CONTOH SOAL 2 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 6.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{2x+4}{1+\sqrt{x}}\: ,\: \textrm{maka}\: \: \displaystyle {f}\, '(4) =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4} &&\textrm{d}.\quad 1 \\ \textrm{b}.\quad \displaystyle \frac{3}{7} \quad &\textrm{c}.\quad \displaystyle \frac{3}{5} \quad &\textrm{e}.\quad 4 \end{array}\\ \end{array}$.

Jawaban: $\begin{array}{ll} &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{2x+4}{1+\sqrt{x}}\\ &=\displaystyle \frac{U}{V}\\ {f}\, '(x)&=\displaystyle \frac{{U}\, '.V-U.{V}\, '}{V^{2}}\\ &=\displaystyle \frac{(2)\left ( 1+\sqrt{x} \right )-(2x+4).\left ( 1.\left ( 1+\sqrt{x} \right )^{0} .\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )}{\left ( 1+\sqrt{x} \right )^{2}} \\&\textrm{ingat}\: \: \sqrt{x}=x^{^{\frac{1}{2}}} \\ {f}\, '(4) &=\displaystyle \frac{2\left ( 1+\sqrt{4} \right )-(2.4+4).\frac{1}{2}. 4^{^{-\frac{1}{2}}} }{\left ( 1+\sqrt{4} \right )^{2}}\\ &=\displaystyle \frac{2(1+2)-(12).\frac{1}{2}. \frac{1}{2}}{(1+2)^{2}}\\ &\textrm{ingat juga}\: \: 4^{^{-\frac{1}{2}}}=\left ( 2^{2} \right )^{^{-\frac{1}{2}}}=2^{^{-1}}=\displaystyle \frac{1}{2^{1}}=\frac{1}{2}\\ &=\displaystyle \frac{6-3}{9}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: \: f(x)=\displaystyle 3x^{2}-2ax+7\\ &\textrm{dan}\: \: {f}\, '(1)=0 \: ,\: \textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1 &&\textrm{d}.\quad 6 \\ \textrm{b}.\quad \displaystyle 2 \quad &\textrm{c}.\quad \displaystyle 4 \quad &\textrm{e}.\quad 8 \end{array}\\ \end{array}$.

Jawaban: $\begin{array}{ll} &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle 3x^{2}-2ax+7\\ {f}\, '(x)&=6x-2a\\ {f}\, '(1)&=0\\ 6(1)-2a&=0\\ 6&=2a\\ 3&=a\\ \textrm{sehingga}\, &\: \\ {f}\, '(x)&=6x-6\\ \textrm{maka}\, ,\: \quad &\\ {f}\, '(2)&=6.2-6\\ &=12-6\\ &=6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \: f(x)=\displaystyle (6x-3)^{3}(2x-1)\\ &\textrm{maka}\: \: {f}\, '(1)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 162 \\\\ \textrm{b}.\quad \displaystyle 24 \quad &\textrm{c}.\quad \displaystyle 54 \quad &\textrm{e}.\quad 216 \end{array}\\\end{array}$.

Jawaban: $\begin{array}{ll} &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle (6x-3)^{3}(2x-1)\\ &=(3.(2x-1))^{3}(2x-1)^{1}\\ &=3^{3}.(2x-1)^{3+1}\\ &=27(2x-1)^{4}\\ {f}\, '(x)&=4.27(2x-1)^{4-1}.2\\ &=216.(2x-1)^{3}\\ {f}\, '(1)&=216.(2.1-1)^{3}\\ &=216.1\\ &=216 \end{aligned} \end{array}$.

CONTOH SOAL 1 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: {f}\, '(2)=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2},\\ & \textrm{maka fungsi}\: \: f(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 12&&\textrm{d}.\quad x^{3}\\ \textrm{b}.\quad 2\quad&\textrm{c}.\quad x\quad&\textrm{e}.\quad x-8 \end{array}\\ \end{array}$.

