Mathematics Tricks & Tips of Relations for NDA

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Tricks & Tips of Relations for NDA

Finding Domain and range of relation

- A relation `R` from a non-empty set `A` to a non-empty set `B` is a subset of the cartesian product `A xx B`.

`"Domain : -"` The set of all first elements of the ordered pairs in a relation `R` from a set `A` to a set `B` is called the domain of the relation `R`.

`"Codomain & Range :-"` The set of all second elements in a relation `R` from a set `A` to a set `B` is called the range of the relation R. The whole set B is called the codomain of the relation `R`. Note that range `⊆ `codomain.

Q 2671234126

If `R = {(x, y) : x + 2y = 8}` is a relation on `N`, write the range of `R`.

CBSE-12th 2014

Solution:

The set of natural numbers, `N = 1,2,3,4,5,6....`

The relation is given as

`R = { (x , y) : x + 2y = 8 }`

Thus, `R = { (6,1) , ( 4,2) , (2,3) }`

Domain `= { 6,4,2 }`

Range `= { 1,2,3 }`

Q 1985745667

Let `A= { 1, 2, 3, 4, 6}`. Let `R` be the relations on

`A` defined by `{(a, b)}: a, b in A, a` divides `b}`.

(i) Write `R` in roster form

(ii) Find the domain of `R`

(iii) Find the range of `R`

Class 11 Exercise 2.2 Q.No. 5

Solution:

(i) Here `A= { 1, 2, 3, 4, 6}` and `R` is a relation on `A` defined by `{(a, b): a in A, b in A, a` divides Clearly,

`R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (3, 3), (3, 6), (4, 4), (6, 6)}`.

(ii) Domain `= {1,2,3,4,6}`.

(iii) Range `= { 1 ,2, 3, 4, 6}`.

Determining type of relation

`"Empty Relation :"` A relation `R` in a set `A` is called empty relation, if no element of `A` is related to any element of `A`, i.e., `R = φ ⊂ A xx A`.

`"Universal relation :"` A relation `R` in a set `A` is called universal relation, if each element of `A` is related to every element of `A`, i.e., `R = A xx A`.

`"(i) Reflexive Relation : "` : A relation in a set A is called reflexive relation if `(a,a) ∈ R` for every element `a ∈ A`. if `(a, a) ∈ R`, for every `a ∈ A`,

`"(ii) Symmetric Relation :"` A relation in a set `A` is called if `(a,b) ∈ R` the `(b,a) ∈ R` for all `a,b ∈ A`
if `(a_1, a_2) ∈ R` implies that `(a_2, a_1) ∈ R`, for all `a_1, a_2 ∈ A`.

`"(iii) Transitive Relation : "` A relation R on a set A is called transitive if whenever (a, b) is in R and (b, c) is in R, then (a, c) is in R. if `(a_1, a_2) ∈ R` and `(a_2, a_3) ∈ R` implies that `(a_1, a_3) ∈ R`, for all `a_1, a_2,a_3 ∈ A.`

`"Equivalence relation : "` A relation `R` in a set `A` is said to be an equivalence relation, If `R` is reflexive, symmetric and transitive.

Q 2612134930

Prove that the relation `R` in the set `A = {1, 2, 3, 4, 5}` given by `R = {(a, b): |a - b|` is even}, is an equivalence relation.
CBSE-12th 2009

Solution:

`A = {1, 2, 3, 4, 5}`

`R = {(a, b): a − b` is even}

For R to be an equivalence relation it must be

(i) Reflexive, `|a- a| = 0`

`:. (a,a) in R` for`AA a in A`

So R is reflexive.

(ii) Symmetric,

if `(a,b) in R => |a- b|` is even

`=> |b- a|` is also even

So `R` is symmetric.

