Question

If $x$ satisfies the inequality ${\log }_{(x+3)}(x^2-x)<1,$ then

• A. $x\in (-3,-2)$
• B. $x\in (-1,3]$
• C. $x\in (1,3)$
• D. $x\in (-1,0)$

Answer 1

Answer: (A), (C), (D)

$x\in (-1,0)$

${\log }_{x+3}(x^2-x)<1$
$x(x-1)>0$
$\Rightarrow x>1\text{or}x<0$

Let $x+3>1\Rightarrow x>-2$
$\Rightarrow x^2-x<x+3$
$\Rightarrow x^2-2x-3<0$
$\Rightarrow (x-3)(x+1)<0$
Hence, $x\in (-1,0)\cup (1,3)\cdots \left(1\right)$

Similarly, let $0<x+3<1$
$\Rightarrow -3<x<-2$
$\Rightarrow x^2-x>x+3$
$\Rightarrow x^2-2x-3>0\Rightarrow (x-3)(x+1)>0$
$\Rightarrow x\in (-3,-2)\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ $x\in (-3,-2)\cup (-1,0)\cup (1,3)$