If $f(x) = \begin{cases}
k\sqrt{x+1}, & 0 \le x \le 3 \
mx+2, & 3

Q: If $f(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx+2, & 3 < x \le 5 \end{cases}$ is differentiable, then $k+m=$

If $f(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx+2, & 3 < x \le 5 \end{cases}$ is differentiable, then $k+m=$

Question

If $f(x) = \begin{cases}
k\sqrt{x+1}, & 0 \le x \le 3 \
mx+2, & 3 < x \le 5
\end{cases}$ is differentiable, then $k+m=$

Accepted Answer

To determine the value of $ k + m $ such that the function $ f(x) $ is differentiable at $ x = 3 $, we need to ensure that both the function itself and its derivative are continuous at this point.

### Step 1: Continuity at $ x = 3 $

For $ f(x) $ to be continuous at $ x = 3 $, the left-hand limit as $ x $ approaches 3 must equal the right-hand limit and the value of the function at $ x = 3 $.

- **Left-hand limit**: As $ x \to 3^- $, $ f(x) = k\sqrt{x+1} $. Thus, $ \lim_{x \to 3^-} f(x) = k\sqrt{3+1} = 2k $.

- **Right-hand limit**: As $ x \to 3^+ $, $ f(x) = mx + 2 $. Thus, $ \lim_{x \to 3^+} f(x) = 3m + 2 $.

For continuity at $ x = 3 $, these limits must be equal:
$$ 2k = 3m + 2 $$

### Step 2: Differentiability at $ x = 3 $

For $ f(x) $ to be differentiable at $ x = 3 $, the derivatives from the left and right must also be equal.

- **Derivative from the left**: For $ 0 \le x \le 3 $, $ f(x) = k\sqrt{x+1} $. The derivative is:
  $$
  f'(x) = \frac{k}{2\sqrt{x+1}}
  $$
  At $ x = 3 $, $ f'(3^-) = \frac{k}{2\sqrt{4}} = \frac{k}{4} $.

- **Derivative from the right**: For $ 3 < x \le 5 $, $ f(x) = mx + 2 $. The derivative is:
  $$
  f'(x) = m
  $$
  At $ x = 3 $, $ f'(3^+) = m $.

For differentiability at $ x = 3 $, these derivatives must be equal:
$$ \frac{k}{4} = m $$

### Solving the System of Equations

We now have two equations:
1. $ 2k = 3m + 2 $
2. $ \frac{k}{4} = m $

Substitute $ m = \frac{k}{4} $ into the first equation:
$$ 2k = 3\left(\frac{k}{4}\right) + 2 $$
$$ 2k = \frac{3k}{4} + 2 $$

Multiply through by 4 to clear the fraction:
$$ 8k = 3k + 8 $$
$$ 5k = 8 $$
$$ k = \frac{8}{5} $$

Substitute $ k = \frac{8}{5} $ back into $ m = \frac{k}{4} $:
$$ m = \frac{\frac{8}{5}}{4} = \frac{8}{20} = \frac{2}{5} $$

Finally, calculate $ k + m $:
$$ k + m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2 $$

Thus, the correct answer is $ k + m = 2 $, which corresponds to **Option 1**.

### Explanation of Incorrect Options

- **Option 2 ($\frac{16}{5}$)**: This value does not satisfy the equations derived from the conditions for continuity and differentiability.
- **Option 3 ($\frac{10}{3}$)**: This value also does not satisfy the necessary conditions for the function to be differentiable at $ x = 3 $.
- **Option 4 (4)**: This value is incorrect as it does not solve the system of equations derived from the continuity and differentiability conditions.

Therefore, the correct answer is indeed **Option 1**.