Functions Part 2

Concept:The domain of a logarithmic function requires its argument to be positive, and the domain of $\sin^{-1}$ requires its argument to be in $[-1,1]$.Explanation:For the first term, let $\lambda = x^2 - 10x + 85$.The condition $\log_3(\log_5(7 - \log_2 \lambda))$ is defined requires:1. $\lambda > 0$ (always true)2. $7 - \log_2 \lambda > 0 \Rightarrow \log_2 \lambda < 7 \Rightarrow \lambda < 2^7$3. $\log_5(7 - \log_2 \lambda) > 0 \Rightarrow 7 - \log_2 \lambda > 1 \Rightarrow \log_2 \lambda < 6 \Rightarrow \lambda < 2^6$So $0 < \lambda < 2^6$, i.e., $0 < x^2 - 10x + 85 < 64$.Solve $x^2 - 10x + 85 < 64$ gives $x^2 - 10x + 21 < 0 \Rightarrow (x-3)(x-7) < 0 \Rightarrow x \in (3,7)$.For the second term, $\left| \frac{3x-7}{17-x} \right| \leq 1$, which is equivalent to $(3x-7)^2 \leq (17-x)^2$ and $x \neq 17$.Simplify: $9x^2 - 42x + 49 \leq 289 - 34x + x^2 \Rightarrow 8x^2 - 8x - 240 \leq 0 \Rightarrow x^2 - x - 30 \leq 0 \Rightarrow (x-6)(x+5) \leq 0 \Rightarrow x \in [-5,6]$.Intersect the two domains: $(3,7) \cap [-5,6] = (3,6]$.Thus $\alpha = 3$, $\beta = 6$.Answer:$\alpha + \beta = 3 + 6 = 9$, which corresponds to option D.