If sets A and B are defined as A

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

(a) $A \cap B=A$

(b) $A \cap B=B$

(c) $A \cap B=\phi$

(d) $A \cup B=A$


$A=\left\{(x, y): y=\frac{1}{x} ; 0 \neq x \in R\right\}$

$B=\{(x, y): y=-x ; x \in R\}$

then $(x, y) \in A \cap B$

i. e $(x, y) \in A$ and $(x, y) \in B$

i. e $y=\frac{1}{x}$ and $y=-x$

i. $\mathrm{e} \frac{1}{x}=-x ; \quad x \neq 0 \in R$

i. e $1=-x^{2} ; \quad x \neq 0 \in R$

i. e $x^{2}+1=0 ; x \neq 0 \in R$

No such real x exist such that x2 + 1 = 0

$\Rightarrow A \cap B=\phi$

Hence, the correct answer is option C.


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