\pi }{2},x=\pi ,x={\displaystyle \frac{3\pi }{2}\text{atau}x=2\pi }}\\ & \bullet f\left({\displaystyle \frac{\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{\pi }{2},0}\right)\\ & \bullet f\left({\displaystyle \pi }\right)=\sin 2\left({\displaystyle \pi }\right)=0\Rightarrow \left({\displaystyle \pi ,0}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{3\pi }{2},0}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 84.& \text{Diketahui fungsi}f(x)=-3\cos 2x+1\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},2}\right)\\ \text{b}.& \left({\displaystyle \frac{2\pi }{3},\frac{5}{2}}\right)\\ \text{c}.& \left({\displaystyle \frac{3\pi }{2},4}\right)\\ \text{d}.& \left({\displaystyle \frac{5\pi }{4},1}\right)\\ \text{e}.& \left({\displaystyle \frac{5\pi }{3},\frac{5}{2}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=-3\cos 2x+1\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=6\sin 2x\Rightarrow f^{″}(x)=12\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 12\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{3\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{3\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{5\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{5\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{7\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{7\pi }{4},1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 85.& \text{Diketahui fungsi}f(x)={\sin }^{2}x+2\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{11}{4}}\right)\\ \text{c}.& \left({\displaystyle \pi ,2}\right)\\ \text{d}.& \left({\displaystyle \frac{4\pi }{3},\frac{11}{4}}\right)\\ \text{e}.& \left({\displaystyle \frac{11\pi }{6},\frac{9}{4}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x+2\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=2\sin x\cos x\Rightarrow f^{\prime }(x)=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{3\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{5\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{5\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{7\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{7\pi }{4},\frac{5}{2}}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 86.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{1}{2}\sin 2x\text{dengan}}\\ & 0^{\circ }<x<{360}^{\circ }.\text{Kurva akan cekung}\\ & \text{ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0^{\circ }<x<{90}^{\circ }\\ \text{b}.& 0^{\circ }<x<{90}^{\circ }\text{atau}{180}^{\circ }<x<{270}^{\circ }\\ \text{c}.& {45}^{\circ }<x<{225}^{\circ }\\ \text{d}.& {90}^{\circ }<x<{180}^{\circ }\text{atau}{270}^{\circ }<x<{360}^{\circ }\\ \text{e}.& {180}^{\circ }<x<{225}^{\circ }\text{atau}{225}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{1}{2}\sin 2x}\\ & f^{\prime }(x)=\cos 2x\Rightarrow f^{″}(x)=-2\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0^{\circ }\\ & \iff 2x=0^{\circ }+k{.360}^{\circ }\text{atau}2x={180}^{\circ }+k{.360}^{\circ }\\ & \iff x=0^{\circ }+k{.180}^{\circ }\text{atau}x={90}^{\circ }+k{.180}^{\circ }\\ & \iff x=0^{\circ },x={90}^{\circ },x={180}^{\circ }\text{dan}x={270}^{\circ }\\ & \text{serta}x={360}^{\circ }\\ & \bullet \text{Selang}{0}^{\circ }<x<{90}^{\circ },\text{misal}x={45}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({45}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{90}^{\circ }<x<{180}^{\circ },\text{misal}x={135}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({135}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{180}^{\circ }<x<{270}^{\circ },\text{misal}x={225}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({225}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{270}^{\circ }<x<{360}^{\circ },\text{misal}x={315}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({315}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 87.& \text{Diketahui fungsi}f(x)={\cos }^{2}x-{\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke bawah}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{2}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{d}.& {\displaystyle \frac{7\pi }{4}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}x-{\sin }^{2}x=\cos 2x\\ & f^{\prime }(x)=-2\sin 2x\Rightarrow f^{″}(x)=-4\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=-4\cos 2\left({30}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=-4\cos 2\left({90}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=-4\cos 2\left({210}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=-4\cos 2\left({300}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=-4\cos 2\left({330}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 88.& \text{Diketahui fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva fungsi tersebut akan}\\ & \text{cekung ke bawah pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{5\pi }{4}<x<\frac{7\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{7\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle 0<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{d}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}}\\ \text{e}.& {\displaystyle 0<x<\frac{\pi }{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=2\cos 2\left({30}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=2\cos 2\left({90}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=2\cos 2\left({210}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=2\cos 2\left({300}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=2\cos 2\left({330}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 89.& \text{Diketahui fungsi}f(x)=2\sin x-2\cos x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke atas}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{3\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<2\pi }\\ \text{d}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{e}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& f(x)=2\sin x-2\cos x\\ & f^{\prime }(x)=2\cos x+2\sin x\\ & f^{″}(x)=-2\sin x+2\cos x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin x+2\cos x=0\iff \sin x=\cos x\\ & \iff \tan x=1\\ & \iff x={\displaystyle \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{5\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \text{Sebagai gambaran saja}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={90}^{\circ }={\displaystyle \frac{\pi }{2}}}}\\ & \Rightarrow f^{″}({90}^{\circ })=-2\sin {90}^{\circ }+2\cos {90}^{\circ }=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 90.& \text{Diketahui fungsi}f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{dengan}0<x<2\pi .\text{Kurva fungsi tersebut}\\ & \text{akan cekung ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0<x<\frac{\pi }{6}\text{atau}\frac{\pi }{2}<x<\frac{5\pi }{6}}\\ \text{b}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{5\pi }{6}<x<\pi }\\ \text{c}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{5\pi }{6}<x<\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{\prime }(x)=3\cos (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{″}(x)=-9\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -9\sin (3x+{\displaystyle \frac{\pi }{2}})=0\iff \sin (3x+{\displaystyle \frac{\pi }{2}})=0\\ & \iff \sin (3x+{\displaystyle \frac{\pi }{2}})=\sin 0\\ & \iff (3x+{\displaystyle \frac{\pi }{2}})=0+k.2\pi \iff (3x+{\displaystyle \frac{\pi }{2}})={\displaystyle \pi +k.2\pi }\\ & \iff 3x=-{\displaystyle \frac{\pi }{2}+k.2\pi \iff 3x={\displaystyle \frac{\pi }{2}+k.2\pi }}\\ & \iff x=-{\displaystyle \frac{\pi }{6}+k.\frac{2\pi }{3}\iff x={\displaystyle \frac{\pi }{6}+k.{\displaystyle \frac{2\pi }{3}}}}\\ & \iff x={\displaystyle \frac{\pi }{6},x={\displaystyle \frac{\pi }{2},x=\frac{5\pi }{6},x=\frac{7\pi }{6},}}\\ & \text{dan}x=\frac{3\pi }{2},\text{serta}x=\frac{11\pi }{6}\\ & \text{Ingat bahwa domain}0\le x\le 2\pi \text{saja}\\ & \text{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{6},\text{misal}x={15}^{\circ }=\frac{\pi }{12}}\\ & \Rightarrow f^{″}({15}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{12}}\right)+{\displaystyle \frac{\pi }{2}})=-{\displaystyle \frac{9}{2}\sqrt{2}<0}\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{6}<x<{\displaystyle \frac{\pi }{2},\text{misal}x={60}^{\circ }={\displaystyle \frac{\pi }{3}}}}\\ & \Rightarrow f^{″}({60}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{3}}\right)+{\displaystyle \frac{\pi }{2}})=9>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 71.& \text{Sebuah mesin diprogram untuk dapat}\\ & \text{begerak tiap waktu mengikuti posisi}\\ & x=2\cos 3t\text{dan}y=2\cos 2t\text{di mana}\\ & x,y\text{dalam}cm,\text{dan}t\text{dalam detik}\\ & \text{Jika kecepatakan dirumuskan dengan}\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2},\text{maka nilai}v\\ & \text{saat}t=30detik\text{adalah}...cm/detik\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{3}\\ \text{b}.& 2\sqrt{11}\\ \text{c}.& 2\sqrt{10}\\ \text{d}.& 6\\ \text{e}.& 4\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui Kecepatan gerak mesin}\\ & \{\begin{array}{ll}x=2\cos 3x& \Rightarrow {\displaystyle \frac{dx}{dt}=-6\sin 3t}\\ y=2\cos 2x& \Rightarrow {\displaystyle \frac{dy}{dt}=-4\sin 2t}\end{array}\\ & \text{Maka kecepatan mesin saat}t=30\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}\\ & v=\sqrt{{(-6\sin 3t)}^2+{(-4\sin 2t)}^2}\\ & =\sqrt{{(-6\sin 3(30))}^2+{(-4\sin 2(30))}^2}\\ & =\sqrt{{(-6(1))}^2+{(-4\left({\displaystyle \frac{1}{2}\sqrt{3}}\right))}^2}\\ & =\sqrt{36+12}=\sqrt{48}=\sqrt{16.3}\\ & =4\sqrt{3}\end{array}\end{array}$

