Question

If the domain of the function
\[ f(x)=\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)+\frac{1}{\log_e(10-x)} \]
is \( (-\infty,\alpha] \cup [\beta,\gamma) - \{\delta\} \), then \( 6(\alpha+\beta+\gamma+\delta) \) is equal to

• A. 68
• B. 66
• C. 70
• D. 67

Hint:

When finding the domain of composite functions, always evaluate each term separately and then take the intersection of all valid intervals.

Answer 1

The Correct Option is C

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 \(\frac{1}{\log_e(10-x)}\).

1. 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 \(x \geq \frac{2}{3}\).
2. 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., \(10-x \neq 1 \Rightarrow x \neq 9\).
3. 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: \(6 \times \frac{35}{3} = 2 \times 35 = 70\).