ALGEBRA 1UNIT
Estimated Teaching Periods : 33
Competency
To solve the problems by demonstrating the relation and function.
To comprehend the concept of sequence and series and solve the related problems.
At the end of this unit, the students will be able to: Learning Outcomes
introduce a relation and a function, differentiate between them and write the relation of function in symbol.
identify the types of function and represent in different methods (table graph, mapping diagram, function machine).
introduce polynomial on the basis of its types (quadratic, cubic and biquadratic) and add and subtract the
polynomials.
evaluate the general term of the sequence and series according to its pattern and express the series in
∑ (sigma) notation.
What do you learn?
Relation and Function
� Ordered pair and Cartesian product
� Introduction to Relation with its domain, range and inverse
� Representations and types of relations
� Introduction to function with its image, pre–image, domain, range and co–domain
� Representations and types of function
� Vertical line test for function and functional value
Polynomial
� Introduction to polynomial, its degree and equal polynomials
� Operations on polynomials (addition, subtraction and multiplication)
Sequence and Series
� Introduction to sequence and its types
� General term of liner or arithmetic sequence
� General term of quadratic sequence and geometric sequence
� Series with its types, sigma notation and partial sum
Are you ready?
� Pair of numbers, coordinate points. � Product of two sets.
� Tree diagram and mapping or arrow diagram. � Relationship between pair of numbers and its components.
� Operating machine or rule. � Image and pre–image in operating system.
� Equation � Polynomial and its degree
� Addition and subtractions of like terms � Set of numbers with definite rule.
� Specific term of sequence or series � Make the pattern of numbers and figures.
Specification Grid
Unit Chapter Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion ability Ques- Marks
tions
Topic 1 mark 2 marks 4 marks 5 marks
1 Algebra From All Topics 2 3 2 1 8 21
ALGEBRA 1.1 ORDERED PAIR AND CARTESIAN
PRODUCT
Learning Objectives
At the end of this topic, the students will be able to:
• define ordered pair and Cartesian product. • represent the Cartesian product in different ways.
Ordered Pair
Let us take any two numbers 2 and 3 which is a pair of natural number greater than 1 and less than 4.
Suppose the number 2 as in first position and 3 as in second position. Then it can be written in ordered
form as (2, 3).
So, a pair of numbers (or objects) arranged in ordered form which is separated by comma (,) and enclosed
by small brackets ( ) is called an ordered pair. The numbers or objects of the ordered pairs are called
components. Also, the component in the first position is called x-component and the component in the
second position is called y-component. So, the ordered pair represents a coordinate point.
In the case of set, the set of the numbers 2 and 3 is written as {2, 3} or {3, 2}. That is {3, 2} and
{2, 3} are the same sets. The order of the components in the set has no meaning in set representation make
the set different. But in the case of ordered pair, the ordered pairs (2, 3) and (3, 2) are not the same because
the positions of 2 and 3 are not the same. So, we say that (2, 3) ≠ (3, 2).
Let us consider the two ordered pairs 42, 12 = (2, 4) and 174, 20 = (2, 4).
3 5
24, 12 174, 20 42, 12 174, 20
This shows that the ordered pairs 3 and 5 are equal, i.e., 3 = 5 .
Two ordered pairs are said to be equal if they have equal corresponding components.
Note:
(i) In two ordered pairs (x, y) and (a, b), if x = a and y = b, then (x, y) = (a, b).
(ii) If (x, y) = (a, b) then we can equate the corresponding components of equal ordered pairs
i.e., x = a and y = b.
(iii) In an ordered pair, the first component represents along x-axis and the second component
along y-axis of the coordinate plane.
Example 1 : Which of the following ordered pairs are equal ?
(a) 54, 2 and 1125, 16 (b) 135, 12 and 360, 30
3 24 6 18
Solution: (a) Here, 54, 2 = 45, 2 and 1125, 16 = 54, 2
3 3 24 3
Here, both corresponding components of the given ordered pairs are equal.
2 Illustrated Optional Mathematics-9
Thus, the ordered pairs 45, 2 and 45, 2 are equal. ALGEBRA
3 3
(b) Here, 135, 12 = (5, 2) and 360, 30 = 5, 5
6 18 3
Here, both corresponding components of the given ordered pairs are not equal.
So, the ordered pairs 135, 12 and 360, 30 are not equal.
6 18
Example 2 : If (3x + 2y, 2) = (– 4, 2x – y), then find the values of x and y.
Solution: Here, (3x + 2y, 2) = (– 4, 2x – y)
Equating the corresponding components, we get
3x + 2y = – 4 ……. (i)
2x – y = 2 …….. (ii) × 2
Multiplying by 2 in eqn (ii) and adding with eqn (i), we get
4x – 2y = 4
3x + 2y = – 4
7x = 0
or, x = 0
Substituting the value of x in eqn (i), we get
3 × 0 + 2y = – 4
or,
y = –4 = –2 ∴ x = 0 and y = – 2.
2
Cartesian Product of Two Sets
Consider two non-empty sets A = {2, 3} and B = {4, 5, 6}
All the possible ordered pairs can be constructed from the set A to the set B as given below:
Set B Ordered pairs
Set A 4 (2, 4)
25 (2, 5)
6 (2, 6)
4 (3, 4)
35 (3, 5)
6 (3, 6)
Here, the set of all possible ordered pairs = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Thus, these all possible ordered pairs of two non-empty sets A and B is called a Cartesian product from A
to B. It is denoted by A × B and is defined by A × B = {(x, y): x∈A, y∈B}
It is also known as the cross product from A to B.
Also, make the Cartesian product from the set B to A as given below:
B × A = {4, 5, 6} × {2, 3} = {(4, 2), (4, 3), (5, 2), (5, 3), (6, 2), (6, 3)}
Thus, we conclude that A × B ≠ B × A.
Ordered Pair and Cartesian Product 3
ALGEBRA Way to Represent Cartesian Product
We can represent the Cartesian product of two non-empty sets in various ways. They are given in the table
below:
SN Name of ways Representation of A × B
1. Set of Ordered pairs (Roster form) A × B = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
2. Standard Description (Using a rule of a formula) A × B = {(x, y): x∈A, y∈B}
3. Tree Diagram Set A Set B Ordered pairs
2 4 (2, 4)
5 (2, 5)
3 6 (2, 6)
4 (3, 4)
5 (3, 5)
6 (3, 6)
Sets B
4. Tabulation Method ×4 56
5. Mapping (or Arrow) Diagram
A 2 (2, 4) (2, 5) (2, 6)
3 (3, 4) (3, 5) (3, 6)
A×B
AB
24
5
36
Y
6. Lattice or Graphical Method B
O Z
A
Example 3 : If P = {4, 5} and Q = {1, 2, 3} then prove that P × Q ≠ Q × R. Show them in the table
and mapping diagram.
Solution: Here, P = {4, 5} and Q = {1, 2, 3}
Now,
P × Q = {4, 5} × {1, 2, 3} = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)} and
Q × P = {1, 2, 3} × {4, 5} = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
4 Illustrated Optional Mathematics-9
Thus, P × Q ≠ Q × R. Proved. ALGEBRA
Showing P × Q and Q × P in the table.
Sets Q Sets P
×1 23 ×4 5
P 4 (4, 1) (4, 2) (4, 3) Q 1 (1, 4) (1, 5)
5 (5, 1) (5, 2) (5, 3) 2 (2, 4) (2, 5)
(3, 5)
3 (3, 4)
Showing P × Q and Q × P in the mapping diagram:
P P×Q Q Q Q×P
P
4 11 4
22
5 3 3 5
Example 4 : If P = {x : 5 ≤ x < 8}, the find P2 and present in the graph.
Solution: Give, P = {x : 5 ≤ x < 8} = {5, 6, 7}
Now,
P2 = P × P
= {5, 6, 7} × {5, 6, 7}
= {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (6, 7), (7, 5), (7, 6), (7, 7)}
Showing P2 in the graph:
Y
P
O X
P
EXERCISE – 1.1
1. (a) What is ordered pair?
(b) In which condition the two ordered pairs are equal?
(c) Define Cartesian product with suitable example.
2. Which of the following ordered pairs are equal? Give reasons.
(a) 2, 4 and 24, 12 (b) (2 + 3, 1 – 2) and (4 + 1, 4 – 5)
3 9
(c) (2 + 1, 3 – 2) and (3 – 1, 4 – 5) (d) 1 + 12, 4 and 23, 2 + 6
3
Ordered Pair and Cartesian Product 5
3. Find the values of x and y, p and q or a and b from the following equal ordered pairs:
ALGEBRA (a) (2x – y, 2) = (5, y + 1) (b) (3p, 2q – p) = (2p – 1, 1)
(c) (4a + 2b, 3) = (2a + b, 3a) (d) (2x – y, 3y) = (3, x + y)
(e) (x2 + 1, x + y) = (2, 2) (f) (2p + q, 3) = (2, 3p + 2q)
4. Find the Cartesian product of the following two sets which are defined as:
(a) A × B = {(x, y) : x∈A = {a, b, c}, y∈B = (4, 5)}
(b) P × Q = {(x, y) : x∈P = {1, 2} y∈Q = (p, q)}
5. (a) If A = {–1, 0, 1} and B = {2, 1}, then find A × B and B × A. Are A × B and B × A equal ?
(b) If M = {Multiple of 4 less than 15} and N = {Factors of 9}, then show that M × N ≠ N × M.
6. If P = {1, 2}, Q = {3, 4} and R = {4, 5}, find the following Cartesian products and show them in
tree diagram and graph:
(a) P × Q (b) P × R (c) Q × R (d) R × P
7. If A = {a, b, c} and B = {1, 2}, then find the following Cartesian products and show them in the
table and mapping diagrams:
(a) B × A (b) A × A (c) B2 (d) A × B
ANSWERS
1. a, b and d
2. (a) 3, 1 (b) – 1, 0 (c) 1, – 2 (d) 2, 1 (e) 1, –1; 1, 3 (f) 1, 0
3. (a) {(a, 4), (a, 5), (b, 4), (b, 5), (c, 4), (c, 5)} (b) {(1, p), (1, q), (2, p), (2, q)}
4. (a) (–1, 2), (–1, 1), (0, 2), (0, 1), (1, 2), (1, 1)}
(2, –1), (2, 0), (2, 1), (1, –1), (1, 0), (1, 1)}, Not equal
5. (a) {(1, 3), (1, 4), (2, 3), (2, 4)} (b) (1, 4), (1, 5), (2, 4), (2, 5)}
(c) {(3, 4), (3, 5), (4, 4), (4, 5)} (d) {(4, 1), (4, 2), (5, 1), (5, 2)}
6. (a) (1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}
(b) (a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
(c) (1, 1), (1, 2), (2, 1), (2, 2)}
(d) (a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
Project Work
Throw two different coloured die three times and record the outcomes in each time in the form of sets
say A and B. Write the possible ordered pairs to make from the set A to the set B and from the set B to
the set A. Write them in Cartesian product form. Are these sets of ordered pairs equal? Show them in
the following methods:
(a) Set-builder Method (b) Tabulation Method
(c) Tree-diagram Method (d) Mapping or Arrow Diagram Method
(e) Graphical or Lattice Diagram Method
6 Illustrated Optional Mathematics-9
1.21 RELATION ALGEBRA
1.2 (A) Introduction to Relation
Learning Objectives
At the end of this topic, the students will be able to:
• define relation. • find the domain, range and inverse of a relation.
