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

$$\begin{array}{rl}f(n)& =f(n-1)+{\displaystyle \frac{n}{2018}}\\ & =f(n-2)+{\displaystyle \frac{n-1}{2018}}+{\displaystyle \frac{n}{2018}}\\ & =f(n-3)+{\displaystyle \frac{n-2}{2018}}+{\displaystyle \frac{n-1}{2018}}+{\displaystyle \frac{n}{2018}}\\ & =\cdots \\ & =f(0)+{\displaystyle \frac{1+2+\cdots +n}{2018}}\\ & ={\displaystyle \frac{2017}{2}}+{\displaystyle \frac{{\displaystyle \frac{1}{2}}n(n+1)}{2018}}\\ & ={\displaystyle \frac{2017}{2}}+{\displaystyle \frac{n(n+1)}{2\cdot 2018}}\\ f(2018)& ={\displaystyle \frac{2017}{2}}+{\displaystyle \frac{2018(2018+1)}{2\cdot 2018}}\\ & ={\displaystyle \frac{2017}{2}}+{\displaystyle \frac{2019}{2}}\\ & =2018\end{array}$$

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
   

$$\begin{array}{rl}f(f(-4))& =f(-f(4))\\ & =f(-(-7))\\ & =-f(-7)\\ & =-5\end{array}$$

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

$$\begin{array}{rl}f\left({\displaystyle \frac{1}{x}}\right)& ={\displaystyle \frac{{\displaystyle \frac{1}{x}}}{1-{\displaystyle \frac{1}{x}}}}\cdot {\displaystyle \frac{x}{x}}\\ & ={\displaystyle \frac{1}{x-1}}\end{array}$$

$$\begin{array}{rl}f(x)+f\left({\displaystyle \frac{1}{x}}\right)& ={\displaystyle \frac{x}{1-x}}+{\displaystyle \frac{1}{x-1}}\\ & ={\displaystyle \frac{x}{1-x}}-{\displaystyle \frac{1}{1-x}}\\ & ={\displaystyle \frac{x-1}{1-x}}\\ & =-1\end{array}$$

$$\begin{array}{rl}f(2016)+f(2015)+\cdots +f(3)+f(2)+f\left({\displaystyle \frac{1}{2}}\right)+f\left({\displaystyle \frac{1}{3}}\right)+\cdots +f\left({\displaystyle \frac{1}{2015}}\right)+f\left({\displaystyle \frac{1}{2016}}\right)& =2015\cdot (-1)\\ & =-2015\end{array}$$

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

$$\begin{array}{rl}f(1-x)& ={\displaystyle \frac{1}{{2021}^{1-x}+\sqrt{2021}}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{2021}{{2021}^x}}+\sqrt{2021}}}\cdot {\displaystyle \frac{{2021}^x}{{2021}^x}}\\ & ={\displaystyle \frac{{2021}^x}{2021+{2021}^x\sqrt{2021}}}\end{array}$$

$$\begin{array}{rl}f(x)+f(1-x)& ={\displaystyle \frac{1}{{2021}^x+\sqrt{2021}}}\cdot {\displaystyle \frac{\sqrt{2021}}{\sqrt{2021}}}+{\displaystyle \frac{{2021}^x}{2021+{2021}^x\sqrt{2021}}}\\ & ={\displaystyle \frac{\sqrt{2021}}{{2021}^x\sqrt{2021}+2021}}+{\displaystyle \frac{{2021}^x}{{2021}^x\sqrt{2021}+2021}}\\ & ={\displaystyle \frac{\sqrt{2021}+{2021}^x}{{2021}^x\sqrt{2021}+2021}}\cdot {\displaystyle \frac{\sqrt{2021}\cdot \sqrt{2021}}{2021}}\\ & ={\displaystyle \frac{2021+{2021}^x\sqrt{2021}}{2021+{2021}^x\sqrt{2021}}}\cdot {\displaystyle \frac{\sqrt{2021}}{2021}}\\ & ={\displaystyle \frac{\sqrt{2021}}{2021}}\end{array}$$

$$\begin{array}{rl}{\displaystyle \sum _{x=-2020}^{2021}f(x)}& =f(-2020)+f(-2019)+\cdots +f(2020)+f(2021)\\ & =(f(-2020)+f(2021))+(f(-2019)+f(2020))+\cdots +(f(0)+f(1))\\ & =2021\cdot {\displaystyle \frac{\sqrt{2021}}{2021}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{2021}}}}}\end{array}$$

Jadi, {\displaystyle\sum_{x=-2020}^{2021}f(x)=\sqrt{2021}}. JAWAB: D