Key Concepts and Formulas

1. Domain of $\log_b(g(x))$: The argument $g(x)$ must be strictly positive, i.e., $g(x) > 0$. The base $b$ must be positive and not equal to 1.
2. Domain of $\sin^{-1}(h(x))$: The argument $h(x)$ must lie within the closed interval $[-1, 1]$, i.e., $-1 \leq h(x) \leq 1$.
3. Domain of a Sum/Difference of Functions: The domain of $f(x) + g(x)$ or $f(x) - g(x)$ is the intersection of the domain of $f(x)$ and the domain of $g(x)$.
4. Solving Rational Inequalities: Critical points are where the numerator or denominator is zero. These points divide the number line into intervals. Test a value from each interval to determine where the inequality holds. Exclude points that make the denominator zero.

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Step-by-Step Solution

The function is given by $f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right)$. To find the domain of $f(x)$, we need to find the values of $x$ for which both terms are simultaneously defined.

Step 1: Determine the domain of the logarithmic term.

For $\log_e\left(\frac{2 x-3}{5+4 x}\right)$ to be defined, its argument must be strictly positive: $\frac{2 x-3}{5+4 x} > 0$ We find the critical points by setting the numerator and denominator to zero:

• Numerator: $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$
• Denominator: $5 + 4x = 0 \Rightarrow x = -\frac{5}{4}$

These critical points divide the number line into three intervals: $\left(-\infty, -\frac{5}{4}\right)$, $\left(-\frac{5}{4}, \frac{3}{2}\right)$, and $\left(\frac{3}{2}, \infty\right)$. We test a value from each interval:

• For $x < -\frac{5}{4}$ (e.g., $x=-2$): $\frac{2(-2)-3}{5+4(-2)} = \frac{-7}{-3} > 0$.
• For $-\frac{5}{4} < x < \frac{3}{2}$ (e.g., $x=0$): $\frac{2(0)-3}{5+4(0)} = \frac{-3}{5} < 0$.
• For $x > \frac{3}{2}$ (e.g., $x=2$): $\frac{2(2)-3}{5+4(2)} = \frac{1}{13} > 0$.

Since the inequality is strict ($>0$), and $x = -\frac{5}{4}$ makes the denominator zero, the domain for the logarithmic term is: $D_1 = \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)$

Step 2: Determine the domain of the inverse sine term.

For $\sin^{-1}\left(\frac{4+3 x}{2-x}\right)$ to be defined, its argument must be between $-1$ and $1$ (inclusive): $-1 \leq \frac{4+3 x}{2-x} \leq 1$ This compound inequality can be split into two separate inequalities:

Inequality 2a: $\frac{4+3 x}{2-x} \leq 1$ $\frac{4+3 x}{2-x} - 1 \leq 0$ $\frac{4+3 x - (2-x)}{2-x} \leq 0$ $\frac{2+4 x}{2-x} \leq 0$ To apply the wavy curve method easily, we can multiply the numerator and denominator by $-1$ and reverse the inequality sign: $\frac{-(2+4 x)}{-(2-x)} \leq 0 \Rightarrow \frac{2+4 x}{x-2} \geq 0$ The critical points are $2+4x=0 \Rightarrow x = -\frac{1}{2}$ and $x-2=0 \Rightarrow x = 2$. Testing intervals:

• For $x < -\frac{1}{2}$ (e.g., $x=-1$): $\frac{2+4(-1)}{-1-2} = \frac{-2}{-3} > 0$.
• For $-\frac{1}{2} < x < 2$ (e.g., $x=0$): $\frac{2+4(0)}{0-2} = \frac{2}{-2} < 0$.
• For $x > 2$ (e.g., $x=3$): $\frac{2+4(3)}{3-2} = \frac{14}{1} > 0$. Since $x \ne 2$ (denominator cannot be zero) and the inequality is $\geq 0$, the solution is: $x \in \left(-\infty, -\frac{1}{2}\right] \cup (2, \infty)$

