Question

The range of values of $x$ satisfying the inequality $(x^2-2x)(2x-2)-9\frac{2x-2}{x^2-2x}\le 0$ is

• A. $(-\infty ,-1]\cup (0,1]\cup (2,3]$
• B. $(-\infty ,-1)\cup (0,1]\cup (2,3]$
• C. $(-\infty ,-1]\cup (0,1)\cup [2,3)$
• D. $(-\infty ,-1)\cup (0,1)\cup [2,3)$

Answer 1

Answer: (A)

$(-\infty ,-1]\cup (0,1]\cup (2,3]$

Given,

$(x^2-2x)(2x-2)-9\frac{2x-2}{x^2-2x}\le 0$
$\Rightarrow (2x-2)\left(\frac{{(x^2-2x)}^2-9}{x^2-2x}\right)\le 0$
$\Rightarrow 2(x-1)\left(\frac{(x^2-2x+3)(x^2-2x-3)}{x(x-2)}\right)\le 0$
$\Rightarrow 2(x-1)\frac{(x^2-2x+3)(x-3)(x+1)}{x(x-2)}\le 0$
$\Rightarrow x\in (-\infty ,-1]\cup (0,1]\cup (2,3]$