Question

The solution set of $\frac{2x-1}{x^2-x-12}>\frac{1}{2}$ is

• A. $x\in (-3,\frac{5-\sqrt{6}5}{2})\cup (4,\frac{5+\sqrt{6}5}{2})$
• B. $\left[-3,-2\right]\cup \left[4,7\right]$
• C. $\left[-\infty ,-2\right]\cup \left[3,2\right]\cup \left[7,\infty \right]$
• D. none of these

Answer 1

The correct option is A $x\in (-3,\frac{5-\sqrt{6}5}{2})\cup (4,\frac{5+\sqrt{6}5}{2})$

$\frac{2x-1}{x^2-x-12}>\frac{1}{2}$

$\frac{2x-1}{x^2-x-12}-\frac{1}{2}>0$

$\frac{x^2-5x-10}{x^2-x-12}<0$

$\frac{(x-\frac{5\pm \sqrt{6}5}{2})}{x^2-4x+3x-12}<0$

$\frac{(x-\frac{5+\sqrt{6}5}{2})(x-\frac{5-\sqrt{6}5}{2})}{(x-4)(x+3)}<0$

From the figure, we get $x\in (-3,\frac{5-\sqrt{6}5}{2})\cup (4,\frac{5+\sqrt{6}5}{2})$