Question

If the domain of the function \[ f(x)=\frac{1}{\ln(10-x)}+\sin^{-1}\!\left(\frac{x+2}{2x+3}\right) \] is \[ (-\infty,-a]\ \cup\ (-1,b)\ \cup\ (b,c), \] then find the value of \( (b+c+3a) \).

• A. \(22\)
• B. \(24\)
• C. \(23\)
• D. \(21\)

Hint:

For domain problems:
Logarithm requires argument \(>0\) and denominator \(\neq 0\) 
\(\sin^{-1}(x)\) requires \(-1\le x\le 1\) 
Final domain is the \textbf{intersection} of all individual domains

Answer 1

The Correct Option is A

Concept: The domain of a function involving logarithmic and inverse trigonometric terms is obtained by finding the intersection of:
the domain of the logarithmic expression, 
the domain of the inverse sine function. 
Step 1: Domain of the logarithmic term. \[ \frac{1}{\ln(10-x)} \text{ is defined when } \ln(10-x)\neq 0 \text{ and } 10-x>0 \] Conditions: \[ 10-x>0 \Rightarrow x<10 \] \[ \ln(10-x)\neq 0 \Rightarrow 10-x\neq 1 \Rightarrow x\neq 9 \] Hence, from the logarithmic term: \[ x\in(-\infty,9)\cup(9,10) \] 
Step 2: Domain of the inverse sine term. \[ \sin^{-1}\!\left(\frac{x+2}{2x+3}\right) \] is defined when: \[ -1\le \frac{x+2}{2x+3}\le 1 \quad \text{and} \quad 2x+3\neq 0 \] 
Step 3: Solve the inequalities. \[ \frac{x+2}{2x+3}\le 1 \Rightarrow x+2\le 2x+3 \Rightarrow x\ge -1 \] \[ \frac{x+2}{2x+3}\ge -1 \Rightarrow x+2\ge -2x-3 \Rightarrow 3x\ge -5 \Rightarrow x\ge -\frac{5}{3} \] Also, \[ 2x+3\neq 0 \Rightarrow x\neq -\frac{3}{2} \] Combining: \[ x\in\left[-\frac{5}{3},-1\right)\ \cup\ (-1,\infty) \] 
Step 4: Find the intersection of both domains. Logarithmic domain: \[ (-\infty,9)\cup(9,10) \] Inverse sine domain: \[ \left[-\frac{5}{3},-1\right)\cup(-1,\infty) \] Intersection: \[ (-\infty,-\tfrac{5}{3}]\ \cup\ (-1,9)\ \cup\ (9,10) \] 
Step 5: Compare with the given domain. \[ (-\infty,-a]\ \cup\ (-1,b)\ \cup\ (b,c) \] Thus, \[ a=\frac{5}{3}, \quad b=9, \quad c=10 \] 
Step 6: Compute the required value. \[ b+c+3a = 9+10+3\left(\frac{5}{3}\right)=19+5=24 \] \[ \boxed{22} \]