Lanjutan 3 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{3. Grafik Fungsi Tangen}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 0& \frac{1}{3}\sqrt{3}& 1& \sqrt{3}& \mathrm{\infty }& -\sqrt{3}& -1& -\frac{1}{3}\sqrt{3}& 0\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& \frac{1}{3}\sqrt{3}& 1& \sqrt{3}& \mathrm{\infty }& -\sqrt{3}& -1& -\frac{1}{3}\sqrt{3}& 0& \\ \hline\end{array}$

Pada fungsi Tangen demikian juga nanti Cotangen ada beberapa nilai yang tida terdefinisi. Dalam fungsi Tangen fungsi, nilai fungsi yang tidak terdefini terdapat saat nilai  $x={\displaystyle \frac{\pi }{2}={90}^{\circ }}$ dan $x={\displaystyle \frac{3\pi }{2}={270}^{\circ }}$. Sehingga pada saat posisi nilai itu, maka dibuatlah garis putus-putus pada grafik yang dan ditampakkan berupa grais vertikal yang selanjutnya garis vertikal itu disebut sebagai asimtot.

LIMIT FUNGSI ALJABAR

 $\text{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{array}{rl}\underset{x\to a}{\text{lim}}f(x)=L& \mathbf{\text{adalah}}\text{Jika}x\text{mendekati}a\\ & \text{dengan tidak sama dengan}a,\\ & \text{maka nilai}f(x)\text{mendekati}L\end{array}$.

Perhatikan definisi di atas istilah  $x\text{mendekati}a$ dituliskan dengan simbol  $(x\to a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{array}{rl}\underset{x\to a^{-}}{\text{lim}}f(x)=\underset{x\to a^{+}}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}f(x)=L& \end{array}$.

$\begin{array}{rl}\text{Perlu di}& \text{perhatikan bahwa didekati dari}\\ \bullet \mathbf{\text{kiri}}& \text{disimbolkan dengan}\underset{x\to a^{-}}{\text{lim}}f(x),\text{dan}\\ \bullet \mathbf{\text{kan}}& \mathbf{\text{an}}\text{disimbolkan dengan}\underset{x\to a^{+}}{\text{lim}}f(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari}f(x)={\displaystyle \frac{x^2-4}{x-2}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika fungsi}{\displaystyle \frac{x^2-4}{x-2}}\\ & \text{di sekitar}x=2\text{sebagaimana dalam tabel}\\ & \text{berikut}\end{array}\end{array}$.

$\begin{array}{rl}.& \text{Jadi, nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}=2\text{atau dapat dikatakan}}\\ & \text{nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}\mathbf{\text{ada}}}\\ & \text{meskipun nilai substitusi langsung}x=2\text{yaitu}\\ & f(0)={\displaystyle \frac{0^2-0}{0-0}=\frac{0}{0}\text{berupa bentuk tak}}\\ & \text{tentu. Berikut ilustrasinya}\end{array}$

$\begin{array}{l}\\ 2.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 5}{\text{lim}}f(x),\text{untuk}f(x)=\{\begin{array}{ll}x& \text{saat}x<5\\ 5-x& \text{saat}x\ge 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 5^{-},\text{maka}\underset{x\to 5^{-}}{\text{lim}}f(x)=x\\ & \text{atau}\underset{x\to 5^{-}}{\text{lim}}f(x)=5\\ & \text{boleh juga dituliskan dengan}\\ & \underset{x\to 5^{-}}{\text{lim}}f(x)=x=5.\text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 5^{+},\\ & \text{maka}\underset{x\to 5^{+}}{\text{lim}}f(x)=\underset{x\to 5^{+}}{\text{lim}}(5-x)=5-5=0.\\ & \text{Karena nilai}\underset{x\to 5^{-}}{\text{lim}}f(x)\ne \underset{x\to 5^{+}}{\text{lim}}f(x),\text{maka}\\ & \text{nilai atau harga}\underset{x\to 5}{\text{lim}}f(x)\mathbf{\text{tidak ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 0}{\text{lim}}f(x),\text{untuk}f(x)=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 0^{-},\text{maka}\underset{x\to 0^{-}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& -0,5& -0,4& -0,3& -0,2& -0,1& 0\\ \cos x& ...& ...& 0,999986& 0,999994& 0,9999985& 1\\ \hline\end{array}\\ & \text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 0^{+},\\ & \text{maka}\underset{x\to 0^{+}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& 0& 0,1& 0,2& 0,3& 0,4& 0,5\\ \cos x& 1& 0,9999985& 0,999994& 0,999986& ...& ...\\ \hline\end{array}\\ & \text{Karena nilai}\underset{x\to 0^{-}}{\text{lim}}f(x)=\underset{x\to 0^{+}}{\text{lim}}f(x)=1,\text{maka}\\ & \text{nilai}\underset{x\to 0}{\text{lim}}\cos x\mathbf{\text{ ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\text{B. Sifat-Sifat Limit Fungsi}$

$\begin{array}{rl}& \text{Misalkan}f\text{dan}g\text{adalah fungsi-fungsi yang}\\ & \text{mempunyai nilai limit di titik sekitar}x=a\\ & \text{atau}(x\to a)\text{dan}c\text{adalah suatu konstanta}\\ & \text{serta}n\text{adalah suatu bilangan bulat positif},\\ & \text{maka berlaku sifat-sifat berikut}:\\ & \begin{array}{l}\\ 1.& \underset{x\to a}{\text{lim}}{\displaystyle c=c}\\ 2.& \underset{x\to a}{\text{lim}}{\displaystyle x^n=a^n}\\ 3.& \underset{x\to a}{\text{lim}}c.f(x)=c.\underset{x\to a}{\text{lim}}f(x)\\ 4.& \underset{x\to a}{\text{lim}}(f(x)\pm g(x))=\underset{x\to a}{\text{lim}}f(x)\pm \underset{x\to a}{\text{lim}}g(x)\\ 5.& \underset{x\to a}{\text{lim}}(f(x)\times g(x))=\underset{x\to a}{\text{lim}}f(x)\times \underset{x\to a}{\text{lim}}g(x)\\ 6.& \underset{x\to a}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}}}\\ 7.& \underset{x\to a}{\text{lim}}{(f(x))}^n={[\underset{x\to a}{\text{lim}}f(x))]}^n\\ 8.& \underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to a}{\text{lim}}f(x)\ge 0\\ & \text{dan}n\text{genap}\end{array}\end{array}$

Lanjutan 2 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{2. Grafik Fungsi Cosinus}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 1& \frac{1}{2}\sqrt{3}& \frac{1}{2}\sqrt{2}& \frac{1}{2}& 0& -\frac{1}{2}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}\sqrt{3}& -1\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& -\frac{1}{2}\sqrt{3}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}& 0& \frac{1}{2}& \frac{1}{2}\sqrt{2}& \frac{1}{2}\sqrt{3}& 1& \\ \hline\end{array}$

Notasi Sigma Lanjutan Induksi Matematika (Matematika Wajib Kelas XI)

$\text{A. Pendahuluan}$

Notasi sigma dari asalnya dari yaitu dari huruf yunani yang memiliki makna jumlah. Dalam matematika lambang notasi sigma $"\sum "$  selanjutnya akan menunjukkan penjumlahan yang teratur sehingga penulisan sebuah deret dari suatu bilangan yang berpola tertentu dapat disederhanakan lebih ringkas.

Sebagai ilustrasinya untuk deretny adalah sebagai berikut

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100\\ \text{b}.1+3+5+7+9+\cdots +199\\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2\end{array}$

Dari bentuk deret di atas jika dimodelkan dengan notasi sigma maka bentuknya akan menjadi lebih sederhana, yaitu:

$\begin{array}{rl}& \sum _{i=1}^n{a}_i=a_1+a_2+a_3+\cdots +a_n\\ & \text{Dibaca}:"\text{Jumlah}\text{dari}{a}_i\text{untuk}i\\ & \text{dari 1 sampai dengan}n"\text{dan}{a}_i\text{adalah suku ke}-i\end{array}$

Sehingga contoh ilustrasi deret di atas jika dinotasikan dengan notasi sigma menjadi

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100=\sum _{i=1}^{100}i\\ \\ \text{b}.1+3+5+7+9+\cdots +199=\sum _{i=1}^{100}(2i-1)\\ \\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2=\sum _{i=1}^{100}{i}^2\end{array}$

$\text{B. Sifat-Sifat Notasi Sigma}$

Misalkan diketahui $a_k$  dan $b_k$  adalah suku ke-k dan C adalah sebuah konstanta, maka

$\begin{array}{l}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C=nC}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C.a_k=C\sum _{k=1}^n{a}_k}\\ & ⧫{\displaystyle \sum _{k=1}^n(a_k+b_k)={\displaystyle \sum _{k=1}^n{a}_k+\sum _{k=1}^n{b}_k}}\\ & ⧫{\displaystyle \sum _{k=1}^n{(a_k+b_k)}^2={\displaystyle \sum _{k=1}^n{a}_k^2+2\sum _{k=1}^n{a}_k{b}_k+\sum _{k=1}^n{b}_k^2}}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^{n-1}{a}_k+a_n}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^m{a}_k+\sum _{k=m+1}^n{a}_k,1<m<n}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Uraikan jumlah berikut dengan lengkap}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k}& \text{f}.{\displaystyle \sum _{k=1}^3{2}^k}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)}& \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k}& \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)}& \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)}& \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)}\end{array}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & \begin{array}{l}\\ 1.& \text{Perhatikanlah,}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k=1+2+3+4=10}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k=5.1+5.2+5.3+5.4=50}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)=30}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)=(2.1+3)+(2.2+3)+(2.3+3)=21}\\ & \text{f}.{\displaystyle \sum _{k=1}^3{2}^k=2^1+2^2+2^3+2^4=2+4+8+16=30}\\ & \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}={\displaystyle \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}={\displaystyle \frac{9+3+1}{27}=\frac{13}{27}}}}\\ & \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)=\cdots +\cdots +\cdots +\cdots +\cdots }\\ & \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k=\cdots +\cdots +\cdots +\cdots }\\ & \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)=\cdots +\cdots +\cdots +\cdots +\cdots }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ & \text{a}.2+4+8+16+32+64\\ & \text{b}.2+6+18+54+162\\ & \text{c}.15+24+35+48\\ & \text{d}.{\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}}\\ & \text{e}.ab+a^2{b}^2+a^3{b}^3+a^4{b}^4\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a})2+4+8+16+32+64={\displaystyle \sum _{k=1}^6{2}^k}\\ & (\text{b})2+6+18+54+162={\displaystyle \sum _{k=1}^5{2.3}^{k-1}}\\ & (\text{c})15+24+35+48={\displaystyle \sum _{k=1}^4(k^2+6k+8)}\\ & (\text{d}){\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}={\displaystyle \sum _{k=1}^5\frac{2^k}{(2k+1)}}}\\ & (\text{e})ab+a^2{b}^2+a^3{b}^3+a^4{b}^4={\displaystyle \sum _{k=1}^4(ab{)}^k}\end{array}$