Consider the two non-empty sets A = {3, 4} and B = {1, 2, 4}.
Then, A × B = {3, 4} × {1, 2, 4}
= {(3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 4)}
We select any subset from A × B as
R1 = {(3, 1), (3, 2), (4, 1), (4, 2)}
R2 = {(3, 4)}
R3 = {(4, 4)}
R4 = {(4, 1), (4, 2), (4, 4)} and so on.
Here, the subset R1 is described as "the x-component is greater than the y-component". The subset
R2 is described as "the x-component is less than the y-component". The subset R3 is described as
"the x-component is equal to the y-component". The subset R4 is described as "the x-component is the
multiple of the y-component."
Thus, the descriptions of the subsets R1, R2, R3 and R4 are the relationship between the x-component and
the y-component. Let's show the above subsets R1, R2, R3 and R4 in the mapping diagrams below:
R1 R2 R3 R4
AB AB AB AB
31 31 31 31
2 222
4 44 44 44 4
"is greater than" "is less than" "is equal to" "is multiple of"
A subset of the Cartesian product of two non-empty sets is said to be a relation. If A and B are two
non-empty sets, then the relation from A to B is denoted by R:A →B where R ⊆ A × B and is defined by
R = {(x, y}: x ∈ A, y ∈ B}. It is written as xRy .
It is read as "x is related to y" and is written as (x, y) ∈ R or xRy.
Relation 7
ALGEBRA Domain and Range of Relation
Consider a relation "is smaller than", R = {2, 6), (3, 7), (4, 8), (5, 9)}
Now separate the x-components and y-components of the relation R in the table below:
Here,
Ordered Pairs x – component y – component
(2, 6) 2 6
(3, 7) 3 7
(4, 8) 4 8
(5, 9) 5 9
The set of x-components of R = {2, 3, 4, 5} is called domain of R.
The set of y-components of R = {6, 7, 8, 9} is called range of R.
The set of x-components of R is considered as the domain of R and the set of y-components of R is
considered as the range of R. They are symbolically denoted as:
the domain, D = {x: (x, y) ∈ R} and the range, R = {y : (x, y) ∈ R}.
Example 1 : If P = {2, 3, 4} and Q = {4, 5, 6}, find the set of ordered pairs in the relation "is less
than" from P to Q.
Solution: Given, P = {2, 3, 4} and Q = {4, 5, 6}
∴ P × Q = {2, 3, 4} × {4, 5, 6}
= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
From the above P × Q, select the ordered pairs whose x-component "is less than y-component"
is {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6)}, which is the required relation.
Example 2 : If A = {1, 0, 2} and B = {1, 2, 3}, find the relation R so that x + y = 3 where x∈A and y∈B.
Solution: Here, A = {1, 0, 2}, B = {1, 2, 3}, and x + y = 3 where x∈ A and y ∈ B
∴ The required relation R = {(1, 2), (0, 3), (2, 1)}.
Example 3 : Find the domain and range of the relation R = {(1, 2), (3, 4), (5, 6), (7, 8)}.
Solution: Given, the relation, R = {(1, 2), (3, 4), (5, 6), (7, 8)}
Now, the domain of R = Set of the x-components of all ordered pairs of R = {1, 3, 5, 7}
The range of R = Set of the y-components of all ordered pairs of R = {2, 4, 6, 8}
Example 4 : If the range of the relation R = {(x, y) : y = 2x – 1} is {3, 5, 7}, find the domain of R.
Solution: Given, the relation R = {(x, y) : y = 2x – 1}, the range of R = {3, 5, 7} i.e. y ∈ {3, 5, 7}
Now, if y = 3, then 3 = 2x – 1, or 2x = 3 + 1, or, 2x = 4, or, x = 2
If y = 5, then 5 = 2x – 1, or 2x = 5 + 1, or, 2x = 6, or, x = 3
If y = 7, then 7 = 2x – 1, or, 2x = 7 + 1, or, 2x = 8, or, x = 4
∴ The domain of R = {2, 3, 4}
8 Illustrated Optional Mathematics-9
EXERCISE – 1.2 (A) ALGEBRA
1. (a) Define relation on the basis of Cartesian product:
(b) How can you determine domain and range from any relation?
2. If A = {1, 2, 9} and B = {3, 6, 9}, find A × B in the set of ordered pairs of the following relations
from A to B:
(a) "is less than" (b) "is greater than"
(c) "is one-third of" (d) "is equal to"
3. If P = {2, 4, 6, 8} and Q = {3, 5, 7, 9}, then find the following relations from P to Q:
(a) "is less by 3 than" (b) "is more by 1 than"
4. If A = {0, 1, 2, 3, 5} and B = {4, 5, 6, 8, 9}, find the following relations from A to B.
(a) x + y = 7 (b) y – x = 3 (c) y = x2.
5. Write down the domain and range of the following relations:
(a) R1 = {(1, 4), (2, 5), (3, 6)} (b) R2 = {(1, 3), (1, 5), (2, 6), (3, 7)}
6. Find the domain of the following relations for the given range:
(a) R1 = {(x, y) : y = 3x + 1} : Range = {1, 7, 10}
(b) R2 = {(x, y) : 2x + 3y = 2} : Range = {0, 2, –2}
ANSWERS (b) {(9, 3), (9, 6)}
(d) {(9, 9)}
2. (a) {(1, 3), (1, 6), (1, 9), (2, 3), (2, 6), (2, 9)} (b) R = {(4, 3), (6, 5), (8, 7)}
(c) {(1, 3), (2, 6)} (b) R = {(1, 4), (2, 5), (3, 6), (5, 8)}
3. (a) R = {(2, 5), (4, 7), (6, 9)}
4. (a) R = {(1, 6), (2, 5), (3, 4)} (b) Domain = {1, 2, 3}; Range = {3, 5, 6, 7}
(c) R = {(2, 4), (3, 9)} (b) Domain = {1, –2, 4}
5. (a) Domain = {1, 2, 3}; Range = {4, 5, 6}
6. (a) Domain = {0, 2, 3}
1.2 (B) Representations of Relation
Learning Objectives
At the end of this topic, the students will be able to:
• represent the relation in different ways.
• identify the types of the relation.
Consider two non-empty sets A = {0, 1} and B = {–1, 0, 1}. Then the relation "is square of" from A to B
can be represented by the following ways:
SN Name of Way Notation
1. Set of ordered pairs (Roster Form) R = {(0, 0), (1, –1), (1, 1)}
Relation 9
ALGEBRA SN Name of Way Notation
2. Set builder Form (Description Method) R = {(x, y) : x ∈ A, y ∈ B and x = y2} or R = (x, y) : x is
the square of y}
3. Tabulation Method A0 11
B0 –1 1
4. Mapping or Arrow Diagram R
AB
0 –1
0
11
Y
5. Graphical Method (Lattice) Diagram B
O X
A
Types of Relation
Consider A = {1, 2} and B = {1, 2, 3}. Then, A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}
(i) Take a relation as
R1 = {(1, 1), (2, 2)} AB
11
It is expressed as 1R1 1 and 2R12. Hence each element of the set A in R 2 2
belongs to the same element of the set B = A. This type of relation is called 3
a reflexible relation. If (x, x) belongs to in R for all x belongs to in A, the
reflexible is denoted by xRx. This is also called an identity relation on a set A and is denoted by IA
and IA = {(x, y) : x = y}
(ii) Take another relation as, AB
R2 = {1, 2), (2, 1)} 12
It is expressed as 1R22 and 2R21. i.e. 1R22 → 2R21. This type of relation 21
is called a symmetric relation. If (x, y) belongs to the relation R in A, then 3
(y, x) also belongs to R. i.e. if xRy → yRx. 1→2→1
(iii) Take next relation as
R3 = {1, 2), (2, 3), (1, 3)} 1→2→3
It is expressed as 1R32 and 2R33 → 1R33. This type of relation is called a transitive relation. If
(x, y) belongs to in R and (y, z) belongs to in R, then (x, z) belongs to in R. i.e. if xRy and yRz → xRz.
Note: If the relation satisfies the reflexible, symmetric and transitive, then it is called an equivalent
relation.
(iv) Take next relation as
R4 = {1, 2)} and R5 = {(2, 1)}
They are expressed as 1R42 and 2R51. In these relations the domain and range are interchanged.
These types of relations are called inverse relations to each other.
10 Illustrated Optional Mathematics-9
Example 1 : P = {1, 2, 3}, Q = {1, 4, 9} and a relation R:P → Q is defined by "is the square root of", ALGEBRA
represent the relation R in the following ways:
(a) Set of ordered pairs (Roster Method) (b) Set builder form
(c) Tabulation Method (d) Lattice Diagram
Solution: Here, P = {1, 2, 3}, Q = {1, 4, 9}
The relation R:P → Q is defined by "is the square root of " is represented below:
(a) Set of ordered pairs (Roster Method) R = {(1, 1), (2, 4), (3, 9)}.
(b) Set-builder form R = {(x, y) : x ∈ P, y ∈ Q and y = x2}.
(c) Tabulation Method:
A1 2 3 Y
B 1 4 9
(d) Lattice diagram:
OX
Example 2 : Find the domain and range of the relation shown in
the given graph: Y
Solution: From the given graph, we have
The relation R = {(0, 2), (2, 0), (4, 3), (7, 3), (8, 4)}
Now, the domain of R, D = {–2, 2, 5, 6}
The range of R, Re = {2, 3, 4} O X
Example 3 : Classify the following relations with arrow diagrams. R
(a) R = {(2, 3), (3, 2)} (b) R = {(3, 3), (4, 4), (8, 8)} 3
Solution: Given, 2 2
(a) R = {(2, 3), (3, 2)} R
Here, 2R3 → 3R2 1 3
4
So, R is a symmetric relation. 8
(b) R = {(3, 3), (4, 4), (8, 8)}
Here, first and second components are same, 3
i.e. 3R3, 4R4 and 8R8. 4
So, R is a reflexible relation.
