NCERT Solutions for Class 12 Maths Chapter 1

NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions all exercises including miscellaneous are in PDF Hindi Medium & English Medium along with NCERT Solutions Apps free download.

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NCERT solutions for Class 12 Maths Chapter 1




Class: 12Maths (English and Hindi Medium)
Chapter 1:Relations and Functions

12th Maths Chapter 1 Solutions

NCERT solutions for Class 12 Maths Chapter 1 all exercises are given below to free download in PDF form. NCERT Books as well as Solutions are available in English and Hindi Medium. Ask your doubts related to NIOS Board and CBSE Board through Discussion Forum. NCERT Solutions and Offline apps are based on latest CBSE Syllabus.




Class 12 Maths Chaper 1 Exercise 1.1 Solution in Videos

Class 12 Maths Exercise 1.1 Solution in English
Class 12 Maths Exercise 1.1 Solution in Hindi

Class 12 Maths Chaper 1 Exercise 1.2 Solution in Videos

Class 12 Maths Exercise 1.2 Solution in English
Class 12 Maths Exercise 1.2 Solution in Hindi




Class 12 Maths Chaper 1 Exercise 1.3 Solution in Videos

Class 12 Maths Exercise 1.3 Solution in English
Class 12 Maths Exercise 1.3 Solution in Hindi

Class 12 Maths Chaper 1 Exercise 1.4 Solution in Videos

Class 12 Maths Exercise 1.4 Solution in English
Class 12 Maths Exercise 1.4 Solution in Hindi

Class 12 Maths Chaper 1 Miscellaneous Exercise 1 in Videos

Class 12 Maths Miscellaneous Ex. 1 Solution in English
Class 12 Maths Miscellaneous Ex. 1 Solution in Hindi

Important Terms related to Chapter 1

Before studying this lesson, you should know:

    1. Concept of set, types of sets, operations on sets
    2. Concept of ordered pair and cartesian product of set.
    3. Domain, co-domain and range of a relation and a function
Relation

Let A and B be two sets. Then a relation R from Set A into Set B is a subset of A × B.
Types of Relations

    • Reflexive Relation
    • Symmetric Relation
    • Transitive Relation



Equivalence Relation

A relation R on a set A is said to be an equivalence relation on A iff

    • it is reflexive
    • it is symmetric
    • it is transitive
CLASSIFICATION OF FUNCTIONS

Let f be a function from A to B. If every element of the set B is the image of at least one element of the set A i.e. if there is no unpaired element in the set B then we say that the function f maps the set A onto the set B. Otherwise we say that the function maps the set A into the set B.
Functions for which each element of the set A is mapped to a different element of the set B are said to be one-to-one.

NCERT-Solutions-Class-12-Maths-Chapter-1

A function can map more than one element of the set A to the same element of the set B. Such a type of function is said to be many-to-one. A function which is both one-to-one and onto is said to be a bijective function.

NCERT solutions for class 12 maths chapter 1-Bijective



BINARY OPERATIONS

Let A, B be two non-empty sets, then a function from A × A to A is called a binary operation on A.
If a binary operation on A is denoted by *, the unique element of A associated with the ordered pair (a, b) of A × A is denoted by a * b.
The order of the elements is taken into consideration, i.e. the elements associated with the pairs (a, b) and (b, a) may be different i.e. a * b may not be equal to b * a.
Let A be a non-empty set and ‘*’ be an operation on A, then

