If domain of the function $$\log _{e}\left(\frac{6 x^{2}+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^{2}-3 x+4}{3 x-5}\right)$$ is $(\alpha, \beta) \cup(\gamma, \delta]$, then $18\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)$ is equal to ______________.
Answer (integer)
20
Solution
Domain of $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)$
<br/><br/>So, $\frac{6 x^2+5 x+1}{2 x-1}>0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{(3 x+1)(2 x+1)}{2 x-1}>0 \\\\
& \Rightarrow x \in\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \infty\right)
\end{aligned}
$$
<br/><br/>Domain of
$\cos ^{-1} x \rightarrow[-1,1]$
<br/><br/>$\text { For domain of } \cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$
<br/><br/>$$
\begin{aligned}
& -1 \leq \frac{2 x^2-3 x+4}{3 x-5} \leq 1 \\\\
& \frac{2 x^2-1}{3 x-5} \geq 0 \text { and } \frac{2 x^2-6 x+9}{3 x-5} \leq 0
\end{aligned}
$$
<br/><br/>$$
\Rightarrow x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup\left(\frac{5}{3}, \infty\right)
$$
<br/><br/>So, common domain is $\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$
<br/><br/>$$
\begin{aligned}
& \therefore 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}\right) \\\\
& =18\left(\frac{9+4+9+18}{36}\right)=\frac{1}{2}(40)=20
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.