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NCERT Solutions for Class 12 Maths Chapter 1: Relation and Functions - Exercise 1.1

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NCERT Solutions for Class 12 Maths Chapter 1 (Ex 1.4)

NCERT Solutions for Class 12 Maths Chapter 1 Ex 1.4 are the best option that every student must consider who is struggling to understand the concept of the Mathematics chapter based on Functions. The simplified version offered by Vedantu has helped numerous students from various academic backgrounds. Exercise 1.4 Class 12 Maths Solutions identify all the critical problems based around Functions as all the steps have been highlighted and discussed in detail by professional teachers. Download the PDF to elevate your mathematical skills.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 1 - Relations and Functions

Exercise:

Exercise - 1.4

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

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

Exercise 1.4

1. Determine whether 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 \[\text{a * b =}\frac{\text{ab}}{\text{2}}\]. 

Ans: On \[\text{Q}\], \[\text{*}\] is defined by \[\text{a * b =}\frac{\text{ab}}{\text{2}}\].

For all \[\text{a, b}\in \text{Q}\], \[\text{ab = ba}\].

\[\Rightarrow \frac{\text{ab}}{\text{2}}=\frac{\text{ba}}{\text{2}}\] 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 \[\text{1, 2, 3}\in \text{Q}\],

\[\text{(a*b)*c=}\left( \frac{\text{ab}}{\text{2}} \right)\text{*c}\]

\[\text{=}\frac{\left( \frac{\text{ab}}{\text{2}} \right)\text{c}}{\text{2}}\]

\[\text{=}\frac{\text{abc}}{\text{4}}\]

And

\[\text{a*(b*c)=}\left( \frac{\text{bc}}{\text{2}} \right)\text{*c}\]

\[\text{=}\frac{\left( \frac{\text{bc}}{\text{2}} \right)\text{a}}{\text{2}}\]

\[\text{=}\frac{\text{abc}}{\text{4}}\]

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 \[\text{a*b = }\frac{\text{a}}{\text{b+1}}\]. 

Ans: On \[\text{R - }\left\{ \text{-1} \right\}\], \[\text{*}\] is defined by \[\text{a*b = }\frac{\text{a}}{\text{b+1}}\].

For \[\text{2, 3}\in \text{R- }\!\!\{\!\!\text{ -1 }\!\!\}\!\!\text{ }\].

\[\text{2*3 = }\frac{\text{2}}{\text{3+1}}\]

\[\text{=}\frac{\text{1}}{\text{2}}\]

\[\text{3*2 = }\frac{\text{3}}{\text{2+1}}\]

\[\text{=1}\]

Therefore, \[\text{2 * 3}\ne \text{3*2}\].

Hence, the operation \[\text{*}\] is not commutative.

For \[\text{2, 3, 4}\in \text{R- }\!\!\{\!\!\text{ -1 }\!\!\}\!\!\text{ }\]

\[\text{(2*3)*4=}\frac{\text{2}}{\text{3+1}}\text{*4}\]

\[\text{=}\frac{\text{1}}{\text{2}}\text{*4}\]

\[\text{=}\frac{\frac{\text{1}}{\text{2}}}{\text{4+1}}\]

\[\text{=}\frac{\text{1}}{\text{10}}\]

\[\text{2*(3*4) = 2*}\frac{\text{3}}{\text{4+1}}\]

\[\text{=2*}\frac{\text{3}}{\text{5}}\]

\[\text{=}\frac{\text{2}}{\frac{\text{3}}{\text{5}}\text{+1}}\]

\[\text{=}\frac{\text{2}}{\frac{\text{8}}{\text{5}}}\]

\[\text{=}\frac{\text{5}}{\text{4}}\]

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 \[\text{*}\] is given by \[\text{a * b = a - b}\].

Ans: On, The binary operation \[\text{*}\] is defined as \[\text{a * b = a - b}\] on set \[\text{Q}\].

.For \[\frac{\text{1}}{\text{2}}\text{,}\frac{\text{1}}{\text{3}}\in \text{Q}\],

\[\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}}\text{=}\frac{\text{1}}{\text{2}}\text{-}\frac{\text{1}}{\text{3}}\]

\[\text{=}\frac{\text{3-2}}{\text{3}}\]

\[\text{=}\frac{\text{1}}{\text{6}}\]

And;

\[\frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{1}}{\text{3}}\text{-}\frac{\text{1}}{\text{2}}\]

\[\text{=}\frac{\text{2-3}}{\text{6}}\]

\[\text{=}\frac{\text{-1}}{\text{6}}\]

Therefore, \[\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}}\ne \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{2}}\].

