Approved by Government of Nepal, Ministry of Education,Science and Technology,
Curriculum Development Center(CDC), Sanothimi, Bhaktapur, Nepal
Infinity
Optional Mathematics
9Grade
Authors
Nil Prasad Ghimire
Shakti Prasad Acharya
Sujit Shrestha
Editors
Ramesh Subedi
Jibnath Sharma
Shubharambha Publication Pvt.Ltd.
Kathmandu, Nepal
Published by:
Shubharambha Publication Pvt. Ltd.
Kathmandu, Nepal
URL: www.shubharambhapublication.com.np
E-mail: [email protected]
www.facebook.com/shubharambhapublication
Book : Infinity Optional Mathematics-Book 9
Authors : Nil Prasad Ghimire
Shakti Prasad Acharya
Sujit Shrestha
Layout Design : Zeeta Computer Service Pvt. Ltd.
Ghantaghar, Kathmandu
Mobile No. 9841418545
Copyright © : Publisher
Edition : First: 2077 B.S.
© : Publisher
No part of this book may be reproduced or transmitted by means (electronic,
photocopying, recording or otherwise) without prior written permission from the
publisher. Any breach of this condition will entail legal action and prosecution.
Printed in Nepal
Preface
Infinity Optional Mathematics is a set of 5 books tailored to teach optional mathematics
to the students in the grades between 6 and 10. It is developed in strict compliance with
the recent school syllabus prescribed by the Curriculum Development Centre (CDC),
Ministry of Education, Science and Technology, Government of Nepal.
The text-books in the series are a result of a painstaking effort from a team of dedicated
and hard-working authors who have devised the lesson units with fitting ingenuity
and the acumen derived from years of experience as teachers.
The chapters in the books feature a fundamental theory and sufficiently-worked-out
examples right at the beginning as a ready reference for students. The classification
and the order of the chapters in each book are systematic and are put in a proper
sequence so that students can learn better and comprehend the progression of topics
and subject-matter with clarity. The authors are also confident that the books in their
present form are comprehensive, and helpful to the students in their preparation,
which is consistent with the exam requirements of the new question model (grid).
We thank the management team of Shubharambha Publication for their help in the
publication of this series. We also remain indebted to Mr Ananda Prasad Poudel,
Managing Director of Zeeta Computer Service, and his team for their expertly typing,
layout and cover design. Last but not least, our gratitude is due to Mr Nil Prasad
Ghimire for his help with the conceptual framework, and motivation throughout this
undertaking.
While every effort has been made to keep the series error-free, it is still possible for
some errors to have unwittingly crept into the work. If the beneficiaries of these books,
the reviewers and the others concerned bring to our notice any discrepancies, or areas
for potential improvement, we shall be extremely grateful to you. We would love to
receive your valuable suggestion, feedback, or queries via email at shubharambha.
[email protected].
Once again, we hope that this new edition of Infinity Optional Mathematics does
more than meet the expectations of students and teachers alike.
Authors
Contents 5
5
Unit 1 30
Algebra 39
1.1 Relation and Function
1.2 Polynomial 50
1.3 Sequence and Series
65
Unit 2
2.1 Limit
Unit 3
3.1 MATRIX
Unit 4 94
94
CO–ORDINATE GEOMETRY 102
4.1 Locus 117
4.2 Section Formulae 158
4.3 Equation of Straight lines
4.4 Area of triangle
Unit 5
TRIGONOMETRY 167
5.1 Measurement of Angle 167
5.2 Trigonometric Ratios 187
5.3 Compound angle 226
Unit 6
6.1 VECTORS 237
Unit 7
7.1 TRANSFORMATION 272
Unit 8 305
305
STATISTICS 316
8.1 Partition Values 335
8.2 Measures of Variability or Dispersion
Answer Sheet
UNIT
1 ALGEBRA
1.1 Relation and Function
Review
Discuss the following questions.
(a) Plot the points (1,3), (–3,2), (5, 3), (3, 5), (–3, –5) and (4, –6) in a graph.
(b) Are (5, 3) and (3, 5) equal? Observe the graph and discuss.
(c) Complete the following tables.
x –2 –1 0 1 2 3 4
y = 2x + 1
x
y = 3x – 5 –14 –11 –8 –5 –2 1 4
Ordered pair
Pair set
Suppose, in a school, two students should be selected for quiz contest. Ram and Sita
are selected for it. Are there any differece if Ram is selected first and then Sita or if
Sita is selected first and then Ram. Of course not, either Ram is selected first and
then Sita or Sita is selected first and then Ram. Both are same as the participants
of the quiz contest that means there is no role of order of selection of Ram and Sita
i.e. {Ram, Sita} = {Sita, Ram}. From the above example, it is concluded that
A set containing of any two elements is called a pair. For example, {p, q} is a pair
and {p, q} = {q, p}.
Ordered pair
Suppose, you have to wear shoes and socks. In this situation, you first wear shoes
and then socks or you wear first socks and then shoes are different and they give
different senses. You wear first socks and then shoes is usual but you wear first
shoes and then socks is unusual. That means the order of wearing shoes and socks
is important.
Infinity Optional Mathematics Book - 9 5
Y
Let's consider two co-ordinates (2, 3) and (3, 2). Are (2,3)
(2, 3) = (3, 2)? Of course not because in the adjoining (3, 2)
graph, it is clear that (2, 3) and (3, 2) are different. In
(2, 3) and (3, 2), the orders of 2 and 3 are different. X
Form the above examples, it is concluded that
O
A set containing of any two elements in definite order is called ordered pair.
In ordered pair, the elements are separated by comma (,) and are enclosed by
paranthesis. For example, (a, b) is an ordered pair and (a, b) ≠ (b, a).
In the ordered pair (a, b), a is called first component or x - coordinate or antecedent
and b is called second component or y-coordinate or consequence.
Equality of ordered pair
Two ordered pairs are said to be equal if corresponding components of both the
ordered pairs are equal. Note:
• In set, {a, b} = {b, a}
That is two ordered pairs (a, b) = (c, d) if and • In ordered pair, (a, b) ≠ (b, a)
only if a = c and b = d. For examples, (2, 8) =
(2, 8), (3, 2 × 3) = 93, 6 etc.
Can you write four differences between set and ordered pair?
WORKED OUT EXAMPLES
1. Find the values of x and y if (b) (2x – 1, x + 4) = (3, 2y)
(a) (x + 3, y – 2) = (1, 2)
Solution:
(a) (x + 3, y – 2) = (1, 2)
By comparing the corresponding elements, we get
x+3=1 y–2=2
or, x = 1 – 3 or, y = 2 + 2
∴ x=–2 ∴ y=4
6 Infinity Optional Mathematics Book - 9
(b) (2x – 1, x + 4) = (3, 2y)
By comparing the corresponding elements, we get,
2x – 1 = 3 x + 4 = 2y
or, 2x = 3 + 1 or, 2 + 4 = 2y
or, x = 4 or, 6 = y
2 2
∴ x=2 ∴ y=3
2. If x ∈ {2, 3, 4} and y ∈ {2, 4, 6, 8}, find all the possible ordered pairs
(x, y) such that
(i) x = y
(ii) y is multiple of x.
(iii) x is half of y.
Solution:
Here, x ∈ {2, 3, 4} and y ∈ {2, 4, 6, 8}
(i) For x = y, the ordered pairs are (2, 2), (4, 4)
(ii) For y is multiple of x, the ordered pairs are (2, 2), (2, 4), (4, 4), (2, 6),
(3, 6), (2, 8), (4, 8)
(iii) For x is half of y, the ordered pairs are (2, 4), (3, 6), (4, 8).
Exercise 1.1
Section 'A'
1. (a) Define ordered pair with an example.
(b) Under what condition, two ordered pairs are equal.
2. Which of the following ordered pairs are equal? Write with reason.
(a) (3, 7) and (7, 3) (b) (–2, 5) and (– 2, 5)
(c) (– 4, 5) and (4, –5) (d) (–5 + 8, 6 – 3) and (3, 3)
(e) (15 ÷ 5, 5 × 2) and (8 – 5, 7 + 2) (f) (4 + 3, 16 ÷ 2) and (14 ÷ 2, 4 × 2)
Section 'B'
3. Find the values of x and y if (b) (x + 2, y – 1) = (6, 2)
(a) (x, 5) = (–3, y) (d) (2x – 5, 4) = (9, y + 3)
(c) (5x – 3, x + y) = (2, –6)
(f) (2x + y, x + 3) = (3, 4y)
(e) x – 1, y + 1 = 2, 4
2 3 3 (h) (3x + y , 72y) = (27, 49)
(g) (2x–1, 5y + 2 ) = (2, 5)
4. If x ∈ {2, 4, 6, 8} and y ∈ {2, 3, 8, 10, 12}, find all the possible ordered
pairs (x, y) such that
(a) x is equal to y. (b) x is half of y.
(c) y is multiple of x. (d) y is factor of x.
Infinity Optional Mathematics Book - 9 7
5. Plot the following ordered pairs in the graph.
(0, 2), (–3, 0), (4, 3), (–2, 4), (–6, –1), (7, –5)
Cartesian Product
Consider two sets A = {Ram, Sita} and B = {12, 14}. Let us form all possible
ordered pairs such that first component of each ordered pair from the set A and
the second component from the set B. The all possible ordered pairs formed are
(Ram, 12), (Ram, 14), (Sita, 12) and (Sita, 14). The set of all possible ordered pairs
i.e. {(Ram, 12), (Ram, 14), (Sita, 12), (Sita, 14)} can be written as product of sets i.e.
A × B.
Let A and B be two given non-empty sets. Then the set of all possible ordered
pairs (a, b) such that first element i.e. a belongs to the set A and second
element i.e. b belongs to the set B is called the Cartesian product of the sets
A and B. It is denoted by A × B. Thus, A × B = {(a, b) : a ∈ A and b ∈ B} and
B × A = {(b, a): b ∈ B and a ∈ A}
For example, if A = {1, 2, 3} and B = {a, b} then
A × B = {1, 2, 3} × {a, b}= {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
B × A = {a, b} × {1, 2, 3} = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
Note
• In general A × B ≠ B × A. If A × B = B × A then A = B.
