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Chapter 1 : Relations and Functions


Fundamentals of relations and functions are already known to us such as their definitions, types, graphs, domain and range, ete. In this chapter, we will study these topics in detail. Once we come and over these topics further chapters of our curriculum will be easily understood. Topics of this chapter are - types of relations : reflexive, symmetric, transitive and equivalence relations, one to one and onto functions, inverse of a function, composite functions, Binary operations.

Exercise 1
Q:
A:

(i)  A = {1, 2, 3 … 13, 14}

R = {(x, y): 3x − y = 0}

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

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 is seen that (1, 1) ∉ R.

∴ R is not reflexive.

(1, 6) ∈R

But,

(1, 6) ∉ R.

∴ R is not symmetric.

Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.

∴ 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.

 (x, x) ∈R

∴ R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

∴ 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.

Now,

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.

Now, let (x, y) ∈ R

⇒ x is the wife of y.

Clearly y is not the wife of x.

∴ (y, x) ∉ R

Indeed if x is the wife of y, then y is the husband of x.

∴ R is not transitive.

Let (x, y), (y, z) ∈ R

⇒ x is the wife of y and y is the wife of z.

This case is not possible. Also, this 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}

Now  (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.


Exercise 1
Q:
A:

R = {(a, b): a ≤ b2}

It can be observed that 

\begin{align} \left(\frac{1}{2},\frac{1}{2}\right) ∉ R , since  \frac{1}{2}>\left(\frac{1}{2}\right)^2 = \frac{1}{4}\end{align}

∴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

(as 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.


Exercise 1
Q:
A:

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.


Exercise 1
Q:
A:

R = {(a, b); a ≤ b}

Clearly (a, a) ∈ R as a = a.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

Now, let (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.


Exercise 1
Q:
A:

R = {(a, b): a ≤ b3}

It is observed that

\begin{align} \left(\frac{1}{2},\frac{1}{2}\right) ∉ R , as  \frac{1}{2}>\left(\frac{1}{2}\right)^3 = \frac{1}{8}\end{align}

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 23 = 8)

But,

(2, 1) ∉ R (as 23 > 1)

∴ R is not symmetric.

We have

\begin{align} \left(3,\frac{3}{2}\right),\left(\frac{3}{2},\frac{6}{5}\right) ∉ R , as  3>\left(\frac{3}{2}\right)^3 and \frac{3}{2}<\left(\frac{6}{5}\right)^3 \end{align}

But

\begin{align} \left(3,\frac{6}{5}\right) ∉ R , as  3>\left(\frac{6}{5}\right)^3 \end{align}

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

 

 

Exercise 1
Q:
A:

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.


Exercise 1
Q:
A:

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.


Exercise 1
Q:
A:

A = {1, 2, 3, 4, 5}

R = { (a,b) ; |a – b| is even}

It is clear that for any element a ∈A, we have |a -a| = 0(which is even).

∴R is reflexive.

Let (a, b) ∈ R.

=> |a –b| is even. 

=> |- (a –b)| = |b - a| is also even.

=> (b, a) ∈ R is even.

A = {1, 2, 3, 4, 5}

R  = { (a, b) : | a – b| is even}

It is clear that for any element a ∈A, we have |a - a | = 0 (which is even).

∴R is reflexive.

Let (a, b) ∈ R.

⇒ |a –b| is even. 

⇒ |- (a –b)| = |b - a| is also even. 

⇒ (b, a) ∈ R is even.

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ |a –b| is even and |(b –c)| is even.

⇒ (a – b) is even and (b –c ) is even.

⇒ (a –c ) = (a – b) + (b – c ) is even.    [ Sum of two even integers is even]

⇒ |a – c | is even.

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} 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 all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

 

⇒ |a –b| is even and |(b –c)| is even.

⇒ (a – b) is even and (b –c ) is even. 

⇒ (a –c ) = (a – b) + (b – c ) is even.                                                                                           [ Sum of two even integers is even] 

⇒ |a – c | is even.

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} 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 all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.


Exercise 1
Q:
A:

A = { x ∈ Z : 0 ≤ x ≤ 12} = {0,1,2,3,4,5,6,7,8,9,10,11,12}

(i) R = { (a,b) : |a - b| is a multiple of 4}

For any element a ∈A, we have (a, a) ∈ R as |a - a = 0|is a multiple of 4.

