We are given a piecewise function $f(x)$ defined as: $f(x) = \begin{cases} 1-\sqrt{-x-1} & \text{if } x \le 0 \\ \frac{1}{[\frac{1}{2x}] - 3} & \text{if } x > 0 \end{cases}$ where $[x]$ is the greatest integer less than or equal to $x$. We are asked to find the domain of $f(x)$, the range of $f(x)$, and discuss the continuity of $f(x)$ at $a=1$.

1. Problem Description

We are given a piecewise function $f(x)$ defined as:

$f(x) = \begin{cases} 1-\sqrt{-x-1} & \text{if } x \le 0 \\ \frac{1}{[\frac{1}{2x}] - 3} & \text{if } x > 0 \end{cases}$

where $[x]$ is the greatest integer less than or equal to $x$.

We are asked to find the domain of $f(x)$, the range of $f(x)$, and discuss the continuity of $f(x)$ at $a=1$.

2. Solution Steps

a) Finding the Domain $D_f$ of $f(x)$.

For $x \le 0$, we have $f(x) = 1 - \sqrt{-x-1}$.

For the square root to be defined, we require $-x-1 \ge 0$, which means $-x \ge 1$, so $x \le -1$.

Thus, for $x \le 0$, we must have $x \le -1$.

For $x > 0$, we have $f(x) = \frac{1}{[\frac{1}{2x}] - 3}$.

For the function to be defined, we must have $[\frac{1}{2x}] - 3 \ne 0$, which means $[\frac{1}{2x}] \ne 3$.

So, we must have $3 \le \frac{1}{2x} < 4$ is false. In other words, $\frac{1}{2x} < 3$ or $\frac{1}{2x} \ge 4$.

If $\frac{1}{2x} < 3$, then $1 < 6x$, so $x > \frac{1}{6}$.

If $\frac{1}{2x} \ge 4$, then $1 \ge 8x$, so $x \le \frac{1}{8}$.

Thus, for $x > 0$, we must have $0 < x \le \frac{1}{8}$ or $x > \frac{1}{6}$.

Therefore, the domain of $f(x)$ is $(-\infty, -1] \cup (0, \frac{1}{8}] \cup (\frac{1}{6}, \infty)$.

b) Finding the Range $R_f$ of $f(x)$.

For $x \le -1$, we have $f(x) = 1 - \sqrt{-x-1}$.

Let $u = -x-1$. Then $u \ge 0$.

$f(x) = 1 - \sqrt{u}$. Since $u \ge 0$, $\sqrt{u} \ge 0$, so $1 - \sqrt{u} \le 1$.

When $x = -1$, $u = -(-1)-1 = 0$, so $f(-1) = 1 - \sqrt{0} = 1$.

As $x \to -\infty$, $u \to \infty$, so $\sqrt{u} \to \infty$, and $1 - \sqrt{u} \to -\infty$.

So, for $x \le -1$, the range of $f(x)$ is $(-\infty, 1]$.

For $x > 0$, we have $f(x) = \frac{1}{[\frac{1}{2x}] - 3}$.

Let $n = [\frac{1}{2x}]$. Then $n$ can be any integer except

3. So $f(x) = \frac{1}{n-3}$ can be any number except when $n=3$.

If $n = 0$, $f(x) = -\frac{1}{3}$. If $n = 1$, $f(x) = -\frac{1}{2}$. If $n = 2$, $f(x) = -1$. If $n = 4$, $f(x) = 1$. If $n = 5$, $f(x) = \frac{1}{2}$.

If $x > 0$, $f(x)$ can take any value of the form $\frac{1}{n-3}$ where $n$ is an integer and $n \ne 3$.

Since $n$ can be any integer except 3, $n-3$ can be any integer except 0, so $\frac{1}{n-3}$ can be any non-zero integer reciprocal.

We have $x > \frac{1}{6}$ or $0 < x \le \frac{1}{8}$.

Consider $0 < x \le \frac{1}{8}$. Then $\frac{1}{2x} \ge 4$. So $[\frac{1}{2x}] \ge 4$. Therefore, $n \ge 4$.

In this case, $f(x)$ can take on the values $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$, i.e., $(0,1]$. But since the domain is $x>\frac{1}{6}$ and $0 < x \le \frac{1}{8}$, the range is not continuous, but discrete values.

If $x > \frac{1}{6}$, then $\frac{1}{2x} < 3$, so $[\frac{1}{2x}] \le 2$. Therefore $n \le 2$.

In this case, $f(x)$ can take on the values $-1, -\frac{1}{2}, -\frac{1}{3}, ...$.

Overall, the range is $(-\infty, 1] \cup \{\frac{1}{n-3} | n \in Z, n\ne 3 \}$

c) Discuss the continuity of $f(x)$ at $a=1$.

Since $1 > 0$, we use the second definition of $f(x)$.

$f(1) = \frac{1}{[\frac{1}{2(1)}] - 3} = \frac{1}{[\frac{1}{2}] - 3} = \frac{1}{0 - 3} = -\frac{1}{3}$.

We need to find $\lim_{x \to 1} f(x)$.

As $x$ approaches 1, $\frac{1}{2x}$ approaches $\frac{1}{2}$. Thus, $[\frac{1}{2x}]$ approaches

0. $\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{1}{[\frac{1}{2x}] - 3} = \frac{1}{0 - 3} = -\frac{1}{3}$.

Since $f(1) = \lim_{x \to 1} f(x)$, $f(x)$ is continuous at $x=1$.

3. Final Answer

a) Domain: $(-\infty, -1] \cup (0, \frac{1}{8}] \cup (\frac{1}{6}, \infty)$

b) Range: $(-\infty, 1] \cup \{\frac{1}{n-3} | n \in Z, n\ne 3 \}$

c) $f(x)$ is continuous at $x=1$.