SBMPTN Zone : Fungsi Komposisi

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Tipe:

Standar SBMPTN HOTS

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
   

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

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

Jadi, \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx=72\ln2-3. JAWAB: A

No.

Jika tabel berikut menyatakan hasil fungsi f dan g

x	0	1	2	3
f(x)	1	3	1	-1
g(x)	2	0	1	2

maka nilai {(f\circ g\circ f)(0)+(g\circ f\circ g)(1)=}

1. 4
2. 3
   
3. 2
   

4. 1
5. 0
   

$$\begin{array}{rl}(f\circ g\circ f)(0)+(g\circ f\circ g)(1)& =f(g(f(0)))+g(f(g(1)))\\ & =f(g(1))+g(f(0))\\ & =f(0)+g(1)\\ & =1+0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$$

Jadi, {(f\circ g\circ f)(0)+(g\circ f\circ g)(1)=1}. JAWAB: D

No.

Diketahui fungsi {f(x) = ax + 6}, dengan a\neq 0 dan {\left(g\circ f\right)(x) = x + \dfrac6a}, maka nilai dari g(6a) =

1. 10
2. 9
3. 8

4. 7
5. 6

SKMP Juli 2020

$$\begin{array}{rl}(g\circ f)(x)& =x+{\displaystyle \frac{6}{a}}\\ g(f(x))& ={\displaystyle \frac{ax}{a}}+{\displaystyle \frac{6}{a}}\\ g(ax+6)& ={\displaystyle \frac{ax+6}{a}}\\ g(x)& ={\displaystyle \frac{x}{a}}\\ 6(6a)& ={\displaystyle \frac{6a}{a}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$$

Jadi, g(6a) ==6. JAWAB: E

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