Relations and Functions.pdf
47 views
239 slides
Feb 03, 2023
Slide 1 of 239
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
About This Presentation
This is very much useful for jee mains, from one of very valuable institutes of INDIA.
Slide Content
Chapter name
Welcome to
Relations & Functions II
Welcome to
Relations & Functions II
Table of contents
Session 01
Cartesian product of sets
Relation
Domain and range of relation
Inverse of a relation
Void relation
Universal Relation
Identity relation
Reflexive relation
Symmetric relation
Transitive relation
Equivalence relation
Composition of a relation
Session 02
Function
Vertical line test
Real valued function
Polynomial function
Identity function
Rational function
Exponential function
Session 03
Logarithmic function
Modulus function
Greatest Integer function
Session 04
Fractional Part function
Signum function
One-One function
Many-one function
Session 05
Number of functions
Number of one-one mappings
Session 06
Onto function (Surjective mapping)
Into function
Bijection function
Principle of inclusion and exculsion
Session 07
Even function
Odd function
Properties of Even/Odd function
Composite function
Session 08
Properties of Composite function
Periodic functions
Properties of Periodic functions
Session 09
Inverse function
Properties of Inverse function
Binary Operation
Properties of Binary Operation
Session 10
Functional Equations
Transformation of graphs
Session 11
Transformation of graphs
03
04
05
06
07
10
11
13
14
18
20
22
29
31
34
36
38
40
47
51
30
53
54
61
66
75
76
82
86
89
110
111
114
117
124
91
101
102
134
135
137
141
150
160
162
163
170
175
180
186
189
190
196
198
203
215
221
Session 01
Cartesian product of sets
Relation
Domain and range of relation
Inverse of a relation
Void relation
Universal Relation
Identity relation
Reflexive relation
Symmetric relation
Transitive relation
Equivalence relation
Composition of a relation
Session 02
Function
Vertical line test
Real valued function
Polynomial function
Identity function
Rational function
Exponential function
Session 03
Logarithmic function
Modulus function
Greatest Integer function
Session 04
Fractional Part function
Signum function
One-One function
Many-one function
Session 05
Number of functions
Number of one-one mappings
Session 06
Onto function (Surjective mapping)
Into function
Bijection function
Principle of inclusion and exculsion
Session 07
Even function
Odd function
Properties of Even/Odd function
Composite function
Session 08
Properties of Composite function
Periodic functions
Properties of Periodic functions
Session 09
Inverse function
Properties of Inverse function
Binary Operation
Properties of Binary Operation
Session 10
Functional Equations
Transformation of graphs
Session 11
Transformation of graphs
03
04
05
06
07
10
11
13
14
18
20
22
29
31
34
36
38
40
47
51
30
53
54
61
66
75
76
82
86
89
110
111
114
117
124
91
101
102
134
135
137
141
150
160
162
163
170
175
180
186
189
190
196
198
203
215
221
Session 1
Introduction to Relations and
Types of Relations
Return To Top
Introduction to Relations and
Types of Relations
Return To Top
Key Takeaways
Let �and �are two non-empty sets. The set of all ordered pairs �,�
[where �∈�and �∈�]is called Cartesian product of sets �and �.
It is denoted by ��.
If �(�)=�,��=�, then the number of elements in cartesian
productof sets is ���=��.
Example:�=�,�,�,�=1,2
⇒�×�={�,1,�,2,�,1,�,2,�,1,�,2}
⇒��×�=6=��×�(�)
•
•
Cartesian product of Sets:
Return To Top
Let �and �are two non-empty sets. The set of all ordered pairs �,�
[where �∈�and �∈�]is called Cartesian product of sets �and �.
It is denoted by ��.
If �(�)=�,��=�, then the number of elements in cartesian
productof sets is ���=��.
Example:�=�,�,�,�=1,2
⇒�×�={�,1,�,2,�,1,�,2,�,1,�,2}
⇒��×�=6=��×�(�)
•
•
Cartesian product of Sets:
Return To Top
Key Takeaways
Relation:
Let �and �be two sets, then a relation�from �to �is a subset of ��.
�⊆�×�
Number of relations =Number of subsets of ��
If ��=�,��=�,and �:�→�,then number of relations =2
��
��
Example: �(�)=6,�(�)=4
⇒��×�=��×��=6×4=24
Number of relations =Number of subsets of ��
=2
24
•
•
•
Return To Top
Relation:
Let �and �be two sets, then a relation�from �to �is a subset of ��.
�⊆�×�
Number of relations =Number of subsets of ��
If ��=�,��=�,and �:�→�,then number of relations =2
��
��
Example: �(�)=6,�(�)=4
⇒��×�=��×��=6×4=24
Number of relations =Number of subsets of ��
=2
24
•
•
•
Return To Top
Domain and range of relation:
The set of all the first components of ordered pairs belonging
to �is called domain of �.
i.e., domain ⊆�
The set of all the second components of ordered pairs
belonging to �is called range of �.
i.e., Range ⊆�
Set �is called the co–domain of �.
Let �be a relation defined from set �to set �.
Let �=�
1,�
1,�
1,�
2,�
2,�
3
BA
�
1
�
2
�
3
�
1
�
2
�
3
�
4
Return To Top
The set of all the first components of ordered pairs belonging
to �is called domain of �.
i.e., domain ⊆�
The set of all the second components of ordered pairs
belonging to �is called range of �.
i.e., Range ⊆�
Set �is called the co–domain of �.
Let �be a relation defined from set �to set �.
Let �=�
1,�
1,�
1,�
2,�
2,�
3
BA
�
1
�
2
�
3
�
1
�
2
�
3
�
4
Return To Top
Inverse of a relation:
Let �and �are two sets and �be a relation from �to �, then the inverse of
�is denoted by �
−1
is a relation from �to �and is defined as:
�
−1
=�,�,�,�∈�
��
��
Domain (�
−1
)=Range of �
Range �
−1
=Domain of �
•
•
Return To Top
Let �and �are two sets and �be a relation from �to �, then the inverse of
�is denoted by �
−1
is a relation from �to �and is defined as:
�
−1
=�,�,�,�∈�
��
��
Domain (�
−1
)=Range of �
Range �
−1
=Domain of �
•
•
Return To Top
If �=�,�:�,�∈ℤ,�
2
+3�
2
≤8is a relation on set of integers ℤ,
then domain of �
−1
.
A
D
B
C
−2,−1,1,2
−1,0,1
{−2,−1,0,1,2}
0,1
Return To Top
2
+3�
2
≤8is a relation on set of integers ℤ,
then domain of �
−1
.
A
D
B
C
−2,−1,1,2
−1,0,1
{−2,−1,0,1,2}
0,1
Return To Top
∴Domain of �
−1
={−1,0,1}
�=�,�:�,�∈ℤ,�
2
+�
2
≤8
Domain of �
−1
=Range of �( values of �)
�=0,�
2
≤8/3
�=1,�
2
≤7/3
�=2,�
2
≤4/3
⇒�∈{−1,0,1}
⇒�∈{−1,0,1}
⇒�∈{−1,0,1}
�=3,�
2
≤−1/3 ⇒�∈??????
If �=�,�:�,�∈ℤ,�
2
+3�
2
≤8is a relation on set of integers ℤ,
then domain of �
−1
.
Solution:
Return To Top
−1
={−1,0,1}
�=�,�:�,�∈ℤ,�
2
+�
2
≤8
Domain of �
−1
=Range of �( values of �)
�=0,�
2
≤8/3
�=1,�
2
≤7/3
�=2,�
2
≤4/3
⇒�∈{−1,0,1}
⇒�∈{−1,0,1}
⇒�∈{−1,0,1}
�=3,�
2
≤−1/3 ⇒�∈??????
If �=�,�:�,�∈ℤ,�
2
+3�
2
≤8is a relation on set of integers ℤ,
then domain of �
−1
.
Solution:
Return To Top
Void relation
A relation �on a set �is called a void or empty relation, if no element
of set�is related to any element of �.
�=??????
Example: �={students in boys' school}
Relation �={(�,�)∶�is sister of �&�,�∈�}
��
•
Return To Top
A relation �on a set �is called a void or empty relation, if no element
of set�is related to any element of �.
�=??????
Example: �={students in boys' school}
Relation �={(�,�)∶�is sister of �&�,�∈�}
��
•
Return To Top
Universal relation
It is a relation in which each element of set �is related to every element
of set�.
�=��
Example: �={set of all the students ofa school}
Relation �={(�,�)∶difference between the heights of �&�
is less than 10meters,where �,�∈�}
It is obvious that the difference between the heights of
any two students ofthe school has tobe less than 10m.
Therefore (�,�)∈�for all �,�∈�.
⇒�=�×�
∴�is the universal-relation on set �.
Explanation:
•
Return To Top
It is a relation in which each element of set �is related to every element
of set�.
�=��
Example: �={set of all the students ofa school}
Relation �={(�,�)∶difference between the heights of �&�
is less than 10meters,where �,�∈�}
It is obvious that the difference between the heights of
any two students ofthe school has tobe less than 10m.
Therefore (�,�)∈�for all �,�∈�.
⇒�=�×�
∴�is the universal-relation on set �.
Explanation:
•
Return To Top
Given: �∈ℝ&�∈ℝ
Since,the difference of two real number is a real number.
�−�→Real number
Absolute value of all real numbers ≥0
�−�≥0
��
1 3
5 2
1−3≥0
5−2≥0
2.5
⋮
1.3
⋮
If �={set of real numbers}, then check whether the relation
�={(�,�)∶�−�≥0,�,�∈�}is a universal relation or not?
Solution:
Return To Top
Since,the difference of two real number is a real number.
�−�→Real number
Absolute value of all real numbers ≥0
�−�≥0
��
1 3
5 2
1−3≥0
5−2≥0
2.5
⋮
1.3
⋮
If �={set of real numbers}, then check whether the relation
�={(�,�)∶�−�≥0,�,�∈�}is a universal relation or not?
Solution:
Return To Top
Key Takeaways
Identity relation:
Relation on set �is identity relation, if each and everyelement of �is
related to itself only.
Example: �={set of integers}
Relation �=�,�:�=�,�,�∈�=??????
??????
��
1
2
3
1
2
3
�=�
�
�
Return To Top
Identity relation:
Relation on set �is identity relation, if each and everyelement of �is
related to itself only.
Example: �={set of integers}
Relation �=�,�:�=�,�,�∈�=??????
??????
��
1
2
3
1
2
3
�=�
�
�
Return To Top
Key Takeaways
Reflexive relation:
A relation �defined on a set �is said to be reflexive if every element
of �is related to itself.
Relation �is reflexive if �,�∈�∀�∈�or ??????⊆�, where ??????is
identity relation on �.
•
Return To Top
Reflexive relation:
A relation �defined on a set �is said to be reflexive if every element
of �is related to itself.
Relation �is reflexive if �,�∈�∀�∈�or ??????⊆�, where ??????is
identity relation on �.
•
Return To Top
�,�→�divides �
For being reflexive following
condition must satisfy:
∴�is a reflexive relation.
�,�⇒�divides �,which is always true.
A relation �defined on set of natural numbers,
�={(�,�)∶�divides�}, then �is a _______
Solution:
Return To Top
For being reflexive following
condition must satisfy:
∴�is a reflexive relation.
�,�⇒�divides �,which is always true.
A relation �defined on set of natural numbers,
�={(�,�)∶�divides�}, then �is a _______
Solution:
Return To Top
�={1,1,1,2,2,2,(3,3)}is:
A
D
B
C
Only identity
Only reflexive
Both �and �
None
Return To Top
A
D
B
C
Only identity
Only reflexive
Both �and �
None
Return To Top
�={1,1,1,2,2,2,(3,3)}is:
Solution:
∴�is a reflexive relation
1
2
3
1
2
3
A
D
B
C
Only identity
Only reflexive
Both �and �
None
Return To Top
Solution:
∴�is a reflexive relation
1
2
3
1
2
3
A
D
B
C
Only identity
Only reflexive
Both �and �
None
Return To Top
Key Takeaways
Symmetric relation:
A relation�on a set �is said to be a symmetric relation,
iff�,�∈�⇒�,�∈�.
Example:
���⇒���,∀�,�∈�
Consider a set �={1,2,3}, which one is symmetric relation
�
1={1,1,1,2,2,1,(1,3),(3,1)}
�
2={(1,1),(1,2),(2,1),(1,3)}
�
3=1,1,2,2,3,3=??????
??????
Symmetric
Not symmetric
Symmetric
Return To Top
Symmetric relation:
A relation�on a set �is said to be a symmetric relation,
iff�,�∈�⇒�,�∈�.
Example:
���⇒���,∀�,�∈�
Consider a set �={1,2,3}, which one is symmetric relation
�
1={1,1,1,2,2,1,(1,3),(3,1)}
�
2={(1,1),(1,2),(2,1),(1,3)}
�
3=1,1,2,2,3,3=??????
??????
Symmetric
Not symmetric
Symmetric
Return To Top
Important Note
Number of Reflexive relation =2
��−1
Number of symmetric relation =2
????????????+1
2
•
•
Return To Top
Number of Reflexive relation =2
��−1
Number of symmetric relation =2
????????????+1
2
•
•
Return To Top
Key Takeaways
Transitive relation
A relation �on set �is said to be a transitive relation,
iff �,�∈�and �,�∈�⇒�,�∈�,∀�,�,�∈�.
Example:Consider a set �=1,2,3
�
1={1,2,2,3,1,3}
�
2={(1,1),(1,3),(3,2)}
�
3=1,1,2,2,3,3=??????
??????
Transitive
Not transitive
Transitive
���and ���⇒���,�,�,�∈�
1�22�31�3
��
1
2
3
1
2
3
Return To Top
Transitive relation
A relation �on set �is said to be a transitive relation,
iff �,�∈�and �,�∈�⇒�,�∈�,∀�,�,�∈�.
Example:Consider a set �=1,2,3
�
1={1,2,2,3,1,3}
�
2={(1,1),(1,3),(3,2)}
�
3=1,1,2,2,3,3=??????
??????
Transitive
Not transitive
Transitive
���and ���⇒���,�,�,�∈�
1�22�31�3
��
1
2
3
1
2
3
Return To Top
Let �,�∈ℝand �,�∈ℝ
So �>�and �>�⇒
Thus �,�∈ℝ
�>�
∴�is a transitive relation.
Show that the relation �defined on the set of real
number such that �=�,�∶�>�is transitive.
Solution:
Return To Top
So �>�and �>�⇒
Thus �,�∈ℝ
�>�
∴�is a transitive relation.
Show that the relation �defined on the set of real
number such that �=�,�∶�>�is transitive.
Solution:
Return To Top
●
Equivalence Relation
A relation �on a set �is said to be equivalence relation on �iff,
If it is reflexive, i.e., �,�∈�,∀�∈�
If it is symmetric, i.e., �,�∈�⇒�,�∈�,∀�,�∈�
If it is transitive, i.e., �,�∈�,�,�∈�⇒�,�∈�,∀�,�,�∈�
Identity Relation is an Equivalence Relation.●
•
•
•
Return To Top
Equivalence Relation
A relation �on a set �is said to be equivalence relation on �iff,
If it is reflexive, i.e., �,�∈�,∀�∈�
If it is symmetric, i.e., �,�∈�⇒�,�∈�,∀�,�∈�
If it is transitive, i.e., �,�∈�,�,�∈�⇒�,�∈�,∀�,�,�∈�
Identity Relation is an Equivalence Relation.●
•
•
•
Return To Top
Key Takeaways
Note:
If a relation is reflexive, symmetric and transitive, then it is
equivalence relation.
Return To Top
Note:
If a relation is reflexive, symmetric and transitive, then it is
equivalence relation.
Return To Top
Let �be the set of all triangles in a plane with �a relation given by
�={�
1,�
2∶�
1iscongruentto�
2}. Show that �is an equivalence relation.
Since every triangle is congruent to itself,⇒�is reflexive
�
1,�
2∈�⇒�
1iscongruentto�
2
⇒�
2is congruent to �
1⇒�is symmetric
Let �
1,�
2∈�and �
2,�
3∈�
⇒�
1is congruent to �
2and �
2is congruent to �
3
⇒�
1is congruent to �
3
⇒�is transitive
Hence, �is an Equivalence Relation.
Solution:
Return To Top
�={�
1,�
2∶�
1iscongruentto�
2}. Show that �is an equivalence relation.
Since every triangle is congruent to itself,⇒�is reflexive
�
1,�
2∈�⇒�
1iscongruentto�
2
⇒�
2is congruent to �
1⇒�is symmetric
Let �
1,�
2∈�and �
2,�
3∈�
⇒�
1is congruent to �
2and �
2is congruent to �
3
⇒�
1is congruent to �
3
⇒�is transitive
Hence, �is an Equivalence Relation.
Solution:
Return To Top
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
A
D
B
C
Statement 1is true, statement 2is true and statement 2is
correct explanation of statement 1.
Statement 1is true, statement 2is true and statement 2is
not correct explanation of statement 1.
Statement 1is true, statement 2is false
Statement 1is false, statement 2is true
JEE Main 2011
Return To Top
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
A
D
B
C
Statement 1is true, statement 2is true and statement 2is
correct explanation of statement 1.
Statement 1is true, statement 2is true and statement 2is
not correct explanation of statement 1.
Statement 1is true, statement 2is false
Statement 1is false, statement 2is true
JEE Main 2011
Return To Top
�,�∈�⇒�−�is an integer
�={�,�∈ℝXℝ∶�−�is an integer}
⇒�−�is an integer⇒�,�∈�
⇒�is reflexive
�,�∈�⇒�−�is an integer⇒�−�is an integer⇒�,�∈�
⇒�is symmetric
�,�∈�and �,�∈�
⇒�−�is an integer and �−�is an integer
⇒�is transitive⇒�−�is an integer⇒�,�∈�
∴�is an equivalence relation.
