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# NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions

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

Last updated date: 24th Mar 2023
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NCERT Solutions for Class 12 Maths Chapter-1 PDF is now available for download on the official website of Vedantu. Students will get the solutions to all the questions given in this NCERT chapter in the relation and function class 12 PDF. All the topics covered in class 12 Maths Chapter 1 are properly discussed here. Subject-matter experts at Vedantu have prepared these Relations and Functions Class 12 NCERT Solutions. Every solution has been worked out in a simple step by step manner so that students can understand it easily.

Download and go through the CBSE NCERT Solutions for Class 12 Maths Chapter 1 PDF for a better understanding of the sums. If you have doubts concerning CH 1 Maths class 12, you can go to the website and drop in your queries. Our experts will assist you with your doubts, in the best possible way.

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions provides solutions for all the questions covered in Chapter 1 Relation and Function. All exercise questions are solved by subject experts using a step by step approach. Solving the questions repeatedly helps in grasping the concepts correctly. Students can download the pdf and go practice these NCERT Solutions for Class 12 Maths to sharpen their Maths skills.

Students are suggested not to understand only the basics of this chapter but also the advanced portions to create a strong command over the Class 12 Maths Relations and Functions. Practicing these NCERT Solutions will help the students in their board and competitive exam preparation in such a way that they would find even the difficult questions from this chapter to be extremely easy. Download the NCERT Solutions PDF of Chapter 1 Maths through the link given below.

### What are Relation and Function?

A relation is defined as a set of inputs and outputs. It is generally written as ordered pairs (input, output). In maths, relations can also be represented by a mapping diagram or a graph. For example, the relation can be represented as:

Function:

In Maths, a function is defined as a  relation in which each input retains only one output. For example, In the relation, b is a function of a, because, for each input a (1, 2, 3, or 0), there is only one output b. a is not a function of b, because the input  b= 3 has several outputs i.e. a = 1 and b = 2.

### Topics Covered In Class 12 Maths Chapter 1

1.1: Introduction To Relation and Function

1.2: Types of Relations

1.3: Types of Function

1.4: Composition of Function And Invertible Function

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

### Exercise 1.1

1. Determine whether each of the following relations are reflexive, symmetric and transitive.

1. Relation $\text{R}$ in the set $\text{A = }\left\{ \text{1, 2, 3}...\text{13, 14} \right\}$ defined as $\text{R = }\left\{ \left( \text{x, y} \right)\text{: 3x - y = 0} \right\}$

Ans: The given relation is: $\text{R = }\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{2, 6} \right)\text{, }\left( \text{3, 9} \right)\text{, }\left( \text{4, 12} \right) \right\}$

Since $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{ }...$ and $\left( \text{14, 14} \right)\notin R$.

We conclude that $\text{R}$ is not reflexive.

Since $\left( \text{1, 3} \right)\in \text{R}$, but $\left( \text{3, 1} \right)\notin \text{R}$. (since $\text{3}\left( \text{3} \right)\text{-1}\ne \text{0}$)

We conclude that $\text{R}$ is not symmetric.

Since $\left( \text{1, 3} \right)$ and $\left( \text{3, 9} \right)\in \text{R}$, but$\left( \text{1, 9} \right)\notin \text{R}\text{. }\left[ \text{3}\left( \text{1} \right)\text{-9}\ne \text{0} \right]$.

We conclude that $\text{R}$ is not transitive.

Therefore, the relation $\text{R}$ is not reflexive, symmetric or transitive.

1. Relation $\text{R}$ in the set $\text{N}$ of natural numbers defined as $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y = x + 5}$ and $\text{x4 }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\left\{ \left( \text{1, 6} \right)\text{, }\left( \text{2, 7} \right)\text{, }\left( \text{3, 8} \right) \right\}$.

Since $\left( \text{1, 1} \right)\notin \text{R}$.

We conclude that $\text{R}$ is not reflexive.

Since $\left( \text{1, 6} \right)\in \text{R}$ but $\left( \text{6, 1} \right)\notin \text{R}$.

We conclude that $\text{R}$ is not symmetric.

In the given relation $\text{R}$ there is not any ordered pair such that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)$ both $\in \text{R}$, therefore we can say that $\left( \text{x, z} \right)$ cannot belong to $\text{R}$.

Therefore $\text{R}$ is not transitive.

Hence, the given relation $\text{R}$ is not reflexive, symmetric or transitive.

1. Relation $\text{R}$in the set $\text{A = }\!\!\{\!\!\text{ 1, 2, 3, 4, 5, 6 }\!\!\}\!\!\text{ }$ as $\text{R= }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y}$is divisible by $\text{x }\!\!\}\!\!\text{ }$

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{ : y}$ is divisible by $\text{x }\!\!\}\!\!\text{ }$

As we know that any number except $\text{0}$ is divisible by itself, therefore $\left( \text{x, x} \right)\in \text{R}$.

We conclude that $\text{R}$ is reflexive.

Since $\left( \text{2, 4} \right)\in \text{R}$ (because $\text{4}$ is divisible by $\text{2}$), but $\left( \text{4, 2} \right)\notin \text{R}$ (since $\text{2}$ is not divisible by $\text{4}$).

We conclude that $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)\in \text{R}$, $\text{y}$ is divisible by $\text{x}$ and $\text{z}$ is divisible by $\text{y}$. Hence $\text{z}$ is divisible by $\text{x}\Rightarrow \left( \text{x, z} \right)\in \text{R}$.

We conclude that $\text{R}$ is transitive.

Hence, the given relation $\text{R}$ is reflexive and transitive but it is not symmetric.

1. Relation $\text{R}$ in the set $\text{Z}$ of all integers defined as $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y }\!\!\}\!\!\text{ }$ is as integer

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y}$ is an integer$\text{ }\!\!\}\!\!\text{ }$

If $\text{x}\in \text{Z, }\left( \text{x, x} \right)\in \text{R}$ because $\text{x-x = 0}$ is an integer.

Hence, we conclude that $\text{R}$ is reflexive.

For $\text{x, y}\in \text{Z}$, if $\left( \text{x, y} \right)\in \text{R}$, then $\text{x - y}$ is an integer and therefore $\left( \text{y-x} \right)$ is also an integer.

Therefore, we conclude that $\left( \text{y, x} \right)\in \text{R}$and hence $\text{R}$ is symmetric.

Assuming that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)\in \text{R}$, where $\text{x, y, z}\in \text{Z}$.

We can say that $\left( \text{x-y} \right)$ and $\left( \text{y-z} \right)$ are integers.

so, $\left( \text{x, z} \right)\in \text{R}$

Hence, we conclude that $\text{R}$ is transitive.

Therefore the given relation $\text{R}$ is reflexive, symmetric, and transitive.

1. Relation $\text{R}$ in the set $\text{A}$ of human beings in a town at a particular time given by

1. The relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ work at the same place$\}$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ and $\text{y}$ work at the same place$\text{ }\!\!\}\!\!\text{ }$

This implies that $\left( \text{x, x} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is reflexive.

Now, $\left( \text{x, y} \right)\in \text{R}$, then $\text{x}$ and $\text{y}$ work at the same place, which means $\text{y}$ and $\text{x}$ also work at the same place. Therefore, $\left( \text{y, x} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is symmetric.

Let us assume that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$.

Then, we can say that $\text{x}$ and $\text{y}$ work at the same place and $\text{y}$ and $\text{z}$ work at the same place. Which means that $\text{x}$ and $\text{z}$ also work at the same place.

Therefore, $\left( \text{x, z} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive, symmetric and transitive.

1. The relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ live in the same locality$\text{ }\!\!\}\!\!\text{ }$

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ live in the same locality$\text{ }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\in \text{R}$.

Therefore, we conclude that $\text{R}$ is reflexive.

Since $\left( \text{x, y} \right)\in \text{R}$, $\text{x}$ and $\text{y}$ live in the same locality. Therefore, $\text{y}$ and $\text{x}$ also live in the same locality, so, $\left( \text{y, x} \right)\in \text{R}$.

Hence, $\text{R}$ is symmetric.

Let $\left( \text{x, y} \right)\in \text{R}$ and $\left( \text{y, z} \right)\in \text{R}$. Hence $\text{x}$ and $\text{y}$ live in the same locality and $\text{y}$ and $\text{z}$ also live in the same locality. Which means that $\text{x}$ and $\text{z}$ also live in the same locality.

Therefore, $\left( \text{x, z} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive, symmetric and transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is exactly $\text{7}$cm taller than $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is exactly $\text{7}$ cm taller than $\text{y }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\notin \text{R}$.

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is exactly $\text{7}$ cm taller than $\text{y}$, therefore $\text{y}$ is obviously not taller than $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is exactly $\text{7}$ cm taller than $\text{y}$and $\text{y}$ is exactly $\text{7}$ cm taller than $\text{z}$. Which means that $\text{x}$ is exactly $\text{14}$ cm taller than $\text{z}$. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is not reflexive, symmetric or transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is wife of $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is the wife of $\text{y }\!\!\}\!\!\text{ }$.

Since, $\left( \text{x, x} \right)\notin \text{R}$

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is the wife of $\text{y}$, therefore $\text{y}$ is obviously not the wife of $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is the wife of $\text{y}$and $\text{y}$ is the wife of $\text{z}$, which is not possible. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore the given relation $\text{R}$is not reflexive, symmetric or transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is father of $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is the father of $\text{y }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\notin \text{R}$

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is the father of $\text{y}$, therefore $\text{y}$ is obviously not the father of $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is the father of $\text{y}$and $\text{y}$ is the father of $\text{z}$, then $\text{x}$ is not the father of  $\text{z}$. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore the given relation $\text{R}$is not reflexive, symmetric or transitive.

2. Show that the relation $\text{R}$ in the set $\text{R}$ of real numbers, defined $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{2}}} \right\}$ is neither reflexive nor symmetric nor transitive.

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{2}}} \right\}$

Since $\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{\text{2}} \right)\notin \text{R}$. (Since $\dfrac{\text{1}}{\text{2}}\text{}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$)

Therefore, $\text{R}$ is not reflexive.

Since $\left( \text{1, 4} \right)\in \text{R}$ as $\text{1}{{\text{4}}^{\text{2}}}$, but $\left( \text{4, 1} \right)\notin \text{R}$ as ${{\text{4}}^{2}}$ is not less than ${{\text{1}}^{2}}$.

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{3, 2} \right)\text{, }\left( \text{2, 1}\text{.5} \right)\in \text{R}$, so, $\text{3}{{\text{2}}^{\text{2}}}\text{=4}$ and $\text{2}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}$ but $\text{3}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive.

3. Check whether the relation $\text{R}$ defined in the set $\left\{ \text{1, 2, 3, 4, 5, 6} \right\}$ as $\text{R = }\left\{ \left( \text{a, b} \right)\text{: b=a+1} \right\}$ is reflexive, symmetric or transitive.

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: b=a+1} \right\}$ defined in the set $\text{A=}\left\{ \text{1, 2, 3, 4, 5, 6} \right\}$.

So, $\text{R=}\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 3} \right)\text{, }\left( \text{3, 4} \right)\text{, }\left( \text{4, 5} \right)\text{, }\left( \text{5, 6} \right) \right\}$

Since, $\left( \text{a, a} \right)\notin \text{R,a}\in \text{A}$.

$\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{5, 5} \right)\text{, }\left( \text{6, 6} \right)\notin \text{R}$

Therefore, $\text{R}$ is not reflexive

Since, $\left( \text{1, 2} \right)\in \text{R}$, but $\left( \text{2, 1} \right)\notin \text{R}$.

Therefore $\text{R}$ is not symmetric.

Since $\left( \text{1, 2} \right)\text{, }\left( \text{2, 3} \right)\in \text{R}$, but $\left( \text{1, 3} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive

4. Show that the relation $\text{R}$ in $\text{R}$ defined as $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le \text{b} \right\}$ is reflexive and transitive but not symmetric.

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le \text{b} \right\}$.

Since, $\left( \text{a, a} \right)\in R$.

Therefore, $\text{R}$ is reflexive.

Since, $\left( \text{2, 4} \right)\in R$ (as $\text{2,4}$), but $\left( \text{4, 2} \right)\notin R$(as $\text{4,2}$).

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$, $\text{a}\le \text{b}$ and $b\le c$, therefore, $a\le c$.

Hence, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive and transitive but not symmetric.

5. Check whether the relation $\text{R}$ in $\text{R}$ defined as $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{3}}} \right\}$ is reflexive, symmetric or transitive.

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{3}}} \right\}$

Since $\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{8} \right)\notin \text{R}$.(Since $\dfrac{\text{1}}{\text{2}}$ is not less than $\dfrac{\text{1}}{8}$)

Therefore, $\text{R}$ is not reflexive.

Since $\left( \text{1, 4} \right)\in \text{R}$ as $\text{1}{{\text{4}}^{\text{3}}}$, but $\left( \text{4, 1} \right)\notin \text{R}$ as $\text{4}$ is not less than ${{\text{1}}^{\text{3}}}$.

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{3, }\dfrac{\text{3}}{\text{2}} \right)\text{, }\left( \dfrac{\text{3}}{\text{2}}\text{, }\dfrac{\text{6}}{\text{5}} \right)\in \text{R}$, so, $\text{3}{{\left( \dfrac{\text{3}}{\text{2}} \right)}^{\text{3}}}$ and $\dfrac{\text{3}}{\text{2}}\text{}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}$ but $\text{3}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive.

6.  Show that the relation $\text{R}$ in the set $\left\{ \text{1, 2, 3} \right\}$ given by $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}$ is symmetric but neither reflexive nor transitive.

Ans: The given relation is $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}$ on the set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Since $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\notin \text{R}$

Therefore, $\text{R}$ is not reflexive.

Since, $\left( \text{1, 2} \right)\in R$ and $\left( \text{2, 1} \right)\in R$.

Therefore $\text{R}$ is symmetric.

Since, $\left( \text{1, 2} \right)\in R$ and $\left( \text{2, 1} \right)\in R$, but $\left( \text{1, 1} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is symmetric but neither reflexive nor transitive.

7.  Show that the relation $\text{R}$ in the set $\text{A}$ of all the books in a library of a college, given by $\text{R = }\!\!\{\!\!\text{ }\left( \text{x,}\,\,\text{y} \right)\text{: x}$ and $\text{y}$ have same number of pages$\text{ }\!\!\}\!\!\text{ }$ is an equivalence relation.

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x,}\,\,\text{y} \right)\text{: x}$ and $\text{y}$ have the same number of pages$\text{ }\!\!\}\!\!\text{ }$

Since $\left( \text{x, x} \right)\notin \text{R}$ as $\text{x}$ and $\text{x}$ have the same number of pages.

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{x, y} \right)\in R$, so $\text{x}$ and $\text{y}$ have the same number of pages, therefore $\text{y}$ and $\text{x}$ will also have the same number of pages.

Therefore $\text{R}$ is symmetric.

Assuming $\left( \text{x, y} \right)\in R$ and $\left( \text{y, z} \right)\in R$. $\text{x}$ and $\text{y}$ have the same number of pages and $\text{y}$ and $\text{z}$ also have the same number of pages. Therefore, $\text{x}$ and $\text{z}$ will also have the same number of pages. So, $\left( \text{x, z} \right)\in R$.

Hence, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

8.  Show that the relation $\text{R}$ in the set $\text{A=}\left\{ \text{1, 2, }\!\!~\!\!\text{ 3, 4, 5} \right\}$ given by $\text{R = }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{: }\left| \text{a-b} \right|$ is even$\text{ }\!\!\}\!\!\text{ }$, is an equivalence relation. Show that all the elements of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ are related to each other and all the elements of $\left\{ \text{2, 4} \right\}$ are related to each other. But no element of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ is related to any element of $\left\{ \text{2, 4} \right\}$.

Ans: Let $\text{a}\in \text{A}$,

So, $\left| \text{a-a} \right|\text{ = 0}$  (which is an even number).

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{a, b} \right)\in \text{R}$,

Now, $\left| \text{a-b} \right|$ is even,

Hence $\left| \text{a-b} \right|$ and $\left| \text{b-a} \right|$ are both even

Therefore, $\left( \text{b, a} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric

Let $\left( \text{a, b} \right)\in \text{R}$and $\left( \text{b, c} \right)\in \text{R}$,

$\Rightarrow \left| \text{a-b} \right|$ is even and $\left| \text{b-c} \right|$ is even

$\Rightarrow \left| \text{a-c} \right|$ is even.

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

All the elements of the set $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ are all odd. Hence, the modulus of the difference of any two elements will be an even number. So, all the elements of this set are related to each other.

All elements of $\left\{ \text{2, 4} \right\}$ are even while all the elements of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ are odd so no element of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ can be related to any element of$\left\{ 2,\text{ }4 \right\}$.

Therefore, the absolute value of the difference between the two elements (from each of these two subsets) will not be an even value.

9. Show that each of the relation $\text{R}$ in the set $\text{A = }\!\!\{\!\!\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\!\!\}\!\!\text{ }$ , is an equivalence relation. Find the set of all elements related to 1 in each case.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{ : }\left| \text{a -- b} \right|$ is a multiple of $\text{4 }\!\!\}\!\!\text{ }$

Ans: The given set $\text{A = }\!\!\{\!\!\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\!\!\}\!\!\text{ = }\left\{ \text{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} \right\}$

The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{ : }\left| \text{a -- b} \right|$ is a multiple of $\text{4 }\!\!\}\!\!\text{ }$.

Let $a\in A$,

$\left( \text{a, a} \right)\in R$ as $\left| \text{a-a} \right|\text{=0}$ is a multiple of $\text{4}$.

Therefore, $\text{R}$ is reflexive.

Let, $\left( \text{a, b} \right)\in \text{R}\Rightarrow \left| \text{a-b} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left| \text{-}\left( \text{a-b} \right) \right|\text{=}\left| \text{b-a} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{b, a} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

$\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$.

$\Rightarrow \left| \text{a-b} \right|$ is a multiple of $\text{4}$ and $\left| \text{b-c} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a-b} \right)$ is a multiple of $\text{4}$ and $\left( \text{b-c} \right)$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a-c} \right)\text{=}\left( \text{a-b} \right)\text{+}\left( \text{b-c} \right)$ is a multiple of $\text{4}$.

$\Rightarrow \left| \text{a-c} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

The set of elements related to 1 is $\left\{ \text{1, }\!\!~\!\!\text{ 5, 9} \right\}$

1. $\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}$

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}$.

$a\in A,\left( \text{a, a} \right)\in R$, since $\text{a = a}$.

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{a, b} \right)\in \text{R}\Rightarrow \text{a=b}$.

$\Rightarrow b\text{=a}\Rightarrow \left( \text{b, a} \right)\in R$

Therefore $\text{R}$ is symmetric.

$\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$

$\Rightarrow a\text{=b}$ and $b\text{=c}$

$\Rightarrow a\text{=c}$

$\Rightarrow \left( \text{a, c} \right)\in R$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

The set of elements related to $\text{1}$ is $\left\{ \text{1} \right\}$.

10. Give an example of a relation. Which is

1. Symmetric but neither reflexive nor transitive.

Ans: Let us assume the relation $\text{R= }\left\{ \left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right) \right\}$ in set $\text{A= }\left\{ \text{5, 6, 7} \right\}$.

So, the relation $\text{R}$ is not reflexive as $\left( \text{5, 5} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{7, 7} \right)\notin \text{R}$.

The relation $\text{R}$ is symmetric as $\left( \text{5, 6} \right)\in R$ and $\left( \text{6, 5} \right)\in R$.

The relation $\text{R}$ is not transitive as $\left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right)\in R$ , but $\left( \text{5, 5} \right)\notin \text{R}$.

Therefore, the given relation $\text{R}$ is symmetric but not reflexive or transitive.

1. Transitive but neither reflexive nor symmetric.

Ans: Let us assume the relation $\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a b} \right\}$

So, the relation $\text{R}$ is not reflexive because for $a\in R$, $\left( \text{a, a} \right)\notin \text{R}$ since a cannot be strictly less than itself.

Let $\left( {1,2} \right) \in R\left( {as1 < 2} \right)$

Since $\text{2}$ is not less than $\text{1}$, $\left( \text{2, 1} \right)\notin \text{R}$.

Therefore $\text{R}$ is not symmetric.

Let $\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$.

$\Rightarrow \left( \text{a, c} \right)\in R$

Therefore, $\text{R}$ is transitive.

So, the relation $\text{R}$ is transitive but not reflexive and symmetric.

1. Reflexive and symmetric but not transitive.

Ans: Let us assume the relation $R=\left\{ \left( \text{4, 4} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{8, 8} \right),\text{ }\left( \text{4, 6} \right),\text{ }\left( \text{6, 4} \right),\text{ }\left( \text{6, 8} \right),\text{ }\left( \text{8, 6} \right) \right\}$ in set $\text{A= }\left\{ \text{4, 6, 8} \right\}$.

The relation $\text{R}$ is reflexive since for $a\in R$, $\left( \text{a, a} \right)\in R$.

The relation $\text{R}$ is symmetric since $\left( \text{a, b} \right)\in R\Rightarrow \left( \text{b, a} \right)\in R$ for $a,b\in R$.

The relation $\text{R}$ is not transitive since $\left( \text{4, 6} \right)\text{, }\left( \text{6, 8} \right)\in R$, but $\left( \text{4, 8} \right)\notin \text{R}$.

Therefore the relation $\text{R}$ is reflexive and symmetric but not transitive.

1. Reflexive and transitive but not symmetric.

Ans: Let us take the relation $\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : }{{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}} \right\}$.

Since $\left( \text{a, b} \right)\in \text{R}$.

Therefore $\text{R}$ is reflexive.

Since $\left( \text{2, 1} \right)\in \text{R}$, but $\left( \text{1, 2} \right)\notin \text{R}$,

Therefore $\text{R}$ is not symmetric.

Let $\left( {a,b} \right),\left( {b,c} \right) \in R$

$\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}}$ and ${{\text{b}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}$

$\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}$

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is reflexive and transitive but not symmetric.

1. Symmetric and transitive but not reflexive.

Ans: Let us take a relation $\text{R=}\left\{ \left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\text{, }\left( \text{-5, -5} \right) \right\}$ in set $\text{A=}\left\{ \text{-5, -6} \right\}$.

The relation $\text{R}$ is not reflexive as $\left( \text{-6, -6} \right)\notin \text{R}$.

