QJEEM24Ma7a29: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7a29 Solution: One of the points of intersection of the curves and is . Let…

Solving curves $y=1+3 x-2 x^2 ~\& ~y=\frac{1}{x}$

$\begin{aligned}& 2 x^3-3 x^2-x+1=0 \\& \Rightarrow \quad(2 x-1)\left(x^2-x-1\right)=0 \\& \Rightarrow \quad x=\frac{1}{2}, x=\frac{1 \pm \sqrt{5}}{2}\end{aligned}$

Area $= \int\limits_{{1 \over 2}}^{{{1 + \sqrt 5 } \over 2}} {\left( {1 + 3x - 2{x^2} - {1 \over x}dx} \right)}$

$= \left[ {x + {{3{x^2}} \over 2} - {{2x} \over 3} - \ln x} \right]$

$= {{\sqrt 5 + 1} \over 2} + {3 \over 8}{(\sqrt 5 + 1)^2} - {1 \over {12}}(\sqrt 5 + 1) - \ln \left( {{{\sqrt 5 + 1} \over 2}} \right) - \left( {{1 \over 2} + {3 \over 8} - {1 \over {12}} - \ln {1 \over 2}} \right)$

$= {1 \over {24}}\left[ {12(\sqrt 5 + 1) + 9{{(\sqrt 5 + 1)}^2} - 2(\sqrt 5 + 1) - 12 - 9 + 2] - \ln \left( {{{\sqrt 5 + 1} \over 2} \times 2} \right)} \right]$

$\begin{aligned}= & \frac{1}{24}[12(\sqrt{5}+1)+9(6+2 \sqrt{5})- \\& 2(5 \sqrt{5}+1+3 \sqrt{5}(\sqrt{5}+1)-19)]-\ln (\sqrt{5}+1) \\= & \frac{1}{24}[14 \sqrt{5}+15]-\ln (\sqrt{5}+1) \\\therefore \quad & \quad l=14, m=15, n=1 \\& \quad I+m+n=30\end{aligned}$