If the function g(x) = begin{cases} ksqrt{x+1 | vedclass

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(B) Since $g(x)$ is differentiable,it must be continuous at $x=3$.For continuity at $x=3$,$\lim_{x 	o 3^-} g(x) = \lim_{x 	o 3^+} g(x) = g(3)$.$\lim

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$2$

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(B) Since $g(x)$ is differentiable,it must be continuous at $x=3$.For continuity at $x=3$,$\lim_{x 	o 3^-} g(x) = \lim_{x 	o 3^+} g(x) = g(3)$.$\lim_{x 	o 3^-} k\sqrt{x+1} = k\sqrt{3+1} = 2k$.$\lim_{x 	o 3^+} (mx+2) = 3m+2$.Thus,$2k = 3m+2$  --- $(i)$.For differentiability at $x=3$,$g'(3^-) = g'(3^+)$.$g'(x) = \begin{cases} \frac{k}{2\sqrt{x+1}}, & 0 $g'(3^-) = \frac{k}{2\sqrt{3+1}} = \frac{k}{4}$.$g'(3^+) = m$.So,$\frac{k}{4} = m \Rightarrow k = 4m$ --- $(ii)$.Substituting $(ii)$ into $(i)$: $2(4m) = 3m+2 \Rightarrow 8m = 3m+2 \Rightarrow 5m = 2 \Rightarrow m = \frac{2}{5}$.Then $k = 4(\frac{2}{5}) = \frac{8}{5}$.Therefore,$k+m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$.

If the function $g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 < x \le 5 \end{cases}$ is differentiable,then the value of $k+m$ is:

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