`A={(x,y):y=e^(x),x in R},B ={(x,y):y=x,x in R}` then

A ={(x,y):y:y=e^(x), x in R},B ={(x,y):y=e^(-x),x in R} then

Let A={(x,y)//y=e^(x),x in R},B={(x,y)//y=e^(-x),x in R} , then

If A={(x,y):y=e^(x),x in R} and B={(x,y):y=e^(-x),x in R} ,then write A nn B.

If A={(x,y):y=1/x, 0 != x in R}, B={(x,y):y=-x,x in R} then (i) AnnB=A (ii) AnnB=B (iii) AnnB=phi (iv) AuuB=A

If A={(x,y):y=e^(x),x in R} and B={(x,y):y=e^(-2x),x in R}, then

In some question of sets, we have to make the use of graphs For example A={(x,y):y=e^(x), x in R} B={{x,y}: y=-x. x in R} Find n(A cap B) It is clear that y=e^(x) and y=-x intersect at one pont. Hence n(A cap B)=1 A={(x,y):x^(2)+y^(2) le 2, x ,y in R} B={{x,y): y gex^(2),x,y in R} Then domain of A cap B is

In some question of sets, we have to make the use of graphs For example A={(x,y):y=e^(x), x in R} B={{x,y}: y=-x. x in R} Find n(A cap B) It is clear that y=e^(x) and y=-x intersect at one pont. Hence n(A cap B)=1 A={(x,y):x^(2)+y^(2) le 2, x ,y in R} B={{x,y): y gex^(2),x,y in R} Then domain of A cap B is

In some question of sets, we have to make the use of graphs For example A={(x,y):y=e^(x), x in R} B={{x,y}: y=-x. x in R} Find n(A cap B) It is clear that y=e^(x) and y=-x intersect at one pont. Hence n(A cap B)=1 A={(x,y):y = sin pi x,x in R} B={(x,y):y=|In|x"||",x in R-{0}} Then, n(A cap B)