Question

The number of integers lying in the domain of the function $f(x) = \sqrt{\log_{0.5}(\frac{5-2x}{x})}$ is

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (C)

1

Answer 2

The domain of the function $f(x) = \sqrt{\log_{0.5}(\frac{5-2x}{x})}$ is determined by two conditions:

1. The expression under the square root must be non-negative: $\log_{0.5}(\frac{5-2x}{x}) \ge 0$.
2. The argument of the logarithm must be positive: $\frac{5-2x}{x} > 0$.

Let's analyze the second condition first: $\frac{5-2x}{x} > 0$. This inequality holds when the numerator and the denominator have the same sign.

Case 1: $5-2x > 0$ and $x > 0$. $5-2x > 0 \implies 5 > 2x \implies x < 5/2$. So, $x > 0$ and $x < 5/2$. This gives $0 < x < 5/2$.

Case 2: $5-2x < 0$ and $x < 0$. $5-2x < 0 \implies 5 < 2x \implies x > 5/2$. So, $x > 5/2$ and $x < 0$. There is no value of $x$ that satisfies both inequalities simultaneously.

The solution to $\frac{5-2x}{x} > 0$ is $0 < x < 5/2$.

Now let's analyze the first condition: $\log_{0.5}(\frac{5-2x}{x}) \ge 0$. The base of the logarithm is 0.5, which is between 0 and 1. For a logarithm with base $b$ such that $0 < b < 1$, $\log_b(y) \ge 0$ is equivalent to $0 < y \le 1$. In this case, $y = \frac{5-2x}{x}$. So, we must have $0 < \frac{5-2x}{x} \le 1$. This is a compound inequality: $\frac{5-2x}{x} > 0$ and $\frac{5-2x}{x} \le 1$.

We have already solved $\frac{5-2x}{x} > 0$, which gives $0 < x < 5/2$.

Now we solve the second part of the compound inequality: $\frac{5-2x}{x} \le 1$.

$\frac{5-2x}{x} - 1 \le 0$

$\frac{5-2x - x}{x} \le 0$

$\frac{5-3x}{x} \le 0$.

This inequality holds when the numerator and the denominator have opposite signs, or the numerator is zero.

Case A: $5-3x \ge 0$ and $x < 0$. $5-3x \ge 0 \implies 5 \ge 3x \implies x \le 5/3$. So, $x \le 5/3$ and $x < 0$. This gives $x < 0$.

Case B: $5-3x \le 0$ and $x > 0$. $5-3x \le 0 \implies 5 \le 3x \implies x \ge 5/3$. So, $x \ge 5/3$ and $x > 0$. This gives $x \ge 5/3$.

The solution to $\frac{5-3x}{x} \le 0$ is $x < 0$ or $x \ge 5/3$. This can be written as $(-\infty, 0) \cup [5/3, \infty)$.

The domain of $f(x)$ is the set of $x$ values that satisfy both $\frac{5-2x}{x} > 0$ and $\frac{5-2x}{x} \le 1$. This is the intersection of the solution sets of the two inequalities: $(0, 5/2) \cap ((-\infty, 0) \cup [5/3, \infty))$.

Let's find the intersection: The interval $(0, 5/2)$ is $(0, 2.5)$. The set $(-\infty, 0) \cup [5/3, \infty)$ is $(-\infty, 0) \cup [1.666..., \infty)$. The intersection of $(0, 2.5)$ with $(-\infty, 0)$ is empty. The intersection of $(0, 2.5)$ with $[5/3, \infty)$ is the set of $x$ such that $0 < x < 2.5$ and $x \ge 5/3$. This intersection is $[5/3, 5/2)$.

The domain of the function is $[5/3, 5/2)$. We need to find the number of integers lying in this domain. The interval is $[1.666..., 2.5)$. The integers in this interval are the integers $k$ such that $1.666... \le k < 2.5$. The only integer that satisfies this condition is $k=2$.

There is only one integer in the domain of the function.