lf the fundamental period of function `f(x)=sinx + cos(sqrt(4-a^2))x` is `4pi`, then the value of a is/are

Options

A. `(sqrt(15))/(2)`
B. `-(sqrt(15))/(2)`
C. `(sqrt(7))/(2)`
D. `-(sqrt(7))/(2)`

Explanation

To solve the problem, we need to find the value of \( a \) such that the fundamental period of the function \( f(x) = \sin x + \cos(\sqrt{4 - a^2} \cdot x) \) is \( 4\pi \).

### Step-by-Step Solution:

1. **Identify the Period of Each Component**:
- The period of \( \sin x \) is \( 2\pi \).
- The period of \( \cos(kx) \) is given by \( \frac{2\pi}{|k|} \). Here, \( k = \sqrt{4 - a^2} \), so the period of \( \cos(\sqrt{4 - a^2} \cdot x) \) is \( \frac{2\pi}{\sqrt{4 - a^2}} \).

2. **Determine the Fundamental Period**:
- The fundamental period of \( f(x) \) is the least common multiple (LCM) of the periods of its components. Therefore, we need to find the LCM of \( 2\pi \) and \( \frac{2\pi}{\sqrt{4 - a^2}} \).

3. **Set Up the Equation**:
- We know that the fundamental period is given as \( 4\pi \). Thus, we set up the equation:
\[ \text{LCM}(2\pi, \frac{2\pi}{\sqrt{4 - a^2}}) = 4\pi \]

4. **Finding the LCM**:
- The LCM of \( 2\pi \) and \( \frac{2\pi}{\sqrt{4 - a^2}} \) can be expressed as:
\[ \text{LCM}(2\pi, \frac{2\pi}{\sqrt{4 - a^2}}) = 2\pi \cdot \frac{1}{\sqrt{4 - a^2}} \text{ if } \sqrt{4 - a^2} < 2 \]
- For the LCM to equal \( 4\pi \), we have:
\[ 2\pi \cdot \frac{1}{\sqrt{4 - a^2}} = 4\pi \]

5. **Simplifying the Equation**:
- Dividing both sides by \( 2\pi \):
\[ \frac{1}{\sqrt{4 - a^2}} = 2 \]
- Taking the reciprocal:
\[ \sqrt{4 - a^2} = \frac{1}{2} \]

6. **Squaring Both Sides**:
- Squaring both sides gives:
\[ 4 - a^2 = \frac{1}{4} \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ 4 - \frac{1}{4} = a^2 \]
- Converting \( 4 \) to a fraction:
\[ \frac{16}{4} - \frac{1}{4} = a^2 \]
- Thus:
\[ \frac{15}{4} = a^2 \]

8. **Finding the Values of \( a \)**:
- Taking the square root:
\[ a = \pm \sqrt{\frac{15}{4}} = \pm \frac{\sqrt{15}}{2} \]

### Final Answer:
The values of \( a \) are \( a = \frac{\sqrt{15}}{2} \) and \( a = -\frac{\sqrt{15}}{2} \).