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1.Relation and Function
normal
Domain of the definition of function
$f(x) = \sqrt {\frac{{4 - {x^2}}}{{\left[ x \right] + 2}}} $ is $($ where $[.] \rightarrow G.I.F.)$
A
$( - \infty ,2)\, \cup \,[ - 1,2]$
B
$[0,2]$
C
$[-1,2]$
D
$(0,2)$
Solution
$\text { Case }-1: 4-\mathrm{x}^{2} \geq 0 $ and $[\mathrm{x}]+2>0 $
$ \Rightarrow \mathrm{x}^{2}-4 \leq 0 \Rightarrow \mathrm{x}[\mathrm{x}]>-2 $
$ \Rightarrow \mathrm{x} \in[-2,2] $ and $ \mathrm{x} \in[-1, \infty) $
$ \mathrm{x} \in[-1,2] $
$ \text { Case }-2: 4-\mathrm{x}^{2} \leq 0 $ and $[\mathrm{x}]+2<0 $
and $[{\rm{x}}] < – 2$
$x \in(-\infty,-2] \cup[2, \infty) $ and $ x \in(-\infty,-2)$
$\Rightarrow x \in(-\infty,-2]$
$\therefore $ answer $x \in(-\infty,-2] \cup[-1,2]$
Standard 12
Mathematics
