Jawaban:
a + 2b = 2
Pembahasan
Komposisi Fungsi
[tex]\large\text{$\begin{aligned}&f(x)=\frac{1}{x+a}\,,\quad g(x)=x^2+b\\\\&(f\circ g)(1)=\frac{1}{2}\\&\iff\frac{1}{2}=\frac{1}{g(1)+a}\\&\iff\frac{1}{2}=\frac{1}{1+b+a}\\&\iff2=1+b+a\\&\iff1+a=2-b\qquad....(i)\\&\iff1+b=2-a\qquad....(ii)\\\\&\iff a+b=2-1=1\\&\iff a+2b=1+b\\&(ii)\,\!\to a+2b=2-a\qquad....(\star)\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&(g\circ f)(1)=2\\&\iff2=\left(f(1)\right)^2+b\\&\iff2=\left(\frac{1}{1+a}\right)^2+b\\&\iff2=\frac{1}{(1+a)^2}+b\\&\iff2=\frac{1+b(1+a)^2}{(1+a)^2}\\&\iff2(1+a)^2=1+b(1+a)^2\\&\iff2(1+a)^2-b(1+a)^2=1\\&\iff(2-b)(1+a)^2=1\\&(i)\to\:(1+a)(1+a)^2=1\\&\iff(1+a)^3=1\\&\iff1+a=\sqrt[3]{1}=1\\&\iff a=\bf0\\\\&\textsf{Subs. nilai $a$ ke ruas kanan pada $(\star)$}\\&a+2b=2-0\\&\therefore\ \boxed{\ \bf a+2b=2\ }\end{aligned}$}[/tex]
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