• If A × B = f then A = f or B = f.
• If A and B are two finite sets then n(A × B) = n(A) × n(B).
Representation of Cartesian Product
The Cartesian product of two sets can be represented in the different methods.
They are :
Tabulation method
Let A = {1, 2, 3} and B = {a, b} be the two sets. We can find the cartesian products
A × B and B × A by this method as follow.
Sets B Sets A
×a b ×123
A 1 (1, a) (1, b) B a (a, 1) (a, 2) (a, 3)
2 (2, a) (2, b) b (b, 1) (b, 2) (b, 3)
3 (3, a) (3, b)
\A × B = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)} \ B × A = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
8 Infinity Optional Mathematics Book - 9
Tree diagram method
Let A = {1, 2, 3} and B = {a, b} be the two sets. The cartesian product A × B and
B × A can be obtained by this method as follows.
A B A × B B A B×A
a (1, a) (a, 1)
1 (a, 2)
(1, b) (a, 3)
(2, a) (b, 1)
(b, 2)
1 a 2 (b, 3)
b
3
a
1
2
b (2, b) b 2
(3, a) 3
a
3
b (3, b) ∴ B × A = {(a, 1), (a, 2), (a, 3), (b,1)
(b, 2), (b, 3)}
∴ A × B = {(1, a), (1, b), (2, a),
(2,b) (3, a), (3, b)}
Arrow diagram method
Let A = {1, 2, 3} and B = {a, b}be the two sets. The cartesian product A × B and
B × A can be obtained by drawing arrow diagram as follows.
A B B A
a
1 a 1
b 2
2
3 b 3
∴ A × B = {(1, a), (1, b), (2,a), ∴ B × A = {(a, 1), (a, 2), (a, 3), (b,1)
(2, b), (3, a), (3, b)}
(b, 2), (b, 3)}
Graph method
Let A = {1, 2, 3} and B = {4, 5} be the two sets. Then A × B = {(1, 4), (1, 5), (2, 4),
(2, 5), (3, 4), (3, 5)} and B × A = {(4, 1,), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)} can be
represented in the graph as follows.
Infinity Optional Mathematics Book - 9 9
YY
6 A×B 6 B×A
5 5
4 4
B3 A3
2 2
1 1
0 123456 X 0 123456 X
A B
WORKED OUT EXAMPLES
1. If A = {a, b} and B = {5} then find the A × B and B × A.
Solution: Here,
A = {a, b} and B ={5}
\ A × B = {a, b} × {5} = {(a, 5), (b, 5)}
\ B × A = {5} × {a, b} = {(5, a), (5, b)}
2. If P = {x : x < 4, x ∈ N} and Q = {x : x2 – 3x + 2 = 0} then find P × Q and
Q × P.
Solution: Here,
P = {x : x < 4, x ∈N} = {1, 2, 3}
Q = {x : x2 – 3x + 2 = 0}
Now, x2 – 3x + 2 = 0
or, x2 – 2x – x + 2 = 0
or, x(x – 2) – 1(x – 2) = 0
or, (x – 2) (x – 1) = 0
Either,
x – 2 = 0 \ x=2
or, x – 1 = 0 \ x=1
\ Q = {1, 2}
Again, P × Q = {1, 2, 3} × {1, 2}
= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
10 Infinity Optional Mathematics Book - 9
Q × P = {1, 2} × {1, 2, 3}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}
3. If A = {Ashok, Yugesh} and B = {Kathmandu, Lalitpur, Bhaktapur},
find A × B and present it in arrow diagram.
Solution: Here,
A = {Ashok, Yugesh}
B = {Kathmandu, Bhaktapur, Lalitpur}
\ A × B = {(Ashok, Kathmandu), (Ashok, Bhaktapur), (Ashok, Lalitpur),
(Yugesh, Kathmandu), (Yugesh, Bhaktapur), (Yugesh, Lalitpur)}
Again, showing A × B in arrow diagram.
Ashok Kathmandu
Lalitpur
Bhaktapur
Yugesh
4. If A = {1, 2}, B = {1, 2, 3} and C = {2, 3} then show that
A × (B∩C) = (A × B) ∩ (A × C).
Solution: Here, A = {1,2}, B = {1, 2, 3} and C = {2, 3}
LHS = A × (B ∩ C)}
= {1, 2} × {2, 3}
= {(1, 2), (1, 3), (2, 2), (2, 3)}
\ A × (B ∩ C) = {(1, 2), (1, 3), (2, 2), (2, 3)}.
R.H.S. = (A × B) ∩ (A × C)
A × B = {1, 2} × {1, 2, 3}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}
A × C = {1, 2} × {2, 3}
= {(1, 2), (1, 3), (2, 2), (2, 3)}
∴ (A × B) ∩ (A × C) = {(1, 2), {1, 3), (2, 2), (2, 3)}
Hence, L.H.S. = R.H.S
5. If A = {1, 2, 3} and B = {4, 5} then verify that n(A × B) = n(A) × n(B).
Solution: Here, A = {1, 2, 3} and B ={4, 5}
∴ A × B = {1, 2, 3} × {4, 5}
= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Infinity Optional Mathematics Book - 9 11
so, n(A × B) = 6
Again, n(A) × n(B) = 3 × 2 = 6
∴ n (A × B) = n(A) × n(B)
Exercise 1.2
Section 'A'
1. (a) Define cartesian product.
(b) If n(A) = 3 and n(B) = 4, find n(A × B) and n(B × A).
2. (a) If A × B = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} then find A, B, n(A),
n(B), B × A and n(B × A).
(b) If P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)} then
find P, Q, n(P), n(Q), Q × P and n(Q × P).
3. From the following arrow diagram, find A × B.
(a) A B (b) A B
1 a Anju Nepal
b
2
3 India
Manju China
Section 'B'
4. Find A × B, B × A, A × A and B × B from the given sets.
(a) A = {a, b}, B = {m} (b) A = {a, b}, B = {m, n}
(c) A = {a, b, c}, B = {m, n} (d) A = {a, b, c}, B = {m, n, o}
5. (a) If A = {1, 2} and B = {2, 3, 4} then find A × B and B × A and also show
that A × B ≠ B × A.
(b) If P = {x, y, z} and Q = {a, b} then find P × Q and Q × P and also show
that P × Q ≠ Q × P.
6. (a) Given that A = {1, 2, 3} and B = {3, 4}, find A × B and B × A and also
present them (i) on the graph (ii) by arrow diagram.
(b) Given that A = {Bindu, Subi} and B = {Mango, Apple, Banana}. Find
A × B and B × A by drawing arrow diagram and tabulation
method.
7. (a) If A = {3, 4, 5} and B = {6, 7} then verify that n(A × B) = n(A) × n(B).
(b) If Y = {a, b, c} and Z = {x} then verify that n(Y × Z) = n(Y) × n(Z).
12 Infinity Optional Mathematics Book - 9
Section 'C'
8. (a) If A = {x : x = 0, 1, 2} and B = {y : y = 2x} then find A × B and B × A.
(b) If A = {x : 2 ≤ x < 5, x∈N} and B = {x : x2 = 9} then find A × B and B × A.
(c) If A = {x : x < 3, x ∈W} and B = {x:x2 –x –6 = 0} then find A×B and B × A.
9. (a) If A = {a, b}, B = {b, c, d} and C = {c, d}, verify that
(i) A × (B∩C) = (A × B) ∩ (B × C)
(ii) A × (B∪C) = (A × B) ∪ (B × C)
(b) If A = {1, 2, 3}, B = {2, 3} and C = {3, 4} then show that
A × (B – C) = (A × B) – (A × C).
Relation
Let A = {Kathmandu, Tokyo, Delhi} and B = {Nepal, India, Japan}.
Then A × B = {(Kathmandu, Nepal), (Kathmandu, India), (Kathmandu, Japan),
(Tokyo, Nepal), ( Tokyo, India), (Tokyo, Japan), (Delhi, Nepal), (Delhi, India),
(Delhi, Japan)}
Let us consider the relation "is the capital of". So, Kathmandu is the capital of
Nepal, Tokyo is the capital of Japan and Delhi is the capital of India.
∴ R = {(Kath, Nepal), (Tokyo, Japan), (Delhi, India)} which is a subset of A × B. i.e.
R ⊂ A × B. So R is relation from A to B.
Let A and B be the two non empty sets. Then any subset of A × B is a relation 'R'
from A to B. Symbolically, R = {(x, y) : x∈A, y∈B} and R ⊆ A × B
A ralation from A to B is denoted by R : A → B or simply R. If (a, b) ∈R then we
write the relation between a and b as aRb which is read as a is related to b. A
relation from A to A is said to be a relation on A.
Representation of a Relation
If A = {1, 2, 3} and B = {2, 3, 4} then
A × B = {1, 2, 3} × {2, 3, 4}
= {(1, 2), (1, 3,), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}
Let us consider a relation from A to B such that y is greater than x by 1. Such
relation 'R' can be presented in the various ways.
(i) Set of ordered pairs method
R = {(1, 2), (2, 3), (3, 4)}
Infinity Optional Mathematics Book - 9 13
(ii) Tabulation method 3 (iv) Graphical method
x 12 4 Y
y 23
(iii) Arrow diagram method
R
A B
1 2 0 X
2 3
3 4
(v) Set builder method
R = {(x, y) : y = x + 1, x∈A, x∈B}
WORKED OUT EXAMPLES
1. If A = {1, 2, 3} and B = {2, 3, 4} then find A × B. Also find the following
relation in A × B.
(a) is equal to (b) is more than
(c) is half of (d) is square of
Solution: Here, A = {1, 2, 3} and B = {2, 3, 4}
A × B = {1, 2, 3} × {2, 3, 4}
= {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}
(a) Let R1 be the relation from A to B such that x is equal to y.
Then, R1 = {(2, 2), (3, 3)}
(b) Let R2 be the relation from A to B such that x is more than y.
Then, R2 = {(3, 2)}
(c) Let R3 be the relation from A to B such that x is half of y.