Hence, the binary operation \[\text{*}\] is not commutative.

For \[\frac{\text{1}}{\text{2}}\text{,}\frac{\text{1}}{\text{3}}\text{,}\frac{\text{1}}{\text{4}}\in \text{Q}\],

\[\left( \frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}} \right)\text{*}\frac{\text{1}}{\text{4}}\text{=}\left( \frac{\text{1}}{\text{2}}\text{-}\frac{\text{1}}{\text{3}} \right)\text{*}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{1}}{\text{6}}\text{*}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{1}}{\text{6}}\text{-}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{2-3}}{\text{12}}\]

\[\text{=}\frac{\text{-1}}{\text{12}}\]

And;

\[\frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{4}} \right)\text{=}\frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{1}}{\text{3}}\text{-}\frac{\text{1}}{\text{4}} \right)\]

\[\text{=}\frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{4-3}}{\text{12}} \right)\]

\[\text{=}\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{12}}\]

\[\text{=}\frac{\text{1}}{\text{2}}\text{-}\frac{\text{1}}{\text{12}}\]

\[\text{=}\frac{\text{5}}{\text{12}}\]

Therefore, \[\left( \frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}} \right)\text{*}\frac{\text{1}}{\text{4}}\ne \frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{4}} \right)\]. 

Hence, the binary operation \[\text{*}\] is not associative.


  1. \[\text{a * b = }{{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\] 

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. \[\text{a * b = a + ab}\] 

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. \[\text{a * b = }{{\left( \text{a - b} \right)}^{\text{2}}}\] 

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. \[\text{a * b =}\frac{\text{ab}}{\text{4}}\] 

Ans: The binary operation \[\text{*}\] is defined as \[\text{a * b =}\frac{\text{ab}}{\text{4}}\] on set \[\text{Q}\].

For \[\text{a,b}\in \text{Q}\],

\[\text{a * b =}\frac{\text{ab}}{\text{4}}\]

\[\text{=}\frac{\text{ba}}{\text{4}}\]

\[\text{= b * a}\]

Therefore, \[\text{a * b = b * a}\].

Hence, the binary operation \[\text{*}\] is commutative.

For \[\text{a,b,c}\in \text{Q}\],

\[\text{(a*b)*c =}\left( \frac{\text{ab}}{\text{4}} \right)\text{*c}\]

\[\text{=}\frac{\frac{\text{ab}}{\text{4}}\text{.c}}{\text{4}}\]

\[\text{=}\frac{\text{abc}}{\text{16}}\] 

And,

\[\text{a*(b*c) = a*}\left( \frac{\text{ba}}{\text{4}} \right)\text{ }\]

\[\text{=}\frac{\frac{\text{bc}}{\text{4}}\text{a}}{\text{4}}\]

\[\text{=}\frac{\text{abc}}{\text{16}}\]

Therefore, \[\left( \text{a * b} \right)\text{ * c = a * }\left( \text{b * c} \right)\]

Hence, the binary operation \[\text{*}\] is associative.


  1. \[\text{a * b = a}{{\text{b}}^{\text{2}}}\] 

Ans: The binary operation \[\text{*}\] is defined as \[\text{a * b = a}{{\text{b}}^{\text{2}}}\] on set \[\text{Q}\].

For \[\frac{\text{1}}{\text{2}}\text{,}\frac{\text{1}}{\text{3}}\in \text{Q}\],

\[\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}}\text{=}\frac{\text{1}}{\text{2}}.{{\left( \frac{\text{1}}{\text{3}} \right)}^{\text{2}}}\]

\[\text{=}\frac{\text{1}}{\text{2}}.\frac{\text{1}}{\text{9}}\]

\[\text{=}\frac{\text{1}}{\text{18}}\] 

And,

\[\frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{1}}{\text{3}}\text{.}{{\left( \frac{\text{1}}{\text{2}} \right)}^{\text{2}}}\]

\[\text{=}\frac{\text{1}}{\text{3}}\text{.}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{1}}{\text{12}}\]

Therefore, \[\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}}\ne \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{2}}\].

