`sin alphax+cos alphax " and " abs(cosx)+abs(sinx)` are periodic functions of same fundamental period, if `alpha` equals

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the functions \( \sin(\alpha x) + \cos(\alpha x) \) and \( |\cos x| + |\sin x| \) have the same fundamental period.

### Step-by-Step Solution:

1. **Determine the Period of \( \sin(\alpha x) + \cos(\alpha x) \)**:
- The period of \( \sin x \) and \( \cos x \) is \( 2\pi \).
- For \( \sin(\alpha x) \) and \( \cos(\alpha x) \), the period is given by:
\[ T = \frac{2\pi}{\alpha} \]

2. **Determine the Period of \( |\cos x| + |\sin x| \)**:
- The function \( |\cos x| \) has a period of \( \pi \) because \( |\cos(x + \pi)| = |\cos x| \).
- Similarly, \( |\sin x| \) also has a period of \( \pi \).
- Therefore, the combined function \( |\cos x| + |\sin x| \) also has a period of \( \pi \).

3. **Set the Periods Equal**:
- We need to set the two periods equal to each other:
\[ \frac{2\pi}{\alpha} = \pi \]

4. **Solve for \( \alpha \)**:
- To solve for \( \alpha \), we can multiply both sides by \( \alpha \) and then simplify:
\[ 2\pi = \alpha \cdot \pi \]
- Dividing both sides by \( \pi \):
\[ 2 = \alpha \]

5. **Conclusion**:
- Thus, the value of \( \alpha \) is:
\[ \alpha = 2 \]

### Final Answer:
\[ \alpha = 2 \]