RBSE Solutions for Class 12 Maths Chapter 1: Relations and Functions helps students understand the core concepts of relations, functions, their types, and properties. This chapter is fundamental for students aiming to master algebra and calculus concepts, especially in higher studies related to mathematics, engineering, and science.
Key Concepts Covered in Chapter 1: Relations and Functions
- Relations
A relation is a set of ordered pairs of elements. It defines a connection between two sets. Students learn about the types of relations, including reflexive, symmetric, transitive, and equivalence relations. - Functions
A function is a relation that assigns exactly one element of one set to each element of another set. Key concepts such as domain, co-domain, range, and the graphical representation of functions are covered in this section. - Types of Functions
- One-to-One Function (Injective)
- Onto Function (Surjective)
- Bijective Function (Injective and Surjective)
- Constant Function
- Identity Function
- Composite Function
- Inverse of a Function
The inverse of a function is also a function where the elements of the domain and co-domain are swapped. Students learn how to find the inverse of a function and understand the conditions for the existence of an inverse.
Detailed RBSE Class 12 Solutions for Chapter 1: Relations and Functions
The RBSE Class 12 Solutions for Chapter 1 are designed to help students understand the concepts with ease, offering detailed solutions to the exercise problems.
Table of Contents
RBSE Class 12 Maths Solutions Chapter 1 Relations and Functions Ex 1.1
Question 1.
Determine whether each of the following relations are reflexive, symmetric and transitive :
(i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as
R = {(x, y): 3x – y = 0}
Answer:
According to question,
A = {1, 2, 3, ……… 13, 14}
and R = {(x, y): 3x – y = 0}
(a) R = {(x, y):3x – y = 0}
Putting y = x
3x – x = 2x ≠ 0
So, xRx is not true. [∵ x ≠ 0]
i.e., (x, x) ∈ R
Thus, R is not reflexive.
(b) Interchanging x and y,
If 3x – y = 0
then 3y – x ≠ 0
So, xRy ⇏⇏ yRx
i.e., (x, y) ∈ R ⇏⇏ (y, x) ∈ R
This, R is not symmetiric.
(c) If 3x – y = 0 and 3y – z = 0, then
On adding
3x – y + 3y – z = 0 + 0
3x – z = – 2y
So, 3x – z ≠ 0
i.e., (x, y) ∈ R, (y, z) ∈ R ⇏⇏ (x, z) ∈ R
Thus, R is not transitive.
Thus, R is not reflexive, symmetric and transitive.
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
Answer:
Set of natural numbers
N = {1, 2, 3, 4, ……… }
R = {(x, y): y = x + 5 and x < 4}
then R = {(1, 6), (2, 7), (3, 8)}
(a) Here (x, x) ∉ R so xRx is not true.
Thus, P is not reflexive.
(b) Again, interchanging x and y
y = x + 5 ⇏⇏ x = y + 5
(1, 6) ∈ P ⇏⇏ (6, 1) ∈ R
or (x, y) ∈ R ⇏⇏ (y, x) ∈ R
So, R is not symmetric.
(c) In relation P = {(1, 6), (2, 7), (3,8)}
In ordered pair (1, 6) and (2, 7), 6 ≠ 2
In ordered pair (2, 7) and (3,8), 7 ≠ 3
Thus, we see that (x, y) ∈ R but (y, z) ∉ P, then (x, z) ∈ R
⇒ P is not transitive.
Thus, R is not reflexive, symmetric and transitive.
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6) as
R = {(x, y): y is divisible by x}
Answer:
According to question,
A = {1, 2, 3, 4, 5, 6}
and R = {(x, y): y, is divisible by x}
Thus, R = {(1,1), (1,2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}
(a) Here, P is reflexive since xPx is true.
Since, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) all ordered pairs are in R.
(b) Again, y is divisible by x but number x is not divisible by y, as
(2, 4) ∈ R but (4, 2) ∈ R
i.e., (x, y) ∈ ⇏⇏ (y, x) ∈ R
Thus, P is not symmetric.
(c) (1, 2) ∈ R and (2, 4) ∈ R ⇒ (1, 4) ∈ R
Similarly, (1, 3) ∈ R, (3, 6) ∈ R ⇒ (1, 6) ∈ R
i.e., (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R
So, P is transitive.
Thus, we see that P is reflexive and transitive but not symmetric.
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x – y is an integer}
Answer:
Set of all integers
Z = {……… – 3, – 2, – 1, 0, 1, 2, 3, ……….}
and P = {(x, y): x – y is an integer}
(a) In x – y putting x = y
x – x = 0, which is an integer
i.e., xRx is true
or (x, x) ∈ R, ∀ x ∈ Z
Thus, R is reflexive.
(b) Again, x – y is an integer then – (x – y) will also be an integer. [- ve of integer]
or y – x is also an integer
or (x, y) integer ⇒ (y – x) integer
or xRy ⇒ yRx, is true
or (x, y) ∈ R ⇒ (y, x) ∈ R
[since by taking x = 1, y = 2, 1 – 2 = – 1 (integer)
and 2 – 1 = 1 (integer)]
Thus, R is symmetric.
(c) If x – y and y – z both are integers.
So, x – y + y – z = x – z will also be integer.
[Sum of integers is also integer]
or xPy and yRx ⇒ xRz
So, R is transitive.
Thus, R is reflexive, symmetric and transitive.
(v) Relation R in the set A of human beings in a town at a particular time given by.
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {x, y): x is wife of y}
(e) R = {(x, y) 😡 is father of y}
Answer:
(a) R = {(x, y): x and y work at same place}
(i) R is reflexive, since each person work at same place at specific time, i.e., xRx is true.
or (x, x) ∈ R
(ii) R is symmetric, since x, y work at same place and at same time so we can say that y and x are also work at same place.
i.e., xRy ⇒ yRx
(x, y) ∈ R ⇒ (y, x) ∈ R
(iii) R is transitive, since x, y and y, z work at same place and at same time, then x, z work at the same place and at the same time i.e.,
xRy and yRz ⇒ xRz
or (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R
Thus, R is reflexive, symmetric and transitive.
