If N denote the set of all natural numbers and R be the relation on N N defined by (a, b) R (c, d), if ad(b + c) = bc(a + d), then R is
symmetric only
reflexive only
transitive only
an equivalence relation
D.
an equivalence relation
For (a, b), (c, d) N N
(a, b) R (c, d)
ad(b + c) = ba(a + b)
Reflexive: Since, ab(b + a) = ba(a + b), ∀ ab ∈ N
(a, b) R (a, b)
So, R is reflexive.
Symmetric: For (a, b), (c, d)
Let (a, b) R (c, d)
So, R is symmetric.
Transitive: For(a, b), (c, d), (e, f) ∈ N N
Let (a, b) R (c, d), (c, d) R (e, f)
ad(b + c) = bc(a + d) , cf(d + e) = de(c + f)
On multiplying Eq. (i) by ef and Eq. (ii) by ab, then we get
adbef + adcef + cfdab + cfeab
= bcaef + bcdef + decab + defab
So, R is transitive.
Hence R is an equivalence relation.
The function f : R R defined by f (x) = (x - 1) (x - 2) (x - 3) is
one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto
The relation R defined on the set of natural numbers as { (a, b) : a differs from b by 3} is given
{(1, 4), (2, 5), (3, 6), ... }
{(4, 1), (5, 2), (6, 3), ... }
{(1, 3), (2, 6), (3, 9), ... }
None of the above
If R be a realtion from A = {1, 2, 3, 4} to B = {1, 3, 5} such that (a, b) ∈R , then ROR-1 is
{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
{(3, 3), (3, 5), (5, 3), (5, 5)}
{(3, 3), (3, 4), (4, 5)}
If F is function such that F (0) = 2, F(1) = 3, F(x + 2)= 2F(x) - F(x + 1) for x > 0, then F(5) is equal to
- 7
- 3
17
13
If f : [2, 3] R is defined by f(x) = x3 + 3x - 2, then the range of f(x) is contained in the interval
[1, 12]
[12, 34]
[35, 50]
[- 12, 12]
If R be a relation defined as aRb iff > 0, then the relation is
reflexive
symmetric
transitive
symmetric and transitive