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

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 (Not in the current syllabus)

## Relations and Functions Chapter at a Glance - Class 12 NCERT Solutions

### 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.

### 1. Let ${f: R }\to { R}$ be defined as ${f}\left( {x} \right){ = 10x + 7}$. Find the function ${g: R }\to { R}$ such that ${gof = fog = }{{{I}}_{{R}}}$. (not in the current syllabus)Ans: The given function ${f: R }\to { R}$ is defined as ${f}\left( {x} \right){ = 10x + 7}$.For the function ${f}$ to be one – one:${f}\left( {x} \right){ = f}\left( {y} \right)$, where ${x, y}\in {R}$.$\Rightarrow {10x + 7 = 10y + 7}$$\Rightarrow {x = y}So, the given function {f} is a one – one function.For the function {f} to be onto:Let {y = 10x + 7}, for {y}\in {R}.\Rightarrow {x =}\dfrac{{y-7}}{{10}}\in {R}So, for any {y}\in {R}, there exists {x =}\dfrac{{y-7}}{{10}}\in {R} such that:\left( {x} \right){ = f}\left( \dfrac{{y-7}}{{10}} \right)$${= 10}\left( \dfrac{{y-7}}{{10}} \right){+ 7}$${=y- 7+7}$${= y}$So, the given function ${f}$ is onto.Therefore, the function ${f}$ is an invertible function.Defining ${g: R }\to { R}$ as $\left( {y} \right){ =}\dfrac{{y-7}}{{10}}{ }$.Calculating ${gof}$:${go}\left( {x} \right){=g}\left( {f}\left( {x} \right) \right)$${= g}\left( {10x+7} \right)$${=}\dfrac{\left( {10x+7} \right){-7}}{{10}}$${=}\dfrac{{10x}}{{10}}$${= x}$ Calculating ${fog}$:${fo}\left( {y} \right){=f}\left( {g}\left( {y} \right) \right)$${=f}\left( \dfrac{{y-7}}{{10}} \right)$${=10}\left( \dfrac{{y-7}}{{10}} \right){+7}$${=y-7+7}$${=y}$Therefore, ${gof = }{{{I}}_{{R}}}$ and ${fog = }{{{I}}_{{R}}}$.2. Let ${f: W }\to { W}$ be defined as ${f}\left( {n} \right){ = n - 1}$, if ${n}$ is odd and${f}\left( {n} \right){ = n + 1}$, if ${n}$ is even. Show that ${f}$is invertible. Find the inverse of ${f}$. Here, ${W}$ is the set of all whole numbers.Ans: The function ${f: W }\to { W}$ is defined as ${f}\left( {n} \right){ =}\left\{ \begin{matrix}{n-1 if n is odd} \\ {n+1 if n is even} \\\end{matrix} \right\}$ For the function to be one – one:${f}\left( {n} \right){ = f}\left( {m} \right)$If ${n}$ is odd and ${m}$ is even, then we will have ${n - 1 = m + 1}$.