8
EXERCISE – 1.2 (B)
1. (a) Write the ways to represent the relation.
(b) Define reflexible and transitive relation.
(c) What is inverse relation? Define with example.
(d) Write one difference between symmetric and inverse relation.
Relation 11
2. Write down the domain, range and inverse of the relations as shown in the following:
ALGEBRA (a) (b) P Q (c) Y
A 2 –1 –3 0 8
B 2 0 –1 1 10
2 12
3 14 O X
3. If A = {1, 4, 5,}, B = {2, 5, 6, 8} and a relation R : A → B is defined by "y – x = 1, x∈A and y∈B,''
represent the relation R by the following methods:
(a) Set of ordered pairs (Roster Method) (b) Set-builder Method
(c) Tabulation Method (d) Mapping Diagram (e) Lattice Diagram
4. Write down the following relations in the set-builder form:
(a) R1 = {(2, 3), (3, 7), (4, 5), (6, 8), (7, 10)} (b) R2 = A –1 0 1 2
B –1 0 1 8
R3 Y
CD
(c) Rupee USA (d) R4 = X' O X
Japan
Dollar Nepal
Pound UK
Yen India
Y'
5. Classify the given relations. Also, show them in the mapping diagrams:
(a) R1 = {(2, 3), (3, 4), (2, 4)} (b) R2 = {(1, 4), (4, 1)}
(c) R3 = {(4, 4), (8, 8)} (d) R4 = {(1, 2), (2, 3), (3, 4), (1, 4)}
6. Write down the set of all ordered pairs of the relation R = {(x, y)} and its inverse R– 1 from the
following conditions:
(a) y = 3x – 2, x ∈ {0, 2, 4} (b) x2 = 2y + 1, x ∈{1, 3, 5}
ANSWERS
2. (a) Domain = {2, –1, –3}; Range = {2, 0, –1}; R–1 = {(2, 2), (0, –1), (–1, –3)}
(b) Domain = {1, 2, 3}; Range = {10, 12, 14}; R–1 = {(10, 1), (12, 3), (14, 1), (14, 2), (14, 3) }
(c) Domain = {2, 4, 7}; Range = {1, 2, 3}; R–1 = {(1, 2), (1, 4), (3, 4) (2, 7), (3, 7)}
3. (a) {(1, 2), (4, 5), (5, 6)} (b) R = {(x, y) : y – x = 1, x ∈ A, y ∈ B} (c), (d) & (e) show to your teacher.
4. (a) {(x, y) : x < y, x ∈ {2, 3, 4, 6, 7}, y ∈ {3, 5, 7, 8, 10} (b) {(x, y) : x3 = y, x ∈ {–1, 0, 1, 2}, y ∈ {–1, 0, 1, 8}
(c) {(x, y) : x is the currency of y, x ∈C, y ∈D } (d) {(x, y) : x = y + 4, x ∈ {1, 2, 3}, y ∈ {–3, –2, –1}
5. (a) Transitive (b) Symmetric (c) Reflexible (d) Transitive
6. (a) {(0, –2), (2, 4), (4, 10)}; {(–2, 0), (4, 2), (10, 4)} (b) {(1, 0), (3, 4), (5, 12)}; {(0, 1), (4, 3), (12, 5)}
Project Work
Throw two different coloured die four times and record the outcomes in each time in the form of sets
say P and Q. Write the Cartesian product from P to Q. Collect the ordered pairs such that x > y where
x ∈ P and y ∈ Q. What is this collection of ordered pairs? Show them in the following methods:
(a) Set-builder Method (b) Tabulation Method
(c) Mapping or Arrow Diagram Method (d) Graphical or Lattice Diagram Method
12 Illustrated Optional Mathematics-9
1.31 FUNCTION ALGEBRA
1.3 (A) Introduction and Representations of Function
Learning Objectives
At the end of this topic, the students will be able to:
• identify a function and define it. • represent the function in different methods.
Introduction to Function
Let us consider the two non-empty sets A = {1, 2, 3} and B = {2,4, 6}.
Then A × B = {1, 2, 3} × {2, 4, 6} = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6)}
Let us take some subsets as relations of A × B and observe carefully:
(i) R1 = {(1, 2), (2, 4), (3, 6)} (ii) R2 = {(1, 2), (1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)}
(iii) R3 = {(2, 2), (3, 2)} (iv) R4 = {(1, 2), (2, 2), (3, 6)}
For the relation R1
The relation R1 = {(1, 2), (2, 4), (3, 6)} from A to B has the following characteristics:
● Every element of the set A pairs separately with each element of the A R1
set B. There is no element left in A and B. 1 B
2
● Each mapped element of the set A is the half of the every element of 2 4
y 3 6
the set B i.e. The relation R1 describes "is the half of" Then R1 : x → 2
where x∈A and y ∈ B. It has specific rule.
● Every element of the relation R1 is mapped with separately from the set A to the set B. So, it is
one-one relation.
For the relation R2 A R2
1 B
The relation R2 = {(1, 2), (1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)} from A to B has 2 2
the following characteristics: 3 4
6
● Every element of the set A is mapped with more than one element of
the set B. Therefore, R2 is one-many relation. There is no element left
in both A and B.
● Each mapped element of the set A is less than each element of the set B, i.e. the relation R2
describes "is less than". Then R2 = {(x, y) : x < y, x ∈ A and y ∈ B}. It does not have specific rule.
For the relation R3 A R3
1 B
The relation R3 = {(2, 2), (3, 2)} from A to B has the following characteristics: 2 2
3 4
● Only two elements of the set A are matched with one element of the set 6
B. The element 1 of A is left which is not mapped with any element of
the set B. It is many to one relation.
● The relation R3 describes "is equal or greater than" i.e. R3 = {(x, y) : x ∈ A, y ∈ B and x ≥ y},
which does not have specific rule.
Function 13
For the relation R4 R4
B
ALGEBRA The relation R4 = {(1, 2), (2, 2), (3, 6)} from A to B has the following characteristics:A 2
1 4
● Every element of the set is mapped with any element of the set B. 6
2
● Two elements 1 and 2 of A are mapped with only one element of B. 3
So, it is many-one relation.
Now, we conclude from the above discussion that the relations R1 and R4 have every element of the set
A mapped with any element of the set B. None element of A is mapped with more than one element of
B. So, the relations R1 and R4 are functions. But in the relation R2, every element of the set A is mapped
with more than one element of B and in the relation R3, one element is left in the set A. So, the relations
R2 and R3 are not functions.
A relation in which every element of the first set is assigned to a unique element A fB
of the second set is known as function. IfAand B are two non-empty sets, then the x f(x)
function from A to B is denoted by f : A → B and is defined by f = {(x, y) : x ∈ A, y ∈ B
=y
and f(x) = y}. It is usually denoted by g, h, k, etc.
Vertical Line Testing for Identifying Function
We can test a graphical representation of a relation by doing a vertical line test to find whether the relation
is function or not. We have already defined that the function is the set of ordered pairs having no two
ordered pairs but have the same x-component. i.e., it is not one-many relation.
A relationship is a function if no vertical line intersects its graph at more than one point. That is if the
vertical line interests at one point in the graph of the relationship, then it is a function. Otherwise, it is not
a function. For examples:
(i) Y (ii) Y (iii) Y
Vertical line
X' O X
OX X' O X Y'
It is a function It is a function It is not a function
Representation of Function
Consider A = {–1, 0, 1}, B = {0, 1, 2} and the function f as "is square of" from A to B. Then the function
f : A → B can be represented in the following ways:
SN Method Notation
1. Roster form f = {(–1, 1), (0, 0), (1, 1)}
2. Set-builder form f = {(x, y) : x ∈ A, y ∈ B and y = x2}
3. Formula
y = x2
4. Tabulation x –1 0 1
y 1 01
14 Illustrated Optional Mathematics-9
SN Method Notation ALGEBRA
5. Mapping Diagram f
6. Graphical Method AB
–1 0
01
12
Y
X' O X
Y'
Example 1 : Which of the following relations are functions? Give reason.
(a) f = {(1, 2), (4, 5), (7, 8)} (b) g = {(1, 2), (3, 4), (3, 5), (4, 6)}
Solution: Here,
(a) The relation f is a function because every element of the range has single pre-image.
(b) The relation g is not a function because the first components of two ordered pairs (3, 4)
and (3, 5) are the same. i.e., 3 has two images.
Example 2 : Which of the following diagrams represent a function. Give reason.
(a) g B (b) B h C (c) Y (d) Y
A
1 11 2 X' OX
2 42 4 X' O X
3 3 8 Y' Y'
Solution: (a) The relation g is not a function g B hC
because the element 3 is left in the A 12
set A in pairing with element in the B 24
set B. 1 38
2 1
3 4 for (b)
(b) The relation h is a function because for (a)
every element of the set C is mapped
with any element of the set D.
(c) This is not a function because the Y
vertical line cuts the given graph (or Y
curve) at more than one point. X' O X
(d) This is a function because the X' O X
vertical line intersects at only one Y'
Y'
point in the graph.
for (c) for (d)
Function 15
ALGEBRA EXERCISE – 1.3 (A)
1. (a) What is function? Define with example.
(b) In how many ways you can represent the function? What are they?
2. Which of the following relations are functions? Give reasons.
(a) f = {2, 3}, (4, 5}, (6, 7), (8, 9)} (b) g = {(0, 1), (1, 7), (2, 10), (2, 12)}
(c) h = {(1, 5), (4, 7), (5, 7)} (d) k = {(a, 1), ( b, 2), (c, 1), (d, 2)}
3. If A = {x, y, z} and B = {4, 5, 6}, then examine which of the following are functions. Give reasons.
(a) {(x, 4), (y, 4), (z, 4)} (b) {(x, 4), (x, 5), (x, 6)}
(c) {(x, 5), (y, 6), (y, 5), (z, 5)} (d) {(x, 5), (y, 4), (z, 5)}
4. State whether the following diagrams represent the functions. Give reasons.
(a) 1 (b) 0 (c) h (d) k
f 8 g 1
27 4 –1 –1 1
1 –2 –2 –3 a 2
2 0 –3 –5 b 3
3 2 –4
c
(e) (f) Y (g) Y (h)
Y Y
X' O X
OX Y' OX
OX
5. It is given that A = {2, 4, 6} and B = {1, 2, 3}. If the function f:B → A is defined by "is the half of",
then represent the function in the following methods:
(a) Roster form (b) Set-builder form (c) Formula
(d) Mapping diagram (e) Lattice diagram (f) Table
ANSWERS
2. (a), (c), (d) 3. (a), (d) 4. (a), (b), (e), (g)
5. (a) {(1, 2), (2, 4), (3, 6)} (b) {(x, y): x ∈B, y∈A and y = x}
2
(c) x= y (d), (e) & (f) show to your teacher.
