Let S be the area of the region enclosed by y=e−x2,y=0,x=0, and x=1. Then
(A)S≥e1✓ Correct
(B)S≥1−e1✓ Correct
(C)S≤41(1+e1)
(D)S≤21+e1(1−21)✓ Correct
Answer: (A), (B), (D)
Solution
S=∫01e−x2dx. Since e−x2≥e−x for x∈[0,1], we have S≥∫01e−xdx=1−e1, which also gives S>e1. For the upper bound, using rectangles: the area under the curve is less than the sum of rectangles with heights at left endpoints. With partition at x=1/2: S<21⋅1+(1−21)⋅e−1/2=21+e1(1−21). Option (C) is incorrect since 41(1+e1)<1−e1.
JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · One or more correct options