(b) R = {{x, y): x and y live in same colony}
(i) R is reflexive, since each person of this colony live in same colony.
i.e., xRx or (x, x) ∈ R
(ii) R is symmetric, since x and y live in same colony then y and x also live in same colony.
i.e., xRy ⇒ yRx
or (x, y) ∈ R ⇒ (y, x) ∈ R
(iii) R is transitive, since x and y, y, and z live at same place (colony) then x and z also live in same colony, i.e.,
xRy and yRz ⇒ xRz
or (x, y) ∈ R, (y, z) ∈ R ⇒ (X, Z) ∈ R
Thus, R is reflexive, symmetric and transitive.
(c) R = {(x, y): x is 7 cm longer than y}
(i) R is not reflexive, any person cannot be 7 cm longer than himself, i.e., xRx is not true.
or (x, x) ∉ R
(ii) R is not symmetric, since if x is 7 cm longer than y then y cannot be 7 cm longer than x.
i.e., xRy ⇏⇏ yRx
or (x, y) ∈ R ⇏⇏ (y, x) ∈ R
(iii) R is not transitive, since if x is 7 cm longer than y, and y is 7 cm longer than z then x cannot be 7 cm longer than z, i.e.,
xRy and yRz ⇏⇏ xRz
or (x, y) ∈ R, (y, z) ∈ R ⇏⇏ (x, z) ∈ R
(d) R = {(x, y): x is wife of y}
(i) R is not reflexive since any person cannot be wife of himself, i.e.,
xRx, is not true
or (x, x) ∉ R
(ii) R is not symmetric, if x is wife of y then y cannot be wife of x, i.e.,
xRy ⇏⇏ yRx
or (x, y) ∈ R ⇏⇏ (y, x) ∈ R
(iii) R is not transitive, since if x is wife of y then y cannot be wife of anyone, i.e., xRy then yRz then xRz or (x, y) ∈ R then (y, z) ∈ R and (x, z) ∈ R.
Thus, R is not reflexive, symmetric and transitive.
(e) R = {(x, y): x is father of y}
(i) R is not reflexive since no one can be father of himself i.e.,
xRx is not true
or (x, x) ∉ R
(ii) R is not symmetric, since if x is father of y then i/ cannot be father of x, i.e.
xRy ⇏⇏ yRx
or (x, y) ∈ R ⇏⇏ (y, x) ∈ R
(iii) R is not transitive, since x is father of y and y is father of z then x cannot be father of z.
i.e., xRy and yRz ⇏⇏ xRz
or (x, y) ∈ R, (y, z) ∈ R ⇏⇏ (x, z) ∈ R
Thus, R is not reflexive, symmetric and transitive.
Question 2.
Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
Answer:
R = Set of real numbers
R = {(a, b): a ≤ b2}
(i) ∀ a ∈ R (R is a set of real numbers)
a ≰ a2
∴ (a, a) ∉ R
∴ R is not reflexive.
For example: 1313 is a real number and 13≤(13)213≤(13)2 is not true, i.e., 1313 R 1313 is not true or (13,13)(13,13) ∉ R.
(ii) aRb ⇒ a ≤ b2
⇒ b ≰ a2
∴ (a, b) ∈ R ⇒ (b, a) ∉ R
∴ R is not symmetric.
For example:
Thus, R is not symmetric.
(iii) aRb and bRc ⇒ a ≤ b2 and b ≤ c2
⇒ a ≤ c4 ⇒ a ≰ c2
∴ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∉ R
∴ R is not transitive.
For example, consider real numbers 3, -3 and 1. Clearly,
3 ≤ (- 3)2 and – 3 ≤ (1)2,
but 3 ≤ 12, is not true
i.e., (3, – 3) ∈ R and (- 3, 1) ∈ R but (3, 1) ∉ R
Thus, R is not transitive.
Question 3.
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.
Answer:
Let A = {1, 2, 3, 4,5, 6}
and R = {(a, b): b = a + 1}
(i) R is not reflexive, since
a ≠ a + 1
i.e., aRa is not true.
(ii) R is not symmetric, since b = a + 1, then a ≠ b + 1
i.e., aRb ⇏ bRa
or (a, b) ∈ R ⇏ (b, a) ∈ R
(iii) R is not transitive since if b = a + 1, c = b + 1, then c ≠ a + 1
i.e., (a, b) ∈ R, (b, c) ∈ R ⇏ (a, c) ∈ R
or aRb, bRc ⇏ aRc
Thus, R is not reflexive, symmetric and transitive.
Question 4.
Show that the relation R in R defined as R = {(a, b): a ≤ b), is reflexive and transitive but not symmetric.
Answer:
In the set of real numbers relation R is defined as
R = {(a, b): a ≤ b}
(i) Relation R is reflexive, since for any real number.
a ≤ a ⇒ (a, a) ∈ R
Thus, aRa is true.
(ii) Relation R is transitive, since for real numbers a, b and c,
a ≤ b, b ≤ c ⇒ a ≤ c
i.e., (aRb) and (bRc) ⇒ aRc
or (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
(iii) R is not symmetiric, since for any two real numbers a and b,
a ≤ b ⇏ b ≤ a
i.e., aRb ⇏ bRa
Thus, (a, b) ∈ R ⇏ (b, a) ∈ R
As (3, 4) ∈ R ⇒ 3 < 4
but 4 < 3 or (4, 3) ∉ R
Thus, R is reflexive and transitive but not symmetiric.
Questin 5.
Check whether the relation R in R defined by R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive.
Answer:
Given relation, R = {(a, b) : a ≤ b3} where a and b are real numbers.
(i)
(ii) aRb ⇒ a ≤ b3 ⇒ b ≰ a3
∴ (a, b) ∈ R ⇒ (b, a) ∉ R
∴ R is not symmetric.
For example:
For real numbers 1414 and 1
(iii) aRb and bRc
⇒ a ≤ b3 and b ≤ c3
⇒ a ≤ c9 ⇒ a ≰ c3
∴ (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∉ R
∴ R is not transitive.
For example: Consider real numbers 5, 2 and 3√223.
∵ 5 ≤ 23 is true ⇒ (5, 2) ∈ R
Again 2 ≤ (21/3)3 is true ⇒ (2, 21/3) ∈ R
But 5 ≰ (21/3)3 ⇒ (5, 3√223) ∉ R
Thus, given relation is not transitive for all real numbers.
Thus, relation R is not reflexive, symmetric and transitive.
Question 6.
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.
Answer:
(i) Relation R = {(1, 2), (2, 1)}, is not reflexive, since ordered pair (1, 1), (2, 2), (3, 3) does not exist in relation R.
