Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

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

1) S ≥ 1 / e

2) S ≥ 1 – (1 / e)

3) S ≤ (1 / 4) [1 + (1 / √e)]

4) S ≤ (1 / √2) + (1 / √e) [1 – (1 / √2)]

Solution: (2) S ≥ 1 – (1 / e)

Applications of Integrals JEE Questions Q20

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

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