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
JEE Main 2011
Solution:
Return To Top
�={�,�∈ℝXℝ∶�−�is an integer}
⇒�−�is an integer⇒�,�∈�
⇒�is reflexive
�,�∈�⇒�−�is an integer⇒�−�is an integer⇒�,�∈�
⇒�is symmetric
�,�∈�and �,�∈�
⇒�−�is an integer and �−�is an integer
⇒�is transitive⇒�−�is an integer⇒�,�∈�
∴�is an equivalence relation.
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
JEE Main 2011
Solution:
Return To Top
�,�∈�⇒�=��
�={�,�∈ℝXℝ∶�=��for some rational number �}
⇒�=��for �=1⇒�,�∈�⇒�is reflexive
�,�∈�⇒�=��
Let �=0,�=1Thus �=0
⇒�,�∉�
But �≠��for any rational �
⇒�is asymmetric
�,�∈�and �,�∈�
⇒�=��and �=��
⇒�is transitive
⇒�=���
⇒�,�∈�
∴�is not an equivalence relation.
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
JEE Main 2011
Solution:
Return To Top
�={�,�∈ℝXℝ∶�=��for some rational number �}
⇒�=��for �=1⇒�,�∈�⇒�is reflexive
�,�∈�⇒�=��
Let �=0,�=1Thus �=0
⇒�,�∉�
But �≠��for any rational �
⇒�is asymmetric
�,�∈�and �,�∈�
⇒�=��and �=��
⇒�is transitive
⇒�=���
⇒�,�∈�
∴�is not an equivalence relation.
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
JEE Main 2011
Solution:
Return To Top
Let ℝbe the set of real numbers.
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
A
D
B
C
Statement 1is true, statement 2is true and statement 2is
correct explanation of statement 1.
Statement 1is true, statement 2is true and statement 2is
not correct explanation of statement 1.
Statement 1is true, statement 2is false
Statement 1is false, statement 2is true
JEE Main 2011
Return To Top
Statement 1∶�={�,�∈ℝ×ℝ∶�−�is an integer}is an equivalence relation on ℝ.
Statement 2∶�={�,�∈ℝ×ℝ∶�=��for some rational number �}is an
equivalence relation.
A
D
B
C
Statement 1is true, statement 2is true and statement 2is
correct explanation of statement 1.
Statement 1is true, statement 2is true and statement 2is
not correct explanation of statement 1.
Statement 1is true, statement 2is false
Statement 1is false, statement 2is true
JEE Main 2011
Return To Top
Composition of a Relation
The composition of two relations�&����is a binary relation from �to �,if and
only if there is �∈�such that ���&���where �∈�&�∈�
Mathematically,
���=�,�|∃�∈�∶���∧���
���
��
�
�
�
���
Return To Top
The composition of two relations�&����is a binary relation from �to �,if and
only if there is �∈�such that ���&���where �∈�&�∈�
Mathematically,
���=�,�|∃�∈�∶���∧���
���
��
�
�
�
���
Return To Top
Session 2
Introduction to Function and
Types of Functions
Return To Top
Introduction to Function and
Types of Functions
Return To Top
Key Takeaways
Function
A function is a relation defined from set �to set �such that each and
everyelement of set �is uniquely related to an element of set �.
Example:
It is denotedby �:�→�
�
�
�
1
2
3
4
Not a function
�
�
�
1
2
3
4
Function
•
Return To Top
Function
A function is a relation defined from set �to set �such that each and
everyelement of set �is uniquely related to an element of set �.
Example:
It is denotedby �:�→�
�
�
�
1
2
3
4
Not a function
�
�
�
1
2
3
4
Function
•
Return To Top
Answer is No.
For being function, every input should have unique
output, here input �doesn’t have any output.
The following relation is a function. Yes or No?
�
�
�
�
1
2
3
4
Solution:
A BYES NO
Return To Top
For being function, every input should have unique
output, here input �doesn’t have any output.
The following relation is a function. Yes or No?
�
�
�
�
1
2
3
4
Solution:
A BYES NO
Return To Top
Domain, Range and Co-domain of function:
Domain: Values of set �for which function is defined.
Range: All values that �takes (Range ⊆Co –domain).
Co-domain: Set of all elements in set �.
(Set of permissible inputs )
(Set of output generated domain )
Domain=1,2,3,4
Range=1,4,9,16
Example:
��
1
2
3
4
1
4
9
16
25
Co-domain=1,4,9,16,25
Return To Top
Domain: Values of set �for which function is defined.
Range: All values that �takes (Range ⊆Co –domain).
Co-domain: Set of all elements in set �.
(Set of permissible inputs )
(Set of output generated domain )
Domain=1,2,3,4
Range=1,4,9,16
Example:
��
1
2
3
4
1
4
9
16
25
Co-domain=1,4,9,16,25
Return To Top
Key Takeaways
Vertical line test:
If any vertical line parallel to �−axis intersect the curve on only one
point, then it is a function. If it is intersecting more than one points,
then it is not a function.
A function
�=�
2
�
�
�
2
+�
2
=9
Not a function
�
�
�
1
�
1
�
2
�
2� �
�
1
�
1
�
1
′
Return To Top
Vertical line test:
If any vertical line parallel to �−axis intersect the curve on only one
point, then it is a function. If it is intersecting more than one points,
then it is not a function.
A function
�=�
2
�
�
�
2
+�
2
=9
Not a function
�
�
�
1
�
1
�
2
�
2� �
�
1
�
1
�
1
′
Return To Top
Key Takeaways
�
2
=�•
�=�
3
Not a function
•
�
�
�
�
A function
��
1
�
1
�
2
�
1�
Vertical line test:
�
2
�
1
�
2
Return To Top
�
2
=�•
�=�
3
Not a function
•
�
�
�
�
A function
��
1
�
1
�
2
�
1�
Vertical line test:
�
2
�
1
�
2
Return To Top
Key Takeaways
Real valued function:
A function which has either ℝor one of its subsets as its range, is
called a real valued function. Further, if its domain is also either ℝor a
subset of ℝ, is called a real function.
�
�⊆ℝ⇒�is real valued function.
Return To Top
Real valued function:
A function which has either ℝor one of its subsets as its range, is
called a real valued function. Further, if its domain is also either ℝor a
subset of ℝ, is called a real function.
�
�⊆ℝ⇒�is real valued function.
Return To Top
For �=1
�
2
=�
1+1
=�
2
�=±�
We get two values of �for single value of �.
Hence, this is not a function.
Check whether �
2
=�
�
2
+�
is function or not.
Solution:
Return To Top
�
2
=�
1+1
=�
2
�=±�
We get two values of �for single value of �.
Hence, this is not a function.
Check whether �
2
=�
�
2
+�
is function or not.
Solution:
Return To Top
Key Takeaways
●
Polynomial function:
Domain : �∈ℝ
If �=0,we get ��=�
0(Constant Polynomial)●
Domain : ℝ
Range : {�
0}
�=�
0
�
�
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,⋯,�
�∈ℝ,�∈??????
0,�
�
Return To Top
●
Polynomial function:
Domain : �∈ℝ
If �=0,we get ��=�
0(Constant Polynomial)●
Domain : ℝ
Range : {�
0}
�=�
0
�
�
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,⋯,�
�∈ℝ,�∈??????
0,�
�
Return To Top
Key Takeaways
If �=1, we get ��= �
1�+�
0
Domain : ℝ
Range : ℝ
�=�
1�+�
0
�
�
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
(Linear Polynomial)
●Domain : �∈ℝ
●
Polynomial function:
Return To Top
If �=1, we get ��= �
1�+�
0
Domain : ℝ
Range : ℝ
�=�
1�+�
0
�
�
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
(Linear Polynomial)
●Domain : �∈ℝ
●
Polynomial function:
Return To Top
Key Takeaways
Identity function:
�
1=1,�
0=0
Domain : ℝ
Range :ℝ
�=�
�
�
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
��=�
Return To Top
Identity function:
�
1=1,�
0=0
Domain : ℝ
Range :ℝ
�=�
�
�
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
��=�
Return To Top
Key Takeaways
If �=2,we get
�(�)=�
2�
2
+�
1�+�
0
Range ∈
−??????
4�
,∞ Range ∈−∞,−
??????
4�
�
�
�
2>0
−??????
4�
0
�
�
2<0
−??????
4�0
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
(Quadratic Polynomial)
Domain : �∈ℝ●
Polynomial function:
Return To Top
If �=2,we get
�(�)=�
2�
2
+�
1�+�
0
Range ∈
−??????
4�
,∞ Range ∈−∞,−
??????
4�
�
�
�
2>0
−??????
4�
0
�
�
2<0
−??????
4�0
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
(Quadratic Polynomial)
Domain : �∈ℝ●
Polynomial function:
Return To Top
Key Takeaways
If ??????is even, �(�)is called an
even degree polynomial whose
range is always a subset of ℝ.
�=�
2
•
�
�
�→+∞
�→+∞
�→−∞
�→+∞
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
Domain : �∈ℝ●
Polynomial function:
Return To Top
If ??????is even, �(�)is called an
even degree polynomial whose
range is always a subset of ℝ.
�=�
2
•
�
�
�→+∞
�→+∞
�→−∞
�→+∞
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
Domain : �∈ℝ●
Polynomial function:
Return To Top
Key Takeaways
If ??????is odd, �(�)is called an odd
degree polynomial whose range
is ℝ.
�=�
3
•
�
�
�→+∞
�→+∞
�→−∞
�→−∞
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
Domain : �∈ℝ●
Polynomial function:
Return To Top
If ??????is odd, �(�)is called an odd
degree polynomial whose range
is ℝ.
�=�
3
•
�
�
�→+∞
�→+∞
�→−∞
�→−∞
●
��=�
��
�
+�
�−1�
�−1
+⋯+�
0
�
0,�
1,…,�
�∈ℝ,�∈??????
Domain : �∈ℝ●
Polynomial function:
Return To Top
��=sin
2
�+cos
2
�=1
�
��
0,1
�
�:�∈ℝ
�
�:�∈1
�
Find domain and range of function. ��=sin
2
�+cos
2
�
Solution:
Return To Top
2
�+cos
2
�=1
�
��
0,1
�
�:�∈ℝ
�
�:�∈1
�
Find domain and range of function. ��=sin
2
�+cos
2
�
Solution:
Return To Top
Find range of the function��=�
2
+4�+3
A
D
B
C
−�,∞
(�,∞)
�,∞
�,∞
Return To Top
2
+4�+3
A
D
B
C
−�,∞
(�,∞)
�,∞
�,∞
Return To Top
Given function:
�
��
0,−1
��=�
2
+4�+3
�=1,�=4,�=3
�
−
??????
4�
=−
4
2
−413
4×1
=−
4
4
=−1
�
�:�∈ℝ
�
�:�∈−
??????
4�
,∞
Hence, range of the function would be −1,∞
Find range of the function��=�
2
+4�+3
Solution:
Return To Top
�
��
0,−1
��=�
2
+4�+3
�=1,�=4,�=3
�
−
??????
4�
=−
4
2
−413
4×1
=−
4
4
=−1
�
�:�∈ℝ
�
�:�∈−
??????
4�
,∞
Hence, range of the function would be −1,∞
Find range of the function��=�
2
+4�+3
Solution:
Return To Top
Key Takeaways
●
Rational Function:
Domain: Check domain of �(�)and �(�), & �(�)≠0
For ℎ�=
��
��
,where �(�)and �(�)are functions of �
If ��&��is both are polynomials, then ℎ�is rational
polynomial function.
●
●
Return To Top
●
Rational Function:
Domain: Check domain of �(�)and �(�), & �(�)≠0
For ℎ�=
��
��
,where �(�)and �(�)are functions of �
If ��&��is both are polynomials, then ℎ�is rational
polynomial function.
●
●
Return To Top
Find domain and range of ??????(??????) =
�+1
3�−5
.
Given: ??????(??????) =
�+1
3�−5
Domain:
Let ??????(??????) = �=
�+1
3�−5
→
⇒3��−5�=�+1
⇒�3�−1=5�+1
⇒�=
5�+1
3�−1
Since,�must be real.
Range:
3�−5≠0⇒�≠
5
3
⇒�∈ℝ−
5
3
Range : �∈ℝ−
1
3
⇒3�−1≠0⇒�≠
1
3
Convert and make
′
�
′
as a subject
Solution:
Return To Top
�+1
3�−5
.
Given: ??????(??????) =
�+1
3�−5
Domain:
Let ??????(??????) = �=
�+1
3�−5
→
⇒3��−5�=�+1
⇒�3�−1=5�+1
⇒�=
5�+1
3�−1
Since,�must be real.
Range:
3�−5≠0⇒�≠
5
3
⇒�∈ℝ−
5
3
Range : �∈ℝ−
1
3
⇒3�−1≠0⇒�≠
1
3
Convert and make
′
�
′
as a subject
Solution:
Return To Top
Let ??????: ℝ→ ℝbe a function defined by ��=
�
�
2
+1
, �∈ℝ.Then
therangeof??????is:
A
D
B
C
ℝ–−
1
2
,
1
2
ℝ−−1,1
−1,1−0
−
1
2
,
1
2
JEE MAIN JAN 2019
Return To Top
�
�
2
+1
, �∈ℝ.Then
therangeof??????is:
A
D
B
C
ℝ–−
1
2
,
1
2
ℝ−−1,1
−1,1−0
−
1
2
,
1
2
JEE MAIN JAN 2019
Return To Top
Domain of ??????(??????)is ℝ
Let�=
�
�
2
+1
⇒��
2
+�=�
⇒��
2
−�+�=0(∵�∈ℝ)
�≥0
⇒1−4�
2
≥0⇒4�
2
−1≤0
⇒�∈−
1
2
,
1
2
−
1
2
1
2
∴Rangeof??????is−
1
2
,
1
2
Let ??????: ℝ→ ℝbe a function defined by ��=
�
�
2
+1
, �∈ℝ.Then
therangeof??????is:
JEE MAIN JAN 2019
Solution:
Return To Top
Let�=
�
�
2
+1
⇒��
2
+�=�
⇒��
2
−�+�=0(∵�∈ℝ)
�≥0
⇒1−4�
2
≥0⇒4�
2
−1≤0
⇒�∈−
1
2
,
1
2
−
1
2
1
2
∴Rangeof??????is−
1
2
,
1
2
Let ??????: ℝ→ ℝbe a function defined by ��=
�
�
2
+1
, �∈ℝ.Then
therangeof??????is:
JEE MAIN JAN 2019
Solution:
Return To Top
Key Takeaways
Exponential function:
Domain : �∈ℝ•
Range : �∈(0,∞)•
�=�
�
,�>0&�≠1
�>1
�
�
�
�
Example:Find domain and range of ??????(??????), where��=�
2�
We know �>1
Domain: �∈ℝ Range: 0,∞
0,1 0,1
Increasing function
Decreasing function0<�<1
Return To Top
Exponential function:
Domain : �∈ℝ•
Range : �∈(0,∞)•
�=�
�
,�>0&�≠1
�>1
�
�
�
�
Example:Find domain and range of ??????(??????), where��=�
2�
We know �>1
Domain: �∈ℝ Range: 0,∞
0,1 0,1
Increasing function
Decreasing function0<�<1
Return To Top
Range of �
�
:(0,∞)
So, range of �
�
+1:(1,∞)
The range of ��=�
�
+1is
Solution:
A
D
B
C
(�,∞)
[−�,∞)
(�,∞)
[�,∞)
Return To Top
�
:(0,∞)
So, range of �
�
+1:(1,∞)
The range of ��=�
�
+1is
Solution:
A
D
B
C
(�,∞)
[−�,∞)
(�,∞)
[�,∞)
Return To Top
Session 3
Some more types of Functions
Return To Top
Some more types of Functions
Return To Top
Key Takeaways
Logarithmic function:
�
�
�
�
Domain : �∈0,∞orℝ
+ Range : �∈−∞,∞orℝ
�=log
��, �>0&�≠1
1,0
1,0
�>1Increasing function Decreasing function0<�<1
• •
Return To Top
Logarithmic function:
�
�
�
�
Domain : �∈0,∞orℝ
+ Range : �∈−∞,∞orℝ
�=log
��, �>0&�≠1
1,0
1,0
�>1Increasing function Decreasing function0<�<1
• •
Return To Top
Key Takeaways
Logarithmic function:
Domain : �∈0,∞orℝ
+ Range : �∈−∞,∞orℝ
�=log
��, �>0&�≠1
Example:Find domain and range of ��=log�−2.
�>1Increasing function Decreasing function0<�<1
��=log
10�−2;
Domain: �−2>0⇒�>2
�
�=2,∞ Range: �∈ℝ
• •
Solution:
�
�
1,03,0
log�−2
Return To Top
Logarithmic function:
Domain : �∈0,∞orℝ
+ Range : �∈−∞,∞orℝ
�=log
��, �>0&�≠1
Example:Find domain and range of ��=log�−2.
�>1Increasing function Decreasing function0<�<1
��=log
10�−2;
Domain: �−2>0⇒�>2
�
�=2,∞ Range: �∈ℝ
• •
Solution:
�
�
1,03,0
log�−2
Return To Top
The domain of the definition of the function
��=
1
4−�
2
+log
10�
3
−�is :
A
D
B
C
1,2∪(2,∞)
–2,–1∪–1,0∪(2,∞)
–1,0∪1,2∪2,∞
–1,0∪1,2∪3,∞
JEE MAIN APR 2019
Return To Top
��=
1
4−�
2
+log
10�
3
−�is :
A
D
B
C
1,2∪(2,∞)
–2,–1∪–1,0∪(2,∞)
–1,0∪1,2∪2,∞
–1,0∪1,2∪3,∞
JEE MAIN APR 2019
Return To Top
�(�)=
1
4−�
2
+ log
10�
3
−�
4−�
2
≠0
and �
3
−�>0
⇒�∈−1,0∪1,∞⋯(????????????)
