The problem asks us to find the value of $6(\alpha + \beta + \gamma + \delta)$ given the domain of the function $f(x) = \sin^{-1}\left(\frac{5-x}{3+2x}\right) + \frac{1}{\log_e(10-x)}$ is $(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$. We need to determine the domain by considering the conditions for each part of the function.
Domain Conditions
- For $\sin^{-1}(u)$, we require $-1 \le u \le 1$. Thus, $-1 \le \frac{5-x}{3+2x} \le 1$.
- For $\log_e(v)$, we require $v > 0$. Thus, $10-x > 0 \implies x < 10$.
- For $\frac{1}{w}$, we require $w \ne 0$. Thus, $\log_e(10-x) \ne 0 \implies 10-x \ne e^0 \implies 10-x \ne 1 \implies x \ne 9$.
- Also, the denominator $3+2x$ in the argument of $\sin^{-1}$ cannot be zero, so $3+2x \ne 0 \implies x \ne -\frac{3}{2}$.
Solving Inequalities for $\sin^{-1}$
We need to solve $-1 \le \frac{5-x}{3+2x} \le 1$. This involves two inequalities:
- Inequality 1: $\frac{5-x}{3+2x} \le 1$ $ \frac{5-x}{3+2x} - 1 \le 0 $ $ \frac{5-x - (3+2x)}{3+2x} \le 0 $ $ \frac{2-3x}{3+2x} \le 0 $ The critical points are $x = 2/3$ and $x = -3/2$. Analyzing the signs, the solution is $x \in (-\infty, -3/2) \cup [2/3, \infty)$.
- Inequality 2: $\frac{5-x}{3+2x} \ge -1$ $ \frac{5-x}{3+2x} + 1 \ge 0 $ $ \frac{5-x + (3+2x)}{3+2x} \ge 0 $ $ \frac{8+x}{3+2x} \ge 0 $ The critical points are $x = -8$ and $x = -3/2$. Analyzing the signs, the solution is $x \in (-\infty, -8] \cup (-3/2, \infty)$.
The intersection of the solutions for Inequality 1 and Inequality 2 is $(-\infty, -8] \cup [2/3, \infty)$.
Combining All Domain Conditions
We combine the result from the $\sin^{-1}$ argument with the other conditions:
- From $\sin^{-1}$: $(-\infty, -8] \cup [2/3, \infty)$.
- From $\log_e$: $x < 10$.
- From $\log_e \ne 0$: $x \ne 9$.
- From denominator $\ne 0$: $x \ne -3/2$.
Intersecting $(-\infty, -8] \cup [2/3, \infty)$ with $x < 10$ gives $(-\infty, -8] \cup [2/3, 10)$.
Now, excluding $x=9$ (since $9$ is in $[2/3, 10)$), the domain becomes $(-\infty, -8] \cup [2/3, 9) \cup (9, 10)$.
The condition $x \ne -3/2$ does not affect this domain as $-3/2$ is not included.
Identifying $\alpha, \beta, \gamma, \delta$
The calculated domain is $(-\infty, -8] \cup [2/3, 9) \cup (9, 10)$.
The given format for the domain is $(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$.
Comparing the calculated domain with the given format, we identify:
- $\alpha = -8$
- $\beta = 2/3$
- $\gamma = 10$
- $\delta = 9$
The domain $(-\infty, -8] \cup [2/3, 10) - \{9\}$ matches the calculated domain $(-\infty, -8] \cup [2/3, 9) \cup (9, 10)$.
Final Calculation
We need to calculate $6(\alpha + \beta + \gamma + \delta)$.
$ \alpha + \beta + \gamma + \delta = -8 + \frac{2}{3} + 10 + 9 $ $ = (-8 + 10 + 9) + \frac{2}{3} $ $ = 11 + \frac{2}{3} $ $ = \frac{33}{3} + \frac{2}{3} $ $ = \frac{35}{3} $Now, multiply by 6:
$ 6(\alpha + \beta + \gamma + \delta) = 6 \times \frac{35}{3} $ $ = 2 \times 35 $ $ = 70 $Similar Questions
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