$\Rightarrow {n-m = 2}$Which is not possible.Similarly, the possibility that of ${n}$ is even and ${m}$ is odd can also be ignored.Therefore, both ${n}$ and ${m}$ must be either odd or even.Now, if both ${n}$ and ${m}$ are odd,${f}\left( {n} \right){ = f}\left( {m} \right)$$\Rightarrow {n - 1 = m - 1} \Rightarrow {n = m} if both {n} and {m} are even,{f}\left( {n} \right){ = f}\left( {m} \right)$$\Rightarrow {n + 1 = m + 1}$ $\Rightarrow {n = m}$Therefore the given function ${f}$ is one – one.For the function to be onto:Now, for any odd number ${2r + 1}\in {N}$ is the image of ${2r}\in {N}$ and any even number ${2r}\in {N}$ is the image of ${2r + 1}\in {N}$.Hence, the function ${f}$ is onto.Therefore, the function ${f}$ is an invertible function.Defining ${g: W }\to { W}$ as $\left( {m} \right){=}\left\{ \begin{matrix}{m+1 if m is even} \\{m-1 if m is odd} \\\end{matrix} \right\}$ When ${n}$ is odd:${go}\left( {n} \right){ = g}\left( {f}\left( {n} \right) \right){ }$${= g}\left( {n - 1} \right){ }$${= n - 1 + 1 }$${= n}When {n} is even{go}\left( {n} \right){ = g}\left( {f}\left( {n} \right) \right){ }$${= g}\left( {n + 1} \right){ }$${= n + 1 - 1 }$${= n}$When ${m}$ is odd${fo}\left( {m} \right){ = f}\left( {g}\left( {m} \right) \right){ }$${= f}\left( {m - 1} \right){ }$${= m - 1 + 1 }$${= m}When {m} is even{fo}\left( {m} \right){ = f}\left( {g}\left( {m} \right) \right){ }$${= f}\left( {m + 1} \right){ }$${= m + 1 - 1 }$${= m}$ Therefore, ${gof = }{{{I}}_{{W}}}$ and ${fog = }{{{I}}_{{W}}}$. Hence, the function ${f}$ is invertible and the inverse of ${f}$ is given by ${{{f}}^{{-1}}}{=g}$, which is the same as ${f}$.3. If ${f: R }\to { R}$ is defined by ${f}\left( {x} \right){ = }{{{x}}^{{2}}}{ - 3x + 2}$, find ${f}\left( {f}\left( {x} \right) \right)$. Ans: The function ${f: R }\to { R}$ is defined as ${f}\left( {x} \right){ = }{{{x}}^{{2}}}{ - 3x + 2}$.Calculating ${f}\left( {f}\left( {x} \right) \right)$:${f}\left( {f}\left( {x} \right) \right){ = f}\left( {{{x}}^{{2}}}{- 3x + 2} \right)$${= }{{\left( {{{x}}^{{2}}}{- 3x + 2} \right)}^{{2}}}{ - 3}\left( {{{x}}^{{2}}}{- 3x+2} \right){+2}$${= }\left( {{{x}}^{{4}}}{+9}{{{x}}^{{2}}}{+4-6}{{{x}}^{{3}}}{-12x+4}{{{x}}^{{2}}} \right){+}\left( {-3}{{{x}}^{{2}}}{+9x-6} \right){+2}$${= }{{{x}}^{{4}}}{-6}{{{x}}^{{3}}}{+10}{{{x}}^{{2}}}{--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 ${f: N }\to { Z}$ and ${g: Z }\to { Z}$ such that ${gof}$ is injective but ${g}$ is not injective. (Not in the current syllabus)

