NCERT CLASS XII RELATION AND FUNCTION MATHEMATICS SOLUTION

 RELATION AND FUNCTION CLASS XII MATHEMATICS 

Relation definition

Suppose we have two sets, A and B, and the elements of A and B are related to each other in some way, either by a formula or a real relation. The related elements are written in order in such a way that the first element belongs to the first set and the second element belongs to the second set. Such a correspondence is known as "relation." The first set is known as the domain, and the second set is known as the codomain. A relation is usually represented by the letter "R".

In this figure {1, 2, 3} ∈ A

and {2, 3, 4} ∈ B 

Both sets are related by a formula: y = x+1. 

x = {1, 2, 3}

y = {2, 3, 4}

The ordered pair {(1,2), (2,3), (3,4)} shows the relation of set A to set B.

Types of relation

Empty relation:  In this relation, there is no connection between the elements of the first set and the other set. Technically, if there are no elements in set A that are related to set B, then this is called an empty relation. 

φ⋲ R. 

If R is such that x∈ A, but for these x, no y∈ B.

Universal Relation: If each element of set A is related to every aspect of A (itself), it is known as a universal relation.

R = A x A

Example: If A = {1,2,3}, then the elements of the universal relation for A consist

R: AxA = {(1,1), (1,2), (1,3), (2,1), (2, 2), (2, 3), (3,1), (3,2), (3,3)}

Definition

A relation "R" 

(i) is said to be reflexive if (a, a) ∈ A, for a ∈ A

(ii) It is called symmetric if (a, b)∈ R implies (b, a) ∈ R for all a, b ∈ A

(iii) It is said to be transitive if (a,b) ∈ R and (b,c) ∈ R implies that (a, c)∈ R for all a, b, c ∈ A

Equivalence Relation

A relation in the set A is called equivalent if it is reflexive, symmetric, and transitive.

Example

Let A be the set of all students of a boys' school. Show that the relation R in A given by R = {(a, b): a is sister of b} is the empty relation, and R′ = {(a, b): the difference between the heights of a and b is less than 3 meters} is the universal relation.

Solution

 Given that set A = the set of all boys in a school 

Since no boy can be the sister of another, it is an empty relation.

R′ = {(a, b): the difference between the heights of a and b is less than 3 meters}

As the difference in height of any two boys can be 3 meters, R' is a universal relation, as it covers all the height variations among the students.

Example 2

Let T be the set of all triangles in a plane, with R a relation in T given by R = {(T1, T2) | T1 is congruent to T2}. Show that R is an equivalence relation.

Solution

Reflexive

T1 R T1 ⇒ any triangle is congruent to itself.

Symmetric

T1 R T2 ⇒ If the triangle T1 is congruent to T2 

Then T2 RT1 ⇒ T2 is congruent to T1, as all the corresponding parts will be equal.

Transitive

 If T1 is congruent to T2, then T1R T2. 

T2 is congruent to T3. ⇒ Corresponding parts of T2 and T3 will be the same.

⇒ The corresponding parts of T1 equal T2, and the corresponding part of T2 equals T3.

⇒ Corresponding parts of the T1 will be equal to triangle T3

i.e. T1R T2  and T2 R T3, then 

⇒ T1 R T3 

Exercise 1.1 

 Q1: Determine whether each of the following relations is 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)3x - y = 0

reflexivity

xRx ⇒ 3x-x = 0 (not possible for any number ∊A)

Symmetric

xRy = 3x-y = 0 (given)

yRX = 3y-x 

let x = 1 and y = 3; then 3x - y = 3*1 - 3 = 0

but 3y - x = 3*3 - 1 ‡ 0 

Hence, it is not symmetric

Transitive

xRY ⇒ 3x - y = 0 

x = 1, y = 3, then 3x - y = 0

y = 3, z = 9

3y-z = 3*3 -9 = 0

xRz = 3x-z = 3*1 -9 does not equal zero. 

Hence, it is neither symmetric nor transitive.

(ii)  R = {(x, y): y = x + 5 and x < 4}

Reflexive

xRx 

x ‡ x + 5

So, it is not reflexive

Symmetric

xRy

y = x+5 

x = 1, 2, 3

y = 6, 7, 8

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

yRx

 if y = 1, 2, 3

x = y + 5

(1, 6) belongs to R, but (6, 1) does not belong to R 

So it is not symmetric.

