If the function $g\left(x\right)=\{\begin{array}{cc}k\sqrt{x+1}& 0\le x\le 3\\ mx+2& 3<x\le 5\end{array}$ is differentiable on $[0,5]$ then the value of $k+m$ is

 $g\left(x\right)=\{\begin{array}{cc}k\sqrt{x+1},& 0\le x3\\ mx+2,& 3<x\le 5\end{array}$
 $L(g\text{'}\left(3\right))=\underset{x\to 3^{-}}{\lim }\frac{k\sqrt{x+1}-2k}{x-3}=\underset{x\to 3^{-}}{\lim }k\left\{\frac{(x+1-4)}{(x-3)(\sqrt{x+1}+2)}\right\}=\frac{K}{4}$
 $R(g\text{'}\left(3\right))=\underset{x\to 3+}{\lim }\frac{mx+2-2k}{x-3}$
Since this limit exists, $3m+2-2k=0\Rightarrow 2k=3m+2\left(i\right)$   
$SoR(g\text{'}\left(3\right))=mbyL\text{'}Hospital\text{'}srule.$ 
Since $g\left(x\right)$ is differentiable, $k=4m\left(ii\right)$   
Solving (i)  and (ii), we get $m=\frac{2}{5},$ and $n=\frac{8}{5}\Rightarrow k+m=2$