If the sets A and B are defined as A = {(x, y)|y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)|y = -x ∈ R} then

If the sets A and B are defined as A = {(x, y)|y = 1 / x, 0 ≠ x �

1.	If the sets A and B are defined as A = {(x, y)\|y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)\|y = -x ∈ R} then
A.	A ∩ B = ϕ
B.	A ∩ B = B
C.	A ∩ B = A
D.	None of these
Answer» B. A ∩ B = B

Discussion

No Comment Found

Related MCQs

In every (n + 1) - - elementic subset of the set (1, 2, 3, .......2n) which of the following is correct:
Of the members of three athletic teams in a school, 21 are in the cricket team, 26 are in the hockey team and 29 are in the football team. Among them, 14 play hockey and cricket, 15 play hockey and football, 12 play football and cricket and 8 play all the three games. The total number of members in the three athletic teams is:
If x = {4n - 3n - 1: n ϵ N} and Y = {9(n - 1) : n ϵ N}, where N is the set of natural numbers, then
Let f(x) = 15 – |x – 10|; x ∈ R. Then the set of all values of x, at which the function, g(x) = f(f(x)) is not differentiable, is:
A square is a convex set, its exterior points are
In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those who number is divisible by 3 opted Physics course and those who number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is:
Let A and B be two events. If \(P(A) = \dfrac{1}{2}, P(B) = \dfrac{1}{4}, P(A \cap B) = \dfrac{1}{5}\) then \(P({A'\over B'})\) =
Let S be a set of all distinct numbers of the form \(\frac{{\rm{p}}}{{\rm{q}}}\), where p, q ∈ {1, 2, 3, 4, 5, 6}. What is the the cardinality of the set S?
If P and Q are two sets, then (P - Q) ∪ (Q - P) ∪ (P ∩ Q) will be
Considering only the principal values of inverse functions, the set \(A = \left\{ {x \ge 0;ta{n^{ - 1}}\left( {2x} \right) + ta{n^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}} \right\}\)

If the sets A and B are defined as A = {(x, y)|y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)|y = -x ∈ R} then

Discussion

No Comment Found

Related MCQs

Reply to Comment