$\begin{array}{l}\\ 72.& \text{Sebuah benda duhubungkan dengan}\\ & \text{pegas dan bergerak sepanjang sumbu}\\ & \text{X dengan formula persamaan}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dari titik}O\text{yang dapat}\\ & \text{dicapai oleh benda tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui gerak benda yang bergerak}\\ & \text{mengikuti formula}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dicapai saat}{x}^{\prime }={\displaystyle \frac{dx}{dt}=0}\\ & x^{\prime }=2\cos 2t-2\sqrt{3}\sin 2t=0\\ & \iff 2\cos 2t=2\sqrt{3}\sin 2t\\ & \iff {\displaystyle \frac{\sin 2t}{\cos 2t}={\displaystyle \frac{1}{3}\sqrt{3}}}\\ & \iff \tan 2t=\tan {30}^{\circ }\\ & \iff 2t={30}^{\circ }+k{.180}^{\circ }\\ & \iff t={15}^{\circ }+k{.90}^{\circ }\{\begin{array}{ll}k=0,& t={15}^{\circ }\\ k=1,& t={105}^{\circ }\\ k=2,& t={195}^{\circ }\\ k=3,& t={285}^{\circ }\\ k=4,& t={375}^{\circ }\\ & \text{dst}\end{array}\\ & \text{Ambil}t={15}^{\circ },\text{maka nilai}\\ & x-\text{nya adalah}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \iff x=\sin 2({15}^{\circ })+\sqrt{3}\cos 2({15}^{\circ })\\ & \iff x={\displaystyle \frac{1}{2}+\sqrt{3}\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ & \iff x={\displaystyle \frac{1}{2}+\frac{3}{2}=2}\end{array}\end{array}$

$\begin{array}{l}\\ 73.& \text{Pada kurva}y=\sin x\text{dibuat}\\ & \text{garis singgung melalui titik}\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{garis singgung tersebut memotong}\\ & \text{sumbu-X di A dan sumbu-Y di B}.\\ & \text{Luas}\mathrm{△}AOB\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{36}}\\ \text{b}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{36}}\\ \text{c}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{16}}\\ \text{d}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{18}}\\ \text{e}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{18}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikan ilustrasi berikut}\end{array}$

