Answer:
Option A
Explanation:
Since , g(x) is differentiable $\Rightarrow$ g(x) must be continuous.
$\therefore$ $g(x)=\begin{cases}k\sqrt{x+1}, & 0 \leq x \leq3\\mx+2, &3<x \leq 5\end{cases}$
At x=3, RHL=3m+2
and at x=3, LHL=2k
$\therefore$ 2k=m+2 .........(i)
Also,
$g'(x)=\begin{cases}\frac{k}{2\sqrt{k+1}}, & 0\leq x < 3\\m, & 3<x \leq5 \end{cases}$
$\therefore$ $L\left\{g'(3)\right\}=\frac{k}{4}$ and $R\left\{g'(3)\right\}=m$
$\Rightarrow$ $\frac{k}{4}=m $ i.e, k=4m
On , solving Eqs(i) and (ii) , we get
$k=\frac{8}{5},m=\frac{2}{5}$
$\Rightarrow$ k+m=2
