SBMPTN Zone : Fungsi Komposisi (Composite Function)

Berikut ini adalah kumpulan soal mengenai Fungsi Komposisi tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Diketahui fungsi ${f(x) = ax + 6}$, dengan ${a\neq0}$ dan ${\left(g\circ f\right)(x) = x + \dfrac6a}$, maka nilai dari $g(6a) =$

1. $10$
2. $9$
3. $8$

4. $7$
5. $6$

Ganesha Operation

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

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

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

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No. 2

Diketahui fungsi ${f(x)=5x+3}$ dan ${g(x)=x^2+ax+b}$. Jika ${\left(g\circ f\right)(1)=53}$ dan ${\left(g\circ f\right)(0)=8}$, maka nilai ${a+b}$ adalah

1. $2$
2. $3$
3. $4$

4. $5$
5. $6$

$\begin{aligned}\left(g\circ f\right)(1)&=53\\g\left(f(1)\right)&=53\\g\left(5(1)+3\right)&=53\\g(8)&=53\\8^2+a(8)+b&=53\\64+8a+b&=53\\8a+b&=-11\end{aligned}$

$\begin{aligned}\left(g\circ f\right)(1)&=53\\g\left(f(0)\right)&=8\\g\left(5(0)+3\right)&=8\\g(3)&=8\\3^2+a(3)+b&=8\\9+3a+b&=8\\3a+b&=-1\end{aligned}$

$\begin{aligned}8a+b&=-11\\3a+b&=-1\qquad-\\\hline5a&=-10\\a&=-2\end{aligned}$

$\begin{aligned}3a+b&=-1\\3(-2)+b&=-1\\-6+b&=-1\\b&=5\end{aligned}$

$\begin{aligned}a+b&=-2+5\\&=\boxed{\boxed{3}}\end{aligned}$

Jadi, $a+b=3$.
JAWAB: B

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