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 will analyze the function \( f(x) = -x^2 + x + p \) and determine the conditions under which the greatest integer function \( g(x) = [f(x)] \) is discontinuous at \( x = \frac{1}{2} \).

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

First, we need to find the value of \( f\left(\frac{1}{2}\right) \):

\[ f\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + p = -\frac{1}{4} + \frac{1}{2} + p \]

Now, simplifying this expression:

\[ f\left(\frac{1}{2}\right) = -\frac{1}{4} + \frac{2}{4} + p = \frac{1}{4} + p \]

### Step 2: Determine when \( g(x) \) is discontinuous

The function \( g(x) = [f(x)] \) will be discontinuous at \( x = \frac{1}{2} \) if \( f\left(\frac{1}{2}\right) \) is an integer. This means that \( \frac{1}{4} + p \) must be an integer.

Let \( n \) be an integer. Then we can write:

\[ \frac{1}{4} + p = n \implies p = n - \frac{1}{4} \]

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

From the equation \( p = n - \frac{1}{4} \), we can see that \( p \) will take values that are \( \frac{3}{4}, \frac{7}{4}, \frac{11}{4}, \ldots \) when \( n = 1, 2, 3, \ldots \) or \( -\frac{1}{4}, -\frac{5}{4}, -\frac{9}{4}, \ldots \) when \( n = 0, -1, -2, \ldots \).

### Step 4: Identify the integer values of \( p \)

To ensure that \( g(x) \) is discontinuous at \( x = \frac{1}{2} \), \( p \) must not be of the form \( n - \frac{1}{4} \) for any integer \( n \). This means \( p \) cannot be \( \frac{1}{4} \) less than any integer.

### Conclusion

Thus, the value of \( p \) cannot be \( \frac{1}{4} \) because if \( p = \frac{1}{4} \), then \( f\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \), which is not an integer, making \( g(x) \) continuous at that point.

### Final Answer

Therefore, \( p \) cannot be \( \frac{1}{4} \).

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