Since $\left( \text{-5, -6} \right)\in \text{R}$ and $\left( \text{-6, -5} \right)\in \text{R}$.

Therefore $\text{R}$ is symmetric.

Since $\left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\in \text{R}$ and $\left( \text{-5, -5} \right)\in \text{R}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is symmetric and transitive but not reflexive.

11. Show that the relation $\text{R}$ in the set $\text{A}$ of points in a plane given by $\text{R = }\!\!\{\!\!\text{ }\left( \text{P, Q} \right)\text{ :}$ Distance of the point $\text{P}$ from the origin is same as the distance of the point $\text{Q}$ from the origin}, is an equivalence relation. Further, show that the set of all points related to a point $\text{P}\ne \left( \text{0, 0} \right)$ is the circle passing through $\text{P}$ with origin as centre.

Ans: The given relation is $\text{ }\!\!~\!\!\text{ R = }\!\!\{\!\!\text{ }\left( \text{P, Q} \right)\text{ :}$ Distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin}

Since, $\left( \text{P, P} \right)\in \text{R}$.

The relation $\text{R}$ is reflexive.

Let $\left( \text{P, Q} \right)\in \text{R}$, distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin similarly distance of $\text{Q}$ from the origin will be the same as the distance of $\text{P}$ from the origin. So, $\left( \text{Q, P} \right)\in \text{R}$.

Therefore $\text{R}$ is symmetric.

Let $\left( \text{P, Q} \right),\left( \text{Q, S} \right)\in \text{R}$.

Distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin and distance of $\text{Q}$ from the origin is the same as the distance of $\text{S}$ from the origin. So, the distance of $\text{S}$ from the origin will be the same as the distance of $\text{P}$ from the origin. So, $\left( \text{P, S} \right)\in \text{R}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

The set of points related to $\text{P}\ne \left( \text{0, 0} \right)$ will be those points whose distance from origin is same as distance of $\text{P}$ from the origin and will form a circle with the centre as origin and this circle passes through $\text{P}$.

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

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\text{ : }{{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}\text{ }\!\!\}\!\!\text{ }$.

The relation $\text{R}$ is reflexive since every triangle is similar to itself.

If $\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\in \text{R}$, then ${{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}$.

$\Rightarrow {{\text{T}}_{\text{2}}}$ is similar to ${{\text{T}}_{\text{1}}}$.

$\Rightarrow \left( {{\text{T}}_{\text{2}}}\text{, }{{\text{T}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\text{,}\left( {{\text{T}}_{\text{2}}}\text{, }{{\text{T}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}$ and ${{\text{T}}_{\text{2}}}$ is similar to ${{\text{T}}_{\text{3}}}$.

$\Rightarrow {{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{3}}}$.

$\Rightarrow \left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

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

Since, the corresponding sides of triangles ${{\text{T}}_{\text{1}}}$ and ${{\text{T}}_{\text{3}}}$ are in the same ratio, therefore triangle ${{\text{T}}_{\text{1}}}$ is similar to triangle ${{\text{T}}_{\text{3}}}$.

Hence, ${{\text{T}}_{\text{1}}}$ is related to ${{\text{T}}_{\text{3}}}$.

13. Show that the relation $\text{R}$ defined in the set $\text{A}$ of all polygons as $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{: }{{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides$\text{ }\!\!\}\!\!\text{ }$, is an equivalence relation. What is the set of all elements in $\text{A}$ related to the right angle triangle $\text{T}$ with sides $\text{3, 4}$  and $\text{5}$?

Ans: $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{: }{{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides$\text{ }\!\!\}\!\!\text{ }$.

Since $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{1}}} \right)\in \text{R}$ , as same polygon has same number of sides.

The relation $\text{R}$ is reflexive.

Let $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\in \text{R}$ .

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{2}}}$ and ${{\text{P}}_{\text{1}}}$ have same number of sides.

$\Rightarrow \left( {{\text{P}}_{\text{2}}}\text{, }{{\text{P}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{,}\left( {{\text{P}}_{\text{2}}}\text{, }{{\text{P}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{2}}}$ and ${{\text{P}}_{\text{3}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{3}}}$ have same number of sides.

$\Rightarrow \left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

The elements in $A$ related to right-angled triangle $\left( \text{T} \right)$ with sides $\text{3, 4}$ and $\text{5}$ are the polygons having $\text{3}$ sides.

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

Ans: $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\text{: }{{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}\text{ }\!\!\}\!\!\text{ }$.

The relation $\text{R}$ is reflexive as any line ${{\text{L}}_{\text{1}}}$ is parallel to itself, so, $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{1}}} \right)\in \text{R}$.

Let $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\in \text{R}$.

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}$, therefore ${{\text{L}}_{\text{2}}}$ is parallel to ${{\text{L}}_{\text{1}}}$.

$\Rightarrow \left( {{\text{L}}_{\text{2}}}\text{, }{{\text{L}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\text{,}\left( {{\text{L}}_{\text{2}}}\text{, }{{\text{L}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}$

$\Rightarrow {{\text{L}}_{\text{2}}}$ is parallel to ${{\text{L}}_{\text{3}}}$

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{3}}}$

$\Rightarrow \left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

Set of all lines related to the line $\text{y=2x+4}$ is the set of all lines that are parallel to the line $\text{y=2x+4}$.

Slope of line $\text{y=2x+4}$ is $\text{m = 2}$. Therefore, lines parallel to the given line are of the form $\text{y=2x+c}$, where $\text{c}\in \text{R}$.

15. Let $\text{R}$ be the relation in the set $\left\{ \text{1, 2, 3, 4} \right\}$ given by $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{1, 1} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 2} \right) \right\}$. Choose the correct answer.

(A) $\text{R}$ is reflexive and symmetric but not transitive.

(B) $\text{R}$ is reflexive and transitive but not symmetric.

(C) $\text{R}$ is symmetric and transitive but not reflexive.

(D) $\text{R}$ is an equivalence relation

Ans: $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{1, 1} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 2} \right) \right\}$.

Since $\left( \text{a, a} \right)\in \text{R}$, for every $\text{a}\in \left\{ \text{1, 2, 3, 4} \right\}$

The relation $\text{R}$ is reflexive.

Since $\left( \text{1, 2} \right)\in \text{R}$ , but $\left( \text{2, 1} \right)\notin \text{R}$ .

Therefore $\text{R}$ is not symmetric.

$\left( \text{a, b} \right)\text{,}\left( \text{b, c} \right)\in \text{R}\Rightarrow \left( \text{a, c} \right)\in \text{R}$ for all $\text{a, b, c}\in \left\{ \text{1, 2, 3, 4} \right\}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is reflexive and transitive but not symmetric.

The correct answer is ($B$) $\text{R}$ is reflexive and transitive but not symmetric.

16. Let $\text{R}$ be the relation in the set N given by $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a = b - 2, b > 6} \right\}$ Choose the correct answer.

(A) $\left( \text{2, 4} \right)\in \text{R}$

(B) $\left( \text{3, 8} \right)\in \text{R}$

(C) $\left( \text{6, 8} \right)\in \text{R}$

(D) $\left( \text{8, 7} \right)\in \text{R}$

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a = b - 2, b 6} \right\}$

Now,

Considering $\left( \text{2, 4} \right)\in \text{R}$.

Since, $\text{b 6}$, so, $\left( \text{2, 4} \right)\notin \text{R}$.

Considering $\left( \text{3, 8} \right)\in \text{R}$.

Since $\text{3 }\ne \text{ 8 - 2}$, so $\left( \text{3, 8} \right)\notin \text{R}$.

Considering $\left( \text{6, 8} \right)\in \text{R}$.

Since $\text{86}$ and $\text{6=8-2}$, so $\left( \text{6, 8} \right)\in \text{R}$.

Therefore, the correct answer is ($C$)$\left( \text{6, 8} \right)\in \text{R}$.

### Exercise 1.2

1. Show that the function $\text{f: }{{\text{R}}_{\text{*}}}\to {{\text{R}}_{\text{*}}}$ defined by $\text{f}\left( \text{x} \right)\text{ =}\dfrac{\text{1}}{\text{x}}$ is one-one and onto, where ${{\text{R}}_{\text{*}}}$ is the set of all non-zero real numbers. Is the result true, if the domain ${{\text{R}}_{\text{*}}}$ is replaced by $\text{N}$ with co-domain being same as ${{\text{R}}_{\text{*}}}$?

Ans: The function $\text{f: }{{\text{R}}_{\text{*}}}\to {{\text{R}}_{\text{*}}}$ is defined by $\text{f}\left( \text{x} \right)\text{ =}\dfrac{\text{1}}{\text{x}}$.

For $\text{f}$ to be one – one:

$\text{x, y}\in {{\text{R}}_{\text{*}}}$ such that $\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{=}\dfrac{\text{1}}{\text{y}}$

$\Rightarrow \text{x = y}$

Therefore, the given function $\text{f}$ is one – one.

For $\text{f}$ to be onto:

For $\text{y}\in {{\text{R}}_{\text{*}}}$ there exists $\text{x =}\dfrac{\text{1}}{\text{y}}\in {{\text{R}}_{\text{*}}}$ as $\text{y}\ne \text{0}$ such that

$\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\left( \dfrac{\text{1}}{\text{y}} \right)}\text{= y}$

Therefore, the given function $\text{f}$ is onto.

Hence the given function $\text{f}$ is one – one and onto.

Consider a function $\text{g: N }\to {{\text{R}}_{\text{*}}}$ defined by $\text{g}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{x}}$

We have, $\text{g}\left( {{\text{x}}_{\text{1}}} \right)\text{=g}\left( {{\text{x}}_{\text{2}}} \right)\Rightarrow \dfrac{\text{1}}{{{\text{x}}_{\text{1}}}}\text{=}\dfrac{\text{1}}{{{\text{x}}_{\text{2}}}}\Rightarrow {{\text{x}}_{\text{1}}}\text{=}{{\text{x}}_{\text{2}}}$

Therefore the function $\text{g}$ is one – one.

The function $\text{g}$ is not onto as for $\text{1}\text{.2}\in \text{=}{{\text{R}}_{\text{*}}}$ there does not exist any $\text{x}$ in $\text{N}$  such that $\text{g}\left( \text{x} \right)\text{ =}\dfrac{\text{1}}{\text{1}\text{.2}}$.

Therefore, the function $\text{g}$ is one-one but not onto.

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

1. $\text{f: N }\to \text{ N}$ given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}$

Ans: The given function $\text{f: N }\to \text{ N}$ is defined by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}$.

For $\text{x, y}\in \text{N}$,

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ = }{{\text{y}}^{\text{2}}}$

$\Rightarrow \text{x = y}$

Therefore the function $\text{f}$ is injective.

Since $\text{2}\in \text{N}$, but, there does not exist any $\text{x}$ in $\text{N}$ such that $\text{f}\left( \text{x} \right)\text{=2}$.

Therefore the function $\text{f}$ is not surjective.

Hence, the function $\text{f}$ is injective but not surjective.

1. $\text{f: Z }\to \text{ Z}$ given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}$

Ans: The given function $\text{f: Z }\to \text{ Z}$ is defined by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}$.

Since,

$\text{f}\left( \text{-1} \right)\text{ = f}\left( \text{1} \right)$

$\text{= 1}$

But $\text{-1 }\ne \text{ 1}$

Therefore the function $\text{f}$ is not injective.

Since $\text{-2}\in \text{Z}$, but, there does not exist any element $\text{x}\in \text{Z}$ such that

$\text{f}\left( \text{x} \right)\text{ = -2}$

Therefore the function $\text{f}$ is not surjective.

Hence, the function $\text{f}$ is neither injective nor surjective.

1. $\text{f: R }\to \text{ R}$  given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}$

Ans: The given function $\text{f: R }\to \text{ R}$  is given by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}$

Now,

$\text{f}\left( \text{-1} \right)\text{ = f}\left( \text{1} \right)$

$\text{= 1}$

But $\text{-1 }\ne \text{ 1}$.

Therefore the function $\text{f}$ is not injective.

Since $\text{-2}\in \text{R}$, but, there does not exist any element $\text{x}\in \text{R}$ such that

$\text{f}\left( \text{x} \right)\text{ = -2}$.

Therefore the function $\text{f}$ is not surjective.

Hence, the function $\text{f}$ is neither injective nor surjective.

1. $\text{f: N }\to \text{ N}$ given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{3}}}$

Ans: The given function $\text{f: N }\to \text{ N}$  is given by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

For $\text{x, y}\in \text{N}$,

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ = }{{\text{y}}^{\text{3}}}$

$\Rightarrow \text{x = y}$

Therefore the function $\text{f}$ is injective.

Since $\text{2}\in \text{N}$, but, there does not exist any $\text{x}$ in $\text{N}$ such that $\text{f}\left( \text{x} \right)\text{=2}$.

Therefore the function $\text{f}$ is not surjective.

Hence, the function $\text{f}$ is injective but not surjective.

1. $\text{f: Z }\to \text{ Z}$given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{3}}}$

Ans: The given function $\text{f: Z }\to \text{ Z}$  is given by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

For $\text{x, y}\in \text{Z}$,

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ = }{{\text{y}}^{\text{3}}}$

$\Rightarrow \text{x = y}$

Therefore the function $\text{f}$ is injective.

Since $\text{2}\in \text{Z}$, but, there does not exist any $\text{x}$ in $\text{Z}$ such that $\text{f}\left( \text{x} \right)\text{=2}$.

Therefore the function $\text{f}$ is not surjective.

Hence, the function $\text{f}$ is injective but not surjective.

3. Prove that the Greatest Integer Function $\text{f: R }\to \text{ R}$ given by $\text{f}\left( \text{x} \right)\text{ = }\left[ \text{x} \right]$, is neither one – one nor onto, where $\left[ \text{x} \right]$ denotes the greatest integer less than or equal to $\text{x}$.

Ans: The function $\text{f: R }\to \text{ R}$ is defined by $\text{f}\left( \text{x} \right)\text{ = }\left[ \text{x} \right]$.

Now,

$\text{f}\left( \text{1}\text{.2} \right)\text{ = }\left[ \text{1}\text{.2} \right]$

$\text{= 1}$

$\text{f}\left( \text{1}\text{.9} \right)\text{ = }\left[ \text{1}\text{.9} \right]$

$\text{= 1}$

Therefore, $\text{f}\left( \text{1}\text{.2} \right)\text{ = f}\left( \text{1}\text{.9} \right)$, but $\text{1}\text{.2}\ne \text{1}\text{.9}$.

Hence the function $\text{f}$ is not one – one.

Taking $0.7\in R$, $\text{f}\left( \text{x} \right)\text{ = }\left[ \text{x} \right]$ is an integer. There does not exist any element $\text{x}\in \text{R}$ such that $\text{f}\left( \text{x} \right)\text{ = 0}\text{.7}$.

Therefore, the function $\text{f}$ is not onto.

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

4. Show that the Modulus Function $\text{f: R }\to \text{ R}$ given by $\left( \text{x} \right)\text{ = }\left| \text{x} \right|$ is neither one – one nor onto, where $\left| \text{x} \right|$ is$x$ , if $\text{x}$ is positive or $\text{0}$ and $\left| \text{x} \right|$ is$\text{-x}$, if $\text{x}$ is negative.

Ans: $\text{f: R }\to \text{ R}$is $\text{f}\left( \text{x} \right)\text{ = }\left| \text{x} \right|$

=\left\{ \begin{align}& \,\,\text{x;}\,\,\text{x0} \\ & \text{-x;}\,\,\text{x0} \\ \end{align} \right\}

Now,

$\text{f}\left( \text{-1} \right)\text{ = }\left| \text{-1} \right|$

$\text{= 1}$

$\text{f}\left( \text{1} \right)\text{ = }\left| \text{1} \right|$

$\text{= 1}$

Therefore, $\text{f}\left( \text{1} \right)\text{ = f}\left( \text{-1} \right)$, but $\text{-1 }\ne \text{ 1}$.

Hence the function $\text{f}$ is not one – one.

Taking $\text{-1}\in \text{R}$, $\text{f}\left( \text{x} \right)\text{ = }\left| \text{x} \right|$ is non-negative. Hence, there does not exist any element $\text{x}\in \text{R}$ such that $\text{f}\left( \text{x} \right)\text{ = -1}$.

Therefore, the function $\text{f}$ is not onto.

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

5. Show that the Signum Function $\text{f: R }\to \text{ R}$, given by $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{matrix}\text{1 if x0} \\\text{0, if x=0} \\ \text{-1 if x0} \\\end{matrix} \right\}$ is neither one-one nor onto.

Ans: The function $\text{f: R }\to \text{ R}$is given by

$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{matrix} \text{1 if x0} \\ \text{0, if x=0} \\ \text{-1 if x0} \\\end{matrix} \right\}$

Now,

$\text{f}\left( \text{1} \right)\text{ = f}\left( \text{2} \right)$

$\text{= 1}$

But $\text{1 }\ne \text{ 2}$

Hence the function $\text{f}$ is not one – one.

Since $\text{f}\left( \text{x} \right)$ takes only $\text{3}$  values $\left( \text{1, 0, or -1} \right)$, for the element $\text{-2}$ in co-domain

$\text{R}$ , there does not exist any $\text{x}$ in domain $\text{R}$ such that $\text{f}\left( \text{x} \right)\text{ = -2}$.

Therefore, the function $\text{f}$ is not onto.

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

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

Ans: The function $\text{f: A }\to \text{ B}$ is defined as $\text{f = }\left\{ \left( \text{1, 4} \right)\text{, }\left( \text{2, 5} \right)\text{, }\left( \text{3, 6} \right) \right\}$. Where $\text{A = }\left\{ \text{1, 2, 3} \right\}\text{, B = }\left\{ \text{4, 5, 6, 7} \right\}$

Since,

$\text{f }\left( \text{1} \right)\text{ = 4}$

$\text{f }\left( \text{2} \right)\text{ = 5}$

$\text{f }\left( \text{3} \right)\text{ = 6}$

Hence the images of distinct elements of $\text{A}$ under $\text{f}$ are distinct.

Therefore, the function $\text{f}$ is one – one.

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

1. $\text{f: R }\to \text{ R}$ defined by $\text{f}\left( \text{x} \right)\text{ = 3 - 4x}$.

Ans: The function $\text{f: R }\to \text{ R}$ is defined by $\text{f}\left( \text{x} \right)\text{ = 3 - 4x}$.

Taking ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\in \text{R}$ such that $\text{f}\left( {{\text{x}}_{\text{1}}} \right)\text{ = f}\left( {{\text{x}}_{\text{2}}} \right)$,

$\Rightarrow \text{3- 4}{{\text{x}}_{\text{1}}}\text{ = 3- 4}{{\text{x}}_{\text{2}}}$

$\Rightarrow \text{-4}{{\text{x}}_{\text{1}}}\text{ = -4}{{\text{x}}_{\text{2}}}$

$\Rightarrow {{\text{x}}_{\text{1}}}\text{ = }{{\text{x}}_{\text{2}}}$

Hence the function $\text{f}$ is one – one.

For any real number $\left( \text{y} \right)$ in $\text{R}$, there exists $\dfrac{\text{3-y}}{\text{4}}$ in $\text{R}$ such that $\text{f}\left( \dfrac{\text{3-y}}{\text{4}} \right)\text{=3-4}\left( \dfrac{\text{3-y}}{\text{4}} \right)$

$\text{=y}$

So, function $\text{f}$ is onto.

Therefore, function $\text{f}$ is bijective.

1. $\text{f: R }\to \text{ R}$defined by $\text{f}\left( \text{x} \right)\text{ = 1 + }{{\text{x}}^{\text{2}}}$

Ans: The function $\text{f: R }\to \text{ R}$defined as $\text{f}\left( \text{x} \right)\text{ = 1 + }{{\text{x}}^{\text{2}}}$

Taking ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\in \text{R}$ such that $\text{f}\left( {{\text{x}}_{\text{1}}} \right)\text{ = f}\left( {{\text{x}}_{\text{2}}} \right)$

$\Rightarrow\text{1+x}_{\text{1}}^{\text{2}}\text{=1+x}_{\text{2}}^{\text{2}}$

$\Rightarrow\text{x}_{\text{1}}^{\text{2}}\text{=x}_{\text{2}}^{\text{2}}$

$\Rightarrow {{\text{x}}_{\text{1}}}\text{= }\!\!\pm\!\!\text{ }{{\text{x}}_{\text{2}}}$

Hence the function $\text{f}$ is not one – one because $\text{f}\left( {{\text{x}}_{\text{1}}} \right)\text{ = f}\left( {{\text{x}}_{\text{2}}} \right)$ does not mean that ${{\text{x}}_{\text{1}}}\text{=}{{\text{x}}_{\text{2}}}$.

Taking $\text{-2}\in \text{R}$. Since $\text{f}\left( \text{x} \right)\text{ = 1 + }{{\text{x}}^{\text{2}}}$ is positive for all $\text{x}\in \text{R}$, so there does not exist any $\text{x}$ in domain $\text{R}$ such that$\text{f}\left( \text{x} \right)\text{ = -2}$.

Therefore, the function $\text{f}$ is not onto.

Hence, the function $\text{f}$ is neither one – one nor onto.

8. Let $\text{A}$ and $\text{B}$ be sets. Show that $\text{f: A }\!\!\times\!\!\text{ B }\to \text{ B }\!\!\times\!\!\text{ A}$ such that $\left( \text{a, b} \right)\text{=}\left( \text{b, a} \right)$ is bijective function.

Ans: The function $\text{f: A }\!\!\times\!\!\text{ B }\to \text{ B }\!\!\times\!\!\text{ A}$ is defined as $\text{f}\left( \text{a, b} \right)\text{ = }\left( \text{b, a} \right)$.