∴R is reflexive.

Now, let (a, b) ∈ R ⇒ |a - b| is a multiple of 4.

⇒ |-(a - b)| = ⇒ |b - a| is a multiple of 4.

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ |(a - b)| is a multiple of 4 and |(b - c)| is a multiple of 4. 

⇒ (a - b) is a multiple of 4 and (b - c) is a multiple of 4.

⇒ (a - c) = (a – b) + (b – c) is a multiple of 4. 

⇒ |a - c| is a multiple of 4.

⇒ (a, c) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

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

 

|1 - 1| = 0 is a multiple of 4, 

|5 - 1| = 4 is a multiple of 4, and 

|9 - 1| = 8 is a multiple of 4.

(ii) R = {(a, b): a = b}

For any element a ∈A, we have (a, a) ∈ R, since a = a.

∴R is reflexive.

Now, let (a, b) ∈ R.

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.


Exercise 1
Q:
A:

(i) Let A = {5, 6, 7}.

Define a relation R on A as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

=> (5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii) Consider a relation R in R defined as:

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

For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a = a.

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ a < b and b < c

⇒ a < c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii) Let A = {4, 6, 8}.

Define a relation R on A as:

A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in R as:

R = {a, b): a3 ≥ b3}

Clearly (a, a) ∈ R as a3 = a3.

∴ R is reflexive.

Now,

(2, 1) ∈ R (as 23 ≥ 13)

But,

(1, 2) ∉ R (as 13 < 23)

∴ R is not symmetric.

Now,

Let (a, b), (b, c) ∈ R.

⇒ a3 ≥ b3 and b3 ≥ c3

⇒ a3 ≥ c3

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v)  Let A = {−5, −6}.

Define a relation R on A as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

∴ The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.


Exercise 1
Q:
A:

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}

Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

∴R is reflexive.

Now,

Let (P, Q) ∈ R.

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.

⇒ (Q, P) ∈ R

∴R is symmetric.

Now,

Let (P, Q), (Q, S) ∈ R.

⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.

⇒ The distance of points P and S from the origin is the same.

⇒ (P, S) ∈ R

∴R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.


Exercise 1
Q:
A:

R = {(T1, T2): T1 is similar to T2}

R is reflexive since every triangle is similar to itself.

Further, if (T1, T2) ∈ R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T2, T1) ∈R

∴R is symmetric.

Now,

Let (T1, T2), (T2, T3) ∈ R.

⇒ T1 is similar to T2 and T2 is similar to T3.

⇒ T1 is similar to T3.

⇒ (T1, T3) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

\begin{align} \frac {3}{6}=\frac {4}{8}=\frac {5}{10} = \left(\frac {1}{2}\right) \end{align}

∴The corresponding sides of triangles T1 and T3 are in the same ratio.

Then, triangle T1 is similar to triangle T3.

Hence, T1­ is related to T3.


Exercise 1
Q:
A:

R = {(P1, P2): P1 and P2 have same the number of sides}

R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself.

Let (P1, P2) ∈ R.

⇒ P1 and P2­ have the same number of sides.

⇒ P2 and P1 have the same number of sides.

⇒ (P2, P1) ∈ R

∴R is symmetric.

Now,

Let (P1, P2), (P2, P3) ∈ R.

⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.

⇒ P1 and P3 have the same number of sides.

⇒ (P1, P3) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.


Exercise 1
Q:
A:

R = {(L1, L2): L1 is parallel to L2}

R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R.

Now,

Let (L1, L2) ∈ R.

⇒ L1 is parallel to L2.

⇒ L2 is parallel to L1.

⇒ (L2, L1) ∈ R

∴ R is symmetric.

Now,

Let (L1, L2), (L2, L3) ∈R.

⇒ L1 is parallel to L2. Also, L2 is parallel to L3.

⇒ L1 is parallel to L3.

∴R is transitive.

Hence, R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.

Slope of line y = 2x + 4 is m = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c, where c ∈R.

Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.


Exercise 1
Q:
A:

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

It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.

∴ R is reflexive.

It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.