(iii) Transitive

If `(a, b) in R (b, c) in R` then `(a, c) in R`

`(a, b) in R ⇒ |a −b|` is even

`(b, c) in R ⇒ |b−c|` is even

Sum of two even numbers is even

So,
` | a- c| = |(a-b) + (b-c) |` = even number + even number = even number

So `|a-c|` is even since, `|a- b|` and `|b- c|` are even

So `(a ,c) in R`

Hence, `R` is transitive.

Therefore, `R` is an equivalence relation.

Q 2581523427

A relation `rho` on the set of real number `R` is
defined as `{x rho y : xy > 0}`. Then, which of the
following is/are true?
WBJEE 2015

(This question may have multiple correct answers)

(A) `rho` is reflexive and symmetric

(B) `rho` is symmetric but not reflexive

(D) `rho` is an equivalence relatij)n

Solution:

We have, `x rho y : xy > 0`

(i) `text (Reflexive)` Suppose `x rho x in R`

`=> x^2 > 0`

which is not true when; `x = 0`.

Hence, relation is not reflexive.

(ii) `text(Symmetric)` `x rho y in R`

`=> xy > 0`

`=> yx > 0`

`=> y rho x in R`

If `(x, y) in R`, then, relation: is symmetric.

(iii) `text (Transitive)` `x rho y in R`

`=> xy > 0`

and `y rho z in R`

`=> yz > 0`

`=> xy^2 z > 0`

`=> xz > 0`

`=> (x,z) in R`

If `(x, y) in R`, then `(y, z) in R`

`=> (x,z) in R`

Hence, relation is transitive.

Correct Answer is `=>` (B)

Q 2418634509

Let a relation `R` be defined on set of all real
numbers by a R b if and only if 1 + ab > 0.
Then, R is
UPSEE 2009

(A)

reflexive, transitive but not symmetric

(B)

reflexive, symmetric but not transitive

(C)

symmetric, transitive but not reflexive

(D)

an equivalence relation

Solution:

As `1 + a· a = 1 + a^2 > 0`

`:. (a, a) in R` which is reflexive.

Also, `(a,b) in R => 1 + ab > 0 => 1 + ba > 0`

`=> (b,a) in R`

`:. R` is symmetric.

But `(a, b) in R` and `(b, c) in R => (a, c) notin R`

Hence, option (b) is correct.

Correct Answer is `=>` (A) reflexive, transitive but not symmetric

Q 2486101077

The relation `R` in `R` defined by
`R ={(a, b) : a <= b^3 }`, is
UPSEE 2014

(A)

reflexive

(B)

symmetric

(C)

transitive

(D)

None of these

Solution:

Given, `R = {(a, b) : a <= b^3 }`

It is observed that `(1/2 . 1/2) in R` as `1/2 < (1/2)^3 = 1/8`

So, `R` is not reflexive.

Now, `(1, 2) in R` (as `1 < 2^3 = 8`)

But `(2, 1) notin R` (as `2^3 > 1`)

So, `R` is not symmetric.

we have, `( 3, 3/2) , (3/2 , 6/5) in R` as `3 < (3/2)^3`

and `3/2 < ( 6/5)^3`

But `(3,6/5) notin R` as `3 > (6/5)^3`

Therefore, `R` is not transitive.

Hence, `R` is neither reflexive nor symmetric nor

transitive.

Correct Answer is `=>` (B) symmetric

Q 2540291113

The relation `R` defined in the set `{1, 2, 3, 4, 5,6}` as `R ={(a, b) : b = a + 1}` is
BCECE Stage 1 2013

(A)

reflexive

(B)

symmetric

(C)

transitive

(D)

None of these

Solution:

Let `A = { 1,2, 3, 4, 5, 6}`

`A` relation `R` is defined on set `A` is

`R = {(a, b) : b = a+ 1}`, therefore

`R = {(1,2),(2, 3), (3, 4),(4,5),(5, 6)}`

Now, `6 in A` but `(6,6) notin R`

Therefore, R is not reflexive.