$\left( {{\text{a}}_{\text{1}}}\text{, }{{\text{b}}_{\text{1}}} \right)\text{, }\left( {{\text{a}}_{\text{2}}}\text{, }{{\text{b}}_{\text{2}}} \right)\in \text{A }\!\!\times\!\!\text{ B}$ such that $\text{f}\left( {{\text{a}}_{\text{1}}}\text{, }{{\text{b}}_{\text{1}}} \right)\text{ = f}\left( {{\text{a}}_{\text{2}}}\text{, }{{\text{b}}_{\text{2}}} \right)$

$\Rightarrow \left( {{\text{b}}_{\text{1}}}\text{, }{{\text{a}}_{\text{1}}} \right)\text{ = }\left( {{\text{b}}_{\text{2}}}\text{, }{{\text{a}}_{\text{2}}} \right)$

$\Rightarrow {{\text{b}}_{\text{1}}}\text{=}{{\text{b}}_{\text{2}}}$ and ${{\text{a}}_{\text{1}}}\text{= }{{\text{a}}_{\text{2}}}$

$\Rightarrow \left( {{\text{b}}_{\text{1}}}\text{, }{{\text{a}}_{\text{1}}} \right)\text{ = }\left( {{\text{b}}_{\text{2}}}\text{, }{{\text{a}}_{\text{2}}} \right)$

Hence the function $\text{f}$ is one – one.

For $\left( \text{b, a} \right)\in \text{B }\!\!\times\!\!\text{ A}$,

There exists $\left( \text{a, b} \right)\in \text{A }\!\!\times\!\!\text{ B}$ such that $\text{f}\left( \text{a, b} \right)\text{ = }\left( \text{b, a} \right)$

So, function $\text{f}$ is onto.

Therefore, the function $\text{f}$ is bijective.

9. Let $\text{f: N }\to \text{ N}$ be defined by $\text{f}\left( \text{n} \right)\text{ =}\left\{ \begin{matrix}\dfrac{\text{n+1}}{\text{2}}\text{, if n is odd} \\ \dfrac{\text{n}}{\text{2}}\text{, if n is even} \\\end{matrix} \right\}$ for all $\text{n}\in \text{N}$. State whether the function $\text{f}$ is bijective. Justify your answer.

Ans: The function $\text{f: N }\to \text{ N}$ is defined by $\text{f}\left( \text{n} \right)\text{ =}\left\{ \begin{matrix}\dfrac{\text{n+1}}{\text{2}}\text{, if n is odd} \\ \dfrac{\text{n}}{\text{2}}\text{, if n is even} \\\end{matrix} \right\}$ for all $\text{n}\in \text{N}$.

Now,

$\text{f(1)=}\dfrac{\text{1+1}}{\text{2}}$

$\text{=1}$

$\text{f(2)=}\dfrac{\text{2}}{\text{2}}$

$\text{=1}$

Here, $\text{f}\left( \text{1} \right)\text{ = f}\left( \text{2} \right)$, but $\text{1 }\ne \text{ 2}$.

Hence the function $\text{f}$ is not one – one.

Taking $\text{n}\in \text{N}$;

Case I: $\text{n}$ is odd

Hence $\text{n = 2r + 1}$, for some $\text{r}\in \text{N}$ there exists $\text{4r + 1}\in \text{N}$ such that

$\text{f}\left( \text{4r + 1} \right)\text{ =}\dfrac{\text{4r+1+1}}{\text{2}}$

$\text{=2r+1}$

Case II: $\text{n}$ is even

Hence, $\text{n = 2r}$ for some $\text{r}\in \text{N}$ there exists $\text{4r}\in \text{N}$ such that

$\text{f}\left( \text{4r} \right)\text{=}\dfrac{\text{4r}}{\text{2}}$

$\text{= 2r}$

So, function $\text{f}$ is onto.

Therefore, the function $\text{f}$ is not bijective.

10. Let $\text{A=R-}\left\{ \text{3} \right\}$ and $\text{B = R - }\left\{ \text{1} \right\}\text{.}$ Consider the function $\text{f:A}\to \text{B}$ defined by $\text{f}\left( \text{x} \right)\text{=}\left( \dfrac{\text{x-2}}{\text{x-3}} \right)$. Is $\text{f}$one-one and onto? Justify your answer.

Ans: The function $\text{f:A}\to \text{B}$ is defined by $\text{f}\left( \text{x} \right)\text{=}\left( \dfrac{\text{x-2}}{\text{x-3}} \right)$, where $\text{A=R-}\left\{ \text{3} \right\}$ and $\text{B = R - }\left\{ \text{1} \right\}\text{.}$

For $\text{x, y}\in \text{A}$ such that $\text{f}\left( \text{x} \right)\text{=f}\left( \text{y} \right)$

$\Rightarrow\dfrac{\text{x-2}}{\text{x-3}}\text{=}\dfrac{\text{y-2}}{\text{y-3}}$

$\Rightarrow \text{(x--2)(y--3) = (y--2)(x--3)}$

$\Rightarrow \text{xy--3x--2y + 6 = xy--2x--3y + 6}$

$\Rightarrow \text{--3x--2y =}\,\text{--2x--3y}$

$\Rightarrow \text{x = y}$

Hence the function $\text{f}$ is one – one.

If $\text{y}\in \text{B = R-}\left\{ \text{1} \right\}$, then $\text{y}\ne \text{1}$.

The function $\text{f}$ is onto if there exists $\text{x}\in \text{A}$ such that $\text{f}\left( \text{x} \right)\text{ = y}$.

$\Rightarrow \dfrac{\text{x-2}}{\text{x-3}}\text{=y}$

$\Rightarrow \text{x--2 = xy--3y}$

$\Rightarrow \text{x(1--y) =}\,\text{--3y + 2}$

$\Rightarrow \text{x=}\dfrac{\text{2-3y}}{\text{1-y}}\in \text{A }\!\![\!\!\text{ y}\ne \text{1 }\!\!]\!\!\text{ }$

for any $\text{y}\in \text{B}$, there exists $\dfrac{\text{2-3y}}{\text{1-y}}\in \text{A}$ such that,

$\text{f}\left( \dfrac{\text{2-3y}}{\text{1-y}} \right)\text{=}\dfrac{\left( \dfrac{\text{2-3y}}{\text{1-y}} \right)\text{-2}}{\left( \dfrac{\text{2-3y}}{\text{1-y}} \right)\text{-3}}$

$\text{=}\dfrac{\text{2-3y-2+2y}}{\text{2-3y-3+3y}}$

$\text{=}\dfrac{\text{-y}}{\text{-1}}$

$\text{=y}$

So, function $\text{f}$ is onto.

Hence, function f is one – one and onto.

11. Let $\text{f: R }\to \text{ R}$ be defined as $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{4}}}$. Choose the correct answer.

1. $\text{f}$ is one-one onto

2. $\text{f}$ is many-one onto

3. $\text{f}$ is one-one but not onto

4. $\text{f}$ is neither one-one nor onto

Ans: The function $\text{f: R }\to \text{ R}$ is defined as $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{4}}}$.

Taking $\text{x, y}\in \text{A}$ such that $\text{f}\left( \text{x} \right)\text{=f}\left( \text{y} \right)$

$\Rightarrow {{\text{x}}^{\text{4}}}\text{=}{{\text{y}}^{\text{4}}}$

$\Rightarrow \text{x= }\!\!\pm\!\!\text{ y}$

Therefore, $\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$ does not necessarily mean that $\text{x=y}$.

Hence the function $\text{f}$ is not one – one.

For $\text{2}\in \text{R}$, there does not exist any $\text{x}$  in domain $\text{R}$ such that $\text{f}\left( \text{x} \right)\text{ = 2}$.

So, function $\text{f}$ is not onto.

Hence, The correct answer is ($D$) function $f$ is neither one – one nor onto.

12. Let $\text{f: R }\to \text{ R}$ be defined as$\text{ f}\left( \text{x} \right)\text{ = 3x}$. Choose the correct answer.

1. $\text{f}$ is one – one onto

2. $\text{f}$ is many – one onto

3. $\text{f}$ is one – one but not onto

4. $\text{f}$ is neither one – one nor onto

Ans: The function $\text{f: R }\to \text{ R}$ is defined as $\text{f}\left( \text{x} \right)\text{ = 3x}$.

Taking $\text{x, y}\in \text{A}$ such that $\text{f}\left( \text{x} \right)\text{=f}\left( \text{y} \right)$

$\Rightarrow \text{3x = 3y}$

$\Rightarrow \text{x = y}$

Hence the function $\text{f}$ is one – one.

For $\text{y}\in \text{R}$, there exists $\dfrac{\text{y}}{\text{3}}$ in $\text{R}$ such that;

$\text{f}\left( \dfrac{\text{y}}{\text{3}} \right)\text{=3}\left( \dfrac{\text{y}}{\text{3}} \right)$

$\text{=y}$

So, function $\text{f}$ is onto.

Therefore, the correct answer is ($A$) function $\text{f}$ is one – one and onto.

### Exercise 1.3

1. Let $\text{f: }\left\{ \text{1, 3, 4} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 5} \right\}$ and $\text{g: }\left\{ \text{1, 2, 5} \right\}\text{ }\to \text{ }\left\{ \text{1, 3} \right\}$ be given by $\text{f = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{3, 5} \right)\text{, }\left( \text{4, 1} \right) \right\}$ and $\text{g =}\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{2, 3} \right)\text{, }\left( \text{5, 1} \right) \right\}$. Write down $\text{gof}$.

Ans: The function $\text{f: }\left\{ \text{1, 3, 4} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 5} \right\}$ is given by $\text{f = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{3, 5} \right)\text{, }\left( \text{4, 1} \right) \right\}$

The function $\text{g: }\left\{ \text{1, 2, 5} \right\}\text{ }\to \text{ }\left\{ \text{1, 3} \right\}$is given by $\text{g =}\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{2, 3} \right)\text{, }\left( \text{5, 1} \right) \right\}$

Now,

$\text{gof}\left( \text{1} \right)\text{ = g}\left[ \text{f}\left( \text{1} \right) \right]$

$\text{= g}\left( \text{2} \right)$         (as $\text{f}\left( \text{1} \right)\text{=2}$)

$=\text{ }3$             (as $\text{g}\left( \text{2} \right)\text{= 3}$)

$\text{gof}\left( \text{3} \right)\text{ = g}\left[ \text{f}\left( \text{3} \right) \right]$

$\text{= g}\left( \text{5} \right)$         (as $\text{f}\left( \text{3} \right)\text{ = 5}$)

$\text{= 1}$               (as $\text{g}\left( \text{5} \right)\text{ = 1}$)

$\text{gof}\left( \text{4} \right)\text{ = g}\left[ \text{f}\left( \text{4} \right) \right]$

$\text{= g}\left( \text{1} \right)$         (as $\text{f}\left( \text{4} \right)\text{ = 1}$)

$\text{= 3}$              (as $\text{g}\left( \text{1} \right)\text{ = 3}$)

Hence, $\text{gof = }\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)\text{, }\left( \text{4, 3} \right) \right\}$.

2. Let $\text{f,g}$ and $\text{h}$ be functions from $\text{R}$ to $\text{R}$. Show that

$\left( \text{f + g} \right)\text{oh = foh + goh}$

$\left( \text{f}\text{.g} \right)\text{oh = }\left( \text{foh} \right)\text{.}\left( \text{goh} \right)$

Ans: To prove: $\left( \text{f + g} \right)\text{oh = foh + goh}$.

Simplifying LHS:

$\text{LHS= }\left[ \left( \text{f + g} \right)\text{oh} \right]\left( \text{x} \right)$

$\text{= }\left( \text{f + g} \right)\left[ \text{h}\left( \text{x} \right) \right]$

$\text{= f }\left[ \text{h}\left( \text{x} \right) \right]\text{ + g}\left[ \text{h}\left( \text{x} \right) \right]$

$\text{= }\left( \text{foh} \right)\left( \text{x} \right)\text{ + }\left( \text{goh} \right)\left( \text{x} \right)$

$\text{= }\left\{ \left( \text{foh} \right)\text{ + }\left( \text{goh} \right) \right\}\left( \text{x} \right)$

$\text{= RHS}$

Hence proved.

To Prove: $\left( \text{f}\text{.g} \right)\text{oh = }\left( \text{foh} \right)\text{.}\left( \text{goh} \right)$.

Simplifying LHS:

$\text{LHS= }\left[ \left( \text{f }\text{. g} \right)\text{oh} \right]\left( \text{x} \right)$

$\text{= }\left( \text{f }\text{. g} \right)\left[ \text{h}\left( \text{x} \right) \right]\text{ }$

$\text{= f }\left[ \text{h}\left( \text{x} \right) \right]\text{ + g}\left[ \text{h}\left( \text{x} \right) \right]$

$\text{= }\left( \text{foh} \right)\left( \text{x} \right)\text{ }\text{. }\left( \text{goh} \right)\left( \text{x} \right)$

$\text{= }\left[ \left( \text{f}\text{.g} \right)\text{oh} \right]\left( \text{x} \right)$

$\text{=RHS}$

Hence proved.

3. Find $\text{gof}$ and $\text{fog}$, if

1. The functions are $\left( \text{x} \right)\text{ = }\left| \text{x} \right|$ and $\left( \text{x} \right)\text{ = }\left| \text{5x - 2} \right|$

Ans: The given functions are $\text{f}\left( \text{x} \right)\text{ = }\left| \text{x} \right|$ and $\text{g}\left( \text{x} \right)\text{ = }\left| \text{5x - 2} \right|$

Calculating $\text{gof}$:

$\text{go f}\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= g}\left( \text{ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ } \right)$

$\text{= }\left| \text{5} \right|\text{x}\left| \text{-2} \right|$

Calculating $\text{fog}$:

$\text{fog}\left( \text{x} \right)\text{ = f}\left( \text{g}\left( \text{x} \right) \right)$

$\text{= f}\left( \text{ }\!\!|\!\!\text{ 5x-2 }\!\!|\!\!\text{ } \right)$

$\text{= }\left| \left| \text{5x-2} \right| \right|$

$\text{= }\left| \text{5x-2} \right|$

1. The functions are $\left( \text{x} \right)\text{=8}{{\text{x}}^{\text{3}}}$ and $\left( \text{x} \right)\text{=}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}$

Ans: The given functions are $\text{f}\left( \text{x} \right)\text{=8}{{\text{x}}^{\text{3}}}$ and $\text{g}\left( \text{x} \right)\text{=}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}$

Calculating $\text{gof}$:

$\text{go f}\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= g}\left( \text{8}{{\text{x}}^{\text{3}}} \right)$

$\text{= }{{\left( \text{8}{{\text{x}}^{\text{3}}} \right)}^{\dfrac{1}{3}}}$

$\text{= 2x}$

Calculating $\text{fog}$:

$\text{fog}\left( \text{x} \right)\text{ = f}\left( \text{g}\left( \text{x} \right) \right)$

$\text{= f}{{\left( {{\text{x}}^{\dfrac{\text{1}}{\text{3}}}} \right)}^{\text{3}}}$

$\text{= 8}{{\left( {{\text{x}}^{\dfrac{\text{1}}{\text{3}}}} \right)}^{\text{3}}}$

$\text{= 8x}$

4. If $\left( \text{x} \right)\text{=}\dfrac{\left( \text{4x+3} \right)}{\left( \text{6x-4} \right)}\text{,x}\ne \dfrac{\text{2}}{\text{3}}$, show that $\text{fof}\left( \text{x} \right)\text{ = x}$, for all $\text{x}\ne \dfrac{\text{2}}{\text{3}}$. What is the inverse of $\text{f}$?

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{(4x+3)}}{\text{(6x-4)}}\text{,x}\ne \dfrac{\text{2}}{\text{3}}$.

So,

$\left( \text{fof} \right)\left( \text{x} \right)\text{=f}\left( \text{f}\left( \text{x} \right) \right)$

$\text{=f}\dfrac{\text{(4x+3)}}{\text{(6x-4)}}$

$\text{=}\dfrac{\text{4}\left( \dfrac{\text{4x+3}}{\text{6x-4}} \right)\text{+3}}{\text{6}\left( \dfrac{\text{4x+3}}{\text{6x-4}} \right)\text{-4}}$

$\text{=}\dfrac{\text{16x+12+18x-12}}{\text{24x+18-24x+16}}$

$\text{=}\dfrac{\text{34x}}{\text{34}}$

$\text{=x}$

Hence proved.

$\Rightarrow \text{fof = }{{\text{I}}_{\text{x}}}$

Therefore, the given function $\text{f}$ is invertible and the inverse of the function $\text{f}$ is the function $\text{f}$ itself.

5. State with reason whether following functions have inverse

1. The function $\text{f: }\left\{ \text{1, 2, 3, 4} \right\}\text{ }\to \text{ }\left\{ \text{10} \right\}$ defined as $\text{f = }\left\{ \left( \text{1, 10} \right)\text{, }\left( \text{2, 10} \right)\text{, }\left( \text{3, 10} \right)\text{, }\left( \text{4, 10} \right) \right\}$

Ans: The function $\text{f: }\left\{ \text{1, 2, 3, 4} \right\}\text{ }\to \text{ }\left\{ \text{10} \right\}$ defined as $\text{f = }\left\{ \left( \text{1, 10} \right)\text{, }\left( \text{2, 10} \right)\text{, }\left( \text{3, 10} \right)\text{, }\left( \text{4, 10} \right) \right\}$

As we can see, the function $\text{f}$ is a many one function as

$\text{f}\left( \text{1} \right)\text{ = f}\left( \text{2} \right)\text{ = f}\left( \text{3} \right)\text{ = f}\left( \text{4} \right)\text{ = 10}$

Hence the function $\text{f}$ is not one – one.

Therefore, the function $\text{f}$ does not have an inverse.

1. The function $\text{g: }\left\{ \text{5, 6, 7, 8} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 3, 4} \right\}$ defined as $\text{g = }\left\{ \left( \text{5, 4} \right)\text{, }\left( \text{6, 3} \right)\text{, }\left( \text{7, 4} \right)\text{, }\left( \text{8, 2} \right) \right\}$

Ans: The function $\text{g: }\left\{ \text{5, 6, 7, 8} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 3, 4} \right\}$ is defined as $\text{g = }\left\{ \left( \text{5, 4} \right)\text{, }\left( \text{6, 3} \right)\text{, }\left( \text{7, 4} \right)\text{, }\left( \text{8, 2} \right) \right\}$

As we can see, the function $\text{g}$ is a many one function as

$\text{g}\left( \text{5} \right)\text{=g}\left( \text{7} \right)\text{=4}$.

Hence the function $\text{g}$ is not one – one.

Therefore, the function $\text{g}$ does not have an inverse.

1. The function $\text{h: }\left\{ \text{2, 3, 4, 5} \right\}\text{ }\to \text{ }\left\{ \text{7, 9, 11, 13} \right\}$ defined as $\text{h = }\left\{ \left( \text{2, 7} \right)\text{, }\left( \text{3, 9} \right)\text{, }\left( \text{4, 11} \right)\text{, }\left( \text{5, 13} \right) \right\}$

Ans: The function $\text{h: }\left\{ \text{2, 3, 4, 5} \right\}\text{ }\to \text{ }\left\{ \text{7, 9, 11, 13} \right\}$defined as

$\text{h = }\left\{ \left( \text{2, 7} \right)\text{, }\left( \text{3, 9} \right)\text{, }\left( \text{4, 11} \right)\text{, }\left( \text{5, 13} \right) \right\}$

All distinct elements of the set $\left\{ \text{2, 3, 4, 5} \right\}$ have distinct images under $\text{h}$.

Hence the function $\text{h}$ is one – one.

Since for every element $\text{y}$ of the set $\left\{ \text{7, 9, 11, 13} \right\}$ there is an element x in the set $\left\{ \text{2, 3, 4, 5} \right\}$ such that $\text{h}\left( \text{x} \right)\text{ = y}$, therefore the function $\text{h}$ is onto.

Hence. The function $\text{h}$ is a one – one and onto function.

Therefore the function $\text{h}$ has an inverse.

6. Show that $\text{f: }\left[ \text{-1, 1} \right]\to \text{R}$, given by$\left( \text{x} \right)\text{= X}\left( \text{X+2} \right)$ is one – one. Find the inverse of the function $\text{f: }\left[ \text{-1, 1} \right]\text{ }\to \text{ Range f}$.

(Hint: For $\text{y}\in \text{Range f, y=}\left( \text{x} \right)\text{=}\left( \text{X+2} \right)$ , for some $\text{x}$ in $\text{ }\!\![\!\!\text{ -1, 1 }\!\!]\!\!\text{ , i}\text{.e}\text{., x = 2y(1-y)}$)

Ans: The function $\text{f: }\left[ \text{-1, 1} \right]\to \text{R}$ is given as $\left( \text{x} \right)\text{= X}\left( \text{X+2} \right)$

For function $\text{f}$ to be one – one,

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow \left( \text{X+2} \right)\text{ = }\left( \text{Y+2} \right)$

$\Rightarrow \text{xy +2x = xy +2y}$

$\Rightarrow \text{2x = 2y}$

$\Rightarrow \text{x = y}$

Hence the function $\text{f}$ is one – one.

And $\text{f: }\left[ \text{-1, 1} \right]\text{ }\to \text{ Range f}$ is onto.

Therefore, the function $\text{f: }\left[ \text{-1, 1} \right]\text{ }\to \text{ Range f}$ is one – one and onto and therefore, the inverse of the function $\text{f: }\left[ \text{-1, 1} \right]\text{ }\to \text{ Range f}$ exists.

Let us assume that the function $\text{g: Range f}\to \left[ \text{-1, 1} \right]$ be the inverse of $\text{f}$.

Let $\text{y}$ be an arbitrary element of range $\text{f}$.

Since $\text{f: }\left[ \text{-1, 1} \right]\text{ }\to \text{ Range f}$is onto,

$\text{y = f}\left( \text{x} \right)$ for some $\text{x}\in \text{ }\!\![\!\!\text{ -1, 1 }\!\!]\!\!\text{ }$

$\Rightarrow \text{y=}\dfrac{\text{x}}{\text{(x+2)}}$

$\Rightarrow \text{xy+2y=x}$

$\Rightarrow \text{x(1-y)=2y}$

$\Rightarrow \text{x-}\dfrac{\text{2y}}{\text{1-y}}\text{,y}\ne \text{1}$

Let us define $\text{g: Range f }\to \text{ }\left[ \text{-1, 1} \right]$ as:

$\text{g(y)=}\dfrac{\text{2y}}{\text{1-y}}\text{,y}\ne \text{1}$.