Then, R3 = {(1, 2), (2, 4)}
(d) Let R4 be the relation from A to B such that x is square of y.
Then, R4 = { }
2. If A = {2, 3, 4}, find the relation R = {(x, y) : x + y ≥ 6} from A to A.
Solution: Here, A = {2, 3, 4}
∴ A × A = {2, 3, 4} × {2, 3, 4}
14 Infinity Optional Mathematics Book - 9
= {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}
∴ R = {(x, y) : x + y ≥ 6}
= {(2, 4), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}
Exercise 1.3
Section 'A'
1. (a) Define relation.
(b) Write the methods of representing relation.
2. Let A = {1, 2, 3} and B = {a, b, c}. Which of the following set of ordered pairs
represent the relation from A to B?
(a) {(1, a), (2, b), (3, c)} (b) {(2, a), (2, b), (3, c)}
(c) {(1, a), (b, 2), (3, b), (3, c)} (d) {(1, a), (1, b), (1, c), (2, a), (c, 2), (c, 3)}
3. Explain the following relation by the set of ordered pairs.
(a) R1 (b) R2
A B A B
1 4 a x
2 5 b y
3 6 c z
(c) A R 3 B (d) A R4 B
Ashok Pokhara Nepal Deilhi
Chitwan Tokyo
Bindu Kathmandu India
Subi
China Pokhara
Japan Kathmandu
4. Given that A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}.
Write the relation on A × B satisfying the following conditions.
a. is double of b. is less than
c. is equal to d. is square of
Infinity Optional Mathematics Book - 9 15
Section 'B'
5. Represent the following relation by arrow diagram.
a. {(a, x), (b, y), (c, z)}
b. {(1, 2), (1, 3), (2, 3), (2, 4)}
c. {(Ram, 12), (Shyam, 12), (Hari, 13)}
d. {(Sunday, 5°C), (Monday, 7°C), (Tuesday, 7°C), (Wednesday, 4°C),
(Thursday, 5°C), (Friday, 6°C), (Saturday, 5°C)}
6. If A = {2, 3, 4} and B = {1, 2, 3}. Find the relation from A to B satisfying
the following conditions.
a. is double of b. is half of
c. is equal to d. is square of
7. If A = {1, 2, 3}, find the relations on A satisfying the following conditions.
a. y = 2x b. y > x c. x + y > 6
d. x + y ≤ 5 e. 2x + y > 7 f. x is multiple of y
Section 'C'
8. If A = {2, 3, 4} and B = {4, 6, 8, 10}, find the following relations in A × B.
a. R1 = {(x, y) : y = 2x} b. R2 = {(x, y): x is factor of y}
Represent each of the relation in following methods.
i. Set of ordered pairs ii. Arrow diagram
iii. Graph iv. Table
9. Find the cartesian product P × Q if P = {x : x= 1, 2, 3} and Q = {y: y =
2x - 1}. Obtain a relation R from P to Q such that x + y > 4.
Domain and Range of a Relation
Consider a relation "is square of" i.e. R = {(1, 1), (4, 2), (9, 3), (16, 4)}. The set of x
-components of all ordered pairs of R = {1, 4, 9, 16} is called domain of R. The set of
y-components of all ordered pairs of R = {1, 2, 3, 4} is called range of R.
If R is a relation from A to B then the set of x-components of all ordered pairs in R is
called domain of R and the set of y-components of all ordered pairs in R is called range of
R. Symbolically, Domain of R = {x: (x, y) ∈ R}, Range of R = {y : (x, y) ∈ R}
16 Infinity Optional Mathematics Book - 9
Types of Relation
Let A = {1, 2, 3}. Then
A × A = {1, 2, 3} × {1, 2, 3}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
Reflexive relation
A relation R1 = {(1, 1), (2, 2), (3, 3)} on A is called reflexive relation.
If R is a relation on A then R is called reflexive relation if (x, x)∈ R for all x ∈ A. That
is xRx for all x ∈ A.
Symmetric relation
A relation R2 = {(1, 2), (1, 3), (2, 3), (2, 1), (3, 1), (3, 2)} on A is called symmetric
relation.
A relation R on A is called symmetric relation if (x, y)∈ R then (y, x) ∈ R. That is if
xRy then yRx.
Transitive relation
A relation R3 = {(1, 3), (3, 2), (2, 1)} on A is called transitive relation.
A relation R on A is called transitive relation if (x, y)∈ R and (y, z) ∈ R then
(x, z) ∈ R. That is if xRy and yRz then xRz.
Note
A relation which satisfies reflexive, symmetric and transitive is called
an equivalence relation.
Inverse relation
Consider two relation R4 = {(1, 2), (1, 3), (3, 2)} and R5 = {(2, 1), (3, 1), (2, 3)}.
Here, domain of R4 = {1, 3} and range of R4 R5
R4 = {2, 3} domain of R5 = {2, 3} and range of 1 22 1
R5 = {1, 3}. So, the relation R4 and R5 are inverse 3 33 3
to each other.
A new relation obtained by interchanging the ∴ R4–1 = R5
given relation R is called as the inverse relation
of R. It is denoted by R-1.
Symbolically, it is defined as follows,
If R = {(x, y): x ∈ A and y ∈ B} then R-1 = {(y, x) : y ∈ B and x ∈ A}
Infinity Optional Mathematics Book - 9 17
WORKED OUT EXAMPLES
1. If A = {1, 2, 3, 4} then find the relation on A × A determined by the
condition x + y < 5. Also find its domain, range and inverse relation.
Solution:
Here, A = {1, 2, 3, 4}
∴ A × A = {1, 2, 3, 4} × {1, 2, 3, 4}
= {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3),
(3,4), (4,1), (4,2), (4,3), (4,4)}
∴ R(x + y < 5) = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}
Again, domain of R = {1, 2, 3}
range of R = {1, 2, 3}
R-1 = {(1, 1), (2, 1), (3, 1), (1, 2), (2, 2), (1, 3)}
2. Given that A = {1, 2} and B = {1, 2, 3}. Find out reflexive, symmetric and
transitive relations from A to B.
Solution:
Here, A = {1, 2} and B = {1, 2, 3}
∴ A × B = {1, 2} × {1, 2, 3}
= {(1, 1), (1, 2), (1,3 ), (2, 1), (2, 2), (2, 3)}
Now, Reflexive relation = {(1, 1), (2, 2)}
Symmetric relation = {(1, 2), (2, 1)}
Transitive relation = {(1, 2), (2, 3), (3, 1)}
Exercise 1.4
Section 'A'
1. (a) Define relation.
(b) Define domain and range of a relation.
(c) Define the following.
i. Reflexive relation ii. Symmetric relation
iii. Transitive relation iv. Inverse relation
(d) Write one difference between symmetric and inverse relation.
18 Infinity Optional Mathematics Book - 9
2. From the following relations in arrow diagram, find their domain
and range.
(a) R1 (b) R2
AB AB
14 aw
25
36 b x
y
cz
3. From the following relations representing in a graph, find their
domain and range.
(a) Y (b) Y
4 4
3 3
2 2
1 1
0 1234 X 0 1234 X
Section 'B'
4. (a) Given that A × B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3),
(3, 4). List out reflexive, transitive and symmetric relation in A × B.
(b) Given that A × B = {(2, 3), (2, 4), (3, 3), (3, 4), (4, 3), (4, 4)}. List out
reflexive, symmetric and transitive relation in A × B.
5. Find the domain, range and inverse relation of the following relations.
(a) R1 = {(1, 2), (1, 3), (2, 3)} (b) R2 = {(1, 4), (2, 4), (3, 4), (4, 4)}
(c) R3 = {(a, x) (b, y), (c, z)} (d) R4 = {(a, y), (b, y), (c, y), (b, z), (c, z)}
Section 'C'
6. If A = {1, 2, 3} and B = {2, 3, 4} then find the relations from A to B under
the following conditions. Also find their domain and range.
(a) x = y (b) x < y
(c) x + y < 5 (d) x + y = 6
7. If A = {2, 4, 6, 8} then list out reflexive, symmetric and transitive
relations on A.
Infinity Optional Mathematics Book - 9 19
Function
Consider A is a set of name of people i.e. A = {Ram, Hari, Sita, Gita} and B is the
set of days of a week i.e. B = { Sun, Mon, Tue, Wed, Thur, Fri, Sat}. Let's make the
relations from A to B defined as the birthday of people.
R2
R1
A B A B
Ram Ram
Hari Sun Hari Sun
Sita Mon Sita Mon
Gita Tue Gita Tue
Wed Wed
Thur Thur
Fri Fri
Sat Sat
R1 is possible because every R2 is not possible because Gita
people in set A has birthday in in set A has no birthday in set
set B. So. R1 is function. B. So, R2 is not fuction.
R3 R4
A B A B
Ram Ram
Hari Sun Hari Sun
Sita Mon Sita Mon
Gita Tue Gita Tue
Wed Wed
Thur Thur
Fri Fri
Sat Sat
R3 is possible because birthday R4 is not possible because
of Ram, Hari and Sita can be Ram in set A can not have two
same and every people of set birthdays in set B. So, R4 is not
A has birthday in set B. So, R3 a function.
is fuction.
From the above example, it is conducted that function is a relation between a first
set and a second set such that each member of the first set corresponds to exactly
one member of the second set.
Let A and B be two non empty sets. A relation from A to B is said to be a function if every
element of set A is related with unique element of set B. The function from set A to set B
is denoted by f : A → B.
20 Infinity Optional Mathematics Book - 9
Image and pre-image of a function A f
1
Let A = {1, 2, 3} and B = {1, 8, 27, 64}. Consider a function 2 B
f = {(1, 1), (2, 8), (3, 27)} from A to B. In this function, 3 1
1, 2, 3 ∈ A associate with 1, 8, 27 ∈ B respectively. So, 1, 8, 27 8
are the images of 1, 2, 3 and 1, 2, 3 are pre-images of 1, 8, 27 27
respectively. 64
If A and B be two non empty sets and x ∈ A and y ∈ B then y is called image of x
and x is called pre-image of y.
