If the function
`g(x)={{:(,k sqrt(x+1),0 le x le3),(,mx+2,3lt xle5):}` is differentiable, then the value of k+m is

Options

A. 2
B. `(16)/(5)`
C. `(10)/(3)`
D. 4

Answer

To solve the problem, we need to ensure that the function \( g(x) \) is both continuous and differentiable at the point \( x = 3 \). The function is defined as:

\[ g(x) = \begin{cases} k \sqrt{x+1} & \text{for } 0 \leq x \leq 3 \\ mx + 2 & \text{for } 3 < x \leq 5 \end{cases} \]

### Step 1: Ensure Continuity at \( x = 3 \)

For \( g(x) \) to be continuous at \( x = 3 \), the left-hand limit as \( x \) approaches 3 must equal the right-hand limit at \( x = 3 \):

\[ k \sqrt{3 + 1} = m \cdot 3 + 2 \]

Calculating \( \sqrt{3 + 1} \):

\[ k \cdot 2 = 3m + 2 \]

This simplifies to:

\[ 2k = 3m + 2 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 3 \)

Next, we differentiate both parts of the function and set them equal at \( x = 3 \).

1. Differentiate \( g(x) = k \sqrt{x + 1} \):

\[ g'(x) = k \cdot \frac{1}{2\sqrt{x + 1}} \]

2. Differentiate \( g(x) = mx + 2 \):

\[ g'(x) = m \]

Setting these equal at \( x = 3 \):

\[ k \cdot \frac{1}{2\sqrt{3 + 1}} = m \]

Calculating \( \sqrt{3 + 1} \):

\[ k \cdot \frac{1}{2 \cdot 2} = m \]

This simplifies to:

\[ \frac{k}{4} = m \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 2 into Equation 1

From Equation 2, we have \( m = \frac{k}{4} \). Substitute this into Equation 1:

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

Multiplying through by 4 to eliminate the fraction:

\[ 8k = 3k + 8 \]

Rearranging gives:

\[ 8k - 3k = 8 \implies 5k = 8 \implies k = \frac{8}{5} \]

### Step 4: Find \( m \)

Now substitute \( k \) back into Equation 2 to find \( m \):

\[ m = \frac{k}{4} = \frac{\frac{8}{5}}{4} = \frac{8}{20} = \frac{2}{5} \]

### Step 5: Calculate \( k + m \)

Now we can find \( k + m \):

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

Thus, the final answer is:

\[ \boxed{2} \]