PAS MATEMATIKA BLOG

Contoh Soal Distribusi Binomial (2)

$\color{blue}\textbf{Contoh Variabel Acak}$

$\begin{array}{ll}\\ 6.&\textrm{Sebuah uang logam ditos sebanyak 3 kali}\\ &\textrm{Jika}\: \: X\: \: \textrm{sebagai variabel acak dari kejadian}\\ &\textrm{munculnya sisi angka (A), maka peluang}\\ &\textrm{a. kejadian terjadi muncul 0 angka}\\ &\textrm{b. kejadian terjadi muncul 1 angka}\\ &\textrm{c. kejadian terjadi muncul 2 angka}\\ &\textrm{d. kejadian terjadi muncul 3 angka}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned} \color{blue}\textrm{Mula}\: \, &\color{red}(1)\quad (2)\quad (3)\quad \color{blue}\textbf{Ruang sampel}\\ \textbf{Mulai}&\left\{\begin{matrix} A\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow (A,A,A)\\ G\rightarrow (A,A,G) \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (A,G,A)\\ G\rightarrow (A,G,G) \end{matrix}\right. \end{matrix}\right.\\ G\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow (G,A,A)\\ G\rightarrow (G,A,G) \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (G,G,A)\\ G\rightarrow (G,G,G) \end{matrix}\right. \end{matrix}\right. \end{matrix}\right. \end{aligned}\\ &\begin{aligned}P(X=0)&=P((G,G,G))\\ &=\displaystyle \frac{n(X=0)}{n(S)}\\ &=\color{red}\displaystyle \frac{1}{8}\\ P(X=1)&=P((G,G,A),(G,A,G),(A,G,G))\\ &=\displaystyle \frac{n(X=1)}{n(S)}\\ &=\color{red}\displaystyle \frac{3}{8}\\ P(X=2)&=P((G,A,A),(A,G,A),(A,A,G))\\ &=\displaystyle \frac{n(X=2)}{n(S)}\\ &=\color{red}\displaystyle \frac{3}{8}\\ P(X=3)&=P((A,A,A))\\ &=\displaystyle \frac{n(X=3)}{n(S)}\\ &=\color{red}\displaystyle \frac{1}{8} \end{aligned} \end{array}$

Contoh Soal Distribusi Binomial (1)

$\color{blue}\textbf{Contoh Peluang dan Kombinasi}$

$\begin{array}{ll}\\ 1.&\textrm{Seorang melempar sebuah dadu dengan enam muka}\\ &\textrm{Tentukukanlah}\\ &\textrm{a}.\quad \textrm{ruang sampel}\\ &\textrm{b}.\quad \textrm{peluang muncul mata dadu ganjil}\\ &\textrm{c}.\quad \textrm{peluang muncul mata dadu genap}\\ &\textrm{d}.\quad \textrm{peluang muncul mata dadu angka prima}\\ &\textrm{e}.\quad \textrm{peluang muncul mata dadu kurang dari 6}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Mata}&\: \textrm{dadu ada 6, yaitu}:\color{red}1,2,3,4,5,\: \& \: 6\\ \textrm{a}.\quad&\textrm{Raung sampel}\\ &S=\color{red}\left \{ 1,2,3,4,5,6 \right \}\: \color{black}\Rightarrow n(S)=6\\ \textrm{b}.\quad&\textrm{peluang}\: \color{red}\textrm{muncul mata dadu ganjil}\color{black}(J)\\ &\textrm{Mata dadu ganjil}:1,3,5\: \Rightarrow n(J)=3\\ &\textrm{Peluangnya}=\displaystyle \frac{n(J)}{n(S)}=\frac{3}{6}=\frac{1}{2}\\ \textrm{c}.\quad&\textrm{peluang}\: \color{red}\textrm{muncul mata dadu ganap}\color{black}(P)\\ &\textrm{Mata dadu ganap}:2,4,6\: \Rightarrow n(P)=3\\ &\textrm{Peluangnya}=\displaystyle \frac{n(P)}{n(S)}=\frac{3}{6}=\frac{1}{2}\\ \textrm{d}.\quad&\textrm{peluang}\: \color{red}\textrm{muncul mata dadu angka prima}\color{black}(R)\\ &\textrm{Mata dadu angka prima}:2,3,5\: \Rightarrow n(R)=3\\ &\textrm{Peluangnya}=\displaystyle \frac{n(R)}{n(S)}=\frac{3}{6}=\frac{1}{2}\\ \textrm{e}.\quad&\textrm{peluang}\: \color{red}\textrm{muncul mata dadu kurang dari 6}\color{black}(Z)\\ &\textrm{Mata dadu kurang dari 6}:1,2,3,4,5\: \Rightarrow n(Z)=5\\ &\textrm{Peluangnya}=\displaystyle \frac{n(Z)}{n(S)}=\frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Andi akan mengambil 4 buah bola dari}\\ &\textrm{10 warna yang berbeda. Berapakah banyak}\\ &\textrm{kombinasi warna yang berbeda yang diambil}\\ &\textrm{oleh Andi}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}n=10&\: \: \textrm{dan}\: \: r=4\\ C(n,r)&=\displaystyle \frac{n!}{r!(n-r)!}\\ C(10,4)&=\displaystyle \frac{10!}{4!(10-4)!}\\ &=\displaystyle \frac{10!}{4!\times 6!}\\ &=\displaystyle \frac{10\times 9\times 8\times 7\times 6!}{(4\times 3\times 2\times 1)\times 6!}\\ &=420\: \: \textrm{kombinasi warna bola berbeda} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Dua kantong berisi bola merah dan biru}\\ &\textrm{Kantong I memuat 4 bola merah dan }\\ &\textrm{6 bola biru. Sedangkan kantong II memuat}\\ &\textrm{5 bola merah dan 3 bola biru. Jika pada}\\ &\textrm{masing-masing kantong diambil 2 bola}\\ &\textrm{sekaligus, maka peluang terambilnya}\\ &\textrm{1 bola merah dan 1 bola biru pada kantong}\\ &\textrm{I serta 2 bola biru pada kantong II}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Kejadian di atas adalah kejadian saling}\\ &\textrm{bebas karena tidak saling mempengaruhi}\\ &\begin{aligned}\textrm{Misal}&\: X=\textrm{kejadian terambil}\: 1M,1B\\ \bullet \quad&\color{red}\textrm{pada kantong I}\\ P(X)&=\displaystyle \frac{C(4,1)\times C(6,1)}{C(10,2)}\\ &=\displaystyle \frac{4\times 6}{\displaystyle \frac{10\times 9}{2}}=\frac{8}{15}\\ \textrm{Misal}&\: Y=\textrm{kejadian terambil}\: 2B\\ \bullet \quad&\color{red}\textrm{pada kantong II}\\ P(Y)&=\displaystyle \frac{C(3,2)}{C(8,2)}\\ &=\displaystyle \frac{3}{\displaystyle \frac{8\times 7}{2}}=\frac{3}{28}\\ \color{magenta}\textrm{maka}&\: \textrm{peluang dari}\: \: X\: \: \textrm{dan}\: \: Y\\ P(X\cap Y)&=P(X)\times P(Y)\\ &=\displaystyle \frac{8}{15}\times \frac{3}{28}\\ &=\displaystyle \frac{2}{35} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Berapa banyak cara dapat memilih untuk}\\ &\textrm{3 perwakilan dari 10 anggota suatu}\\ &\textrm{kelompok, jika}\\ &\textrm{a. tanpa perlakuan khusus}\\ &\textrm{b. salah seorang harus terpilih}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Dengan tanpa perlakuan}\\ &\textrm{memilih 3 orang dari 10 orang adalah}:\\ &C(10,3)=\displaystyle \frac{10!}{3!(10-3)!}=\frac{10!}{3!\times 7!}=\color{blue}120\\ \textrm{b}.\quad&\textrm{Dengan perlakuan 1 orang terpilih}\\ &\color{red}(\textrm{1 orang ini artinya tidak perlu diperhitungkan})\\ &\textrm{memilih 2 orang dari 9 orang adalah}:\\ &C(9,2)=\displaystyle \frac{9!}{2!(9-2)!}=\frac{9!}{2!\times 8!}=\color{blue}36 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Berapa banyak cara dapat memilih 2 buku}\\ &\textrm{matematika dan 3 buku fisika serta 4 buku}\\ &\textrm{ekonomi pada suatu lemari buku yang}\\ &\textrm{di dalamnya terdapat 10 buku matematika,}\\ &\textrm{11 buku fisika dan 12 buku ekonomi}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Banyak}&\: \textrm{cara pemilihan tersebut adalah}:\\ &=C(10,2)\times C(11,3)\times C(12,4)\\ &=\displaystyle \frac{10!}{2!\times 8!}\times \frac{11!}{3!\times 8!}\times \frac{12!}{4!\times 8!}\\ &=\displaystyle \frac{10\times 9}{1\times 2}\times \frac{11\times 10\times 9}{1\times 2\times 3}\times \frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}\\ &=\color{red}3675375 \end{aligned} \end{array}$

Lanjutan Materi Distribusi Peluang Diskrit (Matematika Peminatan Kelas XII)

$\color{blue}\begin{aligned}\textrm{C. 1}.\quad&\textrm{Distribusi Peluang Diskrit} \end{aligned}$

