Aspire Faculty ID #13051 · Topic: JEE Main 2022 (26 July Morning Shift) · Just now
JEE Main 2022 (26 July Morning Shift)

If $f(x)= \begin{cases} x+a, & x\le 0\\ |x-4|, & x>0 \end{cases} \quad\text{and}\quad g(x)= \begin{cases} x+1, & x<0\\ (x-4)^{2}+b, & x\ge 0 \end{cases}$ are continuous on $\mathbb{R}$, then $(g\circ f)(2)+(f\circ g)(-2)$ is equal to:

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