HOTS Zone : Fungsi
Table of Contents

Tipe:
No.
Jika f adalah fungsi yang memenuhiALTERNATIF 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)) =- −5
- −6
- −7
- −8
- −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
JAWAB: A
No.
Diberikan fungsi f : ℝ⟶ℝ yang memenuhi f(2019x) = x. Nilai dari f(1) adalahALTERNATIF PENYELESAIAN
kita cari nilai x sedemikian sehingga 2019x = 1, didapat x=\dfrac1{2019} .
Jadi, f(1)=\dfrac1{2019} .
No.
Diketahui fungsi bilangan real- −4.034
- −4.032
- −4.030
- −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
JAWAB: TIDAK ADA
No.
Diberikan fungs!-
\dfrac1{2021} -
\dfrac1{2021}\sqrt{2021} - 1
-
\sqrt{2021} - 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
JAWAB: D
No.
Diketahui fungsiALTERNATIF 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 x − m = 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- 15
- 23
- 19
- 27
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f\left(f\left(f(x)\right)\right)&=ax-7\\
f\left(4x-3\right)&=ax-7\\
f\left(4x-3\right)&=\dfrac{a}4(4x-3)+\dfrac{3a}4-7\\
f(x)&=\dfrac{a}4x+\dfrac{3a}4-7\\
f\left(f(x)\right)&=f\left(\dfrac{a}4x+\dfrac{3a}4-7\right)\\
4x-3&=\dfrac{a}4\left(\dfrac{a}4x+\dfrac{3a}4-7\right)+\dfrac{3a}4-7\\
4x-3&=\dfrac{a^2}{16}x+\dfrac{3a^2}{16}-\dfrac{7a}4+\dfrac{3a}4-7\\
4x-3&=\dfrac{a^2}{16}x+\dfrac{3a^2}{16}-a-7\\
\end{aligned}\)
\(\begin{aligned}
\dfrac{a^2}{16}&=4\\
a^2&=64
\end{aligned}\)
\(\begin{aligned} \dfrac{3a^2}{16}-a-7&=-3\\[4pt] \dfrac{3(64)}{16}-a-7&=-3\\[4pt] 12-a-7&=-3\\ a&=8 \end{aligned}\)
\(\begin{aligned} f(10)&=\dfrac84(10)+\dfrac{3(8)}4-7\\ &=20+6-7\\ &=\boxed{\boxed{19}} \end{aligned}\)
\(\begin{aligned} \dfrac{3a^2}{16}-a-7&=-3\\[4pt] \dfrac{3(64)}{16}-a-7&=-3\\[4pt] 12-a-7&=-3\\ a&=8 \end{aligned}\)
\(\begin{aligned} f(10)&=\dfrac84(10)+\dfrac{3(8)}4-7\\ &=20+6-7\\ &=\boxed{\boxed{19}} \end{aligned}\)
Jadi, f (10) = 19.
JAWAB: C
JAWAB: C
No.
Fungsi- 2019
- 3
- 1
- 673
- 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
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- 440 ≤ p < 450
- 450 ≤ p < 460
- 460 ≤ p < 470
- 470 ≤ p < 480
- 480 ≤ p < 490
ALTERNATIF PENYELESAIAN
Misal a adalah bilangan bulat tak-negatif. Untuk 3a ≤ k ≤ 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
\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)
\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
⌊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
JAWAB: D
No.
Jika diketahuiNilai 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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