Let \(\mathrm{S}\) be the area of the region enclosed by \(y=e^{-x^{2}}\), \(y=0, x=0\) and \(x=1\); then

The given curve \(y=e^{-x^{2}}\) Draw a rough sketch of curve at \(x=0, y=1\) and at \(x=1, y=1 / e\) \(\because y=e^{-x^{2}}\) \(\Rightarrow \frac{d y}{d x}=-2 x e^{-x^{2}} < 0 \quad \forall x \in(0,1)\)

\(\therefore y=e^{-x^{2}}\) is decreasing on \((0,1)\)

Hence its graph is as shown in figure given below


Now, \(\mathrm{S}=\) area exclosed by curve \(=\mathrm{XYCO}\)

and area of rectangle \(\mathrm{OCYL}=\frac{1}{e}\)

Clearly \(S>\frac{1}{e} \quad \therefore\) a is true.

For \(x \in[0,1] \Rightarrow x^{2} < x\)

\(\begin{array}{l}

\Rightarrow-x^{2}>-x \quad \Rightarrow e^{-x^{2}} \geq e^{-x} \forall x \in[0,1] \\

\Rightarrow \int_{0}^{1} e^{-x^{2}} d x>\int_{0}^{1} e^{-x} d x=1-\frac{1}{e} \\

\Rightarrow S>1-\frac{1}{e} \quad \therefore \text { (b) is true. }

\end{array}\)

Now \(S < \) area of rectangle \(\mathrm{XADO}+\) area of rectangle \(\mathrm{ZDCN}\) \(\Rightarrow S < \frac{1}{\sqrt{2}} \times 1+\left(1-\frac{1}{\sqrt{2}}\right) \frac{1}{\sqrt{e}}\) \(\therefore S < \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right) \quad \because(\mathrm{d})\) is true

Also as \(\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right) < 1-\frac{1}{e} \quad \therefore\) (c) is incorrect.