If domain of the function _e(6x^2+5x+12x-1)+ ^-1(2x^2-3x+43x-5) is ( , ) ( , ], then 18( ^…

Question

If domain of the function ${\log }_{e}(\frac{6{x}^{2}+5x+1}{2x-1})+{\cos }^{-1}(\frac{2{x}^{2}-3x+4}{3x-5})$ is $(\alpha ,\beta )\cup (\gamma ,\delta ]$, then $18({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}+{\delta }^{2})$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 20

Accepted Answer

Domain of ${\log }_{e}(\frac{6{x}^{2}+5x+1}{2x-1})$

So, $\frac{6{x}^{2}+5x+1}{2x-1}>0$

$\begin{gathered} \Rightarrow \frac{(3x+1)(2x+1)}{2x-1}>0 \ \Rightarrow x\in (\frac{-1}{2},\frac{-1}{3})\cup (\frac{1}{2},\infty) \end{gathered}$

Domain of
${\cos }^{-1}x	o [-1,1]$

${\cos }^{-1}(\frac{2{x}^{2}-3x+4}{3x-5})$

$\begin{gathered} -1\le \frac{2{x}^{2}-3x+4}{3x-5}\le 1 \ \frac{2{x}^{2}-1}{3x-5}\ge 0\frac{2{x}^{2}-6x+9}{3x-5}\le 0 \end{gathered}$

$\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\cup (\frac{5}{3},\infty)$

So, common domain is $(\frac{-1}{2},\frac{-1}{3})\cup [\frac{1}{2},\frac{1}{\sqrt{2}}]$

$\begin{gathered} 	herefore 18({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}+{\delta }^{2})=18(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}) \ =18(\frac{9+4+9+18}{36})=\frac{1}{2}(40)=20 \end{gathered}$

Solutions · 1