Let `f(x) = (sin (pi [ x - pi]))/(1+[x^2])` where [] denotes the greatest integer function then f(x) is

Options

A. continuous at integer points
B. continuous everywhere
C. differentiable once but f''(x) and f''' (x) do not exist
D. differentiable for all x

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{\sin(\pi [x - \pi])}{1 + [x^2]} \), where \([ \cdot ]\) denotes the greatest integer function (also known as the floor function).

### Step 1: Simplifying the function
We start by rewriting the function in terms of the greatest integer function. The expression \([x - \pi]\) can be expressed as:
\[ [x - \pi] = [x] - \pi \]
where \([x]\) is the greatest integer less than or equal to \(x\). Therefore, we can rewrite \(f(x)\) as:
\[ f(x) = \frac{\sin(\pi ([x] - \pi))}{1 + [x^2]} \]

### Step 2: Analyzing the sine function
The sine function, \(\sin(\pi n)\), where \(n\) is an integer, equals zero for all integer values of \(n\). Since \([x]\) is an integer, we can conclude that:
\[ f(x) = \frac{\sin(\pi n)}{1 + [x^2]} = \frac{0}{1 + [x^2]} = 0 \]
for all \(x\) in the real numbers.

### Step 3: Conclusion about the function
Since \(f(x) = 0\) for all \(x\), we can conclude that \(f(x)\) is a constant function.

### Step 4: Continuity and Differentiability
A constant function is continuous and differentiable everywhere in its domain. Therefore, we can state:
- \(f(x)\) is continuous for all \(x \in \mathbb{R}\).
- \(f(x)\) is differentiable for all \(x \in \mathbb{R}\).

### Final Answer
Thus, the correct options regarding the function \(f(x)\) are:
- \(f(x)\) is continuous for all \(x\).
- \(f(x)\) is differentiable for all \(x\).