If R be the set of all real numbers and f : R ➔ R is given by f(x) = 3x2 + 1. Then, the set f-1([1, 6]) is
Let Xn = for all integers is
a singleton set
not a finite set
an empty set
a finite set with more than one element
If f(x) = 2100x + 1, g(x) = 3100x + 1, then the set ofreal numbers x such that f{g(x)} = x is
empty
a singleton
a finite set with more than one element
infinite
If A = {x : x2 - 5x + 6 = 0}, B={2, 4}, C = {4, 5}, then A x (B ∩ C) is
{(2, 4), (3, 4)}
{(4, 2), (4, 3)}
{(2, 4), (3, 4), (4, 4)}
{(2, 2), (3, 3), (4, 4), (5, 5)}
Let A and B be two non-empty sets having n elements in common. Then, the number of elements common to A x B and B x A is
2n
n
n2
None of these
If a set A contains n elements, then which of the following. cannot be the number of reflexive relations on the set A?
2n + 1
2n - 1
2n
A.
2n + 1
A relation on set A is a subset of A A.
Let A= {a1, a2, a3, ..., an}. Then, a reflexive relation on A must contain atleast n elements
(a1,a1), (a2, a2) ... (an, an).
Number of reflexive relations on A is .
Clearly, n2 - n = n, n2 - n = n - 1, n2 - n = n2 - 1 have solutions in N but n2 - n = n + 1 does not have solutions in N.
So, 2n + 1 cannot be the number of reflexive relations on A.
If A and B be two sets such that A x B consists of 6 elements. If three elements A x B are (1, 4), (2, 6) and (3, 6), find B x A.
{(1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)}
{(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)}
{(4, 4), (6, 6)}
{(4, 1), (6, 2), (6, 3)}