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

The function $$g(x)$$ is defined piecewise and is differentiable. For differentiability, the function must be continuous at every point, including at $$x = 3$$ where the definition changes. Additionally, the derivatives from the left and right must be equal at $$x = 3$$.

First, ensure continuity at $$x = 3$$. The left-hand limit (LHL) as $$x$$ approaches 3 from the left uses the first piece: $$g(x) = k\sqrt{x+1}$$. So, LHL = $$k\sqrt{3+1} = k\sqrt{4} = 2k$$. The right-hand limit (RHL) as $$x$$ approaches 3 from the right uses the second piece: $$g(x) = mx + 2$$. So, RHL = $$m \cdot 3 + 2 = 3m + 2$$. The value of the function at $$x = 3$$ is $$g(3) = k\sqrt{3+1} = 2k$$. For continuity, LHL = RHL = $$g(3)$$, so:

$$2k = 3m + 2 \quad \text{(equation 1)}$$

Next, ensure differentiability at $$x = 3$$. The left-hand derivative (LHD) at $$x = 3$$ comes from the first piece. Differentiate $$g(x) = k\sqrt{x+1} = k(x+1)^{1/2}$$:

$$g'(x) = k \cdot \frac{1}{2} (x+1)^{-1/2} \cdot 1 = \frac{k}{2\sqrt{x+1}}$$

So, LHD at $$x = 3$$ is:

$$\frac{k}{2\sqrt{3+1}} = \frac{k}{2 \cdot 2} = \frac{k}{4}$$

The right-hand derivative (RHD) at $$x = 3$$ comes from the second piece. Differentiate $$g(x) = mx + 2$$:

$$g'(x) = m$$

So, RHD at $$x = 3$$ is $$m$$. For differentiability, LHD = RHD:

$$\frac{k}{4} = m \quad \text{(equation 2)}$$

Now solve the system of equations. Substitute equation 2 into equation 1:

$$2k = 3\left(\frac{k}{4}\right) + 2$$

Simplify:

$$2k = \frac{3k}{4} + 2$$

Multiply both sides by 4 to eliminate the denominator:

$$4 \cdot 2k = 4 \cdot \left(\frac{3k}{4} + 2\right)$$

$$8k = 3k + 8$$

Subtract $$3k$$ from both sides:

$$8k - 3k = 8$$

$$5k = 8$$

Divide both sides by 5:

$$k = \frac{8}{5}$$

Now substitute $$k = \frac{8}{5}$$ into equation 2 to find $$m$$:

$$m = \frac{k}{4} = \frac{\frac{8}{5}}{4} = \frac{8}{5} \cdot \frac{1}{4} = \frac{8}{20} = \frac{2}{5}$$

Finally, compute $$k + m$$:

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

Hence, the correct answer is Option B.