The number of points of discontinuity of `f(x)=[2x]^(2)-{2x}^(2)` (where [ ] denotes the greatest integer function and { } is fractional part of x) in the interval `(-2,2)`, is

Options

A. 1
B. 6
C. 2
D. 5

Explanation

To find the number of points of discontinuity of the function \( f(x) = [2x^2] - \{2x^2\} \) in the interval \((-2, 2)\), where \([ ]\) denotes the greatest integer function and \(\{ \}\) denotes the fractional part, we can follow these steps:

### Step 1: Understand the Components of the Function
The function \( f(x) \) consists of two parts:
- The greatest integer function \( [2x^2] \)
- The fractional part function \( \{2x^2\} \)

The fractional part can be expressed as:
\[ \{x\} = x - [x] \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = [2x^2] - (2x^2 - [2x^2]) = 2[x^2] - 2x^2 \]

### Step 2: Identify Points of Discontinuity
The function \( f(x) \) will be discontinuous at points where \( [2x^2] \) changes its value. This occurs at points where \( 2x^2 \) is an integer.

### Step 3: Determine Where \( 2x^2 \) is an Integer
Set \( 2x^2 = n \) where \( n \) is an integer. Thus:
\[ x^2 = \frac{n}{2} \]
This implies:
\[ x = \pm \sqrt{\frac{n}{2}} \]

### Step 4: Find Integer Values of \( n \) in the Interval
We need to find integer values of \( n \) such that \( x \) lies in the interval \((-2, 2)\).

Calculating the bounds:
- For \( x = 2 \), \( 2x^2 = 8 \) (not included)
- For \( x = -2 \), \( 2x^2 = 8 \) (not included)

Thus, \( n \) can take values from \( 0 \) to \( 7 \) (since \( 2x^2 < 8 \) when \( x \) is in the interval \((-2, 2)\)).

### Step 5: Calculate Possible Values of \( n \)
The possible integer values of \( n \) are:
- \( n = 0 \) gives \( x = 0 \)
- \( n = 1 \) gives \( x = \pm \frac{1}{\sqrt{2}} \)
- \( n = 2 \) gives \( x = \pm 1 \)
- \( n = 3 \) gives \( x = \pm \sqrt{3} \)
- \( n = 4 \) gives \( x = \pm \sqrt{2} \)
- \( n = 5 \) gives \( x = \pm \sqrt{\frac{5}{2}} \)
- \( n = 6 \) gives \( x = \pm \sqrt{3} \)
- \( n = 7 \) gives \( x = \pm \sqrt{\frac{7}{2}} \)

### Step 6: Count the Points of Discontinuity
From the above calculations, we find the following points of discontinuity in the interval \((-2, 2)\):
- \( x = 0 \)
- \( x = \pm \frac{1}{\sqrt{2}} \)
- \( x = \pm 1 \)
- \( x = \pm \sqrt{2} \)
- \( x = \pm \sqrt{3} \)
- \( x = \pm \sqrt{\frac{5}{2}} \)
- \( x = \pm \sqrt{\frac{7}{2}} \)

### Final Count
Counting these points gives us a total of **7 points of discontinuity**.

### Conclusion
Thus, the number of points of discontinuity of \( f(x) \) in the interval \((-2, 2)\) is **7**.

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