(Hint: Consider ${f}\left( {x} \right){ = x}$ and ${g }\left( {x} \right){ = }\left| {x} \right|$)

Ans: Taking the function ${f: N }\to { Z}$ as ${f}\left( {x} \right){ = x}$ and the function ${g: Z }\to { Z}$ as ${g }\left( {x} \right){ = }\left| {x} \right|$.

Showing that ${g}$ is not injective:

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

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

Therefore $\left( {-1} \right){ = g}\left( {1} \right)$ but ${-1}\ne { 1}$.

Therefore the function ${g}$ is not injective.

Showing that ${gof}$ is injective:

The function ${gof: N }\to { Z}$ is defined as

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

${= g}\left( {x} \right){ }$

${= }\left| {x} \right|$

Taking ${x, y}\in {N}$ such that ${gof}\left( {x} \right){ = gof}\left( {y} \right)$.

$\Rightarrow \left| {x} \right|{ = }\left| {y} \right|$

Since ${x, y}\in {N}$, both are positive.

$\Rightarrow {x = y}$

Therefore the function ${gof}$ is injective.

7. Given examples of two functions ${f: N }\to { N}$ and ${g: N }\to { N}$ such that ${gof}$ is onto but ${f}$ is not onto.

(Hint: Consider ${f}\left( {x} \right){ = x+1}$ and $\left( {x} \right){ =}\left\{ \begin{matrix}{x-1, if x1} \\{1, if x1} \\\end{matrix} \right\}$)

Ans: Taking the function ${f: N }\to { N}$ as ${f}\left( {x} \right){ = x+1}$

And the function ${g: N }\to { N}$ by ${g}\left( {x} \right){ =}\left\{ \begin{matrix}{x-1, if x1} \\ {1, if x1} \\\end{matrix} \right\}$

Showing that ${g}$ is not onto.

Taking an element ${1}$ in a co-domain ${N}$.

Since this element is not an image of any of the elements in domain ${N}$, therefore ${g}$ is not onto.

Showing that ${gof}$ is onto:

The function ${gof: N }\to { N}$ is defined as ${go}\left( {x} \right){ = g}\left( {f}\left( {x} \right) \right){ }$.

${= g}\left( {x + 1} \right){ }$

${= x + 1 - 1 }$                (as ${x}\in {N}\Rightarrow {x + 1 1}$)

${= x}$

So, for ${y}\in {N}$, there exists ${x = y}\in {N}$ such that ${gof}\left( {x} \right){ = y}$.

Therefore, the function ${gof}$ is onto.

8. Given a non-empty set ${X}$, consider ${P}\left( {X} \right)$ which is the set of all subsets of ${X}$. Define the relation ${R}$ in ${P}\left( {X} \right)$ as follows:

For subsets ${A,B}$ in ${P}\left( {X} \right)$, ${ARB}$ if and only if ${A}\subset {B}$. Is ${R}$ an equivalence relation on ${P}\left( {X} \right)$? Justify you answer.

Ans: We know that every set is a subset of itself, ${ARA}$ for all ${A}\in {P}\left( {X} \right)$

Therefore ${R}$ is reflexive.

Let ${ARB}\Rightarrow {A}\subset {B}$.

This does not mean that ${B}\subset {A}$.

If ${A = }\left\{ {1, 2} \right\}$ and ${B = }\left\{ {1, 2, 3} \right\}$, then it cannot be implied that ${B}$ is related to ${A}$.

Therefore ${R}$ is not symmetric.

If ${ARB}$ and ${BRC}$ , then;

${A}\subset {B}$ and ${B}\subset {C}$

$\Rightarrow {A}\subset {C}$

$\Rightarrow {ARC}$

Therefore ${R}$ is transitive.

Hence, ${R}$ is not an equivalence relation as it is not symmetric.

9. Given a non-empty set ${X}$, consider the binary operation ${*: P}\left( {X} \right){ }\!\!\times\!\!{ P}\left( {X} \right){ }\to { P}\left( {X} \right)$ given by ${A*B = A}\cap {B}\forall {A, B}$ in ${P}\left( {X} \right)$ is the power set of ${X}$. Show that ${X}$ is the identity element for this operation and ${X}$ is the only invertible element in ${P}\left( {X} \right)$ with respect to the operation ${*}$.

Ans: The binary operation ${*: P}\left( {X} \right){ }\!\!\times\!\!{ P}\left( {X} \right){ }\to { P}\left( {X} \right)$ is given by ${A*B = A}\cap {B}\forall {A, B}$ in ${P}\left( {X} \right)$.

${A}\cap {X = A = X }\cap { A}$ for all ${A}\in {P}\left( {X} \right)$

$\Rightarrow {A * X = A = X * A}$ for all ${A}\in {P}\left( {X} \right)$

${X}$ is the identity element for the given binary operation ${*}$.

An element ${A}\in {P}\left( {X} \right)$ is invertible if there exists ${B}\in {P}\left( {X} \right)$ such that

${A * B = X = B * A}$          (As ${X}$ is the identity element)

Or

${A }\cap { B = X = B }\cap { A}$

This is only possible when ${A = X = B}$.

