Question

The range of values of $x$ which satisfies the inequation ${\log }_{1/6}(x^2-3x+2)+1<0$ is

• A. $x\in (-\infty ,-1)\cup (4,\infty )$
• B. $x\in (-\infty ,1)\cup (2,\infty )$
• C. $x\in (2,4)$
• D. $x\in (-\infty ,-1)\cup (2,\infty )$

Answer 1

The correct option is A $x\in (-\infty ,-1)\cup (4,\infty )$

${\log }_{1/6}(x^2-3x+2)+1<0$
$\log$ function is defined when $x^2-3x+2>0$
$\Rightarrow (x-1)(x-2)>0$
$\Rightarrow x\in (-\infty ,1)\cup (2,\infty )...\left(1\right)$

Now, $-{\log }_6(x^2-3x+2)+1<0$
$\Rightarrow {\log }_6(x^2-3x+2)>1$
$\Rightarrow x^2-3x+2>6$
$\Rightarrow x^2-3x-4>0$
​​​​​​​$\Rightarrow (x+1)(x-4)>0$
​​​​​​​$\Rightarrow x\in (-\infty ,-1)\cup (4,\infty )...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$x\in (-\infty ,-1)\cup (4,\infty )$