$\begin{array}{l}\\ 3.& \text{Dengan menggunakan kaidah notasi sigma},\\ & \text{tunjukkan bahwa}\\ & \text{a}.{\displaystyle \sum _{k=1}^6(2k+3)=2\sum _{k=1}^{6}k+18}\\ & \text{b}.{\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=1}^{6}k+30}\\ & \text{c}.{\displaystyle \sum _{k=2}^5(2k^2+3k+3)=2\sum _{k=1}^4{k}^2+7\sum _{k=1}^{4}k+32}\\ & \text{d}.{\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=1}^6{k}^2-2\sum _{k=1}^{6}k+6}\\ & \text{e}.{\displaystyle \sum _{k=3}^6(k^2+2k-3)=\sum _{k=1}^6{k}^2+6\sum _{k=1}^4+20}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a}){\displaystyle \sum _{k=1}^6(2k+3)=\sum _{k=1}^{6}2k+\sum _{k=1}^{6}3}\\ & =\sum _{k=1}^{6}2k+6.3=2\sum _{k=1}^{6}k+18\\ & (\text{b}){\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=3-2}^{8-2}((k+2)+3)}\\ & =\sum _{k=1}^6(k+5)=\sum _{k=1}^{6}k+\sum _{k=1}^{6}5=\sum _{k=1}^{6}k+6.5\\ & =\sum _{k=1}^{6}k+30\\ & (\text{c}){\displaystyle \sum _{k=2}^5(2k^2+3k+3)}\\ & =\sum _{k=2-1}^{5-1}(2(k+1{)}^2+3(k+1)+3)=\cdots \\ & (\text{d}){\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=0+1}^{5+1}(k-1{)}^2=\cdots }\\ & (\text{e}){\displaystyle \sum _{k=3}^6(k^2+2k-3)=\cdots }\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{l}\\ .& \text{Buktikanlah bahwa}\\ & \text{a}.{\displaystyle \sum _{k=6}^{12}{k}^2=\sum _{k=1}^7{k}^2+10\sum _{k=1}^{7}k+175}\\ & \text{b}.{\displaystyle \sum _{k=1}^n(3k-1{)}^2=9\sum _{k=1}^n{k}^2-6\sum _{k=1}^{n}k+n}\\ & \text{c}.{\displaystyle \sum _{k=m}^n{a}_k=\sum _{k=m+p}^{n+p}{a}_{k-p}}\\ & \text{d}.{\displaystyle \sum _{i=m}^n{a}_i=\sum _{i=1}^n{a}_i-\sum _{i=1}^{m-1}{a}_i}\\ & \text{e}.{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=0}^{n-1}{a}_{k+1}=\sum _{k=2}^{n+2}{a}_{k-1}}\\ & \text{f}.{\displaystyle \sum _{k=1}^{n-5}{a}_k=\sum _{k=1}^n{a}_k-\sum _{k=(n-5)+1}^n{a}_k}\end{array}$

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
2. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.

Lanjutan 1 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{B. Grafik Fungsi Trigonometri}$

Dalam melukis grafik fungsi trigonometri nantinya yang ditampilkan adalah nilai-nilai untuk sudut istimewa saja. Dan selanjutnya besar sudutnya disajikan dalam derajat dan atau radian.

Selanjutnya perhatikanlah uraian berikut

$\text{1. Grafik Fungsi Sinus}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 0& \frac{1}{2}& \frac{1}{2}\sqrt{2}& \frac{1}{2}\sqrt{3}& 1& \frac{1}{2}\sqrt{3}& \frac{1}{2}\sqrt{2}& \frac{1}{2}& 0\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& -\frac{1}{2}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}\sqrt{3}& -1& -\frac{1}{2}\sqrt{3}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}& 0& \\ \hline\end{array}$

Induksi Matematika (Kelas XI Matematika Wajib)

$\text{A. Pendahuluan}$

Misalkan kita menjumlahkan 100 bilangan ganjil pertama (anggap saja sebagai penjumlahan suku pertama sampai suku ke seratus) yaitu : 1+3+5+...+199, maka untuk memudahkannya kita dapat menentukan cara menjumlahkan dengan atau menurut pola tertentu sebagaimana ilstrasi berikut ini

$\begin{array}{rl}1& =1^2=S_1\\ 1+3& =2^2=S_2\\ 1+3+5& =3^2=S_3\\ 1+3+5+7& =4^2=S_4\\ 1+3+5+7+9& =5^2=S_5\\ ⋮& \\ 1+3+5+7+9+\cdots +(2n-1)& =n^2=S_n\\ ⋮& \\ 1+3+5+7+9+\cdots +199& ={100}^2=S_{100}\end{array}$

$\text{B. Induksi Matematika}$

Pola bilangan tertentu dalam matematika sebagaimana misal contoh di atas dapat ditarik suatu bentuk umum. Selanjutnya untuk membuktikan bahwa suatu bentuk umum dari sebuah pernyataan berlaku, kita dapat menggunakan Induksi Matematika ini. Tentunya semunya dari pernyataan tersebut harus memenuhi kriteria tertentu. Sehingga Induksi Matematika dapat juga disebutkan sebagai proses pembuktian pernyataan (teorema) dari kejadian-kejadian khusus yang berlaku untuk setiap bilangan asli.

Dalam pembuktian dengan Induksi Matematika, perhatikanlah beberapa langkah-langkah ini

Misalkan $P(n)$ adalah suatu pernyataan yang akan dibuktikan kebenarannya untuk semua bilangan asli $n$, maka

$\begin{array}{rl}& \mathbf{\text{Langkahnya:}}\\ & (1)\text{buktikan}P(1)\text{benar untuk}n=1\\ & \text{Selanjutnya disebut Langkah Basis}\\ & (2)\text{Asumsikan pernyataan berlaku untuk}n=k,\\ & \text{yaitu}P(k)\text{benar, dengan}k\in A,\\ & \text{maka untuk}n=k+1\text{bahwa}P(k+1)\\ & \text{juga benar}\\ & \text{Selanjutnya disebut Langkah Induksi}\\ & (3)\text{Setelah langkah}(1)\text{dan}(2)\text{terpenuhi}\\ & \text{atau benar, maka dapat disimpulkan bahwa}\\ & P(n)\text{benar untuk setiap}n\\ & \text{Selanjutnya disebut Konkulsi}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{rl}(1)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}1={\displaystyle \frac{1(1+1)}{2}}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+2+3+...+k={\displaystyle \frac{k(k+1)}{2},\text{maka}}\\ & \text{untuk}n=k+1:\\ & 1+2+3+\cdots +(k)+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & \underset{{\displaystyle \frac{k(k+1)}{2}}}{\underbrace{1+2+3+\cdots +(k)}}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & {\displaystyle \frac{k(k+1)}{2}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)(\frac{k}{2}+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)\frac{k+2}{2}={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle \frac{(k+1)(k+2)}{2}={\displaystyle \frac{(k+1)(k+2)}{2}\equiv P(k+1)}}\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(2)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+3+5+7+...+(2n-1)=n^2\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+3+5+...+(2n-1)=n^2\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-1=1^2\iff 1=1\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+3+5+...+(2k-1)=k^2,\text{maka}\\ & \text{untuk}n=k+1:\\ & 1+3+5+\cdots +(2k-1)+(2(k+1)-1)=(k+1{)}^2\\ & \underset{{\displaystyle k^2}}{\underbrace{1+3+5+\cdots +(2k-1)}}+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2k+2-1)=(k+1{)}^2\\ & k^2+2k+1=(k+1{)}^2\\ & (k+1{)}^2=P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(3)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 2n-3<2^{n-2}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 2n-3<2^{n-2}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-3<2^{1-2}\\ & \text{demikian pula untuk}n=2\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (2k-3)<2^{k-2},\text{maka}\\ & \text{untuk}n=k+1:\\ & 2(k+1)-3=2k+2-3=(2k-3)+2<2^{k-2}+2\\ & \text{sehingga}\\ & (2k-3)+2<2^{k-2}+2<2^{k-2}+2^{k-2},\text{untuk}k\ge 3\\ & (2k-3)+2<{2.2}^{k-2}\\ & (2k-3)+2<2^{(k+1)-2}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(4)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & (1+h{)}^n\ge 1+nh\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv (1+h{)}^n\ge 1+nh\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}(1+h{)}^1\ge 1+1h\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (1+h{)}^k\ge 1+kh,\text{maka}\\ & \text{untuk}n=k+1:\\ & (1+h{)}^{k+1}\ge (1+kh)(1+h)\\ & (1+h{)}^{k+1}\ge (1+(k+1)h+k{h}^2)\\ & (1+h{)}^{k+1}\ge 1+(k+1)h\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}.& \mathbf{\text{Catatan}}\\ & \text{untuk}k=2\\ & (1+h{)}^2=1+2h+h^2\ge 1+2h,\text{maka}\\ & (1+h{)}^n\ge 1+nh\end{array}$