From equation (??????)and (????????????)
⇒��
2
−1>0
⇒�≠±2⋯(??????)
�∈(–1,0)∪(1,2)∪(2,∞)
0 1−1 2
The domain of the definition of the function
��=
1
4−�
2
+log
10�
3
−�is :
JEE MAIN APR 2019
Solution:
Return To Top
1
4−�
2
+ log
10�
3
−�
4−�
2
≠0
and �
3
−�>0
⇒�∈−1,0∪1,∞⋯(????????????)
From equation (??????)and (????????????)
⇒��
2
−1>0
⇒�≠±2⋯(??????)
�∈(–1,0)∪(1,2)∪(2,∞)
0 1−1 2
The domain of the definition of the function
��=
1
4−�
2
+log
10�
3
−�is :
JEE MAIN APR 2019
Solution:
Return To Top
Key Takeaways
For ℎ(�)=�(�)
�(�)
, to be defined for �(�)>0, and normal condition for �(�).
Note:
●
Return To Top
For ℎ(�)=�(�)
�(�)
, to be defined for �(�)>0, and normal condition for �(�).
Note:
●
Return To Top
Find domain of function ��=1+
3
�
1
�−2
��=1+
3
�
1
�−2
1+
3
�
>0and �−2≠0
⇒�∈(−∞,−3)∪(0,∞)and�≠2
⇒�∈(−∞,−3)∪(0,2)∪(2,∞)
−3 0
−∞
2
−3 0
+
+
−
∞
Solution:
Return To Top
3
�
1
�−2
��=1+
3
�
1
�−2
1+
3
�
>0and �−2≠0
⇒�∈(−∞,−3)∪(0,∞)and�≠2
⇒�∈(−∞,−3)∪(0,2)∪(2,∞)
−3 0
−∞
2
−3 0
+
+
−
∞
Solution:
Return To Top
�∈[4,∞)
��=�
4
+�
2
+4=�
Since ��is a polynomial, it’sdomain is ℝ.
For range, �=�
4
+�
2
+4=�
22
+2×
1
2
�
2
+
1
4
−
1
4
+4
Since,�
2
≥0⇒�
2
+
1
2
≥
1
2
∴�≥
1
2
2
+
15
4
=�
2
+
1
2
2
+
15
4
Alternate Method:
We know that,�
2
,�
4
≥0
⇒�≥4
Find domain and range of ��, where ��= �
4
+�
2
+4.
JEE MAIN JAN 2019
Solution:
Return To Top
��=�
4
+�
2
+4=�
Since ��is a polynomial, it’sdomain is ℝ.
For range, �=�
4
+�
2
+4=�
22
+2×
1
2
�
2
+
1
4
−
1
4
+4
Since,�
2
≥0⇒�
2
+
1
2
≥
1
2
∴�≥
1
2
2
+
15
4
=�
2
+
1
2
2
+
15
4
Alternate Method:
We know that,�
2
,�
4
≥0
⇒�≥4
Find domain and range of ��, where ��= �
4
+�
2
+4.
JEE MAIN JAN 2019
Solution:
Return To Top
Key Takeaways
●
Modulus function
�=|�|=ቐ
�,�≥0
−�,�<0
Domain ∶�∈ℝ
Range ∶�∈[0,∞)
�=−�
�=�
�
�
Return To Top
●
Modulus function
�=|�|=ቐ
�,�≥0
−�,�<0
Domain ∶�∈ℝ
Range ∶�∈[0,∞)
�=−�
�=�
�
�
Return To Top
��=
�
2
�
=
�
�
=1Where �≠0
Domain : �∈ℝ−0
Range : ��∈1
∵��
2
=|��|
Find the domain and the range of��=
�
2
�
.
Solution:
Return To Top
�
2
�
=
�
�
=1Where �≠0
Domain : �∈ℝ−0
Range : ��∈1
∵��
2
=|��|
Find the domain and the range of��=
�
2
�
.
Solution:
Return To Top
Find the range of the function ��=1−�−2.
A
D
B
C
−∞,1
−∞,2
1,∞
−∞,
1
2
Return To Top
A
D
B
C
−∞,1
−∞,2
1,∞
−∞,
1
2
Return To Top
�
�
�=�
�
�
�=�−2
2,02,0�
�
�
�=−�−2
2,0
�
�
�=1−�−2
2,1
�
�=−∞,1
Find the range of the function ��=1−�−2.
Solution:
Return To Top
�
�=�
�
�
�=�−2
2,02,0�
�
�
�=−�−2
2,0
�
�
�=1−�−2
2,1
�
�=−∞,1
Find the range of the function ��=1−�−2.
Solution:
Return To Top
Find the range of the function ��=1−�−2.
A
D
B
C
−∞,1
−∞,2
1,∞
−∞,
1
2
Return To Top
A
D
B
C
−∞,1
−∞,2
1,∞
−∞,
1
2
Return To Top
Key Takeaways
●
Greatest integer function(Step function)
Domain ∶�∈ℝ Range ∶�∈ℤ
�=[�]=Greatest Integer less than or equal to �
�
�
1 2 3−1−2
−1
−2
1
2
0
Return To Top
●
Greatest integer function(Step function)
Domain ∶�∈ℝ Range ∶�∈ℤ
�=[�]=Greatest Integer less than or equal to �
�
�
1 2 3−1−2
−1
−2
1
2
0
Return To Top
If �≤−2,then �∈
A
D
B
C
(−∞,1)
−∞,1
−2,−1
−∞,−2
Return To Top
A
D
B
C
(−∞,1)
−∞,1
−2,−1
−∞,−2
Return To Top
�≤−2
�
�
1 2 3−1−2
−1
−2
1
2
0
⇒�∈−∞,−2
If �≤−2,then �∈
Solution:
Return To Top
�
�
1 2 3−1−2
−1
−2
1
2
0
⇒�∈−∞,−2
If �≤−2,then �∈
Solution:
Return To Top
Key Takeaways
●
Greatest integer function
�=[�]
Domain ∶�∈ℝ
Range ∶�∈ℤ
=Greatest Integer less than or equal to �
�
�
1 2 3−1−2
−1
−2
1
2
0
�=��=�−1
Properties:
�≤��−1<
�+�=�+�;for �∈??????.
�+−�=ቐ
0,�∈??????
−1,�∉??????
•
•
•
Return To Top
●
Greatest integer function
�=[�]
Domain ∶�∈ℝ
Range ∶�∈ℤ
=Greatest Integer less than or equal to �
�
�
1 2 3−1−2
−1
−2
1
2
0
�=��=�−1
Properties:
�≤��−1<
�+�=�+�;for �∈??????.
�+−�=ቐ
0,�∈??????
−1,�∉??????
•
•
•
Return To Top
Find the domain and range of the function :
��=�+1+1,(where .denotes G.I.F)
��=�+1+1⇒��=�+2
�=�
�+�=�+�;for �∈??????.
�
�
1 2 3−1−2
−1
−2
1
2
0
Solution:
Return To Top
��=�+1+1,(where .denotes G.I.F)
��=�+1+1⇒��=�+2
�=�
�+�=�+�;for �∈??????.
�
�
1 2 3−1−2
−1
−2
1
2
0
Solution:
Return To Top
�
�
1 2 3−1−2
−1
1
2
0
3
4
Domain : �∈ℝ
Range : ℤ
Find the domain and range of the function :
��=�+1+1,(where .denotes G.I.F)
��=�+1+1⇒��=�+2 �=�+2Solution:
Return To Top
�
1 2 3−1−2
−1
1
2
0
3
4
Domain : �∈ℝ
Range : ℤ
Find the domain and range of the function :
��=�+1+1,(where .denotes G.I.F)
��=�+1+1⇒��=�+2 �=�+2Solution:
Return To Top
Find the domain of��=1−�
2
,where .denotes G.I.F.
A
D
B
C
1,2
−1,2
1,2
−1,0
Return To Top
2
,where .denotes G.I.F.
A
D
B
C
1,2
−1,2
1,2
−1,0
Return To Top
��=1−�
2
1−�
2
≥0
⇒�
2
−1≤0
⇒�
2
≤1
�
�
1 2 3−1−2
−1
−2
1
2
0
⇒�∈−1,2
⇒−1≤�≤1
Find the domain of��=1−�
2
,where .denotes G.I.F.
Solution:
Return To Top
2
1−�
2
≥0
⇒�
2
−1≤0
⇒�
2
≤1
�
�
1 2 3−1−2
−1
−2
1
2
0
⇒�∈−1,2
⇒−1≤�≤1
Find the domain of��=1−�
2
,where .denotes G.I.F.
Solution:
Return To Top
��=�
�
,�∈1,3(where �denotes G.I.F.).
��=�
�
,�∈1,3
Case 1: Case 2: Case 3:�∈1,2
��=�∵�=1
��∈1,2⋯??????
��=�
2
∵�=2
��∈4,9⋯????????????
��=�
3
∵�=3
��∈27⋯??????????????????
�∈2,3 �=3
??????∪????????????∪??????????????????
��∈1,2∪4,9∪27
Find the range of the function :
��=�
�
,�∈1,3 (where �denotes G.I.F.).
Solution:
Return To Top
�
,�∈1,3(where �denotes G.I.F.).
��=�
�
,�∈1,3
Case 1: Case 2: Case 3:�∈1,2
��=�∵�=1
��∈1,2⋯??????
��=�
2
∵�=2
��∈4,9⋯????????????
��=�
3
∵�=3
��∈27⋯??????????????????
�∈2,3 �=3
??????∪????????????∪??????????????????
��∈1,2∪4,9∪27
Find the range of the function :
��=�
�
,�∈1,3 (where �denotes G.I.F.).
Solution:
Return To Top
Session 4
Fractional part function, Signum
function and One –one and Many-one
function
Return To Top
Fractional part function, Signum
function and One –one and Many-one
function
Return To Top
Key Takeaways
●
Fractional Part Function
�=�
Domain ∶�∈ℝRange ∶�∈0,1
=�−�
�
�
1 2 3−1−2 0
1 �=1
Return To Top
●
Fractional Part Function
�=�
Domain ∶�∈ℝRange ∶�∈0,1
=�−�
�
�
1 2 3−1−2 0
1 �=1
Return To Top
What is the fractional part of 1.53?
Solution:�=�=�−�
=1.53−1=0.53
A
D
B
C
1
0.53
0.47
−0.53
Return To Top
Solution:�=�=�−�
=1.53−1=0.53
A
D
B
C
1
0.53
0.47
−0.53
Return To Top
Key Takeaways
●
Fractional Part Function
�=�
Domain ∶�∈ℝRange ∶�∈0,1
=�−�
Properties:
�+�=�,�∈??????
�+−�=ቐ
0,�∈??????
1,�∉??????
Examples:
−1.25=−1.25−−1.25
=−1.25−−2
=−1.25+2=0.75
1.25=1.25−1.25
=−1.25−1
=0.25
1.25+−1.25=1
•
•
Return To Top
●
Fractional Part Function
�=�
Domain ∶�∈ℝRange ∶�∈0,1
=�−�
Properties:
�+�=�,�∈??????
�+−�=ቐ
0,�∈??????
1,�∉??????
Examples:
−1.25=−1.25−−1.25
=−1.25−−2
=−1.25+2=0.75
1.25=1.25−1.25
=−1.25−1
=0.25
1.25+−1.25=1
•
•
Return To Top
��=2�+1+3⇒��=2�+3
�+�=�,�∈??????
Domain : �∈ℝ
Range : ��∈3,5
0≤�<1
0≤2�<2
0+3≤2�+3<2+3
3≤��<5
Find the domain and range of the function :
��=2�+1+3,(where {.}denotes fractional part function).
Solution:
Return To Top
�+�=�,�∈??????
Domain : �∈ℝ
Range : ��∈3,5
0≤�<1
0≤2�<2
0+3≤2�+3<2+3
3≤��<5
Find the domain and range of the function :
��=2�+1+3,(where {.}denotes fractional part function).
Solution:
Return To Top
�=��=
�
1+�
Let
On cross multiplying,
�1+�=�⇒�+��=�
⇒�=
�
1−�
∵�∈0,1⇒0≤
�
1−�
<1
0 1
++
−
�∈0,1
�
1−�
≥0⇒
�
�−1
≤0
(??????)
Find the range of the function : ��=
�
1+�
,(where .denotes
fractional part function).
Solution:
Return To Top
�
1+�
Let
On cross multiplying,
�1+�=�⇒�+��=�
⇒�=
�
1−�
∵�∈0,1⇒0≤
�
1−�
<1
0 1
++
−
�∈0,1
�
1−�
≥0⇒
�
�−1
≤0
(??????)
Find the range of the function : ��=
�
1+�
,(where .denotes
fractional part function).
Solution:
Return To Top
0≤
�
1−�
<1
1
2
1
++
−
�∈−∞,
1
2
∪(1,∞)
⇒
�
1−�
<1
(????????????)
⇒
�
1−�
−1<0
⇒
2�−1
1−�
<0⇒
2�−1
�−1
>0
By ??????∩(????????????)we get:
�∈0,
1
2
Find the range of the function : ��=
�
1+�
,(where .denotes
fractional part function).
Solution:
Return To Top
�
1−�
<1
1
2
1
++
−
�∈−∞,
1
2
∪(1,∞)
⇒
�
1−�
<1
(????????????)
⇒
�
1−�
−1<0
⇒
2�−1
1−�
<0⇒
2�−1
�−1
>0
By ??????∩(????????????)we get:
�∈0,
1
2
Find the range of the function : ��=
�
1+�
,(where .denotes
fractional part function).
Solution:
Return To Top
Key Takeaways
●
Signum Function
�=sgn�
Domain ∶�∈ℝRange ∶�∈−1,0,1
=൞
�
�
,�≠0
0,�=0
=
1,�>0
−1,�<0
0,�=0
1
−1
�
�
0
sgn(sgn(sgn⋯⋯⋯sgn�=sgn�
●
●
Return To Top
●
Signum Function
�=sgn�
Domain ∶�∈ℝRange ∶�∈−1,0,1
=൞
�
�
,�≠0
0,�=0
=
1,�>0
−1,�<0
0,�=0
1
−1
�
�
0
sgn(sgn(sgn⋯⋯⋯sgn�=sgn�
●
●
Return To Top
Domain : �∈ℝ−{−1}
��=sgn
�
3
+�
2
�+1
⇒��=sgn
�
2
�+1
�+1
⇒��=sgn�
2
Thus, ��∈0,1∵�
2
≥0
Range : ��∈0,1
If �
2
>0⇒��=sgn�
2
=1
If �
2
=0⇒��=sgn�
2
=0
Find the domain and range of the function : ��=sgn
�
3
+�
2
�+1
Solution:
Return To Top
��=sgn
�
3
+�
2
�+1
⇒��=sgn
�
2
�+1
�+1
⇒��=sgn�
2
Thus, ��∈0,1∵�
2
≥0
Range : ��∈0,1
If �
2
>0⇒��=sgn�
2
=1
If �
2
=0⇒��=sgn�
2
=0
Find the domain and range of the function : ��=sgn
�
3
+�
2
�+1
Solution:
Return To Top
Name Kishor
Name Arya Name Ayan Name Alia
Roll no. BYJUS01
Score 92%
Roll no. BYJUS02
Score 93%
Roll no. BYJUS03
Score 95%
Roll no. BYJUS04
Score 92%
Roll no. BYJUS05
Score 93%
Student Roll No.
BYJUS01
BYJUS02
BYJUS05
BYJUS03
BYJUS04
One input -one output
Name Roohi
Return To Top
Name Arya Name Ayan Name Alia
Roll no. BYJUS01
Score 92%
Roll no. BYJUS02
Score 93%
Roll no. BYJUS03
Score 95%
Roll no. BYJUS04
Score 92%
Roll no. BYJUS05
Score 93%
Student Roll No.
BYJUS01
BYJUS02
BYJUS05
BYJUS03
BYJUS04
One input -one output
Name Roohi
Return To Top
Student Score
92%
93%
95%
Many inputs -one output
Name Kishor
Name Arya Name Ayan Name Alia
Roll no. BYJUS01
Score 92%
Roll no. BYJUS02
Score 93%
Roll no. BYJUS03
Score 95%
Roll no. BYJUS04
Score 92%
Roll no. BYJUS05
Score 93%
Name Roohi
Return To Top
92%
93%
95%
Many inputs -one output
Name Kishor
Name Arya Name Ayan Name Alia
Roll no. BYJUS01
Score 92%
Roll no. BYJUS02
Score 93%
Roll no. BYJUS03
Score 95%
Roll no. BYJUS04
Score 92%
Roll no. BYJUS05
Score 93%
Name Roohi
Return To Top
Key Takeaways
One –one function (Injective function/ Injective mapping) :
A function �:�→�is said to be a one-one function if different
elements of set �have different �images in set �.
� � � �
Both are example of one-one function
Return To Top
One –one function (Injective function/ Injective mapping) :
A function �:�→�is said to be a one-one function if different
elements of set �have different �images in set �.
� � � �
Both are example of one-one function
Return To Top
Key Takeaways
Methods to determine whether a function is ONE-ONE or NOT:
For �
1,�
2∈�and��
1,��
2∈�
��
1=��
2⇔�
1=�
2or �
1≠�
2⇔��
1≠��
2
Example:
A function �:ℝ→ℝsuch that
��=3�+5
��=�
2
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒3�
1+5=3�
2+5
⇒�
1=�
2
∴��is one-one.
