# NCERT Solutions For Class 12 Maths Chapter 1

## NCERT Solutions Class 12 Maths Relations and Functions

### Ncert Solutions For Class 12 Maths Chapter 1 PDF Free Download

NCERT solutions for class 12 maths chapter 1 – relations and functions are given here in detail. The relations and functions chapter is one of the most important topics for the students studying in CBSE class 12. While solving the questions from this chapter, one might face several doubts and at that time, these  NCERT maths solutions can help a lot. At BYJU’S, students are provided with the NCERT solutions for all the classes and here, the detailed solutions for class 12 students are provided below.

These solutions are very easy to understand and can help the students to clear all their doubts instantly. These solutions are provided in a detailed manner, where one can find step-by-step NCERT maths solutions for class 12 Relations and Functions (chapter 1). Apart from that, students can also download these solutions in PDF format. NCERT solutions are prepared by the subject experts under the guidelines of CBSE to assists students in their class 12 examination. These solutions not only help the students to clear their doubts but can also help them to have an in-depth understanding of the concepts and let them know the best answers to the different exercise questions from the relations and functions chapter.

### EXERCISE – 1.1

Q-1: Check each and every relation whether the following are symmetric, reflexive and transitive:

(i) The relation R of the set S = {2, 3, 4, 5, 6, 7, 8, 9.  . . . . . . . 15} is defined as

R = {(a, b): 2a – b = 0}

(ii) The relation R of the set S having natural numbers is defined as

R = {(a, b): b = 2a + 6 and a < 5}

(iii) The relation R of the set S = {2, 3, 4, 5, 6, 7} is defined as

R = {(a, b): b is divisible by a}

(iv) The relation R of the set S having only the integers is defined as

R = {(a, b): a – b is an integer}

(v) The relation R from the set H having human beings at a particular time in the town is given by:

(a) R = {(a, b): a and b is working at the same place}

(b) R = {(a, b): a and b are living in the same society}

(c) R = {(a, b): a is exactly 6 cm taller than b}

(d) R = {(a, b): b is husband of a}

(e) R = {(a, b): a is the father of b}

Sol:

(i) S = {2, 3, 4, 5, 6 . . . . . . . 13, 14, 15}

R = {(a, b) : 2a – b = 0}

Therefore, R = {(2, 4), (3, 6), (4, 8), (5, 10), (6, 12), (7, 14)}

Now,

(2, 4) ɛ R, but (4, 2) $\notin$ R, i.e.,

[2 (4) – 2 ≠ 0]

So, R is not symmetric.

(2, 2), (3, 3)………..,(15, 15) $\notin$ R.

So, R is not reflexive.

Since, (2, 4), (4, 8) ɛ R, but (2, 4) $\notin$ R

As, [2(2) – 8 ≠ 0]

So, R is not even transitive.

Therefore, R, in this case, is neither reflexive nor transitive and nor symmetric.

(ii) R = {(a, b) : b = 2a + 6 and a < 5}

Therefore, R = {(1, 8), (2, 10), (3, 12), (4, 14)}

Now,

(1, 8) ɛ R, but (8, 1) $\notin$ R, i.e.,

So, R is not symmetric.

(1, 1) , (2, 2), . . . . . . . . . . . . . . . $\notin$ R.

So, R is not reflexive.

Since, here there isn’t any pair in the form of (a, b) and (b, c) ɛ R, then (a, b) $\notin$ R

So, R is not even transitive.

Therefore, R in this case is neither reflexive, nor transitive and nor symmetric.

(iii) S = {2, 3, 4, 5, 6, 7}

R = {(a, b):  b is divisible by a}

(2, 4) ɛ R      [as 4 is divided by 2 completely]

(4, 2) $\notin$ R      [as 2 can’t be divided by 4]

So, R is not symmetric.

Each and every number is divisible at least by itself.

So, (a, a) ɛ R.

So, R is reflexive.

Now,

As (a, b) and (b, c) ɛ R.

Then clearly, ‘b’ will be divisible by ‘a’ and ‘c’ will be divisible by ‘b’.

Therefore, c is divisible by b.

So, (a, b) ɛ R.

Hence, R is transitive.

Therefore, R in this case is reflexive and transitive, but not symmetric.

(iv) R = {(a, b) : a – b will be any integer}

Now,

For each a, b ɛ S, if (a, b) ɛ R, then

a – b is an integer.

$\boldsymbol{\Rightarrow }$ – (a – b) will also be an integer.

$\boldsymbol{\Rightarrow }$ (b – a) will also be an integer.

Therefore, (a, b) ɛ R.

Hence, R will be symmetric to each other.

For each a ɛ S, (a, a) ɛ R as, a – a = 0 which is also an integer.

Hence, R is reflexive.

Let, a pair in the form of (a, b) and (b, c)  ɛ R, then (a, b, c) ɛ Z.

$\boldsymbol{\Rightarrow }$ (a – b) and (b – c) is an integer.

$\boldsymbol{\Rightarrow }$ a – c = (x – y) + (y – z) will also be an integer.

Hence, R is transitive.

Therefore, R in this case is reflexive, transitive and symmetric.

(v)

(a) R = {(a, b) : a and b is working at the same place}

Now,

If (a, b) ɛ R, then a and b works at the same place

$\boldsymbol{\Rightarrow }$ b and a will also work at the same place.

$\boldsymbol{\Rightarrow }$ (b, a) ɛ R

Hence, R will be symmetric to each other.

For each a ɛ S, (a, a) ɛ R as, ‘a’ and ‘a’ is working at the same place.

Hence, R is reflexive.

Let, a pair in the form of (a, b) and (b, c) ɛ R.

$\boldsymbol{\Rightarrow }$ a and b is working at the same place, also b and c is working at the same place.

$\boldsymbol{\Rightarrow }$ a and c is also working at the same place.

$\boldsymbol{\Rightarrow }$ (a, c) ɛ R

Hence, R is transitive.

Therefore, R in this case is reflexive, transitive and symmetric.

(b) R = {(a, b) : a and b both are living in the same society}

Now,

If (a, b) ɛ R, then ‘a’ and ‘b’ is living in the same society.

$\boldsymbol{\Rightarrow }$ b and a will also live in the same society.

$\boldsymbol{\Rightarrow }$ (b, a) ɛ R

Hence, R will be symmetric to each other.

For each a ɛ S, (a, a) ɛ R as, ‘a’ and ‘a’ is the same human being.

Hence, R is reflexive.

Let, a pair in the form of (a, b) and (b, c) ɛ R.

$\boldsymbol{\Rightarrow }$ a and b is living in the same society, also b and c is living in the same society.

$\boldsymbol{\Rightarrow }$ a and c is also living in the same society.

$\boldsymbol{\Rightarrow }$ (a, c) ɛ R

Hence, R is transitive.

Therefore, R in this case is reflexive, transitive and symmetric.

(c) R = {(a, b) : a is 6 cm taller than b}

Now,

If (a, b) ɛ R, then a is 6 cm taller than b.

$\boldsymbol{\Rightarrow }$ b cannot ever will be taller than a.

$\boldsymbol{\Rightarrow }$ (b, a) $\notin$ R

Hence, R is not symmetric to each other.

For each a ɛ S, (a, a) $\notin$ R.

As, the same human being can’t be taller than himself.

So, ‘a’ can’t be taller than ‘a’.

Hence, R is not reflexive.

Let, a pair in the form of (a, b) and (b, c) ɛ R.

$\boldsymbol{\Rightarrow }$ a is 6 cm taller than b, then b will be 6 cm taller than c.

$\boldsymbol{\Rightarrow }$ a is atleast 12 cm taller than c.

$\boldsymbol{\Rightarrow }$ (a, c) $\notin$ R

Hence, R is not transitive.

Therefore, R in this case neither reflexive, nor transitive and nor symmetric.

(d) R = {(a, b) : b is the husband of a}

Now,

If (a, b) ɛ R,

$\boldsymbol{\Rightarrow }$ b is the husband of a.

$\boldsymbol{\Rightarrow }$ a can’t be husband of b.

$\boldsymbol{\Rightarrow }$ (b, a) $\notin$ R

Instead of this, if ‘b’ is the husband of a, then ‘a’ is the wife of b.

Hence, R is not symmetric to each other.

(a, a) $\notin$ R

Since, b cannot be the husband of himself.

Hence, R is not reflexive.

If (a, b), (b, c) ɛ R

$\boldsymbol{\Rightarrow }$ b is the husband of a and a is the husband of c.

Which can’t ever be possible? Also, it is not so that b is the husband of c.

$\boldsymbol{\Rightarrow }$ (a, c) $\notin$ R

Hence, R is not transitive.

Therefore, R in this case neither reflexive, nor transitive and nor symmetric.

(e) R = {(a, b) : a is the father of b}

Now,

If (a, b) ɛ R,

$\boldsymbol{\Rightarrow }$ a is the father of b.

$\boldsymbol{\Rightarrow }$ b can’t be the father of a.

Thus, b is either the son or daughter of a.

$\boldsymbol{\Rightarrow }$ (b, a) $\notin$ R

Hence, R is not symmetric to each other.

(a, a) $\notin$ R

Since, b cannot be the father of himself.

Hence, R is not reflexive.

Since, (a, b) ɛ R and (b, c) $\notin$ R

$\boldsymbol{\Rightarrow }$  a is the father of b, then b is the father of c.

$\boldsymbol{\Rightarrow }$  a is not the father of c.

Indeed, a will be the grandfather of c.

$\boldsymbol{\Rightarrow }$ (a, c) $\notin$ R

Hence, R is not transitive.

Therefore, R in this case neither reflexive, nor transitive and nor symmetric.

Q-2: Prove that the relation M in the set M of the real numbers which is defined as

M = {(x, y): x ≤ y2} which is neither reflexive, nor transitive, nor symmetric.

Sol:

M = {(x, y): x ≤ y2}

We can see that ($\frac{1}{2}, \; \frac{1}{2}$) $\notin$ R

As, $\frac{1}{2} > \left (\frac{1}{2} \right)^{2}$

Hence, M is not reflexive.

(1, 4) ɛ M as 1 < 42

But, 4 is not lesser than 1

So, (4, 1) $\notin$ M

Hence, M is not symmetric.

(3, 2) and (2, 1.5) ɛ M              [as 3 < 22 = 4 and 2 < (1.5)2 = 2.25]

But, 3 > (1.5)2 = 2.25

So, (3, 1.5) $\notin$ M

Hence, M is not transitive.

Therefore, M is neither transitive, nor symmetric, nor reflexive.

