If S is the area of the region enclosed by y = e^-x2, y = 0, x = 0 and x = 1. Then,

\(\begin{array}{l}S \geq \frac{1}{e}\\ \text { (Since area of a rectangle OCDS =} 1 / \mathrm{e} ),\\ \text \ Since \ e^{-x^{2}} \geq e^{-x}, \forall x \in[0,1]\\ \Rightarrow S \geq \int_{0}^{1} e^{-x} d x=\left(1-\frac{1}{e}\right)\\ \text { Area of a rectangle OAPQ + Area of a rectangle QBRS } > \mathrm{S}\\ \Rightarrow S \leq \frac{1}{\sqrt{2}}(1)+\left(\frac{1-1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{e}}\right)\\ \text \ Since \ \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right) \leq 1-\frac{1}{e}\\ \text { Thus, option (c) is incorrect.}\end{array} \)