JEE Questions for Maths Sets Relations And Functions Quiz 12 - MCQExams.com

On the set N of natural numbers,

R = {(x, y) : x, y ϵ N, 2x + y = 41}

Since (1,1) ∉ R as 2.1 + 1 = 3 ≠ 41. So R is not reflexive.

(1, 39) ϵ R. But (20, 39) ∉ R. So R is not transitive.

Since n | n for all n ϵ N, therefore R is reflexive. Since 2 | 6 but 6 | 2, therefore R is not symmetric.

Let n R m and m R p ⇒ n |m and m| p ⇒ n |p⇒ nRp

So, R is transtive.

We first find R^–1, we have

R^–1 = {(5, 4); (4,1); (6, 4); (6, 7); (7,3)},

We know obtain the elements of R^-1 oR we first pick the element of R and then of R^-1.

Since (4, 5) ϵ R and (5, 4) ϵ R^-1, we have

(4, 4) ϵ R^-1 oR

Similarly (1, 4) ϵ R, (4, 1) ϵ R^-1 ⇒ (1, 1) ϵ R^-1 oR

(4, 6) ϵ R, (6, 4) ϵ R^-1 ⇒ (4, 4) ϵ R^-1 oR

(4, 6) ϵ R, (6, 7) ϵ R^-1 ⇒ (4, 7) ϵ R^-1 oR

(7, 6) ϵ R, (6, 7) ϵ R^-1 ⇒ (7, 7) ϵ R^-1 oR

(3, 7) ϵ R, (7, 3) ϵ R^-1 ⇒ (3, 3) ϵ R^-1 oR

Hence, R^-1 oR = {(1,1); (4, 4); (4, 7); (7, 4);(7,7); (3,3)}.

Given A = {2,4,6,8}; R = {(2,4), (4, 2), (4, 6), (6, 4)}

(a, b) ϵ R ⇒ (b, a) ϵ R and also R^–1 = R

Hence, R is symmetric

For (a, b), (c, d) ϵ N × N

(a, b) R (c, d) ⇒ ad(b + c) = bc(a + d)

Reflexive : Since  ab (b + a) = ba (a + b ) ∀ ab ϵ N,

∴ (a, b) R (a, b), ∴ R is reflexive

Symmetric : For (a, b), (c, d) ϵ N × N,

Let (a, b) R (c, d)

∴ ab(b + c) = bc (a + d) ⇒ bc (a + d) = ad (b + c)

⇒cd (d + a) = da(c + b) ⇒ (c, d) R (a, b)

∴ R is symmetric

Transitive : For (a, b), (c, d), (e, f) ϵ 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)

⇒ adb + adc = bca + bcd  → (1)

and cfd + cfe = dec + def  → (2)

(1)×  ef + (2) × ab gives

abdef + adcef + cfdab + cfeab = bcaef + bcdef + decab + defab

⇒adcf (b + e) = bcde (a + f) ⇒ af(b + e) = be (a + f)

⇒ (a, b) R (e, f)

∴ R is transitive. Hence R is an equivalence relation.

Here R is a relation A to B defined by x is greater than y

R = {(2,1); (3,1)}. Hence, range of R = {1}

Here l₁R₁, l₂, l₁ is parallel l₂ and also l₂ is parallel to l₁, so it is symmetric

Clearly, it is also reflexive and transitive. Hence, it is equivalence relation.

B ∩ C = {4}, A ∪ (B ∩ C) = {1, 2, 3 ∪ {4}

               = {1, 2, 3, 4}

Every element has 3 option. Either set Y or set Z or none.

So, number of ordered paires = 3⁵

Since R is an equivalence relation on set A, therefore (a, a) ϵ R, for all a ϵ A. Hence, R has at least n ordered pairs

bN = The set of positive integral multiples of b, cN = The set of positive integral multiplies of c.

bN ∩ cN = The set of positive integral multiples of

bc = dN = The set of positive integral multiples of d

d = bc

∴ A is not a subset of B

∴ Some point of A will not be a point of B, so that point will being to B^c. Hence A and complement of B are always non- disjoint.

R is reflexive if it is contains (1,1), (2,2), (3,3)

∵ (1, 2) ϵ R, (2, 3) ϵ R

∴ R is symmetric if (2, 1), (3, 2) ϵ R,

Now,

R = {(1,1), (2, 2), (3,3), (2,1), (3,2), (2,3), (1, 2)}

R will be transitive if, (3, 1); (1, 3) ϵ R. Thus R becomes and equivalence

Relation by adding

(1,1), (2, 2), (3,3), (2, 1), (3,2), (1, 3), (3,1). Hence, the total number of orderd pairs is 7.