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒�
1
2
=�
2
2
⇒�
1
2
−�
2
2
=0
⇒(�
1−�
2)(�
1+�
2)=0
⇒�
1=�
2or �
1=−�
2
∴��is not one-one.
Return To Top
Methods to determine whether a function is ONE-ONE or NOT:
For �
1,�
2∈�and��
1,��
2∈�
��
1=��
2⇔�
1=�
2or �
1≠�
2⇔��
1≠��
2
Example:
A function �:ℝ→ℝsuch that
��=3�+5
��=�
2
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒3�
1+5=3�
2+5
⇒�
1=�
2
∴��is one-one.
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒�
1
2
=�
2
2
⇒�
1
2
−�
2
2
=0
⇒(�
1−�
2)(�
1+�
2)=0
⇒�
1=�
2or �
1=−�
2
∴��is not one-one.
Return To Top
⇒�
1
2
+�
1+2=�
2
2
+�
2+2
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒�
1
2
−�
2
2
+�
1−�
2=0
⇒(�
1+�
2)(�
1−�
2)+�
1−�
2=0
⇒�
1−�
2�
1+�
2+1=0
⇒�
1=�
2or �
1+�
2=−1
We get two conclusions here
Which indicates that many such �
1&�
2are possible
∴��is many-one function
Check whether the given function ��is one-one or
many one: ��=�
2
+�+2
Solution:
Return To Top
1
2
+�
1+2=�
2
2
+�
2+2
Suppose for some �
1,�
2∈ℝ
��
1=��
2
⇒�
1
2
−�
2
2
+�
1−�
2=0
⇒(�
1+�
2)(�
1−�
2)+�
1−�
2=0
⇒�
1−�
2�
1+�
2+1=0
⇒�
1=�
2or �
1+�
2=−1
We get two conclusions here
Which indicates that many such �
1&�
2are possible
∴��is many-one function
Check whether the given function ��is one-one or
many one: ��=�
2
+�+2
Solution:
Return To Top
Key Takeaways
Many one function :
A function �:�→�is said to be a many-one function if there exist at
least two or more elements of set �that have same �image in �.
� �
��
Both are example of many one function
Return To Top
Many one function :
A function �:�→�is said to be a many-one function if there exist at
least two or more elements of set �that have same �image in �.
� �
��
Both are example of many one function
Return To Top
Key Takeaways
Methods to determine whether a function is ONE-ONE or MANY ONE :
A function �:�→�is many one iffthere exists atleasttwo elements
�
1,�
2∈�such that ��
1=��
2
(��
1,��
2∈�but �
1≠�
2)
��
Many one function
Return To Top
Methods to determine whether a function is ONE-ONE or MANY ONE :
A function �:�→�is many one iffthere exists atleasttwo elements
�
1,�
2∈�such that ��
1=��
2
(��
1,��
2∈�but �
1≠�
2)
��
Many one function
Return To Top
Session 5
Methods to Find Whether a Function is One -One
or not, Number of Functions and Number of
One-One mappings
Return To Top
Methods to Find Whether a Function is One -One
or not, Number of Functions and Number of
One-One mappings
Return To Top
Key Takeaways
�=3�+5
�=�
3
�
� �
�
Methods to determine whether a function is ONE-ONE or MANY ONE :
Horizontal line test : If we draw straight lines parallel to �−axis, and
they cut the graph of the function at exactly one point, then the
function is ONE-ONE.
Return To Top
�=3�+5
�=�
3
�
� �
�
Methods to determine whether a function is ONE-ONE or MANY ONE :
Horizontal line test : If we draw straight lines parallel to �−axis, and
they cut the graph of the function at exactly one point, then the
function is ONE-ONE.
Return To Top
Key Takeaways
Horizontal line test : If there exists a straight lines parallel to �−axis,
which cuts the graph of the function at atleast two points, then the
function is MANY-ONE.
�
�
�
�
Methods to determine whether a function is ONE-ONE or MANY ONE :
Return To Top
Horizontal line test : If there exists a straight lines parallel to �−axis,
which cuts the graph of the function at atleast two points, then the
function is MANY-ONE.
�
�
�
�
Methods to determine whether a function is ONE-ONE or MANY ONE :
Return To Top
Choose the correct option:
�.�=sin��.�=�
�
�.�=log
2� �.�=��.�=�??????��
A
D
B
C
(a), (b) & (e) are one-one mapping (a) & (e) are many-one mapping
None(a)& (c) are one-one mapping
Return To Top
�.�=sin��.�=�
�
�.�=log
2� �.�=��.�=�??????��
A
D
B
C
(a), (b) & (e) are one-one mapping (a) & (e) are many-one mapping
None(a)& (c) are one-one mapping
Return To Top
�.�=log
2�
�.�=sin�
�.�=�
�
�.�=� �.�=�??????��
Exponents and logarithmic functions are one-one.
�
1�
2�
3
One-One One-One
Many-One
Many-One
Many-One
Choose the correct option:
Solution:
Return To Top
2�
�.�=sin�
�.�=�
�
�.�=� �.�=�??????��
Exponents and logarithmic functions are one-one.
�
1�
2�
3
One-One One-One
Many-One
Many-One
Many-One
Choose the correct option:
Solution:
Return To Top
Choose the correct option:
A
D
B
C
(a), (b) & (e) are one-one mapping (a) & (e) are many-one mapping
None(a)& (c) are one-one mapping
�.�=sin��.�=�
�
�.�=log
2� �.�=��.�=�??????��
Return To Top
A
D
B
C
(a), (b) & (e) are one-one mapping (a) & (e) are many-one mapping
None(a)& (c) are one-one mapping
�.�=sin��.�=�
�
�.�=log
2� �.�=��.�=�??????��
Return To Top
⇒1−�
1
�1
−1
=1−�
1
�2
−1
Suppose for some �
1,�
2∈ℝ
��
1=��
2
Hence, One-one
⇒1−�
1
�
1
−1
=1−�
1
�
2
−1
On squaring both sides:
⇒�
1
�
1
−1
=�
1
�
2
−1
⇒�
1
�
1=�
1
�
2
⇒�
1=�
2
�=�
�
�
1=�
2
Identify the following functions as One-one or Many-one: ��=1−�
1
�
−1
Solution:
Return To Top
1
�1
−1
=1−�
1
�2
−1
Suppose for some �
1,�
2∈ℝ
��
1=��
2
Hence, One-one
⇒1−�
1
�
1
−1
=1−�
1
�
2
−1
On squaring both sides:
⇒�
1
�
1
−1
=�
1
�
2
−1
⇒�
1
�
1=�
1
�
2
⇒�
1=�
2
�=�
�
�
1=�
2
Identify the following functions as One-one or Many-one: ��=1−�
1
�
−1
Solution:
Return To Top
Key Takeaways
Any function which is either increasing or decreasing in given
domainis one-one, otherwise many Many-one.
�
′
�>0
�
1 �
2
�
′
�<0
�
1 �
2
�
′
�>0
�
′
�<0
or
�
1
�
2
Methods to determine whether a function is ONE-ONE or MANY ONE :
Many-OneOne-OneOne-One
Return To Top
Any function which is either increasing or decreasing in given
domainis one-one, otherwise many Many-one.
�
′
�>0
�
1 �
2
�
′
�<0
�
1 �
2
�
′
�>0
�
′
�<0
or
�
1
�
2
Methods to determine whether a function is ONE-ONE or MANY ONE :
Many-OneOne-OneOne-One
Return To Top
��=sin�+5�
�
′
�=cos�−5<0
⇒Always decreasing→one-one
Determine whether a function ��=sin�+5�
is ONE-ONE or MANY-ONE
Solution:
Return To Top
�
′
�=cos�−5<0
⇒Always decreasing→one-one
Determine whether a function ��=sin�+5�
is ONE-ONE or MANY-ONE
Solution:
Return To Top
��=�
3
+�
2
+�+1
�
′
�=3�
2
+2�+1
⇒�(�)is always increasing→one-one
�=2
2
−43×1=−8<0
Hence �
′
�>0always
Determine whether a function ��=�
3
+�
2
+�+1
is ONE-ONE or MANY-ONE
Solution:
Return To Top
3
+�
2
+�+1
�
′
�=3�
2
+2�+1
⇒�(�)is always increasing→one-one
�=2
2
−43×1=−8<0
Hence �
′
�>0always
Determine whether a function ��=�
3
+�
2
+�+1
is ONE-ONE or MANY-ONE
Solution:
Return To Top
Key Takeaways
Number of functions :
Let a function �:�→�
��=4,��=5
If ��=�,��=��<�
Thus, total number of functions from �to �
=�⋅�⋅�⋯�(�times)=�
�
Thus, total number of function from �to �
⇒5⋅5⋯5(4times)=5
4
�
1
�
2
�
�
⋅
⋅
�
1
�
�
⋅
�
2
⋅
Return To Top
Number of functions :
Let a function �:�→�
��=4,��=5
If ��=�,��=��<�
Thus, total number of functions from �to �
=�⋅�⋅�⋯�(�times)=�
�
Thus, total number of function from �to �
⇒5⋅5⋯5(4times)=5
4
�
1
�
2
�
�
⋅
⋅
�
1
�
�
⋅
�
2
⋅
Return To Top
Key Takeaways
Number of ONE-ONE Mappings :
Let a function �:�→�
��=4,��=5
Thus, number of mappings
⇒��−1�−2⋯(�−�+1)=�
�
Thus total number of function from �to �
⇒55−15−2⋯(5−4+1)=
5
�
4
�
1
�
2
�
�
⋅
⋅
�
1
�
�
⋅
�
2
⋅
�
�
�,if �≥�0,if �<�
��
Number of Many-ONE Function
=(Total Number of Functions) −(Number of One-One Functions)
Number of MANY-ONE mappings :
Return To Top
Number of ONE-ONE Mappings :
Let a function �:�→�
��=4,��=5
Thus, number of mappings
⇒��−1�−2⋯(�−�+1)=�
�
Thus total number of function from �to �
⇒55−15−2⋯(5−4+1)=
5
�
4
�
1
�
2
�
�
⋅
⋅
�
1
�
�
⋅
�
2
⋅
�
�
�,if �≥�0,if �<�
��
Number of Many-ONE Function
=(Total Number of Functions) −(Number of One-One Functions)
Number of MANY-ONE mappings :
Return To Top
If �=1,2,3,4, then the number of functions
on set �,which are not ONE-ONE is:
A
D
B
C
240
232
248
256
Return To Top
on set �,which are not ONE-ONE is:
A
D
B
C
240
232
248
256
Return To Top
Number of many one functions
=Total number of functions−Number of ONE-ONE functions
=4
4
−
4
P
4⋅4
4
=256−24
=232
If �=1,2,3,4, then the number of functions
on set �,which are not ONE-ONE is:
Solution:
Return To Top
=Total number of functions−Number of ONE-ONE functions
=4
4
−
4
P
4⋅4
4
=256−24
=232
If �=1,2,3,4, then the number of functions
on set �,which are not ONE-ONE is:
Solution:
Return To Top
If �=1,2,3,4, then the number of functions
on set �,which are not ONE-ONE is:
A
D
B
C
240
232
248
256
Return To Top
on set �,which are not ONE-ONE is:
A
D
B
C
240
232
248
256
Return To Top
Let �=�,�,�and �=1,2,3,4.Then the number of elements
in the set �={�:�→�|2∈��and �is not one-one}is _________.
JEE Main Sept 2020
Only one Image
�
�
�
1
4
3
2
��
�
1
Only two Image and 2has to be there
�
�
�
1
4
3
2
��
3
�
12
3
−2
To select one more
image From 1,3,4
When all element �,�,�are
related to only one image
Solution:
Return To Top
in the set �={�:�→�|2∈��and �is not one-one}is _________.
JEE Main Sept 2020
Only one Image
�
�
�
1
4
3
2
��
�
1
Only two Image and 2has to be there
�
�
�
1
4
3
2
��
3
�
12
3
−2
To select one more
image From 1,3,4
When all element �,�,�are
related to only one image
Solution:
Return To Top
Let �=�,�,�and �=1,2,3,4.Then the number of elements
in the set �={�:�→�|2∈��and �is not one-one}is _________.
JEE Main Sept 2020
The number of elements in set �=1+18=19
Only one Image-1
Only two Image and 2has to be there-
3
�
12
3
−2=18
Solution:
Return To Top
in the set �={�:�→�|2∈��and �is not one-one}is _________.
JEE Main Sept 2020
The number of elements in set �=1+18=19
Only one Image-1
Only two Image and 2has to be there-
3
�
12
3
−2=18
Solution:
Return To Top
Determine whether the following function is ONE-ONE or MANY-ONE:
��=ln�
Any function which is either increasing or decreasing in
the whole domain is one-one, otherwise many-one.
��=ln�
0
1
Solution:
Return To Top
��=ln�
Any function which is either increasing or decreasing in
the whole domain is one-one, otherwise many-one.
��=ln�
0
1
Solution:
Return To Top
Identify the following function as One-One or Many-One:
��=2tan�;
??????
2
,
3??????
2
→�
0
??????
2
3??????
2
5??????
2
−
??????
2
??????
2??????
One-One Function.
Solution:
Return To Top
��=2tan�;
??????
2
,
3??????
2
→�
0
??????
2
3??????
2
5??????
2
−
??????
2
??????
2??????
One-One Function.
Solution:
Return To Top
Session 6
Onto & Into Functions
Return To Top
Onto & Into Functions
Return To Top
Key Takeaways
Onto function (surjective mapping)
If the function �:�→�is such that each element in �(co-domain) must
have at least one pre-image in �, then we say that �is a function of �‘onto’ �.
Or, if range of �=Co –domain of �.
�∶�→�is surjective iff∀�∈�,there
exists some �∈�such that ��=�.
If not given, co-domain of function is taken as �
�� ��
●
●
●
Return To Top
Onto function (surjective mapping)
If the function �:�→�is such that each element in �(co-domain) must
have at least one pre-image in �, then we say that �is a function of �‘onto’ �.
Or, if range of �=Co –domain of �.
�∶�→�is surjective iff∀�∈�,there
exists some �∈�such that ��=�.
If not given, co-domain of function is taken as �
�� ��
●
●
●
Return To Top
Key Takeaways
��=sin�: �→[−1,1]
(0,1)
Example:
�
(0,−1)
�
Onto Function
Range :[−1,1]
Onto function (surjective mapping)
If the function �:�→�is such that each element in �(co-domain) must
have at least one pre-image in �, then we say that �is a function of �‘onto’ �.
Return To Top
��=sin�: �→[−1,1]
(0,1)
Example:
�
(0,−1)
�
Onto Function
Range :[−1,1]
Onto function (surjective mapping)
If the function �:�→�is such that each element in �(co-domain) must
have at least one pre-image in �, then we say that �is a function of �‘onto’ �.
Return To Top
��=cos�: �→[−1,2]
�
�
(0,−1)
0,1
Hence ��is not onto Function
Here, Range ⊂Co-domain
⇒−1,1⊂−1,2
Check �:�→−1,2given by �=cos�is onto function or not.
Range of ��=cos�is −1,1
But given co-domain is −1,2
Solution:
Return To Top
�
�
(0,−1)
0,1
Hence ��is not onto Function
Here, Range ⊂Co-domain
⇒−1,1⊂−1,2
Check �:�→−1,2given by �=cos�is onto function or not.
Range of ��=cos�is −1,1
But given co-domain is −1,2
Solution:
Return To Top
Key Takeaways
If the function �:�→�is such that there exists at least one element in �
(co-domain) which is not the image of any element in domain (�), then �is ‘into’.
Into function
For an into function range of �≠Co –domain of �and Range of �⊂Co –domain of �.
�� ��
If a function is onto, it cannot be into and vice –versa.
●
●
●
Return To Top
If the function �:�→�is such that there exists at least one element in �
(co-domain) which is not the image of any element in domain (�), then �is ‘into’.
Into function
For an into function range of �≠Co –domain of �and Range of �⊂Co –domain of �.
�� ��
If a function is onto, it cannot be into and vice –versa.
●
●
●
Return To Top
Key Takeaways
Solution:
��=�
2
+�−2,�∈ℝ
Range of ��=−
9
4
,∞
Thus, range ≠co-domain
Example:
∴INTO Function
�
0,−
9
4
−
1
2
,0
Into function
Return To Top
Solution:
��=�
2
+�−2,�∈ℝ
Range of ��=−
9
4
,∞
Thus, range ≠co-domain
Example:
∴INTO Function
�
0,−
9
4
−
1
2
,0
Into function
Return To Top
Check whether the following functions are into function or not
(??????)��=�,where []denotes greatest integer function
(????????????)�:ℝ→0,1given by ��=�where {}represents
fractional part function
(??????)Here, co-domain =�
−1 210−2
�
�
Range =�
Range ⊆Co-domain
⇒��is into function
−1 210−2
�
(????????????)Here, Range of ��=0,1
Range =Co-domain
��is onto function
Solution:
Return To Top
(??????)��=�,where []denotes greatest integer function
(????????????)�:ℝ→0,1given by ��=�where {}represents
fractional part function
(??????)Here, co-domain =�
−1 210−2
�
�
Range =�
Range ⊆Co-domain
⇒��is into function
−1 210−2
�
(????????????)Here, Range of ��=0,1
Range =Co-domain
��is onto function
Solution:
Return To Top
Bijection Function
If �:�→�is both an injective and a surjective function, then �is
said to be bijection or one to one and onto function from �to �.