So,

$\text{(gof)(x)=g(f(x))}$

$\text{=g}\left( \dfrac{\text{x}}{\text{x+2}} \right)$

$\text{=2}\left( \dfrac{\text{2}\left( \dfrac{\text{x}}{\text{x+2}} \right)}{\text{1-}\left( \dfrac{\text{x}}{\text{x+2}} \right)} \right)$

$\text{=}\dfrac{\text{2x}}{\text{x+2-x}}$

$\text{=}\dfrac{\text{2x}}{\text{2}}$

$\text{=x}$

And

$\text{(fog)(y)=f(g(y))}$

$\text{=f}\left( \dfrac{\text{2y}}{\text{1-y}} \right)$

$\text{=2}\left(\dfrac{\dfrac{\text{2y}}{\text{1-y}}}{\dfrac{\text{2y}}{\text{1-y}}} \right)$

$\text{=}\dfrac{\text{2y}}{\text{2y+2-2y}}$

$\text{=}\dfrac{\text{2y}}{\text{2}}$

$\text{=y}$

Therefore, $\text{gof=x=}{{\text{I}}_{\left[ \text{-1,1} \right]\text{ }}}$ and $\text{fog=y=}{{\text{I}}_{\text{Range f}}}$.

$\Rightarrow {{\text{f}}^{\text{-1}}}\text{=g}$

Therefore, ${{\text{f}}^{\text{-1}}}\text{(y)=}\dfrac{\text{2y}}{\text{1-y}}\text{,y}\ne \text{1}$.

7. Consider $\text{f: R }\to \text{ R}$ given by $\text{f}\left( \text{x} \right)\text{=4x+3}$. Show that $\text{f}$ is invertible. Find the inverse of $\text{f}$.

Ans: The given function $\text{f: R }\to \text{ R}$ is given by, $\text{f}\left( \text{x} \right)\text{=4x+3}$

For function $\text{f}$ to be one – one

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

$\Rightarrow \text{4x +3 = 4y +3}$

$\Rightarrow \text{4x = 4y}$

$\Rightarrow \text{x = y}$

Hence the function $\text{f}$ is one – one.

For function $\text{f}$ to be onto

$\text{y}\in \text{R}$, let $\text{y = 4x + 3}$.

$\Rightarrow \text{x=}\dfrac{\text{y-3}}{\text{4}}\in \text{R}$

For any $\text{y}\in \text{R}$, there exists $\text{x=}\dfrac{\text{y-3}}{\text{4}}\in \text{R}$, such that

$\text{f}\left( \text{x} \right)\text{=f}\left( \dfrac{\text{y-3}}{\text{4}} \right)$

$\text{=4}\left( \dfrac{\text{y-3}}{\text{4}} \right)\text{+3}$

$\text{=y}$

Hence, the function $\text{f}$ is onto.

Therefore, the inverse of the function $\text{f}$ exists.

Now,

Defining $\text{g: R }\to \text{ R}$ by $\text{g}\left( \text{x} \right)\text{=}\dfrac{\text{y-3}}{\text{4}}$

$\left( \text{gof} \right)\left( \text{x} \right)\text{=g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= g}\left( \text{4x + 3} \right)$

$\text{=}\dfrac{\text{(4x + 3)-3}}{\text{4}}$

$\text{=}\dfrac{\text{4x}}{\text{4}}$

$\text{= x}$

and

$\left( \text{fog} \right)\left( \text{y} \right)\text{ = f}\left( \text{g}\left( \text{y} \right) \right)$

$\text{= f}\left( \dfrac{\text{y-3}}{\text{4}} \right)$

$\text{= 4}\left( \dfrac{\text{y-3}}{\text{4}} \right)\text{+3}$

$\text{=y-3+3}$

$\text{=y}$

Therefore, $\text{gof=fog=}{{\text{I}}_{\text{R}}}$. The function $\text{f}$ is invertible and the inverse of $\text{f}$ is given by ${{\text{f}}^{\text{-1}}}\left( \text{y} \right)\text{= g}\left( \text{y} \right)\text{ =}\dfrac{\text{y-3}}{\text{4}}$.

8. Consider $\text{f: R+ }\to \text{ }\left[ \text{4, }\infty \right)$ given by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}\text{+4}$. Show that $\text{f}$ is invertible with the inverse ${{\text{f}}^{\text{-1}}}$ of given $\text{f}$ by ${{\text{f}}^{\text{-1}}}\text{(y)=}\sqrt{\text{Y-4}}$, where $\text{R+}$ is the set of all non-negative real numbers.

Ans: The given function $\text{f: R+ }\to \text{ }\left[ \text{4, }\infty \right)$ is defined as $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}\text{+4}$.

For function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$

${{\text{x}}^{\text{2}}}\text{ + 4 = }{{\text{y}}^{\text{2}}}\text{ + 4}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ = }{{\text{y}}^{\text{2}}}$

$\Rightarrow \text{x = y }\!\![\!\!\text{ as x = y}\in {{\text{R}}_{\text{+}}}\text{ }\!\!]\!\!\text{ }$

Hence the function $\text{f}$ is one – one.

For function $\text{f}$ to be onto:

For $\text{y}\in \text{ }\!\![\!\!\text{ 4,}\infty \text{)}$, let $\text{y = }{{\text{x}}^{\text{2}}}\text{+4}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ = y - 4 }\ge \text{ 0 }\left[ \text{as y }\ge \text{ 4} \right]$

$\Rightarrow \text{x=}\sqrt{\text{Y - 4}}\ge \text{ 0}$

Hence, for any $\text{y}\in \text{ }\!\![\!\!\text{ 4,}\infty \text{)}$ there exists $\text{x =}\sqrt{\text{Y - 4}}\in {{\text{R}}_{\text{+}}}$, such that

$\text{f}\left( \text{x} \right)\text{ = f}\left( \sqrt{\text{Y - 4}} \right)$

$\text{=}{{\left( \sqrt{\text{Y - 4}} \right)}^{\text{2}}}\text{+4}$

$\text{=y-4+4}$

$\text{= y}$

Hence the function $\text{f}$ is onto.

Since the function $\text{f}$ is one – one and onto and therefore, ${{\text{f}}^{\text{-1}}}$ exists.

Let us define $\text{g: }\left[ \text{4, }\infty \right)\text{ }\to \text{ R+}$ by $\text{g}\left( \text{y} \right)\text{ =Y - 4}$.

So,

$\left( \text{gof} \right)\left( \text{x} \right)\text{= g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= }\sqrt{\left( {{\text{x}}^{\text{2}}}\text{+4} \right)\text{-4}}$

$\text{= }\sqrt{{{\text{x}}^{\text{2}}}}$

$\text{=x}$

$\left( \text{fog} \right)\left( \text{y} \right)\text{ = f}\left( \text{g}\left( \text{y} \right) \right)$

$\text{= f}\left( \sqrt{\text{Y - 4}} \right)$

$\text{= }{{\left( \sqrt{\text{Y - 4}} \right)}^{\text{2}}}\text{ + 4}$

$\text{= y - 4 + 4}$

$\text{= y}$

Therefore, $\text{gof = fog = }{{\text{I}}_{\text{R}}}$.

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

9. Consider $\text{f: R+ }\to \text{ }\left[ \text{-5, }\infty \right)$ given by $\text{f}\left( \text{x} \right)\text{ = 9}{{\text{x}}^{\text{2}}}\text{ + 6x - 5}$. Show that $\text{f}$ is invertible with ${{\text{f}}^{\text{-1}}}\text{ }\left( \text{y} \right)\text{ =}\left( \dfrac{\text{(}\sqrt{\text{y+6}}\text{)-1}}{\text{3}} \right)$.

Ans: The function $\text{f: R+ }\to \text{ }\left[ \text{-5, }\infty \right)$ is given as $\text{f}\left( \text{x} \right)\text{ = 9}{{\text{x}}^{\text{2}}}\text{ + 6x - 5}$.

Let $\text{y}$ be an arbitrary element of $\left[ \text{-5, }\infty \right)$.

$\text{y=9}{{\text{x}}^{\text{2}}}\text{+6x--5}$

$\Rightarrow \text{y = }{{\left( \text{3x + 1} \right)}^{\text{2}}}\text{-1-5=}{{\left( \text{3x + 1} \right)}^{\text{2}}}\text{-6}$

$\Rightarrow \text{y + 6 = }{{\left( \text{3x + 1} \right)}^{\text{2}}}$

$\Rightarrow \text{3x+1 =}\sqrt{\text{Y+6}}\text{ }\!\![\!\!\text{ as y}\ge \text{-5}\Rightarrow \text{y + 6 0 }\!\!]\!\!\text{ }$

$\Rightarrow \text{x =}\left( \dfrac{\text{(}\sqrt{\text{y+6}}\text{)-1}}{\text{3}} \right)$

Therefore, function $\text{f}$ is onto, range $\text{f = }\!\![\!\!\text{ -5,}\infty \text{)}$.

Defining $\text{g: }\left[ \text{-5, }\infty \right)\text{ }\to \text{ R+}$ as $\text{g(y) =}\left( \dfrac{\text{(}\sqrt{\text{y+6}}\text{)-1}}{\text{3}} \right)$

$\left( \text{gof} \right)\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= g}\left( \text{9}{{\text{x}}^{\text{2}}}\text{ + 6x-5} \right)$

$\text{= g}\left( {{\left( \text{3x + 1} \right)}^{\text{2}}}\text{-6} \right)$

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

$\text{=}\dfrac{\text{3x+1-1}}{\text{3}}\text{=}\dfrac{\text{3x}}{\text{3}}$

$\text{=x}$

$\left( \text{fog} \right)\left( \text{y} \right)\text{ = f}\left( \text{g}\left( \text{y} \right) \right)$

$=\left( \dfrac{(\sqrt{y+6})-1}{3} \right)$

$=\left[ 3\left( \dfrac{(\sqrt{y+6})-1}{3} \right)+{{1}^{2}} \right]-6$

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

$\text{=y+6-6}$

$\text{=y}$

Therefore,

$\text{gof = x = }{{\text{I}}_{\text{R}}}$

$\text{fog = y = }{{\text{I}}_{\text{Range f}}}$

So, the function $\text{f}$ is invertible and the inverse of $\text{f}$ is given by:

${{\text{f}}^{\text{-1}}}\text{ }\left( \text{y} \right)\text{=g(y)=}\left( \dfrac{\text{(}\sqrt{\text{y+6}}\text{)-1}}{\text{3}} \right)$.

10. Let $\text{f: X }\to \text{ Y}$ be an invertible function. Show that $\text{f}$ has a unique inverse.

(Hint: suppose ${{\text{g}}_{\text{1}}}$ and ${{\text{g}}_{\text{2}}}$ are two inverses of $\text{f}$. Then for all$\text{y}\in \text{Y, fo}{{\text{g}}_{\text{1}}}\left( \text{y} \right)\text{ = }{{\text{I}}_{\text{Y}}}\left( \text{y} \right)\text{ = fo}{{\text{g}}_{\text{2}}}\left( \text{y} \right)$. Use one – oneness of $\text{f}$).

Ans: Let $\text{f: X }\to \text{ Y}$ be an invertible function.

Assuming that $\text{f}$ has two inverses (${{\text{g}}_{\text{1}}}$ and ${{\text{g}}_{\text{2}}}$)

For all $\text{y}\in \text{Y}$,

$\text{fo}{{\text{g}}_{\text{1}}}\left( \text{y} \right)\text{= }{{\text{I}}_{\text{Y}}}\left( \text{y} \right)$

$\text{= fo}{{\text{g}}_{\text{2}}}\left( \text{y} \right)$

$\Rightarrow \text{f}\left( {{\text{g}}_{\text{1}}}\text{ }\left( \text{y} \right) \right)\text{ = f}\left( {{\text{g}}_{\text{2}}}\text{ }\left( \text{y} \right) \right)$

$\Rightarrow {{\text{g}}_{\text{1}}}\text{ }\left( \text{y} \right)\text{ = }{{\text{g}}_{\text{2}}}\text{ }\left( \text{y} \right)$                   (as $\text{f}$ is invertible $\Rightarrow \text{f}$ is one – one)

$\Rightarrow {{\text{g}}_{\text{1}}}\text{ = }{{\text{g}}_{\text{2}}}$                                  (as $\text{g}$ is one – one)

Therefore, the function $\text{f}$ has a unique inverse.

11. Consider $\text{f: }\left\{ \text{1, 2, 3} \right\}\text{ }\to \text{ }\left\{ \text{a, b, c} \right\}$ given by $\text{f}\left( \text{1} \right)\text{ = a, f}\left( \text{2} \right)\text{ = b}$ and $\text{f}\left( \text{3} \right)\text{ = c}$. Find ${{\text{f}}^{\text{-1}}}$ and show that ${{\left( {{\text{f}}^{\text{-1}}} \right)}^{\text{-1}}}\text{ = f}$.

Ans: Function $\text{f: }\left\{ \text{1, 2, 3} \right\}\text{ }\to \text{ }\left\{ \text{a, b, c} \right\}$ is given by $\text{f}\left( \text{1} \right)\text{ = a, f}\left( \text{2} \right)\text{ = b}$ and$\text{f}\left( \text{3} \right)\text{ = c}$.

If we define $\text{g: }\left\{ \text{a, b, c} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 3} \right\}$ as $\text{g}\left( \text{a} \right)\text{ = 1, g}\left( \text{b} \right)\text{ = 2, g}\left( \text{c} \right)\text{ = 3}$.

We have

$\left( \text{fog} \right)\left( \text{a} \right)\text{ = f}\left( \text{g}\left( \text{a} \right) \right)$

$\text{= f}\left( \text{1} \right)$

$\text{= a}$

$\left( \text{fog} \right)\left( \text{b} \right)\text{ = f}\left( \text{g}\left( \text{b} \right) \right)$

$\text{= f}\left( \text{2} \right)$

$\text{= b}$

$\left( \text{fog} \right)\left( \text{c} \right)\text{ = f}\left( \text{g}\left( \text{c} \right) \right)$

$\text{= f}\left( \text{3} \right)$

$\text{= c}$

And,

$\left( \text{gof} \right)\left( \text{1} \right)\text{ = g}\left( \text{f}\left( \text{1} \right) \right)\text{ }$

$\text{= f}\left( \text{a} \right)$

$\text{= 1}$

$\left( \text{gof} \right)\left( \text{2} \right)\text{ = g}\left( \text{f}\left( \text{2} \right) \right)\text{ }$

$\text{= f}\left( \text{b} \right)$

$\text{= 2}$

$\left( \text{gof} \right)\left( \text{3} \right)\text{ = g}\left( \text{f}\left( \text{3} \right) \right)\text{ }$

$\text{= f}\left( \text{c} \right)$

$\text{= 3}$

So, $\text{gof=}{{\text{I}}_{\text{X}}}$ and $\text{fog = }{{\text{I}}_{\text{Y}}}$, where $\text{X = }\left\{ \text{1, 2, 3} \right\}$ and $\text{Y= }\left\{ \text{a, b, c} \right\}$.

Therefore, the inverse of $\text{f}$ exists and ${{\text{f}}^{\text{-1}}}\text{=g}$.

So, function ${{\text{f}}^{\text{-1}}}\text{:}\left\{ \text{ a, b, c} \right\}\text{ }\to \text{ }\left\{ \text{1, 2, 3} \right\}$ is given by ${{\text{f}}^{\text{-1}}}\left( \text{a} \right)\text{ = 1, f-1 }\left( \text{b} \right)\text{ = 2, f--1 }\left( \text{c} \right)\text{ = 3}$

If we define $\text{h: }\left\{ \text{1, 2, 3} \right\}\text{ }\to \text{ }\left\{ \text{a, b, c} \right\}$ as $\text{h }\left( \text{1} \right)\text{ = a, h }\left( \text{2} \right)\text{ = b, h }\left( \text{3} \right)\text{ = c}$.

$\left( \text{goh} \right)\left( \text{1} \right)\text{ = g}\left( \text{h}\left( \text{1} \right) \right)\text{ }$

$\text{= g}\left( \text{a} \right)$

$\text{= 1}$

$\left( \text{goh} \right)\left( \text{2} \right)\text{ = g}\left( \text{h}\left( \text{2} \right) \right)\text{ }$

$\text{= g}\left( \text{b} \right)$

$\text{= 2}$

$\left( \text{goh} \right)\left( \text{3} \right)\text{ = g}\left( \text{h}\left( \text{3} \right) \right)\text{ }$

$\text{= g}\left( \text{c} \right)$

$\text{= 3}$

And

$\left( \text{hog} \right)\left( \text{a} \right)\text{ = h}\left( \text{g}\left( \text{a} \right) \right)$

$\text{= h}\left( \text{1} \right)$

$\text{= a}$

$\left( \text{hog} \right)\left( \text{b} \right)\text{ = h}\left( \text{g}\left( \text{b} \right) \right)$

$\text{= h}\left( \text{2} \right)$

$\text{= b}$

$\left( \text{hog} \right)\left( \text{c} \right)\text{ = h}\left( \text{g}\left( \text{c} \right) \right)$

$\text{= h}\left( \text{3} \right)$

$\text{= c}$

So, $\text{goh=}{{\text{I}}_{\text{X}}}$ and $\text{hog = }{{\text{I}}_{\text{Y}}}$, where $\text{X = }\left\{ \text{1, 2, 3} \right\}$ and $\text{Y= }\left\{ \text{a, b, c} \right\}$.

The inverse of $\text{g}$ exists and ${{\text{g}}^{\text{-1}}}\text{ = h}\Rightarrow {{\text{(}{{\text{f}}^{\text{-1}}}\text{)}}^{\text{-1}}}\text{ = h}$.

So,$\text{ }\!\!~\!\!\text{ h = f}$.

Hence, ${{\left( {{\text{f}}^{\text{-1}}} \right)}^{\text{-1}}}\text{ = f}$.

12. Let $\text{f: X }\to \text{ Y}$ be an invertible function. Show that the inverse of ${{\text{f}}^{\text{-1}}}$ is $\text{f}$, i.e., ${{\left( {{\text{f}}^{\text{-1}}} \right)}^{\text{-1}}}\text{ = f}$.

Ans: Let $\text{f: X }\to \text{ Y}$ be an invertible function.

Then there exists a function $\text{g: Y }\to \text{ X}$ such that $\text{gof = }{{\text{I}}_{\text{X}}}$ and $\text{fog = }{{\text{I}}_{\text{Y}}}$.

So, ${{\text{f}}^{\text{-1}}}\text{=g}$.

$\Rightarrow \text{gof = }{{\text{I}}_{\text{X}}}$ and $\text{fog = }{{\text{I}}_{\text{Y}}}$.

Hence, ${{\text{f}}^{\text{-1}}}\text{: Y }\to \text{ X}$ is invertible and $\text{f}$ is the inverse of ${{\text{f}}^{\text{-1}}}$ i.e., ${{\left( {{\text{f}}^{\text{-1}}} \right)}^{\text{-1}}}\text{ = f}$.

13. If $\text{f: R }\to \text{ R}$ be given by $\left( \text{x} \right)\text{=}{{\left( \text{3-}{{\text{x}}^{\text{3}}} \right)}^{\dfrac{\text{1}}{\text{3}}}}$, then $\text{fof }\left( \text{x} \right)$ is

1. ${{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}$

2. ${{\text{x}}^{\text{3}}}$

3. $\text{x}$

4. $\left( \text{3 - }{{\text{x}}^{\text{3}}} \right)$

Ans: The function $\text{f: R }\to \text{ R}$ is defined as $\text{f}\left( \text{x} \right)\text{=}{{\left( \text{3-}{{\text{x}}^{\text{3}}} \right)}^{\dfrac{\text{1}}{\text{3}}}}$

$\text{fof }\left( \text{x} \right)\text{=f(f(x))}$

$\text{=f(3-}{{\text{x}}^{\text{3}}}{{\text{)}}^{\dfrac{\text{1}}{\text{3}}}}$

$\text{= }\!\![\!\!\text{ 3-(3-}{{\text{x}}^{\text{3}}}\text{)}{{\text{ }\!\!]\!\!\text{ }}^{\dfrac{\text{1}}{\text{3}}}}$

$\text{=(}{{\text{x}}^{\text{3}}}{{\text{)}}^{\dfrac{\text{1}}{\text{3}}}}$

Therefore, $\text{fof}\left( \text{x} \right)\text{ = x}$.

The correct answer is ($C$) $\text{x}$.

14. Let $\text{f: R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}\text{ }\to \text{ R}$ be a function as $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{4x}}{\text{3x+4}}$. The inverse of $\text{f}$ is map $\text{g: Range f}\to \text{R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}$ given by

1. $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{3y}}{\text{3-4y}}$.

2. $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{4y}}{\text{4-3y}}$.

3. $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{4y}}{\text{4-3y}}$.

4. $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{3y}}{\text{4-3y}}$.

Ans: The function $\text{f: R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}\text{ }\to \text{ R}$ is dined as a function $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{4x}}{\text{3x+4}}$.

Let $\text{y}$ be an arbitrary element of Range $\text{f}$.

There exists $\text{x}\in \text{R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}$ such that $\text{y = f}\left( \text{x} \right)$

$\text{y=}\dfrac{\text{4x}}{\text{3x+4}}$

$\Rightarrow \text{3xy+4y=4x}$

$\Rightarrow \text{x(4-3y)=4y}$

$\Rightarrow \text{x=}\dfrac{\text{4y}}{\text{4-3y}}$

Defining $\text{g: Range f}\to \text{R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}$ as $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{4y}}{\text{4-3y}}$.

Calculating $\text{gof(x)}$:

$\Rightarrow \left( \dfrac{\text{4x}}{\text{3x+4}} \right)\text{=}\dfrac{\text{4}\left( \dfrac{\text{4x}}{\text{3x+4}} \right)}{\text{4-3}\left( \dfrac{\text{4x}}{\text{3x+4}} \right)}$

$\text{=}\dfrac{\text{16x}}{\text{12x+16-12x}}$

$\text{=}\dfrac{\text{16x}}{\text{16}}$

$\text{=x}$

Therefore, $\text{gof(x)= g(f(x)) = g}$.

Calculating $\text{fo}\left( \text{y} \right)$:

$\text{fo(y)=f(g(y))}$

$\text{=f(}\dfrac{\text{4y}}{\text{4-3y}}\text{)}$

$\text{=}\dfrac{\text{4}\left( \dfrac{\text{4y}}{\text{4-3y}} \right)}{\text{3}\left( \dfrac{\text{4y}}{\text{4-3y}} \right)\text{+4}}$

$\text{=}\dfrac{\text{16y}}{\text{12y+16-12x}}$

$\text{=}\dfrac{\text{16y}}{\text{16}}$

$\text{=y}$

Therefore, $\text{fo(y) = f(g(y)) = f}$.

So, $\text{gof=}{{\text{I}}_{\text{R-}\left\{ \dfrac{\text{4}}{\text{3}} \right\}}}$ and $\text{fog=}{{\text{I}}_{\text{Range}\,\text{f}}}$

Hence, $\text{g}$ is the inverse of $\text{f}$ i.e., ${{\text{f}}^{\text{-1}}}\text{=g}$.

The inverse of $\text{f}$ is the map $\text{g: Range f}\to \text{R-}\left\{ \text{-}\dfrac{\text{4}}{\text{3}} \right\}$which is given by $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{4y}}{\text{4-3y}}$.

So, the correct answer is (B) $\text{g}\left( \text{y} \right)\text{=}\dfrac{\text{4y}}{\text{4-3y}}$.

### Exercise 1.4

1. Determine whether or not each of the definitions given below gives a binary operation.

In the event that $\text{*}$ is not a binary operation, give justification for this.

1. On ${{\text{Z}}^{\text{+}}}$, define $\text{*}$ by $\text{a * b = a-b}$.

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = a-b}$.

So, $\text{*}$ is not a binary operation as the image of $\left( \text{1, 2} \right)$ under $\text{*}$ is $\text{1 * 2=1-}\,\text{2=}\,\text{-1}\notin {{\text{Z}}^{\text{+}}}$.

1. On, ${{\text{Z}}^{\text{+}}}$ define $\text{*}$ by $\text{a * b = ab}$.

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = ab}$.

For each $\text{a, b}\in {{\text{Z}}^{\text{+}}}$ there is a unique element $\text{ab}$ in ${{\text{Z}}^{\text{+}}}$.

So, we can say that $\text{*}$ carries each pair $\left( \text{a, b} \right)$ to a unique element $\text{a * b = ab}$ in ${{\text{Z}}^{\text{+}}}$. Therefore, $\text{*}$ is a binary operation.

1. On $\text{R}$, define $\text{*}$ by $\text{a * b = a}{{\text{b}}^{\text{2}}}$.

Ans: On $\text{R}$, $\text{*}$ is defined by $\text{a * b = a}{{\text{b}}^{\text{2}}}$.

For each $a,b \in R$, there is a unique element $\text{a}{{\text{b}}^{\text{2}}}$ in $\text{R}$.

Therefore, $\text{*}$ carries each pair $\left( \text{a, b} \right)$ to a unique element $\text{a * b = a}{{\text{b}}^{\text{2}}}$ in $\text{R}$.

Hence, $\text{*}$ is a binary operation.

1. On, ${{\text{Z}}^{\text{+}}}$ define $\text{*}$ by $\text{a*b=}\left| \text{a-b} \right|$

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = }\left| \text{a - b} \right|$.

For each $\text{a, b}\in {{\text{Z}}^{\text{+}}}$ there is a unique element $\left| \text{a - b} \right|$ in ${{\text{Z}}^{\text{+}}}$.

Therefore, $\text{*}$ carries each pair $\left( {a,b} \right)$ to a unique element $\text{a * b = }\left| \text{a - b} \right|$ in ${{\text{Z}}^{\text{+}}}$. Hence, $\text{*}$ is a binary operation.

1. On ${{\text{Z}}^{\text{+}}}$, define $\text{*}$ by $\text{a * b = a}$

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = a}$.

For each $\text{a, b}\in {{\text{Z}}^{\text{+}}}$ there is a unique element $\text{a}$ in ${{\text{Z}}^{\text{+}}}$.

Therefore, $\text{*}$ carries each pair $\left( \text{a , b} \right)$ to a unique element $\text{a * b = a}$ in ${{\text{Z}}^{\text{+}}}$. Hence, $\text{*}$ is a binary operation.

2. For each binary operation $\text{*}$ defined below, determine whether $\text{*}$ is commutative or associative.

1. On $\text{Z}$, define $\text{a * b = a-b}$

Ans: On $\text{Z}$, $\text{*}$ is defined by $\text{a}$.

For $\text{1, 2}\in \text{Z}$,

$\text{1*2=1-2}$

$=-1$

$\text{2*1=2-1}$

$\text{=1}$

Therefore $\text{1 * 2}\ne \text{2 * 1}$.

Hence, the operation $\text{*}$ is not commutative.

For $\text{1, 2, 3}\in \text{Z}$,

$\left( \text{1*2} \right)\text{*3=}\left( \text{1-2} \right)\text{*3}$

$\text{=-1*3}$

$\text{=-1-3}$

$\text{=}\,\text{-4}$

$\text{1*}\left( \text{2*3} \right)\text{=1*}\left( \text{2-3} \right)$

$=1*-1$

$=1-\left( -1 \right)$

$=2$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

Hence, the operation $\text{*}$ is not associative.

1. On $\text{Q}$, define $a * b = ab + 1$.

Ans: On $\text{Q}$, $\text{*}$ is defined by $\text{a*b=ab+1}$.

For all $\text{a, b}\in \text{Q}$, $\text{ab = ba}$.

$\Rightarrow \text{ab + 1 = ba + 1}$ for all $\text{a, b}\in \text{Q}$

$\Rightarrow \text{a * b = a * b}$ for all $\text{a, b}\in \text{Q}$

Hence, operation $\text{*}$ is commutative.

For $\text{1, 2, 3}\in \text{Q}$,

$\left( \text{1 * 2} \right)\text{ * 3 = }\left( \text{1 }\!\!\times\!\!\text{ 2 + 1} \right)\text{ * 3 }$

$\text{= 3 * 3 }$

$\text{= 3 }\!\!\times\!\!\text{ 3 + 1 }$

$\text{= 10}$

$\text{1 * }\left( \text{2 * 3} \right)\text{ = 1 * }\left( \text{2 }\!\!\times\!\!\text{ 3 + 1} \right)\text{ }$

$\text{= 1 * 7 }$

$\text{= 1 }\!\!\times\!\!\text{ 7 + 1 }$

$\text{= 8}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

So, the operation $\text{*}$ is not associative.

1. On $\text{Q}$, define $\text{a * b =}\dfrac{\text{ab}}{\text{2}}$.

Ans: On $\text{Q}$, $\text{*}$ is defined by $\text{a * b =}\dfrac{\text{ab}}{\text{2}}$.

For all $\text{a, b}\in \text{Q}$, $\text{ab = ba}$.

$\Rightarrow \dfrac{\text{ab}}{\text{2}}=\dfrac{\text{ba}}{\text{2}}$ for all $\text{a, b}\in \text{Q}$

$\Rightarrow \text{a * b = b * a}$ for all $\text{a, b}\in \text{Q}$

Hence, operation $\text{*}$ is commutative.

For $\text{1, 2, 3}\in \text{Q}$,

$\text{(a*b)*c=}\left( \dfrac{\text{ab}}{\text{2}} \right)\text{*c}$

$\text{=}\dfrac{\left( \dfrac{\text{ab}}{\text{2}} \right)\text{c}}{\text{2}}$

$\text{=}\dfrac{\text{abc}}{\text{4}}$

And

$\text{a*(b*c)=}\left( \dfrac{\text{bc}}{\text{2}} \right)\text{*c}$

$\text{=}\dfrac{\left( \dfrac{\text{bc}}{\text{2}} \right)\text{a}}{\text{2}}$

$\text{=}\dfrac{\text{abc}}{\text{4}}$

Therefore, $\left( \text{a*b} \right)\text{*c = a*}\left( \text{b*c} \right)$.

So, the operation $\text{*}$ is associative.

1. On ${{\text{Z}}^{\text{+}}}$, define $\text{a * b = }{{\text{2}}^{\text{ab}}}$.

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = }{{\text{2}}^{\text{ab}}}$.

For all $\text{a, b}\in {{\text{Z}}^{\text{+}}}$, $ab=ba$.

$\Rightarrow {{\text{2}}^{\text{ab}}}\text{ = }{{\text{2}}^{\text{ba}}}$ for all $\text{a, b}\in {{\text{Z}}^{\text{+}}}$

$\Rightarrow \text{a * b = b * a}$ for all $\text{a, b}\in {{\text{Z}}^{\text{+}}}$

Hence, operation $\text{*}$ is commutative.

For $\text{1, 2, 3}\in {{\text{Z}}^{\text{+}}}$,

$\text{(1*2)*3=}{{\text{2}}^{\text{1 }\!\!\times\!\!\text{ 2}}}\text{*3}$

$\text{=4*3}$

$\text{=}{{\text{2}}^{\text{4 }\!\!\times\!\!\text{ 3}}}$

$\text{=}{{\text{2}}^{\text{12}}}$

$\text{1*(2*3) = 1*}{{\text{2}}^{\text{2 }\!\!\times\!\!\text{ 3}}}\text{ }$

$\text{= 1*}{{\text{2}}^{\text{6}}}\text{ }$

$\text{= 1*64 }$

$\text{= }{{\text{2}}^{\text{1 }\!\!\times\!\!\text{ 64}}}\text{ }$

$\text{= }{{\text{2}}^{\text{64}}}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 *3} \right)$.

So, the operation $\text{*}$ is not associative.

1. On ${{\text{Z}}^{\text{+}}}$, define $\text{a*b=}{{\text{a}}^{\text{b}}}$.

Ans: On ${{\text{Z}}^{\text{+}}}$, $\text{*}$ is defined by $\text{a * b = }{{\text{a}}^{\text{b}}}$.

For $\text{1, 2}\in {{\text{Z}}^{\text{+}}}$,

$\text{1*2 = }{{\text{1}}^{\text{2}}}\text{ }$

$\text{=1}$

$\text{2*1=}{{\text{2}}^{\text{1}}}$

$\text{=2}$

Therefore, $\text{1 * 2}\ne \text{2 * 1}$.

Hence, the operation $\text{*}$ is not commutative.

For $\text{2, 3, 4}\in {{\text{Z}}^{\text{+}}}$

$\text{(2*3)*4 = }{{\text{2}}^{\text{3}}}\text{*4}$

$\text{=8*4}$

$\text{=}{{\text{8}}^{\text{4}}}$

$\text{= }{{\text{2}}^{\text{12}}}$

$\text{2*(3*4)=2*}{{\text{3}}^{\text{4}}}$

$\text{=2*81}$

$\text{=}{{\text{2}}^{\text{81}}}$

Therefore, $\left( \text{2 * 3} \right)\text{ * 4}\ne \text{2 * }\left( \text{3 * 4} \right)$.

So, the operation $\text{*}$ is not associative.

1. On $\text{R-}\left\{ \text{1} \right\}$ define $\text{a*b = }\dfrac{\text{a}}{\text{b+1}}$.

Ans: On $\text{R - }\left\{ \text{-1} \right\}$, $\text{*}$ is defined by $\text{a*b = }\dfrac{\text{a}}{\text{b+1}}$.

For $\text{2, 3}\in \text{R- }\!\!\{\!\!\text{ -1 }\!\!\}\!\!\text{ }$.

$\text{2*3 = }\dfrac{\text{2}}{\text{3+1}}$

$\text{=}\dfrac{\text{1}}{\text{2}}$

$\text{3*2 = }\dfrac{\text{3}}{\text{2+1}}$

$\text{=1}$

Therefore, $\text{2 * 3}\ne \text{3*2}$.

Hence, the operation $\text{*}$ is not commutative.

For $\text{2, 3, 4}\in \text{R- }\!\!\{\!\!\text{ -1 }\!\!\}\!\!\text{ }$

$\text{(2*3)*4=}\dfrac{\text{2}}{\text{3+1}}\text{*4}$

$\text{=}\dfrac{\text{1}}{\text{2}}\text{*4}$

$\text{=}\dfrac{\dfrac{\text{1}}{\text{2}}}{\text{4+1}}$

$\text{=}\dfrac{\text{1}}{\text{10}}$

$\text{2*(3*4) = 2*}\dfrac{\text{3}}{\text{4+1}}$

$\text{=2*}\dfrac{\text{3}}{\text{5}}$

$\text{=}\dfrac{\text{2}}{\dfrac{\text{3}}{\text{5}}\text{+1}}$

$\text{=}\dfrac{\text{2}}{\dfrac{\text{8}}{\text{5}}}$

$\text{=}\dfrac{\text{5}}{\text{4}}$

Therefore, $\left( \text{2 * 3} \right)\text{ * 4}\ne \text{2 * }\left( \text{3 * 4} \right)$.

So, the operation $\text{*}$ is not associative.

3. Consider the binary operation $\hat{\ }$ on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ defined by $\text{a}\wedge \text{b = min }\left\{ \text{a, b} \right\}$. Write the operation table of the operation $\wedge$.

Ans: The binary operation $\wedge$ on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ is defined by $\text{a}\wedge \text{b = min }\left\{ \text{a, b} \right\}$ for all $\text{a, b}\in \left\{ \text{1,2, 3, 4, 5} \right\}$.

Operation table for the given binary operation $\wedge$ is:

 $\wedge$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{2}$ $\text{1}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{3}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{3}$ $\text{3}$ $\text{4}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$

4. Consider a binary operation $\text{*}$ on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ given by the following multiplication table.

(Hint: use the following table)

 $\text{*}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{2}$ $\text{1}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{3}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{3}$ $\text{3}$ $\text{4}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$

1. Compute $\left( \text{2 * 3} \right)\text{ * 4}$ and $\text{2 * }\left( \text{3 * 4} \right)$.

Ans: Calculating $\left( \text{2 * 3} \right)\text{ * 4}$:

$\left( \text{2 * 3} \right)\text{ * 4 = 1 * 4 }$

$\text{= 1}$

Calculating $\text{2 * }\left( \text{3 * 4} \right)$:

$\text{2 * }\left( \text{3 * 4} \right)\text{ = 2 * 1 }$

$\text{= 1}$

1. Is $\text{*}$ commutative?

Ans: The operation $\text{*}$ is commutative because for every $\text{a, b}\in \left\{ \text{1,2, 3, 4, 5} \right\}$, we have $\text{a * b = b * a}$.

1. Compute $\left( \text{2*3} \right)\text{*}\left( \text{4*5} \right)$.

Ans: Calculating $\left( \text{2 * 3} \right)\text{ * }\left( \text{4 * 5} \right)$:

Since,

$\left( \text{2 * 3} \right)\text{ = 1}$

$\left( \text{4 * 5} \right)\text{ = 1}$

So,

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

$\text{=1}$

5. Let $\text{* }\!\!'\!\!\text{ }$ be the binary operation on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ defined by $\text{a *}\prime \text{ b = H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b}$. Is the operation $\text{* }\!\!'\!\!\text{ }$ same as the operation $\text{*}$ defined in Exercise $4$ above? Justify your answer.

Ans: The binary operation $\text{* }\!\!'\!\!\text{ }$ on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ is defined by $\text{a *}\prime \text{ b = H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b}$.

Operation table for the operation $\text{* }\!\!'\!\!\text{ }$ can be given as:

 $\text{* }\!\!'\!\!\text{ }$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{1}$ $\text{2}$ $\text{1}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{3}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{3}$ $\text{3}$ $\text{4}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$

As we can see, the operation tables for the operations $\text{*}$ and $\text{* }\!\!'\!\!\text{ }$ are the same. Therefore, operation $\text{* }\!\!'\!\!\text{ }$ is same as the operation $\text{*}$.

6. Let $\text{*}$ be the binary operation on $\text{N}$ given by $\text{a * b = L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$. Find

1. Calculate $\text{5 * 7}$ and $\text{20 * 16}$.

Ans: Calculating $\text{5 * 7}$:

Since, $\text{a * b = L}\text{.C}\text{.M}$ of $a$ and $\text{b}$.

$\text{5 * 7 = L}\text{.C}\text{.M}$ of $5$ and $\text{7}$.

$\text{=35}$

Calculating $\text{20 * 16}$:

Since, $\text{a * b = L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$.

$\text{20 * 16= L}\text{.C}\text{.M}$ of $\text{20}$ and $\text{16}$.

$\text{=80}$

1. Is $\text{*}$ commutative?

Ans: Since $\text{L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$ is equal to $\text{L}\text{.C}\text{.M}$ of $\text{b}$ and $\text{a}$ for all $\text{a, b}\in \text{N}$.

So, $\text{a * b = b * a}$.

Therefore, the operation $\text{*}$ is commutative.

1. Is $\text{*}$ associative?

Ans: For $a,b,c \in N$,

$\left( \text{a * b} \right)\text{ * c =}$ ($\text{L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$) $\text{* c =}$ $\text{L}\text{.C}\text{.M}$ of $\text{a,b}$ and $\text{c}$.

$\text{a * }\left( \text{b * c} \right)\text{ = a *}$ ($\text{L}\text{.C}\text{.M}$ of $\text{b}$ and $\text{c}$ $=$ $\text{L}\text{.C}\text{.M}$ of $\text{a,b}$ and $\text{c}$.

So, $\left( \text{a * b} \right)\text{ * c = a * }\left( \text{b * c} \right)$.

Therefore operation $\text{*}$ is associative.

1. Find the identity of  $\text{*}$ in $\text{N}$.

Ans: The binary operation $\text{*}$ on $\text{N}$ is defined as $\text{L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$.

Now,

$\text{L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{1}$ is equal to $\text{L}\text{.C}\text{.M}$ of $1$ and $a$ for all $\text{a}\in \text{N}$.

So, $\text{a * 1 = a = 1 * a}$ for all $\text{a}\in \text{N}$.

Therefore, $\text{1}$ is the identity of $\text{*}$ in $\text{N}$.

1. Which elements of $\text{N}$ are invertible for the operation $\text{*}$?

Ans: For $\text{a, b}\in \text{N}$, the elements in $\text{N}$ are invertible with respect to the operation $\text{*}$ only for the condition $\text{a * b = e = b * a}$.

$e=1$

$\text{L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b =1= L}\text{.C}\text{.M}$ of $\text{b}$ and $\text{a}$ for all $\text{a, b}\in \text{N}$.

This is only possible when $\text{a}$ and $\text{b}$ are equal to $\text{1}$.

Therefore, $\text{1}$ is the only invertible element of $\text{N}$ with respect to the operation $\text{*}$.

7. Is $\text{*}$ defined on the set $\left\{ \text{1, 2, 3, 4, 5} \right\}$ by $\text{a * b = L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$ a binary operation? Justify your answer.

Ans: The operation $\text{*}$ on the set $\text{A=}\left\{ \text{1, 2, 3, 4, 5} \right\}$ is defined as $\text{a * b = L}\text{.C}\text{.M}$ of $\text{a}$ and $\text{b}$.

The operation table for $\text{*}$ is as follows:

 $\text{*}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$ $\text{1}$ $\text{1}$ $\text{2}$ $\text{3}$ $\text{4}$ $\text{5}$ $\text{2}$ $\text{2}$ $\text{2}$ $\text{6}$ $\text{4}$ $\text{10}$ $\text{3}$ $\text{3}$ $\text{6}$ $\text{3}$ $\text{12}$ $\text{15}$ $\text{4}$ $\text{4}$ $\text{4}$ $\text{12}$ $\text{4}$ $\text{20}$ $\text{5}$ $\text{5}$ $\text{10}$ $\text{15}$ $\text{20}$ $\text{5}$

$\text{3 * 2 = 2 * 3 = 6}\notin \text{A}$

$\text{5 * 2 = 2 * 5 = 10}\notin \text{A}$

$\text{3 * 4 = 4 * 3 = 12}\notin \text{A}$.

$\text{3 * 5 = 5 * 3 = 15}\notin \text{A}$

$\text{4 * 5 = 5 * 4 = 20}\notin \text{A}$

Hence, the given operation $\text{*}$ is not a binary operation.

8. Let $\text{*}$ be the binary operation on $\text{N}$ defined by $\text{a * b = H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b}$. Is $\text{*}$ commutative? Is $\text{*}$ associative? Does there exist an identity for this binary operation on $\text{N}$?

Ans: The binary operation $\text{*}$ on $\text{N}$ is defined as: $\text{a * b = H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b}$.

Since, $\text{H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b=H}\text{.C}\text{.F}$ of $\text{b}$ and $\text{a}$ for all $\text{a, b}\in \text{N}$.

Therefore, $\text{a * b = b * a}$.

Hence, operation $\text{*}$ is commutative.

$\left( \text{a * b} \right)\text{ * c =}$ ($\text{H}\text{.C}\text{.F}$ of $\text{a}$ and $\text{b}$) $\text{* c =}$ $\text{H}\text{.C}\text{.F}$ of $\text{a,b}$ and $\text{c}$.

$\text{a * }\left( \text{b * c} \right)\text{ = a *}$ ($\text{H}\text{.C}\text{.F}$ of $\text{b}$ and $\text{c}$ $=$ $\text{H}\text{.C}\text{.F}$ of $\text{a,b}$ and $\text{c}$.

So, $\left( \text{a * b} \right)\text{ * c = a * }\left( \text{b * c} \right)$.

Therefore operation $\text{*}$ is associative. $\text{a * 1 = a = 1 * a}$ for all $\text{a}\in \text{N}$

So, $\text{e}\in \text{N}$ will be the identity for the operation $\text{*}$ if $\text{a * e = a = e * a}$ for all $\text{e}\in \text{N}$. But this relation is not true for any $\text{e}\in \text{N}$.

Therefore, operation $\text{*}$ does not have any identity in $\text{N}$.

9. Let $\text{*}$ be a binary operation on the set $\text{Q}$ of rational numbers. Find which of the given binary operations are commutative and which are associative.

1. The binary operation $\mathbf{\text{*}}$ is given by $\mathbf{\text{a * b = a - b}}$.

Ans: On, The binary operation $\text{*}$ is defined as $\text{a * b = a - b}$ on set $\text{Q}$.

.For $\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\in \text{Q}$,

$\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{\text{1}}{\text{2}}\text{-}\dfrac{\text{1}}{\text{3}}$

$\text{=}\dfrac{\text{3-2}}{\text{3}}$

$\text{=}\dfrac{\text{1}}{\text{6}}$

And;

$\dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{3}}\text{-}\dfrac{\text{1}}{\text{2}}$

$\text{=}\dfrac{\text{2-3}}{\text{6}}$

$\text{=}\dfrac{\text{-1}}{\text{6}}$

Therefore, $\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}}\ne \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{2}}$.

Hence, the binary operation $\text{*}$ is not commutative.

