Question

lf $x$ is real and $(x^2-3x+2)(x^2-x+7)<0$, then $x$ lies in the interval

• A. $(-\infty ,1)\cup (2,\infty )$
• B. $(1,2)$
• C. $[1,2]$
• D. $(3,4)$

Answer 1

The correct option is B $(1,2)$

Given that $(x^2-3x+2)(x^2-x+7)<0$
The $\Delta$ of $x^2-x+7$ is $1-28<0$
Hence, $x^2-x+7>0$ for all $x$.
$\therefore (x^2-3x+2)<0$

$\Rightarrow (x^2-2x-x+2)<0$

$\Rightarrow (x-2)(x-1)<0$
$x\in (1,2)$
Hence, option B is the correct answer.