1. For the piecewise function

to be differentiable, it must be both continuous and have equal left and right derivatives at the joining point $x=3$.

2. Continuity at $x=3$

From the first part, $g(3)=k\sqrt{3+1}=2k.$

From the second part, the right-hand value at $x=3$ would be $3m+2.$

So continuity gives $2k=3m+2. \qquad (1)$

3. Derivative matching at $x=3$

For $0\le x\le 3$, $g(x)=k\sqrt{x+1}=k(x+1)^{1/2}.$ So, $g'(x)=\frac{k}{2\sqrt{x+1}}.$ Hence left derivative at $x=3$ is $g'_-(3)=\frac{k}{2\sqrt{4}}=\frac{k}{4}.$

For $3<x\le 5$, $g(x)=mx+2,$ so $g'(x)=m.$ Thus right derivative at $x=3$ is $g'_+(3)=m.$

Differentiability gives $\frac{k}{4}=m. \qquad (2)$

4. Substitute (2) into (1): $2k=3\left(\frac{k}{4}\right)+2.$ Multiply by $4$: $8k=3k+8.$ So, $5k=8 \Rightarrow k=\frac{8}{5}.$

Then $m=\frac{k}{4}=\frac{8/5}{4}=\frac{2}{5}.$

5. Therefore, $k+m=\frac{8}{5}+\frac{2}{5}=\frac{10}{5}=2.$

6. Option check

• A: $\frac{10}{3}$ ❌
• B: $4$ ❌
• C: $2$ ✅
• D: $\frac{16}{5}$ ❌

So the correct answer is Option C.