It can be observed that `( 1, 2 ) in R` but `(1 ,2)`

`notin R`. Therefore, R is not symmetric.

Now, `(1, 2), (2, 3) in R` but `(1, 3) notin R`. Therefore R

is not transitive.

Hence, R is neither reflexive nor symmetric

nor transitive.

Correct Answer is `=>` (D) None of these

Cartesian Product of sets of Real `( R × R )`

See examples

Q 2268845705

If `A = {1, 2, 3}, B = {3, 8}`, then

`(A cup B) xx (A cap B) =`
BITSAT Mock

(A)

`{(3, 1), (3, 2), (3, 3), (3, 8)}`

(B)

`{(1, 3), (2, 3), (3, 3), (8, 3)}`

(C)

`{(1, 2), (2, 2), (3, 3), (8, 8)}`

(D)

`{(8, 3), (8, 2), (8, 1), (8, 8)}`

Solution:

`A cup B = {1, 2, 3, 8}` and `A cap B = {3}`

`(A cup B) xx (A cap B) = {(1, 3), (2, 3), (3, 3), (8, 3)}`

Correct Answer is `=>` (B) `{(1, 3), (2, 3), (3, 3), (8, 3)}`

Q 2386823777

Let `A = { x in W`, the set of whole numbers and
`x < 3}, B = {x in N`, the set of natural numbers and
`2 <= x < 4}` and `C = {3, 4}`, then how many elements
will `(A cup B) xx C` contain?
NDA Paper 1 2007

(A)

`6`

(B)

`8`

(C)

`10`

(D)

`12`

Solution:

Given, `A= {x in W`, the set of whole numbers and

`X < 3} = {0, 1, 2}`

`B = {x in N`, the set of natural numbers and `2 <= x < 4}`

`= { 2, 3}`

`C = {3, 4}`

`A cup B = {0, 1, 2, 3}`

`(A cup B) xx C = { 0, 1, 2, 3} xx {3, 4}`

`= {(0, 3), (0, 4), (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}`

So, required number of elements containing by `(A cup B) xx C` is `8`.

Correct Answer is `=>` (B) `8`

Q 2386823777

(A)

`6`

(B)

`8`

(C)

`10`

(D)

`12`

Correct Answer is `=>` (B) `8`

Finding inverse of relation and composite relation

For any relation R

If `R={ (1,3) , (1, 5) , (2,3) , (2,5) , (3,5) , (4,5)}`

then, `R^(-1) ={ (3,1) , (5,1) , (3,2) , (5,2) , (5,3) , (5,4)}`

Q 2731256122

Let R be a relation from `A = [ 1,2 ,3 ,4 ]` to `B = [ 1,3,5]` such that `R = [( a,b ) : a < b ,a in A , b in B ]`

What is `RoR^-1` equal to ?
NDA Paper 1 2016

(A)

`{(1,3), (1,5), (2,3) , (2,5 ), ( 3,5) , (4, 5 ) }`

(B)

`{ (3,1), (5,1),( 3,2),(5,2) , (5,3 ) , (5,4)}`

(C)

`{(3,3),(3,5), (5,3), (5,5)}`

(D)

`{(3,3) ,(3,4), (4,5) }`

Solution:

`R={ (1,3) , (1, 5) , (2,3) , (2,5) , (3,5) , (4,5)}`

`R^(-1) ={ (3,1) , (5,1) , (3,2) , (5,2) , (5,3) , (5,4)}`

`RoR^(-1) ={ (3,3) , (3,5) , (5,3) , (5,5)}`

Correct Answer is `=>` (C) `{(3,3),(3,5), (5,3), (5,5)}`

Max Number of relation

- No. of subsets of A are `2^n`

- No. of proper subsets are `2^(n) -1`

- If A and B are finite set having `n ` and `m` element having relation R, No. of relation which can be defined in A = `2^(nxxm)`

-`(AxxB)cup(BxxA) = (AcapB)xx(BcapA)`