Domain, co-domain and range of a function
In the above examples, the set of all pre-images of set A i.e. {1, 2, 3} is called domain
of function f and the set of all images i.e. {1, 8, 27} is called range of function f. The
set B i.e. {1, 8, 27, 64} is called co-domain of function f.
If f: A → B is a function from A to B then set A is called the domain and set B is called
co-domain of f. The set of elements of set B which are associated with the elements of set
A is called range of f.
Vertical line test
A relation can be presented in a graph. But all graphs may not be a function. To
examine whether the given graph is a function or not, we use vertical line test. For
this a vertical line is drawn. If it cuts the graph at one point only, the graph is a
function but if it cuts the graph at two or more points, the graph is not a function.
Consider the following two graphs.
YY
A
X' A O X
X' O X
B Y'
Y' In this figure, a vertical line
In this figure, a vertical line drawn cuts the graph at one
drawn cuts the graph at two point A. So, it is a function.
points A and B. So, it is not a
function.
Infinity Optional Mathematics Book - 9 21
WORKED OUT EXAMPLES
1. Identify the following relations are function or not.
f (b) g
(a) AB
AB ax
14
25 by
36
7 cz
Also, find the domain, Co-domain and range if it is a function.
Solution:
(a) f is a function because every element of set A associates with unique element
of set B. So,
Domain of f = {1, 2, 3}
Co-domain of f = {4, 5, 6, 7}
Range of f = {4, 5, 6}
(b) g is not a function because the element a of set A associates with two elements
x and y of B.
Exercise 1.5
Section 'A'
1. (a) Define function.
(b) Define domain of a function.
(c) Write the relation between range and co-domain of a function.
2. Identify the following relations are functions or not.
(a) R1 (b) A R2 (c) R3
A B BA B
11 14 14
24 25 25
39 36
6
37
22 Infinity Optional Mathematics Book - 9
(d) R4 (e) R5 (f) R6
A BA B AB
ax aw ax
by bx by
cz cz
y
cz
If it is a function, find
(i) function in set of ordered pairs (ii) its domain
(iii) its range (iv) its co-domain.
3. Let P = {2, 3, 4} and Q = {5, 6, 7}. Examine whether the following
relations define a function from P to Q.
(a) {(2, 5), (3, 6), (4, 7)} (b) {(2, 5 ), (2, 3), (3, 5), (4, 5)}
(c) {(3, 5), (4, 6)} (d) {(2, 5), (3, 5), (4, 6)}
If it is a function then
(i) represent in arrow diagram (ii) find its domain
(iii) find its range (iv) find its co-domain
4. Using vertical line test, examine which of the following graphs
represent function.
(a) Y (b) Y (c) Y
X' O X X' O X X' O X
Y' Y' Y'
(d) Y (e) Y (f) Y
X' O X X' O X X' O X
Y' Y' Y'
Infinity Optional Mathematics Book - 9 23
Types of Function f
AB
Onto function 1
In the figure alongside, f is a function. 24
The range of f = {4, 5} 35
The co-domain of f = {4, 5}
Here, range of f = co-domain of f. So, f is onto function.
A function f: A → B is said to be onto if every element of set B has at least one
pre-image in set A. In this case, range of f = B.
Into function f
In the figure alongside, f is a function. AB
The range of f = {4, 5} 14
The co-domain of f = {4, 5, 6} 25
Here, range of f ⊂ co-domain of f. 36
So, f is into function.
A function: f:A → B is said to be into if there is at least one element in B which
has no pre-image in A. In this case, range of f ⊂ B.
One to one function f f
In the figure alongside, f is a function. AB
14 AB
In this function, every element of set A has 25 14
36 25
different image in set B. That is 36
One to one and onto
image of 1 = 4 function 7
image of 2 = 5
image of 3 = 6 One to one and into
So, f is one to one function. function
A function f: A → B is said to be one to one function if different elements of A
have different images in B.
A f f
Many to one function 1 B AB
In the figure alongside, f is a function 2 4 14
from A to B. In this function, at least 3 25
two elements of set A associate with 6 36
single element of set B. That is
image of 1 = 4 Many to one and onto Many to one and into
image of 2 = 4 function function
image of 3 = 6
So, f is many to one function.
24 Infinity Optional Mathematics Book - 9
A function f : A → B is said to be many to one function if element of set B have
more than one pre-images in set A. In this case, f is not one to one function.
Additional Functions A f
B
Constant function 1
In the figure alongside, f is a function from A to B. In this function, 2 3
every element of set A i.e. 1, 2, 3 and 4 associates with constant 3
element 3 of set B. So, f is constant function. 4
A function f : A → B is said to constant function if f(x) = c where c is a constant.
Identity function A f
In the adjoining figure, f is a function from A to B. In this function, 1 B
images of 1, 2 and 3 are 1, 2 and 3 respectively. So, f is identity 2 1
function. 3 2
3
A function f : A → B is said to be identity function if f(x) = x. A
0 f
Linear function 1 B
Consider a function f : A → B defined as f(x) = 2x + 1. 2 1
3
When x = 0, f(0) = 2 × 0 + 1 = 0. 5
When x = 1, f(1) = 2 × 1 + 1 = 3.
When x = 2, f(2) = 2 × 2 + 1 = 5.
So, f is a linear function.
A function f : A → B is said to be linear function if f(x) = ax + b, where a and b
are constants.
WORKED OUT EXAMPLES
1. Is function f : R → R defined by f (x) = x2 for x ∈ R an onto function?
Give reason.
Solution:
Here, f(x) = x2, x ∈ {............ –2, –1, 0, 1, 2, ...........}.
Put x = – 2, f(–2) = (–2)2 = 4
Infinity Optional Mathematics Book - 9 25
Put x = – 1, f(– 1) = (–1)2 = 1 f
R
Put x = 0, f(0) = 02 = 0 R
–2
Put x = 1, f(1) = 12 = 1 –2 –1
Put x = 2, f(2) = 22 = 4 –1 0
1
Therefore, f is not onto function because all real numbers 0 2
1 3
4
in co-domain do not have corresponding pre-images. 2
Exercise 1.6
Section 'A'
1. (a) Define one to one function.
(b) Write one difference between onto and into function.
(c) Define identity function.
(d) Write one difference between constant and identity function.
2. Write the type of function (with reason) from the following arrow
diagrams and graphs.
(a) f1 (b) f2 (c) f3
A BA
a p1 BA B
b a
c q2 42 b
c
r3 5 3 d
6 4
7
(d) f4 (e) Y (f) Y
A B O X
1 1 2 X
2 1
3 2 O 12
3
Section 'B'
3. Mention the types of function by drawing arrow diagrams.
(a) f1 = {(1, a), (2, a), (3, b), (4, c)}
(b) f2 = {(1, 3), (2, 3), (3, 3)}
(c) g1 = {(a, x), (b, y), (c, z)}
(d) g2 = {(4, 4), (5, 5), (6, 6), (7, 7)}
4. (a) Is function f : R → R defined by f(x) = x2 – 1 into function? Give reasons.
(b) Is function f(x) = x2 + 1, x ∈ R onto function? Give reasons.
26 Infinity Optional Mathematics Book - 9
5. (a) Given that A = {1, 2, 3, 4} and B = {4, 5, 8, 10, 11, 12, 14, 15}. Is the
function f : A → B defined by f(x) = 3x + 2 one to one and into? Give
reasons.
(b) If a function f : A → B is defined by g(x) = 4x2 + 1 with A = {–2, 2, –3, 3}
and B = {17, 37, 65}, find the range of g. Is the function g many to one and
onto? Give reason.
Value of a Function
Consider a function f(x) = 3x – 2. Y
Put x = 1, f(1) = 3 × 1 - 2 = 1 9 X
The value of f(1) = 1 i.e. 1 is the image of 1. 8
Put x = 2, f(2) = 3 × 2 - 2 = 4 7
The value of f(2) = 4 i.e. 4 is the image of 2. 6
Put x = 3, f(3) = 3 × 3 – 2 = 7 5
The value of f(3) = 7 i.e. 7 is the image of 3. 4
3
2
So, x 1 2 3 1
f(x) 1 4 7
0 12 34
∴ f = {(1, 1), (2, 4), (3, 7)} f
AB
Here, the images f(1) i.e. 1, f(2) i.e. 4 and f(3) i.e. 7 are the 1 1
values of the given function. 24
37
For every x ∈ A, there exists y ∈ B such that y is the image of x. Then y is
the value of a function i.e. f(x) = y.
WORKED OUT EXAMPLES
1. If f(x) = 5x – 1 then find f(0), f(–2) and f(3).
Solution: Here, f(x) = 5x – 1
Put x = 0, f(0) = 5 × 0 – 1 = – 1
Put x = – 2, f(–2) = 5 × (–2) – 1 = – 11
Put x = 3, f(3) = 5 × 3 – 1= 14.
Infinity Optional Mathematics Book - 9 27
2. If f(x + 2) = 5x – 7 then find the values of f(x) and f(3).
Solution: Here, f(x + 2) = 5x – 7
Let, x + 2 = a
∴ x = a – 2
So, f(a) = 5(a – 2) – 7 = 5a – 10 – 7 = 5a – 17
Replacing a by x, we get,
f(x) = 5x – 17
Again, put x = 3, f(3) = 5 × 3 – 17 = 15 – 17 = – 2
{3. 2x – 1 for x > 0
Given that f(x) = x + 2 for x < 0
Find the values of f (–2), f 1 , f(0), f(3)
3
Solution: Here,
{ f(x) = 2x – 1 for x > 0
x + 2 for x < 0
For x = –2, f(x) = x + 2
∴ f(–2) = – 2 +2 = 0
For x = 13, f(x) = x + 2
1 1 7
∴ f 3 = 3 + 2 = 3
For x = 0, f(x) = 2x – 1
∴ f(0) = 2 × 0 – 1 = –1
For x = 3, f(x) = 2x – 1
∴ f(3) = 2 × 3 – 1 = 5
3. If f(x) = ax + b, f(–2) = 1 and f(3) = 9, find the values of a and b. Also find f(x).
Solution: Here, f(x) = ax + b, f(–2) = 1 and f(3) = 9
Put x = –2,
f(–2) = –2a + b
or, 1 = – 2a + b
∴ b = 2a + 1 ...(i)
Put x = 3,
f(3) = 3a + b
∴ 9 = 3a + b ... (ii)
Solving equations (i) and (ii), we get,
9 = 3a + 2a – 1
or, 10 = 5a
∴ a = 2
Put value of a in equation (i), we get,
b = 2 × 2 – 1 = 3
∴ f(x) = ax + b = 2x + 3
28 Infinity Optional Mathematics Book - 9
Exercise 1.7
Section 'A' f
1. (a) If f = {(1, 1), (2, 4), (3, 9)} then find AB
aw
(i) Pre-image of 4. (ii) Image of 3. bx
(b) From the given arrow diagram, find
cy
(i) Image of a and c. (ii) Pre-image of w. z
Section 'B'
2. ( a) If f(x) = 4x – 5, find the values of f(–3), f(0), f(2).
