Find the set {x in R : frac{sqrt{|x|^2-2|x|-8 | vedclass

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Question

(B) For the expression to be a real number,the following conditions must be satisfied: $1$. The expression inside the square root must be non-negative

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(B) For the expression to be a real number,the following conditions must be satisfied: $1$. The expression inside the square root must be non-negative: $|x|^2 - 2|x| - 8 \geq 0$. Let $|x| = t$,then $t^2 - 2t - 8 \geq 0 \Rightarrow (t-4)(t+2) \geq 0$. Since $t = |x| \geq 0$,we have $t \geq 4$,which means $|x| \geq 4$,so $x \in (-\infty, -4] \cup [4, \infty)$. $2$. The argument of the logarithm must be positive: $2 - x - x^2 > 0 \Rightarrow x^2 + x - 2 $(x+2)(x-1) $3$. The denominator must not be zero: $\log(2 - x - x^2) 
eq 0$ $\Rightarrow 2 - x - x^2 
eq 1$ $\Rightarrow x^2 + x - 1 
eq 0$. $x 
eq \frac{-1 \pm \sqrt{5}}{2}$. Combining the conditions: $x \in ((-\infty, -4] \cup [4, \infty)) \cap (-2, 1)$. Since there is no intersection between these sets,the solution is the empty set $\phi$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Find the set $\{x \in R : \frac{\sqrt{|x|^2-2|x|-8}}{\log(2-x-x^2)} \text{ is a real number}\}$.

Similar Questions

Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=|[x]|, x \in R$ $2$. $Z$
$C$. $h(x)=|x-[x]|, x \in R$ $3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$ $4$. $[0, 1)$
$5$. $\{-1, 1\}$

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