If �,�are finite sets and ��=��then number of bijective functions
defined from �to �is ��!
If �,�are finite sets and �:�→�is a bijective function, then ��=��•
•
Return To Top
If �:�→�is both an injective and a surjective function, then �is
said to be bijection or one to one and onto function from �to �.
If �,�are finite sets and ��=��then number of bijective functions
defined from �to �is ��!
If �,�are finite sets and �:�→�is a bijective function, then ��=��•
•
Return To Top
Note:
A function can be of one of these four types :
One-one, onto (injective and surjective ) also called as Bijective functions.
One –one, into (injective but not surjective)
Many –one, onto (surjective but not injective)
Many –one, into (neither surjective nor injective)
●
●
●
●
Return To Top
A function can be of one of these four types :
One-one, onto (injective and surjective ) also called as Bijective functions.
One –one, into (injective but not surjective)
Many –one, onto (surjective but not injective)
Many –one, into (neither surjective nor injective)
●
●
●
●
Return To Top
If the function �:ℝ−{−1,1}→�, defined by , is
surjective, then �is equal to :
��=
�
2
1−�
2
=
�
2
1−�
2
⇒�−��
2
=�
2
⇒�
2
=
�
1+�
∵�
2
≥0
��=� ⇒
�
1+�
≥0
⇒�∈−∞,−1∪0,∞
∴�= ℝ−−1,0
A
B
C
D
ℝ−−1,0
ℝ−−1
ℝ−−1,0
0,∞
JEE MAIN APRIL 2016
Solution:
Return To Top
surjective, then �is equal to :
��=
�
2
1−�
2
=
�
2
1−�
2
⇒�−��
2
=�
2
⇒�
2
=
�
1+�
∵�
2
≥0
��=� ⇒
�
1+�
≥0
⇒�∈−∞,−1∪0,∞
∴�= ℝ−−1,0
A
B
C
D
ℝ−−1,0
ℝ−−1
ℝ−−1,0
0,∞
JEE MAIN APRIL 2016
Solution:
Return To Top
If�:ℝ→[�,�],��=2sin�−23cos�+1is onto function,
then the value of �−�is
⇒��∈[−3,5]
��=2sin�−23cos�+1
∵�cos??????+�sin??????∈−�
2
+�
2
,�
2
+�
2
Thus, �=[−3,5]
�−�=8
Solution:
Return To Top
then the value of �−�is
⇒��∈[−3,5]
��=2sin�−23cos�+1
∵�cos??????+�sin??????∈−�
2
+�
2
,�
2
+�
2
Thus, �=[−3,5]
�−�=8
Solution:
Return To Top
A
B
C
D
One-one, ontoFunction
Many-one, ontoFunction
One-one, intoFunction
Many-one, into Function
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Return To Top
B
C
D
One-one, ontoFunction
Many-one, ontoFunction
One-one, intoFunction
Many-one, into Function
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Return To Top
1
�
�
−1
−1 1
�=1
�=−1
Range =Co-domain ⇒Onto
∴One-one, onto
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Solution:
Return To Top
�
�
−1
−1 1
�=1
�=−1
Range =Co-domain ⇒Onto
∴One-one, onto
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Solution:
Return To Top
A
B
C
D
One-one, ontoFunction
Many-one, ontoFunction
One-one, intoFunction
Many-one, into Function
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Return To Top
B
C
D
One-one, ontoFunction
Many-one, ontoFunction
One-one, intoFunction
Many-one, into Function
��=sin
??????�
2
:−1,1→[−1,1]is ______.
Return To Top
Key Takeaways
= + −
��∪� �(�) �(�) �(�∩�)
include exclude
��∪�=��+��−��∩�
Principle of inclusion and exclusion
Return To Top
= + −
��∪� �(�) �(�) �(�∩�)
include exclude
��∪�=��+��−��∩�
Principle of inclusion and exclusion
Return To Top
Key Takeaways
��∪�∪�
�(�) �(�)�(�)�(�∩�)�(�∩�)�(�∩�)�(�∩�∩�)
= + −+ − − +
include include
exclude
�
�
�
Principle of inclusion and exclusion
Return To Top
��∪�∪�
�(�) �(�)�(�)�(�∩�)�(�∩�)�(�∩�)�(�∩�∩�)
= + −+ − − +
include include
exclude
�
�
�
Principle of inclusion and exclusion
Return To Top
Key Takeaways
�(�)
�(�)
�(�)
�(�∩�)
�(�∩�)
�(�∩�)
�(�∩�∩�)
��∪�∪�
=
�
�
�
Principle of inclusion and exclusion
Return To Top
�(�)
�(�)
�(�)
�(�∩�)
�(�∩�)
�(�∩�)
�(�∩�∩�)
��∪�∪�
=
�
�
�
Principle of inclusion and exclusion
Return To Top
Key Takeaways
=��
�−��
�∩�
�+��
�∩�
�∩�
�−⋯
=
�
�
1�−1
�
−
�
�
2�−2
�
+
�
�
3�−3
�
−⋯
⋯+−1
�
��
1∩�
2∩�
3∩⋯∩�
�
�(�
1∪�
2∪�
3∪⋯�
�)
=Total functions where atleast one of element excluded
��
�=Total functions when �
�excluded
��
�
1
�
2
�
�
�
3
⋅
⋅
⋅
�
1
�
2
�
�
�
3
⋅
⋅
⋅
Principle of inclusion and exclusion
Return To Top
=��
�−��
�∩�
�+��
�∩�
�∩�
�−⋯
=
�
�
1�−1
�
−
�
�
2�−2
�
+
�
�
3�−3
�
−⋯
⋯+−1
�
��
1∩�
2∩�
3∩⋯∩�
�
�(�
1∪�
2∪�
3∪⋯�
�)
=Total functions where atleast one of element excluded
��
�=Total functions when �
�excluded
��
�
1
�
2
�
�
�
3
⋅
⋅
⋅
�
1
�
2
�
�
�
3
⋅
⋅
⋅
Principle of inclusion and exclusion
Return To Top
Key Takeaways
=��
�−��
�∩�
�+��
�∩�
�∩�
�−⋯
=
�
�
1�−1
�
−
�
�
2�−2
�
+
�
�
3�−3
�
−⋯
⋯+−1
�
��
1∩�
2∩�
3∩⋯∩�
�
�(�
1∪�
2∪�
3∪⋯�
�)=Total functions where atleast one of element excluded
=Total functions −�(�
1∪�
2∪�
3∪⋯�
�)Number of
onto functions
=�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯
��
�=Total functions when �
�excluded
Principle of inclusion and exclusion
Return To Top
=��
�−��
�∩�
�+��
�∩�
�∩�
�−⋯
=
�
�
1�−1
�
−
�
�
2�−2
�
+
�
�
3�−3
�
−⋯
⋯+−1
�
��
1∩�
2∩�
3∩⋯∩�
�
�(�
1∪�
2∪�
3∪⋯�
�)=Total functions where atleast one of element excluded
=Total functions −�(�
1∪�
2∪�
3∪⋯�
�)Number of
onto functions
=�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯
��
�=Total functions when �
�excluded
Principle of inclusion and exclusion
Return To Top
In how many ways can 5distinct balls be distributed into 3distinct
boxes such that
(??????)any number of balls can go in any number of boxes
•Total ways in which all balls can go in boxes:
=3
5
=3×3×3×3×3
Solution:
Return To Top
boxes such that
(??????)any number of balls can go in any number of boxes
•Total ways in which all balls can go in boxes:
=3
5
=3×3×3×3×3
Solution:
Return To Top
In how many ways can 5distinct balls be distributed into 3distinct
boxes such that
????????????Each box has atleastone ball in it.
3
5Number of
onto functions
=
=150
−
3
�
12
5
+
3
�
21
5
Solution:
Return To Top
boxes such that
????????????Each box has atleastone ball in it.
3
5Number of
onto functions
=
=150
−
3
�
12
5
+
3
�
21
5
Solution:
Return To Top
=Total functions −�(�
1∪�
2∪�
3∪⋯�
�)
Number of
onto functions
=�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯
Number of
onto
functions
�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯,(�>�)
�!,(�=�)
0,(�<�)
(Number of onto functions)
(Total number of functions)
−=
Number of
into functions
��
�
1
�
2
�
�
�
3
⋅
⋅
⋅
�
1
�
2
�
�
�
3
⋅
⋅
⋅
Principle of inclusion and exclusion
=
Return To Top
1∪�
2∪�
3∪⋯�
�)
Number of
onto functions
=�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯
Number of
onto
functions
�
�
−
�
�
1�−1
�
−
�
�
2�−2
�
+⋯,(�>�)
�!,(�=�)
0,(�<�)
(Number of onto functions)
(Total number of functions)
−=
Number of
into functions
��
�
1
�
2
�
�
�
3
⋅
⋅
⋅
�
1
�
2
�
�
�
3
⋅
⋅
⋅
Principle of inclusion and exclusion
=
Return To Top
A
B
C
D
3
5
150
5
3
93
Number of Into functions that can be defined from �to �
if ��=5and ��=3is
Return To Top
B
C
D
3
5
150
5
3
93
Number of Into functions that can be defined from �to �
if ��=5and ��=3is
Return To Top
Solution:
Number of Into functions that can be defined from �to �
if ��=5and ��=3is
��,��
Number of functions from �to �=3
5
=243
Number of onto functions from �to �=3
5
=243
=3
5
−
5
�
12
5
+
5
�
21
5
=150
∴Total number of into functions
=243−150=93
Return To Top
Number of Into functions that can be defined from �to �
if ��=5and ��=3is
��,��
Number of functions from �to �=3
5
=243
Number of onto functions from �to �=3
5
=243
=3
5
−
5
�
12
5
+
5
�
21
5
=150
∴Total number of into functions
=243−150=93
Return To Top
Session 7
Even-Odd Functions and
Composite Functions
Return To Top
Even-Odd Functions and
Composite Functions
Return To Top
Key Takeaways
Even Function
If �−�=��∀�in domain of ‘�’, then �is said to be an even function.
Example:��=cos�
�−�=cos−�=cos�=�(�)
��=|�|
�−�=−�=−1×�=−1×�=�=�(�)
��=�
2
+3
�−�=(−�)
2
+3=�
2
+3=�(�)
Example:
Example:
•
Return To Top
Even Function
If �−�=��∀�in domain of ‘�’, then �is said to be an even function.
Example:��=cos�
�−�=cos−�=cos�=�(�)
��=|�|
�−�=−�=−1×�=−1×�=�=�(�)
��=�
2
+3
�−�=(−�)
2
+3=�
2
+3=�(�)
Example:
Example:
•
Return To Top
Key Takeaways
If �−�=��∀�in domain of ‘�’, then �is said to be an even function. •
The graph of every even function is symmetric about the �–axis.•
��=cos�
��=�
2
�
�
�
�
−
??????
2
,0
??????
2
,0
3??????
2
,0
−3??????
2
,0
0,0
Even Function
Example: Example:
Return To Top
If �−�=��∀�in domain of ‘�’, then �is said to be an even function. •
The graph of every even function is symmetric about the �–axis.•
��=cos�
��=�
2
�
�
�
�
−
??????
2
,0
??????
2
,0
3??????
2
,0
−3??????
2
,0
0,0
Even Function
Example: Example:
Return To Top
Key Takeaways
If �−�=−��∀�in domain of ‘�′, then �is said to be an odd function. •
��=�
��=sin�
��=tan�
�−�=−�=−�(�)
�−�=sin−�=−sin�=−�(�)
�−�=tan−�=−tan�=−�(�)
Odd Function
Example:
Example:
Example:
Return To Top
If �−�=−��∀�in domain of ‘�′, then �is said to be an odd function. •
��=�
��=sin�
��=tan�
�−�=−�=−�(�)
�−�=sin−�=−sin�=−�(�)
�−�=tan−�=−tan�=−�(�)
Odd Function
Example:
Example:
Example:
Return To Top
Key Takeaways
•
If �−�=−��∀�in domain of ‘�′, then �is said to be an odd function. •
The graph of an odd function is symmetric about the origin.•
��=�
If an odd function is defined at �=0, then �0=0.
��=�
3
�
�
�
�
Odd Function
Example: Example:
Return To Top
•
If �−�=−��∀�in domain of ‘�′, then �is said to be an odd function. •
The graph of an odd function is symmetric about the origin.•
��=�
If an odd function is defined at �=0, then �0=0.
��=�
3
�
�
�
�
Odd Function
Example: Example:
Return To Top
�−�=
−�
�
−�
−1
−
�
2
+1=
−��
�
1−�
�
−
�
2
+1
��=
�
�
�
−1
+
�
2
+1
=
��
�
�
�
−1
−
�
2
+1
=
�(�
�
−1)+�
�
�
−1
−
�
2
+1
=�+
�
�
�
−1
−
�
2
+1
=
�
�
�
−1
+
�
2
+1
⇒��=�(−�)
Identify whether the function ��=
�
�
�
−1
+
�
2
+1,is even or not ?
∴Even function
Solution:
Return To Top
−�
�
−�
−1
−
�
2
+1=
−��
�
1−�
�
−
�
2
+1
��=
�
�
�
−1
+
�
2
+1
=
��
�
�
�
−1
−
�
2
+1
=
�(�
�
−1)+�
�
�
−1
−
�
2
+1
=�+
�
�
�
−1
−
�
2
+1
=
�
�
�
−1
+
�
2
+1
⇒��=�(−�)
Identify whether the function ��=
�
�
�
−1
+
�
2
+1,is even or not ?
∴Even function
Solution:
Return To Top
Find whether the following function is even / odd or none :��=ln
1+�
1−�
,�<1
��=ln
1+�
1−�
,�<1
��=ln
1+�
1−�
�−�=ln
1−�
1+�
=−ln
1+�
1−�
⇒�−�=−��
Hence the function is odd
Solution:
Return To Top
1+�
1−�
,�<1
��=ln
1+�
1−�
,�<1
��=ln
1+�
1−�
�−�=ln
1−�
1+�
=−ln
1+�
1−�
⇒�−�=−��
Hence the function is odd
Solution:
Return To Top
Key Takeaways
Properties of Even/Odd Function
The only function which is defined on the entire
number line and is even as well as odd is ��=0.
�
�
�
�=0
Some functions may neither be even nor odd.
Example: �(�)=3�+2
●
●
Return To Top
Properties of Even/Odd Function
The only function which is defined on the entire
number line and is even as well as odd is ��=0.
�
�
�
�=0
Some functions may neither be even nor odd.
Example: �(�)=3�+2
●
●
Return To Top
Key Takeaways
All functions (whose domain is symmetric about
origin)can be expressed as sum of an even and
an odd function
��=
��+�(−�)
2
+
��−�(−�)
2
even odd
Let a function ��=�+�
�
, express it as sum of an
even and an odd function
Example:
��=�+�
�
∴��=
�+�
�
+−�+�
−�
2
+
�+�
�
−−�+�
−�
2
●
Properties of Even/Odd Function
Return To Top
All functions (whose domain is symmetric about
origin)can be expressed as sum of an even and
an odd function
��=
��+�(−�)
2
+
��−�(−�)
2
even odd
Let a function ��=�+�
�
, express it as sum of an
even and an odd function
Example:
��=�+�
�
∴��=
�+�
�
+−�+�
−�
2
+
�+�
�
−−�+�
−�
2
●
Properties of Even/Odd Function
Return To Top
A
B
C
D
2�
1�+��
1�−�
2�
1��
1�
2�
1��
2�
2�
1�+��
2�−�
Let ��=�
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
JEE MAIN 2019
Return To Top
B
C
D
2�
1�+��
1�−�
2�
1��
1�
2�
1��
2�
2�
1�+��
2�−�
Let ��=�
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
JEE MAIN 2019
Return To Top
Let ��=�
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
��=�
�
��=�
1�+�
2�
�
1�+�+�
1�−�=
�
�+�
+�
−(�+�)
2
+
�
�−�
+�
−(�−�)
2
=
�
�
(�
�
+�
−�
)+�
−�
�
�
+�
−�
2
=
�
�
+�
−�
�
�
+�
−�
2
=
2�1�.2�1�
2
=2�
1�.�
1�
JEE MAIN 2019
∴�
1�+�+�
1�−�=2�
1��
1�
Solution:
Return To Top
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
��=�
�
��=�
1�+�
2�
�
1�+�+�
1�−�=
�
�+�
+�
−(�+�)
2
+
�
�−�
+�
−(�−�)
2
=
�
�
(�
�
+�
−�
)+�
−�
�
�
+�
−�
2
=
�
�
+�
−�
�
�
+�
−�
2
=
2�1�.2�1�
2
=2�
1�.�
1�
JEE MAIN 2019
∴�
1�+�+�
1�−�=2�
1��
1�
Solution:
Return To Top
Let ��=�
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
A
B
C
D
2�
1�+��
1�−�
2�
1��
1�
2�
1��
2�
2�
1�+��
2�−�
JEE MAIN 2019
Return To Top
�
(�>0)be written as ��=�
1�+�
2�,where �
1�is an even
function and �
2�is an odd function. Then �
1�+�+�
1�−�equals :
A
B
C
D
2�
1�+��
1�−�
2�
1��
1�
2�
1��
2�
2�
1�+��
2�−�
JEE MAIN 2019
Return To Top
Key Takeaways
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
Odd Odd Odd Even Even
Even Odd NENO Odd Odd
Properties of Even/Odd Function
Return To Top
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
Odd Odd Odd Even Even
Even Odd NENO Odd Odd
Properties of Even/Odd Function
Return To Top
Key Takeaways
��=�
2
,��=� �
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
ℎ�=��+��=�
2
+�
ℎ−�=(−�)
2
+−�
=(�)
2
+�=ℎ�→Even
ℎ�=��×��=�
2
�
ℎ−�=(−�)
2
×−�
=(�)
2
×�=ℎ�→Even
Even Even
Properties of Even/Odd Function
•
Return To Top
��=�
2
,��=� �
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
ℎ�=��+��=�
2
+�
ℎ−�=(−�)
2
+−�
=(�)
2
+�=ℎ�→Even
ℎ�=��×��=�
2
�
ℎ−�=(−�)
2
×−�
=(�)
2
×�=ℎ�→Even
Even Even
Properties of Even/Odd Function
•
Return To Top
Key Takeaways
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
��=�,��=sin�
Odd Odd Odd Even Even
ℎ�=��+��=�+sin�
ℎ−�=−�−sin�
=−ℎ�→odd
odd odd
��=��×��=�×sin�
�−�=−�×(−sin�)
=��→even
Properties of Even/Odd Function
•
Return To Top
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
��=�,��=sin�
Odd Odd Odd Even Even
ℎ�=��+��=�+sin�
ℎ−�=−�−sin�
=−ℎ�→odd
odd odd
��=��×��=�×sin�
�−�=−�×(−sin�)
=��→even
Properties of Even/Odd Function
•
Return To Top
Key Takeaways
��=�
2
,��=�
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
Odd Odd Odd Even Even
Even Odd NENO Odd Oddℎ�=��+��=�
2
+�
ℎ−�=−�
2
−�
even odd
=�
2
−�≠ℎ(�)
≠−ℎ(�)
Neither even
nor odd
��=����=�
2
�
�−�=−�
2
×(−�)
=−��→odd
Properties of Even/Odd Function
•
Return To Top
��=�
2
,��=�
�
Even
�
Even
�±�
Even
�.�
Even
�/�(�≠0)
Even
Odd Odd Odd Even Even
Even Odd NENO Odd Oddℎ�=��+��=�
2
+�
ℎ−�=−�
2
−�
even odd
=�
2
−�≠ℎ(�)
≠−ℎ(�)
Neither even
nor odd
��=����=�
2
�
�−�=−�
2
×(−�)
=−��→odd
Properties of Even/Odd Function
•
Return To Top
Key Takeaways
Composite Functions
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
Here ���•
���
���
���
� �
�
�
���
�
�
�
�
=�
=�
=�(1)=not defined
=�(5)=notdefined
�∶�→�
1 �:�
2→�
Return To Top
Composite Functions
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
Here ���•
���
���
���
� �
�
�
���
�
�
�
�
=�
=�
=�(1)=not defined
=�(5)=notdefined
�∶�→�
1 �:�
2→�
Return To Top
Key Takeaways
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
∴�:�→�
1and �:�
2→�be two functions and �is set of �such that if �∈�,
then �(�)∈�
2
�
�⊆�
�
So, �(��)is defined for only those values of �for which range of �is a subset
of domain of �.