Q-3: Check whether the relation given below is reflexive, symmetric and transitive:

The relation M is defined in the set {2, 3, 4, 5, 6, 7} as M = {(x, y): y = x + 1}.

Sol:

Let us assume that:

S {2, 3, 4, 5, 6, 7}.

The relation M is defined on set S as:

M = {(x, y): y = x + 1}

Therefore, M = {(2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}

We can observe that (2, 3) $\notin$ M, but (3, 2)  ɛ M.

Hence, M is not symmetric.

We need to find (x, x) $\notin$ M, where x  ɛ M.

(2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7) ɛ M

Hence, M is not reflexive.

(2, 3), (3, 4) ɛ M

But, (2, 4) $\notin$ M

Hence, M is not transitive.

Therefore, M is neither symmetric, nor reflexive, nor symmetric.

Q-4: Prove that the relation M in M which is defined as M = {(x, y): x ≤ y} is transitive and reflexive, but not symmetric.

Sol:

M = {(x, y): x ≤ y}

So, (x, x) ɛ M                   [as x = x]

Hence, M is reflexive.

Then, (4, 6) ɛ M (as 4 < 6)

But, (4, 6) $\notin$ M since, 6 is greater than 4.

Hence, M is not symmetric.

Let, (x, y), (y, z) ɛ M

Then, x ≤ y and y ≤ z

$\boldsymbol{\Rightarrow }$ x ≤ z

$\boldsymbol{\Rightarrow }$ (x, z) $\notin$ M

Hence, M is transitive.

Therefore, M is reflexive and transitive, but not symmetric.

Q-5: Check that whether the relation M in M which is defined as M = {(x, y): x ≤ y3} is transitive, reflexive and symmetric.

Sol:

M = {(x, y): x ≤ y}

We can see that ($\frac{1}{2}, \; \frac{1}{2}$) $\notin$

As, $\frac{1}{2} > \left (\frac{1}{2} \right)^{3}$

Hence, M is not reflexive.

(1, 2) ɛ M as 1 < 23 = 8

But, (2, 1) $\notin$ M (as 23 > 1)

Hence, M is not symmetric.

We have,

(3, $\frac{3}{2}$), ($\frac{3}{2}, \; \frac{6}{5}$), as 3 < $\left(\frac{3}{2} \right)^{3}$) and ($\frac{3}{2} < \left(\frac{6}{5} \right)^{3}$

But, (3, $\frac{6}{5}$) $\notin$ M as 3 > $\left(\frac{6}{5} \right)^{3}$

Hence, M is not transitive.

Therefore, M is neither symmetric, nor reflexive, nor transitive.

Q-6: Prove that the relation M from the set {2, 3, 4} which is given by M = {(2, 3), (3, 2)} is not reflexive nor transitive, but it is symmetric.

Sol:

As per the data given in the question, we have

M = {(2, 3), (3, 2)}

The set, say S, = {2, 3, 4}

Any relation M on the set S will be defined as M = {(2, 3), (3, 2)}

Thus, (2, 2), (3, 3), (4, 4) $\notin$ M

Hence, M is not reflexive.

We know that,

(2, 3) ɛ M and (3, 2) ɛ M.

Hence, M is symmetric.

Now,

Since, (2, 3) as well as (3, 2) ɛ M

But, (2, 2) $\notin$ M

Hence, M is not transitive.

Therefore, M is symmetric, but it is neither reflexive nor transitive.

Q-7: Prove that the relation M in the set S for all the books in a library of the college BET, the relation given for it is M = {(a, b): a and b have the same number of pages in the book} which is the equivalence relation.

Sol:

Let, the set S be the set of all the books in the library of the college BET.

Relation M = {(a, b): a and b have equal number of pages in the book}

M is reflexive as (a, a) ɛ M as x and x will have the equal number of pages in the book.

Let, (a, b) ɛ M

$\boldsymbol{\Rightarrow }$ a and b have the same/ equal number of pages.

$\boldsymbol{\Rightarrow }$ b and a will also have the equal number of pages in the book.

$\boldsymbol{\Rightarrow }$ As, (a, b) ɛ M. So, (b, a) ɛ M

Hence, M is symmetric.

Let, (a, b) ɛ M and (b, c) ɛ M

$\boldsymbol{\Rightarrow }$  Since, a and b have equal number of pages in the book so, b and c will also have the equal number of pages in the book.

$\boldsymbol{\Rightarrow }$  a and b will also have the equal number of pages.

$\boldsymbol{\Rightarrow }$  (a, b) ɛ M.

Hence, M is transitive.

Thus, M is symmetric, transitive and also, reflexive.

Therefore, M is an equivalence relation.

Q-8: Prove that the relation M of the set S = {2, 3, 4, 5, 6} which is given by M = {(x, y): | x – y | is even}, is an equivalence relation. Also, prove that all the elements are related to each other of the set {3, 5} and the elements of (2, 4, 6} are inter- related with each other. But, elements of {3, 5} and {2, 4, 6} are not related to each other nor their any of the elements are interlinked.

Sol:

As per the data given in the question, we have

S = {2, 3, 4, 5, 6, 7} and M = {(x, y): |x – y | is even}

So,

For any of the element x ɛ S, we have | x – x | = 0 and we know that 0 is an even number.

Hence, R is reflexive.

Let us assume that, (x, y) ɛ M.

$\boldsymbol{\Rightarrow }$  | x – y | is even

$\boldsymbol{\Rightarrow }$  – | -(x – y) | = | y – a | will also be even.

$\boldsymbol{\Rightarrow }$  (y, x) ɛ M.

Hence, M is symmetric.

Let us now assume that, (x, y) ɛ M and (y, z) ɛ M.

$\boldsymbol{\Rightarrow }$  | x – y | is even and | y – z | is even

$\boldsymbol{\Rightarrow }$  (x – y) is even and also, (y – z) is even.

$\boldsymbol{\Rightarrow }$  (x – z) = (x – y) + (y – z) will also be even.   [As we know that the sum of the two integers is even]

$\boldsymbol{\Rightarrow }$  | x – y | is even.

$\boldsymbol{\Rightarrow }$  (x, z) ɛ M

Hence, M is transitive.

Therefore, M is an equivalence relation.

As per the condition given in the question, all the elements are related to each other in the set {3, 5} as each of the elements in the set is odd. Therefore, the mod of the difference of any of the two elements will always be even.

Also, all the elements of the set {2, 4, 6} are inter- related as each and every element is an even number in this subset.

Also, elements of the subset {3, 5} and {2, 4, 6} are not related to each other in any way as all of the elements of {3, 5} are odd and all the elements of {2, 4, 6} are even. Therefore, the modulus of the difference of the elements (for each of the two subsets) won’t be even always, such that 2 – 3, 3 – 4, 2 – 5, 3 – 6, 4 – 3, 4 – 5, 5 – 2, 5 – 4, 5 – 6, 6 – 3 and 6 – 5 all are an odd numbers.

Q-9:  Prove that all the relation M of the set S = {a ɛ P : 0 ≤ a ≤ 12}, which is given by

(a) M = {(x, y): | x – y | is a multiple of 3}

(b) M = {(x, y): x = b} is an equivalence relation. Get all the sets of elements which are related to 1 in every case.

Sol:

S = {a ɛ P: 2 ≤ a ≤ 14} = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

(i)

M = {(x, y): | x – y | will be the multiple of 3}

Let, (x, y) ɛ M

$\boldsymbol{\Rightarrow }$  | x – y | will be the multiple of 3.

$\boldsymbol{\Rightarrow }$  |- (x – y)| = | y – x | will also be the multiple of 3.

$\boldsymbol{\Rightarrow }$  (y, x) ɛ M

Hence, M is symmetric.

For any of the element x ɛ S, we have (x, x) ɛ M as | x – x | = 0 which is a multiple of 3.

Hence, M is reflexive.

Let, (x, y), (y, z) ɛ M

$\boldsymbol{\Rightarrow }$  | x – y | will be the multiple of 3 and | y – z | will be the multiple of 3.

$\boldsymbol{\Rightarrow }$  (x – y) will be the multiple of 3 and (y – z) will be the multiple of 3.

$\boldsymbol{\Rightarrow }$  (x – z) = (x – y) + (y – z) will be the multiple of 3.

$\boldsymbol{\Rightarrow }$  | x – z | will be the multiple of 3.

$\boldsymbol{\Rightarrow }$  (x, z) ɛ M

Hence, M is transitive.

Therefore, M is an equivalence relation.

The set of elements related to 1 is {1, 4, 7, 10, 13} as:

| 1 -1 | = 0 which is a multiple of 0.

| 4 – 1 | = 0 which is a multiple of 3.

| 7 – 1 | = 6 which is a multiple of 3.

| 10 – 1 | = 9 which is a multiple of 3.

| 13 – 1 | = 12 which is a multiple of 3.

(b)

M = {(x, y): x = y}

Let, (x, y) ɛ M

$\boldsymbol{\Rightarrow }$  x = y

$\boldsymbol{\Rightarrow }$  y = x

$\boldsymbol{\Rightarrow }$  (y, x) ɛ M

Hence, M is symmetric.

For every element x ɛ M, as we have (x, x) ɛ M, as x = x.

Hence, M is reflexive.

Let, (x, y) Let, (x, y) ɛ M and (y, z) ɛ M

$\boldsymbol{\Rightarrow }$  x = y and y = z

$\boldsymbol{\Rightarrow }$  x = z

$\boldsymbol{\Rightarrow }$  (x, z) ɛ M

Hence, M is transitive.

Therefore, M is an equivalence relation.

All the elements in M which are related to 1 can be those elements from the set M which is equal to 1.

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

Q-10: Give an example for each of the relation, which is

(a) Symmetric but, neither transitive nor reflexive.

(b) Transitive but, neither reflexive nor symmetric.

(c) Symmetric and reflexive but, not transitive.

(d) Transitive and reflexive but, not symmetric.

(e) Transitive and symmetric but, not reflexive.

Sol:

(a) Let, S = {7, 8, 9}

A relation M in S as S = {(7, 8), (8, 7)}

The relation M is not reflexive because (7, 7),(8, 8), (9, 9) $\notin$ M.

(7, 8) ɛ M and also, (8, 7) ɛ M

Hence, M is symmetric.

Now,

(7, 8), (8, 7) ɛ M but, (7, 7) $\notin$ M

Hence, M is not transitive.

Therefore, the relation M is symmetric but, neither reflexive nor transitive.

(b) Consider a relation M which is defined as:

M = {(x, y): x < y}

For a ɛ M, we have (x, x) $\notin$ M as x won’t be less than itself ever.

