The correct option is
C A∩B=ϕGiven A={(x,y):y=ex,x∈R}
Thus, in roaster form, set A can be written as,
A={.............(−4,1e4),(−3,1e3),(−2,1e2),(−1,1e1),(0,1),(1,e1),(2,e2),(3,e3),(4,e4)...........} (1)
Given B={(x,y):y=x,x∈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.
∴A∩B=ϕ