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

Options

A. `underset(n inI)uuu[n + 1/3,n + 1/2]`
B. `(-oo,0)`
C. `underset(n in I)uuu[ n + 1/6n +1]`
D. None of these

Answer

To find the domain of the function \( f(6\{x\}^2 - 5\{x\} + 1) \) given that the domain of \( f(x) \) is \( (-\infty, 0) \), we need to ensure that the expression \( 6\{x\}^2 - 5\{x\} + 1 \) falls within the domain of \( f(x) \).

### Step-by-Step Solution:

1. **Understanding the Fractional Part Function**:
The fractional part function \( \{x\} \) is defined as \( \{x\} = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). The output of \( \{x\} \) lies in the interval \( [0, 1) \).

2. **Setting Up the Inequality**:
We want to find when \( 6\{x\}^2 - 5\{x\} + 1 < 0 \). This will give us the values of \( \{x\} \) for which the output falls within the domain of \( f \).

3. **Rearranging the Inequality**:
Let's denote \( y = \{x\} \). We need to solve the inequality:
\[ 6y^2 - 5y + 1 < 0 \]

4. **Finding the Roots**:
To solve the quadratic inequality, we first 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 two roots:
\[ y_1 = \frac{6}{12} = \frac{1}{2}, \quad y_2 = \frac{4}{12} = \frac{1}{3} \]

5. **Analyzing the Sign of the Quadratic**:
The quadratic \( 6y^2 - 5y + 1 \) opens upwards (since the coefficient of \( y^2 \) is positive). The quadratic will be negative between its roots:
\[ \frac{1}{3} < y < \frac{1}{2} \]

6. **Mapping Back to \( x \)**:
Since \( y = \{x\} \), we have:
\[ \frac{1}{3} < \{x\} < \frac{1}{2} \]
This means that \( x \) can be expressed as \( n + \{x\} \) where \( n \) is an integer. Therefore, the possible values of \( x \) can be represented as:
\[ n + \frac{1}{3} < x < n + \frac{1}{2} \]
for any integer \( n \).

7. **Final Domain**:
The overall domain of \( f(6\{x\}^2 - 5\{x\} + 1) \) is the union of all intervals of the form:
\[ (n + \frac{1}{3}, n + \frac{1}{2}) \quad \text{for all integers } n \]

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