Composite Functions �∶�→�
1 �:�
2→�
Return To Top
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
∴�:�→�
1and �:�
2→�be two functions and �is set of �such that if �∈�,
then �(�)∈�
2
�
�⊆�
�
So, �(��)is defined for only those values of �for which range of �is a subset
of domain of �.
Composite Functions �∶�→�
1 �:�
2→�
Return To Top
Key Takeaways
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
If �≠∅, then the function ℎdefined by ℎ�=�(��)is called composite function
of �and �and is denoted by �o�. It is also called as function of a function.
Composite Functions
Return To Top
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
If �≠∅, then the function ℎdefined by ℎ�=�(��)is called composite function
of �and �and is denoted by �o�. It is also called as function of a function.
Composite Functions
Return To Top
Key Takeaways
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
Note :Domain of �o�is �which is subset of �(the domain of �).
Range of �o�is a subset of range of �. If �=�, then ��⊆Y
2
Pictorially , ���(�)can be viewed as -
�
��(�)
�
���
�
���∶�,��
���∶�,�Composite Functions
Return To Top
��
1
�
�
�
�
1
3
5
9
�
�
2 �
0
3
9
13
�
�
�
�
�
� �
�
�
���
�
�
�
�
Note :Domain of �o�is �which is subset of �(the domain of �).
Range of �o�is a subset of range of �. If �=�, then ��⊆Y
2
Pictorially , ���(�)can be viewed as -
�
��(�)
�
���
�
���∶�,��
���∶�,�Composite Functions
Return To Top
(??????)Two functions �and�defined from ℝ→ℝsuch that ��=�+1,
��=�+2, then find a) ���b) ���
(????????????)Two functions �and�defined from ℝ→ℝsuch that ��=�
2
,
��=�+1, then show that ���≠���
(??????)���=��+2=�+1+2=�+3
���=��+1=�+2+1=(�+3)
⇒����=���(�)
(????????????)���=��+1=�
2
+1
���=��
2
=�+1
2
=�
2
+2�+1
⇒����≠���(�)
Solution:
Return To Top
��=�+2, then find a) ���b) ���
(????????????)Two functions �and�defined from ℝ→ℝsuch that ��=�
2
,
��=�+1, then show that ���≠���
(??????)���=��+2=�+1+2=�+3
���=��+1=�+2+1=(�+3)
⇒����=���(�)
(????????????)���=��+1=�
2
+1
���=��
2
=�+1
2
=�
2
+2�+1
⇒����≠���(�)
Solution:
Return To Top
The composition of functions are not commutative in general i.e.,
two functions �and �are such that if ���and ���are both
defined, then in general ���≠���.
Note :
Composite Functions
Return To Top
two functions �and �are such that if ���and ���are both
defined, then in general ���≠���.
Note :
Composite Functions
Return To Top
If ��=log
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
A
B
C
D
2�(�)
��
2
2�(�
2
)
−2��
JEE MAIN 2019
Return To Top
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
A
B
C
D
2�(�)
��
2
2�(�
2
)
−2��
JEE MAIN 2019
Return To Top
If ��=log
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
��=log
�
1−�
1+�
��=
2�
1+�
2
log
�
1−�(�)
1+�(�)
=log
�
1−
2�
1+�
2
1+
2�
1+�
2
=log
�
1−�
2
1+�
2
Let
Then��(�)=
JEE MAIN 2019
∴��(�)=2log
�
1−�
1+�
=2��
Solution:
Return To Top
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
��=log
�
1−�
1+�
��=
2�
1+�
2
log
�
1−�(�)
1+�(�)
=log
�
1−
2�
1+�
2
1+
2�
1+�
2
=log
�
1−�
2
1+�
2
Let
Then��(�)=
JEE MAIN 2019
∴��(�)=2log
�
1−�
1+�
=2��
Solution:
Return To Top
If ��=log
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
A
B
C
D
2�(�)
��
2
2�(�
2
)
−2��
JEE MAIN 2019
Return To Top
�
1−�
1+�
, �<1, then �
2�
1+�
2
is equal to :
A
B
C
D
2�(�)
��
2
2�(�
2
)
−2��
JEE MAIN 2019
Return To Top
Key Takeaways
Let ��=�,��=sin�, ℎ�=�
�
, domain of �,�,ℎis ℝExample:
����ℎ�=���ⅇ
�
=�(sin�
�
)=sin�
�
����ℎ(�)=(sinℎ�=sin�
�
∴����ℎ=����ℎ
The composition of functions are associative i.e.if three functions �,�,ℎare
such that ����ℎand ����ℎare defined , then ����ℎ=����ℎ
Composite Functions
Return To Top
Let ��=�,��=sin�, ℎ�=�
�
, domain of �,�,ℎis ℝExample:
����ℎ�=���ⅇ
�
=�(sin�
�
)=sin�
�
����ℎ(�)=(sinℎ�=sin�
�
∴����ℎ=����ℎ
The composition of functions are associative i.e.if three functions �,�,ℎare
such that ����ℎand ����ℎare defined , then ����ℎ=����ℎ
Composite Functions
Return To Top
Session 8
Composite Functions and
Periodic Functions
Return To Top
Composite Functions and
Periodic Functions
Return To Top
If��=ቊ
1−�,�≤0
�
2
,�>0
��=ቊ
−�,�<1
1−�,�≥1
,thenfind���(�)
����=ቊ
����=ቐ
1−�(�),�(�)≤0
(��)
2
,�(�)>0
1−−�,�∈0,1
1−(1−�),�≥1
(−�)
2
,�<0
�
�
�
�=−�
(1,0)
(0,−1)
�=�(�)
�=1−�
∴����=൞
(�)
2
,�∈−∞,0
1+�,�∈[0,1)
�,�∈1,∞
Solution:
Return To Top
1−�,�≤0
�
2
,�>0
��=ቊ
−�,�<1
1−�,�≥1
,thenfind���(�)
����=ቊ
����=ቐ
1−�(�),�(�)≤0
(��)
2
,�(�)>0
1−−�,�∈0,1
1−(1−�),�≥1
(−�)
2
,�<0
�
�
�
�=−�
(1,0)
(0,−1)
�=�(�)
�=1−�
∴����=൞
(�)
2
,�∈−∞,0
1+�,�∈[0,1)
�,�∈1,∞
Solution:
Return To Top
Key Takeaways
Properties of Composite Function
•
If �and �are one –one , then ���if defined will be one –one.
•If �and �are bijections and ���is defined , then ���will be a bijection iff
range of �is equal to domain of �.
Return To Top
Properties of Composite Function
•
If �and �are one –one , then ���if defined will be one –one.
•If �and �are bijections and ���is defined , then ���will be a bijection iff
range of �is equal to domain of �.
Return To Top
Key Takeaways
Periodic Functions :
Mathematically, a function ��is said to be periodic function
if ∃a positive real number �,such that
●
Here �is called period of function �and smallest value of �is
called fundamental period.
●
Note :
��+�=��,∀�∈domain of ′�
′
;�>0
Domain of periodic function should not be restricted (bounded).
Return To Top
Periodic Functions :
Mathematically, a function ��is said to be periodic function
if ∃a positive real number �,such that
●
Here �is called period of function �and smallest value of �is
called fundamental period.
●
Note :
��+�=��,∀�∈domain of ′�
′
;�>0
Domain of periodic function should not be restricted (bounded).
Return To Top
Key Takeaways
Example :● ��=sin�
−8??????−7??????−6??????−5??????−4??????−3??????−2??????−?????? �??????2??????3??????4??????5??????6??????7??????8??????
�
�
1
−1
��+�=��⇒sin�+�=sin�
⇒sin�+�−sin�=0⇒2sin
??????
2
cos
2�+??????
2
=0
⇒
�
2
=�??????⇒�=2�??????,�∈??????Thus, fundamental period =2??????
Periodic Functions :
Return To Top
Example :● ��=sin�
−8??????−7??????−6??????−5??????−4??????−3??????−2??????−?????? �??????2??????3??????4??????5??????6??????7??????8??????
�
�
1
−1
��+�=��⇒sin�+�=sin�
⇒sin�+�−sin�=0⇒2sin
??????
2
cos
2�+??????
2
=0
⇒
�
2
=�??????⇒�=2�??????,�∈??????Thus, fundamental period =2??????
Periodic Functions :
Return To Top
Note :
If a function is dis-continuous, it’s discontinuity should repeat
after a particular interval for the function to be periodic.
Periodic Functions
Return To Top
If a function is dis-continuous, it’s discontinuity should repeat
after a particular interval for the function to be periodic.
Periodic Functions
Return To Top
�
??????
??????
2
−
??????
2
−??????
−
3??????
2
3??????
2
Find the period of function.
??????)��=tan�????????????)��={�}where .denotes fractional part function.
−2 −1 0 1 2
??????)��+�=��tan�+??????=tan�????????????)��=�
Period is ??????
Period is 1
Solution:
Return To Top
??????
??????
2
−
??????
2
−??????
−
3??????
2
3??????
2
Find the period of function.
??????)��=tan�????????????)��={�}where .denotes fractional part function.
−2 −1 0 1 2
??????)��+�=��tan�+??????=tan�????????????)��=�
Period is ??????
Period is 1
Solution:
Return To Top
Key Takeaways
Properties of Periodic Functions :
●If a function ��has a period �,then
1
��
,��
�
�∈ℕ,��,��
also has a period �(�may or may not be fundamental period.)
Example :�=cosⅇc�
Fundamental period =2??????
Return To Top
Properties of Periodic Functions :
●If a function ��has a period �,then
1
��
,��
�
�∈ℕ,��,��
also has a period �(�may or may not be fundamental period.)
Example :�=cosⅇc�
Fundamental period =2??????
Return To Top
Key Takeaways
Example :
Fundamental period =??????
�=sin�
Return To Top
Example :
Fundamental period =??????
�=sin�
Return To Top
Key Takeaways
Example :�=cos
2
�
Fundamental period =??????
Return To Top
Example :�=cos
2
�
Fundamental period =??????
Return To Top
Key Takeaways
Properties of Periodic Functions :
●If a function ��has a period �,
then ���+�has the period
??????
�
.
●For y=sin�,fundamental period =2??????
●For y=sin2�,fundamental period =??????
Return To Top
Properties of Periodic Functions :
●If a function ��has a period �,
then ���+�has the period
??????
�
.
●For y=sin�,fundamental period =2??????
●For y=sin2�,fundamental period =??????
Return To Top
Properties of Periodic Functions
●
●
Every constant function defined for unbounded domain is always periodic with
no fundamental period.
Example :
��=sin
2
�+cos
2
�,domain is ℝ
Periodic with no fundamental period.
⇒��=1
Return To Top
●
●
Every constant function defined for unbounded domain is always periodic with
no fundamental period.
Example :
��=sin
2
�+cos
2
�,domain is ℝ
Periodic with no fundamental period.
⇒��=1
Return To Top
Find the period of function. ??????)��=�⋅
1
�
????????????)�(�)=cos�⋅sⅇc�
??????)��=�⋅
1
�
(domain �∈ℝ−0)
????????????)��=cos�⋅sⅇc�(domain�∈ℝ−2�+1
??????
2
,�∈ℤ)
Not periodic
Period : ??????
Solution:
Return To Top
1
�
????????????)�(�)=cos�⋅sⅇc�
??????)��=�⋅
1
�
(domain �∈ℝ−0)
????????????)��=cos�⋅sⅇc�(domain�∈ℝ−2�+1
??????
2
,�∈ℤ)
Not periodic
Period : ??????
Solution:
Return To Top
Fundamental period of �=
�
3
,where ⋅denotes fractional part function is
A
B
C
D
�
�
�
�
�
�
Return To Top
�
3
,where ⋅denotes fractional part function is
A
B
C
D
�
�
�
�
�
�
Return To Top
Fundamental period of �=
�
3
,where ⋅denotes fractional part function is
If a function ��has a period �,
then ���+�has the period
??????
�
.
For �,fundamental period =1
For
�
3
,fundamental period =3
Solution:
Return To Top
�
3
,where ⋅denotes fractional part function is
If a function ��has a period �,
then ���+�has the period
??????
�
.
For �,fundamental period =1
For
�
3
,fundamental period =3
Solution:
Return To Top
Session 9
Inverse Functions & Binary
operations
Return To Top
Inverse Functions & Binary
operations
Return To Top
Key Takeaways
Properties of Periodic Functions :
●If ��has a period �
1and ��has a period �
2,then
��±��,��⋅��or
��
��
is L.C.M of �
1and �
2(provided L.C.M exists).
L.C.M of
�
�
,
�
�
=
L.C.M�,�
H.C.F�,�
However, L.C.M need not be fundamental period.
If L.C.M does not exists, then ��±��,��⋅��or
��
��
is non-periodic or aperiodic.
●
Return To Top
Properties of Periodic Functions :
●If ��has a period �
1and ��has a period �
2,then
��±��,��⋅��or
��
��
is L.C.M of �
1and �
2(provided L.C.M exists).
L.C.M of
�
�
,
�
�
=
L.C.M�,�
H.C.F�,�
However, L.C.M need not be fundamental period.
If L.C.M does not exists, then ��±��,��⋅��or
��
��
is non-periodic or aperiodic.
●
Return To Top
Find the period of function. ??????)��=sin
3�
2
+cos
9�
4
????????????)��=sin�+cos�
cos
9�
4
→�
2=
2??????
9
4
=
8??????
9
??????)���+�has the period
??????
�
L.C.M of
4??????
3
,
8??????
9
⇒L.C.M of
4
3
,
8
9
??????
L.C.M4,8
H.C.F3,9
??????=
8
3
??????
sin
3�
2
→�
1=
2??????
3
2
=
4??????
3
??????
2
may also be period.
????????????)Period of ��±��is L.C.M of �
1,�
2
��+
??????
2
=sin�+
??????
2
+cos�+
??????
2
=cos�+−sin�
L.C.M of ??????,??????=??????
=��
Period is
??????
2
.
Solution:
Return To Top
3�
2
+cos
9�
4
????????????)��=sin�+cos�
cos
9�
4
→�
2=
2??????
9
4
=
8??????
9
??????)���+�has the period
??????
�
L.C.M of
4??????
3
,
8??????
9
⇒L.C.M of
4
3
,
8
9
??????
L.C.M4,8
H.C.F3,9
??????=
8
3
??????
sin
3�
2
→�
1=
2??????
3
2
=
4??????
3
??????
2
may also be period.
????????????)Period of ��±��is L.C.M of �
1,�
2
��+
??????
2
=sin�+
??????
2
+cos�+
??????
2
=cos�+−sin�
L.C.M of ??????,??????=??????
=��
Period is
??????
2
.
Solution:
Return To Top
Key Takeaways
Properties of Periodic Functions :
●
●
If �is a function such that ���is defined on the domain of �and �is periodic
with �,then ���is also periodic with �as one of its period.
