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,
$$\begin{array}{rl}2f(0+2)+f(-0)& =0-3\\ 2f(2)+f(0)& =-3\\ 4f(2)+2f(0)& =-6\end{array}$$

Untuk x=-2,
$$\begin{array}{rl}2f(-2+2)+f(-(-2))& =-2-3\\ 2f(0)+f(2)& =-5\\ f(2)+2f(0)& =-5\end{array}$$

$$\begin{array}{rl}4f(2)+2f(0)& =-6\\ f(2)+2f(0)& =-5-\\ \\ 3f(2)& =-1\\ f(2)& =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{3}}}}}}\end{array}$$

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

$$\begin{array}{rl}2x-1& =3\\ 2x& =4\\ x& =2\end{array}$$

$$\begin{array}{rl}{x}^2+3ax+1& =5\\ 2^2+3a(2)+1& =5\\ 4+6a+1& =5\\ 6a+5& =5\\ 6a& =0\\ a& =0\end{array}$$

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

$$\begin{array}{rl}2f(3)\cdot g(2)-f^2(3)-g^2(2)& =-{(f(3)-g(2))}^2\\ & =-{(3h(3)+4-{\displaystyle \frac{\sqrt{f(2)}}{9}})}^2\\ & =-{(3(3(3{)}^2+2(3)-1)+4-{\displaystyle \frac{\sqrt{3h(2)+4}}{9}})}^2\\ & =-{(3(27+6-1)+4-{\displaystyle \frac{\sqrt{3(3(2{)}^2+2(2)-1)+4}}{9}})}^2\\ & =-{(3\left(32\right)+4-{\displaystyle \frac{\sqrt{3(12+4-1)+4}}{9}})}^2\\ & =-{(96+4-{\displaystyle \frac{\sqrt{3\left(15\right)+4}}{9}})}^2\\ & =-{(100-{\displaystyle \frac{\sqrt{45+4}}{9}})}^2\\ & =-{(100-{\displaystyle \frac{\sqrt{49}}{9}})}^2\\ & =-{(100-{\displaystyle \frac{7}{9}})}^2\\ & =-{\left({\displaystyle \frac{893}{9}}\right)}^2\\ & =-{\displaystyle \frac{797449}{81}}\end{array}$$

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)

$$\begin{array}{rl}f(x)f(y)& =a^x{a}^y\\ & =a^{x+y}\\ & =f(x+y)\end{array}$$

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

$$\begin{array}{rl}2f(1-x)+f(1-(1-x))& =(1-x{)}^2\\ 2f(1-x)+f(1-1+x)& =1-2x+x^2\\ 2f(1-x)+f(x)& =x^2-2x+1\\ f(x)+2f(1-x)& =x^2-2x+1\end{array}$$

$$\begin{array}{rlr}2f(x)+f(1-x)& =x^2& \times 2\\ f(x)+2f(1-x)& =x^2-2x+1\end{array}$$

$$\begin{array}{rl}4f(x)+2f(1-x)& =2x^2\\ f(x)+2f(1-x)& =x^2-2x+1& -\\ \\ 3f(x)& =x^2+2x-1\\ f(x)& ={\displaystyle \frac{1}{3}}{x}^2+{\displaystyle \frac{2}{3}}x-{\displaystyle \frac{1}{3}}\end{array}$$

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

$$\begin{array}{rl}(f\circ g)(x)& ={\displaystyle \frac{6x+3}{2x-5}}\\ f(g(x))& ={\displaystyle \frac{12x+6}{4x-10}}\\ f(4x-11)& ={\displaystyle \frac{3(4x-11)+39}{4x-11+1}}\\ f(x)& ={\displaystyle \frac{3x+39}{x+1}}\\ f^{-1}(x)& ={\displaystyle \frac{-x+39}{x-3}}\\ f^{-1}(x-1)& ={\displaystyle \frac{-(x-1)+39}{x-1-3}}\\ & ={\displaystyle \frac{-x+1+39}{x-4}}\\ & ={\displaystyle \frac{-x+40}{x-4}}\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{f^{-1}(x-1)}{g(3)}}dx}& ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{{\displaystyle \frac{-x+40}{x-4}}}{4(3)-11}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{{\displaystyle \frac{-x+4+36}{x-4}}}{12-11}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{-1+{\displaystyle \frac{36}{x-4}}}{1}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}(-1+{\displaystyle \frac{36}{x-4}})dx}\\ & ={[-x+36\ln |x-4|]}_5^8\\ & =[-8+36\ln |8-4|]-[-5+36\ln |5-4|]\\ & =[-8+36\ln 4]-[-5+36\ln 1]\\ & =[-8+36\ln 2^2]-[-5-42(0)]\\ & =[-8+72\ln 2]-[-5]\\ & =-8+72\ln 2+5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 72\ln 2-3}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \frac{2}{\sqrt{2x-1}}}& =1\\ \sqrt{2x-1}& =2\\ 2x-1& =4\\ 2x& =5\\ x& ={\displaystyle \frac{5}{2}}\end{array}$$

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
$$\begin{array}{rl}2x& =tx-5t\\ 2x-tx& =-5t\\ tx-2x& =5t\\ (t-2)x& =5t\\ x& ={\displaystyle \frac{5t}{t-2}}\end{array}$$

$$\begin{array}{rl}f\left({\displaystyle \frac{2x}{x-5}}\right)& =2x-1\\ f(t)& =2\left({\displaystyle \frac{5t}{t-2}}\right)-1\\ & ={\displaystyle \frac{10t}{t-2}}-1\\ & ={\displaystyle \frac{10t-(t-2)}{t-2}}\\ & ={\displaystyle \frac{10t-t+2}{t-2}}\\ & ={\displaystyle \frac{9t+2}{t-2}}\end{array}$$

$$\begin{array}{rl}f(3)+f(1)& ={\displaystyle \frac{9(3)+2}{3-2}}+{\displaystyle \frac{9(1)+2}{1-2}}\\ & ={\displaystyle \frac{29}{1}}+{\displaystyle \frac{11}{-1}}\\ & =29-11\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 18}}}}\end{array}$$

CARA 2

$$\begin{array}{rl}{\displaystyle \frac{2x}{x-5}}& =3\\ 2x& =3x-15\\ -x& =-15\\ x& =15\end{array}$$

$$\begin{array}{rl}f(3)& =2(15)-1\\ & =29\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{2x}{x-5}}& =1\\ 2x& =x-5\\ x& =-5\end{array}$$

$$\begin{array}{rl}f(1)& =2(-5)-1\\ & =-11\end{array}$$

$$\begin{array}{rl}f(3)+f(1)& =29+(-11)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 18}}}}\end{array}$$

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,
$$\begin{array}{rl}2f(-(-1))+f(-1+3)& =-1+5\\ 2f(1)+f(2)& =4\\ 4f(1)+2f(2)& =8\end{array}$$

Untuk x=-2,
$$\begin{array}{rl}2f(-(-2))+f(-2+3)& =-2+5\\ 2f(2)+f(1)& =3\\ f(1)+2f(2)& =3\end{array}$$

$$\begin{array}{rl}4f(1)+2f(2)& =8\\ f(1)+2f(2)& =3-\\ \\ 3f(1)& =5\end{array}$$

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

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

$$\begin{array}{rl}f(9)& ={}^3\log (20(1{)}^2+3(1)+4)\\ & ={}^3\log \left(27\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

Jadi, f(9) =3. JAWAB: A

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