વિધેય f(x) = sqrt {frac{{4 - {x^2}}}{{left[ x | vedclass

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Question

$	ext { Case }-1:  4-\mathrm{x}^{2} \geq 0 $ and $[\mathrm{x}]+2>0 $ 

$ \Rightarrow \mathrm{x}^{2}-4 \leq 0 \Rightarrow \mathrm{x}[\m

Vedclass · Accepted Answer

$( - \infty ,2)\, \cup \,[ - 1,2]$

Vedclass · Answer

$ ext { 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] $ $ ext { Case }-2: 4-\mathrm{x}^{2} \leq 0 $ and $[\mathrm{x}]+2<0 $ and $[{ m{x}}] < - 2$ $x \in(-\infty,-2] \cup[2, \infty) $ and $ x \in(-\infty,-2)$ $\Rightarrow x \in(-\infty,-2]$ $ herefore $ answer $x \in(-\infty,-2] \cup[-1,2]$

વિધેય $f(x) = \sqrt {\frac{{4 - {x^2}}}{{\left[ x \right] + 2}}} $ નો પ્રદેશ્ગણ ........... થાય. $($ જ્યા $[.] \rightarrow G.I.F.)$

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