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.