For $\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\text{,}\dfrac{\text{1}}{\text{4}}\in \text{Q}$,

$\left( \dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}} \right)\text{*}\dfrac{\text{1}}{\text{4}}\text{=}\left( \dfrac{\text{1}}{\text{2}}\text{-}\dfrac{\text{1}}{\text{3}} \right)\text{*}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{1}}{\text{6}}\text{*}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{1}}{\text{6}}\text{-}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{2-3}}{\text{12}}$

$\text{=}\dfrac{\text{-1}}{\text{12}}$

And;

$\dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{4}} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{1}}{\text{3}}\text{-}\dfrac{\text{1}}{\text{4}} \right)$

$\text{=}\dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{4-3}}{\text{12}} \right)$

$\text{=}\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{12}}$

$\text{=}\dfrac{\text{1}}{\text{2}}\text{-}\dfrac{\text{1}}{\text{12}}$

$\text{=}\dfrac{\text{5}}{\text{12}}$

Therefore, $\left( \dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}} \right)\text{*}\dfrac{\text{1}}{\text{4}}\ne \dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{4}} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

1. $\mathbf{\text{a * b = }{{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}}$

Ans: On, The binary operation $\text{*}$ is defined as $\text{a * b = }{{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}$ on set $\text{Q}$.

For $\text{a,b}\in \text{Q}$,

$\text{a*b=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}$

$\text{=}{{\text{b}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}$

$\text{=b*a}$

Therefore, $\text{a*b=b*a}$.

Hence, the binary operation $\text{*}$ is commutative.

For $\text{1,2,3}\in \text{Q}$,

$\left( \text{1 * 2} \right)\text{ * 3 = }\left( {{\text{1}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{2}}} \right)\text{ * 3 }$$\text{= 1 * }\left( \text{4 + 9} \right)\text{ }$

$\text{= }\left( \text{1 + 4} \right)\text{ * 3 }$

$\text{= 5 * 3 }$

$\text{= }{{\text{5}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}\text{ }$

$\text{= 34}$

And;

$\text{1 * }\left( \text{2 * 3} \right)\text{ = 1 * }\left( {{\text{2}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}} \right)\text{ }$

$\text{= 1 * }\left( \text{4 + 9} \right)\text{ }$

$\text{= 1 * 13 }$

$\text{= }{{\text{1}}^{\text{2}}}\text{ + 1}{{\text{3}}^{\text{2}}}\text{ }$

$\text{=170}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

1. $\mathbf{\text{a * b = a + ab}}$

Ans: The binary operation $\text{*}$ is defined as $\text{a * b = a + ab}$ on set $\text{Q}$.

For $\text{1,2}\in \text{Q}$,

$\text{1 * 2 = 1 + 1 }\!\!\times\!\!\text{ 2 }$

$\text{= 1 + 2 }$

$\text{= 3}$

$\text{2 * 1 = 2 + 2 }\!\!\times\!\!\text{ 1 }$

$\text{= 2 + 2 }$

$\text{= 4}$

Therefore, $\text{1 * 2}\ne \text{2 * 1}$.

Hence, operation $\text{*}$ is not commutative.

For $\text{1,2,3}\in \text{Q}$,

$\left( \text{1 * 2} \right)\text{ * 3 = }\left( \text{1+ 1 }\!\!\times\!\!\text{ 2 } \right)\text{ * 3 }$

$\text{= }\left( \text{1 + 2} \right)\text{ * 3 }$

$\text{= 3 * 3 }$

$\text{= 3 + 3 }\!\!\times\!\!\text{ 3 }$

$\text{= 3 + 9 }$

$\text{= 12}$

And,

$\text{1 * }\left( \text{2 * 3} \right)\text{ = 1 * }\left( \text{2 + 2 }\!\!\times\!\!\text{ 3 } \right)\text{ }$

$\text{= 1 * }\left( \text{2 + 6} \right)\text{ }$

$\text{= 1 * 8 }$

$\text{= 1 + 1 }\!\!\times\!\!\text{ 8 }$

$\text{= 9}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

1. $\mathbf{\text{a * b = }{{\left( \text{a - b} \right)}^{\text{2}}}}$

Ans: The binary operation $\text{*}$ is defined as $\text{a * b = }{{\left( \text{a - b} \right)}^{\text{2}}}$ on set $\text{Q}$.

For $\text{a,b}\in \text{Q}$,

$\text{a * b = }{{\left( \text{a - b} \right)}^{\text{2}}}$

$\text{b * a = }{{\left( \text{b - a} \right)}^{\text{2}}}$

$\text{=}{{\left[ \text{- }\left( \text{a - b} \right) \right]}^{\text{2}}}$

$\text{=}{{\left( \text{a - b} \right)}^{\text{2}}}$

$\text{a * b = b * a}$

Therefore, the binary operation $\text{*}$ is commutative.

For $\text{1,2,3}\in \text{Q}$,

$\left( \text{1 * 2} \right)\text{ * 3 = }{{\left( \text{1 -- 2} \right)}^{\text{2}}}\text{* 3}$

$\text{=}{{\left( \text{-- 1} \right)}^{\text{2}}}\text{* 3 }$

$\text{= 1 * 3 }$

$\text{= }{{\left( \text{1 -- 3} \right)}^{\text{2}}}\text{ }$

$\text{= }{{\left( \text{-- 2} \right)}^{\text{2}}}\text{ }$

$\text{= 4}$

And,

$\text{1 * }\left( \text{2 * 3} \right)\text{ = 1 * }{{\left( \text{2 -- 3} \right)}^{\text{2}}}$

$\text{=1*}{{\left( \text{-- 1} \right)}^{\text{2}}}$

$\text{=1 * 1}$

$\text{=0}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

1. $\mathbf{\text{a * b =}\dfrac{\text{ab}}{\text{4}}}$

Ans: The binary operation $\text{*}$ is defined as $\text{a * b =}\dfrac{\text{ab}}{\text{4}}$ on set $\text{Q}$.

For $\text{a,b}\in \text{Q}$,

$\text{a * b =}\dfrac{\text{ab}}{\text{4}}$

$\text{=}\dfrac{\text{ba}}{\text{4}}$

$\text{= b * a}$

Therefore, $\text{a * b = b * a}$.

Hence, the binary operation $\text{*}$ is commutative.

For $\text{a,b,c}\in \text{Q}$,

$\text{(a*b)*c =}\left( \dfrac{\text{ab}}{\text{4}} \right)\text{*c}$

$\text{=}\dfrac{\dfrac{\text{ab}}{\text{4}}\text{.c}}{\text{4}}$

$\text{=}\dfrac{\text{abc}}{\text{16}}$

And,

$\text{a*(b*c) = a*}\left( \dfrac{\text{ba}}{\text{4}} \right)\text{ }$

$\text{=}\dfrac{\dfrac{\text{bc}}{\text{4}}\text{a}}{\text{4}}$

$\text{=}\dfrac{\text{abc}}{\text{16}}$

Therefore, $\left( \text{a * b} \right)\text{ * c = a * }\left( \text{b * c} \right)$

Hence, the binary operation $\text{*}$ is associative.

1. $\mathbf{\text{a * b = a}{{\text{b}}^{\text{2}}}}$

Ans: The binary operation $\text{*}$ is defined as $\text{a * b = a}{{\text{b}}^{\text{2}}}$ on set $\text{Q}$.

For $\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\in \text{Q}$,

$\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{\text{1}}{\text{2}}.{{\left( \dfrac{\text{1}}{\text{3}} \right)}^{\text{2}}}$

$\text{=}\dfrac{\text{1}}{\text{2}}.\dfrac{\text{1}}{\text{9}}$

$\text{=}\dfrac{\text{1}}{\text{18}}$

And,

$\dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{3}}\text{.}{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{=}\dfrac{\text{1}}{\text{3}}\text{.}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{1}}{\text{12}}$

Therefore, $\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}}\ne \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{2}}$.

Hence, the binary operation $\text{*}$ is not commutative.

For $\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\text{,}\dfrac{\text{1}}{\text{4}}\in \text{Q}$,

$\left( \dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}} \right)\text{*}\dfrac{\text{1}}{\text{4}}\text{=}\left[ \dfrac{\text{1}}{\text{2}}.{{\left( \dfrac{\text{1}}{\text{3}} \right)}^{\text{2}}} \right]\text{*}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{1}}{\text{18}}\text{*}\dfrac{\text{1}}{\text{4}}$

$\text{=}\dfrac{\text{1}}{\text{18}}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{2}}$

$\text{=}\dfrac{\text{1}}{\text{18}\times \text{16}}$

$\text{=}\dfrac{\text{1}}{\text{288}}$

And;

$\dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{4}} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{*}\left[ \dfrac{\text{1}}{\text{3}}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{2}} \right]$

$\text{=}\dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{48}}$

$\text{=}\dfrac{\text{1}}{\text{2}}{{\left( \dfrac{\text{1}}{\text{48}} \right)}^{\text{2}}}$

$\text{=}\dfrac{\text{1}}{\text{2}\times \text{2304}}$

$\text{=}\dfrac{\text{1}}{\text{4608}}$

Therefore, $\left( \dfrac{\text{1}}{\text{2}}\text{*}\dfrac{\text{1}}{\text{3}} \right)\text{*}\dfrac{\text{1}}{\text{4}}\ne \dfrac{\text{1}}{\text{2}}\text{*}\left( \dfrac{\text{1}}{\text{3}}\text{*}\dfrac{\text{1}}{\text{4}} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

10. Find which of the operations given above has identity.

Ans: For the binary operation $\text{*}$, $\text{e}\in \text{Q}$ will be the identity element only if $\text{a * e = a = e * a}$, for all $\text{a}\in \text{Q}$.

As we can see, there is no such element $\text{e}\in \text{Q}$ for the operations given above satisfying the condition $\text{a * e = a = e * a}$.

Therefore, none of the operations given above has identity.

11. $\text{A = N }\!\!\times\!\!\text{ N}$ and $\text{*}$ be the binary operation on $\text{A}$ defined by

$\left( \text{a, b} \right)\text{ * }\left( \text{c, d} \right)\text{ = }\left( \text{a + c, b + d} \right)$. Show that $\text{*}$ is commutative and associative. Find the identity element for $\text{*}$ on $\text{A}$, if any.

Ans: The binary operation $\text{*}$ on $\text{A = N }\!\!\times\!\!\text{ N}$ a is defined by $\left( \text{a, b} \right)\text{ * }\left( \text{c, d} \right)\text{ = }\left( \text{a + c, b + d} \right)$.

Let $\left( \text{a, b} \right)\text{, }\left( \text{c, d} \right)\in \text{A}$ and $\text{a, b, c, d}\in \text{N}$

$\left( \text{a, b} \right)\text{ * }\left( \text{c, d} \right)\text{ = }\left( \text{a + c, b + d} \right)$

$\left( \text{c, d} \right)\text{ * }\left( \text{a, b} \right)\text{ = }\left( \text{c + a, d + b} \right)\text{ = }\left( \text{a + c, b + d} \right)$

Therefore, $\left( \text{a, b} \right)\text{ * }\left( \text{c, d} \right)\text{ = }\left( \text{c, d} \right)\text{ * }\left( \text{a, b} \right)$.

Hence, the binary operation $\text{*}$ is commutative.

Let $\left( a,\text{ }b \right),\text{ }\left( c,\text{ }d \right),\text{ }\left( e,\text{ }f \right)\in A$ and $a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f\in N$

$\left[ \left( \text{a, b} \right)\text{*}\left( \text{c, d} \right) \right]\text{*}\left( \text{e, f} \right)\text{ = }\left( \text{a + c, b + d} \right)\text{*}\left( \text{e, f} \right)\text{ }$

$\text{= }\left( \text{a + c + e, b + d + f} \right)$

And,

$\left( \text{a, b} \right)\text{*}\left[ \left( \text{c, d} \right)\text{*}\left( \text{e, f} \right) \right]\text{ = }\left( \text{a, b} \right)\text{*}\left( \text{c + e, d + f} \right)\text{ }$

$\text{= }\left( \text{a + c + e, b + d + f} \right)$

Therefore, $\left[ \left( \text{a, b} \right)\text{*}\left( \text{c, d} \right) \right]\text{*}\left( \text{e, f} \right)\text{ = }\left( \text{a, b} \right)\text{*}\left[ \left( \text{c, d} \right)\text{*}\left( \text{e, f} \right) \right]$.

Hence, the binary operation $\text{*}$ is associative.

For the binary operation $\text{*}$, $\text{e = }\left( {{\text{e}}_{\text{1}}}\text{,}{{\text{e}}_{\text{2}}} \right)\in \text{A}$ will be an identity element only if $\text{a * e = a = e * a}$ for all $\text{a = }\left( {{\text{a}}_{\text{1}}}\text{,}{{\text{a}}_{\text{2}}} \right)\in \text{A}$, that is, $\left( {{\text{a}}^{\text{1}}}\text{+}{{\text{e}}^{\text{1}}}\text{, }{{\text{a}}^{\text{2}}}\text{ + }{{\text{e}}^{\text{2}}} \right)\text{=}\left( {{\text{a}}^{\text{1}}}\text{, }{{\text{a}}^{\text{2}}} \right)\text{=}\left( {{\text{e}}^{\text{1}}}\text{+}{{\text{a}}^{\text{1}}}\text{, }{{\text{e}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}} \right)$. Which is not true for any element in $\text{A}$.

Hence the binary operation $\text{*}$ does not have any identity element.

12. State whether the following statements are true or false. Justify.

1. For an arbitrary binary operation $\text{*}$ on a set $\text{N}$, $\text{a * a = a}\forall \text{a}\in \text{N}$.

Ans: Let us define an binary operation $\text{*}$ on $\text{N}$ as $\text{a * a = a+b}\forall \text{a}\in \text{N}$

Taking $\text{b = a = 3}$,

$\text{3 * 3 = 3 + 3 = 6 }\ne \text{ 3}$

Therefore the given statement is false.

1. If $\text{*}$ is a commutative binary operation on $\text{N}$, then $\text{a*}\left( \text{b*c} \right)\text{=}\left( \text{c*b} \right)\text{*a}$.

Ans: It is given that if $\text{*}$ is a commutative binary operation on $\text{N}$, then $\text{a*}\left( \text{b*c} \right)\text{=}\left( \text{c*b} \right)\text{*a}$.

Simplifying $\text{R}\text{.H}\text{.S}$:

$\text{R}\text{.H}\text{.S= }\left( \text{c * b} \right)\text{ * a}$

$\text{= }\left( \text{b * c} \right)\text{ * a }$           (as $\text{*}$ is commutative)

$\text{= a * }\left( \text{b * c} \right)$             (as $\text{*}$is commutative)

$\text{=L}\text{.H}\text{.S}$

Therefore, $\text{a * }\left( \text{b * c} \right)\text{ = }\left( \text{c * b} \right)\text{ * a}$.

Hence, the given statement is true.

13. Consider a binary operation $\text{*}$ on $\text{N}$ defined as $\text{a*b=}{{\text{a}}^{\text{3}}}\text{+ }{{\text{b}}^{\text{3}}}$. Choose the correct answer.

1. The binary operation $\text{*}$ is both associative and commutative.

2. The binary operation $\text{*}$ is commutative but not associative.

3. The binary operation $\text{*}$ is associative but not commutative.

4. The binary operation $\text{*}$ is neither commutative nor associative.

Ans: The binary operation $\text{*}$ is defined as $\text{a*b=}{{\text{a}}^{\text{3}}}\text{+ }{{\text{b}}^{\text{3}}}$ on $\text{N}$.

For $\text{a,b}\in \text{N}$,

$\text{a*b=}{{\text{a}}^{\text{3}}}\text{+ }{{\text{b}}^{\text{3}}}$

$\text{=}{{\text{b}}^{\text{3}}}\text{+ }{{\text{a}}^{\text{3}}}$

$\text{=b*a}$

Hence, the binary operation $\text{*}$ is commutative.

For $\text{1,2,3}\in \text{N}$,

$\left( \text{1 * 2} \right)\text{ * 3=}\left( {{\text{1}}^{\text{3}}}\text{+}{{\text{2}}^{\text{3}}} \right)\text{ * 3}$

$\text{=}\left( \text{1+8} \right)\text{*3}$

$\text{=9*3}$

$\text{=}{{\text{9}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}}\text{ }$

$\text{= 756}$

And,

$\text{1* }\left( \text{2 * 3} \right)\text{ = 1 * }\left( {{\text{2}}^{\text{3}}}\text{+}{{\text{3}}^{\text{3}}} \right)$

$\text{=1*}\left( \text{8 + 27} \right)$

$\text{=1*35}$

$\text{=}{{\text{1}}^{\text{3}}}\text{+3}{{\text{5}}^{\text{3}}}$

$\text{=1+42875}$

$\text{=42876}$

Therefore, $\left( \text{1 * 2} \right)\text{ * 3}\ne \text{1 * }\left( \text{2 * 3} \right)$.

Hence, the binary operation $\text{*}$ is not associative.

Therefore, the correct answer is option (B) the binary operation $\text{*}$ is commutative, but not associative.

### Exercise 1.5

1. Let $\text{f: R }\to \text{ R}$ be defined as $\text{f}\left( \text{x} \right)\text{ = 10x + 7}$. Find the function $\text{g: R }\to \text{ R}$ such that $\text{gof = fog = }{{\text{I}}_{\text{R}}}$.

Ans: The given function $\text{f: R }\to \text{ R}$ is defined as $\text{f}\left( \text{x} \right)\text{ = 10x + 7}$.

For the function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$, where $\text{x, y}\in \text{R}$.

$\Rightarrow \text{10x + 7 = 10y + 7}$

$\Rightarrow \text{x = y}$

So, the given function $\text{f}$ is a one – one function.

For the function $\text{f}$ to be onto:

Let $\text{y = 10x + 7}$, for $\text{y}\in \text{R}$.

$\Rightarrow \text{x =}\dfrac{\text{y-7}}{\text{10}}\in \text{R}$

So, for any $\text{y}\in \text{R}$, there exists $\text{x =}\dfrac{\text{y-7}}{\text{10}}\in \text{R}$ such that:

$\left( \text{x} \right)\text{ = f}\left( \dfrac{\text{y-7}}{\text{10}} \right)$

$\text{= 10}\left( \dfrac{\text{y-7}}{\text{10}} \right)\text{+ 7}$

$\text{=y- 7+7}$

$\text{= y}$

So, the given function $\text{f}$ is onto.

Therefore, the function $\text{f}$ is an invertible function.

Defining $\text{g: R }\to \text{ R}$ as $\left( \text{y} \right)\text{ =}\dfrac{\text{y-7}}{\text{10}}\text{ }$.

Calculating $\text{gof}$:

$\text{go}\left( \text{x} \right)\text{=g}\left( \text{f}\left( \text{x} \right) \right)$

$\text{= g}\left( \text{10x+7} \right)$

$\text{=}\dfrac{\left( \text{10x+7} \right)\text{-7}}{\text{10}}$

$\text{=}\dfrac{\text{10x}}{\text{10}}$

$\text{= x}$

Calculating $\text{fog}$:

$\text{fo}\left( \text{y} \right)\text{=f}\left( \text{g}\left( \text{y} \right) \right)$

$\text{=f}\left( \dfrac{\text{y-7}}{\text{10}} \right)$

$\text{=10}\left( \dfrac{\text{y-7}}{\text{10}} \right)\text{+7}$

$\text{=y-7+7}$

$\text{=y}$

Therefore, $\text{gof = }{{\text{I}}_{\text{R}}}$ and $\text{fog = }{{\text{I}}_{\text{R}}}$.

2. Let $\text{f: W }\to \text{ W}$ be defined as $\text{f}\left( \text{n} \right)\text{ = n - 1}$, if $\text{n}$ is odd and$\text{f}\left( \text{n} \right)\text{ = n + 1}$, if $\text{n}$ is even. Show that $\text{f}$is invertible. Find the inverse of $\text{f}$. Here, $\text{W}$ is the set of all whole numbers.

Ans: The function $\text{f: W }\to \text{ W}$ is defined as $\text{f}\left( \text{n} \right)\text{ =}\left\{ \begin{matrix}\text{n-1 if n is odd} \\ \text{n+1 if n is even} \\\end{matrix} \right\}$

For the function to be one – one:

$\text{f}\left( \text{n} \right)\text{ = f}\left( \text{m} \right)$

If $\text{n}$ is odd and $\text{m}$ is even, then we will have $\text{n - 1 = m + 1}$.

$\Rightarrow \text{n-m = 2}$

Which is not possible.

Similarly, the possibility that of $\text{n}$ is even and $\text{m}$ is odd can also be ignored.

Therefore, both $\text{n}$ and $\text{m}$ must be either odd or even.

Now, if both $\text{n}$ and $\text{m}$ are odd,

$\text{f}\left( \text{n} \right)\text{ = f}\left( \text{m} \right)$

$\Rightarrow \text{n - 1 = m - 1}$

$\Rightarrow \text{n = m}$

if both $\text{n}$ and $\text{m}$ are even,

$\text{f}\left( \text{n} \right)\text{ = f}\left( \text{m} \right)$

$\Rightarrow \text{n + 1 = m + 1}$

$\Rightarrow \text{n = m}$

Therefore the given function $\text{f}$ is one – one.

For the function to be onto:

Now, for any odd number $\text{2r + 1}\in \text{N}$ is the image of $\text{2r}\in \text{N}$ and any even number $\text{2r}\in \text{N}$ is the image of $\text{2r + 1}\in \text{N}$.

Hence, the function $\text{f}$ is onto.

Therefore, the function $\text{f}$ is an invertible function.

Defining $\text{g: W }\to \text{ W}$ as $\left( \text{m} \right)\text{=}\left\{ \begin{matrix}\text{m+1 if m is even} \\\text{m-1 if m is odd} \\\end{matrix} \right\}$

When $\text{n}$ is odd:

$\text{go}\left( \text{n} \right)\text{ = g}\left( \text{f}\left( \text{n} \right) \right)\text{ }$

$\text{= g}\left( \text{n - 1} \right)\text{ }$

$\text{= n - 1 + 1 }$

$\text{= n}$

When $\text{n}$ is even

$\text{go}\left( \text{n} \right)\text{ = g}\left( \text{f}\left( \text{n} \right) \right)\text{ }$

$\text{= g}\left( \text{n + 1} \right)\text{ }$

$\text{= n + 1 - 1 }$

$\text{= n}$

When $\text{m}$ is odd

$\text{fo}\left( \text{m} \right)\text{ = f}\left( \text{g}\left( \text{m} \right) \right)\text{ }$

$\text{= f}\left( \text{m - 1} \right)\text{ }$

$\text{= m - 1 + 1 }$

$\text{= m}$

When $\text{m}$ is even

$\text{fo}\left( \text{m} \right)\text{ = f}\left( \text{g}\left( \text{m} \right) \right)\text{ }$

$\text{= f}\left( \text{m + 1} \right)\text{ }$

$\text{= m + 1 - 1 }$

$\text{= m}$

Therefore, $\text{gof = }{{\text{I}}_{\text{W}}}$ and $\text{fog = }{{\text{I}}_{\text{W}}}$.