Therefore, ${X}$ is the only invertible element in ${P}\left( {X} \right)$ with respect to the given operation ${*}$.

10. Find the number of all onto functions from the set ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself.

Ans: The total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself will be same as the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$.

Since the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$ is ${n}$, thus total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself are ${n}$.

11. Let ${S = }\left\{ {a, b, c} \right\}$ and ${T = }\left\{ {1, 2, 3} \right\}$. Find ${{{F}}^{{-1}}}$ of the following functions ${F}$ from ${S}$ to${T}$, if it exists.

1. The function is given as ${{F = }\left\{ \left( {a, 3} \right){, }\left( {b, 2} \right){, }\left( {c, 1} \right) \right\}}$.

Ans: The function ${F: S }\to { T}$ is defined as ${F = }\left\{ \left( {a, 3} \right){, }\left( {b, 2} \right){, }\left( {c, 1} \right) \right\}$.

$\Rightarrow {F }\left( {a} \right){ = 3, F }\left( {b} \right){ = 2, F}\left( {c} \right){ = 1}$

Therefore, ${{{F}}^{{-1}}}{:T }\to {S}$ is given by ${{{F}}^{{-1}}}{= }\left\{ \left( {3, a} \right){, }\left( {2, b} \right){, }\left( {1, c} \right) \right\}$.

1. The function is given as ${{ }\!\!~\!\!{ F = }\left\{ \left( {a, 2} \right){, }\left( {b, 1} \right){, }\left( {c, 1} \right) \right\}}$.

Ans: The function ${F: S }\to { T}$ is defined as ${F=}\left\{ \left( {a, 2} \right){, }\left( {b, 1} \right){, }\left( {c, 1} \right) \right\}$.

Since ${F}\left( {b} \right){ = F}\left( {c} \right){ = 1}$, the function ${F}$ is not one – one.

Hence, the function ${F}$ is not invertible so its inverse does not exist.

12. Consider the binary operations ${{*: R }\!\!\times\!\!{ R }\to { R}}$ and ${{o}\,{: R }\!\!\times\!\!{ R }\to { R}}$ defined as ${{a*b}\,{=}\,\left| {a-b} \right|}$ and ${{a}\,{o}\,{b}\,{=}\,{a,}\forall {a,}\,{b}\in {R}}$. Show that ${{*}}$ is commutative but not associative, ${{o}}$ is associative but not commutative. Further, show that ${\forall {a,}\,{b,}\,{c}\in {R,}\,{a*}\left( {b}\,{o}\,{c} \right)\,{=}\,\left( {a*b} \right)\,{o}\,\left( {a*c} \right)}$. [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 ${*: R }\!\!\times\!\!{ R }\to { R}$ and ${o}\,{: R }\!\!\times\!\!{ R }\to { R}$is defined as ${a*b}\,{=}\,\left| {a-b} \right|$ and ${a}\,{o}\,{b}\,{=}\,{a,}\forall {a,}\,{b}\in {R}$

For ${a,}\,{b}\in {R}$, we have ${a*b}\,{=}\,\left| {a-b} \right|$

And,

${b*a}\,{=}\,\left| {b-a} \right|\,=\,\left| -\left( {a-b} \right) \right|\,=\,\left| {a-b} \right|$

Therefore, ${a*b}\,{=}\,{b*a}$.

Hence, operation ${*}$ is commutative.

For ${1,}\,{2,}\,{3}\in {R}$,

$\left( {1*2} \right){*3}\,{=}\,\left( \left| {1}\,{-}\,{2} \right| \right){*3}$

${=1*3}$

${=}\left| {1-3} \right|$

${=2}$

And,

${1*}\left( {2*3} \right)\,{=1*}\left( \left| {2}\,{-}\,{3} \right| \right)$

${=1*1}$

${=}\left| {1-1} \right|$

${=0}$

Therefore, $\left( {1*2} \right){*3}\ne {1*}\left( {2*3} \right)$.

