To find the domain of the given function \(f(x)=\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)+\frac{1}{\log_e(10-x)}\), we need to consider the domain restrictions of both components separately: \(\sin^{-1}(u)\) and \(\frac{1}{\log_e(10-x)}\).
- Domain of \(\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)\):
- The argument of \(\sin^{-1}(x)\) should lie between -1 and 1: \(-1 \leq \frac{5-x}{3+2x} \leq 1\).
- Solving \(-\frac{5-x}{3+2x} \leq 1\):
- Simplifying, \(-5+x \geq 3+2x \Rightarrow -5-3 \geq 2x-x\),
- Thus, \(-8 \geq x \Rightarrow x \leq -8\)
- Solving \(\frac{5-x}{3+2x} \leq 1\):
- Expanding, \(5-x \leq 3+2x \Rightarrow 5-3 \leq 2x+x\),
- Results in \(2 \leq 3x \Rightarrow x \geq \frac{2}{3}\).
- Therefore, the domain from sine inverse is \(x \leq -8\) and \(x \geq \frac{2}{3}\).
- Domain of \(\frac{1}{\log_e(10-x)}\):
- The logarithm is defined when \(10-x > 0\) giving \(x < 10\).
- Additionally, we must have \(\log_e(10-x) \neq 0\), i.e., \(10-x \neq 1 \Rightarrow x \neq 9\).
- Combining the results:
- The domain is \((-\infty, -8]\cup[\frac{2}{3}, 10)\)
- Excluding \(x=9\), the domain becomes \((-\infty, -8]\cup[\frac{2}{3},10)-\{9\}\).
- Identifying values: \(\alpha = -8\), \(\beta = \frac{2}{3}\), \(\gamma = 10\), \(\delta = 9\).
Finally, calculate: \(6(\alpha + \beta + \gamma + \delta)\)
- \(6(-8 + \frac{2}{3} + 10 + 9)\)
- \(6(\frac{-24}{3} + \frac{2}{3} + \frac{30}{3} + \frac{27}{3})\)
- \(6(\frac{35}{3})\)
- Evaluating: \(6 \times \frac{35}{3} = 2 \times 35 = 70\).