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Functions

Section: Mathematics

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Question : 4 of 33

Marks: +1, -0

The set of all real values of

x

f(x)=\sqrt{\;\frac{|x|-2}{|x|-3}}

is a well defined function is

[AP EAPCET 22 May 2025 S2]

$\mathbb{R}-[-3,-2)\cup(2,3]$
$\mathbb{R}-[-3,3]$
$(-3,3)$
$(-3,3)$

Solution:

\;f(x)=\sqrt{\;\frac{|x|-2}{|x|-3}}

\;\Rightarrow\;\frac{|x|-2}{|x|-3}\ge 0

Case

I |x|-2\ge 0

and

|x|-3>0

\;\Rightarrow\;\;|x|\ge 2\;\text{and}\;|x|>3

\;\Rightarrow\;\;x>3\;\text{or}\;x<-3

|x|-2\le 0

Case II

|x|-3< 0

and

\;|x|\le 2\;\text{and}\;|x|<3

\;\Rightarrow\;\;|x|\le 2

\;-2\le x\le 2

\;|x|-3\neq 0\Rightarrow|x|\neq 3

\;\Rightarrow x\neq 3\;\text{and}\;x\neq -3

(-\infty,-3)\cup[-2,2]\cup(3,\infty)

Combining these, we get

\mathbb{R}-[-3,-2)\cup(2,3]

Finally we write

(-3,-2]\cup(2,3]

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