# Class 12 Maths NCERT Solutions | Chapter 1, Relations and Functions Exercise 1.1 | CBSE 12 NCERT Maths Exercise 1.1 Solved

## CBSE Class 12 Mathematics NCERT Solutions

### Chapter 1, relations and functions

CBSE Class XII NCERT Mathematics Exercise 1.1 Solved
(Cbse Class 12 Ncert Maths Textbook Page 5, 6, 7)
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}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x y is as integer}
(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): x is father of y}
Solution:
(i) A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0}
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
We can see that 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.
(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It can be seen that (1, 1) ∉ R.
∴ R is not reflexive.
(1, 6) ∈ R But, (1, 6) ∉ R.
∴ R is not symmetric.
Since there is no pair in R such that (x, y) and (y, z) R, then (x, z) cannot belong to R.
Therefore R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(iii) A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
So, (x, x) R
R is reflexive.
Now, (2, 4) R [because 4 is divisible by 2]
But, (4, 2) R. [Because 2 is not divisible by 4]
Therefore, R is not symmetric.
Let (x, y), (y, z) R. Then, y is divisible by x and z is divisible by y.
z is divisible by x.
(x, z) R
R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) R = {(x, y): x − y is an integer}
Now, for every x Z, (x, x) R as x − x = 0 is an integer.
R is reflexive.
Now, for every x, y Z if (x, y) R, then x − y is an integer.
−(x − y) is also an integer.
(y − x) is an integer.
(y, x) R
R is symmetric.
Let (x, y) and (y, z) R, where x, y, z Z.
(x − y) and (y − z) are integers.
x − z = (x − y) + (y − z) is an integer.
(x, z) R
R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(v)
(a) R = {(x, y): x and y work at the same place}
(x, x) R
R is reflexive.
If (x, y) R, then x and y work at the same place.
y and x work at the same place.
(y, x) R.
R is symmetric.
Now, let (x, y), (y, z) R
x and y work at the same place and y and z work at the same place.
x and z work at the same place.
(x, z) R
R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) R as x and x is the same human being.
R is reflexive.
If (x, y) R, then x and y live in the same locality.
y and x live in the same locality.
(y, x) R
R is symmetric.
Now, let (x, y) R and (y, z) R.
x and y live in the same locality and y and z live in the same locality.
x and z live in the same locality.
(x, z) R
R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y) : x is exactly 7 cm taller than y}
Now, (x, x) R
Since human being x cannot be taller than himself.
R is not reflexive.
Now, let (x, y) R.
x is exactly 7 cm taller than y.
Then, y is not taller than x.
(y, x) R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
R is not symmetric.
Now, Let (x, y), (y, z) R.
x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
x is exactly 14 cm taller than z .
(x, z) R
R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y) : x is the wife of y}
Now, (x, x) R
Since x cannot be the wife of herself.
R is not reflexive.
Let (x, y) R
x is the wife of y.
Clearly y is not the wife of x.
(y, x) R
If x is the wife of y, then y is the husband of x.
R is not transitive.
Say (x, y), (y, z) R
x is the wife of y and y is the wife of z which is not possible. Also, it does not imply that x is the wife of z.
(x, z) R
R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
(x, x) R
As x cannot be the father of himself.
R is not reflexive.
Now, let (x, y) R.
x is the father of y.
y cannot be the father of y.
Indeed, y is the son or the daughter of y.
(y, x) R
R is not symmetric.
Now, let (x, y) R and (y, z) R.
x is the father of y and y is the father of z.
x is not the father of z.
Indeed x is the grandfather of z.
(x, z) R
R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor 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.
Solution: R = {(a, b): a ≤ b2}
It can be observed that

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
(since 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.

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.
Solution: 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.

### CBSE XII Maths NCERT Solutions - Relations and Functions Exercise 1.1

Question 4: Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
Solution: R = {(a, b); a ≤ b} Clearly (a, a) R as a = a.
R is reflexive.
Let us take, (2, 4) R (as 2 < 4)
But, (4, 2) R as 4 is greater than 2.
R is not symmetric.
Suppose (a, b), (b, c) R. Then, a ≤ b and b ≤ c
a ≤ c
(a, c) R
R is transitive. Hence R is reflexive and transitive but not symmetric.

Question 5: Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive.
Solution: R = {(a, b): a ≤ b3}
It can be observed that

R is not reflexive.
For example, (1, 2) R (as 1 < 23 = 8)
Since, R is not symmetric.

Since, R is not transitive so, R is neither reflexive, nor symmetric, nor 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.
Solution: Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen 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.

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.
Solution: 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.

Question 8: Show that the relation R in the set A = {1, 2, 3, 4, 5} given by

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}.

Solution: Clearly for any element a A, we have |a – a| = 0 (which is even).
R is reflexive.
Let (a, b) R.

(a, c) R
Since R is transitive so, R is an equivalence relation.
All elements of the set {1, 2, 3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as elements of {1, 3, 5} are odd and elements of {2, 4} are even. Hence, the modulus of the difference between the two elements from each of these two subsets will not be even.