Question

Consider the inequality ${\log }_{(x+3)}(x^2-x)<1$. Then

• A. $x\in (-3,-2)$ if $-3<x<-2$
• B. $x\in (-1,0)\cup (1,3)$ if $-3<x<-2$
• C. $x\in (-1,0)\cup (1,3)$ if $-2<x<0$ and $1<x<\infty$
• D. $x\in (-3,-2)$ if $-2<x<0$ and $1<x<\infty$

Answer 1

Answer: (A), (C)

The correct options are
A $x\in (-3,-2)$ if $-3<x<-2$
C $x\in (-1,0)\cup (1,3)$ if $-2<x<0$ and $1<x<\infty$

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

If $0<x+3<1$
$\text{i.e.,}x\in (-3,-2)\cdots \left(3\right)$
then $x^2-x>x+3$
$\Rightarrow x^2-2x-3>0\Rightarrow (x-3)(x+1)>0\Rightarrow x\in (-\infty ,-1)\cup (3,\infty )\cdots \left(4\right)$
Hence, from $\left(3\right)$ and $\left(4\right)$,
$x\in (-3,-2)$