One of the points of intersection of the curves y=1+3 x-2 x^2 and y=(1)/(x) is ((1)/(2), 2) . Let the area of the region enclosed by these curves be (1)/(24)(l 5 +m)-n _ e (1+ 5 ) , where l, ~m, n ∈ N . Then l+m+n is

Correct answer: C. 30 aligned & A=∫_ (1)/(2) ^ 1+ 5 2 (1+3 x-2 x^2-(1)/(x)) dx & A=[x+(3 x^2)/(2)-(2 x^3)/(3)- x]_ (1)/(2) ^ 1+ 5 & A= 1+ 5 2 +(3)/(2)( 1+ 5 2 )^2-(2)/(3)( 1+ 5 2 )^3- ( 1+ 5 2 ) & -(1)/(2)-(3)/(2)((1)/(4))+(2)/(3)((1)/(8))+ ((1)/(2)) & A=(1)/(2)+ 5 2 +(3)/(8)+(3)/(4) 5 +(15)/(8)-(4)/(3)-(2)/(3) 5 & -(1)/(2)-(3)/(8)+(1)/(12)- (1+ 5 ) & = 5 ((1)/(2)+(3)/(4)-(2)/(3))+(15)/(8)-(4)/(3)+(1)/(12)- (1+ 5 ) & =(14)/(24) 5 +(15)/(24)- (1+ 5 ) aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Area Under Curves

One of the points of intersection of the curves y=1+3 x-2 x^2 and...

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 One of the points of intersection of the curves y=1+3 x-2 x^2 and y=(1)/(x) is ((1)/(2), 2) . Let the area of the

Options

Correct answer

C. 30

Step-by-step solution

\begin{aligned} & A=\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3 x-2 x^2-\frac{1}{x}\right) \mathrm{dx} \\ & A=\left[x+\frac{3 x^2}{2}-\frac{2 x^3}{3}-\ln x\right]_{\frac{1}{2}}^{1+\sqrt{5}} \\ & A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}\left(\frac{1+\sqrt{5}}{2}\right)^2-\frac{2}{3}\left(\frac{1+\sqrt{5}}{2}\right)^3-\ln \left(\frac{1+\sqrt{5}}{2}\right) \\ & -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right) \\ & A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4} \sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3} \sqrt{5} \\ & -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\frac{14}{24} \sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})\end{aligned}

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