On the set of integers Z, define f : Z Z as f(n) = , then 'f' is
injective but not surjective
neither injective nor surjective
surjective but not injective
bijective
C.
surjective but not injective
Here,we see that forevery odd values of z, it will give zero. It means that ite is a many one function.
For every even values of z, we will get a set of integers (). So, it is onto. Hence, it is surjective but not injective
The inverse of 2010 in the group Q* of all positive rational under the binary operation * defined by a * b = is
2009
2011
1
2010
Define a relation R on A = {1, 2, 3, 4} as xRy if x divides y. R is
reflexive and transitive
reflexive and symmetric
symmetric and transitive
equivalence
On the set of all non-zero reals, an operation * is defined as a * b = . In this group, a solution of (2 * x) * 3-1 = 4-1 is
6
1
1/6
3/2
If A and B have n elements in common, then the numberofelements common to A x B and B x A is
n
2n
n2
0
Which of the following is false ?
(N, *) is a group
(N, +) is a semi-group
(Z, +) is a group
Set of even integers is a group under usual addition
Let S be the set of all real numbers. A relation R has been defined on S by aRb , then R is
symmetric and transitive but not reflexive
reflexive and transitive but not symmetrIc
reflexive and symmetric but not transitive
an equivalence relation
For any two real numbers, an operation * defined by a * b = 1 + ab is
neither commutative nor associative
commutative but not associative
both commutative and associative
associative but not commutative
Let f : N N defined by f(n) = , then f is
onto but not one-one
one-one and onto
neither one-one nor onto
one-one but not onto
Suppose f(x) = (x + 1)2 for x - 1. If g(x) is a function whose graph is the reflection of the graph of f(x) in the line y = x, then g(x) is equal to