If the domain of the function f(x) = log_elef | vedclass

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(B) The given function is $f(x) = \log_e\left(\frac{2x-3}{5+4x}\right) + \sin^{-1}\left(\frac{4+3x}{2-x}\right)$.For the function to be defined,we nee

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$4$

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(B) The given function is $f(x) = \log_e\left(\frac{2x-3}{5+4x}\right) + \sin^{-1}\left(\frac{4+3x}{2-x}\right)$.For the function to be defined,we need:$1) \frac{2x-3}{5+4x} > 0$$2) -1 \leq \frac{4+3x}{2-x} \leq 1$Solving condition $(1)$:$\frac{2x-3}{5+4x} > 0 \Rightarrow x \in \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)$.Solving condition $(2)$:$-1 \leq \frac{4+3x}{2-x} \leq 1$$\Rightarrow \frac{4+3x}{2-x} + 1 \geq 0$ and $\frac{4+3x}{2-x} - 1 \leq 0$$\Rightarrow \frac{4+3x+2-x}{2-x} \geq 0$ and $\frac{4+3x-2+x}{2-x} \leq 0$$\Rightarrow \frac{6+2x}{2-x} \geq 0$ and $\frac{2+4x}{2-x} \leq 0$$\Rightarrow \frac{x+3}{x-2} \leq 0$ and $\frac{x+1/2}{x-2} \geq 0$From $\frac{x+3}{x-2} \leq 0$,we get $x \in [-3, 2)$.From $\frac{x+1/2}{x-2} \geq 0$,we get $x \in (-\infty, -1/2] \cup (2, \infty)$.Intersection of these gives $x \in [-3, -1/2]$.Now,taking the intersection with condition $(1)$:$x \in ([-3, -1/2]) \cap ((-\infty, -5/4) \cup (3/2, \infty)) = [-3, -5/4)$.Thus,$\alpha = -3$ and $\beta = -5/4$.Therefore,$\alpha^2 + 4\beta = (-3)^2 + 4(-5/4) = 9 - 5 = 4$.

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

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