Correct Option: B Let's solve the given inequality step-by-step. The inequality is: $$\frac{x^2 - 6x + 5}{x^2 - 1} \leq \frac{1}{2}$$ First, identify the domain restrictions. The denominator cannot be zero: $x^2 - 1 \neq 0 \implies (x-1)(x+1) \neq 0 \implies x \neq 1 \text{ and } x \neq -1$. Now, rearrange the inequality to have 0 on the right side: $$\frac{x^2 - 6x + 5}{x^2 - 1} - \frac{1}{2} \leq 0$$ Find a common denominator, which is $2(x^2 - 1)$: $$\frac{2(x^2 - 6x + 5) - (x^2 - 1)}{2(x^2 - 1)} \leq 0$$ Simplify the numerator: $$\frac{2x^2 - 12x + 10 - x^2 + 1}{2(x^2 - 1)} \leq 0$$ $$\frac{x^2 - 12x + 11}{2(x^2 - 1)} \leq 0$$ Factorize the numerator and the denominator: Numerator: $x^2 - 12x + 11 = (x-1)(x-11)$ Denominator: $2(x^2 - 1) = 2(x-1)(x+1)$ Substitute the factored forms back into the inequality: $$\frac{(x-1)(x-11)}{2(x-1)(x+1)} \leq 0$$ At this point, we must be careful. We have a common factor $(x-1)$ in both the numerator and the denominator. Since we already established that $x \neq 1$, we can cancel this term. If $x$ were allowed to be $1$, the expression would be undefined, not zero. For $x \neq 1$, the inequality simplifies to: $$\frac{x-11}{2(x+1)} \leq 0$$ Since $2$ is a positive constant, it does not affect the direction of the inequality. So, we can write: $$\frac{x-11}{x+1} \leq 0$$ Now, we find the critical points by setting the numerator and denominator to zero: $x-11 = 0 \implies x = 11$ $x+1 = 0 \implies x = -1$ These critical points divide the number line into three intervals: $(-\infty, -1)$, $(-1, 11)$, and $(11, \infty)$. We test a value from each interval: 1. **For $x < -1$** (e.g., $x=-2$): $$\frac{-2-11}{-2+1} = \frac{-13}{-1} = 13$$ Since $13 \not\leq 0$, this interval is not part of the solution. 2. **For $-1 < x < 11$** (e.g., $x=0$): $$\frac{0-11}{0+1} = \frac{-11}{1} = -11$$ Since $-11 \leq 0$, this interval is part of the solution. 3. **For $x > 11$** (e.g., $x=12$): $$\frac{12-11}{12+1} = \frac{1}{13}$$ Since $\frac{1}{13} \not\leq 0$, this interval is not part of the solution. We also need to consider the equality condition, $x-11 = 0$, which gives $x=11$. This value is included because the inequality is 'less than or equal to'. The denominator $x+1$ cannot be zero, so $x \neq -1$. Combining these, the solution to $\frac{x-11}{x+1} \leq 0$ is $-1 < x \leq 11$. Finally, we must recall the original domain restrictions: $x \neq 1$ and $x \neq -1$. The condition $x \neq -1$ is already satisfied by $-1 < x$. However, the solution range $(-1, 11]$ includes $x=1$. Since $x=1$ was excluded from the original domain, it must be removed from the solution set. So, the valid range for $x$ is $(-1, 1) \cup (1, 11]$. We are looking for the number of *integer* values of $x$ in this range. The integers in the interval $(-1, 1)$ are none. The integers in the interval $(1, 11]$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$. To count these integers, we use the formula: Last Integer - First Integer + 1. Number of integers = $11 - 2 + 1 = 10$. Thus, there are 10 integer values of $x$ that satisfy the given inequality. The final answer is $\boxed{10}$.