Jawaban: $\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Turunan fungsi f di}\: \: x=c\: \: \textrm{adalah}\\ &f\, '(c)=\underset{x\rightarrow c}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(c)}{x-c},\: \: \textrm{maka}\\ &\textrm{turunan fungsi f di}\: \: x=2\: \: \textrm{adalah}\\ &f\, '(2)=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2}\\ &\textrm{sehingga akan didapa}\textrm{tkan fungsi}\: \: f\: \: \textrm{nya }\\ &\textrm{yaitu}\: \: f(x)=x^{3} \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a\neq 0\, ,\: \textrm{maka nilai dari}\\&\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3a\displaystyle \sqrt[3]{a}&&\textrm{d}.\quad \displaystyle \frac{1}{2a}\sqrt[3]{a} \\ \textrm{b}.\quad 2a\displaystyle \sqrt[3]{a} \quad&\textrm{c}.\quad 0\quad&\textrm{e}.\quad \displaystyle \frac{1}{3a}\sqrt[3]{a} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll}&\begin{aligned}\textbf{Jaw}&\textbf{ab}:\\\textrm{Alter}&\textrm{natif 1}\\{f}\, '(a)&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{\left ( \sqrt[3]{x}-\sqrt[3]{a} \right )\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{\left ( \sqrt[3]{a^{2}}+\sqrt[3]{a.a}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\times \displaystyle \frac{\sqrt[3]{a}}{\sqrt[3]{a}}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned}\\ &\textrm{Alternatif 2 (dengan aturan L'Hopital)}\\&\begin{aligned}\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x) }{{g}\, '(x) }\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{x^{^{\frac{1}{3}}}-a^{^{\frac{1}{3}}} }{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{\frac{1}{3}-1}}\: -0}{1-0}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{-\frac{2}{3}}}}{1}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3\sqrt[3]{x^{2}}}\times \frac{\sqrt[3]{x}}{\sqrt[3]{x}}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3x}\sqrt[3]{x}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x}}\: \: \textrm{maka nilai dari}\\ & -2{f}\, '(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{x\sqrt{x}} &&\textrm{d}.\quad \displaystyle -\frac{1}{2x\sqrt{x}} \\ \textrm{b}.\quad \displaystyle x\sqrt{x} \quad&\textrm{c}.\quad \displaystyle -\frac{1}{2\sqrt{x}} \quad&\textrm{e}.\quad \displaystyle -2x\sqrt{x} \end{array}\\ \end{array}$.

Jawaban: $\begin{array}{ll}&\begin{aligned}\textbf{Jaw}&\textbf{ab}:\\\textrm{Dike}&\textrm{tahui}\\ f(x)&=\displaystyle \frac{1}{\sqrt{x}}=\displaystyle \frac{1}{x^{^{\frac{1}{2}}}}=x^{^{-\frac{1}{2}}} \end{aligned}\\ &\begin{array}{|c|c|}\hline y=ax^{n}\rightarrow {y}\, '=nax^{n-1}&\begin{aligned}&\\ y&=\displaystyle \frac{U}{V}\rightarrow {y}\, '=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}}\\ & \end{aligned}\\\hline \begin{aligned} {f}\, '(x)&=-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{2}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{2x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{2x^{1}.x^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}&\begin{aligned} {f}\, '(x)&=\displaystyle \frac{0.\sqrt{x}-1.\frac{1}{2}x^{^{\frac{1}{2}-1}}}{\left ( \sqrt{x}\right )^{2}}\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}}{x}\\ &=-\displaystyle \frac{1}{2xx^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Turunan pertama dari}\: \: y=\displaystyle \sqrt[\displaystyle n]{x}\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{n}x^{^{\displaystyle \frac{1}{n}}}&&\textrm{d}.\quad (n-1)\sqrt[ n-1]{x}\\ \textrm{b}.\quad \displaystyle \frac{1}{n}x^{^{\displaystyle \frac{1-n}{n}}}\quad &\textrm{c}.\quad \displaystyle \frac{1}{n-1}x^{^{\displaystyle n-1}}\quad &\textrm{e}.\quad \sqrt[\displaystyle n-1]{x} \end{array}\\ \end{array}$.

