If `f(x)=[x](sin kx)^(p)` is continuous for real x, then (where [.] represents the greatest integer function)

Options

A. `k in [npi, n in I], p gt 0`
B. `k in {2npi, n in I}, p gt0`
C. `k in {npi, n in I}, p in R-{0}`
D. `k in {npi, n I, n ne 0}, p in R-{0}`

Explanation

To determine the conditions under which the function \( f(x) = [x](\sin(kx))^p \) is continuous for all real \( x \), we need to analyze the components of the function.

### Step 1: Understanding the components of the function
The function \( f(x) \) consists of two parts: the greatest integer function \( [x] \) and \( (\sin(kx))^p \).

- The greatest integer function \( [x] \) is discontinuous at integer values of \( x \).
- The function \( (\sin(kx))^p \) is continuous for all real \( x \) as long as \( p > 0 \).

### Step 2: Identifying points of discontinuity
The function \( f(x) \) will be discontinuous at points where \( [x] \) is discontinuous, which occurs at integer values of \( x \). Specifically, the points of discontinuity are at \( x = n \) where \( n \) is an integer.

### Step 3: Analyzing continuity at integer points
To ensure that \( f(x) \) is continuous at integer points, we need to check the limit of \( f(x) \) as \( x \) approaches an integer \( n \).

- As \( x \) approaches \( n \) from the left, \( [x] \) will be \( n-1 \) and \( f(x) \) will approach \( (n-1)(\sin(kn))^p \).
- As \( x \) approaches \( n \) from the right, \( [x] \) will be \( n \) and \( f(x) \) will approach \( n(\sin(kn))^p \).

For \( f(x) \) to be continuous at \( x = n \), we need:
\[ \lim_{x \to n^-} f(x) = \lim_{x \to n^+} f(x) = f(n) \]
This means:
\[ (n-1)(\sin(kn))^p = n(\sin(kn))^p \]

### Step 4: Solving the equation
Rearranging the equation gives:
\[ (n-1)(\sin(kn))^p - n(\sin(kn))^p = 0 \]
Factoring out \( (\sin(kn))^p \):
\[ ((n-1) - n)(\sin(kn))^p = 0 \]
This simplifies to:
\[ -\sin(kn)^p = 0 \]
Thus, for continuity, we require:
\[ \sin(kn) = 0 \]
This occurs when \( kn = n\pi \) for any integer \( n \), which implies \( k \) must be a rational multiple of \( \pi \).

### Conclusion
For \( f(x) = [x](\sin(kx))^p \) to be continuous for all real \( x \), \( k \) must be such that \( \sin(kn) = 0 \) at all integers \( n \). Therefore, \( k \) must be of the form \( \frac{m\pi}{n} \) where \( m \) and \( n \) are integers.

### Final Answer
The function \( f(x) \) is continuous for all real \( x \) if \( k \) is a rational multiple of \( \pi \).

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