HOTS Zone : Fungsi [2]
Table of Contents
Berikut ini adalah kumpulan soal mengenai Fungsi. Jika ingin bertanya soal, silahkan gabung ke grup Matematika Idhamdaz.
Tipe:
No.
JikaALTERNATIF PENYELESAIAN
Jika x = 0,
\(\begin{aligned} f(0\cdot y+1)&=f(0)f(y)-f(y)-0+2\\ f(1)&=1\cdot f(y)-f(y)+2\\ f(1)&=2 \end{aligned}\)Jika y = 0,
\(\begin{aligned} f(x\cdot0+1)&=f(x)f(0)-f(0)-x+2\\ f(1)&=f(x)\cdot1-1-x+2\\ 2&=f(x)-1-x+2\\ f(x)&=x+1\\ f(2023)&=2023+1\\ &=\color{blue}\boxed{\boxed{\color{black}2024}} \end{aligned}\)Jadi, f(2023) = 2024.
No.
Fungsi- 71
- 81
- 91
- 101
- 111
ALTERNATIF PENYELESAIAN
Jika x = y = 1,
\(\begin{aligned} f(1)f(1)-f(1\cdot1)&=\dfrac11+\dfrac11\\ \left(f(1)\right)^2-f(1)&=2\\ \left(f(1)\right)^2-f(1)-2&=0\\ \left(f(1)+1\right)\left(f(1)-2\right)&=0 \end{aligned}\)
101 − 20 = 81
\(\begin{aligned} f(1)f(1)-f(1\cdot1)&=\dfrac11+\dfrac11\\ \left(f(1)\right)^2-f(1)&=2\\ \left(f(1)\right)^2-f(1)-2&=0\\ \left(f(1)+1\right)\left(f(1)-2\right)&=0 \end{aligned}\)
Untuk f(1) = −1
Jika x = 10 dan y = 1,
\(\begin{aligned} f(10)f(1)-f(10\cdot1)&=\dfrac{10}1+\dfrac1{10}\\[4pt] f(10)(-1)-f(10)&=10+\dfrac1{10}\\[4pt] -2f(10)&=\dfrac{101}{10}\\[4pt] f(10)&=-\dfrac{101}{20} \end{aligned}\)Untuk f(1) = 2
Jika x = 10 dan y = 1,
\(\begin{aligned} f(10)f(1)-f(10\cdot1)&=\dfrac{10}1+\dfrac1{10}\\[4pt] f(10)(2)-f(10)&=10+\dfrac1{10}\\[4pt] f(10)&=\dfrac{101}{10} \end{aligned}\)
101 − 20 = 81
Jadi, m − n = 81 .
JAWAB: B
JAWAB: B
No.
Jika fungsi f : ℝ ⟶ ℝ memenuhi- 2
- 3
- 5
- 7
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)+f(x+4)&=f(x+2)+f(x+6)\\
f(x)+f(x+4)&=f(x+2)+f(x+4+2)\\
f(x)&=f(x+2)
\end{aligned}\)
Jadi, periode dari f adalah 2.
JAWAB: A
JAWAB: A
No.
