These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1/

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

**Class 12 Maths Chapter 1 Exercise 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}

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. Reflexive R = {(1, 3),(2, 6), (3, 9), (4, 12)}

(1.1) ∈ R

∴ R is not reflexive.

Symmetric

(1, 3) ∈ R but (3, 1) ∉ R

∴ R is not symmetric.

Transitive

(1, 3) ∈ R and (3,9) ∈ R but (1, 9) ∉ R

∴ R is not transitive.

Hence R is neither reflexive, nor symmetric nor transitive.

ii. R = {(1, 6),(2, 7), (3, 8)}

(1, 1) ∉ R

∴ R is not reflexive.

(1, 6) ∈ R but (6, 1) ∉ R

∴ R is not symmetric. In R there does not exist ordered pairs of the form (x, y) and (y, z)

∴ R is transitive.

iii. 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)}

Reflexive

(x, x) ∈ R for all x ∈ A, since x is divisible by x.

∴ R is reflexive

Symmetric

(1,2) ∈ R but (2, 1) ∈ R

∴R is not symmetric.

Transitive

If y is divisible by x and z is divisible by y, then z is divisible by x. i.e., (x, y) ∈ R

(y, z) ∈R ⇒ (x, z) ∈R for all x, y, z ∈ A

∴R is transitive

iv. Reflexive

x – x = 0, which is an integer

i.e., (x, x)∈ R for every x ∈ Z

∴ R is reflexive

Symmetric

(x, y) ∈ R

⇒ x – y is an integer

⇒ y – x is an integer

⇒ (y, x) ∈ R

∴ R is symmetric

Transitive

(x, y) ∈ R, (y, z) ∈ R ⇒ x – y is an integer and y – z is an integer.

⇒ (x – y) + (y – z) is an integer

⇒ x – z is an integer

⇒ (x, z) ∈ R for all x, y, z ∈ Z

∴ R is transitive.

v. a. Reflexive

(x, x) ∈ R for every x ∈ A, since x and x work at the same place.

∴R is reflexive.

Symmetric

Let (x, y) ∈ R

⇒ x and y work at the same place

⇒ y and x work at the same place

⇒ (y, x) ∈ R for all x, y ∈ A

∴ R is symmetric

Transitive

Let (x, y), (y, z) ∈ R

⇒ x and y work at the same place;

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. Reflexive, symmetric and transitive

(Similar to v.a)

c. Reflexive

Since x cannot 7 cm taller than x, (x, x) ∉ R.

∴ R is not reflexive

Symmetric

(x, y) ∈ R ⇒ x is exactly 7 cm taller than y

⇒ y cannot be exactly 7 cm taller than x.

⇒ (y, x) ∉ R

∴ R is not symmetric

Transitive

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.

d. Reflexive

(x, x) ∉ R as x cannot be the wife of x.

∴ R is not reflexive

Symmetric

Let (x, y) ∈ R ⇒ x is the wife of y

⇒ y is not the wife of x ⇒ (y, x) ∉ R

∴ R is not symmetric

Transitive

Let (x, y) ∈ R ⇒ x is the wife of y ⇒ y

cannot be the wife of any z

⇒ (y, z) ∉ R

That is, there does not exist any (x, y) and (y, z) in R.

∴ R is transitive

e. Reflexive

(x, x) ∉ R as x cannot be the father of x.

∴ R is not reflexive

Symmetric

(x, y) ∈ R ⇒ x is the father of y.

⇒ y is not the father of x

⇒ (y, x) ∉ R

∴ R is not symmetric

Transitive

Let (x, y), (y, z) ∈ R ⇒ x is the father of y

and y is the father of z

⇒ x cannot be the father of z

⇒ (x, z) ∉ R

∴ R is not transitive

**Chapter 1 Math Class 12 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:

Reflexive

Let a = \(\frac { 1 }{ 2 }\). Then a² = \(\frac { 1 }{ 4 }\)

⇒ \(\frac { 1 }{ 2 }\) is not less than or equal to (\(\frac { 1 }{ 2 }\))²

⇒ (\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\)) ∉ R

∴ R is not reflexive.

