Question

If sets $A$ and $B$ are define as
$A=\{(x,y):y=e^x,x\in R\}$
$B=\{(x,y):y=x,x\in R\}$, then

• A. $B\subset A$
• B. $A\subset B$
• C. $A\cap B=\varphi$
• D. $A\cup B=A$

Answer 1

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

Given $A=\{(x,y):y=e^x,x\in R\}$

Thus, in roaster form, set A can be written as,

$A=\{.............(-4,\frac{1}{e^4}),(-3,\frac{1}{e^3}),(-2,\frac{1}{e^2}),(-1,\frac{1}{e^1}),(0,1),(1,e^1),(2,e^2),(3,e^3),(4,e^4)...........\}$ (1)

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

Thus, in roaster form, set B can be written as,

$B=\{...........(-4,-4),(-3,-3),(-2,-2),(-1,-1),(0,0),(1,1),(2,2),(3,3),(4,4).........\}$ (2)

From equations (1) and (2), it is clear that no element is common in both the sets.

$\therefore A\cap B=\varphi$