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

For the function $$f(x)=\log_e\!\left(\dfrac{2x-3}{5+4x}\right)+\sin^{-1}\!\left(\dfrac{4+3x}{2-x}\right)$$ to be defined, two conditions must hold simultaneously:

(i) Logarithm: $$\dfrac{2x-3}{5+4x}\gt 0$$

(ii) Inverse sine: $$-1 \le \dfrac{4+3x}{2-x} \le 1$$ and $$2-x \ne 0$$ (so $$x\ne 2$$).

Condition (i):

The fraction is positive when numerator and denominator have the same sign.

Case 1 $$2x-3\gt 0\;$$ and $$\;5+4x\gt 0$$
  $$x\gt \dfrac32,$$ $$x\gt -\dfrac54 \;\Rightarrow\; x\gt \dfrac32$$

Case 2 $$2x-3\lt 0\;$$ and $$\;5+4x\lt 0$$
  $$x\lt \dfrac32,$$ $$x\lt -\dfrac54 \;\Rightarrow\; x\lt -\dfrac54$$

Hence, from the logarithm we get

$$x\in(-\infty,\,-\dfrac54)\;\cup\;(\dfrac32,\,\infty) \quad -(1)$$

Condition (ii): Let $$y=\dfrac{4+3x}{2-x}.$$ We solve $$-1\le y\le 1.$$ Break into two parts, keeping the sign of $$2-x$$ in mind.

(a)  First inequality $$\dfrac{4+3x}{2-x}\le 1$$

• If $$x\lt 2$$ (denominator positive): multiply directly
  $$4+3x\le 2-x \;\Longrightarrow\; 4x\le -2 \;\Longrightarrow\; x\le -\dfrac12.$$
• If $$x\gt 2$$ (denominator negative): inequality reverses
  $$4+3x\ge 2-x \;\Longrightarrow\; 4x\ge -2 \;\Longrightarrow\; x\ge -\dfrac12.$$ Since this lies entirely above 2, the whole interval $$x\gt 2$$ satisfies part (a).

(b)  Second inequality $$\dfrac{4+3x}{2-x}\ge -1$$

• If $$x\lt 2$$ (denominator positive):
  $$4+3x\ge -\,\bigl(2-x\bigr) \;\Longrightarrow\; 4+3x\ge -2+x \;\Longrightarrow\; 2x\ge -6 \;\Longrightarrow\; x\ge -3.$$
• If $$x\gt 2$$ (denominator negative): inequality reverses
  $$4+3x\le -\,\bigl(2-x\bigr) \;\Longrightarrow\; 2x\le -6 \;\Longrightarrow\; x\le -3,$$ which is impossible because here $$x\gt 2.$$ Thus part (b) gives $$-3\le x\lt 2.$$

Combining (a) and (b): for $$x\lt 2,$$ both must hold, giving
$$x\in[-3,\,-\dfrac12].$$
For $$x\gt 2,$$ part (b) fails, so no additional points appear.

$$x\in[-3,\,-\dfrac12] \quad -(2)$$

Overall domain: Intersect $$(1)$$ and $$(2)$$.

$$( -\infty,\,-\dfrac54)\cup(\dfrac32,\infty)\; \cap\;[-3,\,-\dfrac12] \;=\;[-3,\,-\dfrac54).$$

Therefore, the domain is $$[\alpha,\beta)=\bigl[-3,\,-\dfrac54\bigr).$$ So $$\alpha=-3,\;\beta=-\dfrac54.$$

Compute $$\alpha^2+4\beta:$$

$$\alpha^2+4\beta = (-3)^2 + 4\!\left(-\dfrac54\right)=9-5=4.$$

Hence, $$\alpha^2+4\beta=4,$$ which matches Option B.