$.\begin{array}{rl}& \text{Misalkan koordinat titik}P\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{maka},\\ & x_p={\displaystyle \frac{2\pi }{3},y_p=k=\sin {\displaystyle \frac{2\pi }{3}={\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & \text{Persamaan garis singgung di titik P}:\\ & y=m_{x_p}(x-x_p)+y_p\\ & \{\begin{array}{ll}\left({\displaystyle \frac{2\pi }{3},k}\right)& =\left({\displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3}}\right)\\ m_{x_p}={\displaystyle \frac{dy}{dx}}& =y^{\prime }=\cos x\\ m_{x_p}& =\cos \left({\displaystyle \frac{2\pi }{3}}\right)=-\frac{1}{2}\end{array}\\ & \text{Sehingga persamaan garis singgungnya}\\ & y=(-{\displaystyle \frac{1}{2}})(x-{\displaystyle \frac{2\pi }{3}})+{\displaystyle \frac{1}{2}\sqrt{3}}\\ & \iff 2y=-x+{\displaystyle \frac{2\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{memotong sumbu-X, maka}{y}_A=0\\ & 2y_A=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 0=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & x_A={\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & \bullet \text{memotong sumbu-Y, maka}{x}_B=0\\ & 2y_B=-x_B+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 2y_B=0+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & y_B={\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{Luas}\mathrm{△}AOB=\left[AOB\right]={\displaystyle \frac{x_A.y_B}{2}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\pi }{3}+\sqrt{3}}\right).\left({\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\right)}{2}}\\ & ={\displaystyle \frac{1}{6}(2\pi +3\sqrt{3}).{\displaystyle \frac{1}{6}(2\pi +3\sqrt{3})}}\\ & ={\displaystyle \frac{1}{36}{(2\pi +3\sqrt{3})}^2}\end{array}$

$\begin{array}{l}\\ 74.& \text{Sebuah wadah penampung air hujan}\\ & \text{memiliki ukuran sisi samping 3 m dan}\\ & \text{sisi horisontal juga 3 m. Sisi samping}\\ & \text{membentuk sudut}\theta (0\le \theta \le {\displaystyle \frac{\pi }{2}})\\ & \text{dengan garis vertikal (lihat gambar)}\\ & \text{Nilai}\theta \text{supaya wadah dapat menampung}\\ & \text{air hujan maksimum adalah}....\end{array}$

$\begin{array}{ll}.& \\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{\pi }{3}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}}\\ \text{c}.& {\displaystyle \frac{\pi }{5}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Supaya memuat dapat maksimum}\\ & \text{maka luas penampang haruslah}\\ & \text{MAKSIMUM, yaitu}\end{array}\end{array}$

gambar 1

gambar 2

$.\begin{array}{rl}& \text{Luas penampang}=\text{Luas Trapesium}\\ & \text{dengan}\{\begin{array}{ll}t& =3\sin \theta \\ n& =3\cos \theta \end{array}\\ & \text{Luas Penampang}={\displaystyle \frac{1}{2}(\sum \text{sisi sejajar})\times t}\\ & \iff L={\displaystyle \frac{1}{2}(6+2n)\times t}\\ & \iff L=(3+n)\times t\\ & \iff L=(3+3\cos \theta )\times 3\sin \theta \\ & \iff L=9\sin \theta +9\sin \theta \cos \theta \\ & \iff L=9\sin \theta +{\displaystyle \frac{9}{2}\sin 2\theta }\\ & \text{Suapa luas penampang}\text{MAKSIMUM}\\ & \text{maka}{L}^{\prime }={\displaystyle \frac{dL}{d\theta }=0}\\ & \iff L^{\prime }=9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9(2{\cos }^2\theta -1)=0\\ & \iff 2{\cos }^2\theta +\cos \theta -1=0\\ & \iff (\cos \theta +1)(2\cos \theta -1)=0\\ & \iff \cos \theta =-1\text{atau}2\cos \theta =1\\ & \iff \cos \theta =-1\text{atau}\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \theta =\cos \pi \text{atau}\cos \theta =\cos {\displaystyle \frac{\pi }{3}}\\ & \iff \theta =\pi \text{atau}\theta ={\displaystyle \frac{\pi }{3}}\end{array}$

$\begin{array}{l}\\ 75.& \text{Seseorang melempar bola dari atap}\\ & \text{sebuah rumah. Ketinggian bola saat}\\ & t(detik)\text{dinyatakan dengan persamaan}\\ & h(t)=5+{\cos }^2\pi t.\text{Kecepatan bola}\\ & \text{ditentukan dengan formula}v={\displaystyle \frac{dh}{dt}}\\ & \text{Besar kecepatan bola saat}t=0,25\\ & \text{detik adalah}....\\ & \begin{array}{ll}\text{a}.& 0\\ \text{b}.& \pi \\ \text{c}.& 2\pi \\ \text{d}.& 3\pi \\ \text{e}.& 4\pi \end{array}\\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}h(t)=5+{\cos }^2\pi t.\text{maka}\\ & v={\displaystyle \frac{dh}{dt}=2\cos \pi t(-\sin \pi t).(\pi )}\\ & \iff v=-\pi \sin 2\pi t\\ & \text{Saat}t=0,25={\displaystyle \frac{1}{4},\text{maka}}\\ & \text{besar kecepatannya adalah}:\\ & \iff v=-\pi \sin 2\pi \left({\displaystyle \frac{1}{4}}\right)\\ & \iff =-\pi \sin {\displaystyle \frac{\pi }{2}}\\ & \iff =-\pi \\ & \text{Tanda negatif menunjukkan}\\ & \text{arah kecepatan ke bawah}\\ & \text{Karena kecepatan merupakan salah}\\ & \text{satu besaran}\text{VEKTOR}\end{array}\end{array}$.