$\begin{aligned} &\textrm{Misalkan}\: \: X\: \: \textrm{adalah variabel acak diskrit}\\ &\textrm{dari nilai}\: :\: \: x_{1},\: x_{2},\: x_{3},\: x_{4},\: \cdots \: ,\: x_{k},\: \textrm{dan}\\ &P\: \textrm{adalah seluruh nilai peluang untuk}\: :\\ &p_{1},\: p_{2},\: p_{3},\: p_{4},\: \cdots \: ,p_{k}, \textrm{maka nilai untuk}\\ &\color{blue}p_{1}+ p_{2}+ p_{3}+ p_{4}+ \cdots +p_{k}=1\\ &\textbf{dan}\\ &\textrm{Fungsi}\: \: f(x) =P(X=x)\: \: \textrm{yang mempunyai}\\ &\textrm{nilai}\: \: p_{1},\: p_{2},\: p_{3},\: p_{4},\: \cdots \: ,p_{k},\: \textrm{pada variabel}\\ &X=x_{1},\: x_{2},\: x_{3},\: x_{4},\: \cdots \: ,\: x_{k},\: \textrm{disebut fungsi}\\ &\textrm{kepekatan peluang dari variabel acak}\: \: X.\\ &\textrm{Selanjutnya jika kita gambar grafik}\: \: f(x)\\ &\textrm{terhadap}\: \: x,\: \textrm{maka kita akan grafik yang}\\ &\textrm{dinamakan dengan}\: \: \color{red}\textbf{grafik peluang} \end{aligned}$

Suatu fungsi $f(x)=P(X=x)$ disebut fungsi peluang (probabilitas) dari $X$, jika memenuhi syarat-syarat:

$\color{blue}\begin{matrix} (\textrm{i})\quad f(x)\geq 0\: \: \: \textrm{untuk semua}\: \: x\qquad\qquad\qquad\qquad\qquad\qquad\: \: \\\\ (\textrm{ii})\quad \sum_{i=1}^{n}f\left ( x_{i} \right )=\color{red}f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n})=\color{black}1 \end{matrix}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Pada percobaan melempar 3 koin identik}\\ &\textrm{sekaligus bersama-sama. Variabel acak}\\ &\textrm{dalam hal ini pada kejadian muncul sisi}\\ &\textrm{gambar, tentukan}\\ &\textrm{a}.\: \: \textrm{distribusi peluangnya}\\ &\textrm{b}.\: \: \textrm{tabel fungsi peluangnya}\\ &\textrm{c}.\: \: \textrm{grafik fungsi peluangnya}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui dari soal}\: \: \color{red}\textrm{variabel acak}\\ &\textrm{pada kejadian di atas adalah munculnya}\\ &\textrm{sisi gambar pada pelemparan 3 koin}\\ &\textrm{maka} \end{aligned}\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya}\\ &\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \textrm{Sampel}&AAA&AA\color{red}G&A\color{red}G\color{black}A&A\color{red}GG&\color{red}G\color{black}AA&\color{red}G\color{black}A\color{red}G&\color{red}GG\color{black}A&\color{red}GGG\\\hline \textrm{Muncul}\: \color{red}(G)&0&1&1&2&1&2&2&3\\\hline \end{array}\\ \textrm{b}.\quad&\textrm{Tabel fungsi peluangnya}\\ &x=\textrm{muncul kejadian sisi gambar}\: \: \color{red}(G)\\ &\begin{array}{|c|c|c|c|c|c|}\hline x&0&1&2&3&\color{red}\textrm{Jumlah}\\\hline f(x)&\displaystyle \frac{1}{8}&\displaystyle \frac{3}{8}&\displaystyle \frac{3}{8}&\displaystyle \frac{1}{8}&\color{red}1\\\hline \end{array}\\ \textrm{c}.\quad&\textrm{Grafik fungsi peluangnya adalah}\\ & \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Pada sebuah kotak terdapat 2 kelereng}\\ &\textrm{biru dan 4 kelereng merah. Tiga kereng}\\ &\textrm{diambil secara acak. Tentukanlah distribusi}\\ &\textrm{peluang}\: \: \color{red}x\: \: \color{black}\textrm{jika}\: \: \color{red}x\: \: \color{black}\textrm{menyatakan banyaknya}\\ &\textrm{terambilnya bola biru}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|c|}\hline \qquad\qquad\textrm{Nama}&\textrm{Perhitungan}\\\hline \textrm{Banyak}&\\ \textrm{titik sampel}&\begin{aligned}C_{3}^{6}&=\displaystyle \frac{6!}{3!(6-3)!}=20 \end{aligned}\\\hline \textrm{Banyak cara}&\\ \textrm{mendapatkan bola biru}&C_{x}^{2}\\\hline \textrm{Banyak cara}&\\ \textrm{mendapatkan bola merah}&C_{3-x}^{4}\\\hline \end{array} \end{array}$

$.\quad \: \begin{array}{|l|l|}\hline \color{red}\textrm{Distribusi peluang}&\qquad\quad\color{red}\textrm{Perhitungan}\\\hline P(X=x)=f(x)&f(x)=\displaystyle \frac{\displaystyle C_{x}^{2}.C_{3-x}^{4}}{\displaystyle C_{3}^{6}},\\ \textrm{untuk}&\begin{aligned}x&=0,1,2 \end{aligned}\\\hline x=0\Rightarrow P(x=0)&f(x)=\displaystyle \frac{\displaystyle C_{0}^{2}.C_{3-0}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{0}^{2}.C_{3}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{0!2!}\times \frac{4!}{3!1!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!3!6!}=0,2\\\hline x=1\Rightarrow P(x=1)&f(x)=\displaystyle \frac{\displaystyle C_{1}^{2}.C_{3-1}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{1}^{2}.C_{2}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{1!1!}\times \frac{4!}{2!2!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!2!6!}=0,6\\\hline x=2\Rightarrow P(x=2)&f(x)=\displaystyle \frac{\displaystyle C_{2}^{2}.C_{3-2}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{2}^{2}.C_{1}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{2!0!}\times \frac{4!}{1!3!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!3!6!}=0,2\\\hline \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa fungsi}\: \: P(x)=\displaystyle \frac{x+2}{12}\\ &\textrm{untuk}\: \: x=1,2,\: \textrm{dan}\: \: 3\: \: \textrm{merupakan fungsi}\\ &\textrm{peluang}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhatik}&\textrm{an bahwa}\\ \bullet \quad P(1)&=\displaystyle \frac{1+2}{12}=\frac{3}{12}=\frac{1}{4}\\ \bullet \quad P(2)&=\displaystyle \frac{2+2}{12}=\frac{4}{12}=\frac{1}{3} \\ \bullet \quad P(3)&=\displaystyle \frac{5+2}{12}=\frac{5}{12} \\ \textrm{Sehing}&\textrm{ga}\: \: \displaystyle \sum_{i=1}^{3}P(i)=\displaystyle \frac{3}{12}+\frac{4}{12}+\frac{5}{12}=\color{red}\frac{12}{12}=1\\ &\begin{cases} (\textrm{i}) & \textrm{Peluangnya berada}\: \: \color{red}0\leq P(i)\leq 1 \\ (\textrm{ii}) & \textrm{dan nilai totolnya}=\displaystyle \color{red}\sum_{i=1}^{3}P(i)=1 \end{cases}\\ \textrm{Jadi},\: &\textrm{fungsi}\: \: P(x)=\displaystyle \frac{x+2}{12}\: \: \textrm{untuk}\: \: x=1,2,\: \textrm{dan}\: \: 3\\ &\textbf{merupakan fungsi peluang} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui fungsi peluang adalah}\: \: P(x)=\displaystyle \frac{\color{red}m}{x+1}\\ &\textrm{untuk}\: \: x=0,1,2,\: \textrm{dan}\: \: 3\: .\: \textrm{Tentukanlah}\\ &\textrm{a}.\: \: \textrm{nilai}\: \: \color{red}m\\ &\textrm{b}.\: \: \textrm{nilai}\: \: \color{red}P(x\leq 2)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \sum_{i=0}^{3}P(i)=\color{blue}1\\ &\Leftrightarrow \displaystyle \frac{\color{red}m}{0+1}+\frac{\color{red}m}{1+1}+\frac{\color{red}m}{2+1}+\frac{\color{red}m}{3+1}=1\\ &\Leftrightarrow \color{red}m\color{black}+\displaystyle \frac{\color{red}m}{2}+\frac{\color{red}m}{3}+\frac{\color{red}m}{4}=1\\ &\Leftrightarrow \left (\displaystyle \frac{12+6+4+3}{12} \right )\color{red}m\color{black}=1\\ &\Leftrightarrow \color{red}m\color{black}=\displaystyle \frac{12}{25}\\ \textrm{b}.\quad&P(x\leq 2)=P(x=0)+P(x=1)+P(x=2)\\ &\Leftrightarrow \color{red}m\color{black}+\displaystyle \frac{\color{red}m}{2}+\frac{\color{red}m}{3}=1\\ &\Leftrightarrow \left ( \displaystyle \frac{6+3+2}{6} \right )\color{red}m\color{black}=\displaystyle \frac{11}{6}\color{red}m\\ &\Leftrightarrow \quad =\displaystyle \frac{11}{6}\left ( \displaystyle \frac{12}{25} \right )=\frac{22}{25} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui fungsi}\: \: f(x)=\begin{cases} \displaystyle \frac{x}{6} &\textrm{untuk}\: \: x=1,2,3 \\\\ 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{adalah suatu fungsi peluang/probabilitas}\\ &\textrm{dari pubah/variabel acak}\: \: X.\: \: \textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{distribusi peluangnya untuk}\: \: X\\ &\textrm{b}.\quad P(X=2),\: P(X< 3),\: \textrm{dan}\: P(X\geq 2)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya adalah}:\\ &\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline X=x&1&2&3&4&5&\cdots &\textrm{Jumlah}\\\hline P(X=x)&\displaystyle \frac{1}{6}&\displaystyle \frac{2}{6}&\displaystyle \frac{3}{6}&\color{red}0&\color{red}0&\color{red}0&\color{blue}1\\\hline \end{array}\\ \textrm{b}.\quad &\textrm{Karena}\: \: f(x)=\begin{cases} \color{red}\displaystyle \frac{x}{6} &\textrm{untuk}\: \: x=1,2,3 \\\\ 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{maka}\\ &\bullet P(X=2)=\color{red}\displaystyle \frac{2}{6}\\ &\bullet P(X<3)=P(X=1)+P(X=2)\\ &\: \quad\qquad \qquad =\displaystyle \frac{1}{6}+\frac{2}{6}\\ &\: \quad\qquad \qquad =\displaystyle \frac{3}{6}=\color{red}\frac{1}{2}\\ &\bullet P(X\geq 2)=P(X=2)+P(X=3)\\ &\: \quad\qquad \qquad =\displaystyle \frac{2}{6}+\frac{3}{6}\\ &\: \quad\qquad \qquad =\color{red}\displaystyle \frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Diketahui fungsi peluang variabel}\: \: X\\ &f(x)=\begin{cases} \displaystyle \frac{x+2}{14} &\textrm{untuk}\: \: x=0,1,2,\: \: \textrm{dan}\: \: 3 \\\\ \quad 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{bahwa}\: \: X\: \: \textrm{merupakan variabel acak diskrit}\\ &\textrm{b}.\quad P(X=4),\: F(2),\: P(1<X\leq 3),\\ &\qquad \textrm{dan}\: P(X\geq 1)\: \: \textrm{serta}\: \: P(\left |X-2 \right |\leq 1)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya adalah}:\\ &\begin{array}{|c|c|c|c|c|c|c}\hline X=x&0&1&2&3&\textrm{Jumlah}\\\hline P(X=x)&\displaystyle \frac{2}{14}&\displaystyle \frac{3}{14}&\displaystyle \frac{4}{14}&\displaystyle \frac{5}{14}&1\\\hline \end{array} \\ &\textrm{Karena}\: \: \displaystyle \sum_{x=0}^{3}f(x)=1,\: \textrm{serta}\\ &0\leq \displaystyle \frac{2}{14},\: \frac{3}{14},\: \frac{4}{14},\: \frac{5}{14}<1.\: \textrm{Sehingga syarat}\\ &0\leq f(x)<1\: \: \textrm{dan}\: \: \sum f(x)=1\: \: \: \color{red}\textbf{terpenuhi}\\ &\textrm{Jadi, terbukti}\: \: X\: \: \textrm{adalah variabel acak diskrit}\\ \textrm{b}.\quad&\bullet P(X=4)=f(4)=\color{red}0\\ &\bullet F(2)=P(X\leq 2)\\ &\quad\qquad=P(X=0)+P(X=1)+P(X=2)\\ &\quad\qquad=f(0)+f(1)+f(2)\\ &\quad\qquad=\displaystyle \frac{2}{14}+\frac{3}{14}+\frac{4}{14}=\color{red}\frac{9}{14}\\ &\bullet P(1<X\leq 3)=P(X=2)+P(X=3)\\ &\quad\qquad =f(2)+f(3)=\displaystyle \frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{9}{14}\\ &\bullet P(X\geq 1)=f(1)+f(2)+f(3)\\ &\quad\qquad =\displaystyle \frac{3}{14}+\frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{12}{14}\\ &\bullet P(\left | X-2 \right |\leq 1)=P(-1\leq X-2\leq 1)\\ &\quad\qquad =P(1\leq X\leq 3)\\ &\quad\qquad =f(1)+f(2)+f(3)\\ &\quad\qquad =\displaystyle \frac{3}{14}+\frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{12}{14} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Distribusipeluang acak X disajikan dalam tabel berikut}\\ &\begin{array}{|c|c|c|c|}\hline x&2&3&4\\\hline f(x)&\displaystyle \frac{1}{8}&k+\displaystyle \frac{1}{8}&2k\\\hline \end{array}\\ &\textrm{Jika X merupakan variabel acak diskret, tentukanlah}\\ &\textrm{a}.\quad \textrm{nilai \textit{k}}\\ &\textrm{b}.\quad \textrm{nilai}\: \: \textrm{P}(\textrm{X}\geq 3)-\textrm{F}(3)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \sum f(x)&=f(2)+f(3)+f(4)=1\\ \Leftrightarrow \quad &\displaystyle \frac{1}{8}+k+\frac{1}{8}+2k=1\\ \Leftrightarrow \quad &3k=1-\displaystyle \frac{2}{8}=\frac{6}{8}\\ \Leftrightarrow \quad &k=\displaystyle \frac{2}{8}=\color{red}\frac{1}{4}\\ \textrm{b}.\quad \textrm{P}(\textrm{X}\, \geq 3&)-\textrm{F}(3)=\textrm{P}(\textrm{X}\geq 3)-\textrm{P}(\textrm{X}\leq 3)\\ &=f(3)+f(4)-\left ( f(2)+f(3) \right )\\ &=f(4)-f(2)\\ &=2\left ( \displaystyle \frac{1}{4} \right )-\frac{1}{8}\\ &=\displaystyle \frac{4}{8}-\frac{1}{8}=\color{red}\frac{3}{8} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Kurnia, N., dkk. 2018. Jelajah Matematika SMA Kelas XII Peminatan MIPA. Bogor: YUDHISTIRA.

