Question

If $x$ is an integer satisfying $x^2-6x+5\le 0$ and $x^2-2x>0$, then the number of possible values of $x$ are

• A. $3$
• B. $4$
• C. $2$
• D. infinite

Answer 1

The correct option is A $3$

$x^2-6x+5\le 0$
$(x-1)(x-5)\le 0$
$\Rightarrow x\in [1,5]$
$\Rightarrow x=1,2,3,4,5$
Also, $x^2-2x>0$
$x(x-2)>0$
$\Rightarrow x\in (-\infty ,0)\cup (2,\infty )$
Hence, the integer values of x satisfying both the inequalities are $x=3,4,5$
Therefore, there are three integral roots.