Question

If the domain of the function \[ f(x) = \cos^{-1} \left( \frac{2 - |x|}{4} \right) + \left( \log_e (3 - x) \right)^{-1} \] is \([-\alpha, \beta) - \{ \gamma \}\), then \( \alpha + \beta + \gamma \) is equal to:

• A. 12
• B. 9
• C. 11
• D. 8

Answer 1

The Correct Option is C

To find the domain of \( f(x) \), analyze each component individually.

For \( \cos^{-1} \left( \frac{2 - |x|}{4} \right) \) to be defined, \( -1 \leq \frac{2 - |x|}{4} \leq 1 \). Solving these inequalities:

\[ -1 \leq \frac{2 - |x|}{4} \leq 1 \]

leads to \( |x| \leq 6 \), so \( x \in [-6, 6] \).

For \( (\log_e (3 - x))^{-1} \) to be defined, \( \log_e (3 - x) \neq 0 \) and \( 3 - x > 0 \).

1. \( 3 - x > 0 \Rightarrow x < 3 \).
2. \( \log_e (3 - x) \neq 0 \Rightarrow x \neq 2 \) (since \( \log_e (3 - x) = 0 \) when \( x = 2 \)).

Combining these conditions, we have:

\[ x \in [-6, 3) - \{2\} \]

Thus, the domain is \( [-\alpha, \beta) - \{\gamma\} \) where \( \alpha = 6 \), \( \beta = 3 \), and \( \gamma = 2 \).

\( \alpha + \beta + \gamma = 11 \)