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

Question

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

Aicharya · Accepted Answer

Given function is
$f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right)$
For domain, the conditions are
$\frac{2 x-3}{5+4 x} \gt 0 \text { and }\left|\frac{4+3 x}{2-x}\right| \leq 1$
Now, $\frac{2 \mathrm{x}-3}{5+4 \mathrm{x}} \gt 0 \Rightarrow \mathrm{x} \in\left(-\infty,-\frac{5}{4}\right) \cup\left[\frac{3}{2}, \infty\right)$
and $\quad-1 \leq \frac{4+3 x}{2-x} \leq 1$
$\Rightarrow\left(-1 \leq \frac{4+3 x}{2-x}\right) \cap\left(\frac{4+3 x}{2-x} \leq 1\right)$
$\Rightarrow\left(\frac{6+2 \mathrm{x}}{2-\mathrm{x}} \geq 0\right) \cap\left(\frac{2+4 \mathrm{x}}{2-\mathrm{x}} \leq 0\right)$
$\begin{aligned} & \Rightarrow \frac{6+2 x}{2-x} \cdot \frac{2+4 x}{2-x} \leq 0 \\ & \Rightarrow x \in\left[-3,-\frac{1}{2}\right]\end{aligned}$
Hence, we get the domain of f as $\mathrm{x} \in\left[-3,-\frac{5}{4}\right)$ This means that $\alpha=-3, \beta=-\frac{5}{4}$ Thus, $\alpha^2+4 \beta=9-5=4$

If the domain of the function $f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3…

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