The function `f(x) =[x]^(2) -[x^(2)]` (where [y] is thegreatest integer less than or equal to y), is discontinuous at:

Options

A. all integers
B. all integrers except 0 and 1
C. all integers except 0
D. all integers except 1

Answer

To determine the points of discontinuity of the function \( f(x) = [x^2] - [x]^2 \), where \([y]\) denotes the greatest integer less than or equal to \(y\), we need to analyze the behavior of the function around integer values.

### Step 1: Identify the function behavior around integers
The function involves the greatest integer function, which can cause discontinuities at integer points. We will check the continuity at integers \( n \).

### Step 2: Evaluate the left-hand limit at integer \( n \)
To find the left-hand limit as \( x \) approaches \( n \) from the left (\( n - h \) where \( h \to 0^+ \)):
\[ f(n - h) = [ (n - h)^2 ] - [n - h]^2 \]
Calculating \( (n - h)^2 \):
\[ (n - h)^2 = n^2 - 2nh + h^2 \]
Since \( n - h \) is slightly less than \( n \), we have:
\[ [n - h] = n - 1 \]
Thus:
\[ f(n - h) = [n^2 - 2nh + h^2] - (n - 1)^2 \]
Calculating \( (n - 1)^2 \):
\[ (n - 1)^2 = n^2 - 2n + 1 \]
Now, we need to find \( [n^2 - 2nh + h^2] \). As \( h \to 0 \), \( n^2 - 2nh + h^2 \) approaches \( n^2 \), so:
\[ [n^2 - 2nh + h^2] = n^2 - 1 \quad \text{(for small } h\text{)} \]
Thus:
\[ f(n - h) = (n^2 - 1) - (n^2 - 2n + 1) = 2n - 2 \]

### Step 3: Evaluate the right-hand limit at integer \( n \)
Now, consider the right-hand limit as \( x \) approaches \( n \) from the right (\( n + h \)):
\[ f(n + h) = [ (n + h)^2 ] - [n + h]^2 \]
Calculating \( (n + h)^2 \):
\[ (n + h)^2 = n^2 + 2nh + h^2 \]
Since \( n + h \) is slightly greater than \( n \), we have:
\[ [n + h] = n \]
Thus:
\[ f(n + h) = [n^2 + 2nh + h^2] - n^2 \]
As \( h \to 0 \), \( n^2 + 2nh + h^2 \) approaches \( n^2 \), so:
\[ [n^2 + 2nh + h^2] = n^2 \]
Thus:
\[ f(n + h) = n^2 - n^2 = 0 \]

### Step 4: Compare left-hand and right-hand limits
For continuity at \( n \):
\[ \text{Left-hand limit} = 2n - 2 \]
\[ \text{Right-hand limit} = 0 \]
Setting them equal for continuity:
\[ 2n - 2 = 0 \implies n = 1 \]

### Step 5: Identify points of discontinuity
The function is continuous at \( n = 1 \) but discontinuous at all other integers. Therefore, the points of discontinuity are all integers except \( n = 1 \).

### Final Answer
The function \( f(x) \) is discontinuous at all integers except \( 1 \).