Question

If the domain of the function \( \log_5 (18x - x^2 - 77) \) is \( (\alpha, \beta) \) and the domain of the function \[ \log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right) \] is \( (\gamma, \delta) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

• A. 195
• B. 174
• C. 186
• D. 179

Hint:

For logarithmic domains, remember to satisfy both the base conditions and the argument conditions simultaneously for accurate results.

Answer 1

The Correct Option is C

Step 1: Determine the domain of \( f_1(x) = \log_2 (18x - x^2 - 77) \). For the logarithm to be defined, the argument must be positive: \[ 18x - x^2 - 77>0 \] Rearranging the inequality yields: \[ x^2 - 18x + 77<0 \] Factoring the quadratic expression: \[ (x - 7)(x - 11)<0 \] The inequality holds for the interval: \[ x \in (7, 11) \] Therefore, \( \alpha = 7 \) and \( \beta = 11 \).

Step 2: Determine the domain of \( f_2(x) = \log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right) \). For this logarithm to be defined, two conditions must be met: - The base must be positive and not equal to 1: \(x - 1>0 \implies x>1\). - The argument must be positive: \[ \frac{2x^2 + 3x - 2}{x^2 - 3x - 4}>0 \] Factoring the numerator and denominator: \[ \frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)}>0 \] Applying the sign chart method to this inequality, the solution is: \[ x \in (4, \infty) \] Considering both conditions (\(x>1\) and \(x \in (4, \infty)\)), the domain for \(f_2(x)\) requires \(x \in (4, \infty)\). Thus, \( \gamma = 4 \) and \( \delta = \infty \) (though \( \delta \) is not required for the final calculation).

Step 3: Calculate the expression \( \alpha^2 + \beta^2 + \gamma^2 \). Substitute the determined values: \[ \alpha^2 + \beta^2 + \gamma^2 = 7^2 + 11^2 + 4^2 = 49 + 121 + 16 = 186 \]