Also, x = x

Hence, M is not reflexive.

As per the given condition, x < y

Let us take an e.g.(2, 3)  ɛ M since, 2 < 3.

But, 3 won’t ever be less than 2.

Therefore, (3, 2) $\notin$ M

Hence, M is not symmetric.

Let, (x, y), (y, x) ɛ M

$\boldsymbol{\Rightarrow }$  x < y and y < z

$\boldsymbol{\Rightarrow }$  x < z

$\boldsymbol{\Rightarrow }$  (x, z)  ɛ M

Hence, M is not transitive.

Therefore, the relation R is neither symmetric nor it is reflexive but, it is transitive.

(c) Let, S = {2, 4, 6}

Let us define the relation M on S as

S = {(2, 2), (4, 4), (6, 6), (2, 4), (4, 2), (4, 6), (6, 4)}

The relation S will be reflexive in this case as for each x ɛ S, (x, x) ɛ M

i.e., {(2, 2), (4, 4), (6, 6)} ɛ M

The relation S will be symmetric in this case as (x, y) ɛ M

$\boldsymbol{\Rightarrow }$  (y, x) ɛ M for each x, y ɛ M

The relation S won’t be transitive in this case as (2, 4), (4, 6) ɛ M, but (2, 6) $\notin$ M.

Therefore, the relation M is not transitive but, it is symmetric and reflexive.

(d) Let us define any relation S in S as,

S = {(x, y): x3 ≥ y3}

(x, x) ɛ M              since, x3 = x3

Hence, M is reflexive.

(3, 2) ɛ M              since, 33 ≥ 23

But, (2, 3) $\notin$ M          since, 23 ≤ 33

Hence, M is not symmetric.

Let, (x, y), (y, z) ɛ M

$\boldsymbol{\Rightarrow }$  x3 ≥ y3  and y3 ≥ z3

$\boldsymbol{\Rightarrow }$  x3 ≥ z3

$\boldsymbol{\Rightarrow }$  (x, z)  ɛ M

Hence, M is transitive.

Therefore, the relation M is not symmetric but, it is transitive and reflexive.

(e) Let, S = {-6, -7}

Let us define a relation M in S as,

S = {(-6, -7), (-7, -6), (-6, -6)}

The relation M is not reflexive as (-7, -7) $\notin$ M

The relation M is symmetric as (-6, -7) ɛ M and (-7, -6) ɛ M

We can see that,

(-6, -7), (-7, -6) ɛ M and also, (-6, -6) ɛ M

Hence, the relation M is transitive also.

Therefore, the relation M is symmetric as well as transitive but, it is not reflexive.

Q-11: Prove that the relation A in the set S for the points in the plane which is given by A = {(M, N): The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0)}, is an equivalence relation. Now, also prove that the set of all the points which is related to the point M ≠ (0, 0) is a circle which is passing from the point P with having centre at origin.

Sol:

A = {(M, N): The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0)}

(M, M) ɛ A, as the distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0).

Hence, R is reflexive.

Let, (M, N) ɛ A

$\boldsymbol{\Rightarrow }$  The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0).

$\boldsymbol{\Rightarrow }$  The distance between the point N from (0, 0) will be the same as the distance between the point M from (0, 0).

$\boldsymbol{\Rightarrow }$  (N, M)  ɛ A

Hence, A is symmetric.

Let, (M, N), (N, P) ɛ A

$\boldsymbol{\Rightarrow }$  The distance between the points M and N from the origin is the same, then also, the distance between the points N and P from the origin is the same.

$\boldsymbol{\Rightarrow }$  The distance of the points M and P is the same from the origin.

$\boldsymbol{\Rightarrow }$  (M, P) ɛ A

Hence, the relation A is transitive.

Thus, the relation A is an equivalence relation.

The set of all the points which are related to M ≠ (0, 0) can be those points which have the same distance from the origin as the distance of the point M from (0, 0).

Simply, If O be the origin and OM = x, then the set of all of the points which are related to the point M is at the distance x from the origin point.

Therefore, the set of points hence forms a circle which having the centre at the origin and this circle is passing through the point M.

Q-12: Prove that the relation A which is defined in the set S for all the triangles as A = {(P1, P2): P1 is similar to P2}, is an equivalence relation. Assume three right angled triangles, say, triangle P1 having sides 4, 5, 6, triangle P2 having sides 6, 14, 15 and triangle P3 having sides 7, 9, 11. Find which triangle among P1, P2 and P3 will be related?

Sol:

M = {(P1, P2): P1 is similar to P2}

M can be reflexive in this case as its obvious that, each triangle is similar to itself.

Now,

If (P1, P2) ɛ A, so P1 is completely similar to P2.

$\boldsymbol{\Rightarrow }$  P2  is similar to P1

$\boldsymbol{\Rightarrow }$  (P2, P1)  ɛ A

Hence, M is symmetric.

Let, (P1, P2), (P2, P3) ɛ A

$\boldsymbol{\Rightarrow }$  P1 is similar to P2 and also, P2 is similar to P3

$\boldsymbol{\Rightarrow }$  P1 is similar to P3

$\boldsymbol{\Rightarrow }$  (P1, P3)  ɛ A

Hence, M is transitive.

Therefore, R is an equivalence relation.

We can see that,

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

Hence, the corresponding sides of the triangle P1 and P3 is in the same proportion (ratio).

Thus, the triangle P1 will be similar to the triangle P3.

Therefore, the triangle P1 is inter- related to the triangle P3

Q-13: Prove that the relation M is defined in the set S of every polygon in such a way that M = {(R1, R2): R1 and R2 must have the equal number of sides}, is an equivalence relations. Find all the sets of all the elements in S which is related to the right angled triangle T having sides 4, 5 and 6.

Sol:

M = {(R1, R2): R1 and R2 must have the equal number of sides}

M will be reflexive in this case, as (R1, R2) ɛ M, since a same polygon has same number of the sides among itself.

Let, (R1, R2) ɛ M

$\boldsymbol{\Rightarrow }$  R1 and R2 has same number of its sides.

$\boldsymbol{\Rightarrow }$  R2 and R1 has same number of its sides.

$\boldsymbol{\Rightarrow }$  (R2, R1)  ɛ M

Hence, M is symmetric.

Let, (R1, R2), (R2, R3) ɛ M

$\boldsymbol{\Rightarrow }$  R1 and R2 has same number of its sides.

Also, R2 and R3 have same number of its sides.

$\boldsymbol{\Rightarrow }$  R1 and R3 has same number of its sides.

$\boldsymbol{\Rightarrow }$  (R1, R3)  ɛ M

Hence, M is transitive.

Therefore, M will be an equivalence relation.

All the elements in the set S is related to the right angled triangle (T) having sides 4, 5 and 6 are the polygon who have exactly 3 sides.

Therefore, all the elements in the set S are related to the right triangle T is the set of all of the triangles.

Q-14: Assume that, Q is the set of all of the lines in the XY- plane and M is the relation in Q which is defined as M = {(P1, P2): P1 is parallel to P2}. Prove that the relation M is an equivalence relation. Hence, find the set of all of the lines which are related to the line b = 2a + 4.

Sol:

M = {(P1, P2): P1 is parallel to P2}

M will be reflexive as any line can always be at least parallel to itself, i.e., (P1, P1) ɛ M.

Let, (P1, P2) ɛ M

$\boldsymbol{\Rightarrow }$  P1 is parallel to P2 and also, P2 will be parallel to P1

$\boldsymbol{\Rightarrow }$  (P2, P1) ɛ M

Hence, M is symmetric.

Let, (P1, P2), (P2, P3) ɛ M

$\boldsymbol{\Rightarrow }$  P1 is parallel to P2 and also, P2 will be parallel to P3

$\boldsymbol{\Rightarrow }$  P1 is parallel to P3

$\boldsymbol{\Rightarrow }$  (P2, P1)  ɛ M

Hence, M is transitive.

Therefore, M is an equivalence relation.

The set of all of the lines which are related to the given line which is b = 2a + 4 will be the set of all of the lines which are parallel to the line b = 2a + 4.

Here, slope of the given line, say m, b = 2a + 4 is 2.

We know that the slope of the parallel lines is the same.

The line which is parallel to the given line is in the form b = 2a + c, where c  ɛ M

Therefore, the set of all of the lines which are related to the given line which is given by b = 2a + c, where c ɛ M.

Q-15: Let, M be the given relation in the set S = {2, 3, 4, 5} which is given by-

M = {(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)}. Select the correct answer:

(a) M is symmetric and reflexive, but it is not transitive.

(b) M is transitive and reflexive, but it is not symmetric.

(c) M is transitive and symmetric, but it is not reflexive.

(d) M is an equivalence relation.

Sol:

Given relation is:

M = {(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)}

We can see that, (x, x) ɛ M, for each x ɛ {2, 3, 4, 5}

Hence, M is reflexive.

We can note that, (2, 3) ɛ M, but (3, 2) $\notin$ M

Hence, M is not symmetric.

We can observe that (x, y), (y, z) ɛ M

$\boldsymbol{\Rightarrow }$  (x, z)  ɛ M for each a, b, c ɛ {2, 3, 4, 5}.

Hence, M is transitive.

Therefore, M is transitive and reflexive, but it is not symmetric.

Q-16: Let us consider M be a relation in any set M which is given by

M = {(x, y): x = y – 2, y > 6}.

Select the correct choice from the following:

(a) (2, 4) ɛ M                  (b) (3, 8) ɛ M

(c) (6, 8) ɛ M                  (d) (8, 7) ɛ M

Sol:

M = {(x, y): x = y – 2, y > 6}

As, y > 6

So, (2, 4) $\notin$ M

3 ≠ 8 – 2 as well,

So, (3, 8) $\notin$ M

8 ≠ 7 – 2 as well,

So, (8, 7) $\notin$ M

Let us consider (6, 8)

6 > 8 and also, we have 6 = 8 – 2

Hence,

(6, 8) ɛ M

Therefore, the correct answer is C.

#### EXERCISE – 1. 2

Q-1: Prove that the function f: $R_{*}\rightarrow R_{*}$ which is defined by f(a) = $\frac{1}{a}$ which is one- one and onto, where $R_{*}$ is the set of all of the non- zero real numbers. Check whether the result is true or not, if the domain, say, $R_{*}$ is replaced by M having co- domain as same as $R_{*}$?

Sol:

As per the data given in the question,

$R_{*}\rightarrow R_{*}$ which is defined by f(a) = $\frac{1}{a}$

For one- one condition:

Let, a and b $\epsilon R_{*}$  such that, f(a) = f(b)

$\Rightarrow \frac{1}{a} = \frac{1}{b}$

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is one- one.

