Let f : R rightarrow R and g : R rightarrow R | vedclass

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(B) The composite function $(g \circ f)(x)$ is defined as:$g(f(x)) = \begin{cases} f(x)+1, & f(x) < 0 \ (f(x)-1)^2+b, & f(x) \geq 0 \end{cases}$Subst

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$1$

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(B) The composite function $(g \circ f)(x)$ is defined as:$g(f(x)) = \begin{cases} f(x)+1, & f(x) Substituting $f(x)$ into the definition:For $x For $x \geq 0$,$f(x) = |x-1|$. Thus,$g(f(x)) = \begin{cases} |x-1|+1, & |x-1| Since $|x-1| \geq 0$ for all $x$,the case $|x-1| $(g \circ f)(x) = \begin{cases} x+a+1, & x For continuity at $x = -a$: $\lim_{x 	o -a^-} (x+a+1) = \lim_{x 	o -a^+} ((x+a-1)^2+b) \implies 1 = (-1)^2 + b \implies b = 0$.For continuity at $x = 0$: $\lim_{x 	o 0^-} ((x+a-1)^2+b) = \lim_{x 	o 0^+} ((|x-1|-1)^2+b) \implies (a-1)^2 + b = (|0-1|-1)^2 + b \implies (a-1)^2 = 0 \implies a = 1$.Thus,$a+b = 1+0 = 1$.

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