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NCERT Exemplar Class 12 Maths solutions chapter 1 is crucial to building a base from ground up for the entire academic session. We here will help you to find the solutions of all NCERT questions of the chapter with complete steps and detailed answers. This chapter of NCERT Class 12 Maths Solutions is divided into two parts; relations and functions. We have solutions for questions from the parts explained and solved in detail. NCERT Exemplar Class 12 Maths solutions chapter 1 PDF download is to be available for students for extending learning of the concepts.
Question:1
Let A = {a, b, c} and the relation R be defined on A as follows:
R = {(a, a), (b, c), (a, b)}.
Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.
Answer:
Here, R = {(a, a), (b, c), (a, b)}
The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.
Therefore, we need 3 ordered pairs to add with R to make it reflexive and transitive.
Question:2
Let D be the domain of the real-valued function f defined by . Then, write D.
Answer:
Therefore, the function can be defined as:
Therefore, the given function has a domain of [-5, 5].
Question:3
Let f, g: be defined by f(x) = 2x + 1 and respectively. Then, find gof.
Answer:
Here,
f(x) = 2x + 1
Therefore,
Question:7
Is a function? If g is described by , then what value should be assigned to .
Answer:
Here,
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also,
Therefore,
Similarly,
By solving (1) and (2), we get
Therefore,
Question:8
Answer:
(i)Here,
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs make a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here,
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.
Question:9
If the mappings f and g are given by and , write
Answer:
Here,
and
Therefore,
So, we can write that:
Question:10
Answer:
Here, is given by
If we assume
Then,
Similarly, for
Therefore, it is clear that f(z) is many-one.
So,
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.
Question:11
Let the function be defined by Show that f is neither one-one nor onto.
Answer:
It is given that,
So, we can write:
Hence,
It is understandable that for any value of , the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.
Question:12
Let . Find whether the following subsets of are functions from X to Y or not.
(i)
(ii)
(iii)
(iv)
Answer:
Here,
Therefore,
(i)
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii)
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii)
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv)
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.
Question:13
If functions , then show that f is one-one and g is onto.
Answer:
Here, it is given:
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.
Question:14
Let be the function defined by . Then, find the range off.
Answer:
Here,
Let’s say,
As the range of cos x is,
Therefore,
Hence, the range of the given function is [1/3, 1].
Question:15
Answer:
Here, we have to a relation R in Z as follows: , aRb if and only if a - b is divisible by n.
Here, aRa is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.
Question:16
If A = {1, 2, 3 }, define relations on A which have properties of being:
(a) reflexive, transitive but not symmetric
(b) symmetric but neither reflexive nor transitive
(c) reflexive, symmetric, and transitive.
Answer:
Here,
(i) Assume
Here, (1, 1), (2, 2) and (3, 3) is reflexive.
. Hence, is transitive.
Now,
Therefore, is not symmetric.
(ii) Let say,
So,
Therefore, is symmetric,
. Therefore, is not reflexive.
, but . Hence, is not transitive.
(iii) Let =
is reflexive as
is symmetric as
Therefore, is reflexive, symmetric and transitive.
Question:17
Answer:
Here,
So, the domain
And the Range
Here, . Therefore, R is not reflexive.
Again, . So, R is not symmetric.
Again, . So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.
Question:18
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
(a) an injective mapping from A to B
(b) a mapping from A to B which is not injective
(c) a mapping from B to A.
Answer:
Here,
(i) Assume,
Then,
, this is an injective mapping.
(ii) Again assume, denotes a mapping: . Hence, it is not an injective mapping.
(iii) Again assume, h: denotes a mapping: . Hence, it is a mapping from B to A.
Question:19
(i) which is one-one but not onto
(ii) which is not one-one but onto
(iii) which is neither one-one nor onto.
Answer:
(i) Say, be a mapping defined by
If,
Then,
So,
, hence, f(x) is one-one.
However, ‘f’ is not onto, as for , therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let , is a mapping that is defined by
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But , so the range is
Therefore, f(x) is onto.
(iii) Assume, , be a mapping which is defined by
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is .
Therefore, f (x) is neither one-one nor onto.
Question:20
Let . Let be defined by . Then show that f is bijective.
Answer:
Here,
be defined by
Hence,
Say
So, f (x) is an injective function.
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.
Question:21
Answer:
Here, A = [-1, 1]
(i)
If,
So, f(x) is one-one.
Also,
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
So, g(x) is not one-one.
Also,
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
Hence, f(x) is one-one.
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
Therefore, k(x) is not one-one.
Question:22
Answer:
(i)Here, x is greater than y;
If , then x > x, that does not satisfy for any .\\
Therefore, R is not reflexive.
Say,
For any , the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
Hence, R is transitive.
(ii) x + y = 10;
Thus,
Therefore,
So, R is not reflexive.
Again,
Therefore, R is symmetric.
And,
Therefore, R is not transitive.
(iii)Here, xy is square of an integer
So,
For any is an integer.
Thus, R is reflexive.
If
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then,
, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10;
Hence, R is not symmetric.
Therefore, R is not transitive.
Question:23
Answer:
This must be true for any
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
Therefore, R is symmetric.
Let
So, R is transitive.
Therefore, R is an equivalence relation.
Question:24
Answer:
Say, be many-one functions.
If,
Then
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, is not onto function.
is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then is invertible.
Question:25
Functions are defined, respectively, by , find
(i) fog (ii) gof (iii) fof (iv) gog
Answer:
Here,
Question:26
Answer:
Here, is a binary operation defined on Q.
(i)
Hence, is not commutative.
(ii)
Therefore, is commutative.
(iii)
Therefore, is not commutative.
(iv)
Therefore, is commutative.