Transitive

xRy

 rule (x, y) ∈ R (y, z) ∈ R

Then (x, z)∈R

Here, 

x = 1, 2, 3

y = 6, 7, 8

z = y + 5

z = 11, 12, 13

(6, 11), (7, 12) and (8, 13) do not belong to R

As the first element should be less than 4

So it is not transitive.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}

Reflexive

x is divisible by x

So, it is reflexive

Symmetric

  xRy 

y is divisible by x = { (1,2), (2, 4), (3, 6)}

but yRx 

{ (2,1), (4,2), (6, 3)} is not possible. 

Transitive

xRy  ⇒ y is divisible by x (let y = cx)

yRz  ⇒ z is divisible by y (let z = ky)

then z = ky = k(cx) = kcx

Hence, z is divisible by x.

So, this relation is transitive.

 (iv) Relation R in the set Z of all integers defined as R = {(x, y): x–y is an integer}

Solution:

Reflexive 

xRx 

x-x = 0 is an integer; hence, it is reflexive.

Symmetric

 xRy = x - y = integer

yRx = y-x = also an integer 

So, this relation is symmetric

Transitive

xRy = x-y = integer

yRz = y-z = integer

xRz = x-z is also an integer

Hence, it is 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}

Reflexive

xRx = one person, x, and x work at the same place.

So, the relation is reflexive.

Symmetric

xRy ➡ x and y work at the same place

yRz ➡ y and z work together

xRz ➡ if x and y and z work together, then x and z work together

Therefore, this relation is reflexive, transitive, and symmetric. Hence, it is an equivalence relation.

 

(c) R = {(x, y) : x is exactly 7 cm taller than y} 

Reflexive

 xRx

A person cannot be taller than himself.

Symmetric

xRy = x is exactly 7 cm taller than y, then 

yRx = y can not be exactly 7 cm taller than x

Transitive

 xRy - x is exactly 7 cm taller than y

yRz - y is exactly 7 cm taller than z

xRz - x is 7 + 7 = 14 cm taller than z

This relation is not reflexive, symmetric, or transitive.

(d) R = {(x, y) : x is wife of y} 

 Reflexivity

xRx - x cannot be the wife of himself

Symmetric

xRy—if x is the wife of y 

yRx - y is the wife of x

This relation can not be possible

Transitive

 xRy = x is the wife of y

yRz = y is the wife of z (not possible)

xRz = x is the wife of z (not possible). 

This is not a kind of symmetric, reflexive, and transitive relation. 

 

(e)  R = {(x, y) : x is father of y

Reflexive

xRx - a person cannot be the father of himself

xRy is not equal to yRx

A son's father relation is not exchangeable. 

Transitive

xRy - x is  the father of y 

yRz - y is the father of z

Then xRz - x is the grandfather of z

Hence, this relation is not reflexive, transitive, or symmetric. 

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

a Ra = a ≤ a² is not correct.

This is not true for all numbers

Symmetric

aRb => a≤ b²

bRa ≤ a²  cannot be true

Transitive

aRb => a≤b²

bRc=> b≤c²  

a≤ (c) = c⁴

 So, this relation is not reflexive, symmetric, or 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.

Reflexive 

 

As R = {(a, b) : b = a + 1}

aRa ➡ a ‡a+1

Symmetric

 aRb ➡ b = a+1

bRa → a = b+1

is not correct

Transitive

aRb 

b = a + 1

bRc 

c = b+1

aRc 

c = a + 1 + 1 = a + 2

is not a 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.

Solution

Reflexive

aRa = any a = a

So, it is reflexive

Synthetic

a, b ∈ R such that a ≤ b

Then b  ≤ a cannot be true

Transitive

a, b such that a  ≤ b

b, c such that b ≤ c

➡ a  ≤ c

Hence, the given relation is reflexive and transitive but not symmetric.

Question 5

Check whether the relation R in R defined by R = {(a, b): a ≤ b 3} is reflexive, symmetric, or transitive.

Solution

Reflexive

 R = {(a, b): a ≤ b^3 

Reflexive

For a belonging to R

The relation 'a ≤a³  'is not true for all numbers. So, it is not reflexive

Symmetric

For a, b belonging to R

aRb => a ≤ b³

bRa => b≤ a³

aRb is not equal to bRa

Transitive

aRb => a ≤ b³

bRc => b ≤ c³

aRc => a ≤ (c)^{3*3  =} c⁹

It is also not true. 

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

Since there are no (1,1) and (2,2) in R, it is not reflexive.

(1, 2) is in R and also (2, 1); hence, it is symmetric.

(1, 2) is in R 

(2,3) is not in R, and so (1,3) is not in R, so this relation 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

If x is in R

 If a comparison of the same book is done, then the number of pages will always be similar.

