If f: `Rrarr R` defined by f(x) = 3x+2a cos x -5 is invertible then 'a' belongs to

Options

A. [-3/2, 3/2 ]
B. `(-oo,3//2] cup [3//2 ,oo]`
C. `(-4,4)
D. R

Explanation

To determine the values of 'a' for which the function \( f(x) = 3x + 2a \cos x - 5 \) is invertible, we need to ensure that the function is either always increasing or always decreasing. This can be checked by analyzing the derivative of the function.

### Step-by-Step Solution:

1. **Find the Derivative**:
We start by differentiating the function \( f(x) \).
\[ f'(x) = \frac{d}{dx}(3x + 2a \cos x - 5) \]
The derivative is:
\[ f'(x) = 3 - 2a \sin x \]

2. **Set the Derivative Not Equal to Zero**:
For the function to be invertible, \( f'(x) \) must not equal zero for any \( x \) in \( \mathbb{R} \).
\[ 3 - 2a \sin x \neq 0 \]
Rearranging gives:
\[ 2a \sin x \neq 3 \]
Thus,
\[ \sin x \neq \frac{3}{2a} \]

3. **Analyze the Range of Sine Function**:
The sine function has a range of \([-1, 1]\). Therefore, for \( \frac{3}{2a} \) to be outside this range, we need:
\[ \frac{3}{2a} > 1 \quad \text{or} \quad \frac{3}{2a} < -1 \]

4. **Solve the Inequalities**:
- For \( \frac{3}{2a} > 1 \):
\[ 3 > 2a \implies a < \frac{3}{2} \]

- For \( \frac{3}{2a} < -1 \):
\[ 3 < -2a \implies a < -\frac{3}{2} \]

5. **Combine the Results**:
From the inequalities, we conclude:
\[ a < -\frac{3}{2} \quad \text{or} \quad a > \frac{3}{2} \]

Thus, the values of \( a \) for which the function \( f(x) \) is invertible are:
\[ a \in (-\infty, -\frac{3}{2}) \cup (\frac{3}{2}, \infty) \]