Relations and Functions Test - 26

Consider that aRb, if a is congruent to b, ∀a, b ∈ T

Then aRa ⇒ a ≅ a,

Which is true for all a ∈ T

So, R is reflexive, ....(i)

Let aRb ⇒ a ≅ b

⇒ b ≅ a

⇒ bRa

So, R is symmetric. ...(ii)

Let aRb and bRc

⇒ a ≅ b and b ≅ c

⇒ a ≅ c ⇒ aRc

So, R is transitive. .....(iii)

Hence, R is equivalence relation.

Given,

aRb ⇒ a is brother of b

∴ aRa ⇒ a is brother of a, which is not true

So, R is not reflexive

aRb ⇒ a is brother of b

This does not mean b is also a brother of a as b can be a sister of a 

Hence, R is not symmetric

aRb ⇒ a is brother of b

and bRc ⇒ b is a brother of c

So, a is brother of c.

Hence, R is transitive.

R on the set {1, 2, 3} be defined by R = {(1,2)}

It is clear that R is transitive.

Given that,

aRb if a ≥ b

⇒ aRa ⇒ a ≥ a which is true

let aRb, a ≥ b, then b ≥≥ a which is not true R is not symmetric.

But aRb and bRc

⇒ a ≥ b and b ≥ c

⇒ a ≥ c

Hence, R is transitive.

We know that, if A and B are two non-empty finite set containing m and n elements respectively, then the number of one-one and onto mapping from A to B is

n! if m = n

0, if m ≠ n

Given that, m = 5 and n = 6

∴ m ≠ n

Number of mapping = 0

Given that, A = {1 , 2, 3,....n} and B = {a, b} If function is surjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a'and b' in such a way that no box remains empty. Now for each object there are two options, either it is put in box 'a' or in box b'. So total number of ways of 'n' different objects = 2 x 2 x 2 ... n times = 2ⁿ  But in one case all the objects are put in box 'a' and in one case all the objects are put in box `b' So, number of surjective functions = 2ⁿ − 2.

Given that,

f(x) = 1/x, ∀ ∈ R

For x = 0, f(x) is not defined

Hence, f(x) is not defined function