The equation `x + cos x = a` has exactly one positive root. Complete set of values of `'a'` is

Options

A. `a in (0,1)`
B. `a in (2,3)`
C. `a in (1,oo)`
D. `a in (-oo,1)`

Answer

To solve the problem of finding the complete set of values of \( a \) for which the equation \( x + \cos x = a \) has exactly one positive root, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x + \cos x - a \). We need to analyze the behavior of this function to determine the conditions under which it has exactly one positive root.

### Step 2: Find the derivative
Calculate the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x + \cos x - a) = 1 - \sin x \]

### Step 3: Analyze the derivative
The function \( \sin x \) oscillates between -1 and 1. Therefore, the derivative \( f'(x) \) can be analyzed:
\[ f'(x) = 1 - \sin x \]
Since \( \sin x \) ranges from -1 to 1, we have:
\[ 0 \leq 1 - \sin x \leq 2 \]
This means \( f'(x) \) is always non-negative, indicating that \( f(x) \) is a non-decreasing function.

### Step 4: Evaluate the function at \( x = 0 \)
Now, evaluate \( f(0) \):
\[ f(0) = 0 + \cos(0) - a = 1 - a \]
For \( f(0) < 0 \) (which indicates that the function starts below the x-axis), we have:
\[ 1 - a < 0 \implies a > 1 \]

### Step 5: Determine the behavior as \( x \to \infty \)
As \( x \) approaches infinity, \( \cos x \) oscillates between -1 and 1, but \( x \) will dominate:
\[ f(x) \to \infty \quad \text{as } x \to \infty \]
This indicates that \( f(x) \) will eventually cross the x-axis.

### Step 6: Conclusion on the number of roots
Since \( f(x) \) is non-decreasing and \( f(0) < 0 \) while \( f(x) \to \infty \) as \( x \to \infty \), there will be exactly one point where \( f(x) = 0 \) for \( a > 1 \).

### Final Answer
The complete set of values of \( a \) for which the equation \( x + \cos x = a \) has exactly one positive root is:
\[ \boxed{(1, \infty)} \]