(1, 1), (2, 2), (3, 3) ∉ R
or 1R1, 2R2 and 3R3 are not true
(ii) R = {(1, 2), (2, 1)} is symmetric, since
(1, 2) ∈ R ⇒ (2, 1) ∈ R
or 1R2 ⇒ 2R1
(iii) R = {(1, 2), (2,1)} is not transitive since R has only two elements (1, 2) and (2,1)
Thus, relation R is neither reflexive nor transitive but it is symmetric.
Hence Proved.
Question 7.
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.
Answer:
Let A = {x : x, books in a library of any college}
Given relation R = {(x, y) : Number of pages in x and y are same}
(i) Given relation R is reflexive since if x is any book then x and x will contain same pages i.e., for all x ∈ A.
xRx is true.
(ii) Relation R is symmetric, since if x, y ∈ A, then (x, y) ∈ R
⇒ Number of pages in x and y are same.
⇒ In x and y number of pages are same and in y and z number of pages are same.
⇒ (y, x) ∈ R
So, (x, y) ∈ R ⇒ (y, x) ∈ R
(iii) Let x, y, z ∈ A and (x, y) ∈ R, (y, z) ∈ R
Then (x, y) ∈ R, (y, z) ∈ R
⇒ [In x and y number of pages are same] and [in y and z number of pages are same.]
⇒ In x and z number of pages are same.
⇒ (x, z) ∈ R
So, (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R ,
Thus, R is an equivalence relation.
Hence Proved.
Question 8.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of [2, 4].
Answer:
Given relation R = {(a, b): |a – b| is even}
where a, b ∈ A and A = {1, 2, 3, 4, 5}
(i) R is reflexive, since |a – a| = 0, which is even.
(a, a) ∈ R
⇒ aRa is true.
(ii) R is symmetric, since (a, b) ∈ R, then
(a, b) ∈ R ⇒ |a – b| is even
⇒ |- (a – b)| is also even
[∵ |x| = |- x|]
⇒ |b – a | is even
⇒ (b, a) ∈ R
i.e., (a, b) ∈ R ⇒ (b, a) ∈ R is true.
(iii) R is transitive, since for real numbers a, b and c, if (a, b) ∈ R and (b, c) ∈ R, then (a, b) ∈ R, (b, c) ∈ R
⇒ |a – b| is even and |b – c| is even.
⇒ |a – b| and |b – c| are even.
⇒ a – b + b – c is also even
[sum of even numbers is also even]
⇒ a – c is also even
⇒ |a – c| is also even
⇒ (a, c) ∈ R
Thus, (a, b) ∈ R, (b, c) ∈ R
⇒ (a, c) ∈ R
⇒ aRb and bRc = aRc
We see that relation R is reflexive, symmetric and transitive. Thus, R is an equivalence relation.
Again, on set [1, 3, 5],
|1 – 3| = | – 2 | = 2, is an even number
|3 – 5| = | – 2 | = 2, is an even number
|1 – 5| = | – 4| = 4, is an even number
Thus, in {1, 3 , 5} all elements are related to each other, since elements of set [1, 3, 5] verify the given condition.
Similarly, on set {2, 4}
|2 – 4| = |- 2| = 2 is even
|4 – 2| = |2| = 2 is even
Thus, all elements of set {2, 4} are related to each other since elements of set {2, 4} verify the given condition.
Now, elements of set {1, 3, 5} and set {2, 4} are not related to each other because their elements did not verify given condition.
i.e., |1 – 2| = |- 1| = 1, which is an not even number
|3 – 4| = | – 1| = 1, which is not an even number
|5 – 2| = | 3 | = 3, which is not an even number
|5 – 4| = |1| = 1, which is not an even number
Hence Proved.
Question 9.
Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} given by
(i) R = {(a, b): |a – b| is a multiple of 4}
(ii) R = {(a, b):a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer:
Given set
A = {x ∈ Z: 0 < x < 12}
or A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(i) R = {(a, b): |a – b|, is multiple of 4}
Thus, R = {(0, 0), (0, 4), (0, 8), (0, 12), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12), (1, 5), (1, 9), (2, 6), (2, 10), (3, 7), (3, 11), (4, 8), (4, 12), (5, 9), (6, 10), (7, 11), (8, 12)}
(a) R is reflexive since for any a e A
|a – a| = 0
which is a multiple of 4 since 0 × 4 = 0
⇒ (a, a) ∈ R
⇒ aRa is true.
(b) R is symmetric, since for a,b ∈ A
(a, b) ∈ R
⇒ |a – b|, is a multiple of 4
⇒ |a – b| = 4k, where k is a natural number
⇒ |- (b – a) | = 4k, where k is a natural number
⇒ |b – a| = 4k, where k is a natural number
⇒ (b, a) ∈ R
Thus, (a, b) ∈ R ⇒ (b, a) ∈ R
⇒ aRb ⇒ bRa.
(c) R is transitive, since for a, b, c ∈ A
If (a, b) ∈ R and (b, c) ∈ R
⇒ |a – b|, is a multiple of 4 and |b – c|, is a multiple of 4.
⇒ | a – b | = 4m and |b – c| = 4n,
where m and n are natural numbers.
⇒ a – b = ± 4 m and b – c = ± 4n
⇒ a – b + b – c = ± 4 m + (± 4n)
⇒ (a – c) = ± 4m + (± 4n)
⇒ ± 4(m + n), is a multiple of 4
⇒ |a – c| = 4k, is a multiple of 4
If k = m + n and k is a natural number.
⇒ (a, c) ∈ R
Hence, (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
⇒ aRb and bRc ⇒ aRc
We see that relation R, according to given condition in given set is reflexive, symmetric and transitive.
Thus, relation R is an equivalence relation on set A.
Again let, |1 – x| ∈ R then |1 – x|, is a multiple of 4.
⇒ |1 – x| = 0, 4, 8, 12 ∈ A
⇒ When x = 1, then |1 – 1 | = 0, which is a multiple of 4.
When x = 5, then |1 – 5| = |- 4|=4, which is a multiple of 4.
When x = 9, then |1 – 9| = |- 8 | = 8, which is a multiple of 4.
∴ x = 1, 5, 9
Thus, set {1, 5, 9} is set of elements of set A which are related to 1 by R.
(ii) R = {(a, b):a = b}
and A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Thus, R = {(0 ,0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12)}
(a) R is reflexive, since
a = a ⇒ (a, a) ∈ R
So, aRa, is true.
