Verified Solution • 2025–26 Session moderate

QJEEM23Ma7ade: JEE Main 2023 Mathematics - Class 11 JEEM23Ma7ade Solution: If domain of the function is , then is equal to ______________.

If domain of the function log⁡e(6x2+5x+12x−1)+cos⁡−1(2x2−3x+43x−5)\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 (α,β)∪(γ,δ](\alpha, \beta) \cup(\gamma, \delta] , then 18(α2+β2+γ2+δ2)18\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right) is equal to ______________.

Step-by-Step Verified Solution

Domain of log⁡e(6x2+5x+12x−1)\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)

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

amp;⇒(3x+1)(2x+1)2x−1gt;0amp;⇒x∈(−12,−13)∪(12,∞)\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}

Domain of cos⁡−1x→[−1,1]\cos ^{-1} x \rightarrow[-1,1]

 For domain of cos⁡−1(2x2−3x+43x−5)\text { For domain of } \cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)

amp;−1≤2x2−3x+43x−5≤1amp;2x2−13x−5≥0 and 2x2−6x+93x−5≤0\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}

⇒x∈[−12,12]∪(53,∞)\Rightarrow x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup\left(\frac{5}{3}, \infty\right)

So, common domain is (−12,−13)∪[12,12]\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]

amp;∴18(α2+β2+γ2+δ2)=18(14+19+14+12)amp;=18(9+4+9+1836)=12(40)=20\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring