HOTS Zone : Fungsi

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Tipe:

Standar SBMPTN HOTS

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

\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, nilai 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
   

4. -8
5. -9
   

\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 : \mathbb{R}\rightarrow\mathbb{R} yang memenuhi f(2019x) = x. Nilai dari f(1) adalah

SUMBER

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\neq-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
   

3. -4.030
4. -4.028
   

SUMBER

\begin{aligned} f\left(\dfrac1x\right)&=\dfrac{\dfrac1x}{1-\dfrac1x}\cdot\dfrac{x}x\\[8pt] &=\dfrac1{x-1} \end{aligned}

\begin{aligned} f(x)+f\left(\dfrac1x\right)&=\dfrac{x}{1-x}+\dfrac1{x-1}\\[8pt] &=\dfrac{x}{1-x}-\dfrac1{1-x}\\[8pt] &=\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 JAWABAN

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

4. \sqrt{2021}
5. 2021

C S C POSI 2021

\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