The domain of function `f (x) = log _([x+(1)/(2)])(2x ^(2) + x-1), ` where `[.]` denotes the greatest integer function is :

Options

A. `[(3)/(2), oo)`
B. `(2,oo)`
C. `(-(1)/(2), oo)-{(1)/(2)}`
D. `((1)/(2), 1) uu (1,oo)`

Answer

To find the domain of the function \( f(x) = \log_{(x + \frac{1}{2})}(2x^2 + x - 1) \), where \([.]\) denotes the greatest integer function, we need to ensure that the logarithm is defined.

### Step 1: Conditions for the logarithm
The logarithm \( \log_b(a) \) is defined under the following conditions:
1. The base \( b \) must be greater than 0 and not equal to 1.
2. The argument \( a \) must be greater than 0.

### Step 2: Applying the conditions to our function
For our function:
- The base is \( x + \frac{1}{2} \)
- The argument is \( 2x^2 + x - 1 \)

We need to satisfy the following conditions:
1. \( x + \frac{1}{2} > 0 \)
2. \( x + \frac{1}{2} \neq 1 \)
3. \( 2x^2 + x - 1 > 0 \)

### Step 3: Solve the inequalities

#### Condition 1: \( x + \frac{1}{2} > 0 \)
\[ x + \frac{1}{2} > 0 \implies x > -\frac{1}{2} \]

#### Condition 2: \( x + \frac{1}{2} \neq 1 \)
\[ x + \frac{1}{2} \neq 1 \implies x \neq \frac{1}{2} \]

#### Condition 3: \( 2x^2 + x - 1 > 0 \)
To solve this quadratic inequality, we first find the roots of the equation \( 2x^2 + x - 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us the roots:
\[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \]

Now we can analyze the sign of \( 2x^2 + x - 1 \) in the intervals determined by these roots: \( (-\infty, -1) \), \( (-1, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).

- For \( x < -1 \): Choose \( x = -2 \)
\[ 2(-2)^2 + (-2) - 1 = 8 - 2 - 1 = 5 > 0 \]
- For \( -1 < x < \frac{1}{2} \): Choose \( x = 0 \)
\[ 2(0)^2 + (0) - 1 = -1 < 0 \]
- For \( x > \frac{1}{2} \): Choose \( x = 1 \)
\[ 2(1)^2 + (1) - 1 = 2 + 1 - 1 = 2 > 0 \]

Thus, \( 2x^2 + x - 1 > 0 \) in the intervals \( (-\infty, -1) \) and \( (\frac{1}{2}, \infty) \).

### Step 4: Combine the conditions
Now we combine the conditions:
1. From \( x + \frac{1}{2} > 0 \): \( x > -\frac{1}{2} \)
2. From \( x + \frac{1}{2} \neq 1 \): \( x \neq \frac{1}{2} \)
3. From \( 2x^2 + x - 1 > 0 \): \( x \in (-\infty, -1) \cup (\frac{1}{2}, \infty) \)

The intersection of these conditions gives us:
- From \( (-\infty, -1) \) and \( x > -\frac{1}{2} \): The valid interval is \( (-\frac{1}{2}, -1) \) (but this is empty since there are no numbers between \(-\frac{1}{2}\) and \(-1\)).
- From \( (\frac{1}{2}, \infty) \) and \( x \neq \frac{1}{2} \): The valid interval is \( (\frac{1}{2}, \infty) \).

### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{(\frac{1}{2}, \infty)} \]