**NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions**

NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions are created here for the students of 11 class in boosting their preparation of the exams, class test, and worksheet. These NCERT solutions are the solutions of all unsolved questions of exercise 2.1 of chapter 2-Relations and Functions of NCERT maths text book for class 11. The NCERT solutions of exercise 2.1 of chapter 2-Relations and Functions are based on the cartesian product of the sets.

**NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions**

**PDF of NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions**

**Q1. If (x/3 + 1, y – 2/3) = (5/3, 1/3), find the value of x and y.**

Ans. We are given the two ordered pair which are equal to each other

(x/3 + 1, y – 2/3) = (5/3, 1/3)

The corresponding elements of LHS and RHS should be equal to each other

x/3 + 1 = 5/3

x/3 = 5/3 – 1

x/3 = (5-3)/3

x = 2

y – 2/3 = 1/3

y = 1/3 + 2/3

y = 1

Therefore the value of x = 2 and y = 1

**Q2. If set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?**

Ans. The number of elements in set A = 3

We are given the set B = {3, 4, 5}

The number of elements in set (A × B) = (The number of elements in set A) × (The number of elements in the set B)= 3 × 3 = 9

Therefore the number of elements in (A × B) are = 9

**Q3. If G = {7,8} and H = {5,4,2} , find G × H and H × G.**

Ans. We are given G = {7,8} and H = {5,4,2}

It is known to us that the cartesian product of the two non-empty sets A and B is defined as follows.

A × B = { (a,b) : a ∈ A, b ∈ B}

Therefore,we have

G × H = {7,8} × {5,4,2} = { (7,5), (7,4),(7,2),(8,5),(8,4),(8,2)}

H × G** = ** {5,4,2}× {7,8} = { (5,7), (5,8),(4,7),(4,8),(2,7),(2,8)}

**Q4. State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.**

**(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}**

**(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.**

**(iii) If A = {1, 2}, B = {3, 4}, then A × (B ****∩ Φ) = Φ **

Ans. (i) If P = {m, n} and Q = {n, m}, then

P × Q = {(m, m), (m, n), (n, m), (n, n)}

Therefore the given ststement is** false**

(ii) The given ststement is** true**

(iii) The given ststement is** true**

**Q5. If A = { -1, 1} , find A × A × A**.

Ans. It is known to us that the cartesian product of the three non-empty sets A , B and C is defined as follows.

A × B × C = {(a,b,c), a ∈ A, b ∈ B, c ∈ C}

Here B = A, C = A

Since A = {-1,1}

Therefore

A × A × A =(-1,1) × (-1,1)× (-1,1) = {(-1,-1),(-1,1),(1,-1),(1,1)}×(-1,1)={(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}

**Q6. If A × B = {(a,x),(a,y),(b,x),(b,y)}. Find A and B.**

Ans. We are given that

A × B = {(a,x),(a,y),(b,x),(b,y)}

It is known to us that the cartesian product of the two non-empty sets A and B is defined as follows.

A × B = { (a,b) : a ∈ A, b ∈ B}

The first element of A× B shows the element of A and the second element shows the element of B.

Therefore

A = {a,b} and B = {x,y}

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**Q7. Let A = {1,2}, B = {1,2,3,4}, C = {5,6} and D ={5,6,7,8}. Verify that.**

(i) A × (B ∩C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

Ans.

(i) We have to verify that

A × (B ∩C) = (A × B) ∩ (A × C)

Taking LHS

A × (B∩C), in which we are given A = {1,2}, B = {1,2,3,4}, C = {5,6}

Here, (B∩C)= {1,2,3,4}∩ {5,6} = Φ,where Φ is empty set

Now, A × (B∩C) = {1,2} ×Φ = Φ

LHS = Φ

RHS is

(A × B) ∩ (A × C)

A × B = {1,2} × {1,2,3,4} = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

A × C = {1,2}×{5,6} = {(1,5),(1,6),(2,5),(2,6)}

(A × B) ∩ (A × C)

{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}∩{(1,5),(1,6),(2,5),(2,6)}

=Φ

RHS is also Φ

Hence A × (B ∩C) = (A × B) ∩ (A × C) , verified

(ii) A × C is a subset of B × D

Computing A × C as following

A × C = {1,2} × {5,6}= {(1,5),(1,6),(2,5),(2,6)}

Computing B × D as following

B × D ={1,2,3,4}× {5,6,7,8}= {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8)(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}

Observing A × C and B × D ,we see that all the elements of A × C are the elements of B × D

Hence A × C ⊂ B × D, verified

**Q8. Let A = {1,2} and B = {3,4},write A × B.How many subsets will A × B have ? List them.**

Ans. A = {1,2} and B ={3,4}

A × B = {1,2} × {3,4} = {(1,3),(1,4),(2,3),(2,4)}

The number of elements in A × B = 4

n(A × B) = 4

Therefore number of subsets of A × B = 2^{n }= 2^{4}= 2× 2× 2×2= 16

16 subsets of A× B = {(1,3),(1,4),(2,3),(2,4)} are computed as following

Φ(empty set) compulsurily is subset of every set

Taking single element

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

Taking two elements

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

Taking three elements

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

Taking four elements

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

Therefore,the total number of subsets of A × B are 16 listed above

**Q9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x,1), (y,2),(z,1) are in A × B ,find A and B,where x,y and z are distinct elements.**

Ans.We are given that (x,1), (y,2),(z,1) are in A × B

First elements of the ordered pair of A×B shows the element of A and second elements shows the elements of B.

Therefore x,y,z are the elements of A and 1,2,1 are the elements of B

Since n(A) = 3 and n(B) = 2

Hence the set A = {x,y,z} and set B ={1,2}

**Q10. The cartesian product A × A has 9 elements among which are found (-1,0) and (0,1). Find the set A and the remaining elements of A × A.**

Ans. Let the number of elements in set A = n(A)

We are given the number of elements in A × A = 9

The number of elements in the cartesian product of two sets A and B is given as following. If n(A) is number of elements in set A , n(B) is the number of elements in set B and n(A × B) is the number of elements in (A × B)

n(A) × n(B) = n(A × B)

n(A) × n(A) = n(A × A)

n(A) × n(A) = 9

n(A) = √9 = 3

Two of the ordered pair of A×A among 9 given to us (-1,0) and (0,1)

The first element of the ordered pair shows the element of A and second element also show the elements of A

Therefore set A = {-1,0,1}

A × A = {-1,0,1} × {-1,0,1}

= {(-1,-1),**(-1,0)**,(-1,1),(0,-1),(0,0),**(0,1)**,(1,-1),(1,0),(1,1)}

Therefore remaining elements of A × A are

{(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)}

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

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Chapter 4-Determinants | Chapter 12-Linear Programming |

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