(b) R is symmetric, since for a, b ∈ A
If (a, b) ∈ R ⇒ a = b ⇒ b = a
⇒ (b, a) ∈ R
Thus, (a, b) ∈ R
⇒ (b, a) ∈ R, is true
So, aRb ⇒ bRa, is true
(c) R is transitive, since for a,b, c ∈ A
(a, b) ∈ R and (b, c) ∈ R
⇒ a = b and b = c
⇒ a = b = c
⇒ a = c
⇒ (a, c) ∈ R
Thus, (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, is true
or aRb and bRc ⇒ aRc, is true.
We see that relation R in given set A, according to condition is reflexive, symmetric and transitive. Thus, R is an equivalence relation on set A. Again, set related to 1 (according to condition) is {1}. Hence Proved.
Question 10.
Give an example of a relation which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Answer:
(i) Let A = Set of straight lines in a plane or A = {x : x, is a line in plane}
and relation R is defined on set A, such that
R = {(a, b): line a is perpendicular to b}
(a) Relation R is not reflexive because no line is perpendicular to itself, i.e.,
(a, a) ∉ R or aRa, is not true
(b) Relation R is symmetric, since line a is perpendicular to b means line b will be perpendicular to line a (since angle between two lines is 90°), i.e.,
(a, b) ∈ R ⇒ (b, a) ∈ R, is true
or aRb ⇒ bRa, is true
(c) Relation R is not transitive, since if line a is perpendicular to b and line b is perpendicular to line c then line a cannot be perpendicular to line c. Line a will be parallel to line c, i.e.,
(a, b) ∈ R, (b, c) ∈ R ⇏ (a, c) ∈ R
⇒ aRb and bRt ⇏ aRc
which is clear by the following figure:
Here a ⊥ b and b ⊥ c
Now a || c
Thus, relation R = {(a, b): line a is perpendicular to line b} is symmetric but not reflexive and transitive.
(ii) Let set A = Set of real numbers
or A = {x : x, is a real number}
and relation R is defined on set A, such that
R = {(a, b): a > b, where a and b are real numbers}
(a) R is transitive, since for a, b, c ∈ A
If a > b and b > c ⇒ a > c i.e.,
(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
or aRb and bRc ⇒ aRc
As 5 > 4, 4 > 3 then 5 > 3
(b) R is not reflexive since no number exceeds itself,
i.e., a > a is not true.
or aRa, is not true
or (a, a) ∉ R
(c) R is not symmetric, since for a, b ∈ A
If number a is greater than b, then number b cannot be greater than a, i.e.,
a > b ⇏ b > a
or (a, b) ∈ R ⇏ (b, a) ∈ R
or aRb ⇏ bRa
Thus, relation R = {(a, b): a > b, where a and b are real numbers} is transitive but not reflexive and symmetric.
(iii) Let A = {1, 2, 3, 4} and relation R on set A is defined such that
R = {(a, b): a + b ≤ 5, when a, b ∈ A}
or R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(a) R is reflexive, since (1, 1), (2, 2) ∈ R
i.e., 1R1,2R2, is true.
(b) R is symmetric, since (1, 2), (2, 1) ∈ R, (1, 3), (3 ,1) ∈ R and (1, 4), (4, 1) ∈ R
i.e., 1R2 ⇒ 2R1
1R3 ⇒ 3R1
1R4 ⇒ 4R1, is true
(c) R is not transitive, since if (2, 1) ∈ R, (1, 4) ∈ R but (2, 4) ∉ R, i.e.,
2R1 and 1R4 ⇏ 2R4
Thus, relation R = {(a, b) : a + b ≤ 5, where a, b e A} is reflexive and symmetric but not transitive.
(iv) Let A = {1, 2, 3, 4} and relation R is defined on set A, such that
R = {(a, b): a ≤ b)
R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}
(a) R is reflexive, since
(1, 1), (2, 2), (3, 3), (4, 4) ∈ R
i.e., 1R1, 2R2, 3R3, 4R4, are true.
(b) R is transitive, since
(1, 2), (2, 3) ∈ R ⇒ (1, 3) ∈ R
(1, 3), (3, 4) ∈ R ⇒ (1, 4) ∈ R
i.e., 1R2, 2R3 ⇒ 1R3
and 1R3 and 3R4 ⇒ 1R4
(c) R is not symmetric, since if number a is smaller than number b, then number b will not be smaller than number a.
As 1 < 2, then 2 ≰ 1
i.e., 1R2 ⇏ 2R1
or (1, 2) ∈ R ⇏ (2, 1) ∈ R
Thus, relation R = [(a, b): a ≤ b] is reflexive and transitive but is not symmetric.
(v) Let A = {1, 2, 3, 4} and relation R is defined on set A, such that for any a and b
R = {(a, b) : 0 < |a – b| ≤ 3}
or R = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
(a) R is symmetric, since .
(1, 2) ∈ R ⇒ (2, 1) ∈ R
Similarly, (1, 3) ∈ R ⇒ (3, 1) ∈ R
(1, 4) ∈ R ⇒ (4, 1) ∈ R
(2, 3) ∈ R ⇒ (3, 2) ∈ R
and (3, 4) ∈ R ⇒ (4, 3) ∈ R, etc.
(b) R is transitive, since
(1, 2) ∈ R, (2, 3) ∈ R ⇒ (1, 3) ∈ R Similarly,
(1, 4) ∈ R, (4, 2) ∈ R ⇒ (1, 2) ∈ R
i.e., 1R4 and 4R2 ⇒ 1R2 etc.
(c) R is not reflexive, since (1, 1), (2, 2), (3, 3), (4, 4) do not exist in R.
Thus, relation R = {(a, b): 0 < |a – b| ≤ 3} is symmetric and transitive but not reflexive.
Question 11.
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 points related to a point P ≠ (0,0) is the circle passing through P with origin as centre.
