Functions Part 2

Concept:The domain is found by ensuring the arguments of inverse sine and logarithm are valid.Explanation:The function is $f(x) = \sin^{-1}\left(\frac{5-x}{3+2x}\right) + \frac{1}{\log_e(10-x)}$.For $\sin^{-1}(y)$, we require $-1 \le y \le 1$. So $-1 \le \frac{5-x}{3+2x} \le 1$.This yields two inequalities:$\frac{5-x}{3+2x} \le 1$ gives $\frac{2-3x}{3+2x} \le 0$, which simplifies to $(3x-2)(2x+3) \ge 0$. So $x \le -\frac{3}{2}$ or $x \ge \frac{2}{3}$.Also $\frac{5-x}{3+2x} \ge -1$ gives $\frac{8+x}{3+2x} \ge 0$, so $x \le -8$ or $x > -\frac{3}{2}$.Intersecting both gives $x \le -8$ or $x \ge \frac{2}{3}$.For the logarithm, $\log_e(10-x)$ requires $10-x > 0$, so $x < 10$. Also denominator $\log_e(10-x) \neq 0$, so $10-x \neq 1$, i.e., $x \neq 9$.Combining: domain is $(-\infty, -8] \cup [\frac{2}{3}, 10) - \{9\}$.Thus $\alpha = -8$, $\beta = \frac{2}{3}$, $\gamma = 10$, $\delta = 9$.Then $6(\alpha+\beta+\gamma+\delta) = 6\left(-8 + \frac{2}{3} + 10 + 9\right) = 6\left(11 + \frac{2}{3}\right) = 6 \times \frac{35}{3} = 70$.Answer:$70$, so option C is correct.