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If $\mathrm{f(x)= \begin{cases}x+a, & x \leq 0 \\ |x-4|, & x>0\end{cases}}$ and $\mathrm{g(x)= \begin{cases}x+1 & , x<0 \\ (x-4)^{2}+b & , \quad x \geqslant 0\end{cases}}$ are continuous on $\mathrm{R},$ then $\mathrm{(g\! \circ\! f)\, (2)+(f\! \circ\! g)\, (-2)}$ is equal to:
Option: 1
$-10$

Option: 2
$10$

Option: 3
$8$

Option: 4
$-8$

Answers (1)

best_answer

$\mathrm{f(x)=\left\{\begin{array}{l} x+a ; x \leq 0 \\ |x-4| ; x>0 \end{array}\right.}$

$\mathrm{g(x)= \begin{cases}x+1 & x<0 \\ (x-4)^{2}+b & x \geq 0\end{cases}}$

For continuity $\mathrm{a=4}$ and $\mathrm{b=-15}$

$\mathrm{g(f(2))+f(g(-2)) }\\$

$\mathrm{=g(2)+f(-1)=-8}$

Hence correct option is 4

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