Jawaban: $\begin{aligned}y&=\displaystyle \sqrt[\displaystyle n]{\displaystyle x}=\displaystyle x^{^{\displaystyle \frac{1}{n}}},\: \: \textrm{maka}\\ {y}\, '&=\displaystyle \frac{1}{n}x^{^{\displaystyle \frac{1}{n}-1}}=\displaystyle \frac{1}{n}x^{^{\displaystyle \frac{1-n}{n}}} \end{aligned} $.

$\begin{array}{ll}\\ 5.&\textrm{Turunan ke}-n\: \: \textrm{dari}\: \: \: y=\displaystyle \frac{1}{x} \: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle n!\: .\: x^{-(n+1)} & \\ \textrm{b}.\quad \displaystyle (n+1)!\: .\: x^{-(n+1)}& \\ \textrm{c}.\quad \displaystyle (-1)^{n}\: n!\: .\: x^{-(n+1)}\\ \textrm{d}.\quad (-1)^{n+1}\: n!\: .\: x^{-(n+1)}\\ \textrm{e}.\quad (-1)^{n+1}\: (n+1)!\: .\: x^{-(n+1)} \end{array}\\ \end{array}$.

Jawaban: $\begin{array}{ll}&\textbf{Jawab}:\textbf{c}\\ &\begin{array}{|c|c|l|}\hline \textrm{Fungsi}&y=\displaystyle \frac{1}{x}&\quad y=x^{-1}\\\hline y'=\displaystyle \frac{dy}{dx}&-x^{-2}&(-1)^{1}.1!.x^{-(1+1)}\\\hline y''=\displaystyle \frac{d^{2}y}{dx^{2}}&2x^{-3}&(-1)^{2}.2!.x^{-(2+1)}\\\hline y'''=\displaystyle \frac{d^{3}y}{dx^{3}}&-6x^{-4}&(-1)^{3}.3!.x^{-(3+1)}\\\hline y^{IV}=\displaystyle \frac{d^{4}y}{dx^{4}}&24x^{-5}&(-1)^{4}.4!.x^{-(4+1)}\\\hline y^{V}=\displaystyle \frac{d^{5}y}{dx^{5}}&\cdots &\quad \cdots \\\hline \vdots &\vdots &\quad \vdots \\\hline y^{n}=\displaystyle \frac{d^{n}y}{dx^{n}}&\cdots &(-1)^{n}.n!.x^{-(n+1)}\\\hline \end{array} \end{array}$.

TURUNAN FUNGSI

 Berikut Kumpulan Materi Arsip Lama

• materi turunan fungsi aljabar
• bentuk umum turunan aljabar
• sifat turunan pertama dan aturan rantai
• aturan rantai pada turunan pertama dan kedua
• penggunaan turunan fungsi
• menyelesaikan masalah terkait keekstriman fungsi
• contoh soal 1, contoh 2, contoh 3, contoh 4

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12, contoh 13.

CONTOH SOAL 9 VISUALISASI BARISAN DAN DERET

$\begin{array}{ll}\\ 17.\quad&\textrm{Perhatikan hubungan berikut}\\ &\qquad\qquad\qquad\begin{aligned}1=&1^{\displaystyle 2}\\ 2+3+4=&3^{\displaystyle 2}\\ 3+4+5+6+7=&5^{2}\\ 4+5+6+7+8+9+10=&7^{\displaystyle 2}\\ \vdots\qquad &\vdots \end{aligned}\\ &\textrm{Tuliskan pola yang ada pada hubungan di atas dan buktikan} \end{array}$.

