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

Answer: (C)

11

Answer 2

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 $\gamma = 2$.

$\alpha + \beta + \gamma = 11$