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:

Updated On: Oct 4, 2024
  • -10
  • 10
  • 8
  • -8
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The Correct Option is D

Solution and Explanation

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

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

The correct option is (D): -8

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