if `f(x) = tan(pi[(2x-3pi)^3])/(1+[2x-3pi]^2) ` ([.] denotes the greatest integer function), then

Options

A. `f(x)` is continuous in R
B. `f(x)` is continuous in R but not differentiable in R
C. `f(x)` exists everywhere but `f'(x)` does not exists at some `x epsilonR`
D. None of these

Answer

To solve the problem, we need to analyze the function given:

\[ f(x) = \frac{\tan(\pi \lfloor 2x - 3\pi \rfloor)}{1 + \lfloor 2x - 3\pi \rfloor^2} \]

where \( \lfloor . \rfloor \) denotes the greatest integer function.

### Step 1: Understanding the Greatest Integer Function
The greatest integer function \( \lfloor x \rfloor \) gives the largest integer less than or equal to \( x \). Therefore, we can denote:

Let \( n = \lfloor 2x - 3\pi \rfloor \), where \( n \) is an integer.

### Step 2: Expressing the Function in Terms of \( n \)
We can rewrite \( f(x) \) as:

\[ f(x) = \frac{\tan(n\pi)}{1 + n^2} \]

### Step 3: Evaluating \( \tan(n\pi) \)
For any integer \( n \):

\[ \tan(n\pi) = 0 \]

This is because the tangent function is zero at all integer multiples of \( \pi \).

### Step 4: Substituting Back into the Function
Substituting \( \tan(n\pi) = 0 \) into the function gives:

\[ f(x) = \frac{0}{1 + n^2} = 0 \]

### Step 5: Conclusion About the Function
Since \( f(x) = 0 \) for all \( x \) where \( n \) is defined, we conclude that:

\[ f(x) = 0 \text{ for all } x \]

### Step 6: Analyzing Continuity and Differentiability
1. **Continuity**: A constant function is continuous everywhere.
2. **Differentiability**: The derivative of a constant function is zero, hence \( f'(x) = 0 \).

### Final Result
Thus, we conclude that:

- \( f(x) \) is a constant function equal to 0.
- It is continuous everywhere in \( \mathbb{R} \).
- It is differentiable everywhere in \( \mathbb{R} \).

### Answer
The correct option is **A**: \( f(x) \) is continuous and differentiable in \( \mathbb{R} \).

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