One of the points of intersection of the curv | vedclass

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(B) The curves are $y=1+3x-2x^2$ and $y=\frac{1}{x}$. The intersection points are found by $1+3x-2x^2 = \frac{1}{x} \implies x+3x^2-2x^3 = 1 \implies

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$30$

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(B) The curves are $y=1+3x-2x^2$ and $y=\frac{1}{x}$. The intersection points are found by $1+3x-2x^2 = \frac{1}{x} \implies x+3x^2-2x^3 = 1 \implies 2x^3-3x^2-x+1=0$. Given one point is $x=\frac{1}{2}$,the other intersection point is $x=\frac{1+\sqrt{5}}{2}$.The area $A$ is given by $\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}} (1+3x-2x^2-\frac{1}{x}) dx$.$A = \left[x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln|x|\right]_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}$.Evaluating at the limits:$A = \left(\frac{1+\sqrt{5}}{2} + \frac{3}{2}(\frac{1+\sqrt{5}}{2})^2 - \frac{2}{3}(\frac{1+\sqrt{5}}{2})^3 - \ln(\frac{1+\sqrt{5}}{2})\right) - \left(\frac{1}{2} + \frac{3}{2}(\frac{1}{4}) - \frac{2}{3}(\frac{1}{8}) - \ln(\frac{1}{2})\right)$.Simplifying the terms:$A = \frac{1+\sqrt{5}}{2} + \frac{3(6+2\sqrt{5})}{8} - \frac{2(16+8\sqrt{5})}{24} - \ln(\frac{1+\sqrt{5}}{2}) - \frac{1}{2} - \frac{3}{8} + \frac{1}{12} + \ln(\frac{1}{2})$.$A = \frac{1}{2} + \frac{\sqrt{5}}{2} + \frac{9}{4} + \frac{3\sqrt{5}}{4} - \frac{4}{3} - \frac{2\sqrt{5}}{3} - \frac{1}{2} - \frac{3}{8} + \frac{1}{12} - \ln(1+\sqrt{5}) + \ln(2) + \ln(2)$.$A = \sqrt{5}(\frac{1}{2} + \frac{3}{4} - \frac{2}{3}) + (\frac{9}{4} - \frac{4}{3} - \frac{3}{8} + \frac{1}{12}) - \ln(1+\sqrt{5}) + 2\ln(2)$.$A = \frac{7}{12}\sqrt{5} + \frac{54-32-9+2}{24} - \ln(1+\sqrt{5}) + 2\ln(2) = \frac{14\sqrt{5}+15}{24} - \ln(1+\sqrt{5}) + 2\ln(2)$.Comparing with $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,we have $\ell=14, m=15, n=1$. Thus $\ell+m+n = 14+15+1 = 30$.

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

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