Symmetric

Since 1 ≤ 2², (1, 2) ∈ R

But 2 is not less than or equal to 1²

⇒ (2, 1) ∈ R

i.e., (1, 2) ∈ R and (2, 1) ∉ R

∴ R is not symmetric

Transitive

1 ≤ (- 2)² and – 2 ≤ (0)²

But 1 is not less than or equal to 0²

i.e., (1, – 2), (- 2, 0) ∈ R but (1, 0) ∉ R

∴ R is not transitive

**Class 12 Maths Ex 1.1 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:

R={(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(1, 1) ∉ R

∴ R is not reflexive

(1, 2) ∈ R and (2, 1) ∉ R

∴ R is not symmetric

(3, 4) ∈ R , (4, 5) ∉ R but (3, 5) ∉ R

∴ R is not transitive

**Class 12 Ex 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:

Reflexive

(a, a) ∈ R as a < a ∴ R is reflexive

Symmetric

(1, 2) ∈ R ⇒ 1 ≤ 2 ⇒ 2 ≤ 1

⇒ (2, 1) ∉ R

∴ R is not symmetric

Transitive

(a, b), (b, c) ∈ R a ≤ b , b ≤ c

⇒ a ≤ c which is true

∴ R is transitive

**Maths Class 12 Chapter 1 Exercise 1.1 Question 5.**

Check whether the relation R in R defined by R = {(a,b); a ≤ b³} is reflexive, symmetric or transitive.

Solution:

Reflexive

Let a = \(\frac { 1 }{ 2 }\). Then \(\frac { 1 }{ 2 }\) is not less than (\(\frac { 1 }{ 2 }\))³.

i.e., (\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\)) ∉ R

∴ R is not reflexive.

Symmetric

Since 1 ≤ 2³, (1, 2) ∈ R

But 2 is not less than or equal to L^{3}.

⇒ (2, 1) ∉ R

i.e., (1, 2) ∈ R and (2, 1) ∉ R

∴ R is not symmetric

Transitive

9 ≤ 3³ and 3 ≤ 2³, i.e., (9, 3), (3, 2) ∈ R

But 9 is not less than or equal to 2³

i.e., (9, 2) ∉ R

i.e., (9, 3),(3, 2) ∈ R but (9, 2) ∉ R

∴ R is not 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:

(1, 1) ∉ R ∴ R is not reflexive

Symmetric

(1, 2), (2, 1) ∈ R ⇒ (2, 1), (1, 2) ∈ R

∴ R is symmetric

Transitive

(1, 2), (2, 1) ∈ R but (1, 1) ∉ R

∴ R is not 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:

Reflexive

(x, x) ∈ R, since book x and x have same number of pages.

∴ R is reflexive

Symmetric

If (x, y) ∈ R, then books x and y have same number of pages i.e., books y and x have same number of pages. Hence (y, x) ∈ R.

∴ R is symmetric

Transitive

Let (x, y, (y, z) ∈ R ⇒ books x, y and books y, z have same number of pages.

⇒ books x and z have same number of pages.

⇒ (x, z) ∈ R

∴ R is symmetric

R is trarisitive R is reflexive, symmetric and 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 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:

R= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3),(5, 5)}

Reflexive

Since |a – a| = 0 which is even (a, a) ∈ R

∴ R is reflexive

Symmetric

Let (d, b) ∈ R ⇒ |a – b| is even

⇒ |b – a| is even

⇒ (b, a) ∈ R for all a, b ∈ A

∴ R is symmetric

Transitive

Let (a, b), (b, c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ |a – c| is even

∴ R is transitive

Hence R is an equivalence relation.

The elements of {1, 3, 5} are related to each other, since |1 – 1|, |1 – 3|, |1 – 5| etc. are even numbers.

The elements of {2, 4} are related to each other, since |2 – 2|, |2 – 4|, |4 – 2|, |4 – 4| are even numbers.

No elements of {1, 3, 5} is related to any element of {2, 4}, since the absolute value of the difference of any element from {1, 3, 5} with any element from {2, 4} is not even.

