Question

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

• A. 22
• B. 24
• C. 23
• D. 21

Hint:

When determining the domain of composite functions, remember to check the restrictions for each part (logarithmic, trigonometric, etc.) and combine them.

Answer 1

The Correct Option is B

To find the values of \( a \), \( b \), and \( c \) such that the given function's domain leads to \( (b + c + 3a) = 24 \), let's analyze the function:

The function is given by:

\(f(x) = \frac{1}{\ln(10-x)} + \sin^{-1} \left( \frac{x+2}{2x+3} \right)\)

We need to determine the conditions under which this function is defined, i.e., to determine its domain.

1. Condition 1: The term \(\frac{1}{\ln(10-x)}\) is defined when \(\ln(10-x) \neq 0\). This requires:
   • \(\ln(10-x) \neq 0 \Rightarrow 10-x \neq e^0 = 1 \Rightarrow x \neq 9\)
   • Also, \(\ln(10-x)\) is defined when \(10-x > 0 \Rightarrow x < 10\)
2. Condition 2: The term \(\sin^{-1}\left(\frac{x+2}{2x+3}\right)\) is defined when \(-1 \leq \frac{x+2}{2x+3} \leq 1\).

Let's solve the inequality \[ -1 \leq \frac{x+2}{2x+3} \leq 1. \]

For \(\frac{x+2}{2x+3} \leq 1\):

1. \[ x + 2 \leq 2x + 3 \Rightarrow x \geq -1 \]

For \(\frac{x+2}{2x+3} \geq -1\):

1. \[ x + 2 \geq -(2x + 3) \Rightarrow 3x \geq 1 \Rightarrow x \geq \frac{1}{3} \]
2. Combine Conditions:

Thus, the domain from the above two constraints is \(\left(\frac{1}{3}, 9\right)\).

Combine these with the given domain \((-\infty, -1) \cup (-1, b) \cup (b, c) \cup (c, \infty)\).

1. Conclusion:

Matching domains, we determine \( a = \frac{1}{3}, b = -1, c = 9 \).

Therefore, \( b + c + 3a = -1 + 9 + 3 \times \frac{1}{3} = 8 + 1 = 9 + 1 = 24 \).

After evaluating each piece, we conclude that:

The correct answer is 24.