(b) If f(x) = 3x + 2, find f(–2), f 1 , f(6).
3
(c) If f(x) = 6x – 1 and f(x) = 5, find the value of x.
(d) Given that g(x) = 2x + 3. Find the pre-image of the image 7.
3. Find the range of the given function if
(a) f(x) = 5x – 1 ; domain = {–1, 0, 1}
(b) g(x) = x2 – 3; domain = {1, 2, 3, 4}
4. Find the domain of the given function if
(a) f(x) = 2x + 1; range = {3, 5, 7}
(b) g(x) = x2 + 1; range = {5, 10, 17}
5. If f(x) = 5x – 8, find the missing element in the following ordered pairs.
(a) (1, ......) (b) (3, ......) (c) (......, 2) (d) (......, 7)
6. (a) If f(x – 3) = 4x + 5 then find f(x) and f(5).
(b) If g(2x + 1) = 4x – 3 then find f(x) and f(–2).
7. (a) If f(x) = x + 3 then find f(h), f(x + h) and f(x + h) – f(h) (h ≠ 0)
h
(b) If f(x + 5) = f(x) + f(5), prove that f(0) = 0 and f(–5) = –f(5)
8. (a) If f(x) = 2x + 5 for x > 2
x – 3 for x < 2
then find the values of f(1), f(2), f(–3), f 5
2
(b) 2x + 1 for – 1 < x < 0
If f(x) = –3x + 1 for 0 < x < 3
–2x – 1 for 3 < x < 6
then find the values of f(-1), f(0), f(2), f(5).
Infinity Optional Mathematics Book - 9 29
Section 'C'
9. (a) If f(x) = ax + b, f(–1) = -1 and f(3) = 1, find the values of a and b. Also find
f(x).
(b) If g(x) = mx + c, g(2) = 5 and g(3) = 7, find the values of m and c. Also find
g(x).
1.2 Polynomials
Review
Discuss the following questions.
(a) Distinguish constant and variable with examples.
(b) In a term 4x2, numerical coefficient is .........., literal coefficient is ............
(c) Define algebraic expression with examples.
(d) Define linear function with examples.
1
(e) What is the power of x in x3 and 4 x3 ?
(f) What is the degree of a term 5x3y2?
(g) The degree of an expression 3x5 – 2x2y2 + 4 is .......
Introduction of Polynomials
Consider some examples of algebraic expression.
(i) 5x2 – 3x + 2
In this expression, the numerical coefficients are 5, –3 and 2 and powers of
variable in each term are 2, 1, 0. So, it is a polynomial.
2 1
(ii) 3x3 – 5 x2 + 4x – 2 –52, –1
2
In this expression, the numerical coefficients are 3, 4 and and the
exponents of variable in each term are 3, 2, 1, 0. So, it is a polynomial.
(iii) 2 x4 + 3x2 + 5 x
In this expression, the numerical coefficients are 2 , 3 and 5 and the
exponents of variable in each term are 4, 2 and 1. So, it is a polynomial.
4
(iv) 2x3 – x + 3x3/2
In this expression, the numerical coefficients are 2, –4 and 3 and the exponents
of variable in each term are 3, –1, 32. So, it is not a polynomial.
30 Infinity Optional Mathematics Book - 9
(v) 8x3 – 3 x2 + 2x 3 +1
2
8S,o,–i2t3i,s
In this expression, the numerical coefficients are 2 and 1 and the
exponents of variable in each term are 3, 2, 3 and 0. not a polynomial.
An algebraic expression in which the power of variable in each term is non
negative integer is called as polynomial. The general form of polynomial is
f(x) = a0 xn + a1xn - 1 + a2 xn - 2 + ........ + an where x is a variable, a0, a1, a2,
........, an are real numbers (constants) and n is non negative integer.
Each term of the polynomial contain numeral and literal coefficients. For example,
f(x) = 4x2 + 5. In this polynomial, 4 is numerical coefficient and x2 is literal coefficient
of a term of 4x2.
Degree of a Polynomial
If the polynomial is of one variable, the highest exponent (power) of the variable
used in the polynomial is called as the degree of the polynomial. For examples, f(x)
= 5x3 – 4x2 + 3 is a polynomial. In this polynomial, the highest power of the variable
is 3. So, the degree of the polynomial f(x) is 3.
If the polynomial with two or more variables then the highest degree of a term
among the terms of a polynomial is called the degree of the polynomial. For example,
f(x, y) = 3x3y2 + 2x2y + 5xy. In this polynomial, the degree of 3x3y2 = 3 + 2 = 5, the
degree of 2x2y = 2 + 1 = 3 and the degree of 5xy = 1 + 1 = 2. The highest degree of
the first term is 5. So, the degree of the polynomial f(x, y) is 5.
If the degree of the terms of a polynomial are arranged either in ascending or in
descending order then the polynomial is said to be in standard form.
For examples,
f(x) = x4 – 3x3 + 2x2 + 6x – 1 (in descending order)
g(x) = 5 – 2x2 + 4x3 – 8x4 (in ascending order)
Types of Polynomial
The types of polynomial based on the number of terms are given below.
Types of polynomials No. of terms Examples
Monomial 1 f(x) = 3x
Binomial 2 f(x) = 4x – 1
Trinomial 3 f(x) = 3x2 – 9x + 3
Infinity Optional Mathematics Book - 9 31
The types of polynomial based on the degree of term are given below.
Types of polynomials Degree of polynomial Examples
Constant polynomial 0 g(x) = 3 = 3.x0
Linear polynomial 1 g(x) = 4x + 3
Quadratic polynomial 2 g(x) = 2x2 – 5x + 2
Cubic polynomial 3 g(x) = x3 + 2x2 + 5
Biquadratic polynomial 4 g(x) = 6x4 – 5x2 + 4
Equal Polynomials
Two polynomials are said to be equal if the coefficients and degree of their
corresponding terms are equal. For example,
f(x) = 4x3 – 9 x2 + 4x + 16 and g(x) = 8 x3 – 3x2 + 4x + 1.6
3 10 2
Here, the coefficient and degree of each corresponding terms of f(x) and g(x) are
equal. So, f(x) = g(x).
WORKED OUT EXAMPLES
1. In the given monomial 5x2y, find
(a) numerical coefficient
(b) literal coefficient of y
(c) its degree
Solution:
The given monomial is 5x2y.
(a) Its numerical coefficient = 5
(b) Literal coefficient of y = x2
(c) Its degree = 2 + 1 = 3
2. Identify which of the following algebraic expressions are polynomial.
Give reasons.
(a) f(x) = 4x3 – 3x2 + 2 (b) g(x) = 3x2 – 1 + 3
x
Solution:
(a) f(x) is a polynomial because the powers of x in each terms of f(x) are
whole numbers.
(b) g(x) is not polynomial because g(x) contains a term 1 i.e. x-1 where – 1 is
x
not a non negative integer.
32 Infinity Optional Mathematics Book - 9
Exercise 1.8
Section 'A'
1. Identify which of the following algebraic expressions are polynomial.
Give reasons. 1
x
(a) 4x2 – 3x + 2 (b) 2x2 – 5x +
(c) 3 x + 4 6 (d) 3x3 + 5x2 – 2x + 1
3 6
(e) x4 – 5 x3 + 2x – (f) 6x5 + 4x3 – 3x 5 + 2x – 9
2. Find the numerical and literal coefficient of the following terms.
(a) 3x3 (b) x2
(c) 3 x (d) –4x2
3. Find the numerical coefficient of the following terms.
(a) 2xy (b) – 3x2y
(c) 5x2yz (d) 58 xyz
Also find the literal coefficient of y in each of the above terms.
4. Find the degree of the following polynomials.
(a) f(x) = 3xy2 (b) f(x) = 4x2yz 1
3
(c) g(x) = 4x5 – 3x3 + 5 (d) g(x) = 6x3 – x2 + 4x6 + 5
(e) h(x, y) = 5x3y3 – 3x2y + 4xy3 (f) h(x, y) = 3x4y + xy3 – 5x2y3
5. Write the following polynomials in descending and ascending order.
(a) 3x3 + 6x4 + 4 – 2x2 (b) 2x2 – 8 + x + 3x3
(c) x3 + 2x(x – x3) – 4x2 (d) 5 + 3x4 – 3x3 + 5x2
6. Classify the following polynomials on the basis of their number of
terms.
(a) 3x (b) 4x2 – 6x + 2
4
(c) 2x – 5 (d) 5 x –3
7. Classify the following polynomials on the basis of the degree of their
terms.
(a) 4x + 1 (b) 3x3 – 2x2 + 5x + 6
(c) 10 (d) 3x2 – 8
Infinity Optional Mathematics Book - 9 33
8. Examine whether the following pair of polynomials are equal.
(a) f(x) = 4x3 – 2x2 + 5x + 2 and g(x) = 5x – 2x2 + 4x3 + 2
8 6
(b) f(x) = 3x2 + 4x3 – 2x + 2 and g(x) = 4x3 – 2 x2 – 2 x + 2
(c) f(x) = 0.25x4 – 6x3 + 3 1 –
4 and g(x) = 4 x4 6x3 + 0.75
Section 'B'
9. Find the value of k if the following polynomials are equal.
(a) f(x) = 3x3 – (k + 1)x2 + 3 and g(x) = 3x3 + 4x2 + 3
(b) f(x) = 6x4 – ax3 + 5x – 4 and g(x) = 6x4 – 2x3 + 5x + (2k – a)
10. (a) If f(x) = 5x + 4, g(x) = 7x and f(x) = g(x), then find the value of x.