Hence, the function $\text{f}$ is invertible and the inverse of $\text{f}$ is given by ${{\text{f}}^{\text{-1}}}\text{=g}$, which is the same as $\text{f}$.

3. If $\text{f: R }\to \text{ R}$ is defined by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}\text{ - 3x + 2}$, find $\text{f}\left( \text{f}\left( \text{x} \right) \right)$.

Ans: The function $\text{f: R }\to \text{ R}$ is defined as $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}\text{ - 3x + 2}$.

Calculating $\text{f}\left( \text{f}\left( \text{x} \right) \right)$:

$\text{f}\left( \text{f}\left( \text{x} \right) \right)\text{ = f}\left( {{\text{x}}^{\text{2}}}\text{- 3x + 2} \right)$

$\text{= }{{\left( {{\text{x}}^{\text{2}}}\text{- 3x + 2} \right)}^{\text{2}}}\text{ - 3}\left( {{\text{x}}^{\text{2}}}\text{- 3x+2} \right)\text{+2}$

$\text{= }\left( {{\text{x}}^{\text{4}}}\text{+9}{{\text{x}}^{\text{2}}}\text{+4-6}{{\text{x}}^{\text{3}}}\text{-12x+4}{{\text{x}}^{\text{2}}} \right)\text{+}\left( \text{-3}{{\text{x}}^{\text{2}}}\text{+9x-6} \right)\text{+2}$

$\text{= }{{\text{x}}^{\text{4}}}\text{-6}{{\text{x}}^{\text{3}}}\text{+10}{{\text{x}}^{\text{2}}}\text{--3x}$

4. Show that function $\text{f: R }\to \text{ }\!\!\{\!\!\text{ x}\in \text{R:-1 x 1 }\!\!\}\!\!\text{ }$ defined by $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}$ is one – one and onto function.

Ans: The function $\text{f: R }\to \text{ }\!\!\{\!\!\text{ x}\in \text{R:-1 x 1 }\!\!\}\!\!\text{ }$ is defined as $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}$.

For the function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$, where $\text{x, y}\in \text{R}$.

$\Rightarrow \dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{=}\dfrac{\text{y}}{\text{1+ }\!\!|\!\!\text{ y }\!\!|\!\!\text{ }}$

Assuming that $\text{x}$ is positive and $\text{y}$ is negative:

$\dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{2xy=x-y}$

Since, $\text{x y}\Rightarrow \text{x-y 0}$.

But $\text{2xy}$ is negative.

Therefore, $\text{2xy }\ne \text{ x - y}$.

Hence, $\text{x}$ being positive and $\text{y}$ being negative is not possible. Similarly $\text{x}$ being negative and $\text{y}$ being positive can also be ruled out.

So, $\text{x}$ and $\text{y}$ have to be either positive or negative.

Assuming that both $\text{x}$ and $\text{y}$ are positive:

$\text{f(x)=f(y)}$

$\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{x+xy=y+xy}$

$\Rightarrow \text{x=y}$

Assuming that both $\text{x}$ and $\text{y}$ are negative:

$\text{f(x)=f(y)}$

$\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{x+xy=y+xy}$

$\Rightarrow \text{x=y}$

Therefore, the function $\text{f}$ is one – one.

For onto:

$\text{y}\in \text{R}$ such that $\text{-1 y 1}$.

If $\text{y}$ is negative, then, there exists $\text{x = }\dfrac{\text{y}}{\text{1+y}}\in \text{R}$ such that

$\text{f}\left( \dfrac{\text{y}}{\text{1+y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1+y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1+y}} \right|}$

$\text{=}\dfrac{\dfrac{\text{y}}{\text{1+y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1+y}} \right)}$

$\text{=}\dfrac{\text{y}}{\text{1+y-y}}$

$\text{=y}$

If $\text{y}$ is positive, then, there exists $\text{x = }\dfrac{\text{y}}{\text{1-y}}\in \text{R}$ such that

$\text{f}\left( \dfrac{\text{y}}{\text{1-y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1-y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1-y}} \right|}$

$\text{=}\dfrac{\dfrac{\text{y}}{\text{1-y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1-y}} \right)}$

$\text{=}\dfrac{\text{y}}{\text{1-y+y}}$

$\text{=y}$

Therefore, the function $\text{f}$ is onto.

Hence the given function $\text{f}$ is both one – one and onto.

5. Show that the function $\text{f: R }\to \text{ R}$ given by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$ is injective.

Ans: The given function $\text{f: R }\to \text{ R}$ is given as $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$.

For the function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$ where $\text{x, y}\in \text{R}$.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{=}{{\text{y}}^{\text{3}}}$           …… (1)

We need to show that $\text{x=y}$.

Assuming that $\text{x}\ne \text{y}$, then,

$\Rightarrow {{\text{x}}^{\text{3}}}\ne {{\text{y}}^{\text{3}}}$

Since this is a contradiction to (1), therefore, $\text{x=y}$.

Hence, the function $\text{f}$ is injective.

6. Give examples of two functions $\text{f: N }\to \text{ Z}$ and $\text{g: Z }\to \text{ Z}$ such that $\text{gof}$ is injective but $\text{g}$ is not injective.

(Hint: Consider $\text{f}\left( \text{x} \right)\text{ = x}$ and $\text{g }\left( \text{x} \right)\text{ = }\left| \text{x} \right|$)

Ans: Taking the function $\text{f: N }\to \text{ Z}$ as $\text{f}\left( \text{x} \right)\text{ = x}$ and the function $\text{g: Z }\to \text{ Z}$ as $\text{g }\left( \text{x} \right)\text{ = }\left| \text{x} \right|$.

Showing that $\text{g}$ is not injective:

$\left( \text{-1} \right)\text{ = }\left| \text{-1} \right|\text{= 1}$

$\left( \text{1} \right)\text{ = }\left| \text{1} \right|\text{= 1}$

Therefore $\left( \text{-1} \right)\text{ = g}\left( \text{1} \right)$ but $\text{-1}\ne \text{ 1}$.

Therefore the function $\text{g}$ is not injective.

Showing that $\text{gof}$ is injective:

The function $\text{gof: N }\to \text{ Z}$ is defined as

$\text{go}\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)\text{ }$

$\text{= g}\left( \text{x} \right)\text{ }$

$\text{= }\left| \text{x} \right|$

Taking $\text{x, y}\in \text{N}$ such that $\text{gof}\left( \text{x} \right)\text{ = gof}\left( \text{y} \right)$.

$\Rightarrow \left| \text{x} \right|\text{ = }\left| \text{y} \right|$

Since $\text{x, y}\in \text{N}$, both are positive.

$\Rightarrow \text{x = y}$

Therefore the function $\text{gof}$ is injective.

7. Given examples of two functions $\text{f: N }\to \text{ N}$ and $\text{g: N }\to \text{ N}$ such that $\text{gof}$ is onto but $\text{f}$ is not onto.

(Hint: Consider $\text{f}\left( \text{x} \right)\text{ = x+1}$ and $\left( \text{x} \right)\text{ =}\left\{ \begin{matrix}\text{x-1, if x1} \\\text{1, if x1} \\\end{matrix} \right\}$)

Ans: Taking the function $\text{f: N }\to \text{ N}$ as $\text{f}\left( \text{x} \right)\text{ = x+1}$

And the function $\text{g: N }\to \text{ N}$ by $\text{g}\left( \text{x} \right)\text{ =}\left\{ \begin{matrix}\text{x-1, if x1} \\ \text{1, if x1} \\\end{matrix} \right\}$

Showing that $\text{g}$ is not onto.

Taking an element $\text{1}$ in a co-domain $\text{N}$.

Since this element is not an image of any of the elements in domain $\text{N}$, therefore $\text{g}$ is not onto.

Showing that $\text{gof}$ is onto:

The function $\text{gof: N }\to \text{ N}$ is defined as $\text{go}\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)\text{ }$.

$\text{= g}\left( \text{x + 1} \right)\text{ }$

$\text{= x + 1 - 1 }$                (as $\text{x}\in \text{N}\Rightarrow \text{x + 1 1}$)

$\text{= x}$

So, for $\text{y}\in \text{N}$, there exists $\text{x = y}\in \text{N}$ such that $\text{gof}\left( \text{x} \right)\text{ = y}$.

Therefore, the function $\text{gof}$ is onto.

8. Given a non-empty set $\text{X}$, consider $\text{P}\left( \text{X} \right)$ which is the set of all subsets of $\text{X}$. Define the relation $\text{R}$ in $\text{P}\left( \text{X} \right)$ as follows:

For subsets $\text{A,B}$ in $\text{P}\left( \text{X} \right)$, $\text{ARB}$ if and only if $\text{A}\subset \text{B}$. Is $\text{R}$ an equivalence relation on $\text{P}\left( \text{X} \right)$? Justify you answer.

Ans: We know that every set is a subset of itself, $\text{ARA}$ for all $\text{A}\in \text{P}\left( \text{X} \right)$

Therefore $\text{R}$ is reflexive.

Let $\text{ARB}\Rightarrow \text{A}\subset \text{B}$.

This does not mean that $\text{B}\subset \text{A}$.

If $\text{A = }\left\{ \text{1, 2} \right\}$ and $\text{B = }\left\{ \text{1, 2, 3} \right\}$, then it cannot be implied that $\text{B}$ is related to $\text{A}$.

Therefore $\text{R}$ is not symmetric.

If $\text{ARB}$ and $\text{BRC}$ , then;

$\text{A}\subset \text{B}$ and $\text{B}\subset \text{C}$

$\Rightarrow \text{A}\subset \text{C}$

$\Rightarrow \text{ARC}$

Therefore $\text{R}$ is transitive.

Hence, $\text{R}$ is not an equivalence relation as it is not symmetric.

9. Given a non-empty set $\text{X}$, consider the binary operation $\text{*: P}\left( \text{X} \right)\text{ }\!\!\times\!\!\text{ P}\left( \text{X} \right)\text{ }\to \text{ P}\left( \text{X} \right)$ given by $\text{A*B = A}\cap \text{B}\forall \text{A, B}$ in $\text{P}\left( \text{X} \right)$ is the power set of $\text{X}$. Show that $\text{X}$ is the identity element for this operation and $\text{X}$ is the only invertible element in $\text{P}\left( \text{X} \right)$ with respect to the operation $\text{*}$.

Ans: The binary operation $\text{*: P}\left( \text{X} \right)\text{ }\!\!\times\!\!\text{ P}\left( \text{X} \right)\text{ }\to \text{ P}\left( \text{X} \right)$ is given by $\text{A*B = A}\cap \text{B}\forall \text{A, B}$ in $\text{P}\left( \text{X} \right)$.

$\text{A}\cap \text{X = A = X }\cap \text{ A}$ for all $\text{A}\in \text{P}\left( \text{X} \right)$

$\Rightarrow \text{A * X = A = X * A}$ for all $\text{A}\in \text{P}\left( \text{X} \right)$

$\text{X}$ is the identity element for the given binary operation $\text{*}$.

An element $\text{A}\in \text{P}\left( \text{X} \right)$ is invertible if there exists $\text{B}\in \text{P}\left( \text{X} \right)$ such that

$\text{A * B = X = B * A}$          (As $\text{X}$ is the identity element)

Or

$\text{A }\cap \text{ B = X = B }\cap \text{ A}$

This is only possible when $\text{A = X = B}$.

Therefore, $\text{X}$ is the only invertible element in $\text{P}\left( \text{X} \right)$ with respect to the given operation $\text{*}$.

10. Find the number of all onto functions from the set $\text{ }\!\!\{\!\!\text{ 1, 2, 3, }...\text{ , n }\!\!\}\!\!\text{ }$ to itself.

Ans: The total number of onto maps from $\text{ }\!\!\{\!\!\text{ 1, 2, 3, }...\text{ , n }\!\!\}\!\!\text{ }$ to itself will be same as the total number of permutations on $\text{n}$ symbols $\text{1, 2, 3, }...\text{ , n}$.

Since the total number of permutations on $\text{n}$ symbols $\text{1, 2, 3, }...\text{ , n}$ is $\text{n}$, thus total number of onto maps from $\text{ }\!\!\{\!\!\text{ 1, 2, 3, }...\text{ , n }\!\!\}\!\!\text{ }$ to itself are $\text{n}$.

11. Let $\text{S = }\left\{ \text{a, b, c} \right\}$ and $\text{T = }\left\{ \text{1, 2, 3} \right\}$. Find ${{\text{F}}^{\text{-1}}}$ of the following functions $\text{F}$ from $\text{S}$ to$\text{T}$, if it exists.

1. The function is given as $\mathbf{\text{F = }\left\{ \left( \text{a, 3} \right)\text{, }\left( \text{b, 2} \right)\text{, }\left( \text{c, 1} \right) \right\}}$.

Ans: The function $\text{F: S }\to \text{ T}$ is defined as $\text{F = }\left\{ \left( \text{a, 3} \right)\text{, }\left( \text{b, 2} \right)\text{, }\left( \text{c, 1} \right) \right\}$.

$\Rightarrow \text{F }\left( \text{a} \right)\text{ = 3, F }\left( \text{b} \right)\text{ = 2, F}\left( \text{c} \right)\text{ = 1}$

Therefore, ${{\text{F}}^{\text{-1}}}\text{:T }\to \text{S}$ is given by ${{\text{F}}^{\text{-1}}}\text{= }\left\{ \left( \text{3, a} \right)\text{, }\left( \text{2, b} \right)\text{, }\left( \text{1, c} \right) \right\}$.

1. The function is given as $\mathbf{\text{ }\!\!~\!\!\text{ F = }\left\{ \left( \text{a, 2} \right)\text{, }\left( \text{b, 1} \right)\text{, }\left( \text{c, 1} \right) \right\}}$.

Ans: The function $\text{F: S }\to \text{ T}$ is defined as $\text{F=}\left\{ \left( \text{a, 2} \right)\text{, }\left( \text{b, 1} \right)\text{, }\left( \text{c, 1} \right) \right\}$.

Since $\text{F}\left( \text{b} \right)\text{ = F}\left( \text{c} \right)\text{ = 1}$, the function $\text{F}$ is not one – one.

Hence, the function $\text{F}$ is not invertible so its inverse does not exist.

12. Consider the binary operations $\mathbf{\text{*: R }\!\!\times\!\!\text{ R }\to \text{ R}}$ and $\mathbf{\text{o}\,\text{: R }\!\!\times\!\!\text{ R }\to \text{ R}}$ defined as $\mathbf{\text{a*b}\,\text{=}\,\left| \text{a-b} \right|}$ and $\mathbf{\text{a}\,\text{o}\,\text{b}\,\text{=}\,\text{a,}\forall \text{a,}\,\text{b}\in \text{R}}$. Show that $\mathbf{\text{*}}$ is commutative but not associative, $\mathbf{\text{o}}$ is associative but not commutative. Further, show that $\mathbf{\forall \text{a,}\,\text{b,}\,\text{c}\in \text{R,}\,\text{a*}\left( \text{b}\,\text{o}\,\text{c} \right)\,\text{=}\,\left( \text{a*b} \right)\,\text{o}\,\left( \text{a*c} \right)}$. [If it is so, we say that the operation * distributes over the operation o. Does o distribute over *? Justify your answer.

Ans: It is given that $\text{*: R }\!\!\times\!\!\text{ R }\to \text{ R}$ and $\text{o}\,\text{: R }\!\!\times\!\!\text{ R }\to \text{ R}$is defined as $\text{a*b}\,\text{=}\,\left| \text{a-b} \right|$ and $\text{a}\,\text{o}\,\text{b}\,\text{=}\,\text{a,}\forall \text{a,}\,\text{b}\in \text{R}$

For $\text{a,}\,\text{b}\in \text{R}$, we have $\text{a*b}\,\text{=}\,\left| \text{a-b} \right|$

And,

$\text{b*a}\,\text{=}\,\left| \text{b-a} \right|\,=\,\left| -\left( \text{a-b} \right) \right|\,=\,\left| \text{a-b} \right|$

Therefore, $\text{a*b}\,\text{=}\,\text{b*a}$.

Hence, operation $\text{*}$ is commutative.

For $\text{1,}\,\text{2,}\,\text{3}\in \text{R}$,

$\left( \text{1*2} \right)\text{*3}\,\text{=}\,\left( \left| \text{1}\,\text{-}\,\text{2} \right| \right)\text{*3}$

$\text{=1*3}$

$\text{=}\left| \text{1-3} \right|$

$\text{=2}$

And,

$\text{1*}\left( \text{2*3} \right)\,\text{=1*}\left( \left| \text{2}\,\text{-}\,\text{3} \right| \right)$

$\text{=1*1}$

$\text{=}\left| \text{1-1} \right|$

$\text{=0}$

Therefore, $\left( \text{1*2} \right)\text{*3}\ne \text{1*}\left( \text{2*3} \right)$.

Hence, the operation $\text{*}$ is not associative.

We can observe that, for $\text{1,}\,\text{2,}\,\text{3}\in \text{R}$, $\text{1}\,\text{o}\,\text{2}\,\text{=1}$ and $\text{2}\,\text{o}\,\text{1}\,\text{=2}$

Therefore, $\text{1}\,\text{o}\,\text{2}\ne \text{2}\,\text{o}\,\text{1}$

Hence, the operation $\text{o}$ is not commutative.

For $\text{a,}\,\text{b,}\,c\in \text{R}$,

$\left( \text{a}\,\text{o}\,\text{b} \right)\,o\,c\,\text{=}\,\text{a}\,\text{o}\,\text{c}\,\text{=}\,\text{a}$

And,

$\text{a}\,\text{o}\,\left( \text{b}\,o\,c \right)\,\text{=}\,\text{a}\,\text{o}\,\text{b}\,\text{=}\,\text{a}$

Therefore, $\left( \text{a}\,\text{o}\,\text{b} \right)\,o\,c\,\ne \,\text{a}\,\text{o}\,\left( \text{b}\,o\,c \right)$ where $\text{a,}\,\text{b,}\,c\in \text{R}$.

Operation $\text{o}$ is associative.

For $\text{a,}\,\text{b,}\,c\in \text{R}$,

$\text{a}\,*\,\left( \text{b}\,o\,c \right)\,\text{=}\,\text{a}\,*\,\text{b}\,\text{=}\,\left| \text{a-b} \right|$

$\left( \text{a}\,\text{*}\,\text{b} \right)\,\text{o}\,\left( \text{a}\,\text{*}\,\text{c} \right)\,\text{=}\,\left( \left| \text{a-b} \right| \right)\,o\,\left( \left| \text{a-c} \right| \right)\,=\,\left| \text{a-b} \right|$

Hence, $\text{a}\,*\,\left( \text{b}\,o\,c \right)\,\text{=}\,\left( \text{a}\,\text{*}\,\text{b} \right)\,\text{o}\,\left( \text{a}\,\text{*}\,\text{c} \right)$.

So, if $\text{1,}\,\text{2,}\,\text{3}\in \text{R}$, then,

$\text{1}\,\text{o}\,\left( \text{2}\,\text{*}\,\text{3} \right)\,\text{=}\,\text{1}\,\text{o}\,\left( \left| \text{2-3} \right| \right)$

$\text{=}\,\text{1}\,\text{o}\,\text{1}$

$\text{=1}$

$\left( \text{1}\,\text{o}\,\text{2} \right)\,\text{*}\,\left( \text{1}\,\text{o}\,\text{3} \right)\,\text{=}\,\text{1}\,\text{*}\,\text{1}\,$

$\text{=}\,\left| \text{1-1} \right|$

$\text{=}\,\text{0}$

Therefore, the operation $\text{o}$ does not distribute over $\text{*}$.

13. Given a non - empty set $\mathbf{\text{X}}$, let $\mathbf{\text{*:P}\left( \text{X} \right)\text{ }\!\!\times\!\!\text{ P}\left( \text{X} \right)\to \text{P}\left( \text{X} \right)}$ be defined as $\mathbf{\text{A*B=(A-B)}\cup \text{(B-A),}\forall \text{A,B}\in \text{P(X)}}$. Show that the empty set $\mathbf{\Phi}$ is the identity for the operation $\mathbf{\text{*}}$ and all the elements $\mathbf{\text{A}}$ of $\mathbf{\text{P(X)}}$ are invertible with $\mathbf{{{\text{A}}^{\text{-1}}}\text{=A}}$.

(Hint: $\mathbf{\left( \text{A- }\!\!\Phi\!\!\text{ } \right)\cup \left( \text{ }\!\!\Phi\!\!\text{ -A} \right)\text{=A}}$ and $\mathbf{\left( \text{A-A} \right)\cup \left( \text{A-A} \right)\text{=A*A}}$).

Ans: the function $\text{*: P}\left( \text{X} \right)\text{ }\!\!\times\!\!\text{ P}\left( \text{X} \right)\text{ }\to \text{ P}\left( \text{X} \right)$ is defined as $\text{A*B=(A-B)}\cup \text{(B-A),}\forall \text{A,B}\in \text{P(X)}$.

For $\text{A}\in \text{P(X)}$,

$\text{A* }\!\!\Phi\!\!\text{ =}\left( \text{A- }\!\!\Phi\!\!\text{ } \right)\cup \left( \text{ }\!\!\Phi\!\!\text{ -A} \right)$

$\text{=A}\cup \text{ }\!\!\Phi\!\!\text{ }$

$\text{=A}$

$\text{ }\!\!\Phi\!\!\text{ *A=}\left( \text{ }\!\!\Phi\!\!\text{ -A} \right)\cup \left( \text{A- }\!\!\Phi\!\!\text{ } \right)$

$\text{= }\!\!\Phi\!\!\text{ }\cup \text{A}$

$\text{=A}$

Therefore, $\text{A* }\!\!\Phi\!\!\text{ = }\!\!\Phi\!\!\text{ *A}$ for all $\text{A}\in \text{P(X)}$.

