## JEE Main Mathematics Sets and Relations Online Test

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JEE Main Mathematics Sets and Relations Online Test. JEE Main Online Test for Mathematics Sets and Relations. JEE Main Full Online Quiz **for Mathematics Sets and Relations**. **JEE Main Free Mock Test Paper 2019.** JEE Main 2019 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Mathematics Sets and Relations. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n **Take JEE Main Mathematics Sets and Relations…**

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- Question 1 of 20
##### 1. Question

Let n(a) = m and b = (n). Then, the total number of non-empty relations that can be defined from A to B is

CorrectGiven , n(A) = m and n(B) = n

∴ Total number of relations from A to B = 2

^{mn}∴ Total number of non-empty relations from A to B = 2

^{mn}–1IncorrectGiven , n(A) = m and n(B) = n

∴ Total number of relations from A to B = 2

^{mn}∴ Total number of non-empty relations from A to B = 2

^{mn}–1 - Question 2 of 20
##### 2. Question

If A and B are two sets then (A – B) ∪ (B – A) ∪ (A ∩ B) is equal to

CorrectFrom Venn-Euler’s diagram,

∴ (A – B) ∪ (B – A) ∪ (A ∩ B) = A ∪ B.

IncorrectFrom Venn-Euler’s diagram,

∴ (A – B) ∪ (B – A) ∪ (A ∩ B) = A ∪ B.

- Question 3 of 20
##### 3. Question

R is a relation from (11, 12, 13) to (8, 10, 12) defined by y = x – 3 The relation R

^{-1}isCorrectLet A = {11, 12, 13}, B = {8, 10, 12}

∴ R = {(11, 8), (13, 10)}

R

^{-1 }= {(8, 11), (10, 13)}IncorrectLet A = {11, 12, 13}, B = {8, 10, 12}

∴ R = {(11, 8), (13, 10)}

R

^{-1 }= {(8, 11), (10, 13)} - Question 4 of 20
##### 4. Question

Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A is

Correctn(A × A) = n(A). n(A) = 3

^{2}= 9So, the total number of subsets of A × A is 2

^{9}and a subset of A × A is a relation over the set A.Incorrectn(A × A) = n(A). n(A) = 3

^{2}= 9So, the total number of subsets of A × A is 2

^{9}and a subset of A × A is a relation over the set A. - Question 5 of 20
##### 5. Question

Let a relation R be defined by R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)}. The relation R

^{-1}or is given byCorrectGiven, R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)}

∴ R

^{-1}= {(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)}⇒ R

^{-1}or = {(4, 4), (1, 1), (4, 7), (7, 4), (7, 7), (3, 3)}IncorrectGiven, R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)}

∴ R

^{-1}= {(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)}⇒ R

^{-1}or = {(4, 4), (1, 1), (4, 7), (7, 4), (7, 7), (3, 3)} - Question 6 of 20
##### 6. Question

The solution set of the equation x

^{3}– 3x + 2 = 0 in roster from isCorrectOn solving x

^{3}– 3x + 2 = 0we get x = 1 & -2

∴ Solution set is {1, -2}

IncorrectOn solving x

^{3}– 3x + 2 = 0we get x = 1 & -2

∴ Solution set is {1, -2}

- Question 7 of 20
##### 7. Question

The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by

CorrectGiven, R = {(a, b) : a – b = 3} = {(4, 1) (5, 2), (6, 3), …………….}

IncorrectGiven, R = {(a, b) : a – b = 3} = {(4, 1) (5, 2), (6, 3), …………….}

- Question 8 of 20
##### 8. Question

Let A = {1, 2, 3, 4} and R be a relation in A given by R =

{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)}. Then R is

Correct(1, 1), (2, 2), (3, 3), (4, 4) ∈ R; ∴ R is reflexive.

∵ (1, 2), (3, 1) ∈ R and also (2, 1), (1, 3) ∈ R.

Hence, R is synmetric. But clearly R is not transitive.

Incorrect(1, 1), (2, 2), (3, 3), (4, 4) ∈ R; ∴ R is reflexive.

∵ (1, 2), (3, 1) ∈ R and also (2, 1), (1, 3) ∈ R.

Hence, R is synmetric. But clearly R is not transitive.

- Question 9 of 20
##### 9. Question

Let R be the relation from A = {2, 3, 4, 5} to B = {3, 6, 7, 10} defined by ‘x devides y’, then R

^{-1 }is equal toCorrectGiven, A = {2, 3, 4, 5} and B = {3, 6, 7, 10}

∴ R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}

⇒ R

^{-1}= {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}IncorrectGiven, A = {2, 3, 4, 5} and B = {3, 6, 7, 10}

∴ R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}

⇒ R

^{-1}= {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)} - Question 10 of 20
##### 10. Question

If R is an equivalence relation on a set A, then R

^{–1}is NOTCorrectIncorrect - Question 11 of 20
##### 11. Question

If R is a relation from a set A to the set B and S is a relation from B to C, then the relation SoR

CorrectGive, R ⊆ A × B and S ⊆ B × B, we have

So r ⊆ A × C

∴ SoR is a relation from A to C.

