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 the function \(f(x) = \cos^{-1} \left( \frac{2 - |x|}{4} \right) + \left( \log_e (3 - x) \right)^{-1}\), we need to consider the domain restrictions of both parts separately.

Domain of \(\cos^{-1} \left( \frac{2 - |x|}{4} \right)\):

The inverse cosine function, \(\cos^{-1}(y)\), is defined for \(-1 \leq y \leq 1\). Thus, we need:

\(-1 \leq \frac{2 - |x|}{4} \leq 1\)

Solving the inequality:

• \(\frac{2 - |x|}{4} \leq 1\):
  • \(2 - |x| \leq 4\)
  • \(-|x| \leq 2\)
  • \(-1 \leq \frac{2 - |x|}{4}\):
    • \(-4 \leq 2 - |x|\)
    • \(-6 \leq x \leq 6\)

Domain of \(\left( \log_e(3 - x) \right)^{-1}\):

The natural logarithm, \(\log_e(3-x)\), is defined and positive if \(3 - x > 1\), i.e., \(x < 3\).

But since \(\left(\log_e(3-x)\right)^{-1}\) should be defined, \(\log_e(3-x) \neq 0\).

This implies \(3 - x \neq 1\), hence, \(x \neq 2\).

Combining all conditions:

\(-6 \leq x \leq 6\), \(x < 3\), and \(x \neq 2\).

This means the domain is \([-6, 3) \setminus \{2\}\), which translates to \([-6, 2) \cup (2, 3)\).

Thus, we have:

• \(\alpha = 6\) (since \([-\alpha, \beta)\) implies \([-6, 3)\))
• \(\beta = 3\)
• \(\gamma = 2\)

Therefore, \(\alpha + \beta + \gamma = 6 + 3 + 2 = 11\).

The correct answer is 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 \( \gamma = 2 \).

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