Answer:
Let point O, is origin in plane then relation R is defined in set A such that
R = {(P, Q): OP = OQ}
(i) Relation R is reflexive, since for any point P, its distance from origin O is x, then
OP = x
i.e., x = x
or xRx is true
or (x, x) ∈ R
(ii) R is symmetric, since if points P and Q are at x and y distance from origin, then
(P, Q) ∈ R ⇒ OP = OQ
⇒ x = y
⇒ y = x
⇒ OQ = OP
⇒ (Q, P) ∈ R
⇒ (y, x) ∈ R
(x, y) ∈ R ⇒ (y, x) ∈ R
(iii) R is transitive, since P, Q and S are three points in plane such that
(P, Q) ∈ R and (Q, S) ∈ R
⇒ OP = OQ and OQ = OS
⇒ OP = OS
⇒ (P, S) ∈ R
Thus (P, Q) ∈ R, (Q, S) ∈ R
⇒ (P, S) ∈ R
We see that according to the given condition, relation R is reflexive, symrnetric and transitive. Thus, given relation R is an equivalence; relation. Again, let P be any fixed point in plane and second point Q is in plane such that (P, Q) ∈ R, then
(P, Q) ∈ R ⇒ OP = OQ
⇒ Point Q moves in plane such that its distance from origin O(0, 0) is equal to the distance of point P from origin.
⇒ Locus of point Q is a circle whose centre is at origin.
[Fixed distance OP = radius of circle]
As shown in following figure:
Hence Proved.
Question 12.
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) is similar to T2} is equivalence relation. Consider three right angled triangles T1 with sides 3, 4, 5; T2 with sides 5, 12, 13 and T3 with sides 6, 8,10. Which triangles among T1, T2 and T3 are related ?
Answer:
According to the question,
A = Set of all triangles in plane
or A = {x : x, is a triangle in plane]
aqd relation R, on set A is defined such that
R = {(T1 , T2 ): T1 , T2 are similar}
(i) R is reflexive, since each triangle is similar to itself, i.e„
T1 RT1 is true
or (T1 , T1) ∈ R
(ii) R is symmetric, since if triangles T1 and T2 are similar then triangle T2 will be similar to triangle T1 i.e.,
T1RT2 ⇒ T2RT1
or (T1, T2) ∈ R ⇒ (T2, T1) ∈R
(iii) R is transitive, since if triangles T1, T2 are similar and T2, T3 are similar then T1 and T3 will be similar, i.e.,
T1RT2 and T2RT3 ⇒ T1RT3
or (T1, T2) ∈ R, (T2, T3) ∈ R ⇒ (T1, T3) ∈ R
We see that relation R on set A is reflexive, symmetric and transitive. Thus, R is equivalence relation on set A.
Again, 3, 4, 5 are sides of triangle T1 5, 12, 13 are sides of triangle T2 and 6, 8, 10 are sides of triangle T3. If we consider sides of triangles T1 and T3 then
36=48=510=1236=48=510=12
It means sides of triangles T1 and T3 are proportional. Hence, T1 and T3 are similar triangles, i.e., T1RT3 is true or (T1, T3) ∈ R.
Therefore, T1 and T3 are related to each other.
Hence Proved.
Question 13.
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 angled triangle T with sides 3, 4 and 5 ?
Answer:
According to the question,
A = Set of all polygons in plane
or A = {x : x, is a polygon in plane}
and relation R on set A is defined as
R = {(P1, P2): P1 and P2 have same number of sides}
(i) Relation R is reflexive, since number of sides of a polygon is similar to number of sides of itself, i.e., If P1 is any polygon then P1RP1 or (P1, P1) ∈ R is true.
(ii) Relation R is symmetric since, if polygon P1 has n sides and polygon P2 has n sides then polygon P2 will have n sides and polygon P1 will have n sides, i.e.,
P1RP2 ⇒ P2RP1
or (P1, P2) ∈ R ⇒ (P2, P1) ∈ R.
(iii) Relation R is transitive, since if polygon P1 and P2 have n sides and polygon P2 and P3 have n sides then polygon P1 and P3 will have n sides each, i.e.,
P1RP2, P2RP3 ⇒ P1RP3
or (P1, P2) ∈ R, (P2, P3) ∈ R ⇒ (P1, P3) ⇒ R
We see that relation R on set A is reflexive, symmetric and transitive. Thus, R is an equivalence relation on set A. Again, according to question, set A is set of all polygons lie in plane, i.e., set A consists of triangle, quadrilateral, pentagon, hexagon, heptagon etc. Since, we have to find set A related to right angle of sides 3,4 and 5, then by given relation, all the triangles having same sides (three sides) are related to right angled triangle of sides 3,4 and 5, i.e., if T is a triangle of sides 3, 4 and 5 then all the triangles of set A will be related to T.
Hence Proved.
Question 14.
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.
Answer:
According to question,
Set L = Set of all lines in XY plane
or L = {x : x, is a line in XY-plane}
R is a’ set defined on L such that
R = {(L1, L2): L1 is parallel to L2}
where L1, L2 ∈ L
(i) R is reflexive, since each line is parallel to itself, i.e.,
L1 || L1 is true.
or L1RL1 or (L1, L1) ∈ R
(ii) R is symmetric, since if line L1 is parallel to line L2 then line L2 will be parallel to line L1.
i.e., L1 || L2 ⇒ L2 || L1
or L1RL2 ⇒ L2RL1
or (L1, L2) ∈ R (L2, L1) ∈ R
(iii) R is transitive, since if line L1 is parallel to line L2 and line L2 is parallel to line L3 then line L1 will be parallel to line L3, i.e.,
L1 || L2, L2 || L3 ⇒ L1 || L3; L1, L2, L3 ∈ L
or (L1, L2) ∈ R, (L2, L3) ∈ R ⇒ (L1, L3) ∈ R
or L1RL2, L2RL3 ⇒ L1RL3
We see that relation R is reflexive, symmetric and transitive on set L.
Again, slope of line y = 2x + 4 is 2.
Thus, set of all the lines related to line y = 2x + 4 will be those lines whose slope will be 2. In this way, set of lines related toy = 2x + 4 is y = 2x + k, where k is an arbitrary constant.
Question 15.
Let R be the relation in the set {1,2,3,4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer:
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric,
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer:
Let A = {1, 2, 3, 4}, then relation R defined on A,
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
We see that for each a ∈ A, aRa is true.
i.e., (a, a) ∈ R, since
(1, 1), (2, 2), (3, 3), (4,4) ∈ R
Thus, R is reflexive.
Again, for a, b ∈ A
V (a, b) ∈ R ⇏ (b, a) ∈ R
Since (1, 2) ∈ R ⇏ (2, 1) ∈ R
(1, 3) ∈ R ⇏ (3, 1) ∈ R
i.e., 1R2 ⇏ 2R1
1R3 ⇏ 3R1
Thus, R is not symmetric.
