Question

The set of values of $x$ for which the function ​​$f\left(x\right)=\log \left(\frac{x^2-5x+6}{x^2+x+1}\right)+\sqrt{\frac{1}{[x^2-1]}}$ is defined, is
(where $[.]$ denotes the greatest integer function)

• A. $(-\infty ,-\sqrt{2}]$
• B. $(\sqrt{2},\infty )$
• C. $(3,\infty )$
• D. $(-\infty ,2)$

Answer 1

Answer: (A), (C)

$(3,\infty )$

$f\left(x\right)=\log \left(\frac{x^2-5x+6}{x^2+x+1}\right)+\sqrt{\frac{1}{[x^2-1]}}$
For $f\left(x\right)$ to be defined,
$\frac{x^2-5x+6}{x^2+x+1}>0$
$\Rightarrow x^2-5x+6>0[\because x^2+x+1>0\forall x\in R]$
$\Rightarrow (x-2)(x-3)>0$
$\Rightarrow x\in (-\infty ,2)\cup (3,\infty )...\left(1\right)$

Also, $\frac{1}{[x^2-1]}\ge 0$ and $[x^2-1]\ne 0$
$\Rightarrow [x^2-1]>0$
$\Rightarrow x^2-1\ge 1$
$\Rightarrow x^2-2\ge 0$
$\Rightarrow x\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$x\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},2)\cup (3,\infty )$