Question

If sets A and B are defined as $A=\left\{\left(x,y\right):y=\frac{1}{x},0\ne x\in R\right\},B=\left\{\left(x,y\right):y=-x,x\in R\right\},$ then
(a) A ∩ B = A
(b) A ∩ B = B
(c) A ∩ B = ϕ
(d) A ∪ B = A

Answer 1

$$\begin{gathered} A=\left\{\left(x,y\right):y=\frac{1}{x};0\ne x\in R\right\} \\ B=\left\{\left(x,y\right):y=-x;x\in R\right\} \\ \mathrm{then}\left(x,y\right)\in A\cap B \\ \mathrm{i}.\mathrm{e}\left(x,y\right)\in A\mathrm{and}\left(x,y\right)\in B \\ \mathrm{i}.\mathrm{e}y=\frac{1}{x}\mathrm{and}y=-x \\ \mathrm{i}.\mathrm{e}\frac{1}{x}=-x;x\ne 0\in R \\ \mathrm{i}.\mathrm{e}1=-x^2;x\ne 0\in R \\ \mathrm{i}.\mathrm{e}{x}^2+1=0;x\ne 0\in R \end{gathered}$$
No such real x exist such that x² + 1 = 0
⇒ A ∩ B = ϕ
Hence, the correct answer is option C.