(b) Find the value of x if f(x) = 5x2 – 3x + 2 and g(x) = x2 + 2x + 8 are equal.
Operations on Polynomials
(a) Identify the like and unlike terms from the following terms.
5x3, –4x2, 8, 3 x3, x3, 5x, 6x0 , 3 x2, – 85, 6x3
2 x
(b) Using xa × xb = xa+b, simplify the following 39
(i) x5 × x4 = ......... (ii) x–3 × x5 = ..........
(iii) x4 × x4 = .........
Two or more polynomials can be added, subtracted and multiplied.
Addition and subtraction of polynomials
Let g(x) = 3x3 – 4x2 – 7x + 2 and h(x) = 4x3 + 2x2 – 2x be the polynomials. The
addition of g(x) and h(x) can be done as follows:
g(x) + h(x) = (3x3 – 4x2 – 7x + 2) + (4x3 + 2x2 – 2x)
= (3 + 4)x3 + (–4 + 2)x2 + (–7 – 2) x + 2
= 7x3 – 2x2 – 9x + 2
Similarly, the subtraction of f(x) and g(x) can be done as follows:
g(x) – h(x) = (3x3 – 4x2 – 7x + 2) – (4x3 + 2x2 – 2x)
= 3x3 – 4x2 – 7x + 2 – 4x3 – 2x2 + 2x
= (3 – 4)x3 + (–4 – 2)x2 + (– 7 + 2) x + 2
= – x3– 6x2 – 5x + 2
34 Infinity Optional Mathematics Book - 9
Laws of Addition of Polynomials
Addition of polynomials satisfies the following properties.
Closure property
Let f(x) and g(x) be two polynomials. Then f(x) + g(x) is also a polynomial. That
is the sum of two polynomials is also a polynomial. For example,
If f(x) = 4x2 + 5x + 3 and g(x) = x2 + 3x + 1
Then, f(x) + g(x) = (4x2 + 5x + 3) + (x2 + 3x + 1)
= (4 + 1)x2 + (5 + 3)x + (3 + 1)
= 5x2 + 8x + 4 is also a polynomial.
Commutative property
Let f(x) and g(x) be two polynomials. Then f(x) + g(x) = g(x) + f(x). For example,
If f(x) = 3x2 + 2x + 1 and g(x) = 4x2 + 2x + 3, then
f(x) + g(x) = (3 + 4)x2 + (2 + 2)x + (1 + 3) = 7x2 + 4x + 4
g(x) + f(x) = (4 + 3)x2 + (2 + 2)x + (3 + 1) = 7x2 + 4x + 4
Hence, f(x) + g(x) = g(x) + f(x).
Associative property
Let p(x), q(x) and r(x) be three polynomials. Then,
[p(x) + q(x)] + r(x) = p(x) + [q(x) + r(x)]. For example,
If p(x) = 2x2 + x + 3, q(x) = x2 + 3x + 5 and h(x) = 3x2 + 5x + 1 then,
[p(x) + q(x)] + h(x) = [(2x2 + x + 3) + (x2 + 3x + 5)] + (3x2 + 5x + 1)
= [(2 + 1)x2 + (1 + 3)x + (3 + 5)] + (3x2 + 5x + 1)
= (3x2 + 4x + 8) + (3x2 + 5x + 1)
= (3 + 3)x2 + (4 + 5) x + (8 + 1)
= 6x2 + 9x + 9
p(x) + [q(x) + h(x)] = (2x2 + x + 3) + [(x2 + 3x + 5) + (3x2 + 5x + 1)]
= (2x2 + x + 3) + [(1 + 3)x2 + (3 + 5) x + (5 + 1)]
= (2x2 + x + 3) + (4x2 + 8x + 6)
= (2 + 4) x2 + (1 + 8) x + (3 + 6)
= 6x2 + 9x + 9
Hence, [p(x) + q(x)] + h(x) = p(x) + [q(x) + h(x)].
Existence of additive identity
Let 0(x) be a zero polynomial and g(x) be a polynomial.
Then, g(x) + 0 = 0 + g(x) = g(x). For example,
If g(x) = x2 + 3x + 2 and 0(x) = 0.x2 + 0.x + 0, then
Infinity Optional Mathematics Book - 9 35
g(x) + 0 = (1 + 0)x2 + (3 + 0) x + (2 + 0) = x2 + 3x + 2 = g(x)
0 +g(x) = (0 + 1) x2 + (0 + 3) x + (0 + 2) = x2 + 3x + 2 = g(x)
Hence, g(x) + 0 = 0 + g(x) = g(x)
Additive inverse
For any polynomial p(x), there exists a polynomial –p(x) such that p(x) +
[–p(x)] = [–p(x)] + p(x) = 0. Here, –p(x) is called the additive inverse of p(x).
For example,
If f(x) = 5x2 + 6x + 2 then –p(x) = –5x2 – 6x – 2
p(x) + [–p(x)] = (5x2 + 6x + 2) + (–5x2 – 6x – 2)
= (5 – 5)x2 + (6 – 6)x + (2 – 2)
= 0.x2 + 0.x + 0
=0
Similarly, [–p(x)] + p(x) = 0
Hence, p(x) + [–p(x)] = [–p(x)] + p(x) = 0
Multiplication of polynomials
Let f(x) = x2 – x + 1 and g(x) = x2 + x + 1 be two polynomials. The multiplication
of f(x) and g(x) can be done as follows.
f(x). g(x) = (x2 + x + 1).(x2 – x + 1)
= x2(x2 – x + 1) + x(x2 – x + 1) + 1(x2 – x + 1)
= x4 – x3 + x2 + x3 – x2 + x + x2 – x + 1
= x4 + x2 + 1
WORKED OUT EXAMPLES
1. If f(x) = 3x2 – 2x +1, g(x) = x2 + 5x + 3 and h(x) = 5x2 – x + 4 then find f(x)
+ g(x) – h(x).
Solution;
Here, f(x) = x2 – 2x + 1
g(x) = x2 + 5x + 3
h(x) = 5x2 – x + 4
f(x) + g(x) – h(x) = (x2 – 2x + 1) + (x2 + 5x + 3) – (5x2 – x + 4)
= x2 – 2x + 1 + x2 + 5x + 3 – 5x2 + x – 4
= (1 + 1 – 5)x2 + (– 2 + 5 + 1) x + (1 + 3 – 4)
= – 3x2 + 4x + 0
= – 3x2 + 4x
36 Infinity Optional Mathematics Book - 9
2. Given that two polynomials p(y) = 3y2 – 2y + 1 and q(y) = 5y2 – y + 4.
What should be added to p(y) to get polynomial q(y)?
Solution:
Here, p(y) = 3y2 – 2y + 1
q(y) = 5y2 – y + 4
Let the polynomial added to p(y) be k. Then,
p(y) + k = q(y)
or, k = q(y) – p(y)
or, k = (5y2 – y + 4) – (3y2 – 2y + 1)
or, k = 5y2 – y + 4 – 3y2 + 2y – 1
or, k = (5 – 3)y2 + (– 1 + 2) y + (4 – 1)
or, k = 2y2 + y + 3
∴ Required polynomial is 2y2 + y + 3.
3. If f(x) = 2x – 1, g(x) = 4x2 + 2x + 1 and h(x) = 10x3 + 2x2 + 3x – 6 then find
h(x) – [ f(x) × g(x)]
Solution:
Here, f(x) = 2x – 1
g(x) = 4x2 + 2x + 1
h(x) = 10x3 + 2x2 + 3x – 6
Now, f(x) × g(x) = (2x – 1) × (4x2 + 2x + 1)
= 2x(4x2 + 2x + 1) – 1(4x2 + 2x + 1)
= 8x3 + 4x2 + 2x – 4x2 – 2x – 1
= 8x3 – 1
Again, h(x) – [f(x) × g(x)] = 10x3 + 2x2 + 3x – 6 – (8x3 – 1)
= 10x3 + 2x2 + 3x – 6 – 8x3 + 1
= 2x3 + 2x2 + 3x – 5
Exercise 1.9
Section 'B'
1. Find f(x) + g(x) from the following polynomials.
(a) f(x) = 4x2 – 5x + 7 and g(x) = 5x2 – 4x + 3
(b) f(x) = 6x3 + 4x2 – 5x + 2 and g(x) = x3 – x2 – x + 1
(c) f(x) = 5 x3 + 2 x2 – x + 1 and g(x) = 12 x3 – 35x2 + 2x – 2
2 5 3 3
3 3 1 2 5 3 1 1
(d) f(x) = 2 x + 4 – 5 x2 + 3 x3 and g(x) = 3 x + 5 x2 + 3 x3 – 2
Infinity Optional Mathematics Book - 9 37
2. Find p(x) – q(x) from the following polynomials.
(a) p(x) = 3x2 + 6x – 8 and q(x) = 5x2 – 4x + 10
(b) p(x) = 4x2 + 4x – 3 and q(x) = 2x2 – 2x + 3
(c) p( x) = 6x3 – 4x2 + 7x – 10 and q(x) = 2x2 – 3x + 3
(d) p(x) = 5x4 – 2x3 + 5x + 3 and q(x) = 4x3 + 5x2 – 10x
3. (a) If f(y) = 2y3 – 3y2 + 7y + 5, g(y) = 4y2 – 4y + 6 + 3y3 and h(y) = 3y2 + y then
find (i) f(y) + g(y) + h(y) (ii) f(y) – g(y) + h(y)
(b) If p(x) = 4x2 – 2x + 8, q(x) = 23x2 + 3 x – 5 and r(x) = 1 x – 5 then find
2 2
(i) p(x) + q(x) – r(x) (ii) p(x) – q(x) – r(x)
4. (a) What should be added to f(x) to get g(x) where f(x) = 2x2 – 9x + 6 and
g(x) = x2 + 6x – 5?