$\begin{array}{rl}(5)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena untuk}n=1\\ & 3^1={\displaystyle \frac{3}{2}(3^1-1)}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv \sum _{h=1}^k{3}^h={\displaystyle \frac{3}{2}(3^k-1),\text{maka}}\\ & \text{untuk}n=k+1:\\ & \sum _{h=1}^{k+1}{3}^h=\sum _{h=1}^k{3}^h+\sum _{h=k+1}^{k+1}{3}^h\\ & ={\displaystyle \frac{3}{2}(3^k-1)+3^{k+1}}\\ & ={\displaystyle \frac{1}{2}(3^{k+1}-3+{2.3}^{k+1})}\\ & ={\displaystyle \frac{1}{2}({3.3}^{k+1}-3)}\\ & ={\displaystyle \frac{3}{2}({.3}^{k+1}-1)}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{rl}.& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ \end{array}$

$.\begin{array}{l}\\ 1.& 2+4+6+8+...+2n=n^2+n\\ 2.& 1^2+2^2+3^2+...+n^2={\displaystyle \frac{1}{6}n(n+1)(2n+1)}\\ 3.& 1^3+2^3+3^3+...+n^3={\displaystyle \frac{1}{4}{n}^2(n+1{)}^2}\\ 4.& 1.2+2.3+3.4+...+n(n+1)={\displaystyle \frac{1}{3}n(n+1)(n+2)}\\ 5.& {\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}={\displaystyle \frac{n}{n+1}}}\\ 6.& {\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2n-1)(2n+1)}={\displaystyle \frac{n}{2n+1}}}\\ 7.& {\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}={\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}}}\\ 8.& 1+{\displaystyle \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n+1}-2}\\ 9.& n^3-n\text{habis dibagi oleh}3\\ 10.& n^5-n\text{habis dibagi oleh}5\\ 11.& 5^n+{6.7}^n+1\text{habis dibagi oleh}4\\ 12.& 5^{2n}-1\text{habis dibagi oleh}3\\ 13.& 3^n-1\ge 2^n\\ 14.& 2n+7<(n+3{)}^2\\ 15.& 2+4+6+8+...+2n\le 2^n\\ 16.& {(3+\sqrt{5})}^n+{(3-\sqrt{5})}^n\text{habis dibagi oleh}{2}^n\end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S., 2018. Bupena Mathematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: Erlangga.
2. Kemendikbud. 2017. Matematika untuk SMA/MA/SMK Kelas XI Edisi Revisi. Jakarta: Kementerian Pendidikan Nasional.
3. Tampomas, H. 1999. SeribuPena Matematika Jilid 3 untuk SMU Kelas 3. Jakarta: Erlangga
4. Tim ITB. 2007. Program Pembinaan Kompetensi Siswa Bidang Matematika Tahap 1. Bandung: LPPM ITB

Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\text{A. Persamaan Trigonometri}$

Ada minimal 3 yang utama untuk persmaan trigonometri sederhana, yaitu:

$\begin{array}{rl}1.\sin x& =\sin \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k{.360}^{\circ }\\ & =({180}^{\circ }-\alpha )+k{.360}^{\circ }\end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\sin x& =\sin \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k.2\pi \\ & =(\pi -\alpha )+k.2\pi \end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\begin{array}{rl}2.\cos x& =\cos \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k{.360}^{\circ }\\ & =-\alpha +k{.360}^{\circ }\end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\cos x& =\cos \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k.2\pi \\ & =-\alpha +k.2\pi \end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\begin{array}{rl}3.\tan x& =\tan \alpha \\ x& =\alpha +k{.180}^{\circ }\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\tan x& =\tan \alpha \\ x& =\alpha +k.\pi \\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari }\\ & \text{persamaan-persamaan trigonometri berikut}\\ & \text{ini untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \begin{array}{l}\\ \text{a}.& \sin x={\displaystyle \frac{1}{2}}& \text{f}.& \tan x=-{\displaystyle \frac{1}{3}\sqrt{3}}& \text{k}.& \sin 2x={\displaystyle \frac{1}{2}}\\ \text{b}.& \cos x={\displaystyle \frac{1}{2}\sqrt{3}}& \text{g}& 2\cos x=-\sqrt{3}& \text{l}.& \cos 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& \tan x=\sqrt{3}& \text{h}& 3\tan x=\sqrt{3}& \text{m}.& \tan 2x=\sqrt{3}\\ \text{d}.& \sin x=-1& \text{i}.& \sin x=\sin {46}^{\circ }& \text{n}.& \sin (2x-{30}^{\circ })=\sin {45}^{\circ }\\ \text{e}.& \cos x=-{\displaystyle \frac{1}{2}\sqrt{2}}& \text{j}.& \cos x=\cos {93}^{\circ }& \text{o}.& \sin (2x+{60}^{\circ })=\sin {90}^{\circ }\end{array}\end{array}$