For onto condition:

Clearly, for b$\epsilon R_{*}$, there must exists a = $\frac{1}{b} \; \epsilon R_{*}$ [since, b ≠ 0], so that

f(x) = $\frac{1}{\frac{1}{b}}$ = y

Hence, the function f is onto.

Therefore, the given function f(a) is one- one as well as onto.

Let us consider a another function g: M → $R_{*}$ which is defined by g(a) = $\frac{1}{a}$.

g(x1) = g(x2)

$\Rightarrow \frac{1}{a_{1}} = \frac{1}{a_{2}}$

$\boldsymbol{\Rightarrow }$  x1 = x2

Hence, the another function g is also one- one.

It is obvious that g is not onto as for 1.2 $\epsilon R_{*}$, there doesn’t exist any  a in M so that:

g(a) = $\frac{1}{1.2}$

Therefore, the function ‘g’ is one- one but, it is not onto.

Q-2: Check the following functions for their injectivity and surjectivity:

(i)  f: N →N which is given by f(a) = a2

(ii)  f: Z →Z which is given by f(a) = a2

(iii) f: R → R which is given by f(a) = a2

(iv) f: N →N which is given by f(a) = a3

(v) f: Z → Z which is given by f(a) = a3

Sol:

(i) f: N → N which is given by f(a) = a2

We can see that for a, b ɛ N

f(a) = f(y)

$\boldsymbol{\Rightarrow }$  a= b2

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the a in N, so that f(a) = b2 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but, it is injective.

(ii) f: Z → Z which is given by f(a) = a2

We can see that for a, b ɛ Z

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

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ Z

But, there doesn’t exist any of the a ɛ Z, so that f(a) = – 2 or a2 = -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

(iii) f: R → R which is given by f(a) = a2

We can see that for a, b ɛ R

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

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ R

But, there doesn’t exist any of the a ɛ R, so that f(a) = – 2 or a2 = -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

(iv) f: N → N which is given by f(a) = a3

We can see that for a, b ɛ N

f(a) = f(y)

$\boldsymbol{\Rightarrow }$  a= b3

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the element of a  ɛ Z, so that f(a) = 2 or a3 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

(v) f: Z → Z which is given by f(a) = a3

We can see that for a, b ɛ Z

f(a) = f(y)

$\boldsymbol{\Rightarrow }$  a= b3

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is injective.

2 ɛ Z

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a3 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

Q-3: Show that the GIF (Greatest Integer Function) f: R →R which is given by f(a) = [a], which is neither one- one nor onto, where [a] notifies the greatest integer which must be less than or equal to a.

Sol:

f: R → R which is given by f(a) = [a]

f(1.4) = [1.4] = 1, f(1.7) = [1.7] = 1

$\boldsymbol{\Rightarrow }$  f(1.4) = f(1.7) = 1, but 1.4 ≠ 1.7

Hence, the function f is not one- one.

Let us consider that, 0.9 ɛ R.

We know that, f(a) = [a] will always be any integer.

So, there doesn’t exist any of the element a ɛ R, such that f(a) = 0.9

Hence, the function f is not onto.

Therefore, the GIF (Greatest Integer Function) is neither onto nor one- one.

Q-4: Prove that the Modulus Function f: R → R which is given by f(a) = [a], which is neither one- one nor it is onto, where |a| is a, if a will be positive or 0 and |a| is –a, if a will be negative.

Sol:

f : R → R which is given by

f(a) = |a| =  a, if a ≥ 0 and (-a), if a ≤ 0

It is obvious that f(-2) = |-2| = 2 and f(2) = |2| = 2

So, f(-2) = f(2), but -2 ≠ 2

Hence, f is not one- one

Let us consider -2 ɛ R.

We know that, f (a) = |a| will always be non- negative.

Therefore, neither of any element a exist in the domain R, so that

f(a) = |a| = -2.

Hence, f is not onto.

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

Q-5: Prove that the Signum function f: R → R which is given by f(a) =

1,   if a > 0

0,   if a = 0

-1,   if x < 0 is neither onto nor one- one.

Sol:

f: R → R which is given by f(a) =

1,   if a > 0

0,   if a = 0

-1,   if x < 0

Since, f(a) can only take 3 values (i.e., 1, 0, -1) for the elements -3 in the co- domain R, there does not exist any of the a in domain R, so that f(a) = -3.

Hence, f is not onto.

Here, we can see that:

f(2) = f(3) = 1, but 2 ≠ 3

Hence, f is not one- one.

Therefore, the Signum function in this case is neither onto nor one- one.

Q-6: Let us take P as {2, 3, 4}, Q as {5, 6, 7, 8} and let f = {(2, 5), (3, 6), (4, 7)} which are the function from P and Q. Prove that the function f is one- one.

Sol:

As per the data given in the question, we have

P = {2, 3, 4} and Q = {5, 6, 7, 8}

f : P → Q is defines as f = {(2, 5), (3, 6), (4, 7)}

Thus, f (2) = 5, f(3) = 6 and f(4) = 7

Now,

Here we can see that the images of the distinct elements of P under f are also distinct.

Therefore, the function f is not one- one.

Q-7. Check whether the function in each of the following conditions is one- one, onto or bijective. Justify your answer in each cases.

(i) f : R → R which is defines by f(a) = 4 – 5a

(ii) f : R → R which is defined by f(a) = 2 + a2

Sol:

(i) f : R → R which is defined as f(a) = 4 – 5a

Let, a1, a2  ɛ so that, f(a1) = f(a2)

$\boldsymbol{\Rightarrow }$  4 – 5a1 = 4 – 5a2

$\boldsymbol{\Rightarrow }$   – 5a1 =  – 5a2

$\boldsymbol{\Rightarrow }$  a1 = a2

Hence, the function f is one- one.

Now,

For any of the real number (b) in R, there exists $\frac{4 – b}{5}$ in R, so that

f ($\frac{4 – b}{5}$) = $4 – 5\left(\frac{4 – b}{5} \right)$

f ($\frac{4 – b}{5}$) = 4 – (4 – b)

f ($\frac{4 – b}{5}$) = b

Hence, the function f is onto.

Therefore, the function f is bijective.

(ii) f : R → R which is defined as f(a) = 2 + a2

Let, a1, a2  ɛ so that, f(a1) = f(a2)

$\boldsymbol{\Rightarrow }$  2 + a12 = 2 + a22

$\boldsymbol{\Rightarrow }$   a12 =  a22

$\boldsymbol{\Rightarrow }$  a1 = ± a2

Thus, f(a1) = f(a2) which doesn’t imply that a1 = a2

Let us illustrate an example, f(2) = f(-2) = 3

Hence, the function f is not one- one.

Now,

Let us take an element -3 in the co- domain R.

We can see that, f(a) = 2 + a2 which is positive for all of the x  ɛ R.

So, there doesn’t exist any a in the domain R so that f(a) = -3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor onto so, it is not bijective.

Q-8: Let us consider P and Q be the two sets. Prove that f : P × Q → Q × P so that (x, y) = (y, x) is bijective function.

Sol:

f : P × Q → Q × P which is defined as f(x, y) = y, x.

Let, (x1, y1), (x2, y2)  ɛ P × Q so that, f(x1, y1)=  f(x2, y2)

$\boldsymbol{\Rightarrow }$  (y1, x1) = (y2, x2)

$\boldsymbol{\Rightarrow }$  y1 =  y2  and x1 =  x2

$\boldsymbol{\Rightarrow }$  (x1, y1) = (x2, y2)

Hence, the function f is one- one.

Let us consider that (y, x) ɛ Q × P be any element.

Thus,

There will exist (x, y) ɛ P × Q so that,

f(x, y) = (y, x).

Hence, f is onto.

Therefore, the function f is bijective.

Q-9: Let f : N → N which will be defined by f(p) =

$\frac{p + 1}{2}$, if p is odd

$\frac{p}{2}$, if p is even, for all p  ɛ N.

Check and justify your answer that whether the function f is bijective or not?

Sol:

f : N → N which is defined as f(p) =

$\frac{p + 1}{2}$, if p is odd

$\frac{p}{2}$, if p is even

for all n ɛ N

We can see that:

f(3) = $\frac{3 + 1}{2}$ = 2 and f(4) = $\frac{4}{2}$ = 2

$\boldsymbol{\Rightarrow }$  f(3) = f(4), but 3 ≠ 4

Hence, f is not one- one.

Let us consider any natural number (p) in the co- domain N.

Case-1:  [ p is odd ]

$\boldsymbol{\Rightarrow }$  p = 2r + 1 for few r ɛ N.

Thus, there exists 4r + 1 ɛ N so that,

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

Case-2:  [p is even]

$\boldsymbol{\Rightarrow }$  p = 2r for few r ɛ N.

Thus, there exists 4r ɛ N so that,

f(4r) = $\frac{4r}{2}$ = 2r

Hence, the function f is onto.

Therefore, the function f is not a bijective function.

Q-10: Let P = A – {3} and Q = A – {1}. Let us consider the function f : P → Q which defined by f(a) = $\left(\frac{a – 3}{a – 4} \right)$. Check and Justify your answer, whether the function  f is onto and one- one?

Sol:

P = A – {3}, Q = A – {1} and f :  P → Q which is defined by

f(a) = $\left(\frac{a – 3}{a – 4} \right)$

Let, a, b  ɛ P so that f(a) = f(b)

$\boldsymbol{\Rightarrow }$  $\frac{a – 3}{a – 4} = \frac{b – 3}{b – 4}$

$\boldsymbol{\Rightarrow }$  (a – 3)(b – 4) = (b – 3)(a – 4)

$\boldsymbol{\Rightarrow }$  ab – 4a – 3b + 12 = ab – 4b – 3a + 12

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is one- one.

Let, b ɛ Q = A – {1}.

Thus, b ≠ 1

Then, the function f will be onto if a ɛ P so that, f(a) = b.

f(a) = b

$\boldsymbol{\Rightarrow }$  $\frac{a – 3}{a – 4}$ = b

$\boldsymbol{\Rightarrow }$  (a – 3) = b(a – 4)

$\boldsymbol{\Rightarrow }$  a – 3 = ab – 4b

$\boldsymbol{\Rightarrow }$  a(1 – b) = 3 – 4b

$\boldsymbol{\Rightarrow }$  a = $\frac{3 – 4b}{1 – b}$  ɛ [b ≠ 1]

Then,

For any value of b  ɛ Q, there will exist $\frac{3 – 4b}{1 – b}$  ɛ P, so that

f($\frac{3 – 4b}{1 – b}$) = $\frac{\left (\frac{3 – 4b}{1 – b} \right) – 3}{\left (\frac{3 – 4b}{1 – b} \right) – 4}$

= $\frac{\left(3 – 4b \right) – 3\left(1 – b \right)}{\left(3 – 4b \right) – 4\left(1 – b \right)}$

= $\frac{3 – 4b – 3 + 3b}{3 – 4b – 4 + 4b}$

= $\frac{-b}{-1}$ = b

Hence, the function f is onto.