Distribusi Binomial (Matematika Peminatan kelas XII)

$\color{blue}\textrm{A. Pendahuluan}$

$\begin{aligned}&\left\{\begin{matrix} (1)\: \textrm{Review}\begin{cases} \textrm{Peluang} \begin{cases} \textrm{Populasi} \\ \textrm{Sampel}\begin{cases} \textrm{Acak} \\ \textrm{Bukan Acak}.\quad \end{cases} \end{cases} \\ \textrm{Kombiasi} & \end{cases}\\ (2)\: \textrm{Variabel Acak}\begin{cases} \textrm{Diskrit} & .\qquad\qquad\qquad\qquad \\ \textrm{Kontinue} & \end{cases}\\ (3)\: \textrm{Distribusi}\begin{cases} \textrm{Distribusi Peluang Variabel Acak} & \\ \textrm{Fungsi Distribusi Kumulatif} & \\ \textrm{Variabel Acak Binomial}&\\ \textrm{Distribusi Binomial} \end{cases}\\ \end{matrix}\right. \end{aligned}$

$\color{blue}\textrm{Penjelasan}$

$\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Istilah}&\qquad\qquad\qquad\textrm{Penjelasan}\\\hline 1&\textrm{Statistika}&\textrm{Ilmu tentang pengumpulan, pengolahan},\\ &&\textrm{penganalisaan serta penarikan kesimpulan}\\ &&\textrm{data. Selanjutnya akan dibagi dua yaitu}\\ &&\color{blue}\textrm{deskriptif dan inferensia}\\\hline 2&\textrm{Statistik}&\color{red}\textrm{Kumpulan data/ukuran sampel}\\\hline 3&\textrm{Parameter}&\textrm{Ukuran populasi}\\\hline 4&\textrm{Populasi}&\color{blue}\textrm{Keseluruhan/semua anggota objek/data}\\\hline 5&\textrm{Sampel}&\color{blue}\textrm{Subjek/Objek yang mewakili populasi}\\\hline 6&\textrm{Sesus}&\textrm{Penelitian seluruh data (populasi)}\\\hline 7&\textrm{Tekik}&\textrm{Cara pengambilan data terbatas pada}\\ &\textrm{Sampling}&\textrm{sebagian saja dari populasi yang diteliti}\\\hline \end{array}$

$\color{blue}\textrm{lanjutan}$

$\begin{array}{|l|l|l|}\hline \textrm{No}&\textrm{Istilah}&\qquad\qquad\qquad\textrm{Penjelasan}\\\hline 8&\textrm{Cara}&\color{blue}\textrm{atau radom}.\: \textrm{yaitu setiap elemen populasi}\\ &\textrm{Acak}&\textrm{memiliki kesempatan yang yang sama}\\ &&\textrm{sehingga bersifat objektif}\\\hline 9&\textrm{Ruang}&\textrm{Himpunan dari semua hasil yang mungkin}\\ &\textrm{Sampel}&\textrm{dari sebuah percobaan}\\\hline 10&\textrm{Variabel}&\textrm{Suatu fungsi (aturan) yang memetakan }\\ &\textrm{Acak}&\textrm{setiap anggota ruang sampel dengan}\\ &(\textrm{VA})&\textrm{sebuah bilangan riil. Biasanya dinotasikan}\\ &&\textrm{dengan huruf besar, sedangkan nilai}\\ &&\textrm{variabel acaknya dinotasikan dengan}\\ &&\textrm{huruf kecil}\\\hline 11&(\textrm{VA})&\textrm{Jika VA tersebut memiliki sejumlah nilai}\\ &\textrm{Diskrit}&\textrm{yang dapat dihitung(berupa bilangan}\\ &&\textrm{bulat positif)}\\\hline 12&\textrm{VA}&\textrm{Sebaliknya yaitu berupa bilangan yang}\\ &\textrm{Kontinu}&\textrm{tidak bulat}\\\hline \end{array}$

$\color{red}\textrm{Sebagai contoh}$

$\begin{aligned}\textbf{a}\quad&\color{blue}\textrm{Variabel Acak Diskrit (Bilangan bulat positif)}\\ &\bullet \: \: \textrm{Jumlah siswa kelas XII MIA MA FUTUHIYAH}\\ &\: \, \quad \textrm{JEKETRO GUBUG}\\ &\bullet \: \: \textrm{Jumlah guru laki-laki di MA FUTUHIYAH}\\ &\: \, \quad \textrm{JEKETRO GUBUG}\\ &\bullet \: \: \textrm{Jumlah guru dan siswa di MA FUTUHIYAH}\\ &\: \, \quad \textrm{JEKETRO GUBUG yang tidak terpapar}\\ &\: \, \quad \textrm{COVID-19}\\ &\bullet \: \: \textrm{Jumlah motor yang terjual dalam sebulan}\\ \textbf{b}\quad&\color{magenta}\textrm{Variabel Acak Kontinu (Bukan bilangan bulat)}\\ &\bullet \: \: \textrm{Jumlah miyak yang tumpah di suatu lantai}\\ &\bullet \: \: \textrm{Ketinggian permukaan air di sebuah waduk}\\ \end{aligned}$