$.\text{Jawab}:$

$\begin{array}{rl}.\text{a}.\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ {150}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{30}^{\circ },{150}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{b}.\cos x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos x& =\cos {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ -{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ -{30}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tidak memenuhi})& \\ -{30}^{\circ }+{360}^{\circ }={330}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{30}^{\circ },{330}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{c}.\tan x& =\sqrt{3}\\ \tan x& =\tan {60}^{\circ }\\ x& ={60}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={60}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={60}^{\circ }+{180}^{\circ }={240}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={60}^{\circ }+{360}^{\circ }={420}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{60}^{\circ },{240}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{d}.\sin x& =-1\\ \sin x& =\sin {270}^{\circ }\\ x& =\{\begin{array}{ll}{270}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{270}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}{270}^{\circ }& \text{memenuhi}\\ -{90}^{\circ }& \text{tidak memenuhi}\end{array}\\ k=1& \text{tidak memenuhi semuanya}\\ \text{HP}& =\left\{{270}^{\circ }\right\}\end{array}$

$\begin{array}{rl}.\text{n}.\sin (2x-{30}^{\circ })& =\sin {45}^{\circ }\\ (2x-{30}^{\circ })& =\{\begin{array}{ll}{45}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{45}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{45}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\\ {135}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}37,5^{\circ }& +k{.180}^{\circ }\\ 82,5^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }& \\ 82,5^{\circ }& \end{array}\\ k=1& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }+{180}^{\circ }& =217,5^{\circ }\\ 82,5^{\circ }+{180}^{\circ }& =262,5^{\circ }\end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{37,5^{\circ },82,5^{\circ },217,5^{\circ },262,5^{\circ }\}\end{array}$

Contoh Soal 10 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 46.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -2& & & \\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Jumlah akar-akar persamaan}\\ & {2020}^{x^2-7x+7}={2021}^{x^2-7x+7}\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -7\\ \text{b}.& -5\\ \text{c}.& -3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{2020}^{x^2-7x+7}& ={2021}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai dari}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 5\\ \text{c}.& 10\\ \text{d}.& 20\\ \text{e}.& 40\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}}& ={\displaystyle \frac{2^4{.2}^{2016}+2^2{.2}^{2016}}{2^2{.2}^{2018}+2^{2016}}}\\ & ={\displaystyle \frac{2^{2016}(2^4+2^2)}{2^{2016}(2^2+1)}}\\ & ={\displaystyle \frac{16+4}{4+1}}\\ & ={\displaystyle \frac{20}{5}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 49.& (\mathbf{\text{UM IPB}})\text{Jika}ab=a^b\text{dan}{\displaystyle \frac{a}{b}=a^{3b}}\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 0,5\\ \text{c}.& 1\\ \text{d}.& 0,25\\ \text{e}.& 0,75\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \\ ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}=a^{b-1}.....\mathbf{\text{1}}}\\ \text{maka}& \\ {\displaystyle \frac{a}{b}}& =a^{3b}...............\mathbf{\text{2}}\\ \mathbf{\text{1}}& ke\mathbf{\text{2}}\\ {\displaystyle \frac{a}{a^{b-1}}}& =a^{3b}\\ a^{2-b}& =a^{3b}\\ 2-b& =3b\\ -4b& =-2\\ b& ={\displaystyle \frac{1}{2}................\mathbf{\text{3}}}\\ \mathbf{\text{3}}& ke\mathbf{\text{1}}\\ a\left({\displaystyle \frac{1}{2}}\right)& =a^{\frac{1}{2}}\\ {\displaystyle \frac{1}{4}{a}^2}& =a\\ a^2-4a& =0\\ a(a-4)& =0\\ a=0& \text{atau}a=4\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}=4,\text{maka}x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1,1}\\ \text{b}.& {\displaystyle 1,2}\\ \text{c}.& 1,3\\ \text{d}.& {\displaystyle 1,4}\\ \text{e}.& 1,5\\ \\ & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}}& =4\\ {\left(3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}\right)}^2& =4^2\\ 3^2.x^3& =4^2\\ x^3& ={\displaystyle \frac{4^2}{3^2}}\\ x^3& ={\displaystyle \frac{4^2}{3^2}\times \frac{3}{3}}\\ x^3& \le {\displaystyle \frac{4^2}{3^2}\times \frac{4}{3}}\\ x^3& \le \left({\displaystyle \frac{4^3}{3^3}}\right)\\ x^3& \le {\left({\displaystyle \frac{4}{3}}\right)}^3\\ x& \le {\displaystyle \frac{4}{3}}\\ x& \le 1,\overline{333}\\ x& \approx 1,3\end{array}\end{array}$

$\begin{array}{l}\\ 51.& \text{Hitunglah}\\ \\ & {\displaystyle \sqrt[8]{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{...}}}}}}}}}}\\ \\ & \text{nyatakan jawabannya dalam bentuk }{\displaystyle \frac{a\pm b\sqrt{c}}{d}}\\ & \text{dengan a, b, c, dan d bilangan-bilangan bulat}\end{array}$

Pembahasan:

$\begin{array}{rl}{x}^8& =2207-{\displaystyle \underset{x^8}{\underbrace{{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}}}\\ x^8& =2207-{\displaystyle \frac{1}{x^8}}\\ x^8+{\displaystyle \frac{1}{x^8}}& =2207\\ {(x^4+{\displaystyle \frac{1}{x^4}})}^2& =2207+2\\ (x^4+{\displaystyle \frac{1}{x^4}})& =\sqrt{2209}=47\end{array}$

$\begin{array}{rl}{x}^4+{\displaystyle \frac{1}{x^4}}& =47\\ {(x^2+{\displaystyle \frac{1}{x^2}})}^2& =47+2\\ x^2+{\displaystyle \frac{1}{x^2}}& =\sqrt{49}=7\\ {(x+{\displaystyle \frac{1}{x}})}^2& =7+2\\ x+{\displaystyle \frac{1}{x}}& =\sqrt{9}=3\\ x^2-3x+1& =0,\\ & \text{persamaan kuadrat dalam x,}\\ & \mathbf{\text{gunakan rumus abc}}\\ x_{1,2}=& {\displaystyle \frac{3\pm \sqrt{5}}{2}={\displaystyle \frac{3\pm 1\sqrt{5}}{2}={\displaystyle \frac{a\pm b\sqrt{c}}{d}}}}\\ & \mathbf{\text{Sehingga}},\{\begin{array}{ll}& a=3\\ & b=1\\ & c=5\\ & d=2\end{array}\end{array}$

DAFTAR PUSTAKA

1. Enung, S., Untung, W. 2009. Mandiri Matematika SMA Jilid I untuk Kelas X. Jakarta: ERLANGGA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
3. Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
4. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO. 

