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)

$x\in (-1,0)$

${\log }_{x+3}(x^2-x)<1$
For the inequality to be defined,
$x^2-x>0$ and $x+3>0$
$\Rightarrow x(x-1)>0$ and $x>-3$
$\Rightarrow x\in (-\infty ,0)\cup (1,\infty )$ and $x\in (-3,\infty )$
$\therefore x\in (-3,0)\cup (1,\infty )$

Case $1:$ When $x+3>1$ i.e., $x\in (-2,0)\cup (1,\infty )$
$x^2-x<x+3\Rightarrow x^2-2x-3<0\Rightarrow (x-3)(x+1)<0\Rightarrow x\in (-1,3)$
$\therefore x\in (-1,0)\cup (1,3)\cdots \left(1\right)$

Case $2:$ When $0<x+3<1$ i.e., $x\in (-3,-2)$
$x^2-x>x+3$
$\Rightarrow x^2-2x-3>0\Rightarrow (x-3)(x+1)>0$
$\Rightarrow x\in (-\infty ,-1)\cup (3,\infty )$
$\therefore x\in (-3,-2)\cdots \left(2\right)$

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