Question

If the domain of the function \( f(x) = \log_e \left( \frac{2x + 3}{4x^2 + x - 3} \right) + \cos^{-1} \left( \frac{2x - 1}{x + 2} \right) \) is \( (\alpha, \beta] \), then the value of \( 5\beta - 4\alpha \) is equal to

• A. 12
• B. 10
• C. 11
• D. 9

Answer 1

The Correct Option is A

Determine the domain for each part of \( f(x) \): Logarithmic Part: \( \log_{e} \left( \frac{2x + 3}{4x^2 + x - 3} \right) \) requires
\[ \frac{2x + 3}{4x^2 + x - 3} > 0. \]
Critical points are \( x = -\frac{3}{2}, x = -1, \) and \( x = \frac{3}{4} \).
Solution: \( x \in \left( -\frac{3}{2}, -1 \right) \cup \left( \frac{3}{4}, \infty \right) \).

Inverse Cosine Part: \( \cos^{-1} \left( \frac{2x - 1}{x + 2} \right) \) requires
\[ -1 \leq \frac{2x - 1}{x + 2} \leq 1. \]
Solution: \( x \in \left[ -\frac{1}{3}, 3 \right] \).
Intersection of Domains:
The combined domain is \( \left( \frac{3}{4}, 3 \right] \), giving \( (\alpha, \beta] = \left( \frac{3}{4}, 3 \right] \).
Calculate \( 5\beta - 4\alpha \): \( \alpha = \frac{3}{4}, \beta = 3 \).

\[ 5\beta - 4\alpha = 5 \times 3 - 4 \times \frac{3}{4} = 15 - 3 = 12. \]