Let A = { (x,,y):y = {e^x},,x ∈ R} , B = { (x,,y):y = {e^{ - x}},,x ∈ R} . Then

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1.Set Theory

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Let $A = \{ (x,\,y):y = {e^x},\,x \in R\} $, $B = \{ (x,\,y):y = {e^{ - x}},\,x \in R\} .$ Then

A

$A \cap B = \phi $

B

$A \cap B \ne \phi $

C

$A \cup B = {R^2}$

D

None of these

Solution

(b) $\because y = {e^x},\,\,y = {e^{ – x}}$ will meet, when ${e^x} = {e^{ – x}}$

==> ${e^{2x}} = 1,\,\,\therefore x = 0,y = 1$

$\therefore A$ and $B$ meet on $(0, 1), $

$\therefore A \cap B = \phi $.

Standard 11

Mathematics

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