Inequality 2b: $\frac{4+3 x}{2-x} \geq -1$ $\frac{4+3 x}{2-x} + 1 \geq 0$ $\frac{4+3 x + (2-x)}{2-x} \geq 0$ $\frac{6+2 x}{2-x} \geq 0$ Multiplying numerator and denominator by $-1$ and reversing the inequality: $\frac{-(6+2 x)}{-(2-x)} \geq 0 \Rightarrow \frac{6+2 x}{x-2} \leq 0$ The critical points are $6+2x=0 \Rightarrow x = -3$ and $x-2=0 \Rightarrow x = 2$. Testing intervals:

• For $x < -3$ (e.g., $x=-4$): $\frac{6+2(-4)}{-4-2} = \frac{-2}{-6} > 0$.
• For $-3 < x < 2$ (e.g., $x=0$): $\frac{6+2(0)}{0-2} = \frac{6}{-2} < 0$.
• For $x > 2$ (e.g., $x=3$): $\frac{6+2(3)}{3-2} = \frac{12}{1} > 0$. Since $x \ne 2$ and the inequality is $\leq 0$, the solution is: $x \in [-3, 2)$

Intersection of Inequality 2a and 2b: To find the domain for the inverse sine term, we intersect the solutions from Inequality 2a and 2b:

• From 2a: $\left(-\infty, -\frac{1}{2}\right] \cup (2, \infty)$
• From 2b: $[-3, 2)$

The intersection is: $D_2 = \left( \left(-\infty, -\frac{1}{2}\right] \cup (2, \infty) \right) \cap [-3, 2)$ $D_2 = [-3, -\frac{1}{2}]$

Step 3: Find the intersection of the domains of both terms.

The domain of $f(x)$ is the intersection of $D_1$ and $D_2$: $D = D_1 \cap D_2 = \left( \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right) \right) \cap [-3, -\frac{1}{2}]$ Let's visualize this intersection:

• $D_1$: $(-\infty, -1.25) \cup (1.5, \infty)$
• $D_2$: $[-3, -0.5]$

The interval $[-3, -0.5]$ overlaps with $(-\infty, -1.25)$ in the range $[-3, -1.25)$. It does not overlap with $(1.5, \infty)$. Therefore, the intersection is: $D = [-3, -\frac{5}{4})$

The domain of the function $f(x)$ is given as $[\alpha, \beta)$. By comparing this with our derived domain $[-3, -\frac{5}{4})$, we have: $\alpha = -3$ $\beta = -\frac{5}{4}$

Step 4: Calculate $\alpha^2 + 4\beta$.

Now we substitute the values of $\alpha$ and $\beta$ into the expression: $\alpha^2 + 4\beta = (-3)^2 + 4\left(-\frac{5}{4}\right)$ $= 9 + (-5)$ $= 9 - 5$ $= 4$

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Common Mistakes & Tips

• Sign Errors in Inequalities: Be extremely careful when manipulating inequalities, especially when multiplying or dividing by expressions involving variables. Always consider the sign of the multiplier.
• Excluding Denominator Zeros: Remember that values of $x$ that make the denominator of a rational expression zero must always be excluded from the domain, even if the inequality is $\leq$ or $\geq$.
• Intersection of Intervals: Clearly visualize the intersection of intervals on a number line. This is crucial for finding the overall domain of composite functions.

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Summary

To find the domain of the given function, we first determined the domain restrictions for the logarithmic term (argument must be positive) and the inverse sine term (argument must be between -1 and 1 inclusive). We solved the inequalities for each term separately. The domain of the function is the intersection of these individual domains. After finding the intersection to be $[-3, -\frac{5}{4})$, we identified $\alpha = -3$ and $\beta = -\frac{5}{4}$. Finally, we calculated $\alpha^2 + 4\beta$ to be $4$.

The final answer is $\boxed{4}$. This corresponds to option (A).