In `x epsilon(0,1),f(x)=[3x^(2)+1]`, where [x] stands for the greatest integer not exceeding x is:

Options

A. continuous
B. continuous except at one point
C. continuous except at two points
D. continuous except at three points

Answer

To solve the problem, we need to analyze the function \( f(x) = \lfloor 3x^2 + 1 \rfloor \) for \( x \) in the interval \( (0, 1) \), where \( \lfloor x \rfloor \) denotes the greatest integer function.

### Step 1: Determine the range of \( 3x^2 + 1 \)
First, we need to find the values of \( 3x^2 + 1 \) as \( x \) varies from \( 0 \) to \( 1 \).

- When \( x = 0 \):
\[ 3(0)^2 + 1 = 1 \]

- When \( x = 1 \):
\[ 3(1)^2 + 1 = 4 \]

Thus, as \( x \) varies from \( 0 \) to \( 1 \), \( 3x^2 + 1 \) varies from \( 1 \) to \( 4 \).

### Step 2: Identify the integer values in the range
The function \( f(x) = \lfloor 3x^2 + 1 \rfloor \) will take integer values based on the output of \( 3x^2 + 1 \).

- For \( 1 \leq 3x^2 + 1 < 2 \):
- This occurs when \( 0 \leq 3x^2 < 1 \) or \( 0 \leq x < \frac{1}{\sqrt{3}} \).
- In this interval, \( f(x) = 1 \).

- For \( 2 \leq 3x^2 + 1 < 3 \):
- This occurs when \( 1 \leq 3x^2 < 2 \) or \( \frac{1}{\sqrt{3}} \leq x < \sqrt{\frac{2}{3}} \).
- In this interval, \( f(x) = 2 \).

- For \( 3 \leq 3x^2 + 1 < 4 \):
- This occurs when \( 2 \leq 3x^2 < 3 \) or \( \sqrt{\frac{2}{3}} \leq x < 1 \).
- In this interval, \( f(x) = 3 \).

- At \( x = 1 \):
- \( f(1) = \lfloor 4 \rfloor = 4 \) (but \( x = 1 \) is not included in the interval).

### Step 3: Identify points of discontinuity
The function \( f(x) \) will be discontinuous at points where \( 3x^2 + 1 \) is an integer.

- The critical points are:
- When \( 3x^2 + 1 = 2 \) (i.e., \( 3x^2 = 1 \) or \( x = \frac{1}{\sqrt{3}} \)).
- When \( 3x^2 + 1 = 3 \) (i.e., \( 3x^2 = 2 \) or \( x = \sqrt{\frac{2}{3}} \)).

### Step 4: Conclusion
The function \( f(x) \) is continuous everywhere in the interval \( (0, 1) \) except at the points \( x = \frac{1}{\sqrt{3}} \) and \( x = \sqrt{\frac{2}{3}} \).

Thus, the function is discontinuous at **two points** in the interval \( (0, 1) \).

### Final Answer
The function \( f(x) \) is discontinuous at **two points** in the interval \( (0, 1) \).

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