Question

The solution set of $\frac{x^2-7x+10}{14x-x^2-45}\ge 0$ is

• A. $[2,9]$
• B. $[2,9)$
• C. $[2,5)\cup (5,9)$
• D. $[2,5)\cup (5,9]$

Answer 1

The correct option is C $[2,5)\cup (5,9)$

Given $\frac{x^2-7x+10}{14x-x^2-45}\ge 0$
$\Rightarrow \frac{x^2-7x+10}{x^2-14x+45}\le 0\Rightarrow \frac{(x-2)(x-5)}{(x-5)(x-9)}\le 0\Rightarrow \frac{x-2}{x-9}\le 0[\because x\ne 5]$

From wavy curve method,
$x\in [2,9)-\left\{5\right\}$