SBMPTN Zone : Fungsi

Berikut ini adalah kumpulan soal mengenai Fungsi. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Diberikan fungsi f memenuhi persamaan 2f(x+2)+f(-x)=x-3 untuk setiap bilangan bulat x. Nilai f(2) adalah ....

2. 
   
3. 
   

5. 
   

Ganesha Operation

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\\[-12pt] 3f(2)&=-1\\ f(2)&=\boxed{\boxed{-\dfrac13}} \end{aligned}

Jadi, f(2)=-\dfrac13. JAWAB: B

No.

f\left(x^2+3ax+1\right)=2x-1 dan f(5)=3. Hitung nilai a

Grup Facebook

\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)=3x^2 + 2x-1, maka nilai dari 2f(3)\cdot g(2)-f^2(3)-g^2(2) adalah ....

2. 
   
3. 
   

5. 
   

Learncy.net

\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)=a^x, 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)
   

4. f(x)+f(y)=f(xy)
5. f(x)+f(y)=f(x+y)
   

SPMB '02 (Regional 1)

\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)=x^2 untuk semua nilai real x, maka f(x) sama dengan

2. 
   
3. 
   

5. 
   

UMB '08 Kode 270

\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\\[-12pt] 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\ln2-3
2. 36\ln 3-2
   
3. 36\ln 2-6
   

4. 36\ln 2 - 3
5. 72\ln 3-2
   

Grup Telegram

\begin{aligned} (f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[10pt] f(g(x))&=\dfrac{12x+6}{4x-10}\\[10pt] f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[10pt] f(x)&=\dfrac{3x+39}{x+1}\\[10pt] f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[10pt] f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[10pt] &=\dfrac{-x+1+39}{x-4}\\[10pt] &=\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

4. \dfrac52
5. \dfrac32

Grup Telegram

\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\neq5}. Nilai dari {f(3) + f(1)} adalah ....

1. -2
2. 8
   

3. 13
4. 18
   

Grup Telegram

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

4. 2
5. 0

Grup Telegram

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\qquad-\\\hline\\[-12pt] 3f(1)&=5 \end{aligned}

Jadi, 3f(1)=5. JAWAB: A

No.

Jika fungsi {f\left(x^2 + 3x + 5\right) = {^3\log}\left(20x^2 + 3x + 4\right)}, dengan {x\geq0}, maka nilai f(9) =

1. 3
2. 4
   
3. 5
   

4. 6
5. 7
   

Grup Telegram

\begin{aligned} x^2 + 3x + 5&=9\\ x^2+3x-4&=0\\ (x+4)(x-1)&=0 \end{aligned}
x=-4 (TM) atau x=\boxed{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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