If the domain of the function log _5(18 x-x^2 | vedclass

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(A) For $f_1(x) = \log _5(18 x-x^2-77)$,the domain requires $18 x-x^2-77 > 0$.$x^2-18 x+77 < 0 \implies (x-7)(x-11) < 0$.Thus,$x \in (7, 11)$,so $\alp

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$186$

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(A) For $f_1(x) = \log _5(18 x-x^2-77)$,the domain requires $18 x-x^2-77 > 0$.$x^2-18 x+77 Thus,$x \in (7, 11)$,so $\alpha = 7$ and $\beta = 11$.For $f_2(x) = \log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}ight)$,the domain requires:$1) \ x-1 > 0 \implies x > 1$$2) \ x-1 
eq 1 \implies x 
eq 2$$3) \ \frac{2 x^2+3 x-2}{x^2-3 x-4} > 0 \implies \frac{(2 x-1)(x+2)}{(x-4)(x+1)} > 0$.Using the sign scheme for $\frac{(2 x-1)(x+2)}{(x-4)(x+1)} > 0$,the solution is $x \in (-\infty, -2) \cup (-1, 1/2) \cup (4, \infty)$.Taking the intersection with $x > 1$ and $x 
eq 2$,we get $x \in (4, \infty)$.Thus,$\gamma = 4$ and $\delta = \infty$.Finally,$\alpha^2+\beta^2+\gamma^2 = 7^2+11^2+4^2 = 49+121+16 = 186$.

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

If the domain of the function $\log _5(18 x-x^2-77)$ is $(\alpha, \beta)$ and the domain of the function $\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right)$ is $(\gamma, \delta)$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

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