$\begin{array}{l}\\ 76.& \text{Turunan kedua dari}f(x)=x^3-\sin 3x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 6x^2+9\sin 3x\\ \text{b}.& 3x^2+6\sin 3x\\ \text{c}.& 3x-9\sin 3x\\ \text{d}.& 6x+9\sin 3x\\ \text{e}.& 9x-6\sin 3x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}f(x)& =x^3-\sin 3x\\ f^{\prime }(x)& =3x^2-3\cos 3x\\ f^{″}(x)& =6x+9\sin 3x\end{array}\end{array}$

$\begin{array}{l}\\ 77.& \text{Diketahui fungsi}g(x)={\displaystyle \frac{1-\cos x}{\sin x}.\text{Nilai}}\\ & \text{turunan kedua saat}x={\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2}+4\\ \text{b}.& 2\sqrt{2}-3\\ \text{c}.& 2\sqrt{2}+3\\ \text{d}.& 3\sqrt{2}-4\\ \text{e}.& 3\sqrt{2}+4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}g(x)& ={\displaystyle \frac{1-\cos x}{\sin x}}\\ g^{\prime }(x)& ={\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{{\sin }^{2}x-\cos x+{\cos }^{2}x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{1-\cos x}{{\sin }^{2}x}}\\ g^{″}(x)& ={\displaystyle \frac{\sin x({\sin }^{2}x)-2\sin x\cos x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin x({\sin }^{2}x)-\sin 2x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}({\sin }^2{\displaystyle \frac{\pi }{4})-\sin 2{\displaystyle \frac{\pi }{4}(1-\cos {\displaystyle \frac{\pi }{4})}}}}}{{\sin }^4{\displaystyle \frac{\pi }{4}}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right){\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2-1.(1-\left({\displaystyle \frac{1}{\sqrt{2}}}\right))}{{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}{\displaystyle \frac{1}{\sqrt{2}}-1+{\displaystyle \frac{1}{\sqrt{2}}}}}}{{\displaystyle \frac{1}{4}}}\times {\displaystyle \frac{4}{4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4}\end{array}\end{array}$

$\begin{array}{l}\\ 78.& \text{Turunan kedua fungsi}f(x)={\sin }^{2}x-{\cos }^{2}x\\ & \text{adalah}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 6\sin 2x\\ \text{b}.& 4\cos 2x\\ \text{c}.& 2\cos 2x\\ \text{d}.& -2\cos 2x\\ \text{e}.& -4\cos 2x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& ={\sin }^{2}x-{\cos }^{2}x\\ f^{\prime }(x)& =2\sin x\cos x-2\cos x(-\sin x)\\ & =2\sin x\cos x+2\sin x\cos x\\ & =2(2\sin x\cos x)=2\sin 2x\\ f^{″}(x)& =2.2\cos 2x=4\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 79.& \text{Diketahui}f(x)=\sqrt{\sin x}.\text{Jika}{f}^{″}(x)\\ & \text{adalah turunan keduafungsi}f,\text{maka}\\ & \text{nilai dari}{f}^{″}\left({\displaystyle \frac{\pi }{2}}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}}\\ \text{b}.& -{\displaystyle \frac{1}{4}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}f(x)& =\sqrt{\sin x}={\sin }^{\frac{1}{2}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x={\displaystyle \frac{\cos x}{2{\sin }^{\frac{1}{2}}x}}}\\ f^{″}(x)& ={\displaystyle \frac{-\sin x(2{\sin }^{\frac{1}{2}}x)-\cos x\left(2.{\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x}\right)}{4\sin x}}\\ & ={\displaystyle \frac{-2\sin x\sqrt{\sin x}-{\displaystyle \frac{{\cos }^{2}x}{\sqrt{\sin x}}}}{4\sin x}}\\ f^{″}\left({\displaystyle \frac{\pi }{2}}\right)& ={\displaystyle \frac{-2\sin {\displaystyle \frac{\pi }{2}.\sqrt{\sin {\displaystyle \frac{\pi }{2}}}-{\displaystyle \frac{{\cos }^2{\displaystyle \frac{\pi }{2}}}{\sin {\displaystyle \frac{\pi }{2}}}}}}{4\sin {\displaystyle \frac{\pi }{2}}}}\\ & ={\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 80.& \text{Jika}f(x)={\tan }^2(3x-2)\text{maka}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & -18{\sec }^4(3x-2)\\ \text{b}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^2(3x-2)\\ \text{c}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\\ \text{d}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +36{\sec }^4(3x-2)\\ \text{e}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\tan }^2(3x-2)\\ f^{\prime }(x)& =2\tan (3x-2){\sec }^2(3x-2)(3)\\ & =6\tan (3x-2){\sec }^2(3x-2)\\ f^{″}(x)& =6{\sec }^2(3x-2).(3){\sec }^2(3x-2)\\ & +6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ & =18{\sec }^4(3x-2)\\ & +36{\tan }^2(3x-2){\sec }^2(3x-2)\end{array}\end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{l}\\ 61.& \text{Fungsi}f(x)=\sin x-\cos x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<{\displaystyle \frac{7\pi }{4}}}\\ \text{d}.& 0<x<{\displaystyle \frac{3\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{e}.& 0<x<{\displaystyle \frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sin x-\cos x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=\cos x+\cos x=0\\ & \sin x=-\cos x\iff {\displaystyle \frac{\sin x}{\cos x}=-1}\\ & \iff \tan x=-1\\ & \iff \tan x=\tan {\displaystyle \frac{3\pi }{4}}\\ & \iff x={\displaystyle \frac{3\pi }{4}\pm k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}}\\ & \iff k=2\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm 2\pi =\text{tm}}\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & --& & & ++& \\ 0& & {\displaystyle \frac{3\pi }{4}}& & & {\displaystyle \frac{7\pi }{4}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{2}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{1}{2}\pi +\sin {\displaystyle \frac{1}{2}\pi =0+1=1(\text{positif})}}\\ & \text{untuk}x={\displaystyle \frac{3}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{3}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{3}{2}\pi +\sin {\displaystyle \frac{3}{2}\pi =0-1=-1(\text{negatif})}}\\ & \text{untuk}x={\displaystyle \frac{11}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{11}{6}\pi }\right)}\\ & =\cos {\displaystyle \frac{11}{6}\pi +\sin {\displaystyle \frac{11}{6}\pi ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}}}\end{array}\end{array}$

