The domain of the function f(x) = sqrt{frac{4 | vedclass

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(A) For the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$ to be defined,we must have $\frac{4-x^2}{[x]+2} \geq 0$ and $[x]+2 
eq 0$.This implies $\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$ to be defined,we must have $\frac{4-x^2}{[x]+2} \geq 0$ and $[x]+2 
eq 0$.This implies $\frac{x^2-4}{[x]+2} \leq 0$ and $[x] 
eq -2$.Case $1$: $x^2-4 \geq 0$ and $[x]+2 $x^2 \geq 4 \Rightarrow x \in (-\infty, -2] \cup [2, \infty)$.$[x] Intersection: $x \in (-\infty, -2)$.Case $2$: $x^2-4 \leq 0$ and $[x]+2 > 0$.$x^2 \leq 4 \Rightarrow x \in [-2, 2]$.$[x] > -2 \Rightarrow x \geq -1$.Intersection: $x \in [-1, 2]$.Combining both cases,the domain is $(-\infty, -2) \cup [-1, 2]$.

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	....	....	....	....	....	....	....	....	....

The domain of the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$,where $[x]$ denotes the greatest integer not more than $x$,is

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