If [x] denotes the greatest integer leq x, th | vedclass

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Question

(A) For the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$ to be defined, the expression inside the square root must be non-negative and the denominator

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, the expression inside the square root must be non-negative and the denominator must not be zero.$1$. Denominator condition: $[x] + 2 
eq 0 \Rightarrow [x] 
eq -2$. This implies $x 
otin [-2, -1)$.$2$. Inequality condition: $\frac{4-x^2}{[x]+2} \geq 0$.Case $I$: If $[x] + 2 > 0$, then $[x] > -2$, which means $x \geq -1$. Then $4 - x^2 \geq 0$ $\Rightarrow x^2 \leq 4$ $\Rightarrow x \in [-2, 2]$. Combining $x \geq -1$ and $x \in [-2, 2]$, we get $x \in [-1, 2]$.Case $II$: If $[x] + 2 Then $4 - x^2 \leq 0$ $\Rightarrow x^2 \geq 4$ $\Rightarrow x \in (-\infty, -2] \cup [2, \infty)$. Combining $x Combining both cases, the domain is $(-\infty, -2) \cup [-1, 2]$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

If $[x]$ denotes the greatest integer $\leq x$, then the domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$ is

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