Question

If $f(x)=\begin{cases} x+a, & x \leq 0 \\\ |x-4|, & x\gt0\end{cases}$ and
$g(x)= \begin{cases}x+1 & , x\lt0 \\\ (x-4)^2+b, & x \geq 0\end{cases}$
are continuous on $R$, then $(gof) (2)+(fog)(-2)$ is equal to:

• A. -10
• B. 10
• C. 8
• D. -8

Answer 1

Answer: (D)

-8

Answer 2

$f(x)=\begin{cases} x+a, & x \leq 0 \\\ |x-4|, & x\gt0\end{cases}$
$g(x)= \begin{cases}x+1 & , x\lt0 \\\ (x-4)^2+b, & x \geq 0\end{cases}$

For continuity $a = 4$ and $b = –15$
$g(f(2)) + f(g(-2)) = g(2) + f(-1) = -8$

The correct option is (D): -8