JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
If $f(x)=\begin{cases}x+a, & x\le0\\ |x-4|, & x>0\end{cases}$ and $g(x)=\begin{cases}x+1, & x<0\\ (x-4)^2+b, & x\ge0\end{cases}$ are continuous on $\mathbf{R}$, then $(g\circ f)(2)+(f\circ g)(-2)$ is equal to :
Limits & Continuity
Solution
Continuity of $f$ at $0$: $a=|0-4|=4$. Continuity of $g$ at $0$: $1=(0-4)^2+b\Rightarrow b=-15$. $f(2)=|2-4|=2$, $g(2)=(2-4)^2-15=-11$, so $(g\circ f)(2)=-11$. $g(-2)=-1$, $f(-1)=-1+4=3$, so $(f\circ g)(-2)=3$. Sum $=-8$. Option (D).