$\color{blue}\textrm{B. Variabel Acak}$

$\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Istilah}&\qquad\qquad\qquad\textrm{Definisi}\\\hline 13&\textrm{Variabel}&\textrm{Suatu variabel}\: \: X\: \: \textrm{adalah variabel acak jika}\\ &\textrm{Acak}&\textrm{nilai-nilai yang dimiliki oleh}\: \: X\: \: \textrm{merupakan}\\ &&\textrm{suatu kemungkinan atau peristiwa acak}.\\ &&\color{blue}\textrm{Selanjutnya variabel acak dibedakan}\\ &&\color{blue}\textrm{menjadi dua, yaitu variabel acak diskrit dan}\\ &&\color{blue}\textrm{variabel acak kontinu sebagaimana pada}\\ &&\color{blue}\textrm{penjelasan sebelumnya di atas}\\\hline \end{array}$

$\color{blue}\textrm{C. Distribusi Peluang}$

$\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Istilah}&\qquad\qquad\qquad\textrm{Definisi}\\\hline 14&\textrm{Distribusi}&\textrm{Sebuah daftar yang berisi seluruh hasil}\\ &\textrm{Peluang}&\textrm{yang mungkin dari suatu percobaan dan}\\ &(\textrm{Probabilitas})&\textrm{probabilitas yang berkaitan dengan setiap}\\ &&\textrm{hasil tersebut}.\\ &&\color{red}\textrm{Nilai probabilitas berada di antara 0 dan 1}\\ &&\color{blue}\textrm{Jumlah dari seluruh probabilitas hasil harus}\\ &&\color{blue}\textrm{harus sama dengan 1}\\\hline \end{array}$

$\LARGE\colorbox{magenta}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Sebuah koin dilempar sebanyak tiga kali}\\ &\textrm{a}.\quad \textrm{tentukan semua titik sampelnya}\\ &\textrm{b}.\quad \textrm{tentukan peluang mendapatkan tepat}\\ &\qquad \textrm{dua gambar}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\: \: \: &\textrm{Sebuah koin hanya memiliki dua muka,}\\ &\textrm{yaitu muka gambar (G) dan muka angka (A)}\\ &\textrm{sehingga setiap pelemparan hanya memiliki}\\ &\textrm{dua kemungkinan, yaitu muncul sisi A atau G}\\ &\textrm{maka ruang sampelnya adalah}:\\ &\begin{aligned} \color{blue}\textrm{Mula}\: \, &(1)\quad (2)\quad (3)\quad \color{blue}\textbf{Ruang sampel}\\ \textbf{Mulai}&\left\{\begin{matrix} A\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow (A,A,A)\\ G\rightarrow (A,A,G) \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (A,G,A)\\ G\rightarrow (A,G,G) \end{matrix}\right. \end{matrix}\right.\\ G\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow (G,A,A)\\ G\rightarrow (G,A,G) \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (G,G,A)\\ G\rightarrow (G,G,G) \end{matrix}\right. \end{matrix}\right. \end{matrix}\right. \end{aligned} \\ &\textrm{Jadi, banyaknya ruang sampel adalah }\color{red}8 \end{aligned}\\ &\begin{aligned}\textrm{b}.\: \: \: &\textrm{Dari ruang sampel yang tepat}\\ &\textrm{ada 2 sisi gambar : AGG,GAG,GGA}\\ &\textrm{sehingga peluangnya}=\displaystyle \frac{3}{\textrm{total ruang sampel}}=\color{red}\frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Misalkan}\: \: X\: \: \textrm{menyatakan sisi angka (A)}\\ &\textrm{pada soal No.1 di atas, tentukanlah nilai}\\ &X\: \: \textrm{yang mungkin}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ilustrasi berikut}\\ &\begin{aligned} \color{blue}\textrm{Mula}\: \, &(1)\quad (2)\quad (3)\quad \color{blue}\textbf{Ruang sampel}\quad \textbf{Nilai}\\ \textbf{Mulai}&\left\{\begin{matrix} A\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow \color{magenta}(A,A,A)\rightarrow \rightarrow \rightarrow X=3\\ G\rightarrow (A,A,G)\rightarrow \rightarrow \rightarrow X=2 \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (A,G,A)\rightarrow \rightarrow \rightarrow X=2\\ G\rightarrow (A,G,G)\rightarrow \rightarrow \rightarrow X=1 \end{matrix}\right. \end{matrix}\right.\\ G\left\{\begin{matrix} A\left\{\begin{matrix} A\rightarrow (G,A,A)\rightarrow \rightarrow \rightarrow X=2\\ G\rightarrow (G,A,G)\rightarrow \rightarrow \rightarrow X=1 \end{matrix}\right.\\ G\left\{\begin{matrix} A\rightarrow (G,G,A)\rightarrow \rightarrow \rightarrow X=1\\ G\rightarrow \color{red}(G,G,G)\rightarrow \rightarrow \rightarrow X=0 \end{matrix}\right. \end{matrix}\right. \end{matrix}\right. \end{aligned}\\ &\textrm{Jadi, nilai}\: \: X\: \: \textrm{yang mungkin}=\color{red}0,1,2,\: \color{black}\textrm{atau}\: \color{red}3 \end{aligned} \end{array}$

Perhatikanlah contoh pada No.2 di atas, nilai X ternyata tidak memiliki nilai tunggal. Karena X tidak memiliki nilai tunggal, maka X selanjutnya disebut dengan variabel. Dan variabel seperti ini yang nilainya ditentukan oleh percobaan sehingga akan mendapatkan beberapa kemungkinan selanjutnya disebut dengan variabel acak. Sehingga X pada No.2 di atas adalah salah satu contoh untuk variabel acak.

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Aturan Sinus dan Aturan Cosinus

A. Pendahuluan

Aturan sinus maupun aturan cosinus keduanya sangat bermanfaat berkaitan dengan unsur segitiga baik siku-siku maupun segitiga bebas dalam penentuan besar sudut dalam segitiga tersebut maupun panjang sisi yang diingin. Dalam hal penentuan besar sudut atau menentukan permasalahan panjang salah satu sisi segitiga jika nantinya sudut diketahui, terkadang besar sudutnya tidak cuma lancip, dibanyak soal dimunculkan sudut tumpul. Oleh karenanya ada baiknya pembaca mengetahui nilai perbandingan trigonometri diberbagai kuadran dan nilai sudut-sudut istimewa dalam trigonometri serta tak lupa juga beberapa identitas trigonometri.

$\begin{matrix} \sin \alpha =\displaystyle \frac{BC}{AB}\qquad\Leftrightarrow\quad \csc \alpha =\displaystyle \frac{AB}{BC}=\color{blue}\displaystyle \frac{1}{\sin \alpha }\\\\ \cos \alpha =\displaystyle \frac{AC}{AB}\qquad\Leftrightarrow \quad \sec \alpha =\displaystyle \frac{AB}{AC}=\color{blue}\displaystyle \frac{1}{\cos \alpha }\\\\ \tan \alpha =\displaystyle \frac{BC}{AC}\qquad\Leftrightarrow \quad \cot \alpha =\displaystyle \frac{AC}{BC}=\color{blue}\displaystyle \frac{1}{\tan \alpha } \end{matrix}$.

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \alpha ^{0}&0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha ^{0}&0&\color{red}\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha ^{0}&1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\color{red}\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha ^{0}&0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array}$.

$\begin{aligned}&\color{red}\textrm{Macam-Macam Identitas Trigonometri Dasar}\\ &1.\quad \csc \alpha =\displaystyle \frac{1}{\sin \alpha }\qquad\qquad 5.\quad \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &2.\quad \sec \alpha =\displaystyle \frac{1}{\cos \alpha }\qquad\qquad 6.\quad \tan^{2} \alpha +1=\sec ^{2}\alpha \\ &3.\quad \cot \alpha =\displaystyle \frac{1}{\tan \alpha }\qquad\qquad 7.\quad \cot^{2} \alpha +1=\csc ^{2}\alpha \\ &4.\quad \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha }\qquad\qquad 8.\quad \sin^{2} \alpha +\cos ^{2}=1\\ \end{aligned}$.

B. Aturan Sinus

$\Large\begin{array}{|c|}\hline \displaystyle \frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}=2R\\\hline \end{array}$.

C. Aturan Cosinus

$\Large\begin{array}{|c|}\hline \begin{aligned}\bullet \: \: &\cos \angle A=\displaystyle \frac{b^{2}+c^{2}-a^{2}}{2bc}\\ \bullet \: \: &\cos \angle B=\displaystyle \frac{a^{2}+c^{2}-a^{2}}{2ac}\\ \bullet \: \: &\cos \angle C=\displaystyle \frac{a^{2}+b^{2}-a^{2}}{2ab} \end{aligned}\\\hline \end{array}$.

D. Luas Segitiga

$\begin{aligned}\textbf{Luas}\: \triangle \: ABC&=\frac{1}{2}bc.\sin \angle A\\ &=\frac{1}{2}ac.\sin \angle B\\ &=\frac{1}{2}ab.\sin \angle C \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \triangle ABC\: \textrm{dengan panjang sisi}\\&AC=10\: cm\: \: \textrm{dan}\: \: BC=16\: cm\: \textrm{serta luas}\\ &\triangle ABC=40\: cm^{2} ,\: \textrm{maka besar} \: \angle ACB\\ &\textrm{jika sudutnya lancip adalah}\: \cdots \\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui}\: \: \left\{\begin{matrix} AC=10\: cm\\ BC=16\: cm\\ L_{\triangle }=40\: cm^{2} \end{matrix}\right.,\: \textrm{maka}\\&\begin{aligned}L_{\triangle ABC}\quad&=\frac{1}{2}.AC.BC.\sin \angle ACB\\ 40&=\frac{1}{2}.10.16.\sin \angle ACB\\ 40&=80.\sin \angle ACB\\ \frac{40}{80}&=\sin \angle ACB\\ \sin \angle ACB&=\frac{1}{2}\\ \sin \angle ACB&=\sin 30^{0}\\ \angle ACB&=30^{0} \end{aligned}. \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Perhatikanlah gambar berikut} \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika }\\ &AB+3=BC+2=CD+1=AD=4\: cm,\\ &\textrm{maka}\: \cos \angle BAD\: \textrm{adalah}\: \cdots\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan kembali ilustrasi berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Langkah awal kita gunakan garis bantu BD}\\ &\textrm{untuk nantinya kita mendapatkan nilai}\: \: cos\\ &\textrm{dari sudut A, yaitu}:\\&\begin{aligned}BD^{2}&=BA^{2}+DA^{2}-2.BA.DA.\cos \angle A\\&=1^{2}+4^{2}-2.1.4.\cos \angle A\\ &=17-8\cos \angle A\\ BD^{2}&=BC^{2}+DC^{2}-2.BC.DC.\cos \angle C\\ &=2^{2}+3^{2}-2.2.3.\cos \angle C\\ &=13-12\cos \angle C \end{aligned}\\ &\textrm{Perlu diketahui bahwa}\\ &\angle A+\angle C=\angle B+\angle C=180^{0}\\ & \textrm{karena ABCD segiempat talibusur, sehingga}\\ &\angle C=180^{0}-\angle A\\&\begin{aligned}BD^{2}&=BD^{2}\\ 17-8\cos \angle A&=13-12\cos \angle C\\ 12\cos \angle C-8\cos \angle A&=13-17\\ 12\left ( \cos \left ( 180^{0}-\angle A \right ) \right )-8\cos \angle A&=-4\\ 12\left ( -\cos \angle A \right )-8\cos \angle A&=-4\\ -12\cos \angle A-8\cos \angle A&=-4\\ -20\cos \angle A&=-4\\ \cos \angle A&=\frac{-4}{-20}\\ \cos \angle A&=\frac{1}{5} \end{aligned} \end{aligned}$.

