Let `f(x)=-x^(2)+x+p`, where p is a real number. If `g(x)=[f(x)] and g(x)` is discontinuous at `x=(1)/(2)`, then p - cannot be (where `[.]` represents the greatest integer function)

Options

A. `(1)/(2)`
B. `(3)/(4)`
C. `(7)/(4)`
D. `-(1)/(4)`

Answer

To solve the problem, we need to analyze the function \( f(x) = -x^2 + x + p \) and the greatest integer function \( g(x) = [f(x)] \), where \( [.] \) denotes the greatest integer function. We are given that \( g(x) \) is discontinuous at \( x = \frac{1}{2} \).

### Step 1: Calculate \( f\left(\frac{1}{2}\right) \)

First, we need to find the value of \( f\left(\frac{1}{2}\right) \):
\[ f\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + p \]
Calculating this gives:
\[ f\left(\frac{1}{2}\right) = -\frac{1}{4} + \frac{1}{2} + p = -\frac{1}{4} + \frac{2}{4} + p = \frac{1}{4} + p \]

### Step 2: Determine the condition for discontinuity of \( g(x) \)

The function \( g(x) \) is discontinuous at \( x = \frac{1}{2} \) if \( f\left(\frac{1}{2}\right) \) is an integer. Therefore, we need:
\[ \frac{1}{4} + p \text{ must be an integer} \]
Let \( k \) be an integer. Then we can write:
\[ \frac{1}{4} + p = k \implies p = k - \frac{1}{4} \]

### Step 3: Analyze values of \( p \)

From the equation \( p = k - \frac{1}{4} \), we can see that \( p \) will be an integer minus \( \frac{1}{4} \). This means \( p \) can take values like \( k - \frac{1}{4} \) for any integer \( k \).

### Step 4: Identify values of \( p \) that cannot satisfy the condition

Now, we need to find a value of \( p \) that cannot be expressed in the form \( k - \frac{1}{4} \).

If we set \( p = \frac{1}{2} \):
\[ \frac{1}{4} + p = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]
Since \( \frac{3}{4} \) is not an integer, \( g(x) \) will be continuous at \( x = \frac{1}{2} \).

### Conclusion

Thus, the value of \( p \) that cannot make \( g(x) \) discontinuous at \( x = \frac{1}{2} \) is:
\[ \boxed{\frac{1}{2}} \]