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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.

Diberikan fungsi f memenuhi persamaan 2f(x + 2) + f(−x) = x − 3 untuk setiap bilangan bulat x. Nilai f(2) adalah ....
  1. \dfrac13
  2. -\dfrac13
  3. −3
  1. 3
  2. −2
Ganesha Operation
ALTERNATIF PENYELESAIAN
Untuk x = 0, \begin{aligned} 2f(0+2)+f(-0)&=0-3\\ 2f(2)+f(0)&=-3\\ 4f(2)+2f(0)&=-6 \end{aligned} Untuk x = −2, \begin{aligned} 2f(-2+2)+f(-(-2))&=-2-3\\ 2f(0)+f(2)&=-5\\ f(2)+2f(0)&=-5 \end{aligned}
\begin{aligned} 4f(2)+2f(0)&=-6\\ f(2)+2f(0)&=-5\qquad-\\\hline 3f(2)&=-1\\ f(2)&=\boxed{\boxed{-\dfrac13}} \end{aligned}
Jadi, f(2)=-\dfrac13.
JAWAB: B

No.

f(x2 + 3ax + 1) = 2x − 1 dan f(5) = 3. Hitung nilai a
Grup Facebook
ALTERNATIF PENYELESAIAN
\begin{aligned} 2x-1&=3\\ 2x&=4\\ x&=2 \end{aligned}
\begin{aligned} x^2+3ax+1&=5\\ 2^2+3a(2)+1&=5\\ 4+6a+1&=5\\ 6a+5&=5\\ 6a&=0\\ a&=0 \end{aligned}
Jadi, a = 0.

No.

Jika f(x) = 3h(x) + 4; g(x)=\dfrac{\sqrt{f(x)}}9; dan h(x) = 3x2 + 2x − 1, maka nilai dari 2f(3) ⋅ g(2) − f2(3) − g2(2) adalah ....
  1. -\dfrac{797449}{81}
  2. \dfrac{797449}{81}
  3. -\dfrac{893}9
  1. \dfrac{893749}9
  2. \dfrac{893749}{81}
Learncy.net
ALTERNATIF PENYELESAIAN
\begin{aligned} 2f(3)\cdot g(2)-f^2(3)-g^2(2)&=-\left(f(3)-g(2)\right)^2\\ &=-\left(3h(3)+4-\dfrac{\sqrt{f(2)}}9\right)^2\\ &=-\left(3\left(3(3)^2+2(3)-1\right)+4-\dfrac{\sqrt{3h(2)+4}}9\right)^2\\ &=-\left(3\left(27+6-1\right)+4-\dfrac{\sqrt{3\left(3(2)^2+2(2)-1\right)+4}}9\right)^2\\ &=-\left(3\left(32\right)+4-\dfrac{\sqrt{3\left(12+4-1\right)+4}}9\right)^2\\ &=-\left(96+4-\dfrac{\sqrt{3\left(15\right)+4}}9\right)^2\\ &=-\left(100-\dfrac{\sqrt{45+4}}9\right)^2\\ &=-\left(100-\dfrac{\sqrt{49}}9\right)^2\\ &=-\left(100-\dfrac79\right)^2\\ &=-\left(\dfrac{893}9\right)^2\\ &=-\dfrac{797449}{81} \end{aligned}
Jadi, 2f(3)\cdot g(2)-f^2(3)-g^2(2)=-\dfrac{797449}{81}.
JAWAB: A

No.

Jika f(x) = ax, maka untuk setiap x dan y berlaku
  1. f(x)f(y) = f(xy)
  2. f(x)f(y) = f(x + y)
  3. f(x)f(y) = f(x) + f(y)
  1. f(x) + f(y) = f(x)f(y)
  2. f(x) + f(y) = f(x + y)
SPMB '02 (Regional 1)
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)f(y)&=a^xa^y\\ &=a^{x+y}\\ &=f(x+y) \end{aligned}
Jadi, berlaku f(x)f(y) = f(x + y).
JAWAB: B

No.

Bila f(x) memenuhi 2f(x) + f(1 − x) = x2 untuk semua nilai real x, maka f(x) sama dengan
  1. \dfrac12x^2-\dfrac32x+\dfrac12
  2. \dfrac19x^2+\dfrac89x-\dfrac13
  3. \dfrac23x^2+\dfrac12x-\dfrac13
  1. \dfrac13x^2+\dfrac23x-\dfrac13
  2. \dfrac19x^2+x-\dfrac49
UMB '08 Kode 270
ALTERNATIF PENYELESAIAN
\begin{aligned} 2f(1-x)+f(1-(1-x))&=(1-x)^2\\ 2f(1-x)+f(1-1+x)&=1-2x+x^2\\ 2f(1-x)+f(x)&=x^2-2x+1\\ f(x)+2f(1-x)&=x^2-2x+1 \end{aligned} \begin{aligned} 2f(x)+f(1-x)&=x^2\qquad&\color{red}{\times 2}\\ f(x)+2f(1-x)&=x^2-2x+1 \end{aligned}
\begin{aligned} 4f(x)+2f(1-x)&=2x^2\\ f(x)+2f(1-x)&=x^2-2x+1&\color{red}{-}\\\hline 3f(x)&=x^2+2x-1\\ f(x)&=\dfrac13x^2+\dfrac23x-\dfrac13 \end{aligned}
Jadi, f(x)=\dfrac13x^2+\dfrac23x-\dfrac13.
JAWAB: D

No.

