For what value of k, then function f(x) = k(x + sin x) + k is increasing ?

Options

A. k = -1
B. `k lt -1`
C. `k gt 0`
D. `k lt 0`

Answer

To determine the value of \( k \) for which the function \( f(x) = k(x + \sin x) + k \) is increasing, we need to analyze the derivative of the function.

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = k(x + \sin x) + k \]

2. **Differentiate the function**:
We need to find the derivative \( f'(x) \) with respect to \( x \):
\[ f'(x) = k \frac{d}{dx}(x + \sin x) + 0 \]
The derivative of \( x \) is \( 1 \) and the derivative of \( \sin x \) is \( \cos x \):
\[ f'(x) = k(1 + \cos x) \]

3. **Determine when the function is increasing**:
The function \( f(x) \) is increasing when \( f'(x) > 0 \):
\[ k(1 + \cos x) > 0 \]

4. **Analyze the expression**:
- The term \( (1 + \cos x) \) is always non-negative since \( \cos x \) ranges from \(-1\) to \(1\). Thus, \( 1 + \cos x \) ranges from \(0\) to \(2\).
- Therefore, \( 1 + \cos x \) is zero when \( \cos x = -1 \), which occurs at odd multiples of \( \pi \) (i.e., \( x = (2n + 1)\pi \) for \( n \in \mathbb{Z} \)).
- For \( f'(x) > 0 \), we need \( k > 0 \) when \( 1 + \cos x \) is positive.

5. **Conclusion**:
Since \( 1 + \cos x \) can be zero at certain points, \( k \) must be strictly greater than \( 0 \) for the function to be increasing in general.

Thus, the value of \( k \) for which the function \( f(x) \) is increasing is:
\[ k > 0 \]