ℎ�=cos�,where ⋅is fractional part function
Let ��=cos�,��=�then ℎ�=���,period 2??????
ℎ�=cos�,where ⋅is fractional part function.
Let ��=cos�,��=�then ℎ�=���,period 1
●
Example:
Return To Top
Properties of Periodic Functions :
●
●
If �is a function such that ���is defined on the domain of �and �is periodic
with �,then ���is also periodic with �as one of its period.
ℎ�=cos�,where ⋅is fractional part function
Let ��=cos�,��=�then ℎ�=���,period 2??????
ℎ�=cos�,where ⋅is fractional part function.
Let ��=cos�,��=�then ℎ�=���,period 1
●
Example:
Return To Top
Key Takeaways
Properties of Periodic Functions :
●
●
If �is a function such that ���is defined on the domain of �and �is periodic
with �,then ���is also periodic with �as one of its period.
Note :
If �is a function such that ���is defined on the domain of �
and�is aperiodic, then ���may or may not be periodic.
Example :
ℎ�=cos�+sin�
ℎ�=ℎ�+2??????
⇒period of ℎ�is 2??????
Return To Top
Properties of Periodic Functions :
●
●
If �is a function such that ���is defined on the domain of �and �is periodic
with �,then ���is also periodic with �as one of its period.
Note :
If �is a function such that ���is defined on the domain of �
and�is aperiodic, then ���may or may not be periodic.
Example :
ℎ�=cos�+sin�
ℎ�=ℎ�+2??????
⇒period of ℎ�is 2??????
Return To Top
Key Takeaways
Let �=��:�→�be a one –one and onto function , i.e.a bijection , then there will
always exist a bijective function �=��:�→�such that if �,�is an element of �,
�,�will be an element of �and the functions �(�)and �(�)are said to be inverse of
each other.
•�=�
−1
:�→�=��,�|�,��∈�
��
��
�
�
−1
�,�∈�
Then
�,�∈�
−1
Inverse Function
Return To Top
Let �=��:�→�be a one –one and onto function , i.e.a bijection , then there will
always exist a bijective function �=��:�→�such that if �,�is an element of �,
�,�will be an element of �and the functions �(�)and �(�)are said to be inverse of
each other.
•�=�
−1
:�→�=��,�|�,��∈�
��
��
�
�
−1
�,�∈�
Then
�,�∈�
−1
Inverse Function
Return To Top
Key Takeaways
•
��
�
�
�
−1
Why function must be bijective for it to be invertible?
�
�
�
1
2
3
�
1
2
3
�
�
�
�
�
��
�
�
−1
�
�
�
�
1
2
3
4
�
1
2
3
4
�
�
�
�
�
Not a function
��
�
�
−1
�
�
�
�
1
2
3
4
�
1
2
3
4
�
�
�
�
�
5
5
Not a function
Into
Bijective
Inverse of a bijection is unique and also a bijection.•
Inverse Function
Return To Top
•
��
�
�
�
−1
Why function must be bijective for it to be invertible?
�
�
�
1
2
3
�
1
2
3
�
�
�
�
�
��
�
�
−1
�
�
�
�
1
2
3
4
�
1
2
3
4
�
�
�
�
�
Not a function
��
�
�
−1
�
�
�
�
1
2
3
4
�
1
2
3
4
�
�
�
�
�
5
5
Not a function
Into
Bijective
Inverse of a bijection is unique and also a bijection.•
Inverse Function
Return To Top
Key Takeaways
•
To find inverse :
(??????)For �=��, express �in terms of �
(????????????)In �=�(�), replace �by �in �to get inverse.
Example :�=�
�
�=ln�
�=ln�=�
−1
(�)
Inverse Function
Return To Top
•
To find inverse :
(??????)For �=��, express �in terms of �
(????????????)In �=�(�), replace �by �in �to get inverse.
Example :�=�
�
�=ln�
�=ln�=�
−1
(�)
Inverse Function
Return To Top
Solution:
��=
2�+3
4
:ℝ→ℝ,then find it’sinverse.
Let ��=�=
2�+3
4
⇒�=
4�−3
2
=��
∴��=�
−1
�=
4�−3
2
:ℝ→ℝ
To find inverse :
For �=��, express �in terms of �
In �=�(�), replace �by �in �to get inverse.
0,−
3
2
3
2
,
3
2
0,
3
4
3
4
,0
−
3
2
,0
��
��
�=�
Function and its inverse
are symmetric about �=�
Return To Top
��=
2�+3
4
:ℝ→ℝ,then find it’sinverse.
Let ��=�=
2�+3
4
⇒�=
4�−3
2
=��
∴��=�
−1
�=
4�−3
2
:ℝ→ℝ
To find inverse :
For �=��, express �in terms of �
In �=�(�), replace �by �in �to get inverse.
0,−
3
2
3
2
,
3
2
0,
3
4
3
4
,0
−
3
2
,0
��
��
�=�
Function and its inverse
are symmetric about �=�
Return To Top
Inverse Function
��=�
�
, ��=ln�
��=�
�
��
=ln�
�=�
(0,1)
(1,0)
Example:
Return To Top
��=�
�
, ��=ln�
��=�
�
��
=ln�
�=�
(0,1)
(1,0)
Example:
Return To Top
Since ��is bijective.
Let �=�
2
+�+1⇒�
2
+�+1−�=0
Solving for �,
⇒�=
−1±1−4(1−�)
2
=
−1±4�−3
2
But since inverse of a function is unique,
⇒�=
−1+4�−3
2
=��
∴�
−1
�=
−1+4�−3
2
∶1,∞→0,∞
0,1
�
�
Solution:
If ��=�
2
+�+1:0,∞→1,∞, find its inverse.
Return To Top
Let �=�
2
+�+1⇒�
2
+�+1−�=0
Solving for �,
⇒�=
−1±1−4(1−�)
2
=
−1±4�−3
2
But since inverse of a function is unique,
⇒�=
−1+4�−3
2
=��
∴�
−1
�=
−1+4�−3
2
∶1,∞→0,∞
0,1
�
�
Solution:
If ��=�
2
+�+1:0,∞→1,∞, find its inverse.
Return To Top
Key Takeaways
Properties of Inverse Function
•
The graphs of �and �are the mirror images of each other about the line �=�.
•
If functions �and �
−1
intersect , then at least one point of intersection lie on the
line �=�.
�(�)=�
3
�
−1
�=
3
�
�=�
f(x)=�
3
⇒�
−1
(�)=
3
�
Return To Top
Properties of Inverse Function
•
The graphs of �and �are the mirror images of each other about the line �=�.
•
If functions �and �
−1
intersect , then at least one point of intersection lie on the
line �=�.
�(�)=�
3
�
−1
�=
3
�
�=�
f(x)=�
3
⇒�
−1
(�)=
3
�
Return To Top
Key Takeaways
•
�
If �and �are inverse of each other , then ���=���=�.
�
1
2
3
�
�
�
�
�
�
1
2
3
�
�
�
�
���=�3=�
However, ���and ���can be equal even if �and �are not
inverse of each other , but in that case ���=���≠�
Properties of Inverse Function
Return To Top
•
�
If �and �are inverse of each other , then ���=���=�.
�
1
2
3
�
�
�
�
�
�
1
2
3
�
�
�
�
���=�3=�
However, ���and ���can be equal even if �and �are not
inverse of each other , but in that case ���=���≠�
Properties of Inverse Function
Return To Top
Key Takeaways
•
If �and �are two bijections, �:�→�,�:�→�,then inverse
of ���exists and
���
−1
=�
−1
��
−1
��=�+2,��=�+1
Then , ����=�+1+2=�+3
And ,����=�+2+1=�+3,
but �and �are non inverse of each other.
Example:
However, ���and ���can be equal even if �and �are not inverse of
each other , but in that case ���=���≠�
⇒���=���≠�
Properties of Inverse Function
Return To Top
•
If �and �are two bijections, �:�→�,�:�→�,then inverse
of ���exists and
���
−1
=�
−1
��
−1
��=�+2,��=�+1
Then , ����=�+1+2=�+3
And ,����=�+2+1=�+3,
but �and �are non inverse of each other.
Example:
However, ���and ���can be equal even if �and �are not inverse of
each other , but in that case ���=���≠�
⇒���=���≠�
Properties of Inverse Function
Return To Top
Binary Operation:
Definition:
A binary operation ∗on a set �is a function ∗:�×�→�.
Denoted as ∗�,�→�∗�
Show that addition is a binary operation on �,
but division is not a binary operation.
Example:
Solution:+:�×�→�is given by +�,�→�+�, is a function on �
÷:�×�→�is given by ÷�,�→
�
�
, is not a function on �
and not a binary operation as for �=0,
�
0
is not defined.
Return To Top
Definition:
A binary operation ∗on a set �is a function ∗:�×�→�.
Denoted as ∗�,�→�∗�
Show that addition is a binary operation on �,
but division is not a binary operation.
Example:
Solution:+:�×�→�is given by +�,�→�+�, is a function on �
÷:�×�→�is given by ÷�,�→
�
�
, is not a function on �
and not a binary operation as for �=0,
�
0
is not defined.
Return To Top
??????Commutative:
A binary operation ∗on a set �is called commutative if �∗�=�∗�for every �,�∈�.
Example: Addition is commutative on �,but subtraction is not.
Solution:�+�=�+�→commutative
but �−�≠�−�→not commutative
????????????Associative:
A binary operation ∗is said to be associative
�∗�∗�=�∗�∗�,∀�,�,�∈�.
Example:8+5+3=8+5+3associative
8−5−3≠8−5−3not associative
Properties of Binary Operation:
Return To Top
A binary operation ∗on a set �is called commutative if �∗�=�∗�for every �,�∈�.
Example: Addition is commutative on �,but subtraction is not.
Solution:�+�=�+�→commutative
but �−�≠�−�→not commutative
????????????Associative:
A binary operation ∗is said to be associative
�∗�∗�=�∗�∗�,∀�,�,�∈�.
Example:8+5+3=8+5+3associative
8−5−3≠8−5−3not associative
Properties of Binary Operation:
Return To Top
??????????????????Identity:
Given a binary operation ∗:�×�→�,an element �∈�,if it exists, is called identity for
the operation if �∗�=�=�∗�,∀�∈�
Note:i.0is identity for addition on �
ii.1is identity for multiplication on �
????????????Inverse:
Given a binary operation ∗:�×�→�,with identity element �in �,an element �∈�,is
said to be invertible w.r.t ∗,if there exists an element �in �such that �∗�=�=�∗�
and �is called inverse of �and is denoted by �
−1
.
Properties of Binary Operation:
Return To Top
Given a binary operation ∗:�×�→�,an element �∈�,if it exists, is called identity for
the operation if �∗�=�=�∗�,∀�∈�
Note:i.0is identity for addition on �
ii.1is identity for multiplication on �
????????????Inverse:
Given a binary operation ∗:�×�→�,with identity element �in �,an element �∈�,is
said to be invertible w.r.t ∗,if there exists an element �in �such that �∗�=�=�∗�
and �is called inverse of �and is denoted by �
−1
.
Properties of Binary Operation:
Return To Top
Note:i.−�is inverse of �for addition operation on �.
�+−�=0=−�+�
ii.
1
�
is inverse of �(�≠0)for multiplication operation on �−0.
�×
1
�
=1=
1
�
�
Properties of Binary Operation:
Return To Top
�+−�=0=−�+�
ii.
1
�
is inverse of �(�≠0)for multiplication operation on �−0.
�×
1
�
=1=
1
�
�
Properties of Binary Operation:
Return To Top
Given�∗�=�+�+��.
(i) First, we must check commutativity of ∗
Let �,�∈�−−1
Then �∗�=�+�+��
=�+�+��
=�∗�
Therefore, �∗�=�∗�,∀�,�∈�−−1
Now, we have toprove associativity of ∗
Let �,�,�∈�−−1,then
�∗�∗�=�∗(�+�+��)=�+�+�+��+��+�+��
=�+�+�+��+��+��+���
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
(i) First, we must check commutativity of ∗
Let �,�∈�−−1
Then �∗�=�+�+��
=�+�+��
=�∗�
Therefore, �∗�=�∗�,∀�,�∈�−−1
Now, we have toprove associativity of ∗
Let �,�,�∈�−−1,then
�∗�∗�=�∗(�+�+��)=�+�+�+��+��+�+��
=�+�+�+��+��+��+���
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
�∗�∗�=�+�+��∗�
=�+�+��+�+�+�+���
=�+�+�+��+��+��+���
Therefore, �∗�∗�=�∗�∗�,∀�,�,�∈�−−1
Thus, ∗is associative on �−−1.
(ii) Let e be the identity element in �−−1with
respect to ∗such that
�∗�=�=�∗�,∀�∈�−−1
�∗�=�and �∗�=�,∀�∈�−−1
�+�+��=�and �+�+��=�,∀�∈�−−1
�+��=0and �+��=0,∀�∈�−−1
�1+�=0and �1+�=0,∀�∈�−−1
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
=�+�+��+�+�+�+���
=�+�+�+��+��+��+���
Therefore, �∗�∗�=�∗�∗�,∀�,�,�∈�−−1
Thus, ∗is associative on �−−1.
(ii) Let e be the identity element in �−−1with
respect to ∗such that
�∗�=�=�∗�,∀�∈�−−1
�∗�=�and �∗�=�,∀�∈�−−1
�+�+��=�and �+�+��=�,∀�∈�−−1
�+��=0and �+��=0,∀�∈�−−1
�1+�=0and �1+�=0,∀�∈�−−1
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
�=0,∀�∈�−−1[because �≠−1]
Thus, 0is the identity element
in �−−1with respect to ∗.
(iii) Let �∈�−−1and �∈�−−1
be the inverse of �.Then,
�∗�=�=�∗�
�∗�=�and �∗�=�
�+�+��=0and �+�+��=0
�1+�=−�,∀�∈�−−1
�=−
�
1+�
∀�∈�−−1[because �≠−1]
�=−
�
1+�
is the inverse of �∈�−−1
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
Thus, 0is the identity element
in �−−1with respect to ∗.
(iii) Let �∈�−−1and �∈�−−1
be the inverse of �.Then,
�∗�=�=�∗�
�∗�=�and �∗�=�
�+�+��=0and �+�+��=0
�1+�=−�,∀�∈�−−1
�=−
�
1+�
∀�∈�−−1[because �≠−1]
�=−
�
1+�
is the inverse of �∈�−−1
Solution:
Let ∗be a binary operation on �−−1,defined by �∗�=�+�+��for all
�,�∈�−−1,then:
(??????)Show that ∗is both commutative and associative on �−{−1}
(????????????)Find the identity element in �−{−1}
(??????????????????)Show that every element of �−{−1}is invertible.
Also, find inverse of an arbitrary element.
Return To Top
Session 10
Functional Equations and
Transformation of Graphs
Return To Top
Functional Equations and
Transformation of Graphs
Return To Top
Let ��=�=�
2
−3�
Let ��=�=�
2
−3�
⇒�=
3−9+4�
2
Then , �
−1
�=
3−9+4�
2
⇒�
2
−3�−�=0
Since ,��=�
−1
�=�
So, �
2
−3�=�
⇒�=0,4
But , acc. to given domain
Find the solution of equation �
2
−3�=
3−9+4�
2
, �∈−∞,1.
(1,0)
�(�)
�=�
�(�)
�
−1
(�)
�=0
Solution:
Return To Top
2
−3�
Let ��=�=�
2
−3�
⇒�=
3−9+4�
2
Then , �
−1
�=
3−9+4�
2
⇒�
2
−3�−�=0
Since ,��=�
−1
�=�
So, �
2
−3�=�
⇒�=0,4
But , acc. to given domain
Find the solution of equation �
2
−3�=
3−9+4�
2
, �∈−∞,1.
(1,0)
�(�)
�=�
�(�)
�
−1
(�)
�=0
Solution:
Return To Top
Key Takeaways
Functional Equations
��+�=��.��⇒��=�
��
,??????∈ℝ.
If �,�are independent real variable, then
���=��+��⇒��=??????l��
��,??????∈ℝ,�>0,�≠1.
��+�=��+��⇒��=??????�,??????∈ℝ.
���=��.��⇒��=�
�
,�∈ℝ.
If ��is a polynomial of degree ‘�’ , such that
��.�
1
�
=��+�
1
�
⇒��=1±�
�
•
•
•
•
•
Return To Top
Functional Equations
��+�=��.��⇒��=�
��
,??????∈ℝ.
If �,�are independent real variable, then
���=��+��⇒��=??????l��
��,??????∈ℝ,�>0,�≠1.
��+�=��+��⇒��=??????�,??????∈ℝ.
���=��.��⇒��=�
�
,�∈ℝ.
If ��is a polynomial of degree ‘�’ , such that
��.�
1
�
=��+�
1
�
⇒��=1±�
�
•
•
•
•
•
Return To Top
If �(�)is a polynomial function such that ��⋅�
1
�
=��+�
1
�
, such that
�3=−26.Then �4=?
��.�
1
�
=��+�
1
�
⇒�3=−26
�4= −63
⇒1±3
�
=−26
⇒ ⇒�=3
∴��=1−�
3
3
�
=−27
−3
�
=−27
⇒��=1±�
�
−3
�
=−27
Solution:
Return To Top
1
�
=��+�
1
�
, such that
�3=−26.Then �4=?
��.�
1
�
=��+�
1
�
⇒�3=−26
�4= −63
⇒1±3
�
=−26
⇒ ⇒�=3
∴��=1−�
3
3
�
=−27
−3
�
=−27
⇒��=1±�
�
−3
�
=−27
Solution:
Return To Top
If a function ��satisfies the relation ��+�=��+��, where �,�∈ℝand
�1=4.Then find the value of σ
�=1
10
��=?