$\begin{array}{l}\\ 62.& \text{Fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}<x<\pi \text{atau}{\displaystyle \frac{3\pi }{2}<x<2\pi }}\\ \text{b}.& {\displaystyle \frac{2\pi }{3}<x<\pi }\\ \text{c}.& 0<x<{\displaystyle \frac{\pi }{2}\text{atau}\pi <x<{\displaystyle \frac{3\pi }{2}}}\\ \text{d}.& {\displaystyle \frac{4\pi }{3}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{\pi }{3}<x<\pi \text{atau}{\displaystyle \frac{4\pi }{3}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)={\sin }^{2}x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x=0\\ & \iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=\pm k.2\pi \text{atau}2x=\pi \pm k.2\pi \\ & \iff x=\pm k.\pi \text{atau}x=\frac{\pi }{2}\pm k.\pi \\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{2}}\\ & \iff k=1\Rightarrow x=\pi \text{atau}x={\displaystyle \frac{\pi }{2}+\pi =\frac{3\pi }{2}}\\ & \iff k=2\Rightarrow x=2\pi \text{atau}x={\displaystyle \frac{\pi }{2}+2\pi ={\displaystyle \frac{5}{2}\pi (\text{tm})}}\\ & \begin{array}{c}\\ & & & & & & & & \\ & ++& & --& & ++& & --& \\ 0& & {\displaystyle \frac{\pi }{2}}& & \pi & & {\displaystyle \frac{3\pi }{2}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{6}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{1}{6}\pi }\right)=\sin {\displaystyle \frac{1}{3}\pi =\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{3}{4}\pi \Rightarrow f\left({\displaystyle \frac{3}{4}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{3}{4}\pi }\right)=-1(\text{negatif})\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Fungsi}f(x)={\cos }^{2}2x\text{untuk}\\ & 0^{\circ }<x<{360}^{\circ }\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {45}^{\circ }<x<{90}^{\circ }\\ \text{b}.& {135}^{\circ }<x<{180}^{\circ }\\ \text{c}.& {225}^{\circ }<x<{270}^{\circ }\\ \text{d}.& {270}^{\circ }<x<{300}^{\circ }\\ \text{e}.& {315}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ & \text{Selanjutnya}\\ & \iff -2\sin 4x=0\iff \sin 4x=0\iff \sin 4x=\sin 0^{\circ }\\ & \iff \{\begin{array}{ll}4x=0^{\circ }+k{.360}^{\circ }& \Rightarrow x=k{.90}^{\circ }\\ 4x={180}^{\circ }+k{.360}^{\circ }& \Rightarrow x={45}^{\circ }+k{.90}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}x={45}^{\circ }\\ & \iff k=1\Rightarrow x={90}^{\circ }\text{atau}x={135}^{\circ }\\ & \iff k=2\Rightarrow x={180}^{\circ }\text{atau}x={225}^{\circ }\\ & \iff k=3\Rightarrow x={270}^{\circ }\text{atau}x={315}^{\circ }\\ & \iff k=4\Rightarrow x={360}^{\circ }\text{atau}x={405}^{\circ }(\text{tm})\\ & \text{Gunakan titik uji pada}x={30}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({30}^{\circ })=-2\sin 4({30}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={60}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({60}^{\circ })=-2\sin 4({60}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{Gunakan titik uji pada}x={120}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({120}^{\circ })=-2\sin 4({120}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={150}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({150}^{\circ })=-2\sin 4({150}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{dan seterusnya}...\\ & \begin{array}{c}\\ & & & & & & & & \\ & --& & ++& & --& & ++\\ 0& & {\displaystyle {45}^{\circ }}& & {90}^{\circ }& & {135}^{\circ }& & {180}^{\circ }\\ & --& & ++& & --& & ++\\ {180}^{\circ }& & {\displaystyle {225}^{\circ }}& & {270}^{\circ }& & {315}^{\circ }& & {360}^{\circ }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 64.& (\mathbf{\text{SBMPTN 2015}})\\ & \text{Fungsi}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{pada}0<x<\pi \text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x<\frac{11\pi }{12}}\\ \text{b}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{c}.& {\displaystyle \frac{2\pi }{3}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{3\pi }{4}<x<\pi }\\ \text{e}.& {\displaystyle \frac{3\pi }{4}<x<\frac{3\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)={\displaystyle \frac{\sin 2x+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}}=0}\\ & \sin 2x+{\displaystyle \frac{1}{2}\sqrt{3}=0\iff \sin 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{4\pi }{3}}\\ & \iff 2x={\displaystyle \frac{4\pi }{3}+k.2\pi \text{atau}2x=\pi -\frac{4\pi }{3}+k.2\pi }\\ & \iff x={\displaystyle \frac{2\pi }{3}+k.\pi \text{atau}x=-\frac{\pi }{6}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x=-{\displaystyle \frac{\pi }{6}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{5\pi }{3}\text{atau}x={\displaystyle \frac{5\pi }{6}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{\pi }{2}}\right)\sqrt{3}}{2}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{3\pi }{4}}\right)\sqrt{3}}{2}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{2\pi }{3}}}& & {\displaystyle \frac{5\pi }{6}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 65.& \text{Fungsi}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{dengan}x>0\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x\le \frac{13\pi }{12}}\\ \text{b}.& {\displaystyle \frac{7\pi }{12}<x<\frac{11\pi }{12}}\\ \text{c}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{d}.& {\displaystyle \frac{7\pi }{6}<x\le {\displaystyle \frac{13\pi }{6}}}\\ \text{e}.& {\displaystyle \frac{7\pi }{6}<x\le \frac{11\pi }{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)={\displaystyle \frac{\sin 2x+{\displaystyle \frac{1}{2}}}{2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}}=0}\\ & \sin 2x+{\displaystyle \frac{1}{2}=0\iff \sin 2x=-{\displaystyle \frac{1}{2}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{7\pi }{6}}\\ & \iff 2x={\displaystyle \frac{7\pi }{6}+k.2\pi \text{atau}2x=\pi -\frac{7\pi }{6}+k.2\pi }\\ & \iff x={\displaystyle \frac{7\pi }{12}+k.\pi \text{atau}x=-\frac{\pi }{12}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{7\pi }{12}\text{atau}x=-{\displaystyle \frac{\pi }{12}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{19\pi }{12}\text{atau}x={\displaystyle \frac{11\pi }{12}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{1}{2}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{\pi }{2}}\right)}{2}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{1}{2}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{3\pi }{4}}\right)}{2}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{7\pi }{12}}}& & {\displaystyle \frac{11\pi }{12}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 66.& \text{Titik stasioner fungsi}f(x)=\cos 3x\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1),\left({\displaystyle \frac{\pi }{4},1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{b}.& (0,1),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{6},-1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{6},1}\right),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{\pi }{2},1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},-1}\right)\\ \text{e}.& (0,1),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{2\pi }{3},1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\cos 3x\Rightarrow f^{\prime }(x)=-3\sin 3x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -\sin 3x=0\iff \sin 3x=0\iff \sin 3x=\sin 0\\ & \iff 3x=0+k.2\pi \text{atau}3x=\pi +k.2\pi \\ & \iff x=k.{\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \frac{\pi }{3}+k.\frac{2\pi }{3}}}\\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{3}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \pi }}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x=0\Rightarrow f(0)=\cos 3(0)=1\to (0,1)\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\cos 3\left({\displaystyle \frac{\pi }{3}}\right)=\cos \pi }\\ & =-1\to \left({\displaystyle \frac{\pi }{3},-1}\right)\\ & \text{dan seterusnya}\end{array}\end{array}$

