consider the function f(X) =x+cosx -a
values of a which f(X) =0 has exactly one positive root are

Options

A. (0,1)
B. `(-oo,1)`
C. (-1,1)
D. `(1,oo)`

Explanation

To solve the problem, we need to analyze the function \( f(x) = x + \cos x - a \) and determine the values of \( a \) for which this function has exactly one positive root.

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = x + \cos x - a \]

2. **Differentiate the Function**:
To understand the behavior of the function, we differentiate it:
\[ f'(x) = 1 - \sin x \]

3. **Analyze the Derivative**:
The derivative \( f'(x) = 1 - \sin x \) is always non-negative because \( \sin x \) ranges from -1 to 1. Therefore:
\[ f'(x) \geq 0 \quad \text{for all } x \]
This implies that \( f(x) \) is a non-decreasing function.

4. **Find Critical Points**:
To find where \( f'(x) = 0 \):
\[ 1 - \sin x = 0 \implies \sin x = 1 \]
The solutions to this are:
\[ x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
However, since \( f(x) \) is non-decreasing, it can only have one root if it crosses the x-axis exactly once.

5. **Evaluate the Function at Specific Points**:
We evaluate \( f(0) \):
\[ f(0) = 0 + \cos(0) - a = 1 - a \]
For \( f(x) \) to have exactly one positive root, we need \( f(0) < 0 \):
\[ 1 - a < 0 \implies a > 1 \]

6. **Behavior as \( x \to \infty \)**:
As \( x \to \infty \), \( \cos x \) oscillates between -1 and 1, but \( x \) dominates. Therefore, \( f(x) \to \infty \).

7. **Conclusion**:
Since \( f(x) \) is continuous and non-decreasing, and \( f(0) < 0 \) when \( a > 1 \), there will be exactly one positive root for \( a > 1 \).

Thus, the values of \( a \) for which \( f(x) = 0 \) has exactly one positive root are:
\[ \boxed{(1, \infty)} \]