(b) Given that f(x) = 5x4 – 3x3 + 6 and g(x) = 3x4 – 4x2 + 2x3 – x. What should
be subtracted from f(x) to get g(x).
5. Find f(x) × g(x) from the following polynomials
(a) f(x) = x + 3 and g(x) = x – 3
(b) f(x) = x2 + 2x + 1 and g(x) = x – 2
(c) f(x) = x2 + x + 1 and g(x) = x2 – x + 1
(d) f(x) = 5x2 – 2x – 3 and g(x) = x2 + 4x – 1
Section 'C'
6. If f(x) = 3x4 – 8x3 + 2x2 – 5x + 1, g(x) = x4 + 3x3 – 4x2 + 2x – 5 and
h(x) = 8x4 – x3 – x2 – 4x + 9 then verify the following.
(a) f(x) + g(x) = g(x) + f(x)
(b) [f(x) + g(x)] + h(x) = f(x) + [g(x) + h(x)]
7. (a) If f(x) = (x – 2), g(x) = (x2 + 5x – 3) and h(x) = (4x3 – 2x2 + 8x + 9) then find
the polynomial [f(x) × g(x)] + h(x)
(b) If p(x) = (3x + 1), q(x) = (9x2 – 3x + 1) and r(x) = (30x3 – 20x2 + 10x – 5 )
then find r(x) – [p(x) × q(x)]
38 Infinity Optional Mathematics Book - 9
1.3 Sequence and Series
Review
Discuss the following questions.
(a) If f(x) = 2x + 1 then find the functional values of f(1), f(2), f(3), f(4), f(5).
(b) If f(x) = x2 – 2x + 5 then find the functional values of f(1), f(2), f(3), f(4).
(c) Observe the pattern of figures given below.
, , , , ...
(i) Add one more figure of the above pattern.
(ii) Express the above pattern in terms of numbers.
Sequence
Let us consider the following set of numbers.
(a) 2, 5, 8, 11, ...., ...., .... (b) 20, 15, 10, 5, , ...., ...., ....
(c) 3, 6, 12, 24, ...., ...., .... (d) 1, 5, 8, 12, , ...., ...., ....
In an example (a), each number (except first) is obtained by adding 3 to the preceding
term. So, the number next to 11 is 11 + 3 = 14.
In the example (b), each number is less than the preceding number by 5. So, the
number next to 5 is 5 – 5 = 0.
In the example (c), each number is double of the preceding number. So, the number
next to 24 is 24 × 2 = 48.
In the example (d), there is no certain rule of the formation of its each element. So,
we can not say the number next to 12.
The set of numbers in (a), (b) and (c) follow certain rule. So, they form sequence. But
the set of numbers in (d) does not follow any rule. So it does not form a sequence.
A set of numbers each of which is formed according to a certain rule is called a
sequence. Each element of a sequence is called its term.
Infinity Optional Mathematics Book - 9 39
Types of Sequence
A sequence may contain finite or infinite number of terms.
Finite sequence
A sequence having finite number of terms is called finite sequence. In the finite
sequence, there is a last term. For example, 2, 4, 6, 8, 10, 12 is a finite sequence.
Infinite sequence
A sequence having infinite number of terms is called infinite sequence. In the
infinite sequence, there is no last term. For example, 1, 3, 9, 27, 81, ........ is an
infinite sequence.
WORKED OUT EXAMPLES
1. Observe the pattern of figures given below.
,, , , ..., ..., ...
(a) Add one more figure with the same pattern.
(b) Write the numerical sequence of the given pattern of figure.
Solution:
(a) The 5th figure according to the given pattern is
(b) The numerical sequence of the given pattern of figures is 1, 4, 7, 10, .....,
....., ......
2. Write down next two terms in the following sequences.
(a) 1, 5, 9, 13, 17, ....., ....., ..... (b) 1 , 1, 1, ...., ....
Solution: 16 8 4
(a) Here, the given sequence is 1, 5, 9, 13, 17, ......, .....
Since each term of the given sequence is obtained by adding 4 to the preceding
term. So, the next two terms of the given sequence are 17+4=21 and 21+4= 25.
(b) Here, the given sequence is 1 , 1, 1, ......, ......
16 8 4
Since each term of the sequence is obtained by multiplying 2 to the preceding
40 Infinity Optional Mathematics Book - 9
term. So, the next two terms of the given sequences are
1 × 2 = 2 and 2 × 2 = 4
4
Exercise 1.10
Section 'A'
1. (a) Define sequence with an example.
(b) Write one example of finite sequence.
(c) Write one example of infinite sequence.
2. Identify the following set of numbers form a sequence or not. Give
reasons.
(a) 2, 4, 6, 8, ....., ..... (b) 81, 27, 9, 3, ....., .....
(c) 5, 8, 10, 15, ....., ..... (d) 10, 5, 0, – 5, ....., .....
(e) 1, 3, 5, 7 , ....., ..... (f) 1, 3, 6, 24, 48, ....., .....
2 4 8 16
3. Identify the following sequences are finite or infinite? Give reasons.
(a) 1, 3, 5, 7, 9 (b) 6, 3, –2, – 6, ......, .......
(c) 5, 10, 15, 20, ....., ..... (d) 16, 8, 4, 2, 1.
Section 'B'
4. Draw one more shape in the following sequence of figures and
represent in a numerical sequence.
(a) , , , , ..., .... (b) , , , ..., ...
(c) , , , ..., ... (d) , , , ..., ....
5. Find the next three terms of each of the following sequences.
(a) 3, 7, 11, 15, ......., ....... (b) 4, 2, 0, –2, ......., .......
(c) 15, 5, 5, 5, ......., ....... (d) 1, 1, 1, 1, ......., .......
39 248
Infinity Optional Mathematics Book - 9 41
General Term of a Sequence
Consider a sequence 3, 5, 7, 9, ....., ...... . The terms 3, 5, 7, 9 ...... are called first,
second, third, fourth terms, ...... of the sequence 3, 5, 7, 9, ......, ...... respectively.
Can you find the 5th term of the given sequence? Can you find 12th term of the
sequence? Can you find 50th term of that sequence? Can you find 80th term of
that sequence? The 5th term and 12th term of the sequence can be obtained just
by counting numbers. But getting 50th term and 80th term of that sequence are
difficult by counting numbers. For these, at first we find the rule of the sequence as
the specific term for nth term.
Let t1, t2, t3, ......, ......, tn, ......, ...... be a sequence then the nth term is denoted by tn.
The nth term of the given sequence is also known as the general term. The general
term of the sequence is the rule by which each term of the sequence can be obtained.
The general term of the above sequence can be obtained as follows:
t1 = 3 = 2 + 1 =2×1+1
t2 = 5 = 4 + 1 =2×2+1
t3 = 7 = 6 + 1 =2×3+1
t4 = 9 = 8 + 1 =2×4+1
....... .......
.......
∴ tn = 2 × n + 1 = 2n + 1
Alternatively,
t2 – t1 = 2 ∴ t2 = t1 + 2
t3 – t2 = 2, ∴ t3 = t2 + 2
t4 – t3 = 2 ∴ t4 = t3 + 2
.............
............
∴ tn = tn – 1 + 2
General term of linear sequence
Let tn = an + b, where a and b are constants.
Put n = 1, t1 = a + b
Put n = 2, t2 = 2a + b
Put n = 3, t3 = 3a + b
Put n = 4, t4 = 4a + b
............ ............
42 Infinity Optional Mathematics Book - 9
i.e., (a + b), (2a + b), (3a + b), (4a + b), ......, ......
First difference a a a ......
In this sequence, the first difference is same i.e. 'a'. So, first difference = a and
first term = a + b. For example, consider a sequence
1, 3, 5, 7, 9, ......., .......
2 22 2
Here, the first difference of the sequence is same i.e. 2.
So, the general term of that sequence is in the from of tn = an + b.
General term of quadratic sequence
Let tn = an2 + bn + c, where a, b and c are constants.
Put n = 1, t1 = a + b + c
Put n = 2, t2 = 4a + 2b + c
Put n = 3, t3 = 9a + 3b + c
Put n = 4, t4 = 16a + 4b + c
.............. ..............
i.e. ( a + b + c), (4a + 2b + c), (9a + 3b + c), (16a + 4b + c), ......., .......
First difference 3a + b 5a + b 7a + b .......
Second difference 2a 2a .......
In this sequence the second difference is same i.e. 2a.
So, second difference = 2a
First term of first difference = 3a + b
First term of the sequence = a + b + c,
For example, consider a sequence 1, 7, 17, 31, ......., .......
Here, 1, 7, 17, 31, ......., .......
First difference 6 10 14 ....
Second difference 4 4 ......
Here, the second difference of the sequence is same. So, the general term of
that sequence is in the form of tn = an2 + bn + c.
Infinity Optional Mathematics Book - 9 43
General term of other sequence
Consider a sequence 1, 8, 27, 64, ........, .........
Let's check the first and second difference of the above sequence.
1, 8, 27, 64, ......., .......
First difference 7 19 37 ......
Second difference 12 16
Here, the first and second differences of the sequence are not same. So, the general
term of that sequence is not in the form of tn = an + b and tn = an2 + bn + c. In such
situation, the general term of the sequence is obtained by observing and guessing.
WORKED OUT EXAMPLES
1. Find the general term of the following sequences.
(a) 1, 3, 5, 7, ......., ........ (b) 1, 7, 17, 31, ......., ........
(c) 1, 8, 27, 64, ......., ........
Solution:
(a) The given sequence is
1, 3, 5, 7, ......., .......
First difference 2 2 2
The general term of the sequence is tn = an + b.
Here, first difference = 2
∴ a = 2
Again, first term = 1
or, a + b = 1
∴ b = –1
Hence, tn = an + b = 2n –1.
(b) The given sequence is
1, 7, 17, 31, ......., .......
First difference 6 10 14 ......
Second difference 4 4
So, the general term of the sequence is tn = an2 + bn + c.