$\begin{array}{l}\\ 67.& \text{Titik stasioner fungsi}f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1)\text{dan}\left({\displaystyle \frac{\pi }{6},-1}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{6},1}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},-1}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},1}\right)\text{dan}\left({\displaystyle \frac{5\pi }{6},-1}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{2},-1}\right)\text{dan}\left({\displaystyle \pi ,1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\Rightarrow f^{\prime }(x)=2\cos (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 2\cos (2x-{\displaystyle \frac{\pi }{6}})=0\iff \cos (2x-{\displaystyle \frac{\pi }{6}})=0\\ & \cos (2x-{\displaystyle \frac{\pi }{6}})=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff (2x-{\displaystyle \frac{\pi }{6}})=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & \iff x={\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}+k.\pi }\\ x& =-{\displaystyle \frac{\pi }{6}+k.\pi }\end{array}}\\ & \iff k=0\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}}\\ x& =-{\displaystyle \frac{\pi }{6}(\text{tm})}\end{array}\\ & \iff k=1\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{4\pi }{3}\text{tm}}\\ x& ={\displaystyle \frac{5\pi }{6}}\end{array}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\sin \left(2.{\displaystyle \frac{\pi }{3}-{\displaystyle \frac{\pi }{6}}}\right)=\sin \frac{\pi }{2}=1}\\ & =1\to \left({\displaystyle \frac{\pi }{3},1}\right)\\ & x={\displaystyle \frac{5\pi }{6}\Rightarrow f\left({\displaystyle \frac{5\pi }{6}}\right)=\sin \left(2.{\displaystyle \frac{5\pi }{6}-{\displaystyle \frac{\pi }{6}}}\right)}\\ & =\sin \frac{3\pi }{2}=-1\to \left({\displaystyle \frac{5\pi }{6},-1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 68.& \text{Nilai}x\text{pada titik stasioner}\\ & \text{fungsi}f(x)=x+\sin x\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^{\circ }\\ \text{b}.& {135}^{\circ }\\ \text{c}.& {150}^{\circ }\\ \text{d}.& {180}^{\circ }\\ \text{e}.& 360\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=x+\sin x\Rightarrow f^{\prime }(x)=1+\cos x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 1+\cos =0\iff \cos x=-1\\ & \iff \cos x=\cos {180}^{\circ }\\ & \iff x=\pm {180}^{\circ }+k{.360}^{\circ }\\ & \iff k=0\Rightarrow x=\{\begin{array}{ll}{180}^{\circ }& \text{mungkin}\\ -{180}^{\circ }& \text{tidak mungkin}\end{array}\\ & \iff k=1\Rightarrow x=\{\begin{array}{ll}{540}^{\circ }& \text{tidak mungkin}\\ {180}^{\circ }& \text{mungkin}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 69.& \text{Nilai}y\text{pada titik stasioner}\\ & \text{fungsi}f(x)=4\cos x+\cos 2x\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5\text{dan}3\\ \text{b}.& -4\text{dan}2\\ \text{c}.& -3\text{dan}5\\ \text{d}.& -2\text{dan}4\\ \text{e}.& 3\text{dan}5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ & \Rightarrow f^{\prime }(x)=-4\sin x-2\sin 2x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -4\sin x-2\sin 2x=0\\ & \iff -4\sin x-4\sin x\cos x=0\\ & \iff -4\sin x(1+\cos x)=0\\ & \iff \sin x(1+\cos x)=0\\ & \iff \sin x=0\text{atau}1+\cos x=0\\ & \iff \sin x=0\text{atau}\cos x=-1\\ & \iff \sin x=\sin 0^{\circ }\text{atau}\cos x=\cos {180}^{\circ }\\ & \iff x=\{\begin{array}{l}{0}^{\circ }+k{.360}^{\circ }\\ {180}^{\circ }+k{.360}^{\circ }\end{array}\text{atau}x=\{\begin{array}{l}{180}^{\circ }+k{.360}^{\circ }\\ -{180}^{\circ }+k{.360}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}{180}^{\circ }\\ & \text{Nilai}y-\text{nya}\\ & x=0^{\circ }\Rightarrow f(0^{\circ })\\ & =4\cos 0^{\circ }+\cos 2(0^{\circ })=4+1=5\\ & x={180}^{\circ }\Rightarrow f({180}^{\circ })\\ & =4\cos {180}^{\circ }+\cos 2({180}^{\circ })=-4+1=-3\end{array}\end{array}$

