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

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Area Under Curves

One of the points of intersection of the curves

y = 1 + 3 x - 2 x^{2}

and

y = \frac{1}{x}

(\frac{1}{2}, 2)

. Let the area of the region enclosed by these curves be

\frac{1}{24} (ℓ 5 + m) - n lo g_{e} (1 + 5)

, where

ℓ, m, n \in N

. Then

ℓ + m + n

is equal to

Answer: (C)

The curves

y = 1 + 3 x - 2 x^{2}

and

y = \frac{1}{x}

meet where

x + 3 x^{2} - 2 x^{3} = 1

, i.e.

(2 x - 1) (x^{2} - x - 1) = 0

, giving

x = \frac{1}{2}

and

x = \frac{1 + 5}{2}

(the positive one).

Area

= \int_{1/2}^{(1 + 5) /2} [(1 + 3 x - 2 x^{2}) - \frac{1}{x}] d x = [x + \frac{3 x ^{2}}{2} - \frac{2 x ^{3}}{3} - ln x]_{1/2}^{(1 + 5) /2} .

Evaluating puts the result in the form

\frac{1}{24} (ℓ 5 + m) - n ln (1 + 5)

with

ℓ + m + n = 30

.
Option C.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option