Bukti: $\begin{aligned}&\textrm{Pola Umum}\\ &\:\:\textrm{Baris ke}-n\:(n=1,2,3,4,...)\\ &\circ \quad \textrm{ruas kanan berupa}:(2n-1)^{\displaystyle 2}\\ &\circ \quad \textrm{ruas kiri berupa penjumlahan dari}\:\:(2n-1)\:\: \textrm{bilangan bulat berurutan}\\ &\qquad\textrm{yang dimulai dari}\:\:n\\ &\qquad n+(n+1)+(n+2)+\cdots +(3n-2)=(2n-1)^{\displaystyle 2}\\ &\qquad \text{Catatan}:U_{\displaystyle n}=n+\: (2n-1-1).1=3n-2\\ &\qquad \textrm{Bentuk umum}:\sum_{_{\displaystyle k=0}}^{^{\displaystyle 2n-2}}(n+k)=(2n-1)^{\displaystyle 2}\\ &\qquad \bullet \quad k=0\longrightarrow U_{\displaystyle 1}=n+0=n\\ &\qquad \bullet \quad k=1\longrightarrow U_{\displaystyle 2}=n+1\\ &\qquad \bullet \quad k=2\longrightarrow U_{\displaystyle 3}=n+2\\ &\qquad \bullet \quad k=3\longrightarrow U_{\displaystyle 4}=n+3\\ &\qquad\qquad \vdots \\&\qquad \bullet \quad k=2n-2\longrightarrow U_{\displaystyle n}=n+(2n-2)=3n-2\\ &\textrm{Pembuktian Pola}\\ &\quad \circ \quad \textrm{Suku pertama}=a=n\\ &\quad \circ \quad \textrm{Beda}=b=1\\ &\quad \circ \quad \textrm{Banyak suku}=2n-1\\ &\quad \circ \quad \textrm{Suku terakhir}=3n-2\\ &\quad \circ \quad \textrm{Rumus Jumlahderet aritmetika }=\displaystyle \frac{2n-1}{2}\left( a+U_{\displaystyle n} \right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:\:\: =\displaystyle \frac{2n-1}{2}\left( n+3n-2 \right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:\:\: =(2n-1)^{\displaystyle 2}\\\end{aligned}$.

.

Visualisasi Geometris $(2n-1)^2$

Nilai n: 3

3 + 4 + 5 + 6 + 7 = 5² = 25

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Persiapan Sumatif MA Futuhiyah Jeketro

Sumatif Matematika Kelas X

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• Alokasi Waktu: 90 Menit

• Cakupan: Pola Bilangan, Barisan & Deret Aritmetika/Geometri, Anuitas & Aplikasi.

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LATIHAN SOAL GOOGLE FORMULIR BARISAN DAN DERET

CONTOH SOAL 8 BARISAN DAN DERET

 

$\begin{aligned}15.\quad &\textrm{Di antara bilangan}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{terdapat tak hingga banyak}\\ &\textrm{bilangan pecah. Tentukan 999 bilangan pecah di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\\&\textrm{sehingga selisih antara bilangan pecah berikutnya dengan bilangan}\\ &\textrm{pecah sebelumnya konstan}\\ &(\textrm{Maksudnya: jika}\:\: x_{\displaystyle 1},x_{\displaystyle 2},\cdots ,x_{\displaystyle 999}\:\:\textrm{bilangan pecah yang dimaksud}\\ &\textrm{maka},\:\:x_{\displaystyle 2}-x_{\displaystyle 1}=x_{\displaystyle 3}-x_{\displaystyle 2}=\cdots =x_{\displaystyle 999}-x_{\displaystyle 998}) \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Untuk menyisipkan 999 bilangan pecahan di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{dengan}\\ &\textrm{selisih tetap, maka kita sama saja membentuk sebuah barisan aritmetika}\\&\textrm{dengan} \\ &\circ \quad \text{Suku pertama}:\:\: a=U_{\displaystyle 1}=\displaystyle \frac{1}{5}\\ &\circ \quad \text{Banyak sisipan}:\:\:999\:\: \:\textrm{bilangan pecahan}\\ &\circ \quad \text{Total suku}:k=n+2=999+2=1001\\ &\circ \quad \text{Suku terakhir}:\:\: U_{\displaystyle 1001}=\displaystyle \frac{1}{4}\\ &\circ \quad \text{Beda antar dua suku}:\:\: U_{\displaystyle n}=a+(n-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow \displaystyle \frac{1}{5}+(1001-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow 1000b=\displaystyle \frac{1}{4}-\frac{1}{5}=\displaystyle \frac{1}{20}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow b=\displaystyle \frac{1}{20.000}\\ &\textrm{Sehingga bilangan pecahan yang dimaksud adalah}:\\&\circ \quad U_{\displaystyle 1}=\displaystyle \frac{1}{5}=\frac{4000}{20000}\\ &\circ \quad U_{\displaystyle 2}=U_{\displaystyle 1}+b=x_{\displaystyle 1}=\frac{4000}{20000}+\frac{1}{20000}=\displaystyle \frac{4001}{20000}\\ &\circ \quad U_{\displaystyle 3}=U_{\displaystyle 2}+b=x_{\displaystyle 2}=\frac{4001}{20000}+\frac{1}{20000}=\displaystyle \frac{4002}{20000}\\ &\qquad\qquad\qquad\vdots\\ &\circ \quad U_{\displaystyle 1000}=U_{\displaystyle 999}+b=x_{\displaystyle 999}=\frac{4998}{20000}+\frac{1}{20000}=\displaystyle \frac{4999}{20000} \end{aligned}$