Ketidaksamaan

Pada sebuah segitiga ABC dengan panjang sisi AB = c , AC = b, dan BC = a, dari ketiga sisi ini maka akan berlaku pertidaksamaan umum yang melibatkan ketiga sisinya sebagai berikut:

$\begin{aligned}&a+b>\color{red}c\\ &a+c>\color{red}b\\ &b+c>\color{red}a \end{aligned}$

Identitas-Identitas Aljabar yang Menakjubkan

Banyak sekali keunikan-keunikan saat kita mencoba melihat identitas-identitas aljabar yang sudah ditemukan sampai saat ini. Tentu semuanya sangat membantu ketika kita menyelesaikan suatu problem yang mengarah ke sana. Kadang sebagian ada yang menyebutkan dengan manipulasi aljabar.

Berikut bentuk dasar dari identitas-identitas aljabar tersebut

$\begin{aligned}&a^{2}-b^{2}=\color{red}(a-b)(a+b)\\ &a^{3}+b^{3}=\color{red}(a+b)(a^{2}-ab+b^{2})\\ &a^{3}-b^{3}=\color{red}(a-b)(a^{2}+ab+b^{2})\\ &(a+b)^{2}=\color{red}a^{2}+2ab+b^{2}\\ &(a-b)^{2}=\color{red}a^{2}-2ab+b^{2}\\ &(a+b)^{3}=\color{red}a^{3}+b^{3}+3ab(a+b)\\ &(a-b)^{3}=\color{red}a^{3}-b^{3}-3ab(a-b)\\ &(a+b+c)^{2}=\color{red}a^{2}+b^{2}+c^{2}+2(ab+ac+bc)\\ &(a+b+c)^{3}=\color{red}a^{3}+b^{3}+c^{3}+3(a+b)(a+c)(b+c)\\ &a^{3}+b^{3}+c^{3}-3abc=\color{red}(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)\\ &a^{3}+b^{3}+c^{3}-3abc=\color{red}\displaystyle \frac{1}{2}(a+b+c)((a-b)^{2}+(a-c)^{2}+(b-c)^{2})\\ &abc=\color{blue}(a+b+c)(ab+ac+bc)-(a+b)(a+c)(b+c)\\ &\textit{Sophie Germain}:a^{4}+4b^{4}=\color{red}(a^{2}-2ab+2b^{2})(a^{2}+2ab+2b^{2}) \end{aligned}$

Contoh Soal 13 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 61.&\textrm{(UM UNBRAW)}\\ &\textrm{Nilai maksimum dari fungsi}\\ &f(x)=4\cos ^{2}x+14\sin ^{2}x+24\sin x\cos x+10\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&6\\ \textrm{b}.&24\\ \textrm{c}.&26\\ \color{red}\textrm{d}.&32\\ \textrm{e}.&92 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{black}\begin{aligned}&f(x)=4\cos ^{2}x+14\sin ^{2}x+24\sin x\cos x+10\\ &f(x)=4\cos ^{2}x+4\sin ^{2}x+10\sin ^{2}x+12\sin 2x+10\\ &f(x)=4+5\left (1-\cos 2x \right )+12\sin 2x+10\\ &f(x)=19-5\cos 2x +12\sin 2x\\ &f(x)=19+12\sin 2x -5\cos 2x\\ &f(x)=19+\sqrt{12^{2}+(-5)^{2}}\cos \left ( 2x-\theta \right )\\ &f(x)=19+13\cos \left (2x -\theta \right )\\ &\textrm{Karena nilai}\: \: \cos \left ( 2x-\theta \right )=\pm 1,\: \textrm{maka}\\ &f(x)_{maks}=\color{red}19+13=32 \end{aligned} \end{array}$

Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 56.&\textrm{Diketahui fungsi}\: \: f(x)=\displaystyle \frac{1}{2}\sin 2x\: \: \textrm{dengan}\\ &0^{\circ}<x<360^{\circ} \: .\: \textrm{Kurva akan cekung}\\ &\textrm{ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0^{\circ}<x<90^{\circ}\\ \textrm{b}.&0^{\circ}<x<90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}<x<270^{\circ}\\ \textrm{c}.&45^{\circ}<x<225^{\circ}\\ \color{red}\textrm{d}.&90^{\circ}<x<180^{\circ}\: \: \textrm{atau}\: \: 270^{\circ}<x<360^{\circ}\\ \textrm{e}.&180^{\circ}<x<225^{\circ}\: \: \textrm{atau}\: \: 225^{\circ}<x<360^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{1}{2}\sin 2x\\ &f'(x)=\cos 2x\Rightarrow f''(x)=-2\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0^{\circ}\\ &\Leftrightarrow 2x=0^{\circ}+k.360^{\circ}\: \: \textrm{atau}\: \: 2x=180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow x=0^{\circ}+k.180^{\circ}\: \: \textrm{atau}\: \: x=90^{\circ}+k.180^{\circ}\\ &\Leftrightarrow \color{red}x=0^{\circ},\: x=90^{\circ} \: ,\: x=180^{\circ}\: \: \textrm{dan}\: \: x=270^{\circ}\\ &\qquad \color{red}\textrm{serta}\: \: x=360^{\circ}\\ &\bullet \color{red}\textrm{Selang}\: \: 0^{\circ}<x<90^{\circ},\: \: \color{black}\textrm{misal}\: \: x=45^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 45^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 90^{\circ}<x<180^{\circ},\: \: \color{black}\textrm{misal}\: \: x=135^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 135^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: 180^{\circ}<x<270^{\circ},\: \: \color{black}\textrm{misal}\: \: x=225^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 225^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 270^{\circ}<x<360^{\circ},\: \: \color{black}\textrm{misal}\: \: x=315^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 315^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Diketahui fungsi}\: \: f(x)=\cos ^{2}x-\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke bawah}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{\pi }{2}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{d}.&\displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{e}.&\displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}\cos ^{2}x-\sin ^{2}x\color{blue}=\cos 2x\\ &f'(x)=-2\sin 2x\Rightarrow f''(x)=-4\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=-4\cos 2\left ( 30^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=-4\cos 2\left ( 90^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=-4\cos 2\left ( 210^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=-4\cos 2\left ( 300^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=-4\cos 2\left ( 330^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Diketahui fungsi}\: \: f(x)=\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi .\: \textrm{Kurva fungsi tersebut akan}\\ &\textrm{cekung ke bawah pada interval}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{7\pi }{4}<x<2\pi\\ \textrm{c}.&\displaystyle 0<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \textrm{d}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\\ \textrm{e}.&\displaystyle 0<x<\frac{\pi }{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{black}\begin{aligned}&f(x)=\color{red}\sin ^{2}x\\ &f'(x)=2\sin x\cos x=\sin 2x\\ & f''(x)=2\cos 2x\\ &\color{purple}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=2\cos 2\left ( 30^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=2\cos 2\left ( 90^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=2\cos 2\left ( 210^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=2\cos 2\left ( 300^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=2\cos 2\left ( 330^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 59.&\textrm{Diketahui fungsi}\: \: f(x)=2\sin x-2\cos x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke atas}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{3\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x <\displaystyle \frac{5\pi }{4}\\ \textrm{c}.& \displaystyle \frac{3\pi }{4}<x<2\pi \\ \textrm{d}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \color{red}\textrm{e}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}2\sin x-2\cos x\color{blue}\\ &f'(x)=2\cos x+2\sin x\\ & f''(x)=-2\sin x+2\cos x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin x+2\cos x=0\Leftrightarrow \sin x=\cos x\\ &\Leftrightarrow \tan x=1\\ &\Leftrightarrow x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{5\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai gambaran saja}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=90^{\circ}=\displaystyle \frac{\pi }{2}\\ &\Rightarrow f''(90^{\circ})=-2\sin 90^{\circ}+2\cos 90^{\circ}=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Diketahui fungsi}\: \: f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\textrm{dengan}\: \: 0<x<2\pi .\: \textrm{Kurva fungsi tersebut}\\ &\textrm{akan cekung ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0<x<\frac{\pi }{6}\: \: \textrm{atau}\: \: \frac{\pi }{2}<x<\frac{5\pi }{6}\\ \color{red}\textrm{b}.&\color{magenta}\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi\\ \textrm{c}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{e}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &f'(x)=3\cos \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ & f''(x)=-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\\ &\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\sin 0\\ &\Leftrightarrow \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0+k.2\pi \: \: \Leftrightarrow \: \: \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\displaystyle \pi+k.2\pi \\ &\Leftrightarrow 3x=-\displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: 3x=\displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=-\displaystyle \frac{\pi }{6}+k.\frac{2\pi}{3} \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{6}+k.\displaystyle \frac{2\pi}{3} \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{6},\: \: x=\displaystyle \frac{\pi }{2},\: x=\frac{5\pi }{6} \: ,\: x=\frac{7\pi }{6},\\ & \color{black}\textrm{dan}\: \: \color{red}x=\frac{3\pi }{2},\: \color{black}\textrm{serta}\: \: \color{red}x=\frac{11\pi}{6} \\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0\leq x\leq 2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{6},\: \: \color{black}\textrm{misal}\: \: x=15^{\circ}=\frac{\pi }{12}\\ &\Rightarrow f''(15^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{12} \right )+\displaystyle \frac{\pi }{2} \right )=-\displaystyle \frac{9}{2}\sqrt{2}<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{6}<x<\displaystyle \frac{\pi }{2},\: \: \color{black}\textrm{misal}\: \: x=60^{\circ}=\displaystyle \frac{\pi }{3}\\ &\Rightarrow f''(60^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{\pi }{2} \right )=9>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten. PT. INTAN PARIWARA.

