Mathematics - Functions Question with Solution | TestHub

Given function is
$f(x)={\log }_e\left(\frac{2x-3}{5+4x}\right)+{\sin }^{-1}\left(\frac{4+3x}{2-x}\right)$
For domain, the conditions are
$\frac{2x-3}{5+4x}>0\text{ and }\left|\frac{4+3x}{2-x}\right|\le 1$
Now, $\frac{2\mathrm{x}-3}{5+4\mathrm{x}}>0\Rightarrow \mathrm{x}\in \left(-\infty ,-\frac{5}{4}\right)\cup \left[\frac{3}{2},\infty \right)$
and $-1\le \frac{4+3x}{2-x}\le 1$
$\Rightarrow \left(-1\le \frac{4+3x}{2-x}\right)\cap \left(\frac{4+3x}{2-x}\le 1\right)$
$\Rightarrow \left(\frac{6+2\mathrm{x}}{2-\mathrm{x}}\ge 0\right)\cap \left(\frac{2+4\mathrm{x}}{2-\mathrm{x}}\le 0\right)$
$\begin{array}{rl}& \Rightarrow \frac{6+2x}{2-x}\cdot \frac{2+4x}{2-x}\le 0\\ & \Rightarrow x\in \left[-3,-\frac{1}{2}\right]\end{array}$
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, ${\alpha }^2+4\beta =9-5=4$