Question

The number of integer(s) $x$ satisfying the inequation $\frac{x(x^2+2x+2)({\log }_{2}x-3)}{(x+3{)}^2(x^2-x-6)}\le 0,$ is

• A. $0$
• B. $5$
• C. $2$
• D. $8$

Answer 1

The correct option is B $5$

Clearly, $x>0\cdots \left(1\right)$
and $x^2-x-6\ne 0,x\ne -3$
$\Rightarrow (x+2)(x-3)\ne 0,x\ne -3$
$\Rightarrow x\ne -3,-2,3\cdots \left(2\right)$

Now, $\frac{x(x^2+2x+2)({\log }_{2}x-3)}{(x+3{)}^2(x^2-x-6)}\le 0$
$\Rightarrow \frac{x({\log }_{2}x-3)}{(x+2)(x-3)}\le 0$
$(\because x^2+2x+2>0,(x+3{)}^2>0,\forall x>0)$

Critical points are
$x=0$
${\log }_{2}x-3\Rightarrow x=8$
$(x+2)(x-3)\Rightarrow x=-2,3$


By wavy curve method,
$x\in (-2,0]\cup (3,8]\cdots \left(3\right)$
$\left(1\right)\cap \left(2\right)\cap \left(3\right),$ we get
$x\in (3,8]$
Integers are $4,5,6,7,8$