The number of integers satisfying `log_((1)/(x))((2(x-2))/((x+1)(x-5)))ge 1` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the inequality \( \log_{\frac{1}{x}}\left(\frac{2(x-2)}{(x+1)(x-5)}\right) \geq 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic inequality
The inequality can be rewritten using the properties of logarithms. The expression \( \log_{\frac{1}{x}}(y) \geq 1 \) implies that:
\[ y \leq \frac{1}{x} \]
Thus, we have:
\[ \frac{2(x-2)}{(x+1)(x-5)} \leq \frac{1}{x} \]

### Step 2: Cross-multiply to eliminate the fraction
To eliminate the fraction, we can cross-multiply (keeping in mind the signs of the terms):
\[ 2(x-2) \leq \frac{(x+1)(x-5)}{x} \]
This simplifies to:
\[ 2(x-2)x \leq (x+1)(x-5) \]

### Step 3: Expand both sides
Now, we expand both sides:
\[ 2x^2 - 4x \leq x^2 - 4x - 5 \]

### Step 4: Rearrange the inequality
Rearranging gives:
\[ 2x^2 - 4x - x^2 + 4x + 5 \leq 0 \]
This simplifies to:
\[ x^2 + 5 \leq 0 \]

### Step 5: Analyze the quadratic inequality
The inequality \( x^2 + 5 \leq 0 \) has no real solutions since \( x^2 \) is always non-negative and adding 5 makes it positive. Therefore, this part does not contribute any solutions.

### Step 6: Consider the case when \( \frac{1}{x} > 1 \)
Next, we consider the case when \( \frac{1}{x} > 1 \), which implies \( 0 < x < 1 \). In this case, we need to solve:
\[ \frac{2(x-2)}{(x+1)(x-5)} \geq 1 \]

### Step 7: Cross-multiply again
Cross-multiplying gives:
\[ 2(x-2) \geq (x+1)(x-5) \]

### Step 8: Expand and rearrange
Expanding both sides:
\[ 2x - 4 \geq x^2 - 4x - 5 \]
Rearranging gives:
\[ 0 \geq x^2 - 6x + 1 \]

### Step 9: Solve the quadratic inequality
We can find the roots of the quadratic \( x^2 - 6x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2} \]

### Step 10: Determine the intervals
The roots are \( 3 - 2\sqrt{2} \) and \( 3 + 2\sqrt{2} \). The quadratic opens upwards, so the inequality \( x^2 - 6x + 1 \leq 0 \) holds between the roots:
\[ 3 - 2\sqrt{2} \leq x \leq 3 + 2\sqrt{2} \]

### Step 11: Find integer solutions
Now we need to find the integer solutions within the interval \( (0, 1) \) and the interval \( [3 - 2\sqrt{2}, 3 + 2\sqrt{2}] \).

Calculating \( 3 - 2\sqrt{2} \approx 0.172 \) and \( 3 + 2\sqrt{2} \approx 5.828 \), we find the integers in the interval are \( 1, 2, 3, 4, 5 \).

### Conclusion
Thus, the total number of integers satisfying the original inequality is **5**.