If the function.
`g(x)={(ksqrt(x+1),0lexle3),(mx+2,3ltxle5):}` is differentiable, then the value of k+m is :

Options

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

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 \( x = 3 \), we need:

\[ g(3^-) = g(3^+) \]

Calculating \( g(3^-) \):

\[ g(3^-) = k \sqrt{3 + 1} = k \cdot 2 = 2k \]

Calculating \( g(3^+) \):

\[ g(3^+) = m \cdot 3 + 2 = 3m + 2 \]

Setting these equal for continuity:

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

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

For \( g(x) \) to be differentiable at \( x = 3 \), we need:

\[ g'(3^-) = g'(3^+) \]

Calculating \( g'(3^-) \):

Using the derivative of \( k \sqrt{x + 1} \):

\[ g'(x) = \frac{k}{2\sqrt{x + 1}} \quad \text{for } 0 \leq x < 3 \]

Thus,

\[ g'(3^-) = \frac{k}{2\sqrt{3 + 1}} = \frac{k}{2 \cdot 2} = \frac{k}{4} \]

Calculating \( g'(3^+) \):

The derivative of \( mx + 2 \) is simply \( m \):

\[ g'(3^+) = m \]

Setting these equal for differentiability:

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

### Step 3: Solve the System of Equations

Now we have two equations:

1. \( 2k = 3m + 2 \)
2. \( k = 4m \)

Substituting Equation 2 into Equation 1:

\[ 2(4m) = 3m + 2 \]

This simplifies to:

\[ 8m = 3m + 2 \]

Subtracting \( 3m \) from both sides:

\[ 5m = 2 \]

So,

\[ m = \frac{2}{5} \]

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

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

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

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

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

Thus, the value of \( k + m \) is:

\[ \boxed{2} \]