The area enclosed by the curves $y=\log _{e}\left(x+e^{2}\right)$, $x=\log _{e}\left(\frac{2}{y}\right)$ and $x=\log _{e} 2$, above the line $y=1$ is

(B) $1+\mathrm{e}-\log _{\mathrm{e}} 2$

(A) $2+\mathrm{e}-\log _{\mathrm{e}} 2$

(C) $\mathrm{e}-\log _{\mathrm{e}} 2$

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Area Under Curves

The area enclosed by the curves

y = lo g_{e} (x + e^{2})

x = lo g_{e} (\frac{2}{y})

and

x = lo g_{e} 2

, above the line

y = 1

Answer: (B)

The region is bounded by three curves above

y = 1

. Convert to appropriate integrals:

First part: integrate

y = ln (x + e^{2})

from

x = e - e^{2}

(where

y = 1

) to

x = 0

\int_{e - e^{2}}^{0} [ln (x + e^{2}) - 1] d x

Second part: from

x = lo g_{e} (\frac{2}{y})

, we get

y = 2 e^{- x}

. Integrate from

x = 0

x = ln 2

\int_{0}^{l n 2} [2 e^{- x} - 1] d x

Evaluating both integrals and summing gives the total area:

1 + e - ln 2

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option