Question

The value of $x$ satisfying $\frac{1}{x^2-5x+6}\ge \frac{1}{2}$ is

• A. $\left[\right(1,2)\cup (3,4\left)\right]$
• B. $[2,4]$
• C. $[1,4]$
• D. $R$

Answer 1

The correct option is A $\left[\right(1,2)\cup (3,4\left)\right]$

$\frac{1}{x^2-5x+6}\ge \frac{1}{2}$

$\frac{1}{x^2-5x+6}-\frac{1}{2}\ge 0$

$\frac{2-x^2+5x-6}{2(x^2-5x+6)}\ge 0$

$\frac{x^2-5x+4}{x^2-5x+6}\le 0$

$\frac{(x-4)(x-1)}{(x-2)(x-3)}\le 0$

so $xϵ\left[\right(1,2)\cup (3,4\left)\right]$

(at x=3,4 in function's denominator makes 0 so it is not defined so it excludes)

So option ${}^{\prime }{A}^{\prime }$ is correct.