Contoh Soal 11 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 51.&\textrm{Diketahui}\: \: f(x)=\cos ^{2}2x\: .\: \textrm{Jika}\\ &f''(x)=a\sin ^{2}bx+c\cos ^{2}dx,\: \textrm{nilai untuk}\\ &\displaystyle \frac{a-b}{c-d}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5}{3}\\ \textrm{b}.&\displaystyle \frac{2}{3}\\ \color{red}\textrm{c}.&-\displaystyle \frac{3}{5}\\ \textrm{d}.&-\displaystyle \frac{6}{5}\\ \textrm{e}.&-\displaystyle \frac{9}{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{purple}\begin{aligned}&f(x)=\cos ^{2}2x\\ &f'(x)=2\cos 2x(-\sin 2x)(2)\\ &\: \qquad =-4\sin 2x\cos 2x\\ &\color{blue}f''(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ &\: \: \quad\quad=8\sin ^{2}2x-8\cos ^{2}2x\\ &\textrm{Bandingkan dengan}\\ &\color{red}f''(x)=a\sin ^{2}bx+c\cos ^{2}dx\\ &\textrm{maka},\: \: a=8,\: b=2,\: c=-8,\: d=2\\ &\textrm{Jadi},\: \displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\: .\: \textrm{Jika}\\ &f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\: ,\: \textrm{nilai dari}\\ &m.n=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\\ &f'(x)=\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &\, \qquad =\displaystyle \frac{-\sin ^{2}x-\cos ^{2}x+0}{\sin ^{2}+2\sin x\cos x+\cos ^{2}x}\\ &\, \qquad=\displaystyle \frac{-1}{1+\sin 2x}\\ &f''(x)=\displaystyle \frac{0-((-1).2\cos 2x)}{\left ( \sin 2x+1 \right )^{2}}=\frac{2\cos 2x}{\left ( \sin 2x+1 \right )^{2}}\\ &\color{red}\textrm{Bandingkan dengan yang diketahui}\\ &\color{black}f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\\ &\begin{cases} m &=2 \\ n &=2 \end{cases}\\ &\textrm{Jadi},\: \: m.n=2.1=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\: \: \textrm{dengan}\: \: 0\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{4},0 \right )\\ \color{red}\textrm{b}.&\left ( \displaystyle \frac{\pi }{2},0 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{2},1 \right )\\ \textrm{e}.&\left ( \pi ,1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin 2x\\ &f'(x)=2\cos 2x\Rightarrow f''(x)=-4\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=0+k.2\pi \: \: \textrm{atau}\: \: 2x=\pi +k.2\pi \\ &\Leftrightarrow x=0+k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} +k.\pi \\ &\Leftrightarrow \color{black}x=0,\: \color{red}x=\displaystyle \frac{\pi }{2},\: x=\pi \: ,\: x=\displaystyle \frac{3\pi }{2}\: \: \textrm{atau}\: \: \color{black}x=2\pi\\ &\bullet f\left ( \displaystyle \frac{\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{2},0 \right )\\ &\bullet f\left ( \displaystyle \pi \right )=\sin 2\left ( \displaystyle \pi \right )=0\Rightarrow \color{red}\left ( \displaystyle \pi ,0 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{3\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{2},0 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Diketahui fungsi}\: \: f(x)=-3\cos 2x+1\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},2 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{2\pi }{3},\frac{5}{2} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{3\pi}{2} ,4 \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{5\pi }{3},\frac{5}{2} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=-3\cos 2x+1\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=6\sin 2x\Rightarrow f''(x)=12\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &12\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{atau}\: \: x=\frac{7\pi }{4}\\ &\bullet f\left ( \displaystyle \frac{\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{3\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{5\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{5\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{7\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{7\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{7\pi }{4},1 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Diketahui fungsi}\: \: f(x)=\sin^{2} x+2\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left ( \displaystyle \frac{\pi }{4},\frac{5}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{11}{4} \right )\\ \textrm{c}.&\left ( \displaystyle \pi ,2 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{4\pi }{3},\frac{11}{4} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{11\pi }{6},\frac{9}{4} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\sin^{2} x+2\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=2\sin x\cos x\Rightarrow f'(x)=\sin 2x\\ &f''(x)=2\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{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 \color{red}\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 \color{red}\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 \color{red}\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 \color{red}\left ( \displaystyle \frac{7\pi }{4},\frac{5}{2} \right ) \end{aligned} \end{array}$

Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 46.&\textrm{Turunan kedua dari}\: \: f(x)=x^{3}-\sin 3x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&6x^{2}+9\sin 3x\\ \textrm{b}.&3x^{2}+6\sin 3x\\ \textrm{c}.&3x-9\sin 3x\\ \color{red}\textrm{d}.&6x+9\sin 3x\\ \textrm{e}.&9x-6\sin 3x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}f(x)&=x^{3}-\sin 3x\\ f'(x)&=3x^{2}-3\cos 3x\\ f''(x)&=6x+9\sin 3x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Diketahui fungsi}\: \: g(x)=\displaystyle \frac{1-\cos x}{\sin x}\: . \textrm{Nilai}\\ &\textrm{turunan kedua saat}\: \: x=\displaystyle \frac{\pi}{4}\: \: \textrm{adalah}\: .... \\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2}+4\\ \textrm{b}.&2\sqrt{2}-3\\ \textrm{c}.&2\sqrt{2}+3\\ \color{red}\textrm{d}.&3\sqrt{2}-4\\ \textrm{e}.&3\sqrt{2}+4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}g(x)&=\displaystyle \frac{1-\cos x}{\sin x}\\ g'(x)&=\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{\sin ^{2}x}\\ &=\displaystyle \frac{\sin ^{2}x-\cos x+\cos ^{2}x}{\sin ^{2}x}\\ &=\displaystyle \frac{1-\cos x}{\sin ^{2}x}\\ g''(x)&=\displaystyle \frac{\sin x(\sin ^{2}x)-2\sin x\cos x(1-\cos x)}{\sin ^{4}x}\\ &=\displaystyle \frac{\sin x(\sin ^{2}x)-\sin 2x(1-\cos x)}{\sin ^{4}x}\\ &=\color{red}\displaystyle \frac{\sin \displaystyle \frac{\pi }{4}(\sin ^{2}\displaystyle \frac{\pi }{4})-\sin 2\displaystyle \frac{\pi }{4}(1-\cos \displaystyle \frac{\pi }{4})}{\sin ^{4}\displaystyle \frac{\pi }{4}}\\ &=\color{black}\displaystyle \frac{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{2}-1.\left ( 1-\left ( \displaystyle \frac{1}{\sqrt{2}} \right ) \right )}{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{4}}\\ &=\color{black}\displaystyle \frac{\displaystyle \frac{1}{2}\displaystyle \frac{1}{\sqrt{2}}-1+\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \frac{1}{4}}\times \displaystyle \frac{4}{4}\\ &=\displaystyle \frac{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}{1}\\ &=\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Turunan kedua fungsi}\: \: f(x)=\sin ^{2}x-\cos ^{2}x\\ &\textrm{adalah}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&6\sin 2x\\ \color{red}\textrm{b}.&4\cos 2x\\ \textrm{c}.&2\cos 2x\\ \textrm{d}.&-2\cos 2x\\ \textrm{e}.&-4\cos 2x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}f(x)&=\sin ^{2}x-\cos ^{2}x\\ f'(x)&=2\sin x\cos x-2\cos x(-\sin x)\\ &=2\sin x\cos x+2\sin x\cos x\\ &=2(2\sin x\cos x)=\color{black}2\sin 2x\\ f''(x)&=\color{red}2.2\cos 2x=4\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin x}\: .\: \textrm{Jika}\: \: f''(x)\\ &\textrm{adalah turunan keduafungsi}\: \: f,\: \textrm{maka}\\ &\textrm{nilai dari}\: \: f''\left ( \displaystyle \frac{\pi }{2} \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\displaystyle \frac{1}{2}\\ \textrm{b}.&-\displaystyle \frac{1}{4}\\ \textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}f(x)&=\color{black}\sqrt{\sin x}=\sin ^{\frac{1}{2}}x\\ f'(x)&=\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x=\displaystyle \frac{\cos x}{2\sin ^{\frac{1}{2}}x}\\ f''(x)&=\color{red}\displaystyle \frac{-\sin x\left ( 2\sin ^{\frac{1}{2}}x \right )-\cos x\left ( 2.\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x \right )}{4\sin x}\\ &=\displaystyle \frac{-2\sin x\sqrt{\sin x}-\displaystyle \frac{\cos ^{2}x}{\sqrt{\sin x}}}{4\sin x}\\ f''\left ( \displaystyle \frac{\pi }{2} \right )&=\color{black}\displaystyle \frac{-2\sin \displaystyle \frac{\pi }{2}.\sqrt{\sin \displaystyle \frac{\pi }{2}}-\displaystyle \frac{\cos ^{2}\displaystyle \frac{\pi }{2}}{\sin \displaystyle \frac{\pi }{2}}}{4\sin \displaystyle \frac{\pi }{2}}\\ &=\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: f(x)=\tan ^{2}(3x-2) \: \: \textrm{maka}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &-18\sec ^{4}(3x-2)\\ \textrm{b}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{2}(3x-2)\\ \color{red}\textrm{c}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2)\\ \textrm{d}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+36\sec ^{4}(3x-2)\\ \textrm{e}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\tan ^{2}(3x-2)\\ f'(x)&=2\tan (3x-2)\sec ^{2}(3x-2)(3)\\ &=6\tan (3x-2)\sec ^{2}(3x-2)\\ f''(x)&=6\sec ^{2}(3x-2).(3)\sec ^{2}(3x-2)\\ &+6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ &=18\sec ^{4}(3x-2)\\ &+36\tan ^{2}(3x-2)\sec ^{2}(3x-2) \end{aligned} \end{array}$