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:

i. Reflexive

|a – a| = 0 is a multiple of 4, a ∈ A

⇒ (a, a) ∈ R

∴ R is reflexive

Symmetric

Let (a, b) ∈R ⇒ |a – b| is a multiple of 4

⇒ |b – a| is a multiple of 4

⇒ (b, a) ∈R

∴ R is symmetric

Transitive

Let (a, b), (b, c) ∈ R

⇒ |a – b| and |b – c| are multiple of 4

⇒ a- b and b – c are multiple of 4

Now a – c = a – b + b – c

= Multiple of 4 + multiple of 4

= Multiple of 4

∴ |a – c| is a multiple of 4 ⇒ (a, c) ∈R

i.e., (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R

∴R is transitive

Hence R is an equivalence relation.

The set of elements related to 1 is {1, 5,9}

ii. Reflexive

Let a ∈ A. Then a = a

⇒ (a, a) ∈R for all a ∈ A

∴ R is reflexive

Symmetric

Let (a, b) ∈ R ⇒ a – b

⇒ b = a

⇒ (b, a) ∈ R

∴ R is symmetric

Transitive

Let {a, b), (b, c) ∈ R ⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence R is an equivalence relation.

The set of elements related to 1 is {1}

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= {1, 2, 3}

i. Let R= {(1, 2), (2, 1)}

Reflexive

R is not reflexive, since (1, 1) ∉ R

Symmetric

R is symmetric, since (1, 2), (2, 1) ∈ R

⇒ (2, 1),(1, 2) ∈ R

Transitive

R is not transitive, since (1, 2), (2, 1) ∈R but (1, 1) ∉ R

∴ R is symmetric, but neither reflexive nor transitive.

ii. Let R= {(1,2), (1,3), (3, 2)}

Reflexive

R is not reflexive, since (1, 1) ∉ R

Symmetric

R is not symmetric, since(1, 2) ∈ R, but (2, 1) ∉ R

Transitive

R is transitive, since (1, 3), (3, 2) ∈ R and

(1, 2) ∈ R

∴ R is transitive but neither reflexive nor symmetric.

iii. Let R = {(1, 1), (2, 2), (3, 3), (1, 2),

(2,1) , (2,3), (3,2)}

Reflexive

R is reflexive, since (1,1), (2,2), (3,3) ∈ R

Symmetric

R is symmetric, since (1, 2) ∈ R

⇒ (2, 1) ∈ R(2, 3) ∈ R ⇒ (3, 2) ∈R

Transitive

R is not transitive, since (1, 2), (2, 3) ∈ R but(1, 3) ∉ R

∴ R is reflexive and symmetric but not transitive.

iv. Let R= {(1, 1), (2, 2), (3, 3), (1, 2)}

Reflexive

R is reflexive, since (1, 1), (2, 2), (3, 3) ∈ R

Symmetric

R is not symmetric, since (1, 2) ∈ R but (2, 1) ∉ R

Transitive

R is transitive, since (1, 1), (1, 2) ∈ R and , (1, 2) ∈ R

v. Let R= {(1, 2), (2, 1), (1, 1), (2, 2)}

Reflexive

R is not reflexive, since (3, 3) ∉ R

Symmetric

R is symmetric, since (1, 2) ∈ R ⇒ (2, 1) ∈ R

Transitive

R is transitive, since (1, 2), (2, 1) ∈ R ⇒ (1, 1) ∈ R

and (2, 1 )(1, 2) ∈ R ⇒ (2, 2) ∈ R

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.

Solution:

Let O be the origin

Then R = {(P, Q) : OP = OQ}

Reflexive

Let P be a point in the plane Then OP = OP

⇒ (P, P) ∈ R for all P

∴ R is reflexive

Symmetric

Let (P, Q) ∈ R

⇒ OP = OQ

⇒ OQ = OP

⇒ (Q, P) ∈ R

∴ R is symmetric.

Transitive

Let (P, Q), (Q, S) ∈ R

⇒ OP = OQ and OQ = OS

⇒ OP = OS

⇒ (P, S) ∈ R

∴ R is transitive

Hence R is an equivalence relation.

