Question

NCERT Solutions Class 12 Maths Chapter 1 Relations and Functions have in-depth solutions to various topics and topics related to concepts of class 12 biology. Our NCERT Solutions is designed according to the latest syllabus proposed by the CBSE.

This NCERT Solutions Class 12 is articulated in a very concise manner so that students can learn all the basic concepts in complete detail such as:

• Relations – relation in mathematics is defined as the relationship between two or more than two sets of different values. It is the organized set of ordered pairs which contains one object from each different set. The relation also helps us establish a connection between two different objects or sets of things. We can represent relations in many different ways such as: in table form, a mapping form, graphical form, etc.
• Types of Relations – there are different types of relations such as:
  1. Empty relation
  2. Universal relation
  3. Identity relation
  4. Inverse relation
  5. Reflexive relation
  6. Symmetric relation
  7. Transitive relation
  8. Equivalence relation
• Functions – in mathematics the function is described as the expression of a rule or the law determining the relationship between two different variables which are independent of each other but the second variable is dependent on the first variable. Functions are ever-present in mathematics and are very essential for establishing a relationship between the two sets in science. The main definition of function was first proposed by the German mathematician Peter Dirichlet.
• Types of Functions – there are different kinds of functions such as
  1. One to one function or the injective function
  2. Many to one function
  3. Onto function or surjective function
  4. One-one correspondence or Bijective function
• Binary Operations – the binary operation is defined as the process of combining two elements to produce another element.

NCERT Solutions Class 12 Maths has a complete discussion of all kinds of easy and difficult concepts of maths.

Answer 1

NCERT Solutions Class 12 Maths Chapter 1 Relations and Functions

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}

Answer:

(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 clear 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. 

So, (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.

Answer 2

(v) 

(a) R = {(x, y): x and y work at the same place} 

⇒ (x, x) ∈ R    [as x and x work at the same place] 

∴ 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. [Since, y is 7 cm smaller 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} 

(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.

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.

Answer:

R = {(a, b): a ≤ b²} 

It can be observed that (\(\frac12\), \(\frac12\)) ∉ R, since, \(\frac12\) > (\(\frac12\))²

∴ 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.

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.

Answer:

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.

4. Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

Answer:

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.

Answer 3

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

Answer:

R = {(a, b): a ≤ b³} 

It is observed that (\(\frac12\), \(\frac12\)) ∉ R, since, \(\frac12\) > (\(\frac12\))³ 

∴ 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 (3, \(\frac32\)) , (\(\frac32\), \(\frac65\)) ∈ R, since 3 < (\(\frac32\)) 3 and \(\frac32\) < (\(\frac65\))³

But (3, \(\frac65\)) ∉ R as 3 > (\(\frac65\))³

∴ R is not transitive. 

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

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.

Answer:

Let A = {1, 2, 3}. 

A relation R on A is defined as R = {(1, 2), (2, 1)}. 

It is clear 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.

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.

Answer:

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.

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}.

Answer:

A = {1, 2, 3, 4, 5} and 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 

∴ 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 - b| is even

⇒ (a, c) ∈ R 

∴ R is transitive. Hence, R is an equivalence relation. 

Now, 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 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. [as 1 – 2, 1 – 4, 3 – 2, 3 – 4, 5 – 2 and 5 – 4 all are odd]

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.

Answer:

= { ∈ : 0 ≤ ≤ 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} as 

|1 – 1| = 0 is a multiple of 4. 

|5 – 1| = 4 is a multiple of 4. 

|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}. 

Answer 4

10. Given 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.

Answer:

(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.

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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Answer:

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.

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?

Answer:

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

\(\frac34 = \frac48 = \frac5{10}\left(= \frac12\right)\)

∴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.

Answer 5

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?

Answer:

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.

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.

Answer:

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.

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.

Answer:

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.

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

Answer:

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.

17. Show that the function f: R_* → R_* defined by () = 1 is one-one and onto, where R_* is the set of all non-zero real numbers. Is the result true, if the domain R_* is replaced by N with co-domain being same as R_*?

Answer:

It is given that f: R_* → R_* is defined by f(x) = \(\frac1x\)

For one – one: 

Let x, y ∈ R_* such that f(x) = f(y) 

⇒ \(\frac1x\) = \(\frac1y\)

⇒ x = y 

∴ f is one – one. 

For onto: 

It is clear that for y ∈ R_* , there exists x = \(\frac1y\) ∈ R_* [as y ≠ 0] such that

\(f(x) = \frac{1}{(\frac1y)}=y\) 

∴ f is onto. 

Thus, the given function f is one – one and onto. N

ow, consider function g: N → R_* defined by g(x) = \(\frac1x\) 

We have, 

\(g(x_1) = g(x_2)\) 

⇒ \(\frac1{x_1} = \frac1{x_2}\) 

⇒ x₁ = x₂

∴ 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) = \(\frac1{1.2}\) . 

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

Answer 6

18. Check the injectivity and surjectivity of the following functions: 

(i) f: N → N given by f(x) = x² 

(ii) f: Z → Z given by f(x) = x² 

(iii) f: R → R given by f(x) = x² 

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

Answer:

(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) = −2 or x 2 = −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) = −2 or 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 ∈ N such that f(x) = 2 or 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 ∈ Z such that f(x) = 2 or x³ = 2. 

∴ f is not surjective. 

Hence, function f is injective but not surjective.

19. Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one – one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer:

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.

20. Show that the Modulus Function f: R → R given by f(x) = |x|, is neither one – one nor onto, where |x| is x, if x is positive or 0 and |x| is − x, if x is negative.

Answer:

f: R → R is given by f(x) = |x| = \(\begin{cases}x,\, \text{if }\,x\ge0\\-x, \,\text{if}\,x<0\end{cases}\)

It is clear that f(-1) = |-1| = 1 and f(1) = |1| = 1 

∴ f(−1) = f(1), but −1 ≠ 1. 

∴ f is not one – one. 

Now, consider −1 ∈ R. 

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

∴ f is not onto. 

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

21. Show that the Signum Function f: R → R, given by

 \(f(x) = \begin{cases}1, \text{if} \,x>0\\0, \text{if} \,x = 0\\-1,\text{if}\,x<0\end{cases}\) 

is neither one-one nor onto.

Answer:

f: R → R is given by \(f(x) = \begin{cases}1, \text{if} \,x>0\\0, \text{if} \,x = 0\\-1,\text{if}\,x<0\end{cases}\)

 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.

22. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one – one.

Answer:

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.

Answer 7

23. In each of the following cases, state whether the function is one – one, onto or bijective. 

Justify your answer. 

(i) f: R → R defined by f(x) = 3 − 4x 

(ii) f: R → R defined by f(x) = 1 + x²

Answer:

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

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂) 

⇒ 3-4x₁ = 3-4x₂ 

⇒ -4x₁ = -4x₂ 

⇒ x₁ = x₂ 

∴ f is one – one. For any real number (y) in R, there exists \(\frac{3-y}4\) in R such that

\(f\left(\frac{3-y}{4}\right) = 3-4\left(\frac{3-y}4\right)=y\)

∴ f is onto. 

Hence, f is bijective. 

(ii) f: R → R is defined as f(x) = 1 + x² 

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂) 

⇒ 1+ x\(_1^2\) = 1 + x\(_2^2\)

⇒ x\(_1^2\) = x\(_2^2\)

⇒ x₁ = ± x₂ 

∴ f(x₁) = f(x₂) does not imply that x₁ = x₂

For example f(1) = f(-1) = 2 

∴ f is not one – one. 

Consider an element −2 in co-domain R. 

It is seen that f(x) = 1 + x 2 is 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.

24. Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.

Answer:

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

Let (a₁, b₁), (a₂, b₂) ∈ A × B such that f(a₁, b₁) = f(a₂, b₂) 

⇒ (b₁, a₁) = (b₂, a₂) 

⇒ b₁ = b₂ and a₁ = a₂ 

⇒ (a₁, b₁) = (a₂, b₂) 

∴ 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.

25. Let f: N → N be defined by \(f(n)= \begin{cases} \frac{n+1}2, \text{if n is odd}\\\frac n2, \text{if n is even} \end{cases}\) for all n ∈ N 

State whether the function f is bijective. Justify your answer.

Answer:

f: N → N is defined as  \(f(n)= \begin{cases} \frac{n+1}2, \text{if n is odd}\\\frac n2, \text{if n is even} \end{cases}\) for all n ∈ N

It can be observed that: 

f(1) = \(\frac{1+1}{2}\) = 1 and f(2) = \(\frac22\) = 1 [By definition of f(n)] 

f(1) = f(2), where 1 ≠ 2 

∴ 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 

f(4r + 1) = \(\frac{4r + 1 + 1}{2}\) = 2r + 1 

Case II: n is even 

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

f(4r) = \(\frac{4r}{2}\) = 2r. 

∴ f is onto.

Hence, f is not a bijective function.

26. Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by \((x) = \left(\frac{x-2}{x-3}\right)\) . Is f one-one and onto? Justify your answer.

Answer:

A = R − {3}, B = R − {1} and f: A → B defined by \(f(x) = \left(\frac{x-2}{x-3}\right)\)

Let x, y ∈ A such that f(x) = f(y) 

⇒ \(\frac{x-2}{x-3}\) = \(\frac{y-2}{y-3}\)

⇒ (x – 2)(y – 3) = (y – 2)(x – 3) 

⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6 

⇒ – 3x – 2y = – 2x – 3y 

⇒ x = y 

∴ 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, 

f(x) = y 

⇒ \(\frac{x-2}{x-3}\) = y 

⇒ x – 2 = xy – 3y 

⇒ x(1 – y) = – 3y + 2 

⇒ x = \(\frac{2-3y}{1-y}\) ∈ A [y ≠ 1] 

Thus, for any y ∈ B, there exists \(\frac{2-3y}{1-y}\) ∈ A such that

\(f\left(\frac{2-3y}{1-y}\right) = \cfrac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2 + 2y}{2-3y-3 +3y}=\frac{-y}{-1}= y\)

∴ f is onto.

Hence, function f is one - one and onto.

Answer 8

27. Let f: R → R be defined as f(x) = x⁴. Choose the correct answer. 

(A) f is one-one onto 

(B) f is many-one onto 

(C) f is one-one but not onto 

(D) f is neither one-one nor onto

Answer:

f: R → R is defined as f(x) = x⁴. 

Let x, y ∈ R such that f(x) = f(y). 

⇒ x⁴ = y⁴ 

⇒ x = ± y 

∴ f(x) = f(y) does not imply that x = y. 

For example 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.

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

(A) f is one – one onto 

(B) f is many – one onto 

(C) f is one – one but not onto 

(D) f is neither one – one nor onto

Answer:

f: R → R is 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 \(\frac y3\) in R such that f (\(\frac y3\)) = 3 (\(\frac y3\)) = y 

∴ f is onto. 

Hence, function f is one – one and onto. 

The correct answer is A.

29. Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

Answer:

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)}. 

gof(1) = g[f(1)] = g(2) = 3      [as f(1) = 2 and g(2) = 3]

gof(3) = g[f(3)] = g(5) = 1      [as f(3) = 5 and g(5) = 1]

gof(4) = g[f(4)] = g(1) = 3      [as f(4) = 1 and g(1) = 3]

∴ gof = {(1, 3), (3, 1), (4, 3)}

30. Let f, g and h be functions from R to R. Show that 

(f + g)oh = foh + goh

(f.g)oh = (foh).(goh)

Answer:

To prove: (f + g)oh = foh + goh 

LHS = [(f + g)oh](x) 

= (f + g)[h(x)] 

= f [h(x)] + g[h(x)] 

= (foh)(x) + (goh)(x) 

= {(foh)(x) + (goh)}(x) = RHS 

∴ {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x ∈R 

Hence, (f + g)oh = foh + goh

To Prove: (f.g)oh = (foh).(goh) 

LHS = [(f.g)oh](x) 

= (f.g)[h(x)] 

= f[h(x)] . g[h(x)] 

= (foh)(x) . (goh)(x) 

= {(foh).(goh)}(x) = RHS 

∴[(f.g)oh](x) = {(foh).(goh)}(x) for all x ∈R 

Hence, (f.g)oh = (foh).(goh)

31. Find gof and fog, if 

(i) f(x) = |x| and g(x) = |5x − 2| 

(ii) f(x) = 8x³ and g(x) = x^\(\frac13\)

Answer:

(i). f(x) = |x| and g(x) = |5x-  2| 

∴ gof(x) = g(f(x)) = g(|x|) = |5|x|-  2| 

fog(x) = f(g(x)) = f(|5x - 2|) = ||5x - 2|| = |5x - 2| 

(ii). f(x) = 8x³ and g(x) = x^\(\frac13\)

∴ gof(x) = g(f(x)) = g(8x³) = (8x³)^\(\frac13\) = 2x 

fog(x) = f(g(x)) = f(x^\(\frac13\)) = 8(x^\(\frac13\))³ = 8x

32. If f(x) = \(\frac{(4x + 3)}{(6x -4)}\), x ≠ \(\frac23\) , show that fof(x) = x, for all ≠ \(\frac23\) . What is the inverse of f ?

Answer:

It is given that f(x) = \(\frac{(4x + 3)}{(6x -4)}\), x ≠ \(\frac23\)

\((fof)(x) = f(f(x)) = f\left(\frac{4x + 3 }{6x - 4}\right) = \cfrac{4\left(\frac{4x + 3}{6x - 4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}\) 

\(= \frac{16x + 12 + 18x - 12}{24 x + 18 - 24x + 16}= \frac{34x}{34}\)

= x

∴ fof(x) = x, for all x ≠ \(\frac23\).

⇒ fof = I_x

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

Answer 9

33. State with reason whether following functions have inverse 

(i) f: {1, 2, 3, 4} → {10} with 

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

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

g = {(5, 4), (6, 3), (7, 4), (8, 2)} 

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

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

Answer:

(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.

34. Show that f: [−1, 1] → R, given by f(x) = \(\frac{x}{(x+2)}\) is one – one. Find the inverse of the function f: [−1, 1] → Range f.

(For y ∈ Range f, y = f(x) = \(\frac{x}{(x+2)}\), for some x in [−1, 1], i.e., \(x = \frac{2y}{(1-y)}\))

Answer:

 f: [−1, 1] → R, given by f(x) = \(\frac{x}{(x+2)}\)  

For one – one 

Let f(x) = f(y) 

⇒ \(\frac{x}{x+2}\) = \(\frac{y}{y+2}\)

⇒ xy +2x = xy +2y

⇒ 2x = 2y 

⇒ x = 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 

y = f(x) for some x ∈ [−1, 1] 

⇒ y = \(\frac{x}{x+2}\)

⇒ xy + 2y = x 

⇒ x(1 - y) = 2y 

⇒ x = \(\frac{2y}{1-y}\) , y ≠ 1 

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

g(y) = \(\frac{2y}{1-y}\), y ≠ 1

Now,

\((gof)(x) = g(f(x)) = g\left(\frac{x}{x+2}\right) = \cfrac{2\left(\frac{x}{x+2}\right)}{1- \left(\frac x {x+2}\right)} = \frac{2x }{x + 2 -x } = \frac{2x}{ 2} = x\)

and

\((fog)(y) = f(g((y)) = f \left (\frac{2y }{1-y}\right) = \cfrac{\frac{2y}{1 -y}}{\frac{2y}{1-y}+2} = \frac{2y}{2y + 2- 2y} = \frac{2y}{2} = y\)

∴ gof = x = I_[-1,1] and fog = y = I_{Range f} 

∴ f^-1 = g 

⇒ f^-1 (y) = \(\frac{2y}{1-y}\), y ≠ 1

35. Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer:

f: R → R is given by, f(x) = 4x + 3 

For one – one

Let f(x) = f(y) 

⇒ 4x + 3 = 4y + 3 

⇒ 4x = 4y 

⇒ x = y 

∴ f is a one – one function. 

For onto

For y ∈ R, let y = 4x + 3. 

⇒ x = \(\frac{y-3}{4}\) ∈ R 

Therefore, for any y ∈ R, there exists x = \(\frac{y-3}{4}\) ∈ R, such that

\(f(x) = f\left(\frac{y -3}{4}\right) = 4 \left(\frac{y -3}{4}\right) + 3 = y\)

∴ f is onto. 

Thus, f is one – one and onto and therefore, f −1 exists. 

Let us define g: R → R by g(x) = \(\frac{y-3}{4}\) 

Now,

\((gof) (x) = g(f(x)) = g(4x +3 ) = \frac{(4x+3)-3}{4}=\frac{4x}{4}= x\)

and

\((fog)(y) = f(g(y)) = f\left(\frac{y -3}{4} \right) = 4\left(\frac{y-3}{4}\right) + 3 = y -3 + 3 = y\)

∴ gof = fog = I_R 

Hence, f is invertible and the inverse of f is given by f^-1 (y) = g(y) = \(\frac{y-3}{4}\). 

Answer 10

36. Consider f: R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f⁻¹ of given f by f⁻¹ (y) = \(\sqrt{y-4}\), where R₊ is the set of all nonnegative real numbers.

Answer:

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

For one – one 

Let f(x) = f(y) 

⇒ x² + 4 = y² + 4 

⇒ x² = y² 

⇒ x = y        [as x =  y ∈ R₊] 

∴ f is a one – one function. 

For onto

For y ∈ [4, ∞), let y = x² + 4 

⇒ x² = y − 4 ≥ 0       [as ≥ 4] 

⇒ x = \(\sqrt{y-4}\) ≥ 0

Therefore, for any y ∈ [4, ∞), there exists x = \(\sqrt{y-4}\) ∈ R+, such that f(x) = f(\(\sqrt{y-4}\)) = (\(\sqrt{y-4}\)) 2 + 4 = y-4  +  4 = y 

∴ f is onto. 

Thus, f is one – one and onto and therefore, f⁻¹ exists.

Let us define g: [4, ∞) → R+ by g(y) = \(\sqrt{y-4}\)

Now, 

(gof)(x) = g(f(x)) = g(x² + 4) = \(\sqrt{(x^2+4)-4}\) = \(\sqrt{x^2}\) = x 

and 

(fog)(y) = f(g(y)) = f(\(\sqrt{y-4}\)) = (\(\sqrt{y-4}\)) 2 + 4 = y-4 + 4 = y

∴ gof = fog = IR 

Hence, f is invertible and the inverse of f is given by f^-1 (y) = g(y) = \(\sqrt{y-4}\) 

37. Consider f: R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with f⁻¹ (y) = \(\left(\frac{(\sqrt{y+6})-1}{3}\right)\) 

Answer:

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

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

Let y = 9x² + 6x – 5 

⇒ y = (3x + 1)² -1-5 = (3x + 1)² -6 

⇒ y + 6 = (3x + 1)² 

⇒ 3x + 1 = \(\sqrt{y +6}\)    [as y ≥ −5 ⟹ y + 6 > 0]

⇒ x = \(\frac{(\sqrt{y+6})-1}{3}\)

∴ f is onto, thereby range f = [−5, ∞). Let us define g: [−5, ∞) → R₊ as g(y) = \(\frac{(\sqrt{y+6})-1}{3}\) 

Now,

\((gof)(x) = g(f(x)) = g(9x^2 + 6x -5) = g((3x + 1)^2 -6)\)

\(= \sqrt{(3x + 1)^2 - 6 + 6 -1}\)

\(= \frac{3x+1-1}{3} = \frac{3x}{3} = x\)

and

\((fog)(y) = f(g(y)) = f\left(\frac{\sqrt{y+6}-1}{3}\right)= \left[3\left(\frac{\sqrt{y +6}-1}{3}\right)+1\right]^2-6\)

\( = (\sqrt{y +6})^2-6= y + 6 -6 =y\)

∴ gof = x = I_R and fog = y = I_{Range f} 

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

\(f^{-1}(y) = g(y) = \left(\frac{(\sqrt{y +6})+1}{3}\right)\)

38. Let f: X → Y be an invertible function. Show that f has unique inverse. 

(Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y, fog₁(y) = Iy() = fog₂(y). Use one – one ness of f ).

Answer:

Let f: X → Y be an invertible function. 

Also, suppose f has two inverses (sayg₁ and g₂) 

Then, for all y ∈Y, we have 

fog₁(y) = IY(y) = fog₂(y) 

⇒ f(g₁ (y)) = f(g₂ (y)) 

⇒ g₁ (y) = g₂ (y)     [as f is invertible ⇒ f is one – one] 

⇒ g₁ = g₂                [as g is one – one] 

Hence, f has a unique inverse.

39. Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f⁻¹ and show that (f⁻¹)⁻¹ = f.

Answer:

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. 

We have 

(fog)(a) = f(g(a)) = f(1) = a 

(fog)(b) = f(g(b)) = f(2) = b 

(fog)(c) = f(g(c)) = f(3) = c 

and 

(gof)(1) = g(f(1)) = f(a) = 1 

(gof)(2) = g(f(2)) = f(b) = 2 

(gof)(3) = g(f(3)) = f(c) = 3

∴ 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 

We have 

(goh)(1) = g(h(1)) = g(a) = 1 

(goh)(2) = g(h(2)) = g(b) = 2 

(goh)(3) = g(h(3)) = g(c) = 3 

and 

(hog)(a) = h(g(a)) = h(1) = a 

(hog)(b) = h(g(b)) = h(2) = b 

(hog)(c) = h(g(c)) = h(3) = c 

∴ goh = IX and hog = IY, 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