HOTS Zone : Fungsi

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Berikut ini adalah kumpulan soal mengenai Fungsi. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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No.

Jika f adalah fungsi yang memenuhi {f(n)=f(n-1)+\dfrac{n}{2018}} untuk setiap n bilangan asli dan {f(0)=\dfrac{2017}2}, maka nilai f(2018) adalah ....
OMVN 2018
ALTERNATIF PENYELESAIAN
\begin{aligned} f(n)&=f(n-1)+\dfrac{n}{2018}\\ &=f(n-2)+\dfrac{n-1}{2018}+\dfrac{n}{2018}\\ &=f(n-3)+\dfrac{n-2}{2018}+\dfrac{n-1}{2018}+\dfrac{n}{2018}\\ &=\cdots\\ &=f(0)+\dfrac{1+2+\cdots+n}{2018}\\ &=\dfrac{2017}2+\dfrac{\dfrac12n(n+1)}{2018}\\ &=\dfrac{2017}2+\dfrac{n(n+1)}{2\cdot2018}\\ f(2018)&=\dfrac{2017}2+\dfrac{2018(2018+1)}{2\cdot2018}\\ &=\dfrac{2017}2+\dfrac{2019}2\\ &=2018 \end{aligned}
Jadi, f(2018) = 2018.

No.

Diketahui suatu fungsi f bersifat f(−x) = −f(x) untuk setiap bilangan real x. Jika f(4) = −7 dan f(−7) = 5, maka nilai f(f(4)) =
  1. −5
  2. −6
  3. −7
  1. −8
  2. −9
ALTERNATIF PENYELESAIAN
\begin{aligned} f(f(-4))&=f(-f(4))\\ &=f(-(-7))\\ &=-f(-7)\\ &=-5 \end{aligned}
Jadi, f(f(4)) = −5.
JAWAB: A

No.

Diberikan fungsi f : ℝ⟶ℝ yang memenuhi f(2019x) = x. Nilai dari f(1) adalah
ALTERNATIF PENYELESAIAN
kita cari nilai x sedemikian sehingga 2019x = 1, didapat x=\dfrac1{2019}.
Jadi, f(1)=\dfrac1{2019}.

No.

Diketahui fungsi bilangan real f(x)=\dfrac{x}{1-x}, untuk x ≠ −1. Nilai dari
f(2016)+f(2015)+\cdots+f(3)+f(2)+f\left(\dfrac12\right)+f\left(\dfrac13\right)+\cdots+f\left(\dfrac1{2015}\right)+f\left(\dfrac1{2016}\right)
adalah
  1. −4.034
  2. −4.032
  1. −4.030
  2. −4.028
ALTERNATIF PENYELESAIAN
\begin{aligned} f\left(\dfrac1x\right)&=\dfrac{\dfrac1x}{1-\dfrac1x}\cdot\dfrac{x}x\\[3.5pt] &=\dfrac1{x-1} \end{aligned} \begin{aligned} f(x)+f\left(\dfrac1x\right)&=\dfrac{x}{1-x}+\dfrac1{x-1}\\[3.5pt] &=\dfrac{x}{1-x}-\dfrac1{1-x}\\[3.5pt] &=\dfrac{x-1}{1-x}\\ &=-1 \end{aligned} \begin{aligned} f(2016)+f(2015)+\cdots+f(3)+f(2)+f\left(\dfrac12\right)+f\left(\dfrac13\right)+\cdots+f\left(\dfrac1{2015}\right)+f\left(\dfrac1{2016}\right)&=2015\cdot(-1)\\ &=-2015 \end{aligned}
Jadi, f(2016)+f(2015)+\cdots+f(3)+f(2)+f\left(\dfrac12\right)+f\left(\dfrac13\right)+\cdots+f\left(\dfrac1{2015}\right)+f\left(\dfrac1{2016}\right)=-2015.
JAWAB: TIDAK ADA

No.

