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Q.
If $f(x)=\begin{cases} x+a, & x \leq 0 \\ |x-4|, & x>0\end{cases}$ and $g(x)= \begin{cases}x+1 & , x<0 \\ (x-4)^2+b, & x \geq 0\end{cases}$ are continuous on $R$, then $(gof) (2)+(f o g)(-2)$ is equal to :
JEE MainJEE Main 2022Relations and Functions - Part 2
Solution:
$f(x)= \begin{cases}x+a ; x \leq 0 \\|x-4| ; x>0\end{cases} ; g(x)= \begin{cases} x+1 ; x<0 \\(x-4)^2+b ; x \geq 0\end{cases}$
For continuity $a=4$ and $b=-15$
$g ( f (2))+ f ( g (-2)) $
$ = g (2)+ f (-1)=-8$