Let the domains of the functions f(x) = log_4 | vedclass

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(A) For $f(x)$,we require $\log_3 \log_7(8 - \log_2(x^2 + 4x + 5)) > 0$. This implies $\log_7(8 - \log_2(x^2 + 4x + 5)) > 1$,so $8 - \log_2(x^2 + 4x +

Vedclass · Accepted Answer

$15$

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(A) For $f(x)$,we require $\log_3 \log_7(8 - \log_2(x^2 + 4x + 5)) > 0$. This implies $\log_7(8 - \log_2(x^2 + 4x + 5)) > 1$,so $8 - \log_2(x^2 + 4x + 5) > 7$. Thus,$\log_2(x^2 + 4x + 5) Factoring gives $(x + 3)(x + 1) For $g(x)$,we require $-1 \leq \frac{7x + 10}{x - 2} \leq 1$. Solving $\frac{7x + 10}{x - 2} \leq 1$ $\Rightarrow \frac{7x + 10 - x + 2}{x - 2} \leq 0$ $\Rightarrow \frac{6x + 12}{x - 2} \leq 0$ $\Rightarrow \frac{x + 2}{x - 2} \leq 0$,so $x \in [-2, 2)$. Solving $\frac{7x + 10}{x - 2} \geq -1$ $\Rightarrow \frac{7x + 10 + x - 2}{x - 2} \geq 0$ $\Rightarrow \frac{8x + 8}{x - 2} \geq 0$ $\Rightarrow \frac{x + 1}{x - 2} \geq 0$,so $x \in (-\infty, -1] \cup (2, \infty)$. The intersection is $x \in [-2, -1]$. Thus,$\gamma = -2$ and $\delta = -1$. Finally,$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2 = 9 + 1 + 4 + 1 = 15$.

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

Let the domains of the functions $f(x) = \log_4 \log_3 \log_7(8 - \log_2(x^2 + 4x + 5))$ and $g(x) = \sin^{-1}(\frac{7x + 10}{x - 2})$ be $(\alpha, \beta)$ and $[\gamma, \delta]$,respectively. Then $\alpha^2 + \beta^2 + \gamma^2 + \delta^2$ is equal to:

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