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

Answer: (A)

12

Answer 2

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 $5\beta - 4\alpha$: $\alpha = \frac{3}{4}, \beta = 3$.

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