Contoh Soal 9 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 41.& \text{Penyelesaian pertidaksamaan eksponen}\\ & {\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}>\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x>{\displaystyle \frac{6}{5}}\\ \text{b}.& x<-{\displaystyle \frac{6}{5}}\\ \text{c}.& x>{\displaystyle \frac{5}{6}}\\ \text{d}.& x<-2\\ \text{e}.& x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}& >\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\\ 3^{-(2x+1)}& >3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)& >{\displaystyle \frac{1}{2}(3-(x-1))}\\ -4x-2& >4-x\\ -4x+x& >4+2\\ -3x& >6\\ 3x& <-6\\ x& <-2\end{array}\end{array}$

$\begin{array}{l}\\ 42.& (\mathbf{\text{UMPTN 01}})\text{Nilai}x\text{yang memenuhi}\\ & 4^{x^2-x-2}{.2}^{x^2+3x-10}<{\displaystyle \frac{1}{16}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-5\text{atau}x>-2\\ \text{b}.& x<-2\text{atau}x>{\displaystyle \frac{5}{3}}\\ \text{c}.& -2<x<-1\\ \text{d}.& -2<x<{\displaystyle \frac{5}{3}}\\ \text{e}.& -5<x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{4}^{x^2-x-2}{.2}^{x^2+3x-10}& <{\displaystyle \frac{1}{16}}\\ 2^{2(x^2-x-2)+(x^2+3x-10)}& <2^{-4}\\ 2(x^2-x-2)+(x^2+3x-10)& <-4\\ 3x^2-2x+3x-4-10+4& <0\\ 3x^2+x-10+& <0\\ (x+2)(3x-5)& <0\\ \therefore -2<x& <{\displaystyle \frac{5}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 43.& (\mathbf{\text{SPMB 04 Mat IPA}})\text{Himpunan Penyelesaian}\\ & \text{pertidaksamaan eksponen}\\ & 2\sqrt{4^{x^2-3x+2}}<\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x>4\}\\ \text{b}.& \{x|x>2\}\\ \text{c}.& \{x|x<1\}\\ \text{d}.& \{x|1<x<4\}\\ \text{e}.& \{x|2\le x\le 3\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}2\sqrt{4^{x^2-3x+2}}& <\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\\ 2^{1+\frac{2}{2}(x^2-3x+2)}& <2^{-\frac{3-6x}{3}}\\ 1+(x^2-3x+2)& <-{\displaystyle \frac{3-6x}{3}}\\ x^2-3x+3& <-1+2x\\ x^2-5x+4& <0\\ (x-1)(x-4)& <0\\ 1<x& <4\end{array}\end{array}$

$\begin{array}{l}\\ 44.& (\mathbf{\text{SBMPTN 2015 Mat IPA}})\\ & \text{Nilai}c\text{yang memenuhi}\\ & (0,12{)}^{4x^2+8x+c}<(0,0144{)}^{x^2+4x+4}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& c>0\\ \text{b}.& c>2\\ \text{c}.& c>4\\ \text{d}.& c>6\\ \text{e}.& c>8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}(0,12{)}^{4x^2+8x+c}& <(0,0144{)}^{x^2+4x+4}\\ (0,12{)}^{4x^2+8x+c}& <(0,12{)}^{2(x^2+4x+4)}\\ 4x^2+8x+c& >2(x^2+4x+4)\\ 2x^2+c-8& >0\text{haruslah definit positif}\\ \text{Syaratnya}& \{\begin{array}{ll}a& =2>0\\ D& =b^2-4ac<0\end{array}\\ \text{Maka}D& =b^2-4ac<0\\ \mathbf{\text{ambil dari}}& 2x^2-c-8=0\{\begin{array}{ll}a& =2\\ b& =0\\ c& =c-8\end{array}\\ & =0^2-4.2(c-8)<0\\ -8c& +64<0\\ -8c& <-64\\ 8c& >64\\ c& >8\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Nilai}x\text{yang memenuhi}\\ & 9^{2x}-{10.9}^x+9>0,x\in \mathbb{R}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<1\text{atau}x>0\\ \text{b}.& x<0\text{atau}x>1\\ \text{c}.& x<-1\text{atau}x>2\\ \text{d}.& x<1\text{atau}x>2\\ \text{e}.& x<-1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{9}^{2x}-{10.9}^x+9& >0\\ {\left(9^x\right)}^2-10.\left(9^x\right)+9& >0\\ (9^x-1)(9^x-9)& >0\\ 9^x<1\text{atau}{9}^x& >9\\ 9^x<9^0\text{atau}{9}^x& >9^1\\ x<0\text{atau}x& >1\end{array}\end{array}$

