NCERT solutions for class 11 exercise 2.1 of chapter 2-Relations and Functions - Future Study Point

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

Relations and functions class 11

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.

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Relations and functions class 11

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

Exercise 2.2

Exercise 2.3

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11-Question paper of maths 2015

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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

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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)}

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= 24= 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 for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
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Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
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Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

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Chapter 1-Relations and FunctionsChapter 9-Differential Equations
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Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
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