Hence, the operation ${*}$ is not associative.

We can observe that, for ${1,}\,{2,}\,{3}\in {R}$, ${1}\,{o}\,{2}\,{=1}$ and ${2}\,{o}\,{1}\,{=2}$

Therefore, ${1}\,{o}\,{2}\ne {2}\,{o}\,{1}$

Hence, the operation ${o}$ is not commutative.

For ${a,}\,{b,}\,c\in {R}$,

$\left( {a}\,{o}\,{b} \right)\,o\,c\,{=}\,{a}\,{o}\,{c}\,{=}\,{a}$

And,

${a}\,{o}\,\left( {b}\,o\,c \right)\,{=}\,{a}\,{o}\,{b}\,{=}\,{a}$

Therefore, $\left( {a}\,{o}\,{b} \right)\,o\,c\,\ne \,{a}\,{o}\,\left( {b}\,o\,c \right)$ where ${a,}\,{b,}\,c\in {R}$.

Operation ${o}$ is associative.

For ${a,}\,{b,}\,c\in {R}$,

${a}\,*\,\left( {b}\,o\,c \right)\,{=}\,{a}\,*\,{b}\,{=}\,\left| {a-b} \right|$

$\left( {a}\,{*}\,{b} \right)\,{o}\,\left( {a}\,{*}\,{c} \right)\,{=}\,\left( \left| {a-b} \right| \right)\,o\,\left( \left| {a-c} \right| \right)\,=\,\left| {a-b} \right|$

Hence, ${a}\,*\,\left( {b}\,o\,c \right)\,{=}\,\left( {a}\,{*}\,{b} \right)\,{o}\,\left( {a}\,{*}\,{c} \right)$.

So, if ${1,}\,{2,}\,{3}\in {R}$, then,

${1}\,{o}\,\left( {2}\,{*}\,{3} \right)\,{=}\,{1}\,{o}\,\left( \left| {2-3} \right| \right)$

${=}\,{1}\,{o}\,{1}$

${=1}$

$\left( {1}\,{o}\,{2} \right)\,{*}\,\left( {1}\,{o}\,{3} \right)\,{=}\,{1}\,{*}\,{1}\,$

${=}\,\left| {1-1} \right|$

${=}\,{0}$

Therefore, the operation ${o}$ does not distribute over ${*}$.

13. Given a non - empty set ${{X}}$, let ${{*:P}\left( {X} \right){ }\!\!\times\!\!{ P}\left( {X} \right)\to {P}\left( {X} \right)}$ be defined as ${{A*B=(A-B)}\cup {(B-A),}\forall {A,B}\in {P(X)}}$. Show that the empty set ${\Phi}$ is the identity for the operation ${{*}}$ and all the elements ${{A}}$ of ${{P(X)}}$ are invertible with ${{{{A}}^{{-1}}}{=A}}$.

(Hint: ${\left( {A- }\!\!\Phi\!\!{ } \right)\cup \left( { }\!\!\Phi\!\!{ -A} \right){=A}}$ and ${\left( {A-A} \right)\cup \left( {A-A} \right){=A*A}}$).

Ans: the function ${*: P}\left( {X} \right){ }\!\!\times\!\!{ P}\left( {X} \right){ }\to { P}\left( {X} \right)$ is defined as ${A*B=(A-B)}\cup {(B-A),}\forall {A,B}\in {P(X)}$.

