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 both the logarithmic function is defined and valid. This requires two conditions to be satisfied:

1. The argument of the logarithm must be positive: \( 2x^2 + x - 1 > 0 \).
2. The base of the logarithm must be positive and not equal to 1: \( [x + \frac{1}{2}] > 0 \) and \( [x + \frac{1}{2}] \neq 1 \).

### Step 1: Solve \( 2x^2 + x - 1 > 0 \)

First, we will factor the quadratic expression \( 2x^2 + x - 1 \).

To factor, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 1, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 \]
Now substituting back into the quadratic formula:
\[ x = \frac{-1 \pm 3}{4} \]
This gives us two roots:
\[ x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-4}{4} = -1 \]

Now we can factor \( 2x^2 + x - 1 \):
\[ 2x^2 + x - 1 = 2(x + 1)(x - \frac{1}{2}) \]

To find where this expression is greater than zero, we analyze the sign of the product \( 2(x + 1)(x - \frac{1}{2}) > 0 \).

The critical points are \( x = -1 \) and \( x = \frac{1}{2} \).

- For \( x < -1 \): Both factors are negative, so the product is positive.
- For \( -1 < x < \frac{1}{2} \): One factor is negative and the other is positive, so the product is negative.
- For \( x > \frac{1}{2} \): Both factors are positive, so the product is positive.

Thus, the solution for \( 2x^2 + x - 1 > 0 \) is:
\[ x \in (-\infty, -1) \cup \left(\frac{1}{2}, \infty\right) \]

### Step 2: Solve \( [x + \frac{1}{2}] > 0 \)

The greatest integer function \([x + \frac{1}{2}]\) is positive when:
\[ x + \frac{1}{2} \geq 1 \implies x \geq \frac{1}{2} \]
Thus, \([x + \frac{1}{2}] > 0\) gives us:
\[ x \geq \frac{1}{2} \]

### Step 3: Solve \( [x + \frac{1}{2}] \neq 1 \)

Next, we need to ensure that \([x + \frac{1}{2}] \neq 1\):
\[ 1 \leq x + \frac{1}{2} < 2 \implies \frac{1}{2} \leq x < \frac{3}{2} \]
This means:
\[ [x + \frac{1}{2}] = 1 \text{ for } x \in \left[\frac{1}{2}, \frac{3}{2}\right) \]

### Step 4: Combine Conditions

Now we combine the conditions we have:

1. From \( 2x^2 + x - 1 > 0 \): \( x \in (-\infty, -1) \cup \left(\frac{1}{2}, \infty\right) \)
2. From \( [x + \frac{1}{2}] > 0 \): \( x \geq \frac{1}{2} \)
3. From \( [x + \frac{1}{2}] \neq 1 \): \( x \notin \left[\frac{1}{2}, \frac{3}{2}\right) \)

The valid intervals are:
- From \( \left(\frac{1}{2}, \infty\right) \) excluding \( \left[\frac{1}{2}, \frac{3}{2}\right) \), which gives us \( x \in \left[\frac{3}{2}, \infty\right) \).

### Final Domain

Thus, the domain of the function \( f(x) \) is:
\[ \boxed{\left[\frac{3}{2}, \infty\right)} \]