Exercise 1
Q:
A:

R = {(a, b): a = b − 2, b > 6}

Now, since b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

∴(8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴(6, 8) ∈ R

The correct answer is C.


Exercise 2
Q:
A:

It is given that f: R* → R* is defined by 

One-one:

∴f is one-one.

Onto: It is clear that for y∈R*, there exists such that

∴f is onto.

Thus, the given function (f) is one-one and onto.

Now, consider function g: N → R* defined by

We have,

∴g is one-one.

Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such that g(x) =.

Hence, function g is one-one but not onto.


Exercise 2
Q:
A:

(i) f: N → N is given by,

f(x) = x2

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

 

(ii) f: Z → Z is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iii) f: R → R is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iv) f: N → N given by,

f(x) = x3

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

 

(v) f: Z → Z is given by,

f(x) = x3

It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.


Exercise 2
Q:
A:

f: R → R is given by,

f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.

∴ f is not one-one.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.


Exercise 2
Q:
A:
f : R → R is given by,

It is seen that.

∴f( - 1) = f(1), but - 1 ≠ 1.

∴ f is not one-one.

Now, consider - 1 ∈ R.

It is known that f(x) = |x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = |x| = - 1.

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.


Exercise 2
Q:
A:

f : R → R, given by 

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.

∴f is not one-one.

Now, as f(x) takes only 3 values (1, 0, or - 1) for the element - 2 in co-domain R, there does not exist any x in domain R such that f(x) = - 2.

∴ f is not onto.

Hence, the signum function is neither one-one nor onto.


Exercise 2
Q:
A:

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.

∴ f (1) = 4, f (2) = 5, f (3) = 6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.


Exercise 2
Q:
A:

(i) f: R → R is defined as f(x) = 3 - 4x.

.

∴ f is one-one.

For any real number (y) in R, there existsin R such that

∴f is onto.

Hence, f is bijective.

 

(ii) f: R → R is defined as

.

.

∴does not imply that x1 = x2

For instance,

∴ f is not one-one.

Consider an element - 2 in co-domain R.

It is seen thatis positive for all x ∈ R.

Thus, there does not exist any x in domain R such that f(x) = - 2.

∴ f is not onto.

Hence, f is neither one-one nor onto.


Exercise 2
Q:
A:

f: A × B → B × A is defined as f(a, b) = (b, a).

.

∴ f is one-one.

Now, let (b, a) ∈ B × A be any element.

Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]

∴ f is onto.

Hence,f is bijective.


Exercise 2
Q:
A:

f: N → N is defined as

It can be observed that:

∴ f is not one-one.

Consider a natural number (n) in co-domain N.

Case I: n is odd

∴n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈N such that

.

Case II: n is even

∴n = 2r for some r ∈ N. Then,there exists 4r ∈N such that.

∴ f is onto.

Hence, f is not a bijective function


Exercise 2
Q:
A:

A = R - {3}, B = R - {1}

f: A → B is defined as.

.

∴ f is one-one.

Let y ∈B = R - {1}. Then, y ≠ 1.

The function f is onto if there exists x ∈A such that f(x) = y.

Now,

Thus, for any y ∈ B, there existssuch that

Hence, function f is one-one and onto.


Exercise 2
Q:
A:

f: R → R is defined as f(x) = x4

Let x, y ∈ R such that f(x) = f(y).

∴does not imply that x1=x2.

For instance,

f(1) = f(-1) = 1

∴ f is not one-one.

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

∴ f is not onto.

Hence, function f is neither one-one nor onto.

The correct answer is D.


Exercise 2
Q:
A:

f: R → R defined as f(x) = 3x.

Let x, y ∈ R such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

∴f is one-one.

Also, for any real number (y) in co-domain R, there exists in R such that.

∴f is onto.

Hence, function f is one-one and onto.

The correct answer is A.


Exercise 3
Q:
A:

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.


Exercise 3
Q:
A:

To prove:

(f + g)oh = foh + goh

To Prove:

(f . g)oh = (foh) . (goh) 

Hence, (f . g)oh = (foh) . (goh)


Exercise 3
Q:
A:

(i) f(x) = | x | and g(x) = | 5x – 2 |

(ii) f(x) = 8x3 and g(x) = x1/3


Exercise 3
Q:
A:

It is given that.