IncorrectGive, R ⊆ A × B and S ⊆ B × B, we have

So r ⊆ A × C

∴ SoR is a relation from A to C.

- Question 12 of 20
##### 12. Question

The number of elements in the power set of {a, b, c}

CorrectNumber of subsets of {a, b, c} = 2

^{3}= 8 ∴ number of elements in the power set = 8IncorrectNumber of subsets of {a, b, c} = 2

^{3}= 8 ∴ number of elements in the power set = 8 - Question 13 of 20
##### 13. Question

The set {0, 2, 6, 12, 20} in the set-builder form is

CorrectWe see that each number in the given set follow the relation x = n

^{2}– 3n + 2 as forn = 1, x = 1 – 3 + 2 = 0

n = 2, x = 4 – 6 + 2 = 0

n = 3, x = 9 – 9 + 2 = 2

n = 4, x = 16 – 12 + 2 = 6

n = 5, x = 25 – 15 + 2 = 12

n = 6, x = 36 – 18 + 2 = 20

IncorrectWe see that each number in the given set follow the relation x = n

^{2}– 3n + 2 as forn = 1, x = 1 – 3 + 2 = 0

n = 2, x = 4 – 6 + 2 = 0

n = 3, x = 9 – 9 + 2 = 2

n = 4, x = 16 – 12 + 2 = 6

n = 5, x = 25 – 15 + 2 = 12

n = 6, x = 36 – 18 + 2 = 20

- Question 14 of 20
##### 14. Question

The set equivalent to the set {1, 2, 3, 4, 5, 6} is

CorrectA set does not change if the order of the elements is changed.

IncorrectA set does not change if the order of the elements is changed.

- Question 15 of 20
##### 15. Question

Let R = {(a, a)} be a relation on a set A, Then R is

CorrectIncorrect - Question 16 of 20
##### 16. Question

The number of proper subsets of the set {1, 2, 3} is

CorrectNumber of proper subsets = 2

^{3}– 2 = 6IncorrectNumber of proper subsets = 2

^{3}– 2 = 6 - Question 17 of 20
##### 17. Question

If A and B are two given sets, then A ∪ (A ∩ B) is equal to

CorrectA ∩ B ∩ A. Hence A ∪ (A ∩ B) = A

IncorrectA ∩ B ∩ A. Hence A ∪ (A ∩ B) = A

- Question 18 of 20
##### 18. Question

If A = {x : x is an integer, x

^{2}__<__1}, then the elements of ‘A’ areCorrectIf x

^{2}< 1 ⇒ –1__<__x__<__1 & x is an integer ∴ x ∈ {–1, 0, 1}IncorrectIf x

^{2}< 1 ⇒ –1__<__x__<__1 & x is an integer ∴ x ∈ {–1, 0, 1} - Question 19 of 20
##### 19. Question

In a town of 10, 000 families it was found that 40% family buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C, 5% families by A and B, 3% buy B and C and 4% buy A and C. If 2%families buy all the three newspapers, then number of families which buy A only is

Correctn(A) = 40% of 10000 = 4000

n(B) = 20% of 10, 000 = 2000

n(C) = 10% of 10, 000 = 1000

n(A ∩ B) = 5% of 10000 = 500

n(B ∩ C) = 3% of 10000 = 300

n(C ∩ A) = 4% of 10000 = 400

n(A ∩ B ∩ C ) = 2% of 10,000 = 200

We want to find n(A ∩ B

^{C}∩ C^{C}) = n[A ∩ (B ∩ C)^{C}] = n(A) – n[A ∩ B ∪ C] = n(A) –n[(A ∩ B) ∪ (A ∩ C)]= n(A) – [n(A ∩ B) + n(A ∩ C) – n(A ∩ B ∩ C)] = 4000 – [500 + 400 – 200] = 4000 – 700 = 3300

or

Incorrectn(A) = 40% of 10000 = 4000

n(B) = 20% of 10, 000 = 2000

n(C) = 10% of 10, 000 = 1000

n(A ∩ B) = 5% of 10000 = 500

n(B ∩ C) = 3% of 10000 = 300

n(C ∩ A) = 4% of 10000 = 400

n(A ∩ B ∩ C ) = 2% of 10,000 = 200

We want to find n(A ∩ B

^{C}∩ C^{C}) = n[A ∩ (B ∩ C)^{C}] = n(A) – n[A ∩ B ∪ C] = n(A) –n[(A ∩ B) ∪ (A ∩ C)]= n(A) – [n(A ∩ B) + n(A ∩ C) – n(A ∩ B ∩ C)] = 4000 – [500 + 400 – 200] = 4000 – 700 = 3300

or

- Question 20 of 20
##### 20. Question

Let R be a relation on N defined by x + 2y = 8. The domain of R is

CorrectIncorrect