Hence, the binary operation \[\text{*}\] is not commutative.

For \[\frac{\text{1}}{\text{2}}\text{,}\frac{\text{1}}{\text{3}}\text{,}\frac{\text{1}}{\text{4}}\in \text{Q}\],

\[\left( \frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}} \right)\text{*}\frac{\text{1}}{\text{4}}\text{=}\left[ \frac{\text{1}}{\text{2}}.{{\left( \frac{\text{1}}{\text{3}} \right)}^{\text{2}}} \right]\text{*}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{1}}{\text{18}}\text{*}\frac{\text{1}}{\text{4}}\]

\[\text{=}\frac{\text{1}}{\text{18}}{{\left( \frac{\text{1}}{\text{4}} \right)}^{2}}\]

\[\text{=}\frac{\text{1}}{\text{18}\times \text{16}}\]

\[\text{=}\frac{\text{1}}{\text{288}}\]

And;

\[\frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{4}} \right)\text{=}\frac{\text{1}}{\text{2}}\text{*}\left[ \frac{\text{1}}{\text{3}}{{\left( \frac{\text{1}}{\text{4}} \right)}^{2}} \right]\]

\[\text{=}\frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{48}}\]

\[\text{=}\frac{\text{1}}{\text{2}}{{\left( \frac{\text{1}}{\text{48}} \right)}^{\text{2}}}\]

\[\text{=}\frac{\text{1}}{\text{2}\times \text{2304}}\]

\[\text{=}\frac{\text{1}}{\text{4608}}\]

Therefore, \[\left( \frac{\text{1}}{\text{2}}\text{*}\frac{\text{1}}{\text{3}} \right)\text{*}\frac{\text{1}}{\text{4}}\ne \frac{\text{1}}{\text{2}}\text{*}\left( \frac{\text{1}}{\text{3}}\text{*}\frac{\text{1}}{\text{4}} \right)\].

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 * is commutative)

\[\text{= a * }\left( \text{b * c} \right)\]               (as * 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.

(A) The binary operation \[\text{*}\] is both associative and commutative.

(B) The binary operation \[\text{*}\] is commutative but not associative.

(C) The binary operation \[\text{*}\] is associative but not commutative.

(D) 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.


NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 1 All Exercises

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)


Overview of Class 12 Maths Chapter 1

The chapter related to the topic of functions can be difficult to understand. However, with proper guidance, this topic can gradually be turned into a strength. This is where you will need the assistance of Vedantu’s Class 12 Maths Exercise 1.4 Solutions for Chapter 1. Functions is a procedure or a connection that partners every component x of a set X, the area of the capacity, to a solitary component y of another set Y (conceivably a similar set), the codomain of the capacity. On the off chance that the capacity is called f, this connection is indicated y = f (x) (which is spoken out loud as f of x), component x is the contention or contribution of the capacity, and y is the estimation of the capacity, the yield, or the picture of x by f.[1] The image that is utilized for speaking to the information is the variable of the capacity (one regularly says that f is an element of the variable x).


Important Topics covered in Exercise 1.4 of NCERT Solutions for Class 12 Maths Chapter 1

Class 12 Maths Chapter 1 Relations and Functions covers some very important concepts. Exercise 1.4 of NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions is mainly based on the concept of binary operations of sets. Class 12 Maths NCERT Solutions Chapter 1 Exercise 1.4 helps students to develop a good understanding of this concept.

 

In general, a binary operation on sets is an operation whose two domains and the codomain are the same set. Addition (+), multiplication (*), subtraction (-), and division (/) are the four basic Mathematical binary operations on sets. The questions provided in these solutions are related to these four basic binary operations and ways to compute them using various properties and functions. Here, we have to check or prove if the binary operation is commutative or associative and the existence of identity elements. 


Key learnings from Exercise 1.4 of NCERT Solutions for Class 12 Maths Chapter 1

Below are some key learnings from this exercise:

  1. A binary operation * on a set A is called a function * from A × A to A.

  2. A binary operation * on X is called to be commutative if a * b = b * a ∀ a, b in X.

  3. A binary operation * on X is called to be associative if (a * b) * c = a * (b * c) ∀ a, b, c in X.

  4. An element e ∈ X is called the identity element for binary operation * : X × X → X, if a * e = a = e * a ∀ a ∈ X.


How NCERT Exercise 1.4 ch 1 Class 12 Maths Can Help You Reach Your Goals?