For ${A}\in {P(X)}$,

${A* }\!\!\Phi\!\!{ =}\left( {A- }\!\!\Phi\!\!{ } \right)\cup \left( { }\!\!\Phi\!\!{ -A} \right)$

${=A}\cup { }\!\!\Phi\!\!{ }$

${=A}$

${ }\!\!\Phi\!\!{ *A=}\left( { }\!\!\Phi\!\!{ -A} \right)\cup \left( {A- }\!\!\Phi\!\!{ } \right)$

${= }\!\!\Phi\!\!{ }\cup {A}$

${=A}$

Therefore, ${A* }\!\!\Phi\!\!{ = }\!\!\Phi\!\!{ *A}$ for all ${A}\in {P(X)}$.

For the given operation ${*}$, ${ }\!\!\Phi\!\!{ }$ is the identity element.

For ${A}\in {P(X)}$, if there exists ${B}\in {P(X)}$ such that ${A*B= }\!\!\Phi\!\!{ =B*A}$, then the element ${A}\in {P(X)}$ will be invertible.

For all ${A}\in {P(X)}$,

${A*A=}\left( {A-A} \right)\cup \left( {A-A} \right)$

${= }\!\!\Phi\!\!{ - }\!\!\Phi\!\!{ }$

${= }\!\!\Phi\!\!{ }$

Therefore, all the elements ${A}\in {P(X)}$ are invertible with ${{{A}}^{{-1}}}{=A}$.

14. Define a binary operation ${{*}}$ on the set ${\left\{ {0, 1, 2, 3, 4, 5} \right\}}$ as {{a*b=}\left\{ \begin{align}& {a+b}\,\,{if}\,\,{a+b6} \\ & {a+b-6}\,\,{if}\,\,{a+b}\ge {6} \\ \end{align} \right\}}.

Show that zero is the identity for this operation and each element ${a}\ne {0}$ of the set is invertible with ${6-a}$ being the inverse of ${a}$.

Ans: Let us take ${X=}\left\{ {0, 1, 2, 3, 4, 5} \right\}$.

The binary operation ${*}$ on ${X}$ is defined as {a*b=}\left\{ \begin{align}& {a+b}\,\,{if}\,\,{a+b6} \\ & {a+b-6}\,\,{if}\,\,{a+b}\ge {6} \\ \end{align} \right\}.

If ${a*e}\,{=a}\,{=}\,{e*a}$ for all ${a}\in {X}$, then element ${e}\in {X}$ is the identity element for the operation ${*}$.

For ${a}\in {X}$, we have

${a*0}\,{=a}$ A Є X ⇒ a+ 0 <6

${0*a}\,{=0}$ A Є X ⇒ a+ 0 <6

Therefore, ${a*0=0*a}$ for all ${a}\in {X}$.

Hence, ${0}$ is the identity element for the given operation ${*}$.

For ${a}\in {X}$, if there exists ${b}\in {X}$ such that ${a*b=0=b*a}$, then the element ${a}\in {X}$ will be invertible. That is,

{a*b=}\left\{ \begin{align}& {a+b=0=b+a,}\,\,{if}\,\,{a+b6} \\ & {a+b-6=0=b+a-6,}\,\,{if}\,\,{a+b}\ge {6} \\ \end{align} \right\}

$\Rightarrow {a=}\,{-b}$ or ${b=6-a}$

Since, ${X=}\left\{ {0, 1, 2, 3, 4, 5} \right\}$ and ${a,b}\in {X}$. So, ${a}\ne {-b}$.

Therefore, ${b=6-a}$ is the inverse of ${a}$ for all ${a}\in {X}$.

15. Let ${{A=}\left\{ {-1, 0, 1, 2} \right\}{,B=}\left\{ {-4, -2, 0, 2} \right\}}$ and ${{f,g:A }\to { B}}$ be functions defined by ${{f}\left( {x} \right){=}{{{x}}^{{2}}}{-x,}\,{x}\in {A}}$ and ${{g}\left( {x} \right){=2}\left| {x-}\dfrac{{1}}{{2}} \right|{-1,x}\in {A}}$. Are ${{f}}$ and ${{g}}$ equal? Justify your answer. (Hint: One may note that two function ${{f:A }\to { B}}$ and ${{g:A }\to { B}}$ such that ${{f}\left( {a} \right){=g}\left( {a} \right)\forall {a}\in {A}}$, are called equal functions)