2
Project Work ..........
Put the following numbers as input in the adjoining function machine.
What do you find as the output? Write the domain and range of f(x). 3x2f–(x4)x=+ 1
(a) 1 (b) – 2 (c) 8 (d) 25
(e) – 40 (f) – 54 (g) 78 (h) 100
(i) h (j) a + 1 (k) p - 1 (l) a – b .............
16 Illustrated Optional Mathematics-9
1.3 (B) Components of Function ALGEBRA
Learning Objectives
At the end of this topic, the students will be able to:
• find the value, image, pre–image, domain, co–domain and range of the function.
Value of a Function 1
Consider a function f(x) = 3x – 1. If we put x = 1, then
f(1) = 3 × 1 – 1 = 3 – 1 = 2 3x – 1 2
The value of f(1) is 2, i.e., 2 is the image of 1.
Shown in the adjoining operating machine, we get f(x) = 3x – 2
Similarly, f(2) = 3 × 2 –1 = 6 – 1 = 5 1 2
i.e. 5 is the image of 2 and so on. 25
In general, f(m) = 3m – 1 = n (suppose) mn
Thus, if f is a function and (m, n) is in f, then we write f(m) = n, where f(m) is called the value of the
function at m.
Image and Pre-image of a Function f
AB
Consider a function f = {(0, 0), (1, 1), (2, 4)} and its mapping diagram 00
given: 11
24
Here, all the elements 0, 1 and 2 of A have the corresponding elements in B
as 0, 1 and 4 respectively, i.e., f(0) = 0, f(1) = 1 and f(2) = 4. 9
Again, shown the above function in operating machine in the adjoining as:
In the function f, every element of A as input is operated by 'square' into the 2, 1, 0
elements of B as output.
Which elements of them are the original (initial) elements and which Square
elements of them are the product elements ? The elements of the set A are
the original element as input (or raw material) and the elements of the set B 4,
are the product (or final element) as output which come after operating by 1,
the function f. So, the productive elements in B are the images of the initial
or original elements in A and the initial elements in A are the pre-image of 0
the productive elements in B.
In the above example, the images of the elements 0, 1 and 2 of A are 0, 1 x f
and 4 in B respectively. Similarly, the pre–image 0, 1 and 4 of B are 0, 1 y
and 2 in A respectively.
If A and B are two non-empty sets and x ∈ A and y ∈ B, then the elements
x of A is called the pre-image of y and y is called the image of x.
f:x → y i.e., f(x) = y
Function 17
ALGEBRA Domain, Co-domain and Range of Function
Again, from the above discussion the set of all elements of A under the function f:A → B is {0, 1, 2}. This
is called domain of the function f. The set of all elements in B mapped with each of A is {0, 1, 4}. This is
called range of the function f. But the set of all the elements of B, which is mapped or not is {0, 1, 4, 9}.
This is called co-domain of the function f. Therefore, we can say that the set of pre-images is the domain
of f and the set of images is the range of f and vice-versa.
If A and B are two non-empty sets and the function f : A → B. Then the set of all elements in A is called
the domain and the set of all elements in B is called co-domain of f. The set of mapped elements with
each element of A in B is called range of f.
Example 1 : If f(1) = 1, f(2) = 3, f(3) = 5, then find the value of f(100).
Solution: Given, f(1) = 1, f(2) = 3, f(3) = 5
∴ The domain of f = {1, 2, 3}
The range of f = {1, 3, 5}
Now, the element of the range as a sequence 1, 3, 5, ..…
Here, the first term (a) = 1
The constant difference (d) = 3 – 1 = 2
∴ The nth term (tn) = dn + (a – d) = 2n + (1 – 2) = 2n – 1.
Suppose x is the element in the range, then f(x) = 2x – 1, x ∈ N.
∴ f (100) = 2 × 100 – 1 = 200 – 1 = 199.
Example 2 : What is the element in the range which has the pre-image 4 under the function
f(x) = 3x2 + 1? Also find the pre-image of 4 under the same function.
Solution: Given, f (x) = 3x2 + 1.
Since the pre-image (x) = 4, or, 4 = 3x2 + 1
or, 4 – 1 = 3x2
So, f(4) = 3 × 42 + 1 = 48 + r1a n=g4e9is 49. or, 33 = x2
∴ The required element in the
Again, since the image (y) = 4, or, x2 = 1
or, x = ±1
i.e., f(x) = 4
∴ The required pre-image of 4 are 1 and –1.
Now, f (x) = 3x2 + 1
Example 3 : If the range of the function h:x → 5x – 3 is {– 8, 2, 7}, then find its domain.
Solution: Given, h : x → 5x – 3 or, h(x) = 5x – 3. The range of h(x) = {–8, 2, 7}
Now, or, – 8 + 3 = 5x or, –5 = x or, x = –1.
If h(x) = – 8, then – 8 = 5x – 3 5 or, x = 1.
or, x = 2.
If h(x) = 2, then 2 = 5x – 3 or, 2 + 3 = 5x or, 5 = x
5
If h(x) = 7, then 7 = 5x – 3 or, 7 + 3 = 5x or, 10 = x
5
∴ The domain of h(x) is {–1, 1, 2}.
18 Illustrated Optional Mathematics-9
Example 4 : If the function f(x + 7) = f(x) + f(7), x ∈ R, show that f(0) = 0 and f(–7) = – f(7). ALGEBRA
Solution: Given: f(x + 7) = f(x) + f(7)
Put x = 0, f(0 + 7) = f(0) + f(7)
or, f(7) = f(0) + (7)
or, f(7) – f(7) = f(0)
or, f(0) = 0 ….. (i)
Again, Put x = –7, f(–7 + 7) = f(–7) + f(7)
or, f(0) = f(–7) + f(7)
or, 0 = f(–7) + f(7) [ From eqn (i)]
or, f(–7) = –f(7) Proved.
Example 5 : If f(x) = 3x2 – 2x – 1, g(x) = 5x – 3 and f(x) = g(x), find the value of x.
Solution: Given, f(x) = 3x2 – 2x – 1 and g(x) = 5x – 3
Now, by question,
f(x) = g(x)
or, 3x2 – 2x – 1 = 5x – 3
or, 3x2 – 2x – 5x – 1 + 3 = 0
or, 3x2 – 7x + 2 = 0
or, 3x2 – 6x –x + 2 = 0
or, 3x(x – 2) – 1 (x – 2) = 0
or, (x – 2) (3x – 1) = 0
Either, x – 2 = 0, or, x = 2
OR, 3x – 1 = 0, or, x = 1
∴ The value of x are 2 and 31. 3
Example 6 : What elements in the domain have the image 0 under the function f(x) = 3x2 – 8x + 5,
x ∈ R?
Solution: Given, f(x) = 3x2 – 8x + 5
f(x) = 3x2 – 8x + 5
The image, f(x) = 0
Now, 0 = 3x2 – 3x – 5x + 5
or, 0 = 3x(x – 1) – 5(x – 1)
or, 0 = (x – 1) (3x – 5)
Either, x – 1 = 0, or, x = 1.
OR, 3x – 5 = 0, or, x = 5
3
∴ The required elements in the domain are 1 and 53.
Function 19
ALGEBRA EXERCISE – 1.3 (B)
1. (a) If f is a function and (m, n) be in f, then what is the value of f (m)?
(b) What is image and pre-image of a function?
(c) Define domain, co-domain and range of any function.
2. (a) If f(x) = x2 – 1, find f(1), f(10), f(–10).
(b) If f(x) = x3 + 1, find f(0), f(–2), f(x + 1).
(c) If g(x) = x3 + x2 – x – 2, find g(–1) and g(2) + g(1).
(d) If h(x) = 2x2 – x + 1, find h(2)and h(–3) – h(–2).
3x – 2
(e) If f(x) = 2x2 + 1, find the value of f(3 + h) – f(3).
h
(f) If g(x) = 2x2 – 1, find g(x – 1) and g(5 – a) – g(5).
a
3. Find the domain and range in each of the following functions:
(a) f = {(2, 4), (5, 10), (8, 16), (11, 22)} (b) g = {(1, 4), (2, 5), (3, 6), (4, 6)}
(c) h = {(x, y) : y = 2x + 1 and 0 ≤ x ≤ 3} (d) k = {(x, y) : y = 2x2 + 1 and –1 ≤ x ≤ 1}
4. Find the domain, range and co-domain of the functions represented by the following diagrams:
(a) (b) (c) (d)
f g Y h
–1 2 1 1 a 1
b 2
2 54 2 x 3
Xy 4
5 26 9 3 X' O 5
6
36 4
Y'
5. Find the range of the following functions for the given domain:
(a) f(x) = |x – 1|, the domain, D = {–2, –1, 0, 1}
(b) f(x) = 8 – 3x, the domain, D = {3, 4, 5, 6}
(c) f : x → 2x2 – x – 3, x ∈ {2, 3, 4}
(d) f(x) = {(x, y) : y = 2x3 – 5} and x ∈ {–1 ≤ x < 3}, x ∈ Z
6. Find the domain of the following functions for the given range:
(a) g(x) = 4x – 3; the range, R = {1, 5, 9}
(b) h(x) = x2 – 3; the range, R = {1, 6, 13}
(c) h:x → 3x – 4; the range, R = {–1, 2, 5}
(d) f(x) = {(x, y) : y = x + 1, y ∈ {1 ≤ y < 4, x ∈ W}
7. (a) If f = {(x, x2 + 1)} is a function, then find the pre-image of 2.
(b) Which element in the domain has the image 7 under the function f:x → 3x – 4 ?
2
(c) The function is defined by f ={(x, y) : 2x – 3y = 5}. If it has pre-image 7, then find its image.
