Let`f(x)={{:(-1+sinK_(1)pix",", "x is rational."), (1+cosK_(2)pix",", x):}`
If f(x) is a periodic function, then

Options

A. either `K_(1),K_(2) in` rational or `K_(1),K_(2) in` irrational
B. `K_(1),K_(2) in " rational only "`
C. `K_(1),K_(2) in " irrational only "`
D. `K_(1),K_(2) in " irrational such that " K_(1)/K_(2)` is rational

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} -1 + \sin(k_1 \pi x) & \text{if } x \text{ is rational} \\ 1 + \cos(k_2 \pi x) & \text{if } x \text{ is irrational} \end{cases} \]

We want to determine the conditions under which \( f(x) \) is a periodic function.

### Step 1: Determine the Range of Each Part of the Function

1. **For \( x \) rational**:
- The function is \( f(x) = -1 + \sin(k_1 \pi x) \).
- The range of \( \sin(k_1 \pi x) \) is from -1 to 1.
- Therefore, the range of \( -1 + \sin(k_1 \pi x) \) is:
\[ [-1 + (-1), -1 + 1] = [-2, 0] \]

2. **For \( x \) irrational**:
- The function is \( f(x) = 1 + \cos(k_2 \pi x) \).
- The range of \( \cos(k_2 \pi x) \) is also from -1 to 1.
- Therefore, the range of \( 1 + \cos(k_2 \pi x) \) is:
\[ [1 + (-1), 1 + 1] = [0, 2] \]

### Step 2: Identify the Conditions for Periodicity

For \( f(x) \) to be periodic, the two parts of the function must have overlapping ranges.

- The range for rational \( x \) is \([-2, 0]\).
- The range for irrational \( x \) is \([0, 2]\).

The overlapping point is \( 0 \). Thus, for \( f(x) \) to be periodic, we need to ensure that both parts can be equal at some points.

### Step 3: Establish the Periodicity Condition

1. **Period of \( -1 + \sin(k_1 \pi x) \)**:
- The period \( T_1 \) of \( \sin(k_1 \pi x) \) is given by:
\[ T_1 = \frac{2}{k_1} \]

2. **Period of \( 1 + \cos(k_2 \pi x) \)**:
- The period \( T_2 \) of \( \cos(k_2 \pi x) \) is given by:
\[ T_2 = \frac{2}{k_2} \]

For \( f(x) \) to be periodic, \( T_1 \) and \( T_2 \) must be rational numbers. This implies that both \( k_1 \) and \( k_2 \) must be rational numbers.

### Conclusion

Thus, for \( f(x) \) to be a periodic function, both \( k_1 \) and \( k_2 \) must be rational numbers.

### Final Answer

**If \( f(x) \) is a periodic function, then \( k_1 \) and \( k_2 \) are rational numbers.**