Let S be the area of the region enclosed by y=e^-x^2 ,y=0,x=0 and x=1…

Clearly, $S = \int\limits_0^1e^{-x^2}dx$

Also,

S > Area of rectangle OPCR

$⇒S>OP ×OR⇒ S>1×\frac{1}{e}⇒ S>\frac{1}{e}$

Now, $e^{-x^2}≥e^{-x}$ for all $x∈[0, 1]$

$⇒\int\limits_0^1e^{-x^2}dx≥\int\limits_0^1e^{-x}dx$

$⇒S≥[-e^{-x}]_0^1⇒S≥(-\frac{1}{e}+1)⇒S≥1-\frac{1}{e}$

Clearly,

S ≤ Area of rectangle OASQ + Area of rectangle PQBU

$⇒S≤\frac{1}{\sqrt{2}}×1(1-\frac{1}{\sqrt{2}})\frac{1}{\sqrt{e}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$

Since $\frac{1}{4}(1+\frac{1}{\sqrt{e}})<1-\frac{1}{e}$. Therefore, $S>\frac{1}{4}(1+\frac{1}{\sqrt{e}})$

So, option (3) is incorrect.