Question

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

• A. 5
• B. 4
• C. 3
• D. 7

Hint:

For logarithmic functions, the argument must be strictly positive. For inverse sine functions, the argument must lie between -1 and 1, inclusive.

Answer 1

The Correct Option is B

The given function is \[f(x) = \log_e \left( \frac{2x - 3}{5 + 4x} \right) + \sin^{-1} \left( \frac{4 + 3x}{2 - x} \right)\]To determine the domain of \(f(x)\), the following conditions must be met:\[\frac{2x - 3}{5 + 4x}>0 \quad \text{and} \quad \left| \frac{4 + 3x}{2 - x} \right| \le 1\]First, consider the logarithmic condition:\[\frac{2x - 3}{5 + 4x}>0\Rightarrow x \in \left( -\infty, -\frac{5}{4} \right) \cup \left( \frac{3}{2}, \infty \right)\]Next, address the inverse sine condition:\[-1 \le \frac{4 + 3x}{2 - x} \le 1\]This can be divided into two separate inequalities:\[\frac{4 + 3x}{2 - x} \ge -1 \quad \text{and} \quad \frac{4 + 3x}{2 - x} \le 1\]Solving the first inequality:\[\frac{4 + 3x}{2 - x} + 1 \ge 0 \Rightarrow \frac{4 + 3x + 2 - x}{2 - x} = \frac{6 + 2x}{2 - x} \ge 0\]Solving the second inequality:\[\frac{4 + 3x}{2 - x} - 1 \le 0 \Rightarrow \frac{4 + 3x - 2 + x}{2 - x} = \frac{2 + 4x}{2 - x} \le 0\]Combining these two results:\[\left( \frac{6 + 2x}{2 - x} \ge 0 \right) \cap \left( \frac{2 + 4x}{2 - x} \le 0 \right)\]Multiplying the two expressions yields:\[\frac{(6 + 2x)(2 + 4x)}{(2 - x)^2} \le 0\]The solution to this inequality is:\[x \in \left[ -3, -\frac{1}{2} \right]\]Finally, find the intersection of the two main conditions:\[x \in \left( -\infty, -\frac{5}{4} \right) \cup \left( \frac{3}{2}, \infty \right)\quad \cap \quad x \in \left[ -3, -\frac{1}{2} \right]\Rightarrow x \in \left[ -3, -\frac{5}{4} \right)\]
The domain of \(f(x)\) is therefore:\[x \in \left[ -3, -\frac{5}{4} \right)\]Let \( \alpha = -3 \) and \( \beta = -\frac{5}{4} \).Then,\[\alpha^2 + 4\beta = (-3)^2 + 4 \cdot \left( -\frac{5}{4} \right) = 9 - 5 = 4\]