NCERT solutions for class 12 maths chapter 1 Relations Functions

Important Questions on 12th Maths Chapter 1

Determine whether each of the following relation are reflexive, symmetric and transitive: Relation R in the set A = {1, 2, 3… 13, 14} defined as R = {(x, y): 3x – y = 0}
A = {1, 2, 3 … 13, 14} R = {(x, y): 3x − y = 0} ∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)} R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R. Also, R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0] Also, R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0] Hence, R is neither reflexive, nor symmetric, nor transitive.
Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b^2} is neither reflexive nor symmetric nor transitive.
R = {(a, b): a ≤ b^2} It can be observed that (1/2,1/2)∉R, since, 1/2>(1/2)^2 ∴ R is not reflexive. Now, (1, 4) ∈ R as 1 < 42 But, 4 is not less than 12. ∴ (4, 1) ∉ R ∴ R is not symmetric. Now, (3, 2), (2, 1.5) ∈ R [as 3 < 22 = 4 and 2 < (1.5)^2 = 2.25] But, 3 > (1.5)^2 = 2.25 ∴ (3, 1.5) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Let A = {1, 2, 3, 4, 5, 6}. A relation R is defined on set A as: R = {(a, b): b = a + 1} ∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} we can find (a, a) ∉ R, where a ∈ A. For instance, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R ∴ R is not reflexive. It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R. ∴ R is not symmetric. Now, (1, 2), (2, 3) ∈ R but, (1, 3) ∉ R ∴ R is not transitive Hence, R is neither reflexive, nor symmetric, nor transitive.
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Let A = {1, 2, 3}. A relation R on A is defined as R = {(1, 2), (2, 1)}. It is clear that (1, 1), (2, 2), (3, 3) ∉ R, ∴ R is not reflexive. Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric. Now, (1, 2) and (2, 1) ∈ R, however, (1, 1) ∉ R, ∴ R is not transitive. Hence, R is symmetric but neither reflexive nor transitive.
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation.
Set A is the set of all books in the library of a college. R = {x, y): x and y have the same number of pages} Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages. Let (x, y) ∈ R ⇒ x and y have the same number of pages. ⇒ y and x have the same number of pages. ⇒ (y, x) ∈ R ∴ R is symmetric. Now, let (x, y) ∈R and (y, z) ∈ R. ⇒ x and y and have the same number of pages and y and z have the same number of pages. ⇒ x and z have the same number of pages. ⇒ (x, z) ∈ R ∴ R is transitive. Hence, R is an equivalence relation.
Show that the relation R in the set A of points in a plane given by R = {(P, Q): Distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
R = {(P, Q): Distance of point P from the origin is the same as the distance of point Q from the origin} Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin. ∴ R is reflexive. Now, Let (P, Q) ∈ R. ⇒ The distance of point P from the origin is the same as the distance of point Q from the origin. ⇒ The distance of point Q from the origin is the same as the distance of point P from the origin. ⇒ (Q, P) ∈ R, ∴ R is symmetric. Now, Let (P, Q), (Q, S) ∈ R. ⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same. ⇒ The distance of points P and S from the origin is the same. ⇒ (P, S) ∈ R ∴ R is transitive. Therefore, R is an equivalence relation. The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin. In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin. Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
R = {(P1, P2): P1 and P2 have same the number of sides} R is reflexive, Since (P1, P1) ∈ R, as the same polygon has the same number of sides with itself. Let (P1, P2) ∈ R. ⇒ P1 and P2 have the same number of sides. ⇒ P2 and P1 have the same number of sides. ⇒ (P2, P1) ∈ R, ∴ R is symmetric. Now, Let (P1, P2), (P2, P3) ∈ R. ⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides. ⇒ P1 and P3 have the same number of sides. ⇒ (P1, P3) ∈ R ∴ R is transitive. Hence, R is an equivalence relation. The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (Since T is a polygon with 3 sides). Hence, the set of all elements in A related to triangle T is the set of all triangles.
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
R = {(L1, L2): L1 is parallel to L2} R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R. Now, let (L1, L2) ∈ R. ⇒ L1 is parallel to L2 ⇒ L2 is parallel to L1. ⇒ (L2, L1) ∈ R ∴ R is symmetric. Now, let (L1, L2), (L2, L3) ∈ R. ⇒ L1 is parallel to L2. Also, L2 is parallel to L3. ⇒ L1 is parallel to L3. ∴ R is transitive. Hence, R is an equivalence relation. The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4. Slope of line y = 2x + 4 is m = 2 It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y = 2x + c, where c ∈ R. Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.
Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. gof(1) = g[f(1)] = g(2) = 3 [as f(1) = 2 and g(2) = 3] gof(3) = g[f(3)] = g(5) = 1 [as f(3) = 5 and g(5) = 1] gof(4) = g[f(4)] = g(1) = 3 [as f(4) = 1 and g(1) = 3] ∴ gof = {(1, 3), (3, 1), (4, 3)}
Show that the function f: R → R given by f(x) = x^3 is injective.
f: R → R is given as f(x) = x^3. For one – one Suppose f(x) = f(y), where x, y ∈ R. ⇒ x^3 = y^3 … (1) Now, we need to show that x = y. Suppose x ≠ y, their cubes will also not be equal. ⇒ x^3 ≠ y^3 However, this will be a contradiction to (1). ∴ x = y Hence, f is injective.
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