Diberikan fungs! f(x)=\dfrac1{2021^x+\sqrt{2021}} untuk setiap bilangan real x. Nllai dari {\displaystyle\sum_{x=-2020}^{2021}f(x)=} ....
  1. \dfrac1{2021}
  2. \dfrac1{2021}\sqrt{2021}
  3. 1
  1. \sqrt{2021}
  2. 2021
ALTERNATIF PENYELESAIAN
\begin{aligned} f(1-x)&=\dfrac1{2021^{1-x}+\sqrt{2021}}\\ &=\dfrac1{\dfrac{2021}{2021^x}+\sqrt{2021}}{\color{red}\cdot\dfrac{2021^x}{2021^x}}\\ &=\dfrac{2021^x}{2021+2021^x\sqrt{2021}}\\ \end{aligned} \begin{aligned} f(x)+f(1-x)&=\dfrac1{2021^x+\sqrt{2021}}{\color{red}\cdot\dfrac{\sqrt{2021}}{\sqrt{2021}}}+\dfrac{2021^x}{2021+2021^x\sqrt{2021}}\\ &=\dfrac{\sqrt{2021}}{2021^x\sqrt{2021}+2021}+\dfrac{2021^x}{2021^x\sqrt{2021}+2021}\\ &=\dfrac{\sqrt{2021}+2021^x}{2021^x\sqrt{2021}+2021}{\color{red}\cdot\dfrac{\sqrt{2021}\cdot\sqrt{2021}}{2021}}\\ &=\dfrac{2021+2021^x\sqrt{2021}}{2021+2021^x\sqrt{2021}}\cdot\dfrac{\sqrt{2021}}{2021}\\ &=\dfrac{\sqrt{2021}}{2021} \end{aligned} \begin{aligned} \displaystyle\sum_{x=-2020}^{2021}f(x)&=f(-2020)+f(-2019)+\cdots+f(2020)+f(2021)\\ &=\left(f(-2020)+f(2021)\right)+\left(f(-2019)+f(2020)\right)+\cdots+\left(f(0)+f(1)\right)\\ &=2021\cdot\dfrac{\sqrt{2021}}{2021}\\ &=\boxed{\boxed{\sqrt{2021}}} \end{aligned}
Jadi, {\displaystyle\sum_{x=-2020}^{2021}f(x)=\sqrt{2021}}.
JAWAB: D

No.

Diketahui fungsi f :ℕ2 ⟶ ℕ yang memenuhi kedua persamaan
f (x, 1) = x dan f (x + 1, y) = f (y, x) + 1
untuk setiap bilangan asli x dan y. Tentukan banyaknya pasangan terurut bilangan asli (a, b) yang memenuhi f (a, b) = 2023.
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x+1,y)&=f(y,x)+1\\ &=f(x,y-1)+2\\ &=f(y-1,x-1)+3 &=\cdots\\ &=f(y-m,x-m)+2m+1 \end{aligned} Jika xm = 1 ⟶ x = m + 1 \begin{aligned} f(m+1+1,y)&=f(y-m,1)+2m+1\\ f(m+2,y)&=y-m+2m+1\\ &=y+m+1\\ f(m,y)&=y+(m-2)+1\\ &=m+y-1\\ f(x,y)&=x+y-1 \end{aligned} \begin{aligned} f(a,b)&=2023\\ a+b-1&=2023\\ a+b&=2024 \end{aligned} Banyaknya pasangan (a, b) ada 2024 − 1 = 2023
Jadi, banyaknya pasangan terurut bilangan asli (a, b) yang memenuhi f (a, b) = 2023 ada 2023.

No.

Diketahui ℝ menyatakan himpunan semua bilangan real. Diberikan fungsi f : ℝ ⟶ ℝ yang memenuhi
f (f (x)) = 4x − 3 dan f (f (f (x))) = ax − 7
untuk setiap bilangan real x. Nilai dari f (10) adalah ....
  1. 15
  2. 23
  1. 19
  2. 27
ALTERNATIF PENYELESAIAN
\(\begin{aligned} f\left(f\left(f(x)\right)\right)&=ax-7\\ f\left(4x-3\right)&=ax-7\\ f\left(4x-3\right)&=\dfrac{a}4(4x-3)+\dfrac{3a}4-7\\ f(x)&=\dfrac{a}4x+\dfrac{3a}4-7\\ f\left(f(x)\right)&=f\left(\dfrac{a}4x+\dfrac{3a}4-7\right)\\ 4x-3&=\dfrac{a}4\left(\dfrac{a}4x+\dfrac{3a}4-7\right)+\dfrac{3a}4-7\\ 4x-3&=\dfrac{a^2}{16}x+\dfrac{3a^2}{16}-\dfrac{7a}4+\dfrac{3a}4-7\\ 4x-3&=\dfrac{a^2}{16}x+\dfrac{3a^2}{16}-a-7\\ \end{aligned}\)

\(\begin{aligned} \dfrac{a^2}{16}&=4\\ a^2&=64 \end{aligned}\)

\(\begin{aligned} \dfrac{3a^2}{16}-a-7&=-3\\[4pt] \dfrac{3(64)}{16}-a-7&=-3\\[4pt] 12-a-7&=-3\\ a&=8 \end{aligned}\)

\(\begin{aligned} f(10)&=\dfrac84(10)+\dfrac{3(8)}4-7\\ &=20+6-7\\ &=\boxed{\boxed{19}} \end{aligned}\)
Jadi, f (10) = 19.
JAWAB: C

No.

