# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1

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## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1

Ex 1.1 Class 12 Maths 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 an 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) Relation R in the set A = {1, 2,….,14} defined as R= {(x,y): 3x  -y = 0}
(a) Put y = x, 3x – x ≠ 0 => R is not reflexive.
(b) If 3x – y = 0, then 3y – x ≠ 0, R is not symmetric
(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0,R is not transitive.

(ii) Relations in the set N of natural numbers in defined by R = {(x, y): y = x + 5 and x < 4}
(a) Putting y = x, x ≠ x + 5, R is not reflexive
(b) Putting y = x + 5, then x ≠ y + 5,R is not symmetric.
(c) If y = x + 5, z = y + 5, then z ≠ x + 5 =>R is not transitive.

(iii) Relation R in the set A = {1,2,3,4,5,6} asR = {(x, y): y is divisible by x}
(a) Putting y = x, x is divisible by x => R is reflexive.
(b) If y is divisible by x, then x is not divisible by y when x ≠ y => R is not symmetric.
(c) If y is divisible by x and z is divisible by y then z is divisible by x e.g., 2 is divisible by 1,4 is divisible by 2.
=> 4 is divisible by 1 => R is transitive.

(iv) Relation R in Z of all integers defined as R = {(x, y): x – y is an integer}
(a) x – x=0 is an integer => R is reflexive
(b) x – y is an integer so is y – x => R is transitive.
(c) x – y is an integer, y- z is an integer and x – z is also an integer => R is transitive.

(v) R is a set of human beings in a town at a particular time.
(a) R = {(x, y)} : x and y work at the same place. It is reflexive as x works at the same place. It is symmetric since x and y or y and x work at same place.
It is transitive since X, y work at the same place and if y, z work at the same place,u then x and z also work at the same place.
(b) R : {(x, y) : x and y line in the same locality}
With similar reasoning as in part (a), R is reflexive, symmetrical and transitive.
(c) R: {(x, y)}: x is exactly 7 cm taller than y it is not reflexive: x cannot 7 cm taller than x. It is not symmetric: x is exactly 7 cm taller than y, y cannot be exactly 7 cm taller than x. It is not transitive: If x is exactly 7 cm taller than y and if y is exactly taller than z, then x is not exactly 7 cm taller than z.
(d) R = {(x, y): x is wife of y}
R is not reflexive: x cannot be wife of x. R is not symmetric: x is wife of y but y is not wife of x.
R is not transitive: if x is a wife of y then y cannot be the wife of anybody else.
(e) R= {(x, y): x is a father of y}
It is not reflexive: x cannot be father of himself. It is not symmetric: x is a father of y but y cannot be the father of x.
It is not transitive: x is a father of y and y is a father of z then x cannot be the father of z.

Ex 1.1 Class 12 Maths 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:
(i) R is not reflexive, a is not less than or equal to a² for all a ∈ R, e.g., $\ frac { 1 }{ 2 }$ is not less than $\ frac { 1 }{ 4 }$.
(ii) R is not symmetric since if a ≤ b² then b is not less than or equal to a² e.g. 2 < 5² but 5 is not less than 2².
(iii) R is not transitive: If a ≤ b², b ≤ c², then a is not less than c², e.g. 2 < (-2)², -2 < (-1)², but 2 is not less than (-1)².

Ex 1.1 Class 12 Maths 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:
(i) R is not reflexive a ≠ a + 1.
(ii) R is not symmetric if b = a + 1, then a ≠ b+ 1
(iii) R is not transitive if b = a + 1, c = b + 1 then c ≠ a + 1.

Ex 1.1 Class 12 Maths Question 4.
Show that the relation R in R defined as R={(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
R = {(a,b):a≤b}
Solution:
(i) R is reflexive, replacing b by a, a ≤ a =>a = a is true.
(ii) R is not symmetric, a ≤ b, and b ≤ a which is not true 2 < 3, but 3 is not less than 2.
(iii) R is transitive, if a ≤ b and b ≤ c, then a ≤ c, e.g. 2 < 3, 3 < 4 => 2 < 4.

