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

Given:
\[ f(x)=\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)+\frac{1}{\ln(10-x)} \] 
Step 1: Domain of \[ \sin^{-1}\!\left(\frac{5-x}{3+2x}\right) \] For sine inverse to be defined: \[ -1 \le \frac{5-x}{3+2x} \le 1 \] Solving: 
\[ \frac{5-x}{3+2x} \le 1 \Rightarrow x \ge \frac{2}{3} \] \[ \frac{5-x}{3+2x} \ge -1 \Rightarrow x \le -8 \] Hence, \[ x \in (-\infty,-8] \cup \left[\frac{2}{3},\infty\right) \] 
Step 2: Domain of \[ \frac{1}{\ln(10-x)} \] Conditions: 
1. Argument of log must be positive: \[ 10-x>0 \Rightarrow x<10 \] 2. Denominator cannot be zero: \[ \ln(10-x)\neq 0 \Rightarrow 10-x\neq 1 \Rightarrow x\neq 9 \] So, \[ x<10,\quad x\neq 9 \] 
Step 3: Combined domain Intersection of both conditions: \[ (-\infty,-8] \cup \left[\frac{2}{3},10\right) - \{9\} \] Comparing with: \[ (-\infty,\alpha] \cup [\beta,\gamma) - \{\delta\} \] We get: \[ \alpha=-8,\quad \beta=\frac{2}{3},\quad \gamma=10,\quad \delta=9 \] 
Step 4: Required value \[ 6(\alpha+\beta+\gamma+\delta) =6\left(-8+\frac{2}{3}+10+9\right) \] \[ =6\left(\frac{35}{3}\right)=70 \] 

Final Answer: \[ \boxed{70} \].