$\begin{aligned}16.\quad&(\textbf{LM UGM ke-32 Th 2021 Tk.SMA})\\&\textrm{Diberikan barisan}\:(a_{\displaystyle n})\: \textrm{yang memenuhi}\\ &\qquad\qquad\qquad a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}\\ &\textrm{Jika}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:\: \textrm{dan}\:\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{nilai dari}\:\: 2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}\:\: \textrm{adalah}\:....\\ &\text{a}.\quad \displaystyle \frac{1}{4}\qquad \qquad \qquad\qquad\qquad\:\: \text{d}.\quad \displaystyle 2\\ &\text{b}.\quad \displaystyle \frac{1}{2}\qquad\qquad \text{c}.\quad 1\qquad\qquad \text{e}.\quad 4\end{aligned}$

Jawaban: $\begin{aligned}&\textbf{Jawab: d}\\&\textrm{Perhatikan bahwa}\\ &a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{2\sqrt{a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2}}}\\ &\Leftrightarrow a_{\displaystyle n+1}^{\displaystyle 2}=\displaystyle \frac{a_{\displaystyle n}^{\displaystyle 2}a_{\displaystyle n-1}^{\displaystyle 2}}{\displaystyle 4\left( a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2} \right)}\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle n+1}^{\displaystyle 2}}=\displaystyle \frac{4}{a_{\displaystyle n}^{\displaystyle 2}}-\frac{4}{a_{\displaystyle n-1}^{\displaystyle 2}}\\ &\textrm{dengan}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:,\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{maka}\:\: a_{\displaystyle 1}^{\displaystyle 2}=\displaystyle \frac{1}{2}\:\: \textrm{dan}\:\: a_{\displaystyle 2}^{\displaystyle 2}=\displaystyle \frac{1}{4}\\ &\textrm{Misalkan saja}\:\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\: \textrm{maka}\:\: b_{\displaystyle n+1}=4\left( b_{\displaystyle n}-b_{\displaystyle n-1} \right)\\ &\textrm{dengan}\:\: b_{\displaystyle 1}=2,\: b_{\displaystyle 2}=4,\: \textrm{maka akan diperoleh}:\\ &\circ \quad b_{\displaystyle 3}=4(4-2)=4.2=8\\ &\circ \quad b_{\displaystyle 4}=4(8-4)=4.4=16\\ &\textrm{Tampak bahwa}\:\: b_{\displaystyle n}=2^{\displaystyle n}\: \textrm{dan}\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\\ &\textrm{maka}\:\: 2^{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}}\Leftrightarrow a_{\displaystyle n}=\displaystyle 2^{\displaystyle -\frac{n}{2}}.\\ &\textrm{Karena}\:\: a_{\displaystyle n}\gt 0,\:\: \textrm{maka}\:\: \prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=\prod_{\displaystyle n=1}^{\displaystyle 100}2^{\displaystyle -\frac{1}{2}n}\\ &=2^{\displaystyle -\frac{1}{2}(1+2+3+4+\cdots +99+100)}=2^{\displaystyle -\frac{1}{2}\left( \frac{100\times 101}{2} \right)}\\ &=2^{\displaystyle -2525}\\ &\textrm{Akibatnya adalah}:\\ &2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=2^{\displaystyle 2526}\times 2^{\displaystyle -2525}=2^{\displaystyle 1}=2 \end{aligned}$

CONTOH SOAL 7 BARISAN DAN DERET

 

$\begin{array}{ll}\\ 13.&\textrm{Syarat untuk deret geometri tak hingga }\\ &\textrm{dengan suku pertama}\: \: a\: \: \textrm{konvergen dengan }\\ &\textrm{jumlah 2 adalah}\: ....\:.\\ &\textrm{a}.\quad -2< a< 0\\ &\textrm{b}.\quad -4< a< 0\\ &\textrm{c}.\quad 0< a< 2\\ &\textrm{d}.\quad 0< a< 4\\ &\textrm{e}.\quad -4< a< 4\\ \end{array}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\\&\textrm{Diketahui bahwa}\: \: S_{\infty }=2,\: \: \textrm{dengan}\\ &S_{\infty }=\displaystyle \frac{a}{1-r}\Leftrightarrow 1-r=\displaystyle \frac{a}{S_{\infty }}\Leftrightarrow r=1-\displaystyle \frac{a}{S_{\infty }}\\ &\Leftrightarrow -1< 1-\displaystyle \frac{a}{S_{\infty }}< 1\Leftrightarrow -2< -\displaystyle \frac{a}{S_{\infty }}< 0\\ &\Leftrightarrow 0< \displaystyle \frac{a}{S_{\infty }}< 2\Leftrightarrow \Leftrightarrow 0< \displaystyle \frac{a}{2}< 2\\ & \Leftrightarrow 0< a<4 \end{aligned}$

$\begin{aligned}14.\quad &\textrm{Didefinisikan}\quad t_{\displaystyle n}=\displaystyle \frac{t_{\displaystyle n-1}-1}{t_{\displaystyle n-1}+1}\quad \textrm{untuk}\quad n\ge n\quad \textrm{dan}\quad t_{\displaystyle 1}=2\\ &\textrm{Berapakah nilai}\quad t_{\displaystyle 2026}? \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Untuk menentukan}\quad t_{\displaystyle 2026}\quad \textrm{kita dapat menghitung beberapa suku}\\ &\textrm{pertama dari}\quad t_{\displaystyle n}\quad \textrm{untuk menentukan pola keberulangannya, yaitu}:\\ &\textrm{Diketahui}\quad t_{\displaystyle 1}=2\quad \textrm{dan relasi rekursif}\quad t_{\displaystyle n}=\displaystyle \frac{t_{\displaystyle n-1}-1}{t_{\displaystyle n-1}+1}\\ &\circ \quad \text{Suku kedua}\:(n=2)\Rightarrow t_{\displaystyle 2}=\displaystyle \frac{t_{1}-1}{t_{1}+1}=\displaystyle \frac{2-1}{2+1}=\frac{1}{3}\\ &\circ \quad \text{Suku ketiga}\:(n=3)\Rightarrow t_{\displaystyle 3}=\displaystyle \frac{t_{2}-1}{t_{2}+1}=\displaystyle \frac{\left( \displaystyle \frac{1}{3} \right)-1}{\left( \displaystyle \frac{1}{3} \right)+1}=-\frac{1}{2}\\ &\circ \quad \text{Suku keempat}\:(n=4)\Rightarrow t_{\displaystyle 4}=\displaystyle \frac{t_{3}-1}{t_{3}+1}=\displaystyle \frac{\left( \displaystyle -\frac{1}{2} \right)-1}{\left( \displaystyle -\frac{1}{2} \right)+1}=-3\\ &\circ \quad \text{Suku kelima}\:(n=5)\Rightarrow t_{\displaystyle 5}=\displaystyle \frac{t_{4}-1}{t_{4}+1}=\displaystyle \frac{-3-1}{-3+1}=2\\ &\textrm{Karena}\quad t_{\displaystyle 5}=t_{\displaystyle 1}=2,\quad \textrm{maka nilai-nilai suku akan berulang setiap}\\ &\textrm{4 periode dengan}:\\ &\qquad\qquad\qquad t_{\displaystyle n}=\begin{cases} 2&\textrm{untuk}\quad n\equiv 1\:(\textrm{mod 4}) \\\displaystyle \frac{1}{3}&\textrm{untuk}\quad n\equiv 2\:(\textrm{mod 4})\\ \displaystyle -\frac{1}{2}&\textrm{untuk}\quad n\equiv 3\:(\textrm{mod 4})\\ -3&\textrm{untuk}\quad n\equiv 0\:(\textrm{mod 4}) \end{cases}\\ &\textrm{Untuk}\quad n=2026=4\times506+2\Rightarrow 2026\equiv 2\: (\textrm{mod 4})\\ &\textrm{Jadi, nilai}\quad t_{\displaystyle 2026}=t_{\displaystyle 2}=\displaystyle \frac{1}{3} \end{aligned}$

CONTOH SOAL 6 BARISAN DAN DERET

 

$\begin{aligned}11.\quad &\textrm{Diberikan}\\ &\quad A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\cdots\\ &\textrm{dan}\\ &\quad B=1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\displaystyle \frac{1}{9^{\displaystyle 4}}+\cdots\\ &\textrm{Nyatakan}\quad \displaystyle \frac{A}{B}\quad \textrm{sebagai pecahan} \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Perhatikan bahwa}\\ &A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots\\ &\textrm{Selanjutnya deret di ataskita bagi menjadi dua bagian, yaitu}:\\&A=\underset{\textrm{suku-suku genap}}{\underbrace{\left( \displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{8^{\displaystyle 4}}+\cdots \right)}}+\underset{\textrm{suku-suku ganjil}}{\underbrace{\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)}}\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}\left( 1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\cdots \right)+\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}A+B\Leftrightarrow A=\displaystyle \frac{1}{16}A+B\Leftrightarrow \displaystyle \frac{15}{16}A=B\Leftrightarrow \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15}\\ &\textrm{Jadi, perbandingan}\quad \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15} \end{aligned}$

$\begin{array}{ll}\\ 12.&\textbf{UM UGM}\\ &\textrm{Jumlah deret geometri tak hingga adalah 6}\\ & \textrm{Jika tiap suku dikuadratkan, maka jumlahnya}\\ &\textrm{adalah}\: \: 4\: .\: \textrm{Suku pertama deret ini adalah}\: ....\\ &\textrm{a}.\quad \displaystyle \frac{2}{5}\: \: \qquad\qquad\qquad\qquad\quad\:   \textrm{d}.\quad \displaystyle \frac{5}{6}\\ &\textrm{b}.\quad \displaystyle \frac{3}{5}\qquad\qquad \textrm{c}.\quad \displaystyle \frac{4}{5}\qquad\quad \textrm{e}.\quad \displaystyle \frac{6}{5}\\ \end{array}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\\ &\textrm{DG}=\textrm{Deret Geometri}\\ &a+ar+ar^{2}+\cdots =S_{\infty }=\displaystyle \frac{a}{1-r}=6\\ &\Leftrightarrow a=6(1-r)=6-6r\: ............(1)\\ &\textrm{Saat dikuadratkan masing-masing sukunya}\\ &a^{2}+a^{2}r^{2}+a^{2}r^{4}+\cdots =S_{\infty }=\displaystyle \frac{a^{2}}{1-r^{2}}=4\\ &\Leftrightarrow a^{2}=4(1-r^{2})=4-4r^{2}\: .......(2)\\ &\textrm{Substitusi (1) ke (2), maka} \\ &a^{2}=a^{2}\\ &\Leftrightarrow (6-6r)^{2}=4-4r^{2}\\ &\Leftrightarrow 36-72r+36r^{2}=4-4r^{2}\\ &\Leftrightarrow 40r^{2}-72r+32=0\\ &\Leftrightarrow (5r-4)(r-1)=0\\ &\Leftrightarrow r=\displaystyle \frac{4}{5}\: (memenuhi)\: \: \textbf{atau}\: \: r=1\: (tidak)\\ &\textrm{Selanjutnya kita tentukan nilai}\: \: a,\\ &a=6-6\left ( \displaystyle \frac{4}{5} \right )=6\left ( \displaystyle \frac{1}{5} \right )=\displaystyle \frac{6}{5} \end{aligned}$