Question

Let the domains of the functions $ f(x) = \log_4 \log_3 \log_7 (8 - \log_2(x^2 + 4x + 5)) $ and $ g(x) = \sin^{-1} \left( \frac{7x + 10}{x - 2} \right)$ be $ (\alpha, \beta) $ and $ [\gamma, \delta] $, respectively. Then $ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 $ is equal to:

• A. 15
• B. 13
• C. 16
• D. 14

Hint:

Always check the domains of logarithmic and trigonometric functions to ensure they are properly defined.

Answer 1

The Correct Option is A

First, analyze the function \( f(x) = \log_4 \log_3 \log_7 (8 - \log_2(x^2 + 4x + 5)) \). 
For this function to be defined, the expression inside the logarithms must be positive. 
Solving the inequalities gives the domain of \( f(x) \) as \( (\alpha, \beta) \). 
Similarly, for the function \( g(x) = \sin(x^2) \), the domain is \( [\gamma, \delta] \). 
After finding the domains, we compute \( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 15 \). 
Thus, the correct answer is \( 15 \).

Answer 2

Step 1: First inequality:

\[ \log_3 \left( \log_7 \left( 8 - \log_2 \left( x^2 + 4x + 5 \right) \right) \right) > 0. \] This implies: \[ \log_7 \left( 8 - \log_2 \left( x^2 + 4x + 5 \right) \right) > 1. \] Simplifying further: \[ 8 - \log_2 \left( x^2 + 4x + 5 \right) > 7. \] Thus: \[ \log_2 \left( x^2 + 4x + 5 \right) < 1. \]

Step 2: Second inequality:

\[ \log_2 \left( x^2 + 4x + 5 \right) < 1 \quad \Rightarrow \quad x^2 + 4x + 5 < 2. \] This simplifies to: \[ x^2 + 4x + 3 < 0. \] The solution is: \[ x \in (-3, -1). \]

Step 3: Third inequality:

\[ -1 \leq \frac{7x + 10}{x - 2} \leq 1. \] The solution to this inequality is: \[ x \in [-2, -1]. \]

Step 4: Values of \( \alpha, \beta, \gamma, \delta \):

\[ \alpha = -3, \, \beta = -1, \, \gamma = -2, \, \delta = -1. \] Now, calculate: \[ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 15. \]