$f(x)=\dfrac{16^x}{16^x+4}$, maka nilai dari $f\left(\dfrac1{2024}\right)+f\left(\dfrac2{2024}\right)+\cdots+f\left(\dfrac{2022}{2024}\right)+f\left(\dfrac{2023}{2024}\right)$ adalah ....- 1
- 1011
- $\dfrac{2023}2$
- 1012
- $\dfrac{2025}2$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(1-x)&=\dfrac{16^{1-x}}{16^{1-x}+4}\\[4pt]
&=\dfrac{\dfrac{16}{16^x}}{\dfrac{16}{16^x}+4}{\color{red}\cdot\dfrac{16^x}{16^x}}\\[4pt]
&=\dfrac{16}{16+4\cdot16^x}\\[4pt]
&=\dfrac{4\cdot4}{4\left(4+16^x\right)}\\[4pt]
&=\dfrac{4}{4+16^x}\\[4pt]
&=\dfrac{4}{16^x+4}
\end{aligned}\)
$f(x)+f(1-x)=\dfrac{16^x}{16^x+4}+\dfrac{4}{16^x+4}=1$
\(\begin{aligned} f\left(\dfrac1{2024}\right)+f\left(\dfrac2{2024}\right)+\cdots+f\left(\dfrac{2022}{2024}\right)+f\left(\dfrac{2023}{2024}\right)&=\left(f\left(\dfrac1{2024}\right)+f\left(\dfrac{2023}{2024}\right)\right)+\left(f\left(\dfrac2{2024}\right)+f\left(\dfrac{2022}{2024}\right)\right)+\cdots+\left(f\left(\dfrac{1011}{2024}\right)+f\left(\dfrac{1013}{2024}\right)\right)+f\left(\dfrac{1012}{2024}\right)\\[4pt] &=1011+f\left(\dfrac12\right)\\[4pt] &=1011+\dfrac{16^{\frac12}}{16^{\frac12}+4}\\[4pt] &=1011+\dfrac{4}{4+4}\\[4pt] &=1011+\dfrac12\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{2023}2}} \end{aligned}\)
$f(x)+f(1-x)=\dfrac{16^x}{16^x+4}+\dfrac{4}{16^x+4}=1$
\(\begin{aligned} f\left(\dfrac1{2024}\right)+f\left(\dfrac2{2024}\right)+\cdots+f\left(\dfrac{2022}{2024}\right)+f\left(\dfrac{2023}{2024}\right)&=\left(f\left(\dfrac1{2024}\right)+f\left(\dfrac{2023}{2024}\right)\right)+\left(f\left(\dfrac2{2024}\right)+f\left(\dfrac{2022}{2024}\right)\right)+\cdots+\left(f\left(\dfrac{1011}{2024}\right)+f\left(\dfrac{1013}{2024}\right)\right)+f\left(\dfrac{1012}{2024}\right)\\[4pt] &=1011+f\left(\dfrac12\right)\\[4pt] &=1011+\dfrac{16^{\frac12}}{16^{\frac12}+4}\\[4pt] &=1011+\dfrac{4}{4+4}\\[4pt] &=1011+\dfrac12\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{2023}2}} \end{aligned}\)
Jadi, $f\left(\dfrac1{2024}\right)+f\left(\dfrac2{2024}\right)+\cdots+f\left(\dfrac{2022}{2024}\right)+f\left(\dfrac{2023}{2024}\right)=\dfrac{2023}2$.
JAWAB: C
JAWAB: C
No.
Diketahui $f(x)=x+\dfrac1{2x+\dfrac1{2x+\dfrac1{2x+\cdots}}}$. Hitung nilai dari $f(99)\cdot f'(99)$ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)&=x+\dfrac1{2x+\dfrac1{2x+\dfrac1{2x+\cdots}}}\\
&=x+\dfrac1{x+f(x)}\\[4pt]
f(x)-x&=\dfrac1{f(x)+x}\\[4pt]
f^2(x)-x^2&=1\\
f^2(x)&=x^2+1\\
2f(x)\cdot f'(x)&=2x\\
f(x)\cdot f'(x)&=x\\
f(99)\cdot f'(99)&=\color{blue}\boxed{\boxed{\color{black}99}}
\end{aligned}\)
Jadi, $f(99)\cdot f'(99)=99$.
No.