For the given operation $\text{*}$, $\text{ }\!\!\Phi\!\!\text{ }$ is the identity element.

For $\text{A}\in \text{P(X)}$, if there exists $\text{B}\in \text{P(X)}$ such that $\text{A*B= }\!\!\Phi\!\!\text{ =B*A}$, then the element $\text{A}\in \text{P(X)}$ will be invertible.

For all $\text{A}\in \text{P(X)}$,

$\text{A*A=}\left( \text{A-A} \right)\cup \left( \text{A-A} \right)$

$\text{= }\!\!\Phi\!\!\text{ - }\!\!\Phi\!\!\text{ }$

$\text{= }\!\!\Phi\!\!\text{ }$

Therefore, all the elements $\text{A}\in \text{P(X)}$ are invertible with ${{\text{A}}^{\text{-1}}}\text{=A}$.

14. Define a binary operation $\mathbf{\text{*}}$ on the set $\mathbf{\left\{ \text{0, 1, 2, 3, 4, 5} \right\}}$ as \mathbf{\text{a*b=}\left\{ \begin{align}& \text{a+b}\,\,\text{if}\,\,\text{a+b6} \\ & \text{a+b-6}\,\,\text{if}\,\,\text{a+b}\ge \text{6} \\ \end{align} \right\}}.

Show that zero is the identity for this operation and each element $\text{a}\ne \text{0}$ of the set is invertible with $\text{6-a}$ being the inverse of $\text{a}$.

Ans: Let us take $\text{X=}\left\{ \text{0, 1, 2, 3, 4, 5} \right\}$.

The binary operation $\text{*}$ on $\text{X}$ is defined as \text{a*b=}\left\{ \begin{align}& \text{a+b}\,\,\text{if}\,\,\text{a+b6} \\ & \text{a+b-6}\,\,\text{if}\,\,\text{a+b}\ge \text{6} \\ \end{align} \right\}.

If $\text{a*e}\,\text{=a}\,\text{=}\,\text{e*a}$ for all $\text{a}\in \text{X}$, then element $\text{e}\in \text{X}$ is the identity element for the operation $\text{*}$.

For $\text{a}\in \text{X}$, we have

$\text{a*0}\,\text{=a}$ A Є X ⇒ a+ 0 <6

$\text{0*a}\,\text{=0}$ A Є X ⇒ a+ 0 <6

Therefore, $\text{a*0=0*a}$ for all $\text{a}\in \text{X}$.

Hence, $\text{0}$ is the identity element for the given operation $\text{*}$.

For $\text{a}\in \text{X}$, if there exists $\text{b}\in \text{X}$ such that $\text{a*b=0=b*a}$, then the element $\text{a}\in \text{X}$ will be invertible. That is,

\text{a*b=}\left\{ \begin{align}& \text{a+b=0=b+a,}\,\,\text{if}\,\,\text{a+b6} \\ & \text{a+b-6=0=b+a-6,}\,\,\text{if}\,\,\text{a+b}\ge \text{6} \\ \end{align} \right\}

$\Rightarrow \text{a=}\,\text{-b}$ or $\text{b=6-a}$

Since, $\text{X=}\left\{ \text{0, 1, 2, 3, 4, 5} \right\}$ and $\text{a,b}\in \text{X}$. So, $\text{a}\ne \text{-b}$.

Therefore, $\text{b=6-a}$ is the inverse of $\text{a}$ for all $\text{a}\in \text{X}$.

15. Let $\mathbf{\text{A=}\left\{ \text{-1, 0, 1, 2} \right\}\text{,B=}\left\{ \text{-4, -2, 0, 2} \right\}}$ and $\mathbf{\text{f,g:A }\to \text{ B}}$ be functions defined by $\mathbf{\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}\text{-x,}\,\text{x}\in \text{A}}$ and $\mathbf{\text{g}\left( \text{x} \right)\text{=2}\left| \text{x-}\dfrac{\text{1}}{\text{2}} \right|\text{-1,x}\in \text{A}}$. Are $\mathbf{\text{f}}$ and $\mathbf{\text{g}}$ equal? Justify your answer. (Hint: One may note that two function $\mathbf{\text{f:A }\to \text{ B}}$ and $\mathbf{\text{g:A }\to \text{ B}}$ such that $\mathbf{\text{f}\left( \text{a} \right)\text{=g}\left( \text{a} \right)\forall \text{a}\in \text{A}}$, are called equal functions)

Ans: Let $\text{A=}\left\{ \text{-1, 0, 1, 2} \right\}\text{,B=}\left\{ \text{-4, -2, 0, 2} \right\}$ and $\text{f,g:A }\to \text{ B}$ are defined by $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{2}}}\text{-x,}\,\text{x}\in \text{A}$ and $\text{g}\left( \text{x} \right)\text{=2}\left| \text{x-}\dfrac{\text{1}}{\text{2}} \right|\text{-1,x}\in \text{A}$.

$\text{f}\left( \text{-1} \right)\text{=}{{\left( \text{-1} \right)}^{\text{2}}}\text{-}\left( \text{-1} \right)$

$\text{=1+1}$

$\text{=2}$

And,

$\text{g}\left( \text{-1} \right)\text{=2}\left| \left( \text{-1} \right)\text{-}\dfrac{\text{1}}{\text{2}} \right|\text{-1}$

$\text{=2}\left( \dfrac{\text{3}}{\text{2}} \right)\text{-1}$

$\text{=3-1}$

$\text{=2}$

$\Rightarrow \text{f}\left( \text{-1} \right)\text{=g}\left( \text{-1} \right)$

$\text{f}\left( \text{0} \right)\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-}\left( \text{0} \right)$

$\text{=0}$

And,

$\text{g}\left( \text{0} \right)\text{=2}\left| \left( \text{0} \right)\text{-}\dfrac{\text{1}}{\text{2}} \right|\text{-1}$

$\text{=1-1}$

$\text{=0}$

$\Rightarrow \text{f}\left( \text{0} \right)\text{=g}\left( \text{0} \right)$

$\text{f}\left( \text{1} \right)\text{=}{{\left( \text{1} \right)}^{\text{2}}}\text{-}\left( \text{1} \right)$

$\text{=1-1}$

$\text{=0}$

And,

$\text{g}\left( \text{1} \right)\text{=2}\left| \left( \text{1} \right)\text{-}\dfrac{\text{1}}{\text{2}} \right|\text{-1}$

$\text{=2}\left( \dfrac{1}{\text{2}} \right)\text{-1}$

$\text{=1-1}$

$\text{=0}$

$\Rightarrow \text{f}\left( \text{1} \right)\text{=g}\left( \text{1} \right)$

$\text{f}\left( \text{2} \right)\text{=}{{\left( \text{2} \right)}^{\text{2}}}\text{-}\left( \text{2} \right)$

$\text{=4-2}$

$\text{=2}$

And,

$\text{g}\left( \text{2} \right)\text{=2}\left| \left( \text{2} \right)\text{-}\dfrac{\text{1}}{\text{2}} \right|\text{-1}$

$\text{=2}\left( \dfrac{\text{3}}{\text{2}} \right)\text{-1}$

$\text{=3-1}$

$\text{=2}$

$\Rightarrow \text{f}\left( \text{2} \right)\text{=g}\left( \text{2} \right)$

Therefore, $\text{f}\left( \text{a} \right)\text{=g}\left( \text{a} \right)\forall \text{a}\in \text{A}$. Hence the functions $\text{f}$ and $\text{g}$ are equal.

16. Let $\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}$ Then number of relations containing $\mathbf{\left( \text{1, 2} \right)}$ and $\mathbf{\left( \text{1, 3} \right)}$  which are reflexive and symmetric but not transitive is

1. $\mathbf{\text{1}}$

2. $\mathbf{\text{2}}$

3. $\mathbf{\text{3}}$

4. $\mathbf{\text{4}}$

Ans: We are given a set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Let us take the relation $\text{R}$, containing $\left( \text{1, 2} \right)$ and $\left( \text{1, 3} \right)$, as $\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\text{, }\left( \text{3, 1} \right) \right\}$.

As we can see that $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\in \text{R}$, therefore relation $\text{R}$ is reflexive.

Since $\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\in \text{R}$, the relation $\text{R}$ is symmetric.

The relation Relation $\text{R}$ is not transitive because $\left( \text{1, 2} \right)\text{,}\,\left( \text{3, 1} \right)\in \text{R}$, but $\left( \text{3, 2} \right)\notin \text{R}$.

The relation Relation $\text{R}$ will become transitive on adding and two pairs $\left( \text{3, 2} \right)\text{, }\left( \text{2, 3} \right)$.

Therefore the total number of desired relations is one.

The correct answer is option $(A)$ $\text{1}$.

17. Let $\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}$ Then number of equivalence relations containing $\mathbf{\left( \text{1, 2} \right)}$ is

1. $\mathbf{\text{1}}$

2. $\mathbf{\text{2}}$

3. $\mathbf{\text{3}}$

4. $\mathbf{\text{4}}$

Ans: We are given a set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Let us take the relation $\text{R}$, containing $\left( \text{1, 2} \right)$ as $\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}$.

Now the pairs left are $\left( \text{2, 3} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)$

In order to add one pair, say $\left( \text{2, 3} \right)$, we must add $\left( \text{3, 2} \right)$ for symmetry. And we are required to add $\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)$ for transitivity.

So, only the equivalence relation (bigger than $\text{R}$) is the universal relation.

Therefore, the total number of equivalence relations containing $\left( \text{1, 2} \right)$ are two.

Hence, the correct answer is (B) $\text{2}$.

18. Let $\mathbf{\text{f:R }\to \text{ R}}$ be the Signum Function defined as \mathbf{\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{1,x0} \\ & \text{0,x=0} \\ & \text{-1,x0} \\ \end{align} \right\}} and $\mathbf{\text{g:R }\to \text{ R}}$ be the Greatest Integer Function given by $\mathbf{\text{g}\left( \text{x} \right)\text{=}\left[ \text{x} \right]}$ where $\mathbf{\left[ \text{x} \right]}$ is greatest integer less than or equal to $\mathbf{\text{x}}$. Then does $\mathbf{\text{fog}}$ and $\mathbf{\text{gof}}$ coincide in $\mathbf{\text{(1, 0 }\!\!]\!\!\text{ }}$?

Ans: It is given that, $\text{f:R }\to \text{ R}$ is defined as \text{R}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{1,x0} \\ & \text{0,x=0} \\ & \text{-1,x0} \\ \end{align} \right\} and $\text{g:R }\to \text{ R}$ is defined as $\text{g}\left( \text{x} \right)\text{=}\left[ \text{x} \right]$ where $\left[ \text{x} \right]$ is the greatest integer less than or equal to $\text{x}$.

Let us take $\text{x}\in \text{(1, 0 }\!\!]\!\!\text{ }$.

So, $\left[ \text{x} \right]\text{=1}$ if $\text{x=1}$ and $\left[ \text{x} \right]\text{=0}$ if $\text{0x1}$.

Now,

$\text{fog}\left( \text{x} \right)\text{ = f}\left( \text{g}\left( \text{x} \right) \right)$

$\text{= f}\left( \left[ \text{x} \right] \right)\text{ }$

\text{= }\left\{ \begin{align}& \text{f}\left( \text{1} \right)\text{,}\,\text{if}\,\text{x=1} \\ & \text{f}\left( \text{0} \right)\text{,}\,\text{if}\,\text{x}\in \left( \text{0,1} \right) \\ \end{align} \right\}

\text{= }\left\{ \begin{align}& \text{1,}\,\text{if}\,\text{x=1} \\ & \text{0,}\,\text{if}\,\text{x}\in \left( \text{0,1} \right) \\ \end{align} \right\}

$\text{go}\left( \text{x} \right)\text{ = g}\left( \text{f}\left( \text{x} \right) \right)\text{ =g}\left( \text{1} \right)$   (as $\text{x0}$)

$\text{=}\left[ \text{1} \right]$

$\text{=1}$

Therefore, when $\text{x}\in \text{(1, 0 }\!\!]\!\!\text{ }$, $\text{fog=0}$ and $\text{gof=1}$, so they do not coincide in $\text{(1, 0 }\!\!]\!\!\text{ }$.

19. Number of binary operations on the set $\mathbf{\left\{ \text{a, b} \right\}}$ are

1. $\mathbf{\text{10}}$

2. $\mathbf{\text{16}}$

3. $\mathbf{\text{20}}$

4. $\mathbf{\text{8}}$

Ans: A binary operation $\text{*}$  on $\left\{ \text{a, b} \right\}$ is a function defined as $\left\{ \text{a, b} \right\}\times \left\{ \text{a, b} \right\}\text{ }\to \text{ }\left\{ \text{a, b} \right\}$.

That is, $\text{*}$ is a function from $\left\{ \left( \text{a, a} \right)\text{, }\left( \text{a, b} \right)\text{, }\left( \text{b, a} \right)\text{, }\left( \text{b, b} \right) \right\}\text{ }\to \text{ }\left\{ \text{a, b} \right\}$.

Therefore the total number of binary operations on the set $\left\{ \text{a, b} \right\}$ is ${{\text{2}}^{\text{4}}}\text{=16}$.

Hence, the correct answer is (B) $\text{16}$.

Students often feel troubled because they cannot find the appropriate solutions for NCERT solutions for class 12 maths chapter 1 relations and functions. Here, the learners will find the proper solutions to the sums of relation and function class 12 chapter. These solutions can be downloaded for free from the official website of Vedantu, so it’s just a click away.

### NCERT Solutions for Class 12 Maths Chapter 1

#### 1.1 Introduction

The NCERT Class 12 Maths Chapter 1 delves deep into the concepts of Relations and Functions. Students will have a recap of their learning from the previous class. The assessment will pave the way to a more in-depth understanding of the fundamental concepts of class 12 Maths Chapter 1. A student will also learn the quantifiable relationship between two objects belonging to the sets. All the key points pertaining to relations and functions can be clearly understood with the given set of instructions. For a good understanding of the NCERT solutions for class 12 Maths Chapter 1, there is a need to revise the topics covered in the previous class.

Students will gain a refined understanding of the topics covered in Chapter 1 by studying the given activities. By going through the activities carefully, and by solving the sums of Chapter 1 they will be able to establish a clear idea of the properties of Relations and Functions. The level of difficulty will be propelled from an easier stage to a difficult one as one goes on solving the sums, one after the other. After gaining an overall idea of Chapter 1, it will be easier for students to answer every question with the utmost clarity.

#### 1.2 Recall

In NCERT Class 12 Maths Chapter 1 Solutions, students will get to revise some topics from the previous class pertaining to Relations and Functions. A reiteration of these topics will help them to gain a deeper understanding of vertible and invertible functions. Comprehensive knowledge of real numbers and the usage of addition, multiplication, division, and subtraction will help them develop a better understanding of this chapter. Along with this, an insight into how the usage of relations accentuates the complimentary integer is also included in this chapter. Furthermore, it will pave the way for understanding the domain and the co-domain that adheres to the principles of functions.

Exercise 1.1 Solutions: 16 Questions (3 Short Questions, 13 Long Questions)

#### 1.3 Types of Functions

The above section explains the various types of Functions like identity function, constant function, polynomial function, rational function, modulus function, signum function, etc. This will help in understanding the relationship between the injective and the surjective function. One will also learn how the elements of three distinctly different numbers are in association with a host element. It furthermore provides a deeper sense of understanding about the finite as well as the infinite sets.

Exercise 1.2 Solutions: 12 Questions (5 Short Questions, 7 Long Questions)

#### 1.3 Composition of Functions and Invertible Function

This section will focus on the composition of the functions and how the inverse quality concerning bijective works. Once you gain an understanding of this, you will be provided with examples exploring the relationship between the set and the codes. This section of the chapter focuses on the combination of two distinctly different functions and how a code can be eventually attached to it.

Exercise 1.3 Solutions: 14 Questions (4 Short Questions, 10 Long Questions)

#### 1.5 Binary Operations

You must be aware of the fundamental concepts of the BODMAS rule right from your junior school days. It will show how operational functions assist in inducing two numbers and therefore, their association as binary operations is used in the merging of two integers into one. You will be able to cover all of the aspects of the operations. You will also learn how a set of numbers will be eventually substituted by an arbitrary set of numbers which concerns the binary operation. Moreover, the formulation of the pair pertaining to the elements will help in developing a deeper sense of understanding of the concepts covered in this chapter.

Exercise 1.4 Solutions: 13 Questions (6 Short Questions, 7 Long Questions)

### Key Features of NCERT Solutions for Class 12 Maths Chapter 1

You must go through every single point stated in this chapter for securing good marks in the exams. If you refer to the relations and functions class 12 NCERT solutions that have been prepared by our experts in NCERT, then you are definitely going to learn the basic, as well as, the core concepts of this topic. The key features of the solutions are listed below.

The relation and function class 12 NCERT solutions are explained in a simple and logical manner to guide and assist the students in gaining an understanding of the chapter.

By practicing the sums of relations and functions class 12 solutions, students will be able to instill better knowledge of the topics covered in the chapter.

If students have any kind of doubts, they can refer to the class 12 Maths NCERT solutions chapter 1 and reach out to our experts for explanations.

By following the instructions provided in the solutions will help students in getting good marks in the exam.

 Chapter 1 - Relations and Functions Exercises in PDF Format Exercise 1.1 16 Questions & Solutions (3 Short Answers, 13 Long Answers) Exercise 1.2 12 Questions & Solutions (5 Short Answers, 7 Long Answers) Exercise 1.3 14 Questions & Solutions (4 Short Answers, 10 Long Answers) Exercise 1.4 13 Questions & Solutions (6 Short Answers, 7 Long Answers)

## FAQs on NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions

1. Why is Vedantu’s NCERT Solutions for Class 12 Maths Chapter 1 very reliable?

Class 12 NCERT solutions Chapter 1 is of utmost importance to enhance your basic Maths knowledge, which will be immensely helpful in your future. You have to understand not just the basic concepts of this chapter but even the advanced portion to hold a grasp over this.

Vedantu’s Relation and Function solutions PDF (you can also refer to our live video courses) will strengthen your fundamentals of the chapter. It will also prepare you in such a way that even the advanced questions from this chapter will be easy for you to solve.

Vedantu’s Relations and Functions questions and answers PDF have both easy-to-understand solutions to let you understand the basics and also advanced solutions to erase any kind of doubts related to the chapter. In our NCERT Solutions PDF for Class 12 Maths Chapter 1, you will find answers for the basic problems like finding if the Relation is reflexive, symmetric or transitive. Intermediate problems etc.

2. How many exercises are there in Class 12 Maths NCERT textbook Chapter 1?

Class 12 Maths Chapter 1 contains four exercises in total and all of these exercises consist of various kinds of questions such as short answer type, multiple choice questions.

• Exercise 1.1: 16 Questions (14 Short Answers, 2 MCQ)

• Exercise 1.2: 12 Questions (10 Short Answers, 2 MCQ)

• Exercise 1.3: 14 Questions (12 Short Answers, 2 MCQ)

• Exercise 1.4: 13 Questions (12 Short Answers, 1 MCQ)

3. Can you give a primary overview of the topics and sub-topics of the chapter?

The main topics and sub-topics covered in this chapter are given below. There is also a Miscellaneous Q&A section at the end of Chapter 1.

• 1.1 - Introduction

• 1.2 - Types of Relations

• 1.3 - Types of Functions

• 1.4 - Composition of Functions and Invertible Function

• 1.5 - Binary Operations

4. Why should I opt for NCERT Solutions for Class 12 Maths Chapter 1?

There are various reasons behind why every Class 12 student should opt for the NCERT Solutions for Class 12 Maths Chapter 1 as the best study guide. A few are given in the following. Take a look:

• All the answers to the questions asked in the textbook exercises are aimed at providing an effortless solution in solving the problems from Chapter 1 of Class 12 Maths named Relation and Function.

• NCERT solutions for Class 12 Maths provide the list of all the important formulas under one roof.

• NCERT Solutions are carefully created by some of the excellent subject matter experts from the relevant industry.

• These solutions can be beneficial while studying for various competitive exams such as JEE Main, Olympiad etc. apart from the board exams.

• These solutions help you to learn the question patterns, marks weightage etc.

5. What are the concepts explained in NCERT Solutions for Chapter 1 of Class 12 Maths?

The first chapter of Class 12 Maths is “Relation and Functions,” and it gives an introduction to new concepts, as well as concepts learned previously that might be used in the Class 12. It covers types of relations, types of functions, the composition of functions and invertible functions, and binary operations. Each exercise in the Chapter has about 12 questions. It provides all the necessary formulas required to understand the concepts with sufficient examples.

6. What is Relation, according to  Chapter 1 of Class 12 Maths?

Relation is the concept that describes the relationship between two sets of quantities with each other. The relations can be described as empty or universal. These concepts along with examples explaining them step-by-step can be found in the NCERT Solutions for Chapter 1 of Class 12 Maths. All the exercises are also solved for the reference of students and extra questions for practice are provided as well.

7. How many chapters are present in Class 12 Maths, including Chapter 1?

The NCERT Maths textbook for Class 12 Maths has a total of 12 chapters including Chapter 1. All these chapters are covered in the NCERT Solutions PDF provided by Vedantu. The solutions book explains all the chapters with enough examples that help students understand the concepts better. Each exercise is solved with step-by-step answers for the students to follow. Extra questions based on previous years' papers and patterns are also provided for practice. These solutions are prepared by subject matter experts and are based on the latest exam guidelines. So these solutions are 100% reliable.

8. Where can I download the NCERT Solutions for Chapter 1 of Class 12 Maths?

The NCERT Solutions for Chapter 1 of Class 12 Maths is available for download free of cost on the Vedantu website and the Vedantu app. The NCERT Solutions PDF is written keeping in mind the easy learning of students and how each student has a different caliber. All the concepts and examples are explained in an easy language for the students to understand. It helps you familiarise with the kind of questions to expect in the exam. The questions and solutions are formulated by professionals.

9. How can I prepare Chapter 1 of Class 12 Maths for the board exam?

Board exams are very important in determining a student’s future and  Chapter 1 of Class 12 Maths is important. Chapter 1 is important not only for the board but also for various competitive exams, including JEE. The students should practice all the questions and examples from the NCERT textbook thoroughly. The students can refer to the solutions PDF for extra questions and explanations. Students should also practice sample papers and papers from previous years for the students to know the kind of questions to expect in the exam and be well-versed with it.