If sets A and B are defined as A - x-y - y-1-x- x -0- x - R-B - x-y - y-x- x - R- then

X ∈ (A ∩ B) ⇒ x ∈ A and x ∈ B i.e. y=1/x and y= x ∴ x ∈ A ∩ B ⇒ x=1/x ⇒ x^2=1 ⇒ x^2= 1, This is not possible for any real value of x. ∴ A ∩ B = ϕ nb ...

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Standard XII

Mathematics

Intersection

If sets A and...

Question

If sets $A$ and $B$ are defined as
$A = {(x, y) | y = \frac{1}{x}, x \neq 0, x \in R},$
$B = {(x, y) | y = - x, x \in R},$ then

$A \cap B = A$

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$A \cup B = B$

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$A \cap B = ϕ$

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$A \cup B = A$

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Solution

The correct option is C
$A \cap B = ϕ$

$x \in (A \cap B) \Rightarrow x \in A and x \in B i . e . y = \frac{1}{x} and y = - x$

$∴ x \in A \cap B \Rightarrow - x = \frac{1}{x}$
$\Rightarrow - x^{2} = 1$
$\Rightarrow x^{2} = - 1,$
This is not possible for any real value of $x$ .
$∴ A \cap B = ϕ$

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If sets A and B are defined as A={(x,y) | y=1x, x≠0, x ∈ R}, B={(x,y) | y=−x, x ∈ R}, then

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