If domain of `f(x)` is`(-oo,0)`, then domain of `f(6{x}^2-5(x)+ 1)` is (where {.} represents fractional part function)

Options

A. `bigcup_(n epsilonI)[n+1/3,n+1/2]`
B. `(-oo,0)`
C. `bigcup_(n epsilonI)[n+1/6,n+1]`
D. `bigcup_(n epsilonI)[n-1/2,n-1/2]`

Explanation

To find the domain of the function \( f(6\{x\}^2 - 5\{x\} + 1) \), where \( \{x\} \) represents the fractional part of \( x \), we start with the given domain of \( f(x) \), which is \( (-\infty, 0] \).

### Step 1: Understand the fractional part function
The fractional part function \( \{x\} \) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
This means \( \{x\} \) always lies in the interval \( [0, 1) \).

### Step 2: Determine the range of \( 6\{x\}^2 - 5\{x\} + 1 \)
Next, we need to analyze the expression \( 6\{x\}^2 - 5\{x\} + 1 \) for \( \{x\} \in [0, 1) \).

Let \( y = \{x\} \). Then we need to evaluate the quadratic function:
\[ g(y) = 6y^2 - 5y + 1 \]
for \( y \in [0, 1) \).

### Step 3: Find the critical points of \( g(y) \)
To find the minimum or maximum values of \( g(y) \), we can calculate its derivative:
\[ g'(y) = 12y - 5 \]
Setting the derivative to zero to find critical points:
\[ 12y - 5 = 0 \implies y = \frac{5}{12} \]
This critical point \( \frac{5}{12} \) lies within the interval \( [0, 1) \).

### Step 4: Evaluate \( g(y) \) at the endpoints and critical points
Now we evaluate \( g(y) \) at the endpoints \( y = 0 \), \( y = 1 \) (not included), and at the critical point \( y = \frac{5}{12} \):
1. At \( y = 0 \):
\[ g(0) = 6(0)^2 - 5(0) + 1 = 1 \]
2. At \( y = \frac{5}{12} \):
\[ g\left(\frac{5}{12}\right) = 6\left(\frac{5}{12}\right)^2 - 5\left(\frac{5}{12}\right) + 1 \]
Calculate:
\[ = 6 \cdot \frac{25}{144} - \frac{25}{12} + 1 = \frac{150}{144} - \frac{300}{144} + \frac{144}{144} = \frac{150 - 300 + 144}{144} = \frac{-6}{144} = -\frac{1}{24} \]
3. As \( y \) approaches \( 1 \):
\[ g(1) = 6(1)^2 - 5(1) + 1 = 6 - 5 + 1 = 2 \]

### Step 5: Determine the range of \( g(y) \)
The minimum value of \( g(y) \) occurs at \( y = \frac{5}{12} \) and is \( -\frac{1}{24} \), and the maximum value as \( y \) approaches \( 1 \) is \( 2 \). Thus, the range of \( g(y) \) is:
\[ [-\frac{1}{24}, 2) \]

### Step 6: Find the domain of \( f(g(y)) \)
Since the domain of \( f(x) \) is \( (-\infty, 0] \), we need \( g(y) \leq 0 \):
\[ -\frac{1}{24} \leq 0 \]
This means \( g(y) \) is valid for \( y \) values that yield outputs in the range \( [-\frac{1}{24}, 0] \).

### Step 7: Solve \( g(y) \leq 0 \)
To find the values of \( y \) such that \( g(y) \leq 0 \):
\[ 6y^2 - 5y + 1 \leq 0 \]
We can find the roots of the equation:
\[ 6y^2 - 5y + 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} \]
This gives us:
\[ y = \frac{6}{12} = \frac{1}{2} \quad \text{and} \quad y = \frac{4}{12} = \frac{1}{3} \]

### Step 8: Determine the intervals
The quadratic opens upwards (since the coefficient of \( y^2 \) is positive), so it is negative between the roots:
\[ \frac{1}{3} \leq y \leq \frac{1}{2} \]

### Step 9: Convert back to \( x \)
Since \( y = \{x\} \), we can express \( x \) as:
\[ \{x\} = y \implies x = n + y \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
\[ n + \frac{1}{3} \leq x < n + \frac{1}{2} \quad \text{for integers } n \]

### Final Answer
The domain of \( f(6\{x\}^2 - 5\{x\} + 1) \) is:
\[ \bigcup_{n \in \mathbb{Z}} \left[n + \frac{1}{3}, n + \frac{1}{2}\right) \]