If the domain of the function f(x)=loge(2x-3/5+4x)+ sin^(-1)(4+3x/2-x) is [α, β}, then α^2+4β is equal to

Question

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

(1) 5

(2) 4

(3) 3

(4) 7

Answer 1

Correct option is: (2) 4   

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

\(\frac{2 x-3}{5+4 x}>0\) and \(\left|\frac{4+3 x}{2-x}\right| \leq 1\)

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 \(-1 \leq \frac{4+3 x}{2-x} \leq 1\)

\(\Rightarrow\left(-1 \leq \frac{4+3 \mathrm{x}}{2-\mathrm{x}}\right) \cap\left(\frac{4+3 \mathrm{x}}{2-\mathrm{x}} \leq 1\right)\)

\(\Rightarrow\left(\frac{6+2 \mathrm{x}}{2-\mathrm{x}} \geq 0\right) \cap\left(\frac{2+4 \mathrm{x}}{2-\mathrm{x}} \leq 0\right)\)      

\(\Rightarrow \frac{6+2 \mathrm{x}}{2-\mathrm{x}} \cdot \frac{2+4 \mathrm{x}}{2-\mathrm{x}} \leq 0\)

\(\Rightarrow \mathrm{x} \in\left[-3,-\frac{1}{2}\right]\)

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, \(\alpha^{2}+4 \beta=9-5=4\)