Let [x] be the greatest integer function `f(x)=(sin(1/4(pi[x]))/([x]))` is

Options

A. (a)Not continuous at any point
B. (b)continuous at `x = (3)/(2)`
C. (c)discontinuous at x = 2
D. (d)differentiable at `x = (4)/(3)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{\sin\left(\frac{1}{4} \pi [x]\right)}{[x]} \), where \([x]\) is the greatest integer function (also known as the floor function).

### Step 1: Identify the points of discontinuity
The greatest integer function \([x]\) changes its value at integer points. Therefore, we need to check the continuity of \( f(x) \) at integer points, specifically at \( x = 0, 1, 2, 3, \ldots \).

### Step 2: Evaluate \( f(x) \) at integer points
1. **At \( x = 0 \)**:
\[ f(0) = \frac{\sin\left(\frac{1}{4} \pi [0]\right)}{[0]} = \frac{\sin(0)}{0} \]
This is undefined.

2. **At \( x = 1 \)**:
\[ f(1) = \frac{\sin\left(\frac{1}{4} \pi [1]\right)}{[1]} = \frac{\sin\left(\frac{1}{4} \pi \cdot 1\right)}{1} = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

3. **At \( x = 2 \)**:
\[ f(2) = \frac{\sin\left(\frac{1}{4} \pi [2]\right)}{[2]} = \frac{\sin\left(\frac{1}{4} \pi \cdot 2\right)}{2} = \frac{\sin\left(\frac{\pi}{2}\right)}{2} = \frac{1}{2} \]

4. **At \( x = 3 \)**:
\[ f(3) = \frac{\sin\left(\frac{1}{4} \pi [3]\right)}{[3]} = \frac{\sin\left(\frac{1}{4} \pi \cdot 3\right)}{3} = \frac{\sin\left(\frac{3\pi}{4}\right)}{3} = \frac{\frac{\sqrt{2}}{2}}{3} = \frac{\sqrt{2}}{6} \]

### Step 3: Check continuity at critical points
To check continuity at \( x = 1, 2, 3 \):

- **At \( x = 1 \)**:
- Left limit as \( x \to 1^- \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 0\right)}{0} \) (undefined)
- Right limit as \( x \to 1^+ \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 1\right)}{1} = \frac{\sqrt{2}}{2} \)
- Since the left limit is undefined, \( f(x) \) is not continuous at \( x = 1 \).

- **At \( x = 2 \)**:
- Left limit as \( x \to 2^- \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 1\right)}{1} = \frac{\sqrt{2}}{2} \)
- Right limit as \( x \to 2^+ \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 2\right)}{2} = \frac{1}{2} \)
- Since the left limit does not equal the right limit, \( f(x) \) is not continuous at \( x = 2 \).

- **At \( x = 3 \)**:
- Left limit as \( x \to 3^- \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 2\right)}{2} = \frac{1}{2} \)
- Right limit as \( x \to 3^+ \): \( f(x) = \frac{\sin\left(\frac{1}{4} \pi \cdot 3\right)}{3} = \frac{\sqrt{2}}{6} \)
- Since the left limit does not equal the right limit, \( f(x) \) is not continuous at \( x = 3 \).

### Conclusion
The function \( f(x) \) is not continuous at \( x = 0, 1, 2, 3 \).

### Final Answer
The function \( f(x) \) is not continuous at any integer point, specifically at \( x = 0, 1, 2, 3 \).