IIT-JEE 2012 · Paper 1, Mathematics question 52

On $[0,1]$, $e^{-x^2}$ decreases from 1 to $\frac1e$ and is concave on $[0,\frac1{\sqrt2}]$ (its inflection point). Numerically $S\approx0.747$.

• (A) $e^{-x^2}\ge\frac1e$ on $[0,1]$, so $S\ge\frac1e$. True.
• (B) $x^2\le x$ on $[0,1]$, so $e^{-x^2}\ge e^{-x}$ and $S\ge\int_0^1e^{-x}dx=1-\frac1e\approx0.632$. True.
• (C) $\frac14\left(1+e^{-1/2}\right)\approx0.40<S$. False.
• (D) Bound $e^{-x^2}$ by 1 on $[0,\frac1{\sqrt2}]$ and by $e^{-1/2}$ on $[\frac1{\sqrt2},1]$ (it is decreasing):
  $$S\le\frac1{\sqrt2}+\frac1{\sqrt e}\left(1-\frac1{\sqrt2}\right)\approx0.885.$$
  True.

(A), (B), (D)