One of the points of intersection of the curves y=1+3 x-2 x^2 and [y=(1)/(x)] is [((1)/(2)…

Question

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

Accepted Answer

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

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

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

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

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

Solutions · 1