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1.Set Theory
medium
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