For a, b, c ∈ A
(a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
Since, (1, 1) ∈ R, (1, 2) ∈ R ⇒ (1, 2) s R, is true.
(1, 3) ∈ R, (3, 3) ∈ R ⇒ (1, 3) ∈ R, is true.
(1, 3) ∈ R, (3, 2) ∈ R ⇒ (1, 2) ∈ R, is true.
Thus, R is transitive.
We see that relation R on set A is reflexive, transitive but not symmetric.
Thus, (B) is correct.
Question 16.
Let R be the relation in the set N given by R = [(a, b): a = b – 2, b > 6}. Choose the correct answer:
(A) (2, 4) ∈R
(B) (3, 8) ∈ R
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R
Answer:
According to the question
N = Set of natural numbers
or N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ………. }
And relation R on set N is defined in the following way:
R = {(a, b): a = b – 2, b > 6}, where a, b ∈ N
a = b – 2, b > 6
Putting, b = 8, we have a = 8 – 2 = 6
Then (6, 8) ∈ R, which is given in (C).
Thus, option (C) is correct answer.
RBSE Class 12 Maths Solutions Chapter 1 Relations and Functions Ex 1.2
Question 1.
Show that the function f: R* → R* defined by f(x) = 1x1x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R* ?
Answer:
Given function f(x) = 1x1x, f:R* → R*
Let x1, x2 ∈ R* (domain)
then f(x1) = f(x2) ⇒ 1×1=1x21x1=1×2
⇒ x1 = x2
Thus, function is one one.
Again, if y is an arbitrary element of co-domain.
If f(x) = y, then
y = 1x1x ⇒ x = 1y1y, y ≠ 0
which is an element of domain.
(∀ y ∈ R*, x ∈ R*)
∴ f(R*) = R*
But each element of co-domain is image of one and only one element of domain.
∴ f is onto.
Thus, the given function is one-one and onto.
(ii) When domain R* is replaced by set of natural numbers N and co-domain remains R* then f: N → R*
and f(x) = 1x1x, where x ∈ N
If x1, x2 ∈ N, then
f(x1) = f(x2) ⇒ 1×1=1x21x1=1×2
⇒ x1 = x2, x1, x2 ∈ N
∴ function f is one-one.
Since, domain is set of natural numbers and co-domain is set of non-zero rea1 “ambers, then
f:N → R* and f(x) = 1x1x
Let y ∈ R* (co-domain) is any arbitrary element, and f(x) = y
y = 1x1x ⇒ x = 1y1y
Thus, range of f ⊂R*
Since, some elements of R* are not images of any element of domain.
∴ f is not onto.
So, f is one-one but not onto. Thus, by replacing R* by N result also changes.
Hence Proved.
Question 2.
Check the injectivity and surjectivity of the following functions:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f:R → R given by f(x) = x2
(iv) f:N → N given by f(x) = x3
(v) f: Z → Z given by f(x) = x3
Answer:
(i) Function f: N → N is given by f(x) = x2
Let x1, x2 ∈ N then
f(x1) = f(x2) ⇒ x12 = x22
⇒ x1 = x2
So, function is (injective).
[x1 = ± x2, x2 cannot be – ve, since x2 ∈ N]
Again, in f(x) = x2 putting x = 1, 2, 3 …….
f(1) = 12 = 1
f(2) = 22 = 4
f(3) = 32 = 9
i.e., There exist some elements in co-domain which are not images of any element of domain. For example, if 5, exist in co-domain, then 5 is not the image of any element of domain. So, f is not onto. (Surjective)
∴ Range of f ⊂ N (co-domain)
∴ f is not onto.
Thus, f is injective but not surjective.
(ii) f: Z → Z, f(x) = x2
Let x1, x2 ∈ Z(domain), then
f(x1) = f(x2)
⇒ x12 = x22 (by definition off)
⇒ (x12 – x22) = 0
⇒ (x1 – x2) (x1 + x2) = 0
⇒ (x1 – x2) = 0 or (x1 + x2) = 0
⇒ x1 = x2 or x1 = – x1
∴ f(x1) = f(x2) ⇒ x1 ≠ x2 (∵ x1 = – x2)
Thus, f is not one-one.
Since x = 1 then f(1) = 12 = 1
then x = – 1 then f(- 1) = (- 1)2 = 1
∴ 1 ≠ – 1 ⇒ f(1) = f(-1) = 1
or f(1) = f(-1) ⇒ 1 ≠ (- 1)
Again, let y ∈ Z (co-domain) is any arbitrary element,
then y = x2 ⇒ x = ± √y (y > 0)
Thus, f is not onto.
Range of f ⊂ Z (co-domain)
∴ f is not onto (Surjective).
Since, in f(x) = x2 putting x = – 1, – 2, – 3, 1, 2, 3
f(1) = 12 = 1, f(- 1) = (- 1)2 = 2,
f(2) = 22 = 4, f(- 2) = (- 2)2 = 4,
f(3) = 32 = 9,
f(- 3) = (-3)2 = 9 etc.
i.e., There exist some elements in co-domain which are not image of any element of domain. For example 7, exist in co-domain but not image of any element of domain. So, f is not onto.
Thus, function is neither injective nor surjective.
(iii) f: R → R is given by f(x) = x3
If 1, – 1 ∈ R, (domain)
Then, f(1) = 12 ⇒ f(-1) = (- 1)2 = 1
Thus, 1 ≠ – 1 ⇒ f(1) = f(- 1)
Here two distinct elements of domain have same image.
∴ Function is not injective.
Again, we see that – 2, 3 are in co-domain but are not the images of any element of domain. Thus, f is not surjective.
∴ f is neither injective nor surjective.
(iv) f: N → N is given by f(x) = x3
If x1, x2 ∈ N domain, then
f(x1) = f(x2) ⇒ x13 = x23,
⇒ x1 = x2
We see that if images of two elements are same, then two elements are same.
Thus, f is injective.
Again, in N, co-domain of function/, there exist many elements which are not image of any one element of domain like 2, 3, 4, 5 etc.
Range of function f = {1, 8, 27, 64,…} ⊂ N
i.e. f(N) ⊂ N
Function is not surjective.
Thus, function is injective but not surjective.
(v) f: Z → Z is given by f(x) = x3
If x1, x2 ∈ Z (domain), then
f(x1) = f(x2) ⇒ x13 = x23
⇒ x1 = x2
Thus, f is injective.