$\begin{array}{l}\\ 70.& \text{Nilai stasioner fungsi}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}}\\ & \text{untuk}0\le x\le 2\pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},\frac{1}{2}}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-\frac{1}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3}}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow f^{\prime }(x)={\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}}}\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & {\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}=0\iff 2\cos x-1=0}\\ & \iff \cos x={\displaystyle \frac{1}{2}\iff \cos x=\cos {\displaystyle \frac{\pi }{3}}}\\ & \iff x=\pm {\displaystyle \frac{\pi }{3}+k.2\pi }\\ & \iff k=0\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}\iff x=\{\begin{array}{ll}{\displaystyle \frac{\pi }{3}}& \text{memenuhi}\\ -{\displaystyle \frac{\pi }{3}}& \text{tidak memenuhi}\end{array}}\\ & \iff k=1\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}+2\pi \iff x=\{\begin{array}{ll}{\displaystyle \frac{7\pi }{3}}& \text{tidak memenuhi}\\ {\displaystyle \frac{5\pi }{3}}& \text{memenuhi}\end{array}}\\ & \text{Titiknya adalah}\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{3}}}{2-\cos {\displaystyle \frac{\pi }{3}}}={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{\pi }{3},{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\\ & x={\displaystyle \frac{5\pi }{3}\Rightarrow f\left({\displaystyle \frac{5\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{5\pi }{3}}}{2-\cos {\displaystyle \frac{5\pi }{3}}}={\displaystyle \frac{-{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{5\pi }{3},-{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\end{array}\end{array}$.

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

 $\begin{array}{l}\\ 51.& \text{Turunan pertama dari fungsi}\\ & g(x)={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\\ \text{b}.& {\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}}\\ \text{c}.& {\displaystyle \frac{1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{d}.& {\displaystyle \frac{-1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{e}.& {\displaystyle {\sin }^{2}x{\cos }^{2}x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ g(x)& ={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}\\ & =\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}={\displaystyle \frac{1}{\sin x\cos x}}\\ \text{maka}& \\ g^{\prime }(x)& ={\displaystyle \frac{0.(\sin x\cos x)-1.({\cos }^{2}x-{\sin }^{2}x)}{(\sin x\cos x{)}^2}}\\ & ={\displaystyle \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\end{array}\end{array}$

