f(x)= _e((2x-3)/(5+4x))+ ^-1((4+3x)/(2-x))[ , ) ^2+4

Question

$f(x)={\log }_{e}(\frac{2x-3}{5+4x})+{\sin }^{-1}(\frac{4+3x}{2-x})[\alpha ,\beta ){\alpha }^{2}+4\beta $

Accepted Answer

$\frac{2x-3}{4x+5}>0$

$\begin{gathered} 	herefore \, x\in (-\infty,-\frac{5}{4})\cup (\frac{3}{2},\infty)........(i) \ -1\le \frac{3x+4}{2-x}\le 1 \ \frac{3x+4}{2-x}\le 1 \ \Rightarrow \, \frac{3x+4}{2-x}-1\le 0 \end{gathered}$

$\begin{gathered} \Rightarrow \, \frac{3x+4-2+x}{x-2}\ge 0 \ \Rightarrow \, \frac{4x+2}{x-2}\ge 0 \ \Rightarrow \, x\in (-\infty,-\frac{1}{2}]\cup (2,\infty)........(ii) \end{gathered}$

$\begin{gathered} \frac{3x+4}{2-x}\ge -1 \ \Rightarrow \frac{3x+4}{2-x}+1\ge 0 \ \Rightarrow \frac{3x+4+2-x}{2-x}\ge 0 \ \Rightarrow \frac{2x+6}{x-2}\le 0 \ 	herefore x\in [-3,2)........(iii) \end{gathered}$

Taking intersection of (i), (ii) and (iii)

$\begin{gathered} x\in [-3,-\frac{5}{4}) \ \alpha =-3,\beta =-\frac{5}{4} \ {\alpha }^{2}+4\beta =4 \end{gathered}$

Solutions · 1