Question

$\text{The range of values of}x\text{which satisfies}$
$\text{the inequality}{\log }_{(x+3)}(x^2-x)<1\text{is}$

Answer 1

Given:
${\log }_{(x+3)}(x^2-x)$
$\text{For}\log \text{to be defined},$
$x+3>0,x+3\ne 1\text{and}{x}^2-x>0$
$\Rightarrow x>-3,x\ne -2\text{and}x\in (-\infty ,0)\cup (1,\infty )$
$\Rightarrow x\in (-3,-2)\cup (-2,0)\cup (1,\infty )$

Now we solve,
${\log }_{(x+3)}(x^2-x)<1$

Case 1:
$\text{If}x+3>1$
$\text{i.e.,}x>-2$
$\text{i.e.,}x\in (-2,0)\cup (1,\infty )\cdots \left(1\right)$
$\text{then}{x}^2-x<x+3$
$\Rightarrow x^2-2x-3<0\Rightarrow (x-3)(x+1)<0$
$\Rightarrow x\in (-1,3)\cdots \left(2\right)$
$\text{Hence, from equation}\left(1\right)\text{and}\left(2\right),$
$x\in (-1,0)\cup (1,3)$

Case 2:
$\text{If}0<x+3<1$
$\text{i.e.,}x\in (-3,-2)\cdots \left(3\right)$
$\text{then}{x}^2-x>x+3$
$\Rightarrow x^2-2x-3>0\Rightarrow (x-3)(x+1)>0\Rightarrow x\in (-\infty ,-1)\cup (3,\infty )\cdots \left(4\right)$
$\text{Hence, from equation}\left(3\right)\text{and}\left(4\right),$
$x\in (-3,-2)$
Hence the correct answer are Option A and Option C.