Question

The set of the solutions for $\frac{2x-5}{(x-1)(x-7)}\le 0$ is

• A. $(-\infty ,1)\cup [\frac{5}{2},7)$
• B. $(-\infty ,1)\cup (\frac{5}{2},7)$
• C. $(-\infty ,1]\cup (\frac{5}{2},7)$
• D. $(-\infty ,1)\cup [\frac{5}{2},7]$

Answer 1

The correct option is A $(-\infty ,1)\cup [\frac{5}{2},7)$

$\frac{2x-5}{(x-1)(x-7)}\le 0$


Since, $x\ne 1,7$
$\Rightarrow x\in (-\infty ,1)\cup [\frac{5}{2},7)$