If `f:R->R` where `f(x) = ax + cosx` is an invertible function, then `(a).(-2, -1]uu [1,2)`; `(b).[-1,1]`; `(c).(-oo, -1]uu [1,oo)`; `(d).(-oo, -2]uu [2,oo)`.

Options

A. `(-2, -1]uu [1,2)`
B. `[-1,1]`
C. `(-oo, -1]uu [1,oo)`
D. `(-oo, -2]uu [2,oo)`

Answer

To determine the values of \( a \) for which the function \( f(x) = ax + \cos x \) is invertible, we need to ensure that the function is one-to-one. A function is one-to-one if it is either strictly increasing or strictly decreasing.

### Step 1: Find the derivative of the function
To analyze the increasing or decreasing nature of the function, we first calculate the derivative:

\[ f'(x) = \frac{d}{dx}(ax + \cos x) = a - \sin x \]

### Step 2: Determine conditions for \( f'(x) \) to be non-negative
For the function to be increasing (and thus one-to-one), we require:

\[ f'(x) \geq 0 \implies a - \sin x \geq 0 \implies a \geq \sin x \]

Since \( \sin x \) varies between -1 and 1, we can express this condition as:

\[ a \geq -1 \quad \text{(minimum value of } \sin x\text{)} \]

### Step 3: Determine conditions for \( f'(x) \) to be non-positive
For the function to be decreasing (and thus one-to-one), we require:

\[ f'(x) \leq 0 \implies a - \sin x \leq 0 \implies a \leq \sin x \]

This gives us:

\[ a \leq 1 \quad \text{(maximum value of } \sin x\text{)} \]

### Step 4: Combine the conditions
From the above steps, we have two inequalities:

1. \( a \geq -1 \)
2. \( a \leq 1 \)

Thus, combining these gives us:

\[ -1 \leq a \leq 1 \]

### Step 5: Identify the intervals for \( a \)
The values of \( a \) that satisfy the condition for the function \( f(x) \) to be invertible are:

\[ a \in [-1, 1] \]

### Conclusion
The only option that matches this interval is:

**(b) \([-1, 1]\)**