(d) Which element in the range has the pre-image 4 under the function f(x) = 2x + 1 ?
3
20 Illustrated Optional Mathematics-9
8. If g(x) = 2x3 + 1, find the missing element in the following ordered pairs:
(a) (…., 3) (b) (–1, ….) (c) (…, 17) (d) (100, …..) ALGEBRA
9. (a) If f(x + 5) = f(x) + f(5), prove that f(0) = 0 and f(–5) = – f(5).
(b) If f(x + 5) = f(x) + f(7), then show that f(2) = 0 and f(7) = – f(–3).
(c) If g(x + 2) = g(x) + g(6) then verify that g(4) = 0 and g(6) = – f(2).
(d) h(x – 3) = h(x) – h(3), then prove that h(0) = 0 and h(6) = 2h(3).
10. (a) If f(1) = 3, f(2) = 4, f(3) = 5, then find the value of f(20).
(b) If f(1) = 4, f(2) = 7, f(3) = 10, then find the value of f(–10).
11. (a) Let the function f : R → R be defined by, (b) If the function g:W → W is defined by
2x + 1 for x > 5, x2 – 1 for x < 2,
f(x) = 3x – 1 for –4 ≤ x ≤ 3, g(x) = 2x3 – 1 for 2 ≤ x ≤ 7,
5x – 3 for x < –4. x4 – x2 + x – 2 for x > 7.
Find g(0), g(5), and g(10).
Find f(10), f(0), and f(–7).
12. (a) If f(x) = ax – b, f(–2) = –4 and f(2) = 4, find the value of a and b.
(b) If g(x) = mx + c, g(–7) = –5 and g(4) = 6, find the values of m and c.
13. (a) If f(x) = 2x2 – x + 1, g(x) = x2 + 3x – 3 and f(x) = g(x), then find the possible value of x.
(b) If f(x) = 3x2 + x – 4 and g(x) = x2 – 2x + 1, for what value of x, f(x) = 2g(x) ?
(c) The functions f and g are defined as f:x → 2x + 1 and g:x → 2x2 – 4x + 5. If f(x) + 2 = 2g(x) – 1,
find the possible value of x.
(d) If f(x) = x2 + 2x + 1, then for what values of x, f(x) = f(x + 2)?
ANSWERS
2. (a) 0, 99, 99 (b) 1, – 7; x3 + 3x2 + 3x + 2 (c) – 1, 7 (d) 7 , – 5
(e) 2h + 12 (f) 2x2 – 4x + 1; 2a – 20 4 8
3. (a) Domain = {2, 5, 8, 11}, Range = {4, 10, 16, 22} (b) Domain = {1, 2, 3, 4}, Range = {4, 5, 6}
(c) Domain = {0, 1, 2, 3}, Range = {1, 3, 5, 7} (d) Domain = {– 1, 0, 1}, Range = {3, 1}
4. (a) Domain = {– 1, 2, 5}, Range = {2, 5, 26}, Co–domain = {2, 5, 26, 36}
(b) Domain = {1, 4, 9}, Range = {1, 3}, Co-domain = {1, 2, 3, 4}
(c) Domain = {– 2, 0, 2, 4, 6, 8}, Range = {–2, 0, 2, 4} = Co-domain
(d) Domain = {a, b, x, y}, Range = {1, 6} = Co-domain = {1, 2, 3, 4, 5, 6}
5. (a) {3, 2, 1, 0} (b) {– 1, – 4, – 7, – 10} (c) {3, 12, 25} (d) {– 7, – 5, – 3, 11}
6. (a) {1, 2, 3} (b) {±2, ±3, ±4} (c) {1, 2, 3} (d) {0, 1, 2}
7. (a) 1 or –1 (b) 6 (c) 3 (d) 3
8. (a) 1 (b) – 1 (c) 2 (d) 2000001
10. (a) 22 (b) – 29
11. (a) 21, – 1, – 38 (b) – 1, 249, 9908
12. (a) 2, 0 (b) 1, 2 (c) 1, 3 (d) – 2
13. (a) 2, 2 (b) 1, –6 2
Function 21
ALGEBRA 1.3 (C) Types of Function
Learning Objectives
At the end of this topic, the students will be able to:
• identify the types of function.
Observe the following mapping diagrams of the functions: h k
fg
1 a –1 0 1 a –1 0
2 b0 12 b0 1
3 c1 c1 2
In the function f, there is not any element left in the second set. The range and the co-domain are equal to
{a, b, c}. Therefore, it is onto function. Again, every element of range has single pre-image. Therefore, it
is also one-one function.
In the function g, there is also no any element left in the second set unmapped. The range and the
co-domain are equal to {0, 1}. Therefore, it is also onto function. But the element 1 has double pre-images
–1 and 1. So, it is also many–one function.
In the function h, there is one element left in the second set (or the co-domain) unmapped and therefore,
the range = {a, b} and the co–domain = {a, b, c}. So, the range is the proper subset of the co-domain, i.e,,
range ⊂ co-domain. Therefore, it is into function. Again, every element of range has single pre-image.
Therefore, it is also one–one function.
In the function k, there is one element 2 left in the second set unmapped and then the range ={0, 1} and the
co-domain = {0, 1, 2}. So, the range is proper subset of the co-domain i.e., Range ⊂ Co-domain. Therefore, it
is into function. But the element 1 has two pre-images –1 and 1. So, it is also many-one function.
Let us now summarize the above description in the table:
Func- One-one Definition Definition Many-one Definition
tions Definition Example Example
At least one
The range f Every element The range g element of range
has more than
and codomain of range has of the –1 0 one pre-image in
01 a function.
of a function 1 a single pre- function is 1
Onto are equal to 2 equal to its
3 b
c image in the
each other. function. codomain.
One-one and Onto Function Many-One and Onto Function
The range of Every element The range of At least one
a function h of range has a function k element of
is proper 1 a single pre- is proper –1 0 the range has
2 b image in the subset of its 0 1 more than one
Into subset of its c function. co-domain. 1
codomain. 2 pre-image in the
function.
One-one and Into Function Many-One and Into Function
22 Illustrated Optional Mathematics-9
Algebraic Functions
Consider some algebraic expressions or polynomials as 4, 3x, 2x + 1, x2 + 2x – 1, x3 – 1, etc. ALGEBRA
For any value of x, they give unique value as y or f(x). So, they can be expressed as f(x) = 4,
f(x) = 3x, f(x) = 2x + 1, f(x) = x2 + 2x – 1, f(x) = x3 – 1, etc. These equalities are called algebraic functions
or polynomial functions. Hence, the algebraic functions are represented as the algebraic operations of
any algebraic terms. They have different degrees. Therefore, we can classify the following some special
functions based on their degree: f Y
y = 2x – 1
(i) Linear Function: An algebraic function of first 0. .–1
degree in the form of f(x) = mx + c or, f(x) = ax + 1. .1 O X
b, where m, a, b, c are constants, is called a linear 2. .3
function. The linear function always gives a straight Mapping Diagram Graph
line in a graph. The linear function f(x) = 2x – 1, Y
i.e., y = 2x – 1 is shown in the alongside mapping
diagram and graph:
(ii) Constant Function: An algebraic function of zero f
degree in the form of y = c, where c is a constant, –1. .1 y=2 X
is called a constant function. It represents a straight 0. .2
line parallel to x–axis. The constant function y = 1. .3 O
2 is shown in the adjoining mapping diagram and Mapping Diagram Graph
graph: Y
f
(iii) Identity Function: A linear function on the form y y=x
= x in which every element maps into itself is called 0. .0
an identity function. The identity function has the 1. .1 45° X
same domain and range. It also represents a straight 3. .3 O
line making + 45o with x – axis. The graph of the Mapping Diagram
identity function y = x is shown in the adjoining Graph
arrow diagram and graph:
(iv) Quadratic Function: An algebraic function of second degree in the form of y = ax2 + bx + c, where
a ≠ 0 and a, b, c are constants, is called a quadratic function. It represents a curve of conic section
named parabola (U-shaped opened mouth). The quadratic function y = x2 – 2x + 1 is shown in the
adjoining mapping diagram and graph:
f Y y = x2 – 2x + 1 X Cone Parabola
1. .0 Plane
2. X' O
– 1. .1 Y' Conic section of parabola
2.
3. .4 Graph
Mapping Diagram
Function 23
ALGEBRA (v) Cubic Function: An algebraic function of third degree in the form of y = ax3 + bx2 + cx + d, where
a ≠ 0 and a, b, c, d are constants, is called a cubic function. It represents the two half parabolas. The
graph of the cubic function y = x3 is shown in the adjoining mapping diagram and graph.
Trigonometric Functions
A function involving the trigonometric ratios as sine, cosine, tangent, cotangent, secant or cosecant is
called a trigonometric function. For examples; f(θ) = sin θ, g(α) = cos 2α, h(β) = sec2 β + cosec β, etc. are
the trigonometric functions.
Graph of sine function Graph of cosine function
Consider a sine function, y = sin θ
θ 0º 30º 60º 90º 120º 150º Consider a cosine function, y = cos θ
y 0 0.5 0.87 1 0.87 0.5 180º θ 0º 30º 60º 90º 120º 150º 180º
0
θ 210º 240º 270º 300º 330º 360º ….. Y 1 0.87 0.5 0 –0.5 –0.87 –1
…..
y –0.5 –0.87 –1 –0.9 –0.5 0 θ 210º 240º 270º 300º 330º 360º …..
Y X
y = sin θ Y –0.87 –0.5 0 0.5 0.87 1 …..
X' O 30° 60° 90°120°150°180°210°240°270°300°330°360° 1Y y = cos θ X
X' O 30° 60° 90°120°150°180°210°240°270°300°330°360°
Y' –1
Y'
Y
Graph of tangent function y = tan θ
Consider a tangent function, y = tan θ
θ 0º 30º 60º 90º 120º 150º 180º 2 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° X
y 0 0.58 1.7 ∞ –1.73 –0.58 0 1
θ 210º 240º 270º 300º 330º 360º …..
y 0.58 1.73 ∞ –1.73 0.58 0 ….. X' O
–1
–2
Y'
Example 1 : Is f(x) = x2 – 1, x ∈ R onto function? Given reason. h
Solution: Given, ... –3
f(x) = x2 – 1, x ∈ R = {… , –2, –1, 0, 1, 2, 3, …..} –2 –2
Put x = – 2, f(–2) = (–2)2 – 1 = 4 – 1 = 3 –1 –1
Put x = –1, f(x) = (–1)2 – 1 = 1 – 1 = 0 0 0
Put x = 0, f(0) = 02 – 1 = 0 – 1 = – 1 1 1
Put x = 1, f(1) = 12 – 1 = 1 – 1 = 0 2 2
... 3
Put x = 2, f(2) = 22 – 1 = 4 – 1 = 3
Showing the above rules in the mapping diagram.
