Question

The set of all real values of $x$ for which $\frac{x^2-1}{(x-4)(x-3)} \ge 1$ is

• A. $[-1,1] \cup (3,4)$
• B. $[1,\frac{7}{3}] \cup (4,\infty)$
• C. $(-\infty, \frac{-13}{7}] \cup (3,4)$
• D. $\mathbb{R} - [3,4]$

Hint:

Simplify the inequality to a single fraction. Find the critical points and analyze the sign of the expression in the intervals determined by the critical points.

Answer 1

The Correct Option is B

$\frac{x^2-1}{(x-4)(x-3)} - 1 \ge 0$ $\frac{x^2-1-(x^2-7x+12)}{(x-4)(x-3)} \ge 0$ $\frac{7x-13}{(x-4)(x-3)} \ge 0$ Critical points are $x=3$, $x=4$, and $x=\frac{13}{7}$. Consider the intervals:

• $x<3$: All three factors are negative, so the expression is negative.
• $3<x<\frac{13}{7}$: Numerator is negative, $(x-3)$ is positive, and $(x-4)$ is negative, so the expression is positive.
• $\frac{13}{7}<x<4$: Numerator is positive, $(x-3)$ is positive, and $(x-4)$ is negative, so the expression is negative.
• $x>4$: All three factors are positive, so the expression is positive.

Thus, the solution is $(3, \frac{13}{7}] \cup (4, \infty)$. Since $x=3$ and $x=4$ make the denominator zero, they are excluded. Since the inequality is $\ge 0$, $x=\frac{13}{7}$ is included.