Again, f(1) = 13 = 1, f(-1) = (-1)3 = – 1
f2) = 23 = 8, f(- 2) = (-2)3 = – 8
f(3) = 33 = 27, f(- 3) = (- 3)3 = – 27
f(4) = 43 = 64, f(- 4) = (- 4)3 = – 64
Thus, range of/.
= {….. – 64, – 27, -8, -1, 0,1, 8, 27, 64, ……} ⊂ Z
i.e. f(Z) ⊂ Z (co-domain)
∴ Function f is not surjective.
Thus, function f is injective but not surjective.
Question 3.
Prove that the Greatest Integer Function f:R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answer:
Given, function f: R → R and f(x) = [x] then,
f(1.4) = 1 and f(1.7) = 1
1.4 ≠ 1.7 ⇒ f(1.4) = f(1.7) = 1
Here, 1.4 and 1.7 both have image 1.
∴ f is not injective.
Since, co-domain of/is set of real numbers R and in the co-domain, all numbers are not integers. But image of x ∈ R (domain) is integer.
∴ The element of co-domain, which is not an integer, is not the image of any element of domain, i.e.,
f(R) ⊂ R (co-domain)
∴ f is not surjective.
Thus, f is neither injective nor surjective.
Hence proved.
Question 4.
Show that the Modulus Function f: R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.
Answer:
Here f: R → R and f(x) = |x|, then
f(1) = |1| = 1 and f(- 1) = |- 1| = 1
Here 1 ≠ – 1 ⇒ f(1) = f(- 1) = 1
i.e., Image of two elements 1 and -1 of domain is same as 1.
∴ f is not injective.
f(o) = o
Again f(1) = | 1 | = 1,
f(- 1) = |- 1| = 1
f(2) = |2| = 2,
f(3) = |3|= 3,
f(- 3) = |- 3| = 3
f(4) = |4| = 4,
f(- 4) = |- 4| = 4
Since, there are negative numbers also in co-domain of f but any negative number of co-domain is not the image of any element of domain of f.
∴ f is not surjective.
Thus, function f is neither injective nor surjective.
Hence Proved.
Question 5.
Show that the Signum Function f: R → R, given by
is neither one-one nor onto.
Answer:
Here f: R → R, and
Here, f(1) = 1, f(2) = 1, since 1 > 0, 2 > 0
∴ 1 ≠ 2 ⇒ f(1) = f(2)
i.e., two elements of domain 1 and 2, have same image 1.
Function is not injective.
Since, co-domain of function is set of real numbers and – 1, 0, 1 is only image of elements of domain, then
f(R) = { -1, 0, 1} ⊂ R (Co-domain)
Range of f = {- 1, 0, 1} ⊂ R (Domain)
∴ f is not surjective.
Thus, function f is neither injective nor surjective.
Hence Proved.
Question 6.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = (1, 4,), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Answer:
According to the questions,
A = {1, 2, 3}, B = {4, 5, 6, 7}
f: A → B is such that
f = {(1, 4), (2, 5), (3, 6)}
Then f(1) = 4, f(2) = 5, f(3) = 6
i.e., Each elements of A has distinct image.
Thus, f is injective.
This can be shown as follows:
Hence Proved.
Question 7.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 – 4x
(ii) f: R → R defined by f(x) = 1 + x2
Answer:
(i) Here, f: R → R and f(x) = 3 – 4x
If x1, x2 ∈ R domain, then
f(x1) = f(x2) ⇒ 3 – 4x1 = 3 – 4x2
⇒ – 4x1 = – 4x2
⇒ x1 = x2
∴ f is injective.
Again, if y is any arbitrary element of co-domain such that f(x) = y
then y = 3 – 4x ⇒ y – 3 = – 4x
Thus, each element of co-domain is image of any element of domain necessarily.
∴ f is surjective.
Thus, f is one-one and onto.
(ii) f: R → R and f(x) = 1 + x2
Let x1, x2 ∈ R (domain), then,
f(x1) = f(x2) ⇒ 1 + x12 = 1 + x22
⇒ x12 = x22
⇒ (x12 – x22) = 0
⇒ (x1 – x2) (x1 + x2) = 0
⇒ x1 = x2 or x1 = – x2
i.e,, Being images of elements equal, elements are not equal.
Example: f(1) = 1 + 12 = 2
f(- 1) = 1 + (- 1)2
= 1 + 1 = 2
i.e„ 1 ≠ – 1
⇒ f(1) = f(- 1) = 2
∴ f is not injective.
Co-domain of function f has negative numbers also which are not image of any element of domain.
∴ f is not surjective.
Let y ∈ R (co-domain), then
f(x) = y = 1 + x2
⇒ x2 = y – 1
then x = ± √y – 1
If y = 0, then x = ± √-1 ∉ R
So, f is not surjective.
Thus, function f is neither injective nor surjective.
Question 8.
Let A and B be sets. Show that f: A × B → B × A such that f(a, b) = (b, a) is bijective function.
Answer:
According to question,
f: A × B → B × A if f(a, b) = (b, a)
Let (a1, b1) and (a2, b2) ∈ A × B, then
f(a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
⇒ b1 = b2 , a1 = a2
⇒ (b1, a1) = (b2, a2)
Thus, f(a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
where (a1, b1), (a2, b2) ∈ A × B
∴ Function f is injective.
Again, let (b, a) be an arbitrary element of set B × A, then (F, a) ∈ B × A.
⇒ b ∈ B and a ∈ A
⇒ a ∈ A and b ∈ B
⇒ (a, b) ∈ A × B
Thus, ∀ (b, a) ∈ B × A ∃ (a, b) ∈ A × B is such that
f(a, b) = {a, b)
∴ f: A × B → B × A is surjective.
Thus, given function is bijective function.
Hence Proved.
Question 9.
Let f: N → N be defined by
for all n ∈ N. State whether the function f is bijective. Justify your answer.
Answer:
f: N → N and
i.e., Image of two distinct elements 1 and 2 is same as element 2.
Thus, function is not one-one, i.e., function is many-one. Let n be any arbitrary constant of set N, i.e., n ∈ N.
If n is odd then (2n – 1) will also be odd, then
f(2n – 1) = (2n−1)+12(2n−1)+12 = 2n22n2 = n
f(n) = (n−1)2(n−1)2, when n is odd
and when n is even then 2n will also be even.