$\begin{array}{l}\\ 52.& \text{Diketahui}h(x)=\cos \left({\displaystyle \frac{3}{x}}\right),\\ & \text{maka}{\displaystyle \frac{dh}{dx}}\\ & \begin{array}{l}\\ \text{a}.& -3\sin {\displaystyle \frac{3}{x}}\\ \text{b}.& -{\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{c}.& -{\displaystyle \frac{3}{x}\sin \frac{3}{x}}\\ \text{d}.& {\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{e}.& {\displaystyle \frac{3}{x}\sin \frac{3}{x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos {\displaystyle \frac{3}{x}}& =-\sin {\displaystyle \frac{3}{x}\left({\displaystyle \frac{0.(x)-3.1}{x^2}}\right)}\\ & ={\displaystyle \frac{-(-3)}{x^2}\sin \frac{3}{x}}\\ & ={\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Turunan pertama dari}\tan (\cos x),\\ & \text{terhadap}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\sec }^2(\cos x)\sin x\\ \text{b}.& {\sec }^2(\cos x)\sin x\\ \text{c}.& {\sec }^2(\sin x)\cos x\\ \text{d}.& {\displaystyle \sin x}\\ \text{e}.& {\displaystyle -\sin x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ y& =\tan x(\cos x)\\ y^{\prime }& ={\sec }^2(\cos x)\times (-\sin x)\\ & =-{\sec }^2(\cos x).\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 54.& (\mathbf{\text{UN 2005}})\text{Turunan pertama dari}\\ & f(x)=\sqrt[3]{{\cos }^2(3x^2+5x)}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{b}.& {\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)}\\ \text{c}.& -{\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{d}.& -{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\\ \text{e}.& {\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ f(x)& =\sqrt[3]{{\cos }^2(3x^2+5x)}\\ f^{\prime }(x)& ={\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)\\ & ={\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{2}}}(3x^2+5x)\times (-\sin (3x^2+5x))}\\ & \times (6x+5)\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)}\\ & \times {\cos }^{-1}(3x^2+5x)\times \sin (3x^2+5x)\\ & =-{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\end{array}$.

$\begin{array}{l}\\ 55.& \text{Persamaan garis singgung pada kurva}\\ & y=3\sin x\text{pada titik yang berabsis}{\displaystyle \frac{\pi }{3}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})-\frac{2\sqrt{2}}{3}}\\ \text{b}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})+\frac{2\sqrt{2}}{3}}\\ \text{c}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{3}}{2}}\\ \text{d}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})+\frac{3\sqrt{3}}{2}}\\ \text{e}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{2}}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& y=3\sin x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0=3\sin \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)=\frac{3\sqrt{3}}{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=3\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=3\cos \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}}\right)=\frac{3}{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y={\displaystyle \frac{3}{2}(x-{\displaystyle \frac{\pi }{3}})+{\displaystyle \frac{3\sqrt{3}}{2}}}\end{array}\end{array}$

$\begin{array}{l}\\ 56.& \text{Kurva}y=\sin x+\cos x\text{untuk}\\ & 0<x<\pi \text{memotong sumbu X}\\ & \text{di titik A. Persamaan garis}\\ & \text{singgung di titik A adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{b}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{2}})\\ \text{c}.& y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\\ \text{d}.& y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{e}.& y=\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Kurva memotong sumbu X}\\ & \text{di titik A, berarti}y=0\\ & \sin x+\cos x=0\\ & \sin x=-\cos x\\ & {\displaystyle \frac{\sin x}{\cos x}=-1\iff \tan x=-1}\\ & \tan x=\tan \left({\displaystyle \frac{3\pi }{4}}\right)\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \text{Jadi, titik A-nya}:\left({\displaystyle \frac{3\pi }{4},0}\right)\\ & \text{dan nilai gradien}m=y^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{3\pi }{4}}\right)-\sin \left({\displaystyle \frac{3\pi }{4}}\right)\\ & m=-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Persamaan garis singgung di A}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})+0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y={\sec }^{2}x\text{pada titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{b}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{c}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{d}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{e}.& y=4\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& y={\sec }^{2}x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0={\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)=(2{)}^2=4\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=2{\sec }^{2}x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=2{\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)\tan \left({\displaystyle \frac{\pi }{3}}\right)\\ & =2(4)\sqrt{3}=8\sqrt{3}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\end{array}\end{array}$

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

$\begin{array}{l}\\ 59.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sec x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=\sqrt{3}x-{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{b}.& y=\sqrt{3}x+{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{c}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{d}.& y=\sqrt{2}x+{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{e}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& y=\sec x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & y_0=\sec \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=\sec x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & m=\sec \left({\displaystyle \frac{\pi }{4}}\right)\tan \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}.1=\sqrt{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})+\sqrt{2}\\ & \iff y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sin x+\cos x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{2}\text{akan memotong sumbu}}\\ & \text{Y dengan ordinatnya berupa}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}+1}\\ \text{b}.& {\displaystyle \frac{\pi }{2}-1}\\ \text{c}.& 1-{\displaystyle \frac{\pi }{2}}\\ \text{d}.& 2+{\displaystyle \frac{\pi }{2}}\\ \text{e}.& 2-{\displaystyle \frac{\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& y=\sin x+\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{2}}\\ & y_0=\sin \left({\displaystyle \frac{\pi }{2}}\right)+\cos \left({\displaystyle \frac{\pi }{2}}\right)=1+0=1\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{\pi }{2}}\right)-\sin \left({\displaystyle \frac{\pi }{2}}\right)\\ & m=0-1=-1\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-1(x-{\displaystyle \frac{\pi }{2}})+1\\ & \iff y=-x+{\displaystyle \frac{\pi }{2}+1}\\ & \text{Ordinat garis singgungnya saat}\\ & \text{memotong sumbu-Y adalah}:x=0,\\ & \text{maka}\\ & y=-0+{\displaystyle \frac{\pi }{2}+1={\displaystyle \frac{\pi }{2}+1}}\end{array}\end{array}$