Diberikan fungsi f dengan- 32
- 33
- 34
- 35
- 36
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
a-f^{-1}(a)&=9\\
a-9&=f^{-1}(a)\\
f(a-9)&=a\\
(a-9)^3+a-9+1&=a\\
(a-9)^3&=8\\
a-9&=2\\
a&=11
\end{aligned}\)
\(\begin{aligned} f^{-1}(a-8)&=f^{-1}(11-8)\\ c&=f^{-1}(3)\\ f(c)&=3\\ c^3+c+1&=3\\ c&=1 \end{aligned}\)
\(\begin{aligned} f(a-8)+f^{-1}(a-8)&=f(3)+f^{-1}(3)\\ &=3^3+3+1+1\\ &=\color{blue}\boxed{\boxed{\color{black}32}} \end{aligned}\)
\(\begin{aligned} f^{-1}(a-8)&=f^{-1}(11-8)\\ c&=f^{-1}(3)\\ f(c)&=3\\ c^3+c+1&=3\\ c&=1 \end{aligned}\)
\(\begin{aligned} f(a-8)+f^{-1}(a-8)&=f(3)+f^{-1}(3)\\ &=3^3+3+1+1\\ &=\color{blue}\boxed{\boxed{\color{black}32}} \end{aligned}\)
Jadi, f(a − 8) + f−1(a − 8) = 32 .
JAWAB: A
JAWAB: A
No.
Didefinisikan fungsi f yang memenuhi \[f(x,y,z)=\dfrac1x+\dfrac1y+\dfrac1z+\dfrac1{xy}+\dfrac1{yz}+\dfrac1{zx}+\dfrac1{xyz}\] untuk setiap bilangan real tak nol x, y, dan z. JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x,y,z)&=\dfrac1x+\dfrac1y+\dfrac1z+\dfrac1{xy}+\dfrac1{yz}+\dfrac1{zx}+\dfrac1{xyz}\\[4pt]
&=\left(\dfrac1x+1\right)\left(\dfrac1y+1\right)\left(\dfrac1z+1\right)-1\\[4pt]
f(2019,2020,2021)&=\left(\dfrac1{2019}+1\right)\left(\dfrac1{2020}+1\right)\left(\dfrac1{2021}+1\right)-1\\[4pt]
&=\left(\dfrac{2020}{2019}\right)\left(\dfrac{2021}{2020}\right)\left(\dfrac{2022}{2021}\right)-1\\[4pt]
&=\dfrac{2022}{2019}-1\\[4pt]
&=\dfrac{3}{2019}\\[4pt]
&=\dfrac1{673}
\end{aligned}\)
1 + 673 =674
1 + 673 =
Jadi, a + b = 674.
No.
Untuk setiap bilangan asli n didefinisikan S(n) yang merupakan jumlahan digit-digit n dan- 0
- 1
- 3
- 6
- 2
ALTERNATIF PENYELESAIAN
Perhatikan bahwa
n mod 9 ≡ S(n) mod 9
n − S(n) mod 9 ≡ 0 mod 9
f(n) mod 9 ≡ 0 mod 9
$\underbrace{(f\circ f\circ f \circ \cdots \circ f)}_{sebanyak\ 2017}\left(2017^{2017}\right)\mod9\equiv0\mod9$
n mod 9 ≡ S(n) mod 9
n − S(n) mod 9 ≡ 0 mod 9
$\underbrace{(f\circ f\circ f \circ \cdots \circ f)}_{sebanyak\ 2017}\left(2017^{2017}\right)\mod9\equiv0\mod9$
Jadi, sisa pembagian dari $\underbrace{(f\circ f\circ f \circ \cdots \circ f)}_{sebanyak\ 2017}\left(2017^{2017}\right)$ oleh 9 adalah 0.
JAWAB: A
JAWAB: A
No.
Diketahui suatu fungsif(f(x + 1)) = x + 1
Jika
ALTERNATIF PENYELESAIAN
$\begin{aligned}
f(2024)&=2025\\
f\left(f(2024)\right)&=f(2025)\\
2024&=f(2025)
\end{aligned}$
$\begin{aligned} f(x)+x&=4049\\ f(x)&=4049-x\\ f(-1)&=4049-(-1)\\ &=\boxed{\boxed{4050}} \end{aligned}$
$\begin{aligned} f(x)+x&=4049\\ f(x)&=4049-x\\ f(-1)&=4049-(-1)\\ &=\boxed{\boxed{4050}} \end{aligned}$
Jadi, f(−1) = 4050.
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