The complete set of values of x in the domain of function `f (x)= sqrt(log _(x+2{x})([x] ^(2)-5[x] +7))` where [.] denote greatest integer functioon and `{.}` denote fraction pert function) is :

Options

A. `(-(1)/(3), 0) uu ((1)/(3), 1) uu (2,oo)`
B. `(0,1)uu(1,oo)`
C. `(-(2)/(3), 0) uu((1)/(3), 1) uu (1,oo)`
D. `(-(1)/(3),0) uu((1)/(3), 1) uu (1,oo)`

Answer

To find the complete set of values of \( x \) in the domain of the function

\[ f(x) = \sqrt{\log_{(x + 2 \{x\})}([x]^2 - 5[x] + 7)} \]

where \([x]\) denotes the greatest integer function and \(\{x\}\) denotes the fractional part function, we need to ensure that the expression inside the square root is defined and non-negative.

### Step 1: Determine the conditions for the logarithm

The logarithm is defined only when its base is positive and not equal to 1, and its argument must be positive.

1. **Base Condition**:
\[ x + 2\{x\} > 0 \quad \text{and} \quad x + 2\{x\} \neq 1 \]

2. **Argument Condition**:
\[ [x]^2 - 5[x] + 7 > 0 \]

### Step 2: Analyze the base condition

The base \( x + 2\{x\} \) can be rewritten as:
\[ x + 2\{x\} = x + 2(x - [x]) = x + 2x - 2[x] = 3x - [x] \]

Since \([x]\) is the greatest integer less than or equal to \(x\), we have:
\[ 3x - [x] > 0 \implies 3x > [x] \]
This is always true since \([x] \leq x\).

Next, we check when \(3x - [x] \neq 1\):
\[ 3x - [x] \neq 1 \implies 3x \neq [x] + 1 \]

### Step 3: Analyze the argument condition

Now, we analyze the quadratic \( [x]^2 - 5[x] + 7 \):
The discriminant of this quadratic is:
\[ D = (-5)^2 - 4 \cdot 1 \cdot 7 = 25 - 28 = -3 \]
Since the discriminant is negative, the quadratic is always positive. Therefore, this condition is satisfied for all \( x \).

### Step 4: Solve the base condition

Now, we need to solve the inequalities derived from the base condition:
1. \( 3x > [x] \)
2. \( 3x \neq [x] + 1 \)

#### Case 1: \( x \) in the interval \( [n, n+1) \) where \( n \) is an integer.

In this interval, \([x] = n\). Therefore:
1. \( 3x > n \) implies \( x > \frac{n}{3} \).
2. \( 3x \neq n + 1 \) implies \( x \neq \frac{n + 1}{3} \).

Thus, the valid range for \( x \) in this interval is:
\[ \left(\frac{n}{3}, n+1\right) \quad \text{and} \quad x \neq \frac{n + 1}{3} \]

### Step 5: Combine intervals

We need to consider all integers \( n \):
1. For \( n = 0 \): \( x \in \left(0, 1\right) \)
2. For \( n = 1 \): \( x \in \left(\frac{1}{3}, 2\right) \) excluding \( \frac{2}{3} \)
3. For \( n = 2 \): \( x \in \left(\frac{2}{3}, 3\right) \)
4. For \( n = 3 \): \( x \in \left(1, 4\right) \)
5. For \( n \geq 4 \): \( x \in \left(n/3, n+1\right) \)

### Final Solution

Combining all these intervals, we get:
\[ (-\frac{1}{3}, 0) \cup \left(\frac{1}{3}, 1\right) \cup (1, \infty) \]