Symmetric

xRy => x and y are two books having the same number of pages. 

Then, yRx = y and x have the same number of pages.

Transitive

xRy => x and y have the same number of pages

yRz => y and z have the same number of pages, where z is also in R

xRz => x and z will also have the same number of pages

So, this relation is reflexive, symmetric, and transitive.

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}.

{1,3,5} R {2,4}

Reflexive

We can check for each number that |a – a| = 0, 

0 is divisible by 2.

Symmetric

1R 3 = |1– 3| = even

3R1 = |3– 1| = even

1R5 = 5R1

2R4 = 4R2

3R5 = 5R3 

Hence, this relation is symmetric

Transitive

 1R3 =|1 – 3| = even

3R5 = |3 – 5| = even

1R5 = |1 – 5| = even 

Similarly, we can check for the other realtions. This is transitive

Hence, this relation is equivalent. 

{1,3, 5} and {2,4} are related within themselves 

|1 – 3| = |1-5|=|3 – 5| = |2 – 4| all differences are even

but 1R2 => |1 – 2| = not even, 

3R2 = |3– 2| = odd

5R4 = |5– 4|= odd 

So, {1, 3, 5} is not related to {2,4}

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

A = { 1,2,3,4,5,6,7,8,9,10,11,12}

Reflexive

aRa => |a – a| for all a belonging to A give results 0

Hence, it is reflexive

Symmetric

aRb => |a – b| 

for 1R5 = |1 – 5| and 5R1 = |5 – 1|

for 2R6 => |2 – 6| and 6R2 => |6 – 2|

3R9 =>9R3 

4R12 => 12R4 

Similarly, more relations can be explored, which give multiples of 4

Hence, this relation is symmetric

Transitive

1R5 =|1– 5| =multiple of 4

5R9 = |1 – 9| = multiple of 4

1R9 => |1 – 9| = multiple of 4

Similarly, these relations can be explored 

This relation is an equivalence. 

(ii) R = {(a, b): a = b}

for aRa => a=a

For aRb => a =b

bRa => b =a

aRb = a=b

bRc = >b=c

=> a=c

=>aRc

This is true for a=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

(i) Symmetric but neither reflexive nor transitive: R = {(1, 2), (2, 1)}
(ii) Transitive but neither reflexive nor symmetric: R = {(1, 2), (2, 3), (1, 3)}
(iii) Reflexive and symmetric but not transitive: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
(iv) Reflexive and transitive but not symmetric: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(v) Symmetric and transitive but not reflexive: R = {(1, 2), (2, 1), (2, 3), (3, 2)} 

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 the 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

Reflexive

PRP => A point P will always be of its distance from the origin

Let two points P and Q be in R

PRQ = distance of P from origin = distance of Q from origin

QRP = distance of Q from origin = distance of P from origin

This relation is symmetric

Transitive 

PRQ = distance of P from origin = distance of Q from origin

Q and S are in R

QRS = distance of Q from origin = distance of S from origin

=> distance of P from origin = distance of S from origin

This relation is transitive. 

So, this relation is an equivalence. 

The distances of P, Q, and R from the origin (0,0) are equal

Hence, these points will form a circle. 

Question 12

Let f: R → R be defined as f(x) = 3x. Choose the correct answer. 

(A) f is one-to-one onto

 (B) f is many-one onto 

(C) f is one-one but not onto

 (D) f is neither one-one nor onto.

Solution 

One- one

let f(x1) = f(x2) where x1 and x2 belongs to R

Then 3x1 = 3x2

It is possible only when x1 = x2

That means this function is one-to-one

On to 

Let y belong to R

y = 3x 

then x = y/3

f(x) = f(y/3) = 3*y/3

                    = y 

f(x) = y 

hence the function is on to 

(A) is correct

Composition of Function

 Let f: A ➡B  

g: B➡C are two functions such that when we operate function g on f then we write 

gof(x) or g(f(x) 

step- first operate f on x, then operate function g on the f(x) 

So as, 

f➡(x) ➡f(x) ➡g on f(x) ➡ g(f(x))

Similarly, we can obtain fo(g(x))

g➡(x) ➡g(x) ➡f on g(x) ➡ f(g(x))

Example1: 

Let f(x) = x+2 and g(x) = 2x ; where x is any real number then

f(gx)) = f(2x) = 2x+2

g(f(x)) = g(2x) = 2(2x) = 4x

to be continued...

To encorage for writing more useful contents you can donate as per your wish