Question

The set of all real values of $x$ which satisfies $(x-2)(x-3)\le 2\le (x-1)(x-2)$, is

• A. $[3,4]$
• B. $[1,4]$
• C. $[1,3]$
• D. $[0,3]$

Answer 1

The correct option is A $[3,4]$

$(x-2)(x-3)\le 2\le (x-1)(x-2)$

$\left(i\right)(x-2)(x-3)\le 2$
$\Rightarrow x^2-5x+6\le 2\Rightarrow x^2-5x+4\le 0\Rightarrow (x-1)(x-4)\le 0\Rightarrow x\in [1,4]\cdots \left(1\right)$

$\left(ii\right)2\le (x-1)(x-2)$
$\Rightarrow x^2-3x+2\ge 2\Rightarrow x^2-3x\ge 0\Rightarrow x(x-3)\ge 0\Rightarrow x\in (-\infty ,0]\cup [3,\infty )\cdots \left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,
$x\in [3,4]$