Contoh Soal 8 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 36.& \text{Bentuk sederhana dari}\\ & \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\ \\ & (\mathbf{\text{SIMAK UI 2012 Mat IPA}})\\ & \begin{array}{l}\\ \text{a}.& 2-\sqrt{2}\\ \text{b}.& 8-\sqrt{2}\\ \text{c}.& -2+\sqrt{2}\\ \text{d}.& 2+5\sqrt{2}\\ \text{e}.& 8+5\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{misalkan},\\ & \begin{array}{rl}x& =\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ & =(\sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}})\\ & =\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^2& =33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{(33+20\sqrt{2})(27-18\sqrt{2})}\\ & =60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ & =60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2(\sqrt{162}-\sqrt{9})\\ & =60+2\sqrt{2}-2(9\sqrt{2}-3)\\ x^2& =66-16\sqrt{2}\\ x& =\sqrt{66-2.8\sqrt{2}}\\ & =\sqrt{64+2-2\sqrt{64.2}}\\ & =\sqrt{64}-\sqrt{2}\\ & =8-\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Jika}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}={\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12},}}\\ \\ & \text{maka}a+b+c=....\\ & (\mathbf{\text{UM UGM 2016 Mat Das}})\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}& ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times {\displaystyle \frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}-\sqrt{5})}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}}\\ & \{\begin{array}{ll}a& =3\\ b& =2\\ c& =-1\end{array}\\ a+b+c& =3+2+(-1)\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Tunjukkan bahwa}\\ & \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\ \\ & \text{Bukti}\\ & \begin{array}{rl}{x}^2& =x^2\\ x^2& =1+(x^2-1)\\ & =1+(x-1)(x+1)\\ & =1+(x-1)\sqrt{{(x+1)}^2}\\ & =1+(x-1)\sqrt{1+{(x+1)}^2-1}\\ & =1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & =1+(x-1)\sqrt{1+x(x+2)}\\ & =1+(x-1)\sqrt{1+x\sqrt{{(x+2)}^2}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+{(x+2)}^2-1}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{{(x+3)}^2}}}\\ x^2& =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\\ x& =\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Jika bentuk}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}\\ & \text{dinyatakan dalam pangkat positif}=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a^2}{a-b}}& & & \\ \\ \text{b}.& {\displaystyle \frac{a^2}{a-1}}\\ \\ \text{c}.& {\displaystyle \frac{b-a}{ab}}\\ \\ \text{d}.& {\displaystyle \frac{a^2}{b-a}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a-b}}\\ \\ & & & & (\mathbf{\text{textit{SAT Test Math Level 2}}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}& ={\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}}\\ & ={\displaystyle \frac{a^2}{b-a}}\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Jika terdapat hubungan berikut}\\ & \text{a}.2^p=3^q=6^r,\text{tunjukkan bahwa}pr+qr-pq=0\\ & \text{b}.2^x=3^{2y}=6^z,\text{tunjukkan bahwa }2xy-2yz-xz=0\\ & \text{c}.3^{15a}=5^{5b}={15}^{3c},\text{tunjukkan bahwa }5ab-bc-3ac=0\end{array}$

$\mathbf{\text{bukti}}$

Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:

$\begin{array}{rl}{3}^{15a}& =5^{5b}={15}^{3c}\{\begin{array}{ll}3=5^{\frac{5b}{15a}}& \\ 3^{\frac{15a}{5b}}=b& (a^b=c^d\to a=c^{\frac{d}{b}}\text{atau}{a}^{\frac{b}{d}}=c)\end{array}\\ 3^{15a}& ={15}^{3c}\\ 3^{15a}& =(3\times 5{)}^{3c}\\ 3^{15a}& =(3\times 3^{\frac{15a}{5b}}{)}^{3c}\\ 3^{15a}& =3^{3c+\frac{9c}{b}}\\ a^{f(x)}& =a^{g(x)}\\ f(x)& =g(x)\\ 15a& =3c+\frac{9ac}{b}\\ 15ab& =3bc+9ac\\ 5ab& =bc+3ac\\ 5ab-bc-3ac& =0◼\end{array}$

Contoh Soal 7 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misalkan}A& =\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =3\\ \text{dikuadratkan}& \\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =9\\ x+3& =9\\ x& =9-3\\ x& =6\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi}\\ & x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 7\sqrt[7]{7}\\ \text{b}.& 7\\ \text{c}.& 14\\ \text{d}.& 49\\ \text{e}.& \sqrt[3]{81}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x& =\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49x\\ x^2& =49\\ x& =\sqrt{49}\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Nilai}x\text{yang memenuhi}\\ & x^{x^{x^{x^{x^{\cdots }}}}}=2020\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2020}\\ \text{b}.& \sqrt[2020]{2020}\\ \text{c}.& {2020}^{\sqrt{2020}}\\ \text{d}.& {\sqrt{2020}}^{\sqrt{2020}}\\ \text{e}.& \sqrt{2020\sqrt{2020}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^{x^{x^{x^{x^{\cdots }}}}}& =2020\\ x^{2020}& =2020\\ x& =\sqrt[2020]{2020}\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai dari}\\ & {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& \sqrt[3]{2}+1\\ \text{c}.& \sqrt[3]{2}-1\\ \text{d}.& \sqrt[3]{4}+1\\ \text{e}.& \sqrt[3]{4}-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}}\\ & ={\displaystyle \frac{{\left(\sqrt[3]{2}\right)}^2-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{2-1}}\\ & =\sqrt[3]{4}-1\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai dari}\\ & {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\sqrt{2}-1\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{d}.& \sqrt{{\displaystyle \frac{5}{3}}}\\ \text{e}.& \sqrt{{\displaystyle \frac{2}{5}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-(\sqrt{2}-1)}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}\right)+\left({\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}\right)}{\sqrt{5}+1}+1-\sqrt{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}}{1+\sqrt{5}}+1-\sqrt{2}}\\ & ={\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}}\\ & =\sqrt{2}+1-\sqrt{2}\\ & =1\end{array}\end{array}$