Therefore, the function f is onto as well as it is one- one.

Q-11: Let us assume, f : R → R will be defined as g(a) = a5. Select the correct answer from the following:

(a) f is many- one onto                                                        (b) f is one- one onto

(c) f is neither one- one nor onto                                      (d) f is one- one but, not onto

Sol:

f : R → R which is defined as g(a) = a4.

Let, a, b  ɛ R so that f(a) = f(b).

$\boldsymbol{\Rightarrow }$  a4 = b4

$\boldsymbol{\Rightarrow }$  a = ± b

Thus, f(a) = f(b) doesn’t imply that a = b.

Example:

f(3) = f(-3) = 3

Hence, the function f is not one- one.

Let us consider an element 3 in the co- domain R. It’s obvious that there doesn’t exist any a in the domain R so that, f(a) = 3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor it is onto.

Thus, the correct answer is option (c).

Q-12: Let us assume, f : R → R will be defined as g(a) = 4a . Select the correct answer from the following:

(a) f is many- one onto                                                        (b) f is one- one onto

(c) f is neither one- one nor onto                                      (d) f is one- one but, not onto

Sol:

f : R → R which is defined as g(a) = 4a.

Let, a, b ɛ R so that f(a) = f(b).

$\boldsymbol{\Rightarrow }$  4a = 4b

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is one- one.

For any of the real number (b) in the co- domain R, there must exist $\frac{b}{3}$ in R so that, f($\frac{b}{3} = 3\left(\frac{b}{3} \right) = y$)

Hence, the function f is onto.

Therefore, the function f is one- one and also, it is onto.

Thus, the correct answer is option (b).

#### EXERCISE – 1.3

Q-1: Let us consider, g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will begiven by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}. Find fog.

Sol:

The given functions g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will be defined by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}

fog(3) = f[g(3)] = f(4) = 5                                        [as, g(3) = 4 and f(4) = 5]

fog(5) = f[g(5)] = f(7) = 3                                        [as, g(5) = 7 and f(7) = 3]

fog(6) = f[g(6)] = f(3) = 5                                        [as, g(6) = 3 and f(3) = 5]

Hence,

fog = {(3, 5), (5, 3), (6, 5)}

Q-2: Let us consider

f, g and h be the functions from R to R. Prove that:

(g + f)oh = goh + foh

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

Sol:

We need to prove that,

(g + f)oh = foh + goh

$\boldsymbol{\Rightarrow }$  {(g + f)oh}(x) = {goh + foh}(x)

Now,

LHS = {(g + f)oh}(x)

= (g + f)[h(x)]

= g[h(x)] + f[h(x)]

= (goh)(x) + (foh)(x)

= {(goh) + (foh)}{x)

= RHS

Therefore, {(g + f)oh} = goh + foh

Hence, proved.

We need to prove:

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

LHS = [(g.f)oh](x)

= (g.f)[h(x)]

= g[h(x)]. f[h(x)]

= (goh)(x) . (foh)(x)

= {(goh).(foh)}(x)

= (goh).(foh)

Hence, LHS = RHS

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

Q-3: Find fog and gof, if

(i) g(a) = |a| and f(a) = |5a – 2|

(ii) g(a) = 8a3 and f(a) = $x^{\frac{1}{3}}$

Sol:

(i) g(a) = |a| and f(a) = |5a – 2|

Therefore, the binary operation * is not associative.

fog(a) = f(g(a)) = f(|a|) = $\left | 5\left | x \right | – 2 \right |$

Therefore,  gof(a) = g(f(a)) = g(|5x – 2|) = $\left |\left | 5x – 2 \right | \right |$ = $\left | 5x – 2 \right |$

(ii) g(a) = 8a3 and f(a) = a$\frac{1}{3}$

Therefore, fog(a) = f(g(a)) = f(8a3) = (8a3)$\frac{1}{3}$ = (23 a3) $\frac{1}{3}$ = 2a

Therefore, gof(a) = g(f(a)) = g($a^{\frac{1}{3}}$) = 8($a^{\frac{1}{3}}$)3 = 8a

Q-4: If g(a) = $\frac{\left(4a + 3 \right)}{\left(6a – 4 \right)}$, a ≠ $\frac{2}{3}$. Prove that gog(a) = a, for every a ≠ $\frac{2}{3}$. What will be the inverse of the function g?

Sol:

As per the data given in the question, we have

g(a) = $\frac{4a + 3}{6a – 4}$, a ≠ $\frac{2}{3}$

(gog)(x) = g(g(x)) = g($\frac{4a + 3}{6a – 4}$) = $\frac{4\left (\frac{4a + 3}{6a – 4} \right) + 3}{6 \left (\frac{4a + 3}{6a – 4} \right) – 4}$

= $\frac{16a + 12 + 18a – 12}{24a + 18 – 24a + 16}$

= $\frac{34a}{34}$

Therefore, gog(a) = a, for all of the a ≠ $\frac{2}{3}$

So, gog = Ia

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

Q-5: Explain with suitable reason that, whether the following functions has any inverse

(i) g : {2, 3, 4, 5} → {11} with

g = {(2, 11), (3, 11), (4, 11), (5, 11)}

(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} with

f = {(6, 5), (7, 4), (8, 5), (9, 3)}

(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} with

h = {(3, 8), (4, 10), (5, 12), (6, 14)}

Sol:

(i) g : {2, 3, 4, 5} → {11} which is defined as g = {(2, 11), (3, 11), (4, 11), (5, 11)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

g(2) = g(3) = g(4) = g(5) = 11

Hence, the function g is not one- one.

Therefore, in this case, the function g doesn’t have any inverse.

(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} which is defined as f = {(6, 5), (7, 4), (8, 5), (9, 3)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

f(6) = f(7) = 4

Hence, the function f is not one- one.

Therefore, in this case, the function f doesn’t have any inverse.

(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} which is defined as

h = {(3, 8), (4, 10), (5, 12), (6, 14)}

Here, we can see that, all the distinct elements from the set {3, 4, 5, 6} have distinct images under h.

Hence, the function h is one- one.

Also, the function h is onto as each element y from the set {8, 10, 12, 14}, there exists an element, say a, in the set {3, 4, 5, 6} so that h(a) = b.

Hence, h is one- one as well as onto function.

Therefore, the function h will have an inverse.

Q-6: Prove that:

g : [-2, 2] → R,

which is given by g(a) = $\frac{a}{\left(a + 2 \right)}$ will be one- one. What will be the inverse of the function g : [-2, 2] → Range g.

Sol:

g : [-2, 2] → R which is given by

g(a) = $\frac{a}{\left(a + 2 \right)}$

For one- one:

Let, g(a) = g(b)

$\boldsymbol{\Rightarrow }$  $\frac{a}{\left(a + 2 \right)}$ = $\frac{b}{\left(b + 2 \right)}$

$\boldsymbol{\Rightarrow }$  a(b + 2) = b (a + 2)

$\boldsymbol{\Rightarrow }$  ab + 2a = ab + 2b

$\boldsymbol{\Rightarrow }$  2a = 2b

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function g is one- one function.

We can see that, g : [-2, 2] → Range g is onto.

Hence, g : [-2, 2] → Range g  here is one- one and onto, and thus, the inverse of the function f : [-2, 2] → Range g exists.

Let us consider f : Range g → [-2, 2] be the inverse for g.

Let, b be any arbitrary element of the range f.

As, f : [-2, 2] → Range g is onto, here we have

b = g(a) for some values of a  ɛ [-2, 2]

$\boldsymbol{\Rightarrow }$  b = $\frac{a}{\left(a + 2 \right)}$

$\boldsymbol{\Rightarrow }$  b(a + 2) = a

$\boldsymbol{\Rightarrow }$  ab + 2b = a

$\boldsymbol{\Rightarrow }$  a(1 – b) = 2b

$\boldsymbol{\Rightarrow }$  a = $\frac{2b}{\left(1 – b \right)}$, b ≠ 1

Then,

Let we define f: Range g → [-2, 2] as

f(b) = $\frac{2b}{\left(1 – b \right)}$, b ≠ 1

Thus,

(fog)(a) = f(g(a)) = f$\left(\frac{a}{a + 2} \right)$= $\frac{2\left (\frac{a}{a + 2} \right)}{1 – \left (\frac{a}{a + 2} \right)} = \frac{2a}{a + 2 – a} = \frac{2a}{2} = a$

and,

(gof)(a) = g(f(a)) = g $\left (\frac{2b}{1 – b} \right) = \frac{\left (\frac{2b}{1 – b} \right)}{\left (\frac{2b}{1 – b} \right) + 2} = \frac{2b}{2b + 2 – 2b} = \frac{2b}{2} = b$

Hence,

fog = a = $I_{\left \lfloor -1, 1 \; \right \rfloor}$ and gof = b = IRange f

Thus,

g-1 = f

g-1(b) = $\frac{2b}{2}$, b ≠ 1.

Q-7: Let us consider g : R → R which is given by g(a) = 4a + 3. Prove that the function g is invertible. Also find the inverse of g.

Sol:

g : R → R which is given by,

g(a) = 4a + 3

For one- one function,

Let, g(a) = g(b)

$\boldsymbol{\Rightarrow }$  4a + 3 = 4b + 3

$\boldsymbol{\Rightarrow }$  4a = 4b

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function g is a one- one function.

For onto function,

For b  ɛ R, let b = 4a + 3.

$\boldsymbol{\Rightarrow }$  a = $\frac{b – 3}{4}$  ɛ R.

Hence, for every b  ɛ R, there does exist a = $\frac{b – 3}{4}$ ɛ R, so that

g(a) = g $\left (\frac{b – 3}{4} \right)$ = 4 $\left (\frac{b – 3}{4} \right)$ + 3 = b

Hence, the function g is onto.

Therefore, the function g is one- one and onto and hence, g-1 will exist.

