Question

The set of real values of $x$ for which expression
${\log }_{12}(x^2-x)<1$ holds true is

Answer 1

Given :
${\log }_{12}(x^2-x)<1$

For log to be defined $x^2-x>0$

$x(x-1)>0$

$x\in (-\infty ,0)\cup (1,\infty )\cdots \left(1\right)$

Base of logarithm is greater than 1, then inequality is equivalent to

$\Rightarrow x^2-x<(12{)}^1$

$\Rightarrow (x-4)(x+3)<0$

Using wavy curve method
$\Rightarrow x\in (-3,4)\cdots \left(2\right)$

$\text{From equation}1\text{and}2,\text{common region is}$
$x\in (-3,0)\cup (1,4)$

Hence the correct answer is Option A.