Contoh Soal 6 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 05}})\text{Hasil dari}\\ & \sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b},\text{maka}{\displaystyle \frac{1}{a}+\frac{1}{b}=....}\\ & \begin{array}{l}\\ \text{a}.& 25\\ \text{b}.& 20\\ \text{c}.& 15\\ \text{d}.& 10\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{0,3+\sqrt{0,08}}& =\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ & =\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ & =\sqrt{0,2}+\sqrt{0,1}\\ \text{maka},a=0,2& ,b=0,1\\ \text{sehingga}{\displaystyle \frac{1}{a}+\frac{1}{b}}& ={\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{SPMB 06}})\text{Jika bilangan bulat}a\text{dan}b\text{memenuhi}\\ & {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30},\text{maka}ab=....}\\ & \begin{array}{l}\\ \text{a}.& -22\\ \text{b}.& -11\\ \text{c}.& -9\\ \text{d}.& 2\\ \text{e}.& 13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}}& ={\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}}}\\ & ={\displaystyle \frac{5-2\sqrt{30}+6}{5-6}}\\ & ={\displaystyle \frac{11-2\sqrt{30}}{-1}}\\ & =-11+2\sqrt{30}\\ \text{sehingga}& a=-11,b=2,\text{maka}\\ ab& =(-11)\times 2\\ & =-22\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{OSK 2013}})\text{Misal}a\text{dan}b\text{bilangan asli}\\ & \text{dengan}a>b.\text{Jika}\sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ & \text{maka nilai}a-b\text{adalah... .}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\sqrt{94+2\sqrt{2013}}& =\sqrt{61+33+2\sqrt{61\times 33}}\\ & =\sqrt{61}+\sqrt{33}\\ & =\sqrt{a}+\sqrt{b}\\ \text{Sehingga}a& =61,b=33,\text{maka}\\ a-b& =61-33\\ & =28\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Daerah hasil dari fungsi eksponen}y=x^{-\frac{2}{3}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y<0\\ \text{b}.& y>0\\ \text{c}.& y\ge 0\\ \text{d}.& y\le 0\\ \text{e}.& \text{Semua bilangan real}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikanlah gambar berikut}\end{array}$

$.\begin{array}{rl}\text{diketahui}& \\ y& =x^{-{\displaystyle \frac{2}{3}}}\\ y^3& =x^{-2}\\ y^3& ={\displaystyle \frac{1}{x^2},\text{atau}}\\ y^3\times x^2& =1,\\ \text{sehingga}& y\text{tidak mungkin berharga}0\end{array}$

$\begin{array}{l}\\ 30.& \text{Jika}f(x)=b^x,\text{di mana konstan positif},\\ \\ & {\displaystyle \frac{f(x^2+x)}{f(x+1)}=....}\\ & \begin{array}{l}\\ \text{a}.& f\left(x^2\right)& & & \\ \text{b}.& f(x+1)f(x-1)\\ \text{c}.& f(x+1)+f(x-1)\\ \text{d}.& f(x+1)-f(x-1)\\ \text{e}.& f(x^2-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{f(x^2+x)}{f(x+1)}}& =\frac{b^{x^2+x}}{b^{x+1}}\\ & =b^{x^2+x-(x+1)}\\ & =b^{x^2-1}\\ & =f(x^2-1)\end{array}\end{array}$

Contoh Soal 5 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{l}\\ 21.& \text{Nilai}p-q^{p-q}\text{untuk}p=2\text{dan}q=-2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -18\\ \text{b}.& -14\\ \text{c}.& 1\\ \text{d}.& 18\\ \text{e}.& 256\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}p-q^{p-q}& =(2-(-2){)}^{2-(-2)}\\ & =4^4\\ & =256\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 9\\ \text{c}.& 7\\ \text{d}.& 7^{2x+2}\\ \text{e}.& 3^{2x-1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}}& ={\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}}\\ & ={\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}}\\ & =3^0\times 7^2\\ & =1\times 49\\ & =49\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Jika}{\displaystyle \frac{(0,24{)}^3(0,243{)}^5}{(3,6{)}^7}=\frac{2^p{3}^q}{5^r},}\\ & \text{maka nilai}p+q+r\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 7\\ \text{b}.& 8\\ \text{c}.& 9\\ \text{d}.& 10\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{(0,24{)}^3(0,243{)}^5}{(3,6{)}^7}}& ={\displaystyle \frac{{\left(\frac{24}{100}\right)}^3\times {\left(\frac{243}{1000}\right)}^5}{{\left(\frac{36}{10}\right)}^7}}\\ & ={\displaystyle \frac{(8\times 3{)}^3\times {\left(3^5\right)}^5}{(2^2\times 3^2{)}^7}\times {\displaystyle \frac{{10}^7}{{100}^3\times {1000}^5}}}\\ & ={\displaystyle \frac{(2^3\times 3{)}^3\times 3^{25}}{(2^{14}\times 3^{14})}\times {\displaystyle \frac{{10}^7}{{\left({10}^2\right)}^3\times {\left({10}^3\right)}^5}}}\\ & =2^{9-14}{.3}^{3+25-14}{.10}^{7-6-15}\\ & =2^{-5}{.3}^{14}{.10}^{-14}=2^{-5}{.3}^{14}.(2.5{)}^{-14}\\ & =2^{-5-14}{.3}^{21}{.5}^{-14}\\ & ={\displaystyle \frac{2^{-19}{.3}^{14}}{5^{14}}}\\ \text{Sehingga}& p+q+r=-19+14+14=9\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai eksak dari}\\ & {\displaystyle \frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+{\displaystyle \frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}}}\\ & \begin{array}{l}\\ \text{a}.& 2020& & & \\ \text{b}.& 2020,5\\ \text{c}.& 2021\\ \text{d}.& 2021,5\\ \text{e}.& 2022\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misal},x& =\frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ \text{maka},x& =\frac{1}{\frac{1}{{10}^{2020}}+1}+\frac{1}{\frac{1}{{10}^{2019}}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{{10}^{2019}}{1+{10}^{2019}}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{1}{{10}^{2020}+1}+\frac{{10}^{2019}}{1+{10}^{2019}}+\frac{1}{{10}^{2019}+1}+...+\frac{1}{1^0+1}\\ & =\frac{{10}^{2020}+1}{{10}^{2020}+1}+\frac{{10}^{2019}+1}{{10}^{2019}+1}+\frac{{10}^{2018}+1}{{10}^{2018}+1}+...+\frac{{10}^1+1}{{10}^1+1}+\frac{1}{1^0+1}\\ & =\underset{\text{sebanyak}2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ & =2020,5\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai dari}\\ & \sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10\\ \text{b}.& 11\\ \text{c}.& 12\\ \text{d}.& 5\sqrt{6}\\ \text{e}.& 6\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misal diketah}& \text{ui}\\ x& =\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \text{untuk}& \\ \sqrt{54+14\sqrt{5}}& =\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}& =\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}& =\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}+\\ & ---------------\\ & =7+5\\ & =12\end{array}\end{array}$