Question

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

Answer 1

$\bigcup_{n \in \mathbb{Z}} \left[n + \frac{1}{3}, n + \frac{1}{2}\right]$

Answer 2

The problem asks for the domain of the composite function $f(g(x))$, where $f(x)$ has a domain of $(-\infty, 0]$ and $g(x) = 6\{x\}^2 - 5\{x\} + 1$. Here, $\{\cdot\}$ represents the fractional part function.

For $f(g(x))$ to be defined, the argument of $f$ must be within its domain. Given that the domain of $f(x)$ is $(-\infty, 0]$, we must have $g(x) \le 0$. So, we need to solve the inequality: $6\{x\}^2 - 5\{x\} + 1 \le 0$

Let $y = \{x\}$. We know that the range of the fractional part function is $0 \le \{x\} < 1$. Substitute $y$ into the inequality: $6y^2 - 5y + 1 \le 0$

To solve this quadratic inequality, we first find the roots of the corresponding quadratic equation $6y^2 - 5y + 1 = 0$. We can factorize the quadratic expression: $6y^2 - 3y - 2y + 1 = 0$ $3y(2y - 1) - 1(2y - 1) = 0$ $(3y - 1)(2y - 1) = 0$

The roots are $y = \frac{1}{3}$ and $y = \frac{1}{2}$.

Since the coefficient of $y^2$ (which is 6) is positive, the parabola $6y^2 - 5y + 1$ opens upwards. Therefore, the quadratic expression is less than or equal to zero between its roots. So, the inequality $6y^2 - 5y + 1 \le 0$ holds when: $\frac{1}{3} \le y \le \frac{1}{2}$

Now, substitute back $y = \{x\}$: $\frac{1}{3} \le \{x\} \le \frac{1}{2}$

We need to find the values of $x$ for which its fractional part lies in the interval $[\frac{1}{3}, \frac{1}{2}]$. The fractional part of $x$, denoted by $\{x\}$, is defined as $\{x\} = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Let $n = \lfloor x \rfloor$, where $n$ is an integer ($n \in \mathbb{Z}$). Then the inequality becomes: $\frac{1}{3} \le x - n \le \frac{1}{2}$

To find $x$, we add $n$ to all parts of the inequality: $n + \frac{1}{3} \le x \le n + \frac{1}{2}$

Since this condition must hold for any integer $n$, the domain of the function $f(6\{x\}^2 - 5\{x\} + 1)$ is the union of all such intervals for all integers $n$.

Domain $= \bigcup_{n \in \mathbb{Z}} \left[n + \frac{1}{3}, n + \frac{1}{2}\right]$

The domain consists of an infinite number of disjoint closed intervals. For example, for $n=0$, $x \in [\frac{1}{3}, \frac{1}{2}]$; for $n=1$, $x \in [\frac{4}{3}, \frac{3}{2}]$; for $n=-1$, $x \in [-\frac{2}{3}, -\frac{1}{2}]$, and so on.