Lanjutan Materi (10) Turunan Kedua Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\color{blue}\textrm{H. Turunan Kedua Fungsi Trigonometri}$

Definisi dari bahasan ini adalah jika turunan pertama dari suatu fungsi $f$ dan dinyatakan dengan $f'$ ada dan terdefinisi untuk setiap nilai $x$ dalam daerah terdefinisi $f$, maka turunan kedua dari fungsi $f$ dinyatakan dengan $f''$ adalah:

$\color{blue}f''(x)=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{f'(x+h)-f'(x)}{h}=\displaystyle \frac{d}{dx}\left ( f'(x) \right )$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukan turunan kedua dari}\\ &y=\sin x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\sin x\\ y'&=\cos x\\ y''&=-\sin x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukan turunan kedua dari}\\ &y=\sin 2x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\sin 2x\\ y'&=2\cos 2x\\ y''&=-4\sin 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tentukan turunan kedua dari}\\ &y=\sin^{2} x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\sin^{2} x\\ y'&=2\sin x(-\cos x)=-\sin 2x\\ y''&=-2\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Tentukan turunan kedua dari}\\ &y=\cos x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\cos x\\ y'&=-\sin x\\ y''&=-\cos x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Tentukan turunan kedua dari}\\ &y=\tan x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\tan x\\ y'&=\sec^{2} x\\ y''&=2\sec x(\sec x \tan x)=2\sec ^{2}x\tan x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Tentukan turunan kedua dari}\\ &y=\cot x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\cot x\\ y'&=-\csc ^{2}x\\ y''&=-2\csc x(-\csc x\cot x)=2\csc ^{2}x\cot x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Tentukan turunan kedua dari}\\ &y=\sec x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\sec x\\ y'&=\sec x\tan x\\ y''&=\sec x\tan x(\tan x)+\sec x\left ( \sec ^{2}x \right )\\ &=\sec x\tan ^{2}x+\sec ^{3}x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Tentukan turunan kedua dari}\\ &y=\csc x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}y&=\csc x\\ y'&=-\csc x\cot x\\ y''&=-(-\csc x\cot x)\cot x+(-\csc x)(-\csc ^{2}x)\\ &=\csc x\cot ^{2}x+\csc ^{3}x \end{aligned} \end{array}$

$\color{blue}\textrm{I. Fungsi Naik dan Fungsi Turun}$

Sebelumnya telah diketahui bahwa pada selang terbuka

$\begin{array}{ll}\\ &\bullet \: \textrm{untuk}\: \: \color{blue}f'(x)>0\: \: \textrm{maka fungsi naik}\\ &\bullet \: \textrm{untuk}\: \: \color{red}f'(x)<0\: \: \textrm{maka fungsi turun} \end{array}$

$\begin{array}{ll}\\ &\textrm{Misalkan}\: \: f'\: \: \textrm{dan}\: \: f''\: \: \textrm{ada untuk setiap}\\ &\textrm{titik pada suatu interval yang memuat}\\ &c\: \: \textrm{dengan}\: \: f'(c)=0\\ &\bullet \quad \textrm{jika}\: \: \color{blue}f''(c)>0\: \: \textrm{maka}\: \: f(c)\: \: \textrm{adalah}\\ &\: \, \quad \textrm{nilai minimum lokal (titik minimum)}\\ &\bullet \quad \textrm{jika}\: \: \color{red}f'(c)<0\: \: \textrm{maka}\: \: f(c)\: \: \textrm{adalah}\\ &\, \: \quad \textrm{nilai maksimum lokal (titik maksimum)}\\ &\bullet \quad \color{purple}\textrm{jika}\: \: f''(c)=0\: \: \textrm{maka nilai stasioner}\\ &\, \: \quad \textrm{belum dapat ditentukan} \end{array}$

$\color{black}\begin{array}{ll}\\ &\textrm{Titik Belok}\\\\ &\textrm{Jika}\: \: (c,f(c))\: \: \textrm{adalah titik belok grafik}\\ &f,\: \: \textrm{maka}\: \: f''(x)=0\: \: \textrm{atau}\: \: f''\: \: \textrm{tidak ada}\\ &\textrm{pada} \: \: x=c \end{array}$

$\LARGE\color{black}\fbox{CONTOH SOAL}$

Perhatikan lagi contoh pada bagian ini LANJUTAN MATERI 8 berikut

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah semua titik stasioner}\\ &\textrm{berikut jenisnya dari fungsi}\\ &f(x)=\sin x+\cos x\: \: \textrm{dengan}\\ &0\leq x\leq 2\pi\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\color{red}\textrm{Dengan Turunan Pertama}\\ &\textrm{Diketahui}\\ &f(x)=\sin x+\cos x\\ &f'(x)=\cos x-\sin x\\ &\textrm{Saat}\quad \color{black}f'(x)=0,\\ &\color{black}f'(x)=\cos x-\sin x=0 \: \: \cos x=\sin x\\ &\cos x=\cos \left ( \displaystyle \frac{\pi }{2}-x \right )\\ &\: \: \: \quad x=\pm \left ( \displaystyle \frac{\pi }{2}-x \right )+k.2\pi \\ &\: \: \: \quad \begin{cases} x+x &=\displaystyle \frac{\pi }{2}+k.2\pi ,\: \: \color{red}\textrm{atau} \\ x-x &=-\displaystyle \frac{\pi }{2}+k.2\pi \end{cases}\\ &\textrm{maka}\\ &\: \: \: \quad \begin{cases} x &=\displaystyle \frac{\pi }{4}+k.\pi ,\: \: \color{red}\textrm{atau} \\ 0&=-\displaystyle \frac{\pi }{2}+k.2\pi\: \: (\color{black}\textrm{tidak memenuhi}) \end{cases}\\ &\textrm{Sehingga ada dua absis yang memenuhi}\\ &\color{red}\textrm{sebagai titik STASIONER},\: \: \color{black}\textrm{yaitu}\\ &\color{black}x=\displaystyle \frac{\pi }{4}\: \: \textrm{dan}\: \: \quad x=\frac{5\pi }{4}\\ &\textrm{untuk}\: \: \: \color{black}x=\displaystyle \frac{\pi }{4}\\ &f\left ( \displaystyle \frac{\pi }{4} \right )=\sin \left ( \displaystyle \frac{\pi }{4} \right )-\cos \left (\displaystyle \frac{\pi }{4} \right )\\ &\qquad=\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}=\sqrt{2}\\ &\textrm{untuk}\: \: \: \color{black}x=\displaystyle \frac{5\pi }{4}\\ &f\left ( \displaystyle \frac{5\pi }{4} \right )=\sin \left ( \displaystyle \frac{5\pi }{4} \right )+\cos \left (\displaystyle \frac{5\pi }{4} \right )\\ &\qquad=-\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}\\ &\textrm{Jadi titik stasionernya}:\: \: \left ( \displaystyle \frac{\pi }{4},2 \right )\: \&\: \: \left ( \displaystyle \frac{5\pi }{4},-\sqrt{2} \right )\\ &\color{black}\textrm{Langkah berikutnya gunakanlah titik}\\ &\color{black}\textrm{uji di sekitar nilai stasioner yaitu}:\\ &\begin{array}{ccccccccc} &&&&&&&&\\\hline \color{red}0&&\displaystyle \frac{\pi }{4}&&\color{red}\pi &&\displaystyle \frac{5\pi }{4}&&\color{red}2\pi \end{array}\\ &\textrm{Selanjutnya}\\ &\textrm{Untuk}\: \: f'(x)=\cos x-\sin x\\ &x=0\Rightarrow f'(0)=\cos 0-\sin 0\\ &\quad=1+0=1>0\quad (\color{black}\textrm{positif})\\ &x=\pi \Rightarrow f'(\pi )=\cos \pi -\sin \pi \\ &\quad=-1+0=-1<0\quad (\color{red}\textrm{negatif})\\ &x=0\Rightarrow f'(2\pi )=\cos 2\pi -\sin 2\pi \\ &\quad=1+0=1>0\quad (\color{black}\textrm{positif})\\ &\begin{array}{|c|c|c|c|c|l|}\hline x&0&\displaystyle \frac{\pi }{4}&\pi &\displaystyle \frac{5\pi }{4}&2\pi \\\hline \color{black}f'(x)&+&0&-&0&+\\\hline &&--&&&\\ \color{red}\textrm{Garfik}&/&&\backslash&&/\\ &&&&\_\_\_\_&\\\hline \end{array}\\ &\textrm{Dari tabel di atas didapatkan}\\ &\begin{cases} \color{black}\left ( \displaystyle \frac{\pi }{4},\sqrt{2} \right ) & \color{red}\textrm{titik balik maksimum} \\ \color{black}\left ( \displaystyle \frac{5\pi }{4},-\sqrt{2} \right ) & \color{red}\textrm{titik balik minimum} \end{cases} \end{aligned} \end{array}$