Fungsi f(x) menyatakan hasil penjumlahan semua pembagi positif dari bilangan asli x selain x itu sendiri, dengan kesepakatan f(1) = 1. Contohnya, f(20) = 1 + 2 + 4 + 5 + 10 = 22 dan f(19) = 1. Nilai f(⋯f(f(f(2019)))⋯), dengan fungsi f diterapkan sebanyak 2019 kali, adalah ....
  1. 2019
  2. 3
  3. 1
  1. 673
  2. 91022019
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \underbrace{f(\cdots f(f(f(2019)))\cdots)}_{2019\ f}&=\underbrace{f(\cdots f(f(f(1+3+673)))\cdots)}_{2018\ f}\\[4pt] &=\underbrace{f(\cdots f(f(f(677)))\cdots)}_{2018\ f}\\[4pt] &=\underbrace{f(\cdots f(f(f(1)))\cdots)}_{2017\ f}\\[4pt] &=\underbrace{f(\cdots f(f(f(1)))\cdots)}_{2016\ f}\\ &=\cdots\\ &=f(1)\\ &=\boxed{\boxed{1}} \end{aligned}\)
Jadi, f(⋯f(f(f(2019)))⋯) = 1.
JAWAB: C

No.

Diberikan fungsi \(f(n)=1+\displaystyle\sum_{k=1}^n\left\lfloor{^3\negthinspace\log k}\right\rfloor\) yang terdefinisi pada bilangan asli n. Jika p adalah suatu bilangan sehingga f(p) = 2023, nilai p yang memenuhi berada pada interval ....
  1. 440 ≤ p < 450
  2. 450 ≤ p < 460
  3. 460 ≤ p < 470
  1. 470 ≤ p < 480
  2. 480 ≤ p < 490
ALTERNATIF PENYELESAIAN
Misal a adalah bilangan bulat tak-negatif. Untuk 3ak ≤ 3a + 1 − 1,
3log k⌋ = a

\(\begin{aligned} \displaystyle\sum_{k\ =\ 3^a}^{3^{a+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=a\left(3^{a+1}-1-3^a+1\right)\\ &=a\left(3\cdot3^a-3^a\right)\\ &=2a3^a \end{aligned}\)

\(\begin{aligned} \displaystyle\sum_{k\ =\ 1\ =\ 3^0}^{n\ =\ 3^{b+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2\displaystyle\sum_{a=0}^{b}a3^a\\ &=2\left(\dfrac{3-(b+1)3^{b+1}+b3^{b+2}}{(1-3)^2}\right)\\[4pt] &=2\left(\dfrac{3-(b+1)3^{b+1}+3b3^{b+1}}{4}\right)\\[4pt] &=\dfrac{3+(2b-1)3^{b+1}}2 \end{aligned}\)

Kita car nilai b sehingga \(\dfrac{3+(2b-1)3^{b+1}}2\) mendekati atau sama dengan 2022
\(\begin{aligned} \dfrac{3+(2b-1)3^{b+1}}2&\leq2022\\[4pt] 3+(2b-1)3^{b+1}&\leq4044\\ (2b-1)3^{b+1}&\leq4041\\ (2b-1)3^b&\leq1347 \end{aligned}\)
Jika b = 5,
(2(5) − 1)35 = 2187 > 1347

Jika b = 4,
(2(4) − 1)34 = 567 < 1347

didapat b = 4

\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)

\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
Jadi, nilai p yang memenuhi berada pada interval 470 ≤ p < 480.
JAWAB: D

No.

Jika diketahui f(1) = 2022 dan f(1) + f(2) + f(3) + ⋯ + f(n) = n2f(n).
Nilai f(2022) adalah ...
ALTERNATIF PENYELESAIAN
\(\begin{aligned} f(1)+f(2)+\cdots+f(n-1)+f(n)&=n^2f(n)\\ (n-1)^2f(n-1)+f(n)&=n^2f(n)\\ (n-1)^2f(n-1)&=\left(n^2-1\right)f(n)\\ (n-1)f(n-1)&=(n+1)f(n)\\ f(n)&=\dfrac{n-1}{n+1}f(n-1)\\[4pt] &=\dfrac{n-1}{n+1}\cdot\dfrac{n-2}nf(n-2)\\ &=\cdots\\ &=\dfrac{n-1}{n+1}\cdot\dfrac{n-2}n\cdots\dfrac24f(2)\\[4pt] &=\dfrac{n-1}{n+1}\cdot\dfrac{n-2}n\cdots\dfrac24\cdot\dfrac13f(1)\\[4pt] &=\dfrac{2\cdot1}{(n+1)n}\cdot2022\\[4pt] &=\dfrac{2\cdot2022}{n(n+1)}\\[4pt] f(2022)&=\dfrac{2\cdot2022}{2022(2022+1)}\\ &=\boxed{\boxed{\dfrac2{2023}}} \end{aligned}\)
Jadi, \(f(2022)=\dfrac2{2023}\).


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