Let P ≠ (0,0) be a point in the plane. Consider the circle with centre at origin and radius OP. Then the set of points on this circle are related to P since the distance from the origin to any point on the circle is OP.

Hence the set of points related to P is the circle passing through P with origin as centre.

Question 12.

Show that the relation R defined in the set A of all triangles as R = {(T_{1}, T_{2}) : T_{1} is similar to T_{2}}, is an equivalence relation. Consider three right angled triangles T_{1} with sides 3,4, 5, T_{2} with sides 5, 12, 13 and T_{3} with sides 6, 8,10. Which triangles among T_{1}, T_{1} and T_{3} are related?

Solution:

R = {(T_{1}, T_{2}): T_{1} is similar to T_{2}

Reflexive

Let T ∈ A

Since T is similar to T, (T, T) ∈ R

∴ R is reflexive

Symmetric

Let (T_{1},T_{2}) ∈ R

⇒ T_{1} is similar to T_{2}

⇒ T_{2} is similar to T_{1}

⇒ (T_{2},T_{1}) ∈ R

∴ R is symmetric

Transitive

Let (T_{1}, T_{2}), (T_{2}, T_{3}) ∈ R

⇒ T_{1} is similar to T_{2} and T_{2} is similar to T_{3}

⇒ T_{1} is similar to T_{3}

⇒ (T_{1}, T_{3}) ∈ R

∴ R is transitive

Hence R is an equivalence relation.

Two triangles are similar if their sides are proportional. The sides 3, 4, 5 of triangle T_{1} is proportional to the sides 6, 8, 10 of triangle T_{3}

(\(\frac { 3 }{ 6 }\) = \(\frac { 4 }{ 8 }\) = \(\frac { 5 }{ 10 }\))

i. e., the sides of and T_{1} are proportional. Hence T_{1} is related to T_{3}.

Question 13.

Show that the relation R defined in the set A of all polygons as R = {(P_{1}, P_{2}): P_{1} and P_{2} 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:.

R = {(P_{1}, P_{2}): P_{1} and P_{2} have same number of sides}

Reflexive

Let P ∈ A

Since P and P have same number of sides,

(P, P) ∈ R for all P.

∴ R is reflexive

Symmetric

Let (P_{1}, P_{2}) ∈ R

⇒ Number of sides of P_{1} = Number of sides of P_{2}

⇒ Number of sides of P_{1} = Number of sides of P_{1}

⇒ (P_{2}, P_{1}) ∈ R

∴ R is symmetric

Transitive

Let (P_{1}, P_{2}), (P_{2}, P_{3}) ∈ R

⇒ P_{1}, P_{2} have same number of sides and P_{2}, P_{3} have same number of sides.

⇒ P_{1}, P_{3} have same number of sides

⇒ (P_{1}, P_{3}) ∈ R

∴ R is transitive

Hence R is an equivalence relation.

The right triangle with sides 3, 4, 5 is a polygon having 3 sides.

∴ T he set of elements of A related to T is the set of triangles in the plane.

Question 14.

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L_{1}, L_{2}) : L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Solution:.

Reflexive

(L_{1}, L_{1}) ∈R since the line L_{1} is parallel to itself.

∴ R is reflexive

Symmetric

(L_{1}, L_{2}) ∈ R ⇒ L_{1} is parallel to L_{2}

⇒ L_{2} is parallel to L_{1} ⇒ (L_{2}, L_{1}) ∈ R

∴ R is symmetric.

Transitive

(L_{1}, L_{2}) ∈R, (L_{2}, L_{3}) ∈ R

L_{1}, L_{2} are parallel and L_{2}, L_{3} are parallel

⇒ L_{1} and L_{3} are parallel. ⇒ (L_{1}, L_{3}) ∈ R

∴ R is transitive.

Hence R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the line y = 2x + c where C ∈ R.

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:

Answer:

b. R is reflexive and transitive but not symmetric.

(1, 2) ∈ R but (2, 1) ∉ R

∴ R is not symmetric

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:

Ans. c

If b = 8, a = 6 – 2 = 8 – 2 = 6

∴ (6, 8) ∈ R is correct.