Jika (f\circ g)(x)=\dfrac{6x+3}{2x-5} dan g(x) = 4x − 11, maka hasil dari \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx adalah
  1. 72 ln 2 − 3
  2. 36 ln 3 − 2
  3. 36 ln 2 − 6
  1. 36 ln 2 − 3
  2. 72 ln 3 − 2
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} (f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[3.5pt] f(g(x))&=\dfrac{12x+6}{4x-10}\\[3.5pt] f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[3.5pt] f(x)&=\dfrac{3x+39}{x+1}\\[3.5pt] f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[3.5pt] f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[3.5pt] &=\dfrac{-x+1+39}{x-4}\\[3.5pt] &=\dfrac{-x+40}{x-4} \end{aligned}
\begin{aligned} \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx&=\displaystyle\intop_5^8\dfrac{\dfrac{-x+40}{x-4}}{4(3)-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{\dfrac{-x+4+36}{x-4}}{12-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{-1+\dfrac{36}{x-4}}1\ dx\\ &=\displaystyle\intop_5^8\left(-1+\dfrac{36}{x-4}\right)\ dx\\ &=\left[-x+36\ln|x-4|\right]_5^8\\ &=\left[-8+36\ln|8-4|\right]-\left[-5+36\ln|5-4|\right]\\ &=\left[-8+36\ln4\right]-\left[-5+36\ln1\right]\\ &=\left[-8+36\ln2^2\right]-\left[-5-42(0)\right]\\ &=\left[-8+72\ln2\right]-\left[-5\right]\\ &=-8+72\ln2+5\\ &=\boxed{\boxed{72\ln2-3}} \end{aligned}
Jadi, \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx=72\ln2-3.
JAWAB: A

No.

Jika f\left(\dfrac2{\sqrt{2x-1}}\right)=x dengan x\geq\dfrac12, maka f(1) =
  1. 4
  2. 2
  3. \dfrac12
  1. \dfrac52
  2. \dfrac32
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac2{\sqrt{2x-1}}&=1\\ \sqrt{2x-1}&=2\\ 2x-1&=4\\ 2x&=5\\ x&=\dfrac52 \end{aligned} f(1)=\boxed{\boxed{\dfrac52}}
Jadi, f(1)=\dfrac52.
JAWAB: D

No.

Fungsi f terdefinisi pada bilangan real kecuali 2 sehingga {f\left(\dfrac{2x}{x-5}\right)=2x-1}, x ≠ 5. Nilai dari f(3) + f(1) adalah ....
  1. -2
  2. 8
  1. 13
  2. 18
Grup Telegram
ALTERNATIF PENYELESAIAN

CARA 1

Misal \left(\dfrac{2x}{x-5}\right)=t \begin{aligned} 2x&=tx-5t\\ 2x-tx&=-5t\\ tx-2x&=5t\\ (t-2)x&=5t\\ x&=\dfrac{5t}{t-2} \end{aligned} \begin{aligned} f\left(\dfrac{2x}{x-5}\right)&=2x-1\\ f(t)&=2\left(\dfrac{5t}{t-2}\right)-1\\ &=\dfrac{10t}{t-2}-1\\ &=\dfrac{10t-(t-2)}{t-2}\\ &=\dfrac{10t-t+2}{t-2}\\ &=\dfrac{9t+2}{t-2} \end{aligned} \begin{aligned} f(3)+f(1)&=\dfrac{9(3)+2}{3-2}+\dfrac{9(1)+2}{1-2}\\ &=\dfrac{29}1+\dfrac{11}{-1}\\ &=29-11\\ &=\boxed{\boxed{18}} \end{aligned}

CARA 2

\begin{aligned} \dfrac{2x}{x-5}&=3\\ 2x&=3x-15\\ -x&=-15\\ x&=15 \end{aligned} \begin{aligned} f(3)&=2(15)-1\\ &=29 \end{aligned} \begin{aligned} \dfrac{2x}{x-5}&=1\\ 2x&=x-5\\ x&=-5 \end{aligned} \begin{aligned} f(1)&=2(-5)-1\\ &=-11 \end{aligned} \begin{aligned} f(3)+f(1)&=29+(-11)\\ &=\boxed{\boxed{18}} \end{aligned}
Jadi, f(3) + f(1) = 18.
JAWAB: D

No.

Diberikan fungsi f memenuhi persamaan 2f(−x) + f(x + 3) = x + 5 untuk setiap bilangan real x. Nilai 3f(1) adalah
  1. 5
  2. 3
  3. 4
  1. 2
  2. 0
Grup Telegram
ALTERNATIF PENYELESAIAN
Untuk x = −1, \begin{aligned} 2f(-(-1))+f(-1+3)&=-1+5\\ 2f(1)+f(2)&=4\\ 4f(1)+2f(2)&=8 \end{aligned} Untuk x = −2, \begin{aligned} 2f(-(-2))+f(-2+3)&=-2+5\\ 2f(2)+f(1)&=3\\ f(1)+2f(2)&=3 \end{aligned}
\begin{aligned} 4f(1)+2f(2)&=8\\ f(1)+2f(2)&=3&-\\\hline 3f(1)&=5 \end{aligned}
Jadi, 3f(1) = 5.
JAWAB: A

No.

Jika fungsi f(x2 + 3x + 5) = 3log(20x2 + 3x + 4), dengan x ≥ 0, maka nilai f(9) =
  1. 3
  2. 4
  3. 5
  1. 6
  2. 7
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} x^2 + 3x + 5&=9\\ x^2+3x-4&=0\\ (x+4)(x-1)&=0 \end{aligned} x = −4 (TM) atau x = 1
\begin{aligned} f(9)&={^3\log}\left(20(1)^2 + 3(1) + 4\right)\\ &={^3\log}\left(27\right)\\ &=\boxed{\boxed{3}} \end{aligned}
Jadi, f(9) = 3.
JAWAB: A


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