Thus, f(2n) = 2n22n2 = n [∵ f(n) = n2n2, when n os even]
Thus, we see that either n is even or odd.
Then f(N) = N
∴ f is surjective.
Thus, given function is many-one onto but not one-one onto.
Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by f(x) = (x−2x−3)(x−2x−3). Is f one-one and onto ? Justify your answer.
Answer:
According to the question,
A = R – {3} and B = R – {1} and f: A → B, then
f(x) = x−2x−3x−2x−3
Let x1, x2 ∈ A, then
Let f(x1) = f(x2) ⇒ x1−2×1−3=x2−2×1−3×1−2×1−3=x2−2×1−3
⇒ (x1 – 2) (x2 – 3) = (x2 – 2) (x1 – 3)
⇒ x1x2 – 3x1 – 2x2 + 6 = x1x2 – 3x2 – 2x1 + 6
⇒ 3x2 – 2x2 = 3x1 – 2x1
⇒ x2 = x1
⇒ x1 = x2
∴ Function f is one-one.
Again, let y ∈ B be any arbitrary elements such that
y = f(x)
y = (x−2x−3,x≠3)(x−2x−3,x≠3)
then (x – 3) y = x – 2
⇒ xy – 3y = x – 2
⇒ xy – x = 3y -2
⇒ x(y – 1) = 3y – 2
∴ x = 3y−2y−13y−2y−1
Clearly, x is not defined for y ≠ 1.
Thus, ∀ y ∈ B, ∃ x ∈ A: f(x) = y
∴ Function f is onto.
Thus, given function f is one-one onto.
Question 11.
Let f: R → R be defined as f(x) = x4. Choose the correct answer:
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Answer:
According to the question,
f:R → R, f(x) = x4
Let x1, x2 ∈ R (domain)
∴ f(x1) = f(x2)
⇒ x14 = x24
⇒ x14 – x24 = 0
⇒ (x12 – x22) (x12 + x22) = 0
But x12 + x12 ≠ 0, since sum of squares of two real numbers cannot be zero.
∴ x12 – x22 = 0
⇒ (x1 – x2) (x1 + x2) = 0
⇒ x1 – x2 = 0 or x1 + x2 = 0
⇒ x1 = x2 or x1 = – x2
f(x1) = f(x2) ⇒ x1 ≠ x2,
Thus, f is not one-one.
Since, x = 1 for, f(1) = 14 = 1
and x = – 1 for, f(- 1) = (- 1)4 = 1
and x = 2 for f(2) = 24 = 16
and x = – 2 for, f(- 2) = (- 2)4 = 16
i.e., f(1) = f(-1) and f(2) = f(- 2)
∴ 1 ≠ – 1 ⇒ f(1) = f(- 1) = 1
and 2 ≠ – 2 ⇒ f(2) = f(- 2) = 16
For onto, let y ∈ R (co-domain) be any arbitrary element.
and f(x) = y
then , y = x4 ⇒ x = (y)1/4, y ≥ 0
∵ There are negative numbers also in co-domain. Thus, any negative number (of co-domain) is not the image of any element of domain. For example, element – 2 of co-domain is not the image of any element of domain. In other words, range of f is set of non-negative numbers, i.e.,
Range of f ⊂ R (co-domain)
So, function is not onto.
Thus, given function is neither one-one nor onto and so, (D) is correct.
Question 12.
Let f: R → R be defined as f(x) = 3x. Choose the correct answer:
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Answer:
According to the question,
f: R → R, if f(x) = 3x .
Let x1, x2 ∈ R (domain)
Then f(x1) = f(x2) ⇒ 3x1 = 3x2
⇒ x1 = x2
Thus, f is one-one.
Let y ∈ R co-domain be any arbitrary element such that y = f(x), then
y = f(x) ⇒ y = 3x
Thus, image of y3y3 element of domain R is element y of co-domain.
∴ f is onto
So, function is one-one onto.
Thus, (A) is correct.
Important Formulas and Definitions
- Domain: The set of all possible input values for a function.
- Co-domain: The set of all possible output values that the function can map to.
- Range: The set of all actual output values of a function.
- Injective Function: A function where each element of the domain is mapped to a unique element in the co-domain.
- Surjective Function: A function where every element in the co-domain has a pre-image in the domain.
- Bijective Function: A function that is both injective and surjective.
- Inverse Function: If f:A→Bf: A \to Bf:A→B is a function, its inverse function f−1:B→Af^{-1}: B \to Af−1:B→A satisfies the property that f(f−1(b))=bf(f^{-1}(b)) = bf(f−1(b))=b and f−1(f(a))=af^{-1}(f(a)) = af−1(f(a))=a.
Why Are Relations and Functions Important?
- Foundation for Advanced Studies: Functions and relations form the base for many advanced topics in calculus and algebra.
- Practical Applications: Relations and functions are used in real-life applications like computer science (databases), engineering (signal processing), and economics (supply-demand models).
- Helps in Problem Solving: Understanding these concepts is crucial for solving complex mathematical problems, especially in competitive exams.
FAQs on RBSE Solutions for Class 12 Maths Chapter 1: Relations and Functions
Q1: What is the difference between a relation and a function?
A1: A relation is a set of ordered pairs, whereas a function is a special type of relation where each input (from the domain) maps to exactly one output (in the co-domain).
Q2: How do I know if a function is bijective?
A2: A function is bijective if it is both injective (one-to-one) and surjective (onto), meaning every element of the domain is mapped to a unique element in the co-domain, and every element of the co-domain has a corresponding element in the domain.
Q3: What is the inverse of a function?
A3: The inverse of a function is a function that “reverses” the mapping of the original function. If f(x)=yf(x) = yf(x)=y, then the inverse function f−1(y)=xf^{-1}(y) = xf−1(y)=x.
Q4: Can every function have an inverse?
A4: Not all functions have an inverse. A function will have an inverse if and only if it is bijective (injective and surjective).
Q5: Where can I find detailed solutions for Class 12 Maths Chapter 1?
A5: Detailed solutions for Chapter 1: Relations and Functions are available on websites like rbsesolution.in, where each problem is solved step-by-step for better understanding.
Conclusion
The RBSE Solutions for Class 12 Maths Chapter 1: Relations and Functions provide comprehensive explanations of the essential concepts needed to master this chapter. The solutions are structured to help students understand relations, functions, and their applications with clarity. Practicing these solutions will strengthen your problem-solving skills and prepare you for exams and higher studies.