Let us consider, f: R → R by f(x) = $\frac{b – 3}{4}$

Then,

(fog)(a) = f(g(a)) = f(4a + 3) = $\frac{\left (4a + 3 \right) – 3}{4} = \frac{4x}{4} = x$

And,

(gof)(b) = g(f(x)) = g$g\left (\frac{b – 3}{4} \right) = 4\left (\frac{b – 3}{4} \right) + 3 = b – 3 + 3 = b$

So, fog = gof = IR

Therefore, the function f will be invertible and have inverse which is given by g-1(b) = f(b) = $\frac{b – 3}{4}$

Q-8: Let us consider a function, g : R+ → [4, ) which is given by g(a) = a2 + 4. Prove that the function g is invertible with the inverse g-1 of the given function g by g-1 (b) = $\sqrt{b – 4}$, where R+ will be the set of all the non- negative real number.

Sol:

g : R+ → [4, ∞) which is given by g(a) = a2 + 4

For one- one function,

Let, g(a) = g(b)

$\boldsymbol{\Rightarrow }$  a2 + 4 = b2 + 4

$\boldsymbol{\Rightarrow }$  a2 = b2

$\boldsymbol{\Rightarrow }$  a = b

Hence, the function f is one- one function.

For onto function,

For b  ɛ [4, ∞), let b = a2 + 4

$\boldsymbol{\Rightarrow }$  a2 = b – 4 ≥ 0

$\boldsymbol{\Rightarrow }$  a = $\sqrt{b – 4}$ ≥ 0

Hence, for every b  ɛ [4, ∞), there exists a = $\sqrt{b – 4}$  ɛ R+, so that

G(a) = g($\sqrt{b – 4}$) = ($\sqrt{b – 4}$)2  + 4 = b – 4 + 4 = b

Hence, the function g is onto.

So, the function g is one- one and onto and hence, g-1 exists.

Let we define, f : [4, ∞) → R+ by f(b) = $\sqrt{b – 4}$

Then,

(fog)(a) = f(g(a)) = f($x^{2} + 4$)  = $\sqrt{\left (a^{2} + 4 \right) – 4} = \sqrt{a^{2}}$ = a

And,

(gof)(b) = g(f(b)) = g$\left (\sqrt{b – 4} \right)$ = $\left (\sqrt{b – 4} \right)^{2}$ + 4 = b – 4 + 4 = b

Hence, gof = fog = IR

Therefore, the function g is invertible and it will be the inverse of g which is given by g-1(b) = f(b) = $\sqrt{b – 4}$

Q-9: Let us consider a function, g: R+ → [-5, ∞) which is given by g(a) = 9a2 + 6a – 5. Prove that, the function g is invertible with g-1(b) = $\left (\frac{\left (\sqrt{b + 6} \right) – 1}{3} \right)$.

Sol:

g: R+ → [-5, ∞) which is given by g(a) = $9a^{2} + 6a – 5$

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

Let,

b = $9a^{2} + 6a – 5$

$\boldsymbol{\Rightarrow }$  b = (3a + 1)2 – 1 – 5 = (3a + 1)2 – 6

$\boldsymbol{\Rightarrow }$  b + 6 = (3a + 1)2

$\boldsymbol{\Rightarrow }$  3a + 1 = $\sqrt{b + 6}$   [as b ≥ -5 $\boldsymbol{\Rightarrow }$  b + 6 > 0]

$\boldsymbol{\Rightarrow }$  a = $\frac{\left (\sqrt{b + 6} \right) – 1}{3}$

Hence, the function g is onto, which means having range g = [-5, ∞)

Let us consider, f : [-5, ∞) → R+ as f(b) = $\frac{\left (\sqrt{b + 6} \right) – 1}{3}$

Then,

(fog)(a) = f(g(a)) = f(9a2 + 6a – 5) = g ((3a + 1)2 – 6) = $\frac{\sqrt{\left (3a + 1 \right)^{2} – 6 + 6} – 1}{3}$

=$\frac{3a + 1 – 1}{3}$ = $\frac{3a}{3}$ = a

And,

(gof)(b) = g(f(b)) = g$\left (\frac{\sqrt{b + 6} – 1}{3} \right) = \left [3 \left (\frac{\sqrt{b + 6} – 1}{3} \right) + 1 \right]^{2} – 6$

= $\left (\sqrt{b + 6} \right)^{2} – 6$ = b + 6 – 6 = b

Hence, fog = a = IR and gof = b = IRange g

Therefore, g is invertible and the inverse of g will be given by:

g-1(b) = f(b) = $\left (\frac{\left (\sqrt{y + 6} \right)^{2} – 1}{3} \right)$

Q-10: Let us consider a function g : P → Q be an invertible function. Prove that the function g have a unique inverse.

Sol:

Let, the function, g : P → Q be an invertible function.

Let, function g have two inverses (say, g1 and g2)

Thus, for every b ɛ Q, we have,

gof1 (b) = IY(b) = gof2(b)

$\boldsymbol{\Rightarrow }$  g(f1(b)) = g(f2(b))

$\boldsymbol{\Rightarrow }$  f1(b) = f2(b)     [as g is invertible $\boldsymbol{\Rightarrow }$  g is on- one]

$\boldsymbol{\Rightarrow }$  f1= f2    [as f is one- one]

Therefore, the function g has a unique inverse.

Q-11. Let us consider the function g: {2, 3, 4} → {x, y, z}, generally given by g(2) = x, g(3) = y, g(4) = z. Find g-1 and prove that (g-1)-1 = g.

Sol:

The given function g : {2, 3, 4} → {x, y, z} which is given by g(2) = , g(3) = y, and g(4) = z

If we will define f: {x, y, z} → {2, 3, 4}, since f(x) = 2, f(y) = 3, f(z) = 4.

(gof)(x) = g(f(x)) = g(2) = x

(gof)(y) = g(f(y)) = g(3) = y

(gof)(z) = g(f(z)) = g(4) = z

And,

(fog)(2) = f(g(2)) = f(x) = 2

(fog)(3) = f(g(3)) = f(y) = 3

(fog)(4) = f(g(4)) = f(x) = 4

Thus,

fog = IX  and gof = IY, such that P = {2, 3, 4} and Q = {x, y, z}

Hence, the inverse of the function g exists and g-1 = f

Thus,

g-1 :{x, y, z} → {2, 3, 4} which is given by g-1(x) = 2, g-1(y) = 3 and g-1(z) = 4

Let us now obtain the inverse for g-1, i.e., finding the inverse of the function f.

Let us define h : {2, 3, 4} → {x, y, z} such that h(2) = x, h(3) = y and h(4) = z

Now,

(foh)(2) = f(h(2)) = f(x) = 2

(foh)(3) = f(h(3)) = f(x) = 3

(foh)(4) = f(h(4)) = f(x) = 4

And,

(hof)(x) = h(f(x)) = h(2) = x

(hof)(y) = h(f(y)) = h(3) = y

(hof)(z) = h(f(z)) = h(4) = z

Hence, foh = IX and hof = IY, P = {2, 3, 4} and Q = {x, y, z}.

So, the inverse of the function of f exists and f-1 = h $\boldsymbol{\Rightarrow }$  (g-1)-1 = h

We can see that, h = g

Hence, (f-1)-1 = f.

Q-12: Let us consider g : P → Q be an invertible function. Prove that the inverse of g-1 is g, i.e., (g-1)-1 = g.

Sol:

As per the given condition,

g : P → Q is an invertible function.

Thus, there exists a function say, f : Q → P so that, fog = IP and gof = IQ

So, g-1 = f

Then, fog = IP and gof = IQ

$\boldsymbol{\Rightarrow }$  g-1og= IP and fof-1 = IQ

Therefore, g-1 : Q → P will be invertible and the function g is the inverse of g-1, i.e., (g-1)-1 = g

Q-13: If g : R → R is given by g(a) = $\left (3 – a^{3} \right)^{\frac{1}{3}}$, then gog(a) is

(a) $\frac{1}{a^{3}}$

(b) a3

(c) a

(d) (3 – a3)

Sol:

g : R → R which is given by g(a) = $\left (3 – a^{3} \right)^{\frac{1}{3}}$

Then,

gog(a) = g(g(a)) = g$\left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)$ = $\left [3 – \left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)^{3} \right]^{\frac{1}{3}}$

= $\left [3 – \left (3 – a^{3} \right) \right]^{\frac{1}{3}}$ = $\left (a^{3} \right)^{\frac{1}{3}} = a$

Hence, gog(a) = a

Therefore, the correct answer is C.

Q-14: Let us consider a function g: R – $\left \{-\frac{4}{3} \right \}$ → R as g(a) = $\frac{4a}{3a + 4}$. The inverse of the function g is map f: Range g → R – $\left \{-\frac{4}{3} \right \}$ is given by

(a) f(b) = $\frac{3b}{3 – 4b}$     (b) f(b) = $\frac{4b}{4 – 3b}$

(c) f(b) = $\frac{4b}{3 – 4b}$      (d) f(b) = $\frac{3b}{4 – 3b}$

Sol:

As per the data given in the question, we have

g : R – $\left \{-\frac{4}{3} \right \}$ → R which is a function defined as g(a) = $\frac{4a}{3a + 4}$.

Let, b be the arbitrary element having range g.

Thus, there exists a ɛ R – $\left \{-\frac{4}{3} \right \}$ so that, b = f(a).

$\boldsymbol{\Rightarrow }$  b = $\frac{4a}{3a + 4}$

$\boldsymbol{\Rightarrow }$  b(3a + 4) = 4a

$\boldsymbol{\Rightarrow }$  3ab + 4b = 4a

$\boldsymbol{\Rightarrow }$  a = $\frac{4b}{4 – 3b}$

Let, f : Range g → R – $\left \{-\frac{4}{3} \right \}$ so that, f(b) = $\frac{4b}{4 – 3b}$

Thus,

fog(a) = f(g(a) = f$\left (\frac{4a}{3a + 4} \right) = \frac{4\left (\frac{4a}{3a + 4} \right)}{4 – 3\left (\frac{4a}{3a + 4} \right)}$

= $\frac{16a}{12a + 16 – 12a} = \frac{16a}{16}$ = a

And,

gof(b) = g(f(b)) = g$\left (\frac{4b}{4 – 3b} \right) = \frac{4\left (\frac{4b}{4 – 3b} \right)}{3\left (\frac{4b}{4 – 3b} \right) + 4}$

= $\frac{16b}{12b + 16 – 12b} = \frac{16b}{16} = b$

Thus, fog = IR – $\left \{- \frac{4}{3} \right \}$ and gof = IRange f

Therefore, the inverse of the function g is the map f: Range g → R – $\left \{- \frac{4}{3} \right \}$, which will be given by

g(b) = $\frac{4b}{4 – 3b}$

Hence, the correct answer is b.