Hence, the given function f is invertible and the inverse of f is f itself.

 

 


Exercise 3
Q:
A:

(i) f: {1, 2, 3, 4} → {10} defined as:

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10

∴f is not one-one.

Hence, function f does not have an inverse.

 

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.

∴g is not one-one,

Hence, function g does not have an inverse.

 

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

∴Function h is one-one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.

Thus, h is a one-one and onto function. Hence, h has an inverse.


Exercise 3
Q:
A:

f: [ - 1, 1] → R is given as

Let f(x) = f(y).

∴ f is a one-one function.

It is clear that f: [ - 1, 1] → Range f is onto.

∴ f: [ - 1, 1]→ Range f is one-one and onto and therefore, the inverse of the function:

f: [ - 1, 1] → Range f exists.

Let g: Range f → [ - 1, 1] be the inverse of f.

Let y be an arbitrary element of range f.

Since f: [ - 1, 1] → Range f is onto, we have:

Now, let us define g: Range f → [ - 1, 1] as

∴gof =I[-1, 1]and fog = IRange f

∴ f - 1 = g

⇒ 

  

 

 


Exercise 3
Q:
A:

f: R → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

Therefore, for any y ∈ R,   such that

∴ f is onto.

Thus, f is one-one and onto and therefore, f - 1 exists.

Let us define g: R→ R by.

∴

Hence, f is invertible and the inverse of f is given by

 

 


Exercise 3
Q:
A:

f : R+ → [4, ∞) is given as f(x) = x2 + 4.

One-one:

Let f(x) = f(y).

∴ f is a one-one function.

Onto:

For y ∈ [4, ∞), let y = x2 + 4.

Therefore, for any y ∈ R, there exists  such that

.

∴ f is onto.

Thus, f is one-one and onto and therefore, f - 1 exists.

Let us define g: [4, ∞)  → R+ by,

∴

Hence, f is invertible and the inverse of f is given by


Exercise 3
Q:
A:

f: R+ → [ - 5, ∞) is given as f(x) = 9x2 + 6x - 5.

Let y be an arbitrary element of [ - 5, ∞).

Let y = 9x2 + 6x - 5.

∴f is onto, thereby range f = [ - 5, ∞).

Let us define g: [ - 5, ∞)  → R+ as

We now have:

∴ gof = IR+ and

Hence, f is invertible and the inverse of f is given by


Exercise 3
Q:
A:

Let f: X  → Y be an invertible function.

Also, suppose f has two inverses (say g1 and g2).

Then, for all y ∈Y, we have:

Hence, f has a unique inverse.


Exercise 3
Q:
A:

Function f: {1, 2, 3}  →  {a, b, c} is given by,

f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c}  →  {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:

gof = Ix and fog = Iy where X = {1, 2, 3} and Y= {a, b, c}.

Thus, the inverse of f exists and f - 1 = g.

∴f - 1: {a, b, c} →  {1, 2, 3} is given by,

f - 1(a) = 1, f - 1(b) = 2, f-1(c) = 3

Let us now find the inverse of f - 1 i.e., find the inverse of g.

If we define h: {1, 2, 3}  →  {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we have:

∴, where X = {1, 2, 3} and Y = {a, b, c}.

Thus, the inverse of g exists and g - 1 = h ⇒ (f - 1) - 1 = h.

It can be noted that h = f.

Hence, (f - 1) - 1 = f.


Exercise 3
Q:
A:

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = Ix and fog = Iy

Here, f-1 = g.

Now, gof = Ix and fog = Iy

⇒ f-1of = Ix and fof-1=  Iy

Hence, f-1: Y → X is invertible and f is the inverse of f-1

i.e., (f-1)-1 = f.


Exercise 3
Q:
A:

: R → R is given as.

The correct answer is C.


Exercise 3
Q:
A:

It is given that is defined as 

Let y be an arbitrary element of Range f.

Then, there exists x ∈such that y = f(x)

Let us define g: Rangeas 

      Now,

 

 

∴

Thus, g is the inverse of f i.e., f - 1 = g.

Hence, the inverse of f is the map g: Range, which is given by

.

The correct answer is B.


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