Find the value of 'k' such for `f(x)=k(x+sinx)+k` is increasing in R.

Answer

To find the value of \( k \) such that the function \( f(x) = k(x + \sin x) + k \) is increasing for all \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Find the derivative of the function
To determine if the function is increasing, we need to calculate its derivative \( f'(x) \).

\[ f(x) = k(x + \sin x) + k \]

Differentiating \( f(x) \):

\[ f'(x) = k\left(1 + \cos x\right) \]

### Step 2: Set the condition for increasing function
A function is increasing if its derivative is greater than or equal to zero:

\[ f'(x) \geq 0 \]

Substituting the expression for \( f'(x) \):

\[ k(1 + \cos x) \geq 0 \]

### Step 3: Analyze the term \( 1 + \cos x \)
We know that the cosine function oscillates between -1 and 1. Therefore, \( 1 + \cos x \) will oscillate between:

\[ 1 + (-1) = 0 \quad \text{and} \quad 1 + 1 = 2 \]

Thus, \( 1 + \cos x \) is always non-negative:

\[ 1 + \cos x \geq 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Determine the condition on \( k \)
Since \( 1 + \cos x \) is non-negative for all \( x \), the sign of \( k \) will determine whether \( f'(x) \) is non-negative. For \( f'(x) = k(1 + \cos x) \) to be non-negative for all \( x \), we need:

\[ k \geq 0 \]

### Conclusion
Thus, the value of \( k \) must be:

\[ k \geq 0 \]