Here, all real numbers in the codomain do not have corresponding pre-images.
So, the given function is an into function. No, it is not an onto function.
24 Illustrated Optional Mathematics-9
EXERCISE – 1.3 (C) ALGEBRA
1. (a) Define onto function and into function.
(b) Differentiate between one-one function and many-one function in one sentence.
(c) What is an algebraic function? Write its types.
(d) Define trigonometric function.
2. Identify the types of the following functions:
(a) f (b) g (c) Y (d) h
–1 2 1 1 a 1
2 54 b 2
5 26 9 2 x 3
3 X' O Xy 4
36 4 5
6
Y'
3. (a) If P = {1, 2, 3}, Q = {1, 2, 4, 6, 8, 10} and f = {(x, y) : y = 2x, x∈P, y∈Q}, then find and check
whether it is one–one and into function or not.
(b) If M = { –2, –1, 0, 1, 2}, N = {1, 2, 5} and the function g : x → x2 + 1, then find the function
g:M → N. Is g many–one and onto function?
4. If A = {1, 2}, B = {3, 4} and the function f from A to B, then
(a) how many one-one and onto functions can be defined?
(b) how many one-one and into functions can be defined?
(c) how many many-one and onto functions can be defined?
(d) how many many-one and into functions can be defined?
5. (a) If the function f(x) = 4x – 1 with A = {1, 2, 3, 4} and B = {2, 3, 7, 8, 10, 11, 15, 16}, find the
range of f:A → B. Is the function f one–one and onto? Give reason.
(b) If the function g:P → Q is defined by g : x → 2x2 – 4 with P = {2, 3, 4} and Q = {4, 14, 28},
find the range of g. Is the function g one–one and onto ? Give reason.
6. (a) Is the function f(x) = 2x2 + 1, x ∈ R into function ? Give reason.
(b) Is the function g(x) = x3 + 1, x ∈ W onto function ? Give reason.
7. Identify the types of the following functions:
(a) y = 2x – 3 (b) f(x) = – 5 (c) y = 3x2 – 4 (d) y = 3x3 – 2x + 1
(e) y = x (f) 3x + y = 7 (g) 4x – 3f(x) + 5 = 0 (h) 7x2 – 2y + 2 = 0
ANSWERS
2. (a) One-One and into (b) Many-One and into (c) Many-One and onto (d) Many-One and onto
3. (a) f = {(1, 2), (2, 4), (3, 6)}, f is one–one and onto. (b) g = {(–2, 5), (–1, 2), (0, 1), (1, 2), (2, 5)}, yes
4. (a) two (b) Not defined (c) Not defined (d) two
5. (a) {3, 7, 11, 15}, No, (b) {4, 14, 28}, Yes
6. (a) Yes (b) No
7. (a) linear (b) constant (c) quadratic (d) cubic
(g) linear (h) quadratic
(e) identity (f) linear
Function 25
ALGEBRA 1.41 POLYNOMIAL
1.4 (A) Introduction to Polynomial
Learning Objectives
At the end of this topic, the students will be able to:
• define polynomial. • find the degree of polynomial. • identify the types of polynomial.
Consider an algebraic expression, 3x2 + x + 4. This expression has three algebraic terms 3x2, x and 4.
The term 4 has only one value four. So, 4 is a constant. The term 3x2 has a constant 3, is called numerical
coefficient and the variable x2 is called literal coefficient and has non-negative integer as power 2 and the
term x has the numerical coefficient 1. So, the algebraic expression 3x2 + x + 4 is an example of polynomial.
An algebraic expression having non-negative integer as the power of a variable in each term
is called a polynomial in one variable. Simply, the polynomial in one variable is only called
polynomial. The polynomial may have two or more variables.
For examples;
2, 2x, 3x + 1, 2x2 – 2x + 3, 2x3 – 3x2 + 1 and 3x5 + 4x – 1 etc. are the examples of polynomials.
Standard Form of Polynomial
The polynomial in the variable x of n degree can be generalized in the form anxn + an –1xn –1 + an–2 xn–2 + ….
+ a2x2 + a1x1 + a0 x0 (= a), where an ≠ 0 and an, an–1, … , a1, a0 are rational constants. n is a non-negative
integer. Here, anxn is leading term and an is the leading coefficient. It is denoted by p(x), q(x), r(x) etc. This
form of the polynomial is said to be the standard or general form. It has the following properties:
(i) The power (or exponent) of each variable in each term is always non-negative integers (or
whole numbers).
(ii) No terms of a polynomial has a variable as its denominator.
(iii) The coefficient of each term is as rational number.
(iv) Terms are arranged either descending or ascending order of power of variables.
Degree of Polynomial
The highest power (or exponent) of the variables of the term of a polynomial is called the degree of the
polynomial. For examples;
SN Polynomial Degree
(i) 4 (= 4x0) 0
(ii) 2x 1
(iii) 2x – 1 1
(iv) 3x2 + 3x + 1 2
(v) 3x2y + 3xy + 3y
(vi) 2x5y – 3x2y2 + 5y + 2 2+1=3
5+1=6
26 Illustrated Optional Mathematics-9
Types of Polynomials ALGEBRA
From the above examples mentioned on the degree of polynomials, there are various types of polynomials.
There are commonly two categories of polynomials. They are
(i) Based on the number of terms and
(ii) Based on the degree of terms.
There are mainly four types of polynomials based on the number of terms. They are shown in the table
below:
SN Types of Polynomials No. of Terms Examples
1. Monomial 1 2x, 4x3, 2x2y3, 3, etc.
2. Binomial 2 2x + 1, 3x3 – 2, 4x3 – 2y, etc.
3. Trinomial 3 3x2 + x – 1, 3x2y + xy2 + 2, etc.
4. Multinomial More than 1 2x3 + 2x2 + x – 1, 3x3 + 3x2y + xy2 + y3 – 1, etc.
Note: Multinomial is also polynomial which means that has more terms.
There are mainly five types of polynomials based on the degree of the polynomial. They are shown in the
table below:
SN Types of Polynomial (General form) Degree of Examples
polynomial
1. Constant Polynomial (a) 0 2, 5, 5 , etc.
2
2. Linear Polynomial (ax + b), a ≠ 0 1 2x, 3x + 1, 5x + 2y – 1, etc.
3. Quadratic polynomial (ax2 + bx + c), a ≠ 0 2 2x2 + x – 1, 3x3 – 1, 4x2, etc.
4. Cubic polynomial (ax3 + bx2 + cx + d), a ≠ 0 3 3x3 + 2x2 – x + 4, 3x3 – 1, etc.
5. Bi-quadratic polynomial (ax4 + bx3 + cx2 + dx + e), 4 2x4 – 3x3 + 2x2 + 4x – 5, 2x4 – 1,
a≠0 2x4 – 3x2 – 1, etc.
Note: The polynomial P (x) = 0, which may be considered to have no terms at all, is called zero
polynomial. Its degree is not defined.
Equal Polynomials
Consider the polynomials
P(x) = 6 x3 – 2x + 3 and q(x) = 18 x3 – 6 x + 9
2 6 3 3
After simplifying the polynomial p(x) and q(x), we get
P(x) = 3x3 – 2x + 3 and q(x) = 3x3 – 2x + 3 which have the same coefficient and the same degree on
variable of corresponding terms.
Thus, p(x) and q(x) are equal, i.e., p(x) = q(x). So, they are equal polynomials.
Two polynomials are said to be equal if and only if (iff) the coefficients and the degree of their
corresponding terms are the same.
Polynomial 27
ALGEBRA Example 1 : Find the numerical and literal coefficient of the following polynomials:
(a) 3x2 (b) 4x2y3
Solution: (a) Given, 3x2
Now, the numerical coefficient of 3x2 is 3 and the literal coefficient of 3x2 is x2.
(b) Given, 4x2y3
Now,
The numerical coefficient of 4x2y3 = 4.
The literal coefficient in y3 of 4x2y3 = x2.
The literal coefficient in x2 of 4x2y3 = y3.
Example 2 : Rewrite the following polynomial in the standard form. P(x) = 3x7 + 4x3 + 3x8 – 4x2 + 4 – 3x
Solution: Here, P(x) = 3x7 + 4x3 + 3x8 – 4x2 + 4 – 3x
Now, the standard form of the given polynomial p(x) is
3x8 + 3x7 + 4x3 – 4x2 – 3x+ 4 (Descending Order)
or, 4 – 3x – 4x2 + 4x3 6+ 3x7 + 3x8 (Ascending Order)
Example 3 : Find the degree of the polynomials.
(a) p(x) = 3x3 + 2x2 + 3x – 1 and
(b) q(x) = 3x3y2 + 3x2y + 5x – 4
Solution: Here,
(a) p(x) = 3x3 + 2x2 + 3x – 1
Now, the power of variable x in 3x3 = 3.
The power of variable x in 2x2 = 2.
The power of variable x in 3x = 1.
The power of variable x in 1 = 0.
Here, the highest power of x is 3. So, the degree of p(x) is 3.
(b) q(x) = 3x3y2 + 3x2y + 5x – 4
Now, the total power of the variable x and y in 3x3y2 = 3 + 2 = 5.
The total power of the variable x and y in 3x2y = 2 + 1 = 3.
The total power of the variable x and y in 5x = 1.
The total power of the variable x and y in 4 = 0.
Here, the highest power of the term of q(x) is 5. So, the degree of q(n) is 5.
EXERCISE – 1.4(A)
1. (a) What is polynomial?
(b) Classify the polynomials based on the number of terms in the polynomials.
(c) What are the types of polynomials on the basis of its degree?
28 Illustrated Optional Mathematics-9
2. Which of the following algebraic expressions are polynomials? Give reasons.
(a) x2 + x – 1 (b) 1 + 1 + 12 (c) 3x–3 + 1 ALGEBRA
x2 x
(d) 3t2 – 5 t + 1 (e) 2 x – 1 (f) 8
3. Rewrite the the following polynomials in the standard form:
(a) 2x – 4 + 3x3 – 4x5 (b) 2x3 – 5x + 3x7 – x2 + 5
(c) 4x3 – 5x2 + 1 – 6x5 – x7 – x (d) x2 – x3 + x – 5 + x4
2 3
4. If p(x) = 2x5y + 4x4y2 – 3x3y3 + 5xy + 4, then answer the following questions:
(a) What is the numerical coefficient of xy? (b) Write the constant term of p(x).