44 Infinity Optional Mathematics Book - 9
Here, second difference = 4.
or, 2a = 4.
∴ a = 2.
First term of first difference = 6.
or, 3a + b = 6.
or, 3 × 2 + b = 6.
∴ b = 0.
First term of the sequence = 1
or, a + b + c = 1
or, 2 + 0 + c = 1
∴ c = –1.
Hence, tn = an2 + bn + c = 2n2 + 0 × n – 1 = 2n2 – 1.
(c) The given sequence is
1, 8, 27, 64, ......., .......
H ere,tttt2431====826147====((2((14)3)3))333
...... ......
The general term of the given sequence (tn) = n3
2. Find the general term of the sequence –41, 94, 1–96, 1256, ....., .....
Solution:
The given sequence is -1, 4, -9 , 16, ....., .....
4 9 16 25
Here, t1 = -1 = (–1)1 × 1 2 = (–1)1 × 12
4 2 1+1
t2 = 4 = (–1)2 × 2 2 (–1)2 × 22
9 3 2+1
=
t3 = – 9 = (–1)3 × 3 2 (–1)3 × 32
16 4 3+1
=
t4 = 16 = (–1)4 × 4 2 (–1)4 × 42
25 5 4+1
=
............ ............. .............
∴ The general term of the sequence (tn) = (–1)n n 1 2 (–1)n (n n2
n+ + 1)2
=
Infinity Optional Mathematics Book - 9 45
Exercise 1.11
Section 'A'
1. (a) What is the general term of a linear sequence?
(b) What is the general term of a quadratic sequence?
Section 'B'
2. From the given nth term of the sequences, find first four terms in each
of the following.
(a) tn = 2n + 1 (b) tn = 4n – 3
(c) tn = n2 + 3 (d) tn = n2 + 2n – 1
(e) tn = (–1)n + 1 n +1 (f) tn = n+ 3
n 2n
3. Find the first four terms of the given sequence using the following
relations.
(a) a1 = 2, an + 1 = 3an (b) a1 = 1, an – 1 = 3an
(c) a1 = – 1, an – 1 = 2an – 1 (d) a1 = 1, a2 = 2, an + 2 = an + 1 + an
Section 'C'
4. Find the nth term of the following sequences and use it to find 15th
term.
(a) 2, 6, 10, 14, ......., ....... (b) 8, 3, – 2, –7, ......., .......
(c) 4, 7, 12, 19, ......., ....... (d) 2, 7, 14, 23, ......., .......
(e) 31, 54, 1, 190, ......., ....... (f) 52, -4 , 69, 1-81, ......., .......
7
Series
Consider a sequence 2, 4, 8, 16, 32, ......., ........ Then the symbolic expression
2 + 4 + 8 + 16 + 32 + ........ is corresponding series of the given sequence
2, 4, 8, 16, 32, ......., .......
A series of a sequence is the representation of the terms of a sequence in the sum or
addition.
A series is finite or infinite according as the number of terms in the corresponding
sequence is finite or infinite. In general, if a1, a2, a3, ......, ......, an be a finite sequence
then its corresponding finite series a1 + a2 + a3 + ...... + an. In the same way, if a1,
a2, a3, ......., an, ....... be an infinite sequence, then its corresponding infinite series is
a1 + a2 + a3 + ...... + an + ....... . For example, 1, 3, 5, 7, ......, 15 is a finite sequence.
46 Infinity Optional Mathematics Book - 9
Then 1 + 3 + 5 + 7 + ....... + 15 is the finite series corresponding to 1, 3, 5, 7, ......., 15.
Similarly, 5, 10, 15, 20, ....... is an infinite sequence. Then 5 + 10 + 15 + 20 + ....... is
the infinite series corresponding to 5, 10, 15, 20, ....... .
Sigma notation
Consider a finite series 1 + 2 + 4 + 8 + ....... to 10 terms.
Here, t1 = 1 = 20 = 21 – 1
t2 = 2 = 21 = 22 – 1
t3 = 4 = 22 = 23 –1
t4 = 8 = 23 = 24 – 1
......... ......... .........
∴ tn = 2n – 1
Thus, the sum of first 10 terms is represented as
10
Σ 1 + 2 + 4 + 8 + ....... to 10 terms =
n=1 2n–1
If a1, a2, a3, ........, ......., an be a sequence of n terms then its corresponding series is
a 1 + a2 + a3 + ....... + an. This sum can be written in sigma rotation as:
n
Σ a1 + a2 + a3 + ....... + an = k = 1 ak.
Partial Sum
Consider a series 1 + 3 + 5 + 7 + 9 + 11 + 13. Then
the sum of first term = S1 = 1
the sum of first two terms = S2 = 1 + 3 = 4
the sum of first three terms = S3 = 1 + 3 + 5 = 9
the sum of first four terms = S4 = 1 + 3 + 5 + 7 = 16
the sum of first five terms = S5 = 1 + 3 + 5 + 7 + 9 = 25
the sum of first six terms = S6 = 1 + 3 + 5 + 7 + 9 + 11 = 36
the sum of first seven terms = S7 = 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49.
If a1, a2, a3, ......., an be the n terms of a sequence and Sn be their sum, the sum of n
terms of the given sequence is Σn
Sn = a1 + a2 + a3 + ....... + an = k = 1 ak.
4n
Σ Σ Then, S4 = a1 + a2 + a3 + a4 = k = 1 ak is said to be partial sum of k = 1 ak.
Infinity Optional Mathematics Book - 9 47
Note
Sn – Sn - 1 = (a1 + a2 + a3 + ....... + an-1 + an) – (a1 + a2 + a3 + ....... + an-1) = an
For example, S6 – S5 = a6
WORKED OUT EXAMPLES
1. Find the general term of -2 + 4 – 6 + 8 – 10 + 1125. Also write it with
5 7 9 11 13
sigma rotation.
Solution:
The given series is -2 + 4 – 6 + 8 – 10 + 12
5 7 9 11 13 15
Here, t1 = – 2 = (–1)1 × 2×1
5 2×1+3
t2 = 4 = (–1)2 × 2×2
7 2×2+3
t3 = – 6 = (–1)3 × 2×3
9 2×3+3
t4 = 8 = (–1)4 × 2×4
11 2×4+3
t5 = – 10 = (–1)5 × 2×5
13 2×5+3
............ ................
∴ tn = (–1)n 2×n = (–1)n 2n
2×n+3 2n + 3
Hence, the general term (tn) = (–1)n 2n
2n + 3
66 2n
Σ Σ -2 4 6 8 10 12 2n + 3
Again, 5 + 7 – 9 + 11 – 13 + 15 = n = 1 tn = n = 1 (–1)n .
2. Evaluate: Σ4 3n – 2
n=1
Solution: Σ4 (3n – 2) = (3 × 1 – 2) + (3 × 2 – 2) + (3 × 3 – 2) + (3 × 4 – 2)
n=1
48 Infinity Optional Mathematics Book - 9
= (3 – 2) + (6 – 2) + (9 – 2) + (12 – 2)
= 1 + 4 + 7 + 10
= 22
Exercise 1.12
Section 'A'
1. (a) Define series with an example.
(b) Write one difference between sequence and series.
2. Identify which of the following are sequence or series.
(a) 2 + 4 + 6 + 8 + ....... (b) (3n – 1), n ∈ N
(c) -31 , 25, -3 , 49, -5 (d) Σ6 (5n + 3)
7 11 k=1
3. Identify which of the following series are finite or infinite.
(a) 5 + 8 + 11 + 14 + 17 (b) 1 + 2 + 4 + ...... + 128
(c) Σ10 k 1 (d) – 1 + 2 – 3 + 4 – 5 + 6 – .......
k=3 2k +
4. (a) If S8 = 115 and S7 = 100 then find the value of t8.
(b) If S6 = 56 and S7 = 65 then find the value of t7.
Section 'B'
5. Evaluate: 8
5
(b) Σ (5n – 4)
(a) Σ (2n + 3) n=3
n=1 6
4
(d) Σ k (k + 2)
Σ (c) (k2 + 1) k=3
k=1 4
Σ6 1 Σ(f) 1
n + n n = 1 (–1)n + 1 2n
(e)
n=2
Section 'C'
6. Express the following series by using sigma rotation.
(a) 1 + 3 + 5 + 7 + 9 + 11 + 13 (b) – 15 – 10 – 5 + 0 + 5 + 10
(c) – 2 + 4 – 6 + 8 ....... to 10 terms (d) 3 + 9 + 27 + 81 + ....... to 8 terms
(e) 51 – 3 + 5 – 7 + 291 (f) 2 + 6 + 26 + 126 + 626
9 13 17
Infinity Optional Mathematics Book - 9
49
UNIT
2 LIMIT
2.1 Limit
Review
Discuss the following questions.
(a) Find the domain and range of a function f = {(1, 5), (2, 8), (3, 13)}
(b) If f(x) = 3x + 2 then find f(2), f(5) and f(20).
(c) Give examples of increasing and decreasing sequences.
(d) What is the value of 1 when the value of x is increasing?
(e) x a height of 48m and it rebounds with
A ball is dropped from the height of 1 of
3
previous height. Make a sequence from the given information if the ball is left
until it comes to rest.
Limit of a Sequence of Numbers
Consider the following sequence of numbers
(a) 2, 4, 6, 8, ......, ....... (b) 1, 3, 9, 27, 81, ......, .......
(c) 42, 30, 18, 6, –6, –18, ......, ....... (d) 12, 41, 81, 116, ......, .......
(e) 0.9, 0.99, 0.999, 0.9999, , ......, .......
(f) 0.01, 0.001, 0.0001, 0.00001, ......, .......
Can you find two more terms in each of above sequences?
How many terms are there in each of the above sequence?
Can you find the last term of each of the above sequence?
Consider a general term tn = 3nn. Then the sequence formed from the given general
term is 19, 237, 841, ......, .......
Is this sequence increasing or decreasing? Can you find the last term of this
sequence? Consider a line segment having length 12cm on a number line.
50 Infinity Optional Mathematics Book - 9
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