Let the domain of the function f(x) = log_3 l | vedclass

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(C) For the first term $\log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85))$, we require $\log_5 (7 - \log_2 (x^2 - 10 x + 85)) > 0$, which implies $7 - \log

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$9$

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(C) For the first term $\log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85))$, we require $\log_5 (7 - \log_2 (x^2 - 10 x + 85)) > 0$, which implies $7 - \log_2 (x^2 - 10 x + 85) > 1$, so $\log_2 (x^2 - 10 x + 85) For the second term $\sin^{-1} ( | \frac{3 x - 7}{17 - x} | )$, we require $0 \leq | \frac{3 x - 7}{17 - x} | \leq 1$. The condition $| \frac{3 x - 7}{17 - x} | \leq 1$ implies $(3x - 7)^2 \leq (17 - x)^2$, so $9x^2 - 42x + 49 \leq 289 - 34x + x^2$. This simplifies to $8x^2 - 8x - 240 \leq 0$, or $x^2 - x - 30 \leq 0$. Factoring gives $(x - 6)(x + 5) \leq 0$, so $x \in [-5, 6]$.Combining the domains $(3, 7)$ and $[-5, 6]$, we get $x \in (3, 6]$.Thus, $\alpha = 3$ and $\beta = 6$.Therefore, $\alpha + \beta = 3 + 6 = 9$.

Let the domain of the function $f(x) = \log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85)) + \sin^{-1} ( | \frac{3 x - 7}{17 - x} | )$ be $(\alpha, \beta]$. Then $\alpha + \beta$ is equal to :

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