(c) What is the literal coefficient of y2? (d) What is the literal coefficient of x5?
(e) What are the sums of powers of the variables in each term? (f) Find the degree of p(x).
5. Find the degree of the following polynomials:
(a) 2 x5 + 1 x3 – 1 (b) 5x2y – 3xy + 5
3 2
(c) 2 x5y2 – 2 x4y – 2xy (d) 2x2y(2x3y – 3x2y)
5 3
6. Classify the following polynomials based on their number of terms:
(a) 2x (b) 2x2 – 1 (c) 4x2 – 3x (d) 2x2 – x + 1
(h) x6 – x2 + x – 5
(e) 2 x + 1 (f) 7 (g) 3x3 – x2 + 1
3 (d) 2x2 – x + 1
7. Classify the following polynomials based on the degree of their terms: (h) 34
(a) 2x (b) 2 (c) 4x – 1
(e) 3x4 – x2 + x – 1 (f) 3x3 – 12x + 1 (g) 3x2
8. Which of the following pairs of polynomials are equal?
(a) 2x3 – 2x2 + 3x + 1 and 3x + 2x3 + 1 – 2x2
(b) 62x3 + 12 x2 – 2x – 2 and 4x2 – 8 x3 + 6 x – 2
3 2 2
(c) 8 x4 – 4 x + 6 x2 + 3 and 3x2 – 12 x + 34x4 + 3
6 3 2 9
(d) 2 x3 – 2x2 – 4 x + 4 and 8 x3 – 8 x + 3 – 6 x2 + 1
3 3 12 6 3
ANSWERS
2. (a), (d), (f)
3. (a) – 4x5 + 3x3 + 2x – 4 (b) 3x7 + 2x3 – x2 – 5x + 5
(c) –x7 – 6x5 + 4x3 – 5x2 – x + 1 (d) x4 – x3 + x2 +x–5
(d) y 3 2 (e)
4. (a) 5 (b) 4 (c) x4 6, 6, 6, 2, 0 (f) 6
5. (a) 5 (b) 3 (c) 7 (d) 7
6. (a) Monomial (b) Binomial
(e) Binomial (f) Monomial (c) Binomial (d) Trinomial
7. (a) Linear (b) Constant
(e) Bi-quadratic (f) Cubic (g) Trinomial (h) Multinomial
8. (a), (c), (d)
(c) Linear (d) Quadratic
(g) Quadratic (h) Constant
Polynomial 29
ALGEBRA 1.4 (B) Operations on Polynomials
Learning Objectives
At the end of this topic, the students will be able to:
• add the polynomials. • subtract the polynomials. • multiply the polynomials.
There are four fundamental operations (+, –, ×, ÷) on the polynomials. The operating processes of the
polynomials are as follows:
(a) Addition (or Sum) of Polynomials
Consider two polynomials
p(x) = anxn + an – xn–1 + ….. + a1x1 + a0 and q(x) = bnxn + bn–1 xn–1 + …. + b1x1 + b0.
1
Then the addition (or sum) of p(x) and q(x) is denoted by p(x) + q(x) and is defined by
p(x) + q(x) = (an + bn) xn + (an–1 + bn–1) xn–1 + … + (a1 + b1) x1 + (a0 + b0) which is also polynomial.
Thus, the sum of two polynomials can be obtained by adding the coefficient of the like terms. The degree of
the sum of two polynomials is less than or equal to the degree of the polynomial with higher degree.
Properties of addition of polynomials
In the addition of two polynomials, there exist some special properties as given below:
If p(x), q(x) and r(x) are three polynomials over real numbers, then
SN Property Definition
(i) Closure Property p(x) + q(x) will be also a polynomial over real number.
(ii) Commutative Property p(x) + q(x) = q(x) + p(x)
(iii) Associative Property {p(x) + q(x)} + r(x) = p(x) + {q(x) + r(x)}
(iv) Additive Identity p(x) + 0 = p(x)
(v) Additive Inverse p(x) + {–p(x)} = 0
(b) Subtraction (or Difference) of Polynomials
Consider two polynomials p(x) = anxn + anxn–1 + …. + a1x1 + a0 and q(x) = bnxn + bnxn–1 + … + b1x1 +
b1, where a > b. Then the subtraction (or difference) of p(x) and q(x) is denoted by p(x) – q(x) and
is defined by p(x) – q(x) = (an – bn) xn + (an – 1 – bn–1) xn–1 + ….. + (a1 – b1)x1 + (a0 – b0), which is also
polynomial.
Thus, the difference of two polynomials can be obtained by subtracting the coefficient of the
like terms. The degree of the difference of polynomials is less than or equal to the degree of the
polynomial with higher degree.
(c) Multiplication (Product) of Polynomials
If p(x) and q(x) are two polynomials then their product is denoted by p(x)×q(x) or p(x).q(x), which
is again a polynomial. It is obtained by applying the distributive law repeatedly and then adding
or subtracting the like terms. Its degree is the sum of degree of the given two polynomials. i.e.,
the degree of p(x) × q(x) = {the degree of p(x)} + {the degree of q(x)}, while multiplying the
polynomials, we use the law of index, xm × xn = xm+n.
Example : If p(x) = 2x2 + x – 1 and q(x) = 3x2 – 2x + 3 then
p(x) × q(x) = (2x2 + x – 1) × (3x2 – 2x + 3)
30 Illustrated Optional Mathematics-9
= 2x2(3x2 – 2x + 3) + x(3x2 – 2x + 3) – 1(3x2 – 2x + 3) “Alternatively” ALGEBRA
= 6x4 – 4x3 + 6x2 + 3x3 – 2x2 + 3x – 3x2 + 2x – 3
= 6x4 – x3 + x2 + 5x – 3, 2x2 + x – 1
× 3x2 – 2x + 3
which is the polynomial with the degree 2 + 2 = 4 as the sum of the degree 6x4 + 3x3 – 3x2
of the given polynomials. – 4x3 – 2x2 + 2x
+ + 6x2 + 3x – 3
Properties of Multiplication of Polynomials
6x4 – x3 + x2 + 5x – 3.
Multiplication of polynomials satisfy the properties in the table below.
If p(x), q(x) and 4(x) are three non–zero polynomials then,
SN Property Definition
1. Closure The product p(x).q(x) is also a polynomial.
2. Commutative p(x).q(x) = q(x).p(x).
3. Associative {p(x).q(x)}.r(x) = p(x).{q(x).r(x)}.
P(x)×1 = 1×p(x) = p(x) or, p(x) × I(x) = I(x).p(x) = p(x),
4. Existence of Multiplicative Identity
where I(x) is a unit polynomial.
5. Distributive over Addition p(x){q(x) + r(x)}= p(x).q(x) + p(x).r(x).
6. Cancellation If p(x).q(x) = r(x).q(x), then p(x) = r(x).
Example 1 : Add the following polynomials and Example 2 : Subtract q(x) from p(x): p(x) = 3x2 +
find its degree: p(x) = 2x2 – 3x + 4 4x + 2 and q(x) = 2x2 + 2x + 1
and q(x) = 3x2 + 2x – 3
Solution:
Solution: Here, p(x) = 3x2 + 4x + 2 and q(x) = 2x2 + 2x + 1
Here, p(x) = 2x2 – 3x + 4 and q(x) = 3x2 + 2x – 3 Now, p(x) – q(x)
Now, p(x) + q(x) = 3x2 + 4x + 2 – (2x2 + 2x + 1)
= 2x2 – 3x + 4 + 3x2 + 2x – 3 = 3x2 + 4x + 2 – 2x2 – 2x – 1
= 2x2 + 3x2 – 3x + 2x + 4 – 3 = 3x2 – 2x2 + 4x – 2x + 2 – 1
= 5x2 – x + 1. = (3 – 2)x2 + (4 – 2)x + (2 – 1)
∴ The required degree is 2. = x2 + 2x + 1.
Example 3 : What should be added to p(x) = 3x3 Example 4 : Simplify : f(x) + g(x) – h(x) if
– 2x2 + 4x + 5 to get q(x) = 2x2 – 3? f(x) = 3x3 – 2x + 4, g(x) = 2x2 + 2x – 5
and h(x) = 2x3 – x2 + 3x + 2.
Solution:
Given, p(x) = 3x3 – 2x2 + 4x + 5; q(x) = 2x2 – 3 Solution:
Let r(x) be added to p(x) to get q(x) Here, f(x) = 3x3 – 2x + 4, g(x) = 2x2 + 2x – 5 and
Then, P(x) + r(x) = q(x) h(x) = 2x3 – x2 + 3x + 2
or, r(x) = q(x) – p(x) Now, f(x) + g(x) – h(x)
= 2x2 – 3 – (3x3 – 2x2 + 4x + 5) = (3x3 – 2x + 4) + (2x2 + 2x – 5) –
= 2x2 – 3 – 3x3 + 2x2 – 4x – 5 (2x3 – x2 + 3x + 2)
= – 3x3 + (2 + 2)x2 – 4x – (3 + 5) = 3x3 – 2x + 4 + 2x2 + 2x – 5 – 2x3 + x2 – 3x – 2
= (3 – 2)x3 + (2 + 1)x2 + (–2 + 2 – 3)x + (4 – 5 – 2)
= – 3x3 + 4x2 – 4x – 8. = x3 + 3x2 – 3x – 3.
Polynomial 31
ALGEBRA EXERCISE – 1.4 (B)
1. Add the following polynomials and also find the degree of the sum in each case:
(a) p(x) = 3x2 – 2x + 1 and q(x) = 2x2 + 3x + 4
(b) p(x) = 3x3 – 2x2 + x – 5 and q(x) = 4x – 2x2 – 3x3
2. Subtract q(x) from p(x) and find the degree:
(a) p(x) = 5x2 – 3x + 2 and q(x) = 3x2 + 4x – 2
(b) p(x) = 4x2 + 4x – 3 and q(x) = 2x2 – 2x – 3
3. (a) If p(x) = 3x2 + 2x – 3 and q(x) = 4x2 – 3x + 2, then verify that the commutative property of
addition holds.
(b) If r(x) = 2x3 – 2x + 1 and s(x) = 3x3 – 2x2 + 4x – 3, then show that r(x) + s(x) = s(x) + r(x).