$.\: \quad\begin{array}{|c|}\hline \quad\color{black}\begin{aligned}&\color{red}\textrm{Dengan Turunan Kedua}\\ &f(x)=\sin x+\cos x\\ &f'(x)=\cos x-\sin x\\ &f''(x)=-\sin x-\cos x=-(\sin x+\cos x)\\ &f''\left ( \displaystyle \frac{\pi }{4} \right )=-\left ( \sin \displaystyle \frac{\pi }{4}+\cos \displaystyle \frac{\pi }{4} \right )=\color{blue}-\sqrt{2}<0\\ &\Rightarrow (\color{red}\textrm{maksimum atau cekung ke bawah})\\ &f''\left ( \displaystyle \frac{5\pi }{4} \right )=-\left ( \sin \displaystyle \frac{5\pi }{4}+\cos \displaystyle \frac{5\pi }{4} \right )=\color{blue}\sqrt{2}>0\\ &\Rightarrow (\color{red}\textrm{minimum atau cekung ke atas})\\ &\textrm{Dengan}\: \: f(x)=\sin x+\cos x, \: \: \textrm{maka}\\ &\: \: \bullet \textrm{nilai maksimumnya}:\sin \displaystyle \frac{\pi }{4}+\cos \frac{\pi }{4}=\sqrt{2}\\ &\: \: \bullet \textrm{nilai minimumnya}:\sin \displaystyle \frac{5\pi }{4}+\cos \frac{5\pi }{4}=-\sqrt{2}\\ &\textrm{Jadi, titik maksimumnya}\: \: \color{red}\left ( \displaystyle \frac{\pi }{4},\sqrt{2} \right )\\ &\textrm{dan nilai minimumnya}\: \: \color{red}\left ( \displaystyle \frac{5\pi }{4},-\sqrt{2} \right ) \end{aligned}\\\hline \end{array}$

$.\: \qquad\color{black}\begin{aligned}&\color{red}\textrm{Untuk TITIK BELOK}\\ &\textrm{Syarat titik belok adalah}\: \: f''(x)=0\\ &\color{blue}\textrm{Diketahui}\: \: f(x)=\sin x+\cos x\\ &f''(x)=-(\sin x+\cos x)=0\\ &\Leftrightarrow \: \sin x+\cos x=0\Leftrightarrow \sin x=-\cos x\\ &\Leftrightarrow \: \displaystyle \frac{\sin x}{\cos x}=-1\Leftrightarrow \tan x=-1\\ &\Leftrightarrow \: \tan x=\tan 135^{\circ}\\ &\Leftrightarrow \: x=135^{\circ}+k.180^{\circ}\\ &\Leftrightarrow k=0\Rightarrow x=\color{red}135^{\circ}=\displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow k=1\Rightarrow x=135^{\circ}+180^{\circ}=\color{red}315^{\circ}=\displaystyle \frac{7\pi }{4}\\ &\textrm{Adapun titik beloknya pada fungsi}\: \: f(x)\\ &\textrm{adalah}:\\ &\bullet \: \: x=\displaystyle \frac{3\pi }{4}\\ &\quad f\left ( \displaystyle \frac{3\pi }{4} \right )=\sin \left ( \displaystyle \frac{3\pi }{4} \right )+\cos \left ( \displaystyle \frac{3\pi }{4} \right )=0\\ &\quad \color{red}\textrm{maka titiknya}\: \: \left ( \displaystyle \frac{3\pi }{4},0 \right )\\ &\bullet \: \: x=\displaystyle \frac{7\pi }{4}\\ &\quad f\left ( \displaystyle \frac{7\pi }{4} \right )=\sin \left ( \displaystyle \frac{7\pi }{4} \right )+\cos \left ( \displaystyle \frac{7\pi }{4} \right )=0\\ &\quad \color{red}\textrm{maka titiknya}\: \: \left ( \displaystyle \frac{7\pi }{4},0 \right ) \end{aligned}$

$.\: \qquad \color{purple}\textrm{Berikut Sketsa grafiknya}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah semua titik stasioner}\\ &\textrm{berikut jenisnya dari fungsi}\\ &f(x)=2\sin x\: \: \textrm{dengan}\\ &0\leq x\leq 2\pi\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{red}\textrm{Diketahui}\\ &f(x)=2\sin x\\ &f'(x)=2\cos x\\ &\textrm{Syarat titik stasioner}\: \: f'(x)=0\\ &2\cos x=0\Leftrightarrow \cos x=0\\ &\Leftrightarrow \cos x=\cos 90^{\circ}\Leftrightarrow x=90^{\circ}\pm k.360^{\circ}\\ &\Leftrightarrow k=0\Rightarrow x=90^{\circ}\: \: \color{red}\textrm{yang memenuhi}\\ &\Leftrightarrow k=1\Rightarrow x=270^{\circ}\: \: \color{red}\textrm{yang memenuhi}\\ &\textrm{Turunan kedua fungsi di atas adalah}:\\ &f''(x)=-2\sin x\\ &\color{blue}\textrm{maka},\\ &\begin{array}{|c|l|l|l|}\hline \textrm{Nilai}&\qquad\qquad\quad\quad\textrm{Hasil}&\textrm{Keterangan}&\quad\textrm{Titik}\\\hline x=90^{\circ}&f''(90^{\circ})=-2\sin 90^{\circ}=-2<0&\color{blue}\textrm{Maksimum}&\\ &f(90^{\circ})=2\sin 90^{\circ}=2&&\left ( 90^{\circ},2 \right )\\\hline x=270^{\circ}&f''(270^{\circ})=-2\sin 270^{\circ}=2>0&\color{red}\textrm{Minimum}&\\ &f(270^{\circ})=2\sin 270^{\circ}=-2&&\left ( 270^{\circ},-2 \right )\\\hline \textrm{Syarat}&f''(x)=0&\textrm{Belok}&\\ &\begin{aligned}&-2\sin x=0\Leftrightarrow \sin x=0\\ &\Leftrightarrow \sin x=\sin 0^{\circ}\\ &\Leftrightarrow x=\begin{cases} 0^{\circ} & +k.360^{\circ} \\ 180^{\circ} & +k.360^{\circ} \end{cases}\\ &\textrm{Yang memenuhi}\\ &x=0^{\circ},\: 180^{\circ},\: \: \textrm{dan}\: \: 360^{\circ}\\ &\textrm{Lalu hasilnya disubstitusikan}\\ &\textrm{ke persamaan}\: \: f(x)=2\sin x \end{aligned}&\begin{aligned}&\\ &\textrm{Hasilnya}: \end{aligned}&\begin{aligned}&\left ( 0^{\circ},0 \right ),\\ &\left ( 180^{\circ},0 \right ),\\ &\left ( 360^{\circ},0 \right ) \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$

$\color{blue}\textrm{J. Selang Kecekungan}$

Lihat keterkaitan materi dan contoh di atas berkaitan dengan selang kecekungan kurva fungsi trigonometri

$\color{purple}\begin{aligned}&\textrm{Misalkan pada suatu selang} \: \: (a,b)\\ &\textrm{terdapat sembarang bilangan real}\: \: c\\ &\textrm{serta turunan kedua fungsi}\: \: f\: \: \textrm{ada}\\ &\textrm{pada selang tersebut}\\ &\bullet \quad \color{blue}\textrm{saat}\: \: f''(c)<0\: , \textrm{maka kurva}\\ &\qquad f\: \: \textrm{cekung ke bawah}\\ &\bullet \quad \color{red}\textrm{saat}\: \: f''(c)>0\: , \textrm{maka kurva}\\ &\qquad f\: \: \textrm{cekung ke atas}\\ \end{aligned}$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ &(\color{red}\textrm{Perhatikan lagi contoh soal no.1 di atas})\\ &\textrm{Tentukanlah interval di mana kurva}\\ &\textrm{cekung ke bawah dan atas dari fungsi}\\ &f(x)=\sin x+\cos x\: \: \textrm{dengan}\\ &0< x< 2\pi\\\\ &\textrm{Jawab}:\\ &\color{purple}\begin{aligned}&f(x)=\sin x+\cos x\\ &f'(x)=\cos x-\sin x\\ &f''(x)=-(\sin x+\cos x)\\ &\textrm{Sebelum menentukan batas kecekungan}\\ &\textrm{dengan menentukan titik beloknya dulu}\\ &\textrm{yaitu} :\: \: f''(x)=0\\ &\color{red}\textrm{Sebelumnya telah dibahas titik beloknya}\\ &\textrm{fungsi} \: \: f\: \: \textrm{di atas mempunyai 2 buah}\\ &\textrm{titik belok pada selang}\: \: 0<x<2\pi \\ &\color{blue}x=\displaystyle \frac{3\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{blue}x=\displaystyle \frac{7\pi }{4}\\ &\textrm{Melihat banyaknya titik belok, maka}\\ &\textrm{akan terdapat 3 selang kecekungan, yaitu}:\\ &\begin{cases} 1\: \bullet & 0<x<\displaystyle \frac{3\pi }{4} \\ 2\: \bullet & \displaystyle \frac{3\pi }{4}<x<\frac{7\pi }{4} \\ 3\: \bullet & \displaystyle \frac{7\pi }{4}<x<2\pi \end{cases}\\ &\textrm{Kita ambil titik uji tiap selang di atas}\\ &\textrm{dan substitusikan ke turunan kedua fungsi}\: \: f\\ &f''\left ( \displaystyle \frac{\pi }{2} \right )=-\left ( \sin \displaystyle \frac{\pi }{2} +\cos \displaystyle \frac{\pi }{2} \right )=-1<0\\ &\color{blue}\textrm{Sehingga pada selang ini, kurva cekung ke bawah}\\ &f''\left ( \pi \right )=-\left ( \sin \pi +\cos \pi \right )=1>0\\ &\color{red}\textrm{Sehingga pada selang ini, kurva cekung ke atas}\\ &f''\left ( \displaystyle \frac{11\pi }{6} \right )=-\left ( \sin \displaystyle \frac{11\pi }{6} +\cos \displaystyle \frac{11\pi }{6} \right )=\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}<0\\ &\color{blue}\textrm{Sehingga pada selang ini, kurva cekung ke bawah} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Kurnia, N., dkk. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: YUDHISTIRA
Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA