JEE Mains 28-July-2022 Shift 2 Solved Paper

According to NTA, the required region A2 which is shaded in crossed lines and comes out to be
A2=

2

∫

1

(ln‌

2

y

−ey+e2)dy=1+e−ln‌2
But according to us the required region A1 comes out to be shaded in parallel lines, which can be obtained as
A1=

ln‌2

∫

0

(ln(x+e2)−2e−x)dx
=.{(x+e2)‌ln(x+e2)−x+2e−x}|0ln‌2
=(ln‌2+e2)‌ln(ln‌2+e2)−ln‌2+1
−2e2−2
=(ln‌2+e2)‌ln(ln‌2+e2)−ln‌2−2e2−1
Not given in any option.
The region asked in the question is bounded by three curves
y=ln(x+e2)
x=ln(‌

2

y

)
x=ln‌2
There is only one region which satisfies above requirement and which also lies above line y=1
Line y=1 may or may not be the boundary of the region.