∴
�=1
10
��=
�=1
10
4�
⇒�1=4=??????
=4
�=1
10
�
⇒��=??????���+�=��+��
=220
Solution:
Return To Top
�1=4.Then find the value of σ
�=1
10
��=?
∴
�=1
10
��=
�=1
10
4�
⇒�1=4=??????
=4
�=1
10
�
⇒��=??????���+�=��+��
=220
Solution:
Return To Top
For �∈ℝ−0,the function �(�)satisfies ��+2�1−�=
1
�
. Find the value of �2.
��+2�1−�=
1
�
Put�=2⇒�2+2�−1=
1
2
⋯??????
Put�=−1⇒�−1+2�2=−1⋯????????????
By??????and????????????
�2+2�−1=
1
2
2�−1+4�2=−2
− − −
3�2=−
5
2
⇒�2=−
5
6
A
D
B
C
−
5
6
1
2
−2
3
4
�2=−
5
6
Solution:
Return To Top
1
�
. Find the value of �2.
��+2�1−�=
1
�
Put�=2⇒�2+2�−1=
1
2
⋯??????
Put�=−1⇒�−1+2�2=−1⋯????????????
By??????and????????????
�2+2�−1=
1
2
2�−1+4�2=−2
− − −
3�2=−
5
2
⇒�2=−
5
6
A
D
B
C
−
5
6
1
2
−2
3
4
�2=−
5
6
Solution:
Return To Top
��+�1−�=
4
�
4
�
+2
+
4
1−�
4
1−�
+2
=
4
�
4
�
+2
+
4
4
�
4
4
�
+2
=
4
�
4
�
+2
+
4
4+2⋅4
�
=
4
�
4
�
+2
+
2
4
�
+2
∴��+�1−�=1
⇒�
1
40
+�
2
40
+�
3
40
+⋯+�
20
40
+⋯+�
39
40
=19+�
20
40
⇒�
1
40
+�
2
40
+�
3
40
+⋯+�
39
40
−�
1
2
=19+�
20
40
−�
1
2
Let the function �:[0,1]→�be defined by ��=
4
�
4
�
+2
Then the value of �
1
40
+�
2
40
+�
3
40
+⋯+�
39
40
−�
1
2
is____.
=19
Solution:
Return To Top
4
�
4
�
+2
+
4
1−�
4
1−�
+2
=
4
�
4
�
+2
+
4
4
�
4
4
�
+2
=
4
�
4
�
+2
+
4
4+2⋅4
�
=
4
�
4
�
+2
+
2
4
�
+2
∴��+�1−�=1
⇒�
1
40
+�
2
40
+�
3
40
+⋯+�
20
40
+⋯+�
39
40
=19+�
20
40
⇒�
1
40
+�
2
40
+�
3
40
+⋯+�
39
40
−�
1
2
=19+�
20
40
−�
1
2
Let the function �:[0,1]→�be defined by ��=
4
�
4
�
+2
Then the value of �
1
40
+�
2
40
+�
3
40
+⋯+�
39
40
−�
1
2
is____.
=19
Solution:
Return To Top
Key Takeaways
Let �=�(�)
�=��+??????,??????>0(graph goes to left by
′
??????
′
units)
�,0
�,0
0,�
�
�
Transformation of graphs (horizontal shifts):
�−??????,0
0,�
�
�
•
Return To Top
Let �=�(�)
�=��+??????,??????>0(graph goes to left by
′
??????
′
units)
�,0
�,0
0,�
�
�
Transformation of graphs (horizontal shifts):
�−??????,0
0,�
�
�
•
Return To Top
0,0
�
�
0,0
�
�
−1,0
1units
Plot the following curve:
??????�=�+1
2
????????????�=�−2
2
??????For �=��+??????,??????>0graph shift ??????
units toward left from �=��graph
�=�
2
�=�
2
�=�+1
2
Solution:
Return To Top
�
�
0,0
�
�
−1,0
1units
Plot the following curve:
??????�=�+1
2
????????????�=�−2
2
??????For �=��+??????,??????>0graph shift ??????
units toward left from �=��graph
�=�
2
�=�
2
�=�+1
2
Solution:
Return To Top
Plot the following curve:
??????�=�+1
2
????????????�=�−2
2
????????????�=�−2
2
=�+−2
2
Here graph shift 2units toward right
0,0
�
�
0,0
�
�
2,0
2units
�=�
2
�=�
2
�=�−2
2
Solution:
Return To Top
??????�=�+1
2
????????????�=�−2
2
????????????�=�−2
2
=�+−2
2
Here graph shift 2units toward right
0,0
�
�
0,0
�
�
2,0
2units
�=�
2
�=�
2
�=�−2
2
Solution:
Return To Top
For �=��−??????,??????<0graph shift ??????units
towards right horizontally from �=��graph.
Return To Top
towards right horizontally from �=��graph.
Return To Top
�
�
??????
−1
2??????−??????−2??????
−??????
2
??????
2
3??????
2
−3??????
2
1
�=cos�
�
�
??????
−1
2??????−??????−2??????
−??????
2
??????
2
3??????
2
−3??????
2
1
�=cos�−
??????
2
=sin�
??????
2
units
Plot the curve of function �=cos�−
??????
2
using transformations
Solution:
Return To Top
�
??????
−1
2??????−??????−2??????
−??????
2
??????
2
3??????
2
−3??????
2
1
�=cos�
�
�
??????
−1
2??????−??????−2??????
−??????
2
??????
2
3??????
2
−3??????
2
1
�=cos�−
??????
2
=sin�
??????
2
units
Plot the curve of function �=cos�−
??????
2
using transformations
Solution:
Return To Top
Key Takeaways
Let �=�(�)
�=��+??????,??????>0(graphgoestoupby′??????′units)
Transformation of graphs (Vertical shifts):
0,�
�
�
0,�+??????
�
�
•
Return To Top
Let �=�(�)
�=��+??????,??????>0(graphgoestoupby′??????′units)
Transformation of graphs (Vertical shifts):
0,�
�
�
0,�+??????
�
�
•
Return To Top
0,0
�
�
0,0
�
�
??????For �=��+??????,??????>0graph shift ??????
units toward down from �=��graph
�=�
2
�=�
2
�=�
2
+1
Plot the following curves:
??????�=�
2
+1 ????????????�=�
2
−2
1units
Solution:
Return To Top
�
�
0,0
�
�
??????For �=��+??????,??????>0graph shift ??????
units toward down from �=��graph
�=�
2
�=�
2
�=�
2
+1
Plot the following curves:
??????�=�
2
+1 ????????????�=�
2
−2
1units
Solution:
Return To Top
????????????�=�
2
−2
Here graph shift 2units upward
Plot the following curves:
??????�=�
2
+1 ????????????�=�
2
−2
0,0
�
�
0,0
�
�
�=�
2
�=�
2
�=�
2
−2
2units
Solution:
Return To Top
2
−2
Here graph shift 2units upward
Plot the following curves:
??????�=�
2
+1 ????????????�=�
2
−2
0,0
�
�
0,0
�
�
�=�
2
�=�
2
�=�
2
−2
2units
Solution:
Return To Top
For�=��−??????,??????<0graph of �=��will shift
??????units downwards.
Return To Top
??????units downwards.
Return To Top
Key Takeaways
�=�??????�,??????>1(points on �-axis divided by ′??????′units)
Example:
�
�
??????
−1
2??????
??????
2
3??????
2
1
�=sin�
Two loops in 0to ??????
Let �=�(�)
Transformation of graphs (horizontal stretch):
�
??????
2??????
??????
2
3??????
2
Two loops in 0to 2??????
�=sin2�
−1
1
•
Return To Top
�=�??????�,??????>1(points on �-axis divided by ′??????′units)
Example:
�
�
??????
−1
2??????
??????
2
3??????
2
1
�=sin�
Two loops in 0to ??????
Let �=�(�)
Transformation of graphs (horizontal stretch):
�
??????
2??????
??????
2
3??????
2
Two loops in 0to 2??????
�=sin2�
−1
1
•
Return To Top
Key Takeaways
�=??????⋅��,??????>1(Pointony−axisismultipliedby′??????′units)
0,�
�
�
0,�
�
�
0,??????�
Let �=�(�)
Transformation of graphs (Vertical stretch):
•
Return To Top
�=??????⋅��,??????>1(Pointony−axisismultipliedby′??????′units)
0,�
�
�
0,�
�
�
0,??????�
Let �=�(�)
Transformation of graphs (Vertical stretch):
•
Return To Top
Plot graph of the following functions: �=2sin2�
−1≤sin2�≤1
�
�
??????
−1
2??????
??????
2
1
�=sin�
−2≤2sin2�≤2⇒
Period of sin2�=Period of 2sin2�=??????
�=sin2�
�
�
??????
−1
2??????
??????
2
1
�=2sin2�
−2
2
Solution:
Return To Top
−1≤sin2�≤1
�
�
??????
−1
2??????
??????
2
1
�=sin�
−2≤2sin2�≤2⇒
Period of sin2�=Period of 2sin2�=??????
�=sin2�
�
�
??????
−1
2??????
??????
2
1
�=2sin2�
−2
2
Solution:
Return To Top
Session 11
Playing with Graphs
Return To Top
Playing with Graphs
Return To Top
1.Make the plot of the graph �
�
�
�=�3
2
3
−1
21
1
0−1−2
Solution:??????�=1+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
�
�
�=�3
2
3
−1
21
1
0−1−2
Solution:??????�=1+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
Solution:
�
�
�=1+�
3
2
3
−1
21
1
0−1−2
2.Now, up the graph by 1.
??????�=1+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
�
�
�=1+�
3
2
3
−1
21
1
0−1−2
2.Now, up the graph by 1.
??????�=1+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
Solution:
�∈0,1 �=�+0
�∈1,2 �=�+1
�∈2,3 �=�+2
�∈−1,0 �=�−1
�
�
�−1 123
????????????�=�+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
�∈0,1 �=�+0
�∈1,2 �=�+1
�∈2,3 �=�+2
�∈−1,0 �=�−1
�
�
�−1 123
????????????�=�+�
Plot the following curves for �∈�: ??????�=1+�????????????�=�+�
denotesG.I.F.
Return To Top
??????)�=
1
�+4
Shift �=
1
�
at �=−4
�
�0
�=
1
�+4
�=
1
�
�=−4
Solution:
Plot graph of the following functions. ??????�=
1
�+4
????????????�=
1
�+4
+3
Return To Top
1
�+4
Shift �=
1
�
at �=−4
�
�0
�=
1
�+4
�=
1
�
�=−4
Solution:
Plot graph of the following functions. ??????�=
1
�+4
????????????�=
1
�+4
+3
Return To Top
????????????)�=
1
�+4
+3
Shift �=
1
�
at �=−4
�
�0
�=
1
�+4
+3
�=
1
�
Plot graph of the following functions. ??????�=
1
�+4
????????????�=
1
�+4
+3
Solution:
Return To Top
1
�+4
+3
Shift �=
1
�
at �=−4
�
�0
�=
1
�+4
+3
�=
1
�
Plot graph of the following functions. ??????�=
1
�+4
????????????�=
1
�+4
+3
Solution:
Return To Top
Key Takeaways
●Let �=�(�)
�=�−�,(mirrored about �−axis)
�
�
�
�
�
�
Transformation of graphs
1
1
Return To Top
●Let �=�(�)
�=�−�,(mirrored about �−axis)
�
�
�
�
�
�
Transformation of graphs
1
1
Return To Top
�=−�
�
�
0
Solution:
Plot the curve −�,
Where denotes fractional part function
12 3 4−2−3 −1
Return To Top
�
�
0
Solution:
Plot the curve −�,
Where denotes fractional part function
12 3 4−2−3 −1
Return To Top
Key Takeaways
�,0
�,0
0,�
�
�
●Let �=�(�)
Transformation of graphs:
�,0 �,0
�
�
�=−�(�)
�=�(�)
�=−��,(mirrored about �−axis)
Values of �, multiplied by −1
Return To Top
�,0
�,0
0,�
�
�
●Let �=�(�)
Transformation of graphs:
�,0 �,0
�
�
�=−�(�)
�=�(�)
�=−��,(mirrored about �−axis)
Values of �, multiplied by −1
Return To Top
Key Takeaways
�,0
�,0
0,�
�
�
●Let �=�(�)
�=−�(−�) transformationfrom �=�(�):
�,0�,0
0,�
�
�
�=�−�
Return To Top
�,0
�,0
0,�
�
�
●Let �=�(�)
�=−�(−�) transformationfrom �=�(�):
�,0�,0
0,�
�
�
�=�−�
Return To Top
Key Takeaways
�=−�−�,
�=−�(−�) transformation from �=�(�):
�,0�,0
0,�
�
�
�=−�−�,
Return To Top
�=−�−�,
�=−�(−�) transformation from �=�(�):
�,0�,0
0,�
�
�
�=−�−�,
Return To Top
Key Takeaways
�=��
�,0 �,0
0,�
�
�
−�,0 �,0
�
(image of �for +ve�,about �−axis)
�
At �=1,�=�1
At �=−1,�=�1
At �=2,�=�2
At �=−2,�=�2
�=��is an even function
●Let �=�(�)
Transformation of graphs:
Return To Top
�=��
�,0 �,0
0,�
�
�
−�,0 �,0
�
(image of �for +ve�,about �−axis)
�
At �=1,�=�1
At �=−1,�=�1
At �=2,�=�2
At �=−2,�=�2
�=��is an even function
●Let �=�(�)
Transformation of graphs:
Return To Top
Key Takeaways
�=��
�,0 �,0
�
�
(−ve�−axis portion
flipped about �−axis)
●Let �=�(�)
Transformation of graphs:
�,0 �,0
0,�
�
�
Return To Top
�=��
�,0 �,0
�
�
(−ve�−axis portion
flipped about �−axis)
●Let �=�(�)
Transformation of graphs:
�,0 �,0
0,�
�
�
Return To Top
Key Takeaways
�,0 �,0
�
�
●Let �=�(�)
Transformation of graphs:
�,0 �,0
0,�
�
�
�=�|�|
(+ve�axis portion of ��
flipped about y−axis)
�,0 �,0
�
�
�=��
Return To Top
�,0 �,0
�
�
●Let �=�(�)
Transformation of graphs:
�,0 �,0
0,�
�
�
�=�|�|
(+ve�axis portion of ��
flipped about y−axis)
�,0 �,0
�
�
�=��
Return To Top
Key Takeaways
|�|=�(�)transformation from �=��:
�=�(�)
�,0 �,0
0,�
�
�
�,0 �,0
�
�
�=��
�,0 �,0
�
�
Return To Top
|�|=�(�)transformation from �=��:
�=�(�)
�,0 �,0
0,�
�
�
�,0 �,0
�
�
�=��
�,0 �,0
�
�
Return To Top
�
�
??????�=sin|�|
Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
�
??????�=sin|�|
Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
Shift �=�
1
3at �=2
�
�
�
�=�−2
1
3
2
�=�
1
3
????????????�=�−2
1
3Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
1
3at �=2
�
�
�
�=�−2
1
3
2
�=�
1
3
????????????�=�−2
1
3Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
Now, draw graph for �=�−2
1
3at �=2
�
�
�2
�=|�−2
1
3|
Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
1
3at �=2
�
�
�2
�=|�−2
1
3|
Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
Return To Top
Solution:
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
�
�
1�
??????????????????�=ln�
Return To Top
Plot graphs of the function ??????�=sin�????????????�=�−2
1
3??????????????????�=ln�
�
�
1�
??????????????????�=ln�
Return To Top
Key Takeaways
Return To Top
Return To Top
Key Takeaways
�=�(�)
�=��
�=�(|�|)
�=|��|
Return To Top
�=�(�)
�=��
�=�(|�|)
�=|��|
Return To Top
Number of solutions of two curves �=��&�=��is number of
intersection points for 2curves �=��&��
Return To Top
intersection points for 2curves �=��&��
Return To Top
�
�
�=2
−�
�=ln�
Number of point of intersections
=Number of solutions=2
Find the number of solutions for ln�=2
−�
Solution:
Return To Top
�
�=2
−�
�=ln�
Number of point of intersections
=Number of solutions=2
Find the number of solutions for ln�=2
−�
Solution:
Return To Top
�=sin�
�
�
2????????????−??????−2??????
1
−1
−
??????
2
−
3??????
2
??????
2
3??????
2
in period 2??????
�
2????????????−??????−2??????
1
−1
−
??????
2
−
3??????
2
??????
2
3??????
2
�
�=sin�in period 2??????
Plot the curve of �=sin�:
Solution:
Return To Top
�
�
2????????????−??????−2??????
1
−1
−
??????
2
−
3??????
2
??????
2
3??????
2
in period 2??????
�
2????????????−??????−2??????
1
−1
−
??????
2
−
3??????
2
??????
2
3??????
2
�
�=sin�in period 2??????
Plot the curve of �=sin�:
Solution:
Return To Top
Return To Top
Tags
Categories
GeneralDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 47
- Slides 239
- Age 1036 days
Related Slideshows
22
Pray For The Peace Of Jerusalem and You Will Prosper
RodolfoMoralesMarcuc
33 views
26
Don_t_Waste_Your_Life_God.....powerpoint
chalobrido8
36 views
31
VILLASUR_FACTORS_TO_CONSIDER_IN_PLATING_SALAD_10-13.pdf
JaiJai148317
33 views
14
Fertility awareness methods for women in the society
Isaiah47
30 views
35
Chapter 5 Arithmetic Functions Computer Organisation and Architecture
RitikSharma297999
29 views
5
syakira bhasa inggris (1) (1).pptx.......
ourcommunity56
30 views