If the domain of the function sin^{-1}left(fr | vedclass

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(A) For $\sin^{-1}(u)$,we need $-1 \leq u \leq 1$. So,$-1 \leq \frac{3x-22}{2x-19} \leq 1$. Solving this inequality gives $x \in (5, 8.2]$.For $\log_e

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$97$

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(A) For $\sin^{-1}(u)$,we need $-1 \leq u \leq 1$. So,$-1 \leq \frac{3x-22}{2x-19} \leq 1$. Solving this inequality gives $x \in (5, 8.2]$.For $\log_e(v)$,we need $v > 0$. So,$\frac{3x^2-8x+5}{x^2-3x-10} > 0$,which simplifies to $\frac{(3x-5)(x-1)}{(x-5)(x+2)} > 0$. The solution is $x \in (-\infty, -2) \cup (1, 5/3) \cup (5, \infty)$.Taking the intersection of both conditions,the domain is $(5, 8.2]$,which is $(5, 41/5]$.Here,$\alpha = 5$ and $\beta = 41/5$.Thus,$3\alpha + 10\beta = 3(5) + 10(41/5) = 15 + 82 = 97$.

If the domain of the function $\sin^{-1}\left(\frac{3x-22}{2x-19}\right) + \log_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$ is $(\alpha, \beta]$,then $3\alpha + 10\beta$ is equal to :

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