Question

The solution set of $x^2-7x+12\ge 0$ is

• A. $[3,4]$
• B. $(3,4)$
• C. $(-\infty ,3)\cup (4,\infty )$
• D. $(-\infty ,3]\cup [4,\infty )$

Answer 1

The correct option is D $(-\infty ,3]\cup [4,\infty )$

$x^2-7x+12\ge 0\Rightarrow (x-3)(x-4)\ge 0$


$\therefore x\in (-\infty ,3]\cup [4,\infty )$