The number of values of `x , x ∈ [-2,3]` where `f (x) =[x ^(2)] sin (pix)` is discontinous is (where [.] denotes greatest integer function)

Answer

To determine the number of values of \( x \) in the interval \( [-2, 3] \) where the function \( f(x) = [x^2] \sin(\pi x) \) is discontinuous, we will follow these steps:

### Step 1: Identify points of discontinuity
The function \( f(x) \) can be discontinuous at points where either the greatest integer function \( [x^2] \) changes its value or where \( \sin(\pi x) = 0 \).

### Step 2: Find where \( [x^2] \) changes its value
The greatest integer function \( [x^2] \) changes its value at integer points of \( x^2 \). We need to find the integer values of \( n \) such that \( n = x^2 \) for \( x \in [-2, 3] \).

Calculating the range of \( x^2 \):
- At \( x = -2 \), \( x^2 = 4 \).
- At \( x = 3 \), \( x^2 = 9 \).

Thus, \( x^2 \) can take values from \( 0 \) to \( 9 \). The integer values of \( n \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \).

The corresponding \( x \) values for these integers are:
- \( n = 0 \) gives \( x = 0 \)
- \( n = 1 \) gives \( x = \pm 1 \)
- \( n = 2 \) gives \( x = \pm \sqrt{2} \)
- \( n = 3 \) gives \( x = \pm \sqrt{3} \)
- \( n = 4 \) gives \( x = \pm 2 \)

### Step 3: Identify points where \( \sin(\pi x) = 0 \)
The sine function is zero at integer multiples of \( \pi \):
- \( \sin(\pi x) = 0 \) when \( x \) is an integer.

The integers in the interval \( [-2, 3] \) are \( -2, -1, 0, 1, 2, 3 \).

### Step 4: Find points of discontinuity
The function \( f(x) \) will be discontinuous at points where \( [x^2] \) changes and \( \sin(\pi x) \) is not zero.

From the values we found:
- \( [x^2] \) changes at \( x = 0, \pm 1, \pm \sqrt{2}, \pm \sqrt{3}, \pm 2 \).
- The points where \( \sin(\pi x) = 0 \) are \( -2, -1, 0, 1, 2, 3 \).

Now, we need to exclude the points where both conditions are satisfied (i.e., where \( [x^2] \) changes and \( \sin(\pi x) = 0 \)):
- The points \( -2, -1, 0, 1, 2 \) are both points of discontinuity due to \( [x^2] \) and \( \sin(\pi x) \).

### Step 5: Count the discontinuous points
The points of discontinuity where \( [x^2] \) changes but \( \sin(\pi x) \) is not zero are:
- \( \sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3} \)

Thus, the total points of discontinuity are:
- \( -\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3} \) (4 points).

Adding the points where \( \sin(\pi x) \) is zero and \( [x^2] \) does not change:
- \( -2, -1, 0, 1, 2 \) (5 points).

### Final Count
The total number of points where \( f(x) \) is discontinuous is \( 4 + 5 = 9 \).

### Conclusion
The number of values of \( x \) in the interval \( [-2, 3] \) where \( f(x) \) is discontinuous is **9**.

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