Find the interval in which the function `f(x) = sqrt3 sin 2x - cos 2x + 4` is one - one

Options

A. `[-pi/2,pi/2]`
B. `[-pi/6,pi/2]`
C. `[-pi/6,pi/3]`
D. `[-pi/6,pi/2]`

Answer

To find the interval in which the function \( f(x) = \sqrt{3} \sin(2x) - \cos(2x) + 4 \) is one-to-one, we need to determine where the derivative of the function does not change sign.

### Step-by-step Solution:

1. **Rewrite the Function**:
The given function is:
\[ f(x) = \sqrt{3} \sin(2x) - \cos(2x) + 4 \]

2. **Differentiate the Function**:
We need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sqrt{3} \sin(2x)) - \frac{d}{dx}(\cos(2x)) \]
Using the chain rule:
\[ f'(x) = \sqrt{3} \cdot 2 \cos(2x) + 2 \sin(2x) = 2\sqrt{3} \cos(2x) + 2 \sin(2x) \]
Simplifying, we have:
\[ f'(x) = 2(\sqrt{3} \cos(2x) + \sin(2x)) \]

3. **Set the Derivative Equal to Zero**:
To find critical points, set \( f'(x) = 0 \):
\[ \sqrt{3} \cos(2x) + \sin(2x) = 0 \]
Rearranging gives:
\[ \sin(2x) = -\sqrt{3} \cos(2x) \]
Dividing both sides by \( \cos(2x) \) (assuming \( \cos(2x) \neq 0 \)):
\[ \tan(2x) = -\sqrt{3} \]

4. **Find the General Solution**:
The general solution for \( \tan(2x) = -\sqrt{3} \) is:
\[ 2x = \frac{5\pi}{3} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{5\pi}{6} + \frac{n\pi}{2} \]

5. **Determine Intervals**:
The function \( f(x) \) will be one-to-one where \( f'(x) \) does not change sign. The critical points divide the x-axis into intervals. We need to check the sign of \( f'(x) \) in these intervals.

6. **Testing Intervals**:
Choose test points in the intervals created by the critical points. For example, if we take \( n = 0 \):
- For \( x < \frac{5\pi}{6} \), choose \( x = 0 \):
\[ f'(0) = 2(\sqrt{3} \cdot 1 + 0) > 0 \]
- For \( x > \frac{5\pi}{6} \), choose \( x = \frac{3\pi}{4} \):
\[ f'\left(\frac{3\pi}{4}\right) = 2(\sqrt{3} \cdot 0 - 1) < 0 \]
Thus, \( f'(x) \) changes sign at \( x = \frac{5\pi}{6} \).

7. **Final Interval**:
The function is one-to-one in the interval where \( f'(x) \) does not change sign. Therefore, the function is one-to-one in the interval:
\[ \left(-\frac{\pi}{6}, \frac{\pi}{3}\right) \]