Ex 1.1 Class 12 Maths Question 5.
Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.
Solution:
(i) R is not reflexive.
(ii) R is not symmetric.
(iii) R is not transitive.

Ex 1.1 Class 12 Maths 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:
(i) (1, 1),(2, 2),(3, 3) do not belong to relation R
∴ R is not reflexive.
(ii) It is symmetric (1,2) and (2,1) belong to R.
(iii) there are only two element 1 and 2 in this relation and there is no third element c in it =>R is not transitive

Ex 1.1 Class 12 Maths 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:
(i) The number of pages in a book remain the same
=> Relation R is reflexive.
(ii) The book x has the same number of pages as the book y.
=> Book y has the same number of pages as the book x.
=> The relation R is symmetric.
(iii) Book x and y have the same number of pages. Also book y and z have the same number of pages.
=> Books x and z also have the same number of pages.
R is transitive also Thus, R is an equivalence relation.

Ex 1.1 Class 12 Maths 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}.
Solution:
A= {1,2,3,4,5} and R= {(a,b): |a – b| is even}
R= {(1,3), (1,5), (3,5), (2,4)}
(a) (i) Let us take any element of a set A. then |a – a| = 0 which is even.
=> R is reflexive.
(ii) If |a – b| is even, then |b – a| is also even, where,
R = {(a, b) : |a – b| is even} => R is symmetric.
(iii) Further a – c = a – b + b – c
If |a – b| and |b – c| are even, then their sum |a – b + b – c| is also even.
=> |a – c| is even,
∴ R is transitive. Hence R is an equivalence relation.

(b) Elements of {1,3,5} are related to each other.
Since|1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. All are even numbers.
= Elements of {1, 3, 5} are related to each other. Similarly elements of {2,4} are related to each other. Since |2 – 4| = 2 an even number.
No element of set {1, 3, 5} is related to any element of {2,4}.

Ex 1.1 Class 12 Maths 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.
Solution:
The set A ∴ {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2,…..12}
(i) R= {(a, b): |a – b| is a multiple of 4}
|a – b| = 4k on b = a + 4k.
∴R ={(1,5),(1,9),(2, 6), (2, 10), (3, 7), (3, 11) , (4,8), (4,12), (5,9), (6,10), (7,11), (8.12) ,(0,0),(1,1), (2,2),…..(12,12)
(a) (a – a) = 0 = 4k,where k = 0=>(a,a)∈R
∴ R is reflexive.
(b) If |a – b| = 4k,then|b – a| = 4k i.e. (a,b) and (b, a) both belong to R. R is symmetric.
(c) a – c = a – b + b – c
when a – b and b – c are both multiples of 4 then a – c is also a multiple of 4. This shows if (a,b)(b,c) ∈ R then a – c also ∈ R
∴ R is an equivalence relation. The sets related 1 are {(1,5), (1,9)}.

(ii) R= {(a, b):a = b}
{(0,0), (1,1), (2,2)…..(12,12)}
(a) a = a => (a, a) ∈R
R is reflexive.
(b) Again if (a, b) ∈R, then (b, a) also ∈R
Since a = b and (a, b) ∈ R => R is symmetric.
(c) If(a, b) ∈R,then (b, c) ∈R =>a = b = c
∴ a = c => (a, c) ∈R, Hence, R is transitive set related is {1}.

Ex 1.1 Class 12 Maths 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.
Solution:
Let A = set of straight lines in a plane
(i) R: {(a, b): a is perpendicular to b} let a, b be two perpendicular lines.

(ii) Let A = set of real numbers R= {(a,b):a>b}
(a) An element is not greater than itself
∴R is not reflexive.
(b) If a > b than b is not greater than a => R is not symmetric
(c) If a > b also b > c, then a > c thus R is transitive
Hence, R is transitive but neither reflexive nor symmetric.

(iii) The relation R in the set {1,2,3} is given by R= {(a, b) :a + b≤4}
R= {(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)} (1,1), (2,2) ∈ R => R is reflexive
(1,2), (2,1), (1,3), (3,1) => R is symmetric
But it is not transitive, since (2,1) ∈R, (1,3) ∈R but (2,3)∉R.

