Given f(x)={\begin{matrix}x+a; & x\leq 0 \\ |x-4|; & x>0\end{matrix}and g(x)={\begin{matrix}x+1; & x<0 \\ {(x-4)}^{2}+b; & x\geq 0\end{matrix}
Given f(x)&g(x) are continuous on R
So, for continuity
f(0−)=f(0)=f(0+)
⇒a=∣0−4∣
⇒a=4
And g(0−)=g(0)=g(0+)
⇒1=(0−4)2+b
⇒b=1−16=−15
Then, g(f(2))+f(g(−2))
=g(2)+f(−1)
=(2−4)2−15+(−1+4)
=4−15+3=−8