Question:27
Let be binary operation defined on R by . Then the operation is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative
Answer:
(i) Here,
is a binary operation defined on R by
So,
Therefore, is a commutative binary operation.
Hence, is not associative.
Therefore, is commutative but not associative.
Question:28
Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to . Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these
Answer:
(C) equivalence
Here aRb, if a is congruent to b,
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say,
Therefore, R is symmetric.
Say, aRb and bRc
Therefore, R is transitive.
Therefore, R is an equivalence relation.
Question:29
(A) symmetric but not transitive (B) transitive but not symmetric
(C) neither symmetric nor transitive (D) both symmetric and transitive
Answer:
(B) transitive but not symmetric
If, aRb means a is brother of b.
Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.
Therefore, R is not symmetric.
If, aRb implies that a is the brother of b.
and bRc implies that b is the brother of c.
Therefore, a must be the brother of c.
Hence, R is transitive.
Question:30
The maximum number of equivalence relations on the set A = {1, 2, 3} are
(A) 1 (B) 2
(C) 3 (D) 5
Answer:
(D) 5
Given, set
Now, the number of equivalence relations as follows
Thus, the maximum number of equivalence relations is ‘5’.
Question:31
If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is
(A) reflexive (B) transitive
(C) symmetric (D) none of these
Answer:
(D) none of these
If, R be defined on the set {1, 2, 3} by R = {(1, 2)}
Then, we can say that R is not reflexive, transitive, and symmetric.
Question:32
Let us define a relation R in R as aRb if Then R is
(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.
Answer:
(b).
The defined relation R in R as aRb if
Similarly, aRa implies which is true.
Therefore, it is reflexive.
Let
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, . So, , and this is true.
Therefore, R is transitive.
Question:33
(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.
Answer:
(a)
Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
Therefore, it can be written as: 1R1, 2R2 and 3R3.
Therefore, R is reflexive.
Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.
Therefore, R is not symmetric.
Again, 1R1 and
Therefore, R is transitive.
Question:34
The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is
(a) 1 (b) 0 (c) 2 (d) None of these
Answer:
(c)
Here:
Assume ‘e’ as the identity element.
Therefore,
Question:35
(a) 720 (b) 120 (c) 0 (d) None of these
Answer:
Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping
from A to B is n! when m = n
and, 0 if m ≠ n
It is given that, m = 5 and n = 6.
As,5 ≠ 6 So, from A to B mapping = 0
Question:36
Let and . Then the number of surjections from A to B is
(a)
(b)
(c)
(d) None of these
Answer:
(d)
It is given that, and
Say, the number of elements in set A and B are m and n respectively.
Therefore, is the number of surjections from A to B, for
Given, m=2.
Therefore, the number of surjections from A to B
Question:37
Let be defined by then f is
(a) one-one (b) onto (c) bijective (d) f is not defined
Answer:
(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.
Question:39
Which of the following functions from Z to Z are bijections?
Answer:
(b)
Given that
Say,
For,
or,
Therefore, the function f(x) is one-one.
Say, y = x + 2
or,
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.
Question:43
f(x) = { x, if is rational
{ 1-x , if is irrational
Then (fof)x is
(a) constant
(b) 1 + x
(c) x
(d) None of these
Answer:
(c)
Here,
Question:45
Let be the function defined by and be another function defined by then, gof(3/2) is
Answer:
(d)
Given that,
Therefore,
Hence,
Question:48
Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............
Answer:
Here, aRb : 2a + 3b = 30
Or, 3b = 30 – 2a
Or, b = (30 - 2a)/ 3
for a = 3, b = 8
a = 6, b = 6
a = 9, b = 4
a = 12, b = 2
Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}
Question:49
Let the relation R be defined on the set by . Then R is given by ............
Answer:
Here
Question:53
Answer:
False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}
As (1 ,1) R, R is not a reflexive one.
As (3, 1) R and (1, 3) belongs to R, then R is symmetric.
Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.
Question:54
Let f: R→R be the function defined by , x R. Then f is invertible.
Answer:
False
Here, , x R. This function is not a one-one onto function for all
Again,
.
Hence, the function is not invertible.
Question:55
Every relation which is symmetric and transitive is also reflexive.
Answer:
False
Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}
Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.
Question:56
Answer:
False.
Here, the given relation in the question is reflexive and transitive. However, it is not symmetric
Question:57
Let and N be the set of natural numbers. Then the mapping defined by is onto.
Answer:
True
Here,
Therefore, the range of f is
Hence, mapping is onto.
Question:58
The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}
Answer:
False
As,
Therefore,
Hence, R is not reflexive.
Question:59
Question:62
A binary operation on a set always has the identity element.
Answer:
False
‘+’ is a binary operation on the set N but it has no identity element.
This chapter is divided into various sub-topics that are related to relations and functions and its operations. They are mentioned below:
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· This is the very start of high school algebra, which holds the very root of higher education mathematics. In NCERT Exemplar Class 12 Maths chapter 1 solutions, students will learn about relations and functions, which is a type of relation. The relation is nothing but the relationship between two groups or the relationship between input and output.
· The function is a relation, but is used for obtaining the output from a set input. Learning the relations and functions at the very start of 12 Class will help in setting one's roots in calculus mathematics for cracking entrance exams and higher education.
· Several things are covered in Class 12 Maths NCERT Exemplar solutions chapter 1 which goes into the types of functions and relations. One will learn about various relations that will be used, like universal relations, reflective relations, empty relation, symmetric relation, etc.
Chapter 1 | Relations and Functions |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
Yes, NCERT exemplar solutions for Class 12 Maths chapter 1 are designed well to prepare you for board examinations.
The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.
The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.
There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.
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hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
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