(iv) The relation R in the set {1,2,3} given by R = {(a, b): a≤b} = {(1,2), (2,2), (3,3), (2,3), 0,3)}
(a) (1,1), (2,2), (3,3) ≤R => R is reflexive
(b) (1, 2) ∈R, but (2, 1) ∉R => R is not symmetric
(c) (1,2) ∈R, (2,3) ∈R,Also (1,3) ∈R=>R is transitive.

(v) The relation R in the set {1,2,3} given by R= {(a, b): 0 < |a – b| ≤ 2} = {(1,2), (2,1), (1, 3), (3,1), (2,3), (3,2)}
(a) R is not reflexive
∵ (1,1),(2,2), (3,3) do not belong to R
(b) R is symmetric
∵ (1,2), (2,1), (1,3), (3,1), (2,3), (3,2) ∈R
(c) R is transitive (1,2) ∈R, (2,3) ∈R, Also (1,3) ∈R

Ex 1.1 Class 12 Maths Question 11.
Show that the relation R in the set A of punts in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the punt Q from the origin}, is an equivalence relation. Further, show that the set of all punts related to a point P ≠ (0,0) is the circle passing through P with origin as centre.
Solution:
Let O be the origin then the relation
R={(P,Q):OP=OQ}
(i) R is reflexive. Take any distance OP,
OP = OP => R is reflexive.
(ii) R is symmetric, if OP = OQ then OQ = OP
(iii) R is transitive, let OP = OQ and OQ = OR =>OP=OR
Hence, R is an equivalence relation.
Since OP = K (constant) => P lies on a circle with centre at the origin.

Ex 1.1 Class 12 Maths Question 12.
Show that the relation R defined in the set A of all triangles as R= {(T1, T2): T1 is similar to T2}, is equivalence relation. Consider three right angle 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?
Solution:
(i) In a set of triangles R = {(T1, T2) : T1 is similar T2}
(a) Since A triangle T is similar to itself. Therefore (T, T) ∈ R for all T ∈ A.
∴ Since R is reflexive
(b) If triangle T1 is similar to triangle T2 then T2 is similar triangle T1
∴ R is symmetric.
(c) Let T1 is similar to triangle T2 and T2 to T3 then triangle T1 is similar to triangle T3,
∴ R is transitive.
Hence, R is an equivalence relation.
(ii) Two triangles are similar if their sides are proportional now sides 3,4,5 of triangle T1 are proportional to the sides 6, 8, 10 of triangle T3.
∴ T1 is related to T3.

Ex 1.1 Class 12 Maths 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 angle triangle T with sides 3,4 and 5?
Solution:
Let n be the number of sides of polygon P1.
R= {(P1, P2): P1 and P2 are n sides polygons}
(i) (a) Any polygon P1 has n sides => R is reflexive
(b) If P1 has n sides, P2 also has n sides then if P2 has n sides P1 also has n sides.
=> R is symmetric.
(c) Let P1, P2; P2, P3 are n sided polygons. P1 and P3 are also n sided polygons.
=> R is transitive. Hence R is an equivalence relation.
(ii) The set A = set of all the triangles in a plane.

Ex 1.1 Class 12 Maths 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.
Solution:
L = set of all the lines in XY plane, R= {(L1,L2) : L1 is parallel to L2}
(i) (a) L1 is parallel to itself => R is reflexive.
(b) L1 is parallel to L2 => L2 is parallel to L1 R is symmetric.
(c) Let L1 is parallel to L2 and L2 is parallel to L3 and L1 is parallel to L3 => R is transitive.
Hence, R is an equivalence relation.
(ii) Set of parallel lines related to y = 2x + 4 is y = 2x + c, where c is an arbitrary constant.

Ex 1.1 Class 12 Maths 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.
Solution:
(b)

Ex 1.1 Class 12 Maths 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
Solution:
Option (c) satisfies the condition that a = b – 2
i. e. 6 = 8 – 2 and b > 6, i.e. b = 8
=> option (c) is correct.

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