#### EXERCISE – 1.4

Q-1: Check whether each of the following definitions gives a binary operation or not.

In each of the event, that * will not be a binary operation, give explanation for this condition.

(i) On Z+, define * by x * y = x – y

(ii) On Z+, define * by x * y = xy

(iii) On R, define * by x * y = xy2

(iv) On Z+, define * by x * y = |x – y|

(v) On Z +, define * by x * y = x

Sol:

(i) On Z+, * will be defined by x * y = x – y

It’s not a binary operation,

Such that the image of (2, 3) under * is 2 * 3 = 2 – 3 = -1 $\notin \; Z^{+}$

(ii) On Z+, * will be defined by x * y = xy.

We can see that for every x, y ɛ $\; Z^{+}$, there will be a unique element xy in Z+ .

It means that * carries every pair (x, y) from the unique element x * y = xy in Z+.

Hence, * is the binary operation.

(iii) On R, * will be defined as x * y = xy2.

We can see that for every x, y ɛ R, there will be a unique element xy2 in R.

It means that * will carry each pair (x, y) to a unique element x * y = xy2 in R.

Hence, * is a binary operation.

(iv) On Z+, * will be defined as x * y =|x – y|.

We can see that for every x, y ɛ Z+, there will be a unique element |x – y| in Z+.

It means that * will carry each pair (x, y) to a unique element x * y = |x – y| in Z+.

Hence, * is a binary operation.

(v) On Z+, * will be defined as x * y = x.

We can see that for every x, y ɛ Z+, there will be a unique element x in Z+.

It means that * will carry each pair (x, y) to a unique element x * y = x in Z+.

Hence, * is a binary operation.

Q-2: For every binary operations * which is defined below, check whether * is associative or commutative in the following cases.

(i) On Z, explain x * y = x – y

(ii) On Q, explain x * y = xy + 1

(iii) On Q, explain x * y = $\frac{xy}{2}$

(iv) On Z+, explain x * y = 2xy

(v) On Z+, explain x * y = xy

(vi) On R – {-1}, explain x * y = $\frac{x}{y + 1}$

Sol:

(i) On Z, * will be defined as x * y = x – y

(2 * 3) * 4 = (2 – 3)* 4 = -1 * 4 = -1 – 4 = -5

2*(3 * 4) = 2 *(3 – 4) = 2 * -1 = 2 – (-1) = 3

Thus,

(2 * 3) * 4 ≠ 2*(3 * 4), where 2, 3, 4 ɛ Z.

Therefore, the given operation * is not associative.

Now, also

We can observe that 2 * 3 = 2 – 3 = -1 and 3 * 2 = 3 – 2 = 1

Since, 2 * 3 ≠ 3 * 2, where 2, 3 ɛ Z

Therefore, the given operation * is not associative.

(ii) On Q, * will be defined as x * y = xy + 1

We can observe here that,

(2 * 3)* 4 = (2 × 3 + 1) * 3 = 7 * 3 = 7 × 3 + 1 = 22

2 *(3 * 4) = 2 * (3 × 4 + 1) = 2 * 13 = 2 × 13 + 1 = 27

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the given operation * is not associative.

Now, also

We know that: xy = ya for every x, y ɛ Q.

$\boldsymbol{\Rightarrow }$  xy + 1 = yx + 1 for every x, y  ɛ Q

$\boldsymbol{\Rightarrow }$  x * y = x * y for every x, y  ɛ Q

Therefore, the given operation * is commutative.

(iii) On Q, * will be defined by x * y = $\frac{xy}{2}$

For every x, y, z ɛ Q, we have

(x * y)* z = $\left (\frac{xy}{2} \right) * z = \frac{\left (\frac{xy}{2} \right)z}{2} = \frac{xyz}{4}$

And,

x *(y * z) = $x * \left (\frac{yz}{2} \right) = \frac{x \left (\frac{yz}{2} \right)}{2} = \frac{xyz}{4}$

Thus,

(x * y)* z = x *(y * z), where x, y, z ɛ Q

Hence, the given operation * is associative.

We know that, xy = yx for every x, y ɛ Q

$\boldsymbol{\Rightarrow }$  $\frac{xy}{2} = \frac{yx}{2}$ for every x, y  ɛ Q.

$\boldsymbol{\Rightarrow }$  x * y = y * x for every x, y  ɛ Q.

Hence, the given operation * is commutative.

(iv) On Z+, * will be defined by x * y = 2xy

For every x, y, z ɛ Q, we have

(1 * 3)* 4 = 21 × 3  * 4 = 8 * 4 = 28 × 4

And,

1 *(3 * 4) = 1 * 23 × 4 = 1 * 4096 = 21 × 4096

Thus,

(1 * 3)* 4 ≠ 1 *(3 * 4), where x, y, z ɛ Q

Hence, the given operation * is not associative.

We know that, xy = yx for every x, y ɛ Q

$\boldsymbol{\Rightarrow }$  2xy = 2yx for every x, y  ɛ Q.

$\boldsymbol{\Rightarrow }$  x * y = y * x for every x, y  ɛ Q.

Hence, the given operation * is commutative.

(v) On Z+, * will be defined by x * y = xy

We can see that,

For every x, y, z ɛ Z+, we have

(1 * 3)* 4 = 13 * 4 = 1 * 4 = 1 4 = 1

And,

1 *(3 * 4) = 1 * 3 4 = 1 * 81 = 181 = 1

Thus,

(1 * 3)* 4 = 1 *(3 * 4), where x, y, z ɛ Z+

Hence, the given operation * is associative.

We can also see that,

2 * 3 = 23 = 8 and 3 * 2 = 32 = 9

Hence, 2 * 3 ≠ 3 * 2, where x, y, z ɛ Z+

Therefore, the given operation * is not commutative.

(vi) On R, * – {-1} which is defined as x * y = $\frac{x}{y + 1}$

We can see that

(2 * 3)* 4 = $\left (\frac{2}{3 + 1} \right) * 4 = \left (\frac{2}{4} \right) * 4 = \frac{\frac{1}{2}}{4 + 1} = \frac{1}{2 \times 4} = \frac{1}{8}$

And,

2 *(3 * 4) = $2 * \left (\frac{3}{4 + 1} \right) = 2 * \left (\frac{3}{5} \right) = \frac{2}{\frac{3}{5} + 1} = \frac{1}{\frac{3 + 5}{5}} = \frac{5}{8}$

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ R – {-1}

Hence, the given operation * is not associative.

Q-3: Let us consider a binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.

Draw the operational table for the operation ^.

Sol:

As per the data given in the question, we have

A binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.

Then,

x, y ɛ {2, 3, 4, 5, 6}

Hence, the operational table for the operation ^ given in the question will be given as below:

 ^ 2 3 4 5 6 2 2 2 2 2 2 3 2 3 3 3 3 4 2 3 4 4 4 5 2 3 4 5 5 6 2 3 4 5 6

Q-4: Let us consider a binary operation * for the set {2, 3, 4, 5, 6} which is given by the multiplication table given below:

(i) Find the value of (3 * 4) * 5 and 3 *(4 * 5).

(ii) Check whether * is commutative.

(iii) Find the value of (3 * 4)*(5 * 6)

 * 2 3 4 5 6 2 2 2 2 2 2 3 2 3 2 2 3 4 2 2 4 2 2 5 2 2 2 5 5 6 2 3 2 5 6

Sol:

(i) (3 * 4)* 5 = 2 * 5 = 2

3 *(4 * 5) = 3 * 2 = 2

(ii) For each x, y ɛ {2, 3, 4, 5, 6}, we have x * y = y * x

Hence, the given operation is commutative.

(iii) (3 * 4)*(5 * 6) = 2 * 5 = 2

Q-5: Let us consider *’ be a binary operation for the set {2, 3, 4, 5, 6} which will be defined by x *’ y = HCF of x * y. Check whether the operation *’ is the same as the operation * which is defined and used previously? Also, do justify your answer.

Sol:

A binary operation for the set {2, 3, 4, 5, 6} which will be defined by x * y = HCF of x and y.

The operational table for the operation *’ is given as the table below:

 *’ 2 3 4 5 6 2 2 2 2 2 2 3 2 3 2 2 3 4 2 2 4 2 2 5 2 2 2 5 5 6 2 3 2 5 6

Here, we will observe that the operational table for the operation * and *’ is the same.

Hence, the operation *’ is the same as * operation.

Q-6: Let us consider a binary operation * on N which is given by x * y = LCM of x and y. Find the following:

(i) 6 * 7, 21 * 18                                                (ii) Is the operation * commutative?

(iii) Is the operation * associative?              (iv) Obtain the identity of * in N.

(v) Which of the element of N are invertible for the operation *?

Sol:

As per the data given in the question, we have

A binary operation * on N which is given by x * y = LCM of x and y.

(i) 6 * 7 = LCM of 6 and 7 = 42

21 * 18 = LCM of 21 and 18 = 126

(ii) We know that,

LCM of x and y = LCM of y and x for every x, y  ɛ N.

Hence, x * y = y * x

Therefore, the given operation * is commutative.

(iii) For x, y, z  ɛ N, we have

(x * y) * z = (LCM of x and y) * z = LCM of x, y and z

x * (y * z) = x * (LCM of y and z) = LCM of x, y and z

Hence,

(x * y) * z = x * (y * z)

Therefore, the given operation * is associative.

(iv) We know that:

LCM of x and 1 = LCM of 1 and x = x for all x  ɛ N

$\boldsymbol{\Rightarrow }$  x * 1 = x = 1 * x for all x  ɛ N

Therefore, 1 is an identity of * in N.

(v) The element x in N will be invertible with respect to the operation *, if there will exist an element y in N, so that x * y = e = y * x

Here, e = 1

This means,

LCM of x and y = 1 = LCM of y and x

This case will only be possible when x and y is equal to 1.

Therefore, 1 is the only element which is invertible to N with respect to the operation *

Q-7: Check whether the operation * is defined for the set {1, 2, 3, 4, 5} by x * y = LCM of x and y, is a binary operation? Also, do justify your answer.

Sol:

The operation * for the set R = {1, 2, 3, 4, 5} which is defines as x * y = LCM of x and y.

Thus,

The operational table for the given operation * will be given by:

 * 1 2 3 4 5 1 1 2 3 4 5 2 2 4 6 8 10 3 3 6 9 12 15 4 4 8 12 16 20 5 5 10 15 20 25

We can observe from the table defined above that,

2 * 3 = 3 * 2 = 6 $\notin$ R,

2 * 5 = 5 * 2 = 10 $\notin$ R,

3 * 4 = 4 * 3 = 12 $\notin$ R,

3 * 5 = 5 * 3 = 15 $\notin$ R,

4 * 5 = 5 * 4 = 20 $\notin$ R

Therefore, the operation * is not a binary operation.

Q- 8: Let us consider a binary operation * on N which is defined by x * y = HCF of x and y. Check whether * is commutative? Check whether * is associative? Check whether there exists any identity for the binary operation * on N?

Sol:

A binary operation * on N which is defined by x * y = HCF of x and y

We know that,

HCF of x and y = HCF of y and x for every x, y  ɛ N

Thus,

x * y = y * x

Hence, the binary operation * is commutative.

For x, y, z $\epsilon$, we have

(x * y)* z = (HCF of x and y)* z = HCF of x, y and z

x *(y * z) = x *(HCF of y and z) = HCF of x, y and z.

Hence, (x * y)* z = x *(y * z)

Thus, the binary operation * is associative.

Then,

An element e ɛ N is the identity for the binary operation * if, x * e = x = e * x for every x  ɛ N.

But, this is not true for any x ɛ N.

Therefore, the given binary operation * don’t have any identity in N.

Q-9: Let us consider a binary operation * for the set Q of the rational numbers for the following:

(i) x * y = x – y                                                           (ii) x * y = x2 + y2

(iii) x * y = x + xy                                                       (iv) x * y = (x – y)2

(v) x * y = $\frac{xy}{4}$               (vi) x * y = xy2

Determine which of the binary operations are commutative and which are associative.

Sol:

(i) On Q, the binary operation * is defined by x * y = x – y. It will be observed that:

$\frac{1}{3} * \frac{1}{4} = \frac{1}{3} – \frac{1}{4} = \frac{4 – 3}{12} = \frac{1}{12}$

And,

$\frac{1}{4} * \frac{1}{3} = \frac{1}{4} – \frac{1}{3} = \frac{3 – 4}{12} = -\frac{1}{12}$

Therefore, $\frac{1}{3} * \frac{1}{4} \neq \frac{1}{4} * \frac{1}{3}$, where, $\frac{1}{3}, \; \frac{1}{4}$  ɛ Q.

Hence, the binary operation * is not commutative.

It will be observed that,

$\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{1}{3} – \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{4 – 3}{12} \right) * \frac{1}{5} = \frac{1}{12} * \frac{1}{5} = \frac{1}{12} – \frac{1}{5} = \frac{5 – 12}{60} = \frac{-7}{60}$

And,

$\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{1}{4} – \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{5 – 4}{20} \right) = \frac{1}{3} * \frac{1}{20} = \frac{20 – 3}{60} = \frac{17}{20}$

Thus,

$\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)$, where $\frac{1}{3}, \frac{1}{4} \; and \; \frac{1}{4} \; \epsilon \; Q$

Therefore, the binary operation * is not associative.

(ii) On Q, the binary operation * is defined as x * y = x2 + y2.

For x, y ɛ Q, we have

x * y =  x2 + y2 = y2 + x2 = y * x

$\boldsymbol{\Rightarrow }$  x * y = y * x

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (22 + 32) * 4 = (4 + 9)* 4 = 13 * 4 = 132 + 42 = 185,

And,

2 * (3 * 4) = 2 * (32 + 42) = 2 * (3 + 4) = 2 * 7 = 22 + 72 = 53

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

(iii) On Q, the binary operation * which is defined by x * y = x + xy

It will be observed that,

2 * 3 = 2 + 2 × 3 = 2 + 6 = 8

3 * 2 = 3 + 3 × 2 = 3 + 6 = 9

Thus, the binary operation * is not commutative.

It will be observed that,

(2 * 3)* 4 = (2 + 2 × 3)* 4 = (2 + 6)* 4 = 8 * 4 = 8 + 8 × 4 = 40

And,

2 * (3 * 4) = 2 * (3 + 3 × 4) = 2 * (3 + 12) = 2 * 15 = 2 + 2 × 15 = 32

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the binary operation * is not associative.

(iv) On Q, the binary operation * will be defined by x * y = (x – y)2

For x, y ɛ Q, we have

x * y = (x – y)2

y * x = (y – a)2 = [-(x – y)]2 = (x – y)2

Hence, the binary operation * is commutative.

We can also observe that,

(2 * 3)* 4 = (2 – 3)2 * 4 = 1 * 4 = (1 – 4)2 = (-3)2 = 9

And,

2 * (3 * 4) = 2 * (3 – 4)2 = 2 * 1 = (2 – 1)2 = (-1)2 = 1

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the binary operation * is not associative.

(v) On Q, the binary operation * will be defined as x * y = $\frac{xy}{4}$

For x, y  ɛ Q, we will get

x * y = $\frac{xy}{4}$ = $\frac{yx}{4}$ = y * x

Hence, x * y = y * x

Therefore, the binary operation * is commutative.

For x, y, z $\epsilon$Q, we will get

(x * y) * z = $\left (\frac{xy}{4} \right) * z = \frac{\left (\frac{xy}{4} \right). z}{4} = \frac{xyz}{16}$

And,

x * (y * z) = $x * \left (\frac{yz}{4} \right) = \frac{x * \left (\frac{yz}{4} \right)}{4} = \frac{xyz}{16}$

Hence, (x * y) * z = x * (y * z), where x, y, z $\epsilon$

Therefore, the binary operation * is associative.

(vi) On Q, the binary operation * will be defined as x * y = xy2.

We can observe that,

$\frac{1}{3} * \frac{1}{4} = \frac{1}{3}.\left (\frac{1}{4} \right)^{2} = \frac{1}{3} . \frac{1}{16} = \frac{1}{48}$

And,

$\frac{1}{4} * \frac{1}{3} = \frac{1}{4}.\left (\frac{1}{3} \right)^{2} = \frac{1}{4} . \frac{1}{9} = \frac{1}{36}$

Hence, $\frac{1}{3} * \frac{1}{4}$ ≠ $\frac{1}{4} * \frac{1}{3}$, where $\frac{1}{3} \; and \; \frac{1}{4}$  ɛ Q.

Therefore, the binary operation * is not commutative.

We can also observe that:

$\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left [\frac{1}{3}\left (\frac{1}{4} \right)^{2} \right] * \frac{1}{5} = \frac{1}{48} * \frac{1}{5} = \frac{1}{48} * \left (\frac{1}{5} \right)^{2} = \frac{1}{48 \times 25} = \frac{1}{1200}$

And,

$\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3}\left [\frac{1}{4}\left (\frac{1}{5} \right)^{2} \right] = \frac{1}{3} * \frac{1}{100} = \frac{1}{3} * \left (\frac{1}{100} \right)^{2} = \frac{1}{3 \times 10000} = \frac{1}{30000}$

Hence, $\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} ≠ \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)$, where $\frac{1}{3}, \frac{1}{4}, \frac{1}{5} \; \epsilon \; Q$

Therefore, the binary operation * is not associative.

Thus, the operations which are defined in (ii), (iv), (v) are commutative and the operation defined in (v) is only associative.

Q-10: Check the identity of the operations which are equated above.

Sol:

Let us consider an element e ɛ Q which can be the identity element for the operation * if,

x * e = x = e * x, for every x ɛ Q.

Here, among each of the six operations, there is no such element e ɛ Q, which satisfies the above condition.

Hence, none of the six operations evaluated above have identity.

Q-11: Let us consider M = N × N and * be the binary operation on M which can be defined by

(p, q)* (r, s) = (p + r, q + s).

Prove that the operation * is commutative and associative. If there is any possibility of having an identity element, then what will be the identity element for the operation * on M?

Sol:

M = N × N and * be the binary operation on M which can be defined by

(p, q)* (r, s) = (p + r, q + s)

Let, (p, q), (r, s) ɛ M

Then,

p, q, r, s  ɛ N

Here, we have

(p, q)*(r, s) = (p + r, q + s)

(r, s)*(p, q) = (r + p, s + q) = (p + r, q + s)

[Addition will be commutative in the set of the natural numbers]

Thus, (p, q)*(r, s) = (r, s)*(p, q)

Hence, the operation * is commutative here.

Let, (p, q), (r, s), (t, u) ɛ M

Then, p, q, r, s, t, u ɛ N

Now,

[(p, q)*(r, s)] * (t, u) = (p + r, q + s)*(t, u) = (p + r + t, q + s + u)

And,

(p, q) * [(r, s) * (t, u)] = (p, q)*(r + t, s + u) = (p + r + t, q + s + u)

Thus, [(p, q)*(r, s)] * (t, u) = (p, q) * [(r, s) * (t, u)]

Hence, the operation * is associative.

Let us consider an element e = (e1, e2) ɛ M which can be an identity element for * operation, if

x * e = x = e * x, for every x = (x1, x2)  ɛ M

Then,

(x1 + e1, x2 + e2) = (x1, x2) = (e1 + x1, e2 + x2),

This cannot be true for every element in M.

Hence, the operation * will not have any identity element.

Q-12: Check whether the statements given below are true or false. Also, justify your answer.

(i) For any arbitrary binary operation * for the set N, x * x = x  for every x  ɛ N.

(ii) If the operation * is a commutative binary operation for N, then x *(y * z) = (z * y)* x.

Sol:

(i) Explain a binary operation * on N as x * y = x + y for every x  ɛ N.

Thus, for y = x = 4, we have

4 * 4 = 4 + 4 = 8 ≠ 3.

Hence, the statement (i) is false.

(ii) RHS = (z * y)* x

= (y * z)* x                             [* is commutative]

= x * (y * z)                            [Again, as * is commutative]

= LHS

Therefore, x *(y * z) = (z * y)* x

Hence, statement (ii) is true.

Q-13: Let us consider a binary operation * for N defined as x * y = x3 + y3. Select the correct answer.

(a) Is * commutative but, not associative?

(b) Is * both commutative and associative?

(c) Is * neither commutative nor associative?

(d) Is * associative but, not commutative?

Sol:

On N, the binary operation * will be defined as x * y = x3 + y3.

For x, y  ɛ N, we have

x * y = x3 + y3 =  y3 + x3 = y * x                         [Addition is commutative in N]

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (23 + 33)* 4 = (8 + 27)* 4 = 35 * 4 = 353 + 44 = 42875 + 16 = 42891

And,

2 *(3 * 4) = 2 *(33 + 43) = 2 *(27 + 64) = 2 * 91 = 23 + 913 = 8 + 753571 = 753579

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4  ɛ N

Thus, the operation * is not associative.

Therefore, the operation * is commutative but, not associative.

Hence, the correct statement is a.

#### Practise This Question

The capacity of a soda can be calculated using the formula.