Most of the students generally have a fear of the subject of Mathematics. This is primarily because of the flawed education system in India. People generally believe in the knowledge which is only passed within the limitations of the classroom. However, with the assistance of NCERT, Vedantu now looks to use the tool of technology to break the traditional limitations of the Indian Education system.

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Benefits of Opting for our NCERT Exercise 1.4 ch 1 Class 12 Maths

Importance of NCERT Solutions

NCERT is a council which is responsible for setting rules and standards that aim to make education simple for all the students across the country. Based on these principles Vedantu customizes and publishes its content which you as a student can benefit from. Mathematics as a subject can only be mastered with regular practice. The primary objective of Vedantu is to help students understand the aspects responsible to effectively solve math problems. Chapter 1 Exercise 1.4 Class 12 Solutions thus will guide you into understanding the chapter of Functions is a better way. Hence, you will build confidence and in turn, improve your grades in the long run.


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All the Solutions offered in Vedantu's Exercise 1.4 ch 1 Class 12 Maths are 100% authentic as various methods of solving a particular problem have been duly highlighted. Our company only works with teachers who are highly experienced and also have the required skills in the Mathematics field. Likewise, all the solutions provided are correct as they are formulated under the supervision of expert professionals.


Helps in Scoring Good Marks

If the basic knowledge of a student is strong then all the relative problems can be tackled by applying the fundamental concept. The Solutions offered in Class 12 Maths ch 1 ex 1.1 have been created such that all the fundamentals of Functions are fulfilled. We make sure that you understand the application of each mathematical step. An in-depth analysis of each problem has been carried out to perfection. Therefore, be sure to avail Vedantu's online content to boost your grades in the long run. 


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FAQs on NCERT Solutions for Class 12 Maths Chapter 1: Relation and Functions - Exercise 1.1

1. What is the first chapter of the Class 12 Maths textbook?

The title of the first chapter of the Class 12th Maths textbook is Relations and Functions. The chapter forms an important part of the Maths Class 12th syllabus. The chapter helps you understand different types of relations, composition functions, and invertible functions of different sets of equations. The questions from this chapter are asked in your board exams. So, it is crucial that you get your basics clear. The chapter will also assist you in other chapters. 

2. What are the different types of relations and functions?

The chapter, titled Relations and Functions, helps you to tackle a set of questions based on various relations or functions. The chapter explains to you six different types of relations namely; Empty Relation; Universal Relation; Reflexive Relation; Symmetric Relation; Transitive Relation, and Equivalence relation. Other than the relations, the chapter also describes four types of functions: one-to-one function; onto function; one-to-one and onto functions; and invertible function. 

3. Is it important to solve all questions of Class 12th chapter 1 of Maths textbook?

Maths is one of the few subjects that you learn not by learning or memorizing but by doing. You are bound to forget the equations or their usage if you have not practiced them sufficiently. The purpose of the questions given in chapter 1 of the Class 12th Maths textbook is to help you evaluate whether you have understood the concepts mentioned in the syllabus. As the competition keeps on increasing day-by-day, your practice of Maths must not be confined only to the textbooks. But, it is mandatory to practice all the questions of the chapter. 

4. Where can I find solutions to chapter 1 of Class 12th Maths textbook?

There are numerous websites where you can find step-wise step solutions to all the questions of chapter 1. The correct answers are also given in the textbook to enable you to check if the method used by you gives you the right answer or not. To check whether the steps or the method that you have applied in answering the question is right or not, you can go to the Vedantu website (vedantu.com). The NCERT Solutions for Chapter 1 of Class 12 Maths provided on the website are prepared by the experts after extensive research. Another added advantage is that the answers are as per the revised syllabus. Also, any study material or solutions PDF provided by Vedantu are available free of cost.

5. How should I approach the Relations and Functions chapter of Class 12th Maths?

To understand the basics of the chapter easily and clearly, attend your classes regularly. In your free time, spend some time going through the chapter. Make notes in the language you can understand, read and revise the concept three to four times. Write down the relations, functions and go through them again and again. Practice questions given in the textbook and write tests. Identify your weak areas and learn from your mistakes. Try to answer some of the questions asked in PYQs.