Question

$(x-1)(x^2-5x+7)<(x-,1)$, then x belongs to

• A. $(1,2)\cup (3,\infty )$
• B. $(-\infty ,1)\cup (2,3)$
• C. $(2,3)$
• D. None of these

Answer 1

The correct option is B $(-\infty ,1)\cup (2,3)$

Given that, $(x-1)(x^2-5x+7)<(x-1)$
$\therefore (x-1)(x^2-5x+6)<0$
$\Rightarrow (x-1)(x-2)(x-3)<0$
$\Rightarrow x\in (-\infty ,1)\cup (2,3)$