consider the function f(X) =x+cosx -a
values of a for which f(X) =0 has exactly one negative 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 negative root. Here’s a step-by-step solution:

### Step 1: Analyze the function
We start with the function:
\[ f(x) = x + \cos x - a \]
We need to find the conditions under which \( f(x) = 0 \) has exactly one negative root.

### Step 2: Find the derivative
To understand the behavior of the function, we calculate its derivative:
\[ f'(x) = 1 - \sin x \]
The sine function, \( \sin x \), oscillates between -1 and 1. Therefore, \( f'(x) \) will be:
\[ f'(x) \geq 0 \quad \text{(since } 1 - \sin x \text{ is always non-negative)} \]
This means that \( f(x) \) is a non-decreasing function.

### Step 3: Evaluate \( f(0) \)
Next, we evaluate \( f(0) \):
\[ f(0) = 0 + \cos(0) - a = 1 - a \]
For \( f(0) = 0 \), we need:
\[ 1 - a = 0 \implies a = 1 \]

### Step 4: Behavior of \( f(x) \) as \( x \to -\infty \)
Now, we check the behavior of \( f(x) \) as \( x \) approaches negative infinity:
\[ \lim_{x \to -\infty} f(x) = -\infty \quad \text{(since } x \text{ dominates } \cos x \text{)} \]
This indicates that \( f(x) \) starts from negative infinity and increases.

### Step 5: Condition for exactly one negative root
Since \( f(x) \) is increasing and continuous, it can cross the x-axis at most once. For \( f(x) = 0 \) to have exactly one negative root, the function must cross the x-axis at \( x < 0 \) when \( a < 1 \) and touch the x-axis at \( x = 0 \) when \( a = 1 \).

Thus, \( f(x) = 0 \) has exactly one negative root if:
\[ a < 1 \]

### Conclusion
The values of \( a \) for which \( f(x) = 0 \) has exactly one negative root are:
\[ \boxed{(-\infty, 1)} \]