Ans: Let ${A=}\left\{ {-1, 0, 1, 2} \right\}{,B=}\left\{ {-4, -2, 0, 2} \right\}$ and ${f,g:A }\to { B}$ are defined by ${f}\left( {x} \right){=}{{{x}}^{{2}}}{-x,}\,{x}\in {A}$ and ${g}\left( {x} \right){=2}\left| {x-}\dfrac{{1}}{{2}} \right|{-1,x}\in {A}$.

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

${=1+1}$

${=2}$

And,

${g}\left( {-1} \right){=2}\left| \left( {-1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

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

${=3-1}$

${=2}$

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

${f}\left( {0} \right){=}{{\left( {0} \right)}^{{2}}}{-}\left( {0} \right)$

${=0}$

And,

${g}\left( {0} \right){=2}\left| \left( {0} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=1-1}$

${=0}$

$\Rightarrow {f}\left( {0} \right){=g}\left( {0} \right)$

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

${=1-1}$

${=0}$

And,

${g}\left( {1} \right){=2}\left| \left( {1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

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

${=1-1}$

${=0}$

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

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

${=4-2}$

${=2}$

And,

${g}\left( {2} \right){=2}\left| \left( {2} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

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

${=3-1}$

${=2}$

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

Therefore, ${f}\left( {a} \right){=g}\left( {a} \right)\forall {a}\in {A}$. Hence the functions ${f}$ and ${g}$ 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 ${{f:R }\to { R}}$ be the Signum Function defined as {\left( {x} \right){=}\left\{ \begin{align}& {1,x0} \\ & {0,x=0} \\ & {-1,x0} \\ \end{align} \right\}} and ${{g:R }\to { R}}$ be the Greatest Integer Function given by ${{g}\left( {x} \right){=}\left[ {x} \right]}$ where ${\left[ {x} \right]}$ is greatest integer less than or equal to ${{x}}$. Then does ${{fog}}$ and ${{gof}}$ coincide in ${{(1, 0 }\!\!]\!\!{ }}$? (Not in the current syllabus)

Ans: It is given that, ${f:R }\to { R}$ is defined as {R}\left( {x} \right){=}\left\{ \begin{align}& {1,x0} \\ & {0,x=0} \\ & {-1,x0} \\ \end{align} \right\} and ${g:R }\to { R}$ is defined as ${g}\left( {x} \right){=}\left[ {x} \right]$ where $\left[ {x} \right]$ is the greatest integer less than or equal to ${x}$.

Let us take ${x}\in {(1, 0 }\!\!]\!\!{ }$.

So, $\left[ {x} \right]{=1}$ if ${x=1}$ and $\left[ {x} \right]{=0}$ if ${0x1}$.

Now,

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

${= f}\left( \left[ {x} \right] \right){ }$

{= }\left\{ \begin{align}& {f}\left( {1} \right){,}\,{if}\,{x=1} \\ & {f}\left( {0} \right){,}\,{if}\,{x}\in \left( {0,1} \right) \\ \end{align} \right\}

{= }\left\{ \begin{align}& {1,}\,{if}\,{x=1} \\ & {0,}\,{if}\,{x}\in \left( {0,1} \right) \\ \end{align} \right\}

${go}\left( {x} \right){ = g}\left( {f}\left( {x} \right) \right){ =g}\left( {1} \right)$   (as ${x0}$)

${=}\left[ {1} \right]$

${=1}$

Therefore, when ${x}\in {(1, 0 }\!\!]\!\!{ }$, ${fog=0}$ and ${gof=1}$, so they do not coincide in ${(1, 0 }\!\!]\!\!{ }$.

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)

#### 1.5 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 (Not in the current syllabus)
• 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(not in the current syllabus) , 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.