(c) If f(x) = 3x2 – 2x, g(x) = 2x2 + 2x – 1 and h(x) = 2x3 – 2x2 + 3x – 5, then prove that {f(x) +g(x)}
+ h(x) = f(x) + {g(x) + h(x)}.
(d) Show that the associate property of addition for the polynomials p(x) = 2x2 – 1, q(x) = 3x2 – x
+ 1 and r(x) = 3x – 5 holds.
4. (a) Find the additive inverse of the polynomial p(x) = 3x3 – x2 + 3x + 4.
(b) Find the additive inverse of the polynomial q(x) = 4x4 – x2 + 3x + 2.
5. (a) What should be added to q(x) = 3x2 + 2x – 1 to get p(x) = 3x2 – 3x + 4?
(b) What should be added to r(x) = 2x + 1 to get s(x) = 5x2 – 3x + 3?
(c) What should be subtracted from f(x) = 3x3 + 2x2 – x + 1 to get g(x) = 2x3 + x2 + 3?
(d) What should be subtracted from p(x) = x2 to get q(x) = 3x3 – 4x2 + 2x – 1?
6. (a) Subtract p(x) = 3x2 + 4x – 2 from the sum of q(x) = 2x2 – x + 5 and r(x) = 3x2 + 3x – 2.
(b) Subtract h(x) = 3x3 – 2x2 + 4x + 1 from the sum of f(x) = 3x3 + 4x2 – 3x + 5 and g(x) = 2x3 +
2x2 – x + 1.
(c) Add p(x) = 2x2 – x + 1 with the difference of q(x) = 3x2 – 2x + 2 and r(x) = 4x2 – x – 3.
(d) Add f(x) = 3x3 – x + 1 with the difference of g(x) = 2x3 – 7x2 + 5x – 2 and r(x) = 3x3 – 2x2 + x – 1.
7. Multiply the following polynomials:
(a) p(x) = 2x + 1 and q(x) = 3x2 – x + 4 (b) f(x) = 3x2 – 2 and g(x) = 2x2 + x – 3
(c) p(x) = 3x2 – x + 1 and q(x) = 2x2+ 3x – 2 (d) s(x) = x3 + 2x + 1 and t(x) = 2x2 – x + 3
ANSWERS
1. (a) 5x2 + x + 5; 2 (b) –4x2 + 5x – 5; 2 2. (a) 2x2 – 7x + 4; 2 (b) 2x2 + 6x; 2
4. (a) –3x3 + x2 – 3x – 4 (b) –4x4 + x2 – 3x – 2
5. (a) 5 – 5x (b) 5x2 – 5x + 2 (c) x3 + x2 – x – 2 (d) –3x3 + 5x2 – 2x + 1
6. (a) 2x2 – 2x + 5 (b) 2x3 + 8x2 – 8x + 5 (c) x2 – 2x + 6 (d) 2x3 – 5x2 + 3x
7. (a) 6x3 + x2 + 7x + 4 (b) 6x4 + 3x3 – 13x2 – 2x + 6
(c) 6x4 + 7x3 – 4x2 + 2x – 2 (d) 2x5 – x4 + 7x3 + 5x + 3
Project Work
Verify the properties of polynomial addition and multiplication for the following polynomials:
f(x) = 3x2 + 2x – 5, g(x) = 4x2 + 3x + 1 and h(x) = 2x2 – 5x + 6
32 Illustrated Optional Mathematics-9
1.51 SEQUENCE AND SERIES ALGEBRA
1.5 (A) Introduction and Types of Sequence
Learning Objectives
At the end of this topic, the students will be able to:
• identify and define sequence. • identify the types of sequence.
Review of Sequence
Observe the following patterns of figures:
(i) , , , , _______, _______, ..................
(ii) , , , , _______, _______, ................
Can you draw the next figure that comes immediately to the last figure of the above pattern? Can you find
the rules between every consecutive (or successive) figures? What are the number forms of the patterns
of figures?
The above patterns of figures give the following number forms:
(i) 1, 4, 7, 10, .... (ii) 1, 3, 6, 10, ....
The numeral form (i) has the number in every successive term is
increased by 3, i.e. difference of every successive term is 3. But
in the numeral form (ii) has the number in every successive term and
is increased as a pattern of 2, 3, 4, ..... . Therefore, the next figure
in (i), in which it has 10 + 3 = 13 circles, and (ii) in which it has 10 + 5 = 15 circles, are respectively as:
Now, find the number of small circles of next other two shapes in the sequence, i.e., 6th and 7th shapes in
the same rules in the both patterns. I hope, you can do easily as given below:
Patterns 6th Shape 7th Shape
(i) 13 + 3 = 16 circles 16 + 3 = 19 circles
(ii) 15 + 6 = 21 circles 21 + 7 = 28 circles
Let us consider some more sets of numbers and the rules used to form them shown in the table below:
SN Set of Numbers Rule
(i) 8, 15, 22, 29, ... Increased by 7, i.e. backward difference = 7.
Sequence and Series 33
ALGEBRA (ii) 5, 2, –1, –4, ..... Decreased by 3, i.e. backward difference = –3.
Increased in the a pattern of 4, 6, 8, ......
(iii) 2, 6, 12, 20, .... Multiply by 2, i.e. backward ratio = 2.
(iv) 2, 4, 8, 16, .... 1
Divide by 10, i.e. backward ratio = 10 .
(v) 100, 10, 1, 1 , ...
10
A sequence is the objects (or numbers) of a set with specific rule. A number (or an object) of the sequence
is known as term. Generally, the sequence is written as t1, t2, t3, t4, ......, where, t1 = first term, t2 = second
term, t3 = third term, t4 = fourth term, .... .
Note: the t1 is also denoted by 'a'.
The same (or constant) difference of any two consecutive (successive) terms is known as common difference.
It is also known as backward difference. It is obtained by subtracting the term from just preceding term.
∴ Common difference (d) = Backward term – forward (preceding) term.
Note: d = t2 – t1 = t3 – t2 = .....,
where t1 = first term, t2 = second term, t3 = third term ....
All of the above examples are connected with any rule like as a function. For example,
If N = {1, 2, 3, 4, ... } and the function f(x) = 2x – 1, and x∈N then f(1) = 2 × 1 – 1 = 1, f(2) = 2 × 2 – 1 =
3, f(3) = 2 × 3 – 1 = 5, f(4) = 2 × 4 – 1 = 7 and so on. f
Now, the range of f is {1, 3, 5, 7, ....} under the rule f(x) = 2x – 1. This 1 1
function is showed in mapping diagram. 2 3
3 5
Here, the images of natural numbers under the function f are as the set of 4 7
ordered natural numbers. Thus, the range of the function which has the . .
domain as natural number, is a sequence. . .
. .
Example 1 : Add the next one shape in the following sequences of figures:
(i)
, , , , ........ , .........
(ii)
, , , , ............ , ............
(i) Here,
Solution:
The numeral sequence of the given pattern of figures is
3, 6, 9, 12, ...
Difference: 3 3 3 ... → Common difference
Here, the difference of any two consecutive terms is 3.
Therefore, the number of dots in the next shape is 12 + 3 = 15.
Hence, the required next one shape is as in the adjoining figure:
34 Illustrated Optional Mathematics-9
(ii) Here, the numeral sequence of the given pattern of figures is ALGEBRA
1, 3, 6, 10, ...
1st Difference: 2 3 4 ...
2nd Difference: 1 1 → Common difference
Here, the differences of every term as sequence is 2, 3, 4, .... Then the next difference
will be 5. Therefore, the number of dots in the next one shape is 10 + 5 = 15. Hence,
the required one shape is as in the adjoining figure:
Example 2 : Evaluate the common difference of the sequence 1, 7, 13, 19, .....
Solution: Given, the sequence is 1, 7, 13, 19, ...............
The difference between 7 and 1 = 7 – 1 = 6
The difference between 13 and 7 = 13 – 7 = 6
The difference between 19 and 13 = 19 – 13 = 6
Here, the difference of any two consecutive terms is the same. Therefore, the common
difference of the given sequence is 6.
Type of Sequences
Consider the different types of sequences:
i. 2, 5, 8, 11, 14, 17, 20 ii. 8, 5, 2, –1, –4, .... iii. 5, 8, 12, 17, ... iv. 3, 6, 12, 24, 48, 96
What differences can be observed in the above examples of sequences?
The sequence (i) has finite number of terms, i.e. there are 7 terms in it. Also, its every term is increased one after
another and the difference between any two consecutive terms is 3 in it.
Similarly, the sequence (ii) has infinite number of terms, which cannot be counted. Its every term is
simultaneously decreased by 3. Therefore, the difference between any two consecutive terms is –3 in it.
For the sequence (iii), there are infinite number of terms. Its every following term is increased by 3, 4 and
5 respectively. But its difference is 1.
Similarly, the sequence (iv) has finite number of terms. Its every following term is formed multiplying by
2 respectively. It has the same common ratio. Therefore, its common ratio is 2.
Now, how many types of sequences are there in the above mentioned examples? There are four categories
of sequences. They are as follows:
a. Finite and Infinite Sequences
A sequence having finite number of terms is called a finite sequence and a sequence having infinite
number of terms is called an infinite sequence. The above mentioned examples (i) and (iv) are finite
sequences, but that of the examples (ii) and (iii) are infinite sequences.
b. Ascending and Descending Sequences
A sequence whose every term, except first term, is increased to the previous term is called an
ascending sequence and a sequence whose every term, except first term, is decreased to the previous
term is called a descending sequence. The examples (i), (iii) and (iv) are the ascending sequences,
but, that of the example (ii) is the descending sequence.
Sequence and Series 35
ALGEBRA c. Linear and Quadratic Sequences
A sequence having the same difference at first step is called a linear sequence and a sequence having
the same difference at second step is called a quadratic sequence. The examples (i) and (ii) are the
linear sequences and the example (iii) is the quadratic sequence. The general form of the linear and
quadratic sequences are an + b and an2 + bn + c respectively, where a, b, c ∈ R and n ∈ N.
d. Arithmetic and Geometric Sequences
A sequence having the same difference (backward difference) at first step is called an arithmetic
sequence (AS). The linear sequence is also the arithmetic sequence. A sequence having the same
ratio (backward ratio) at first step is called a geometric sequence (GS). The examples (i) and (ii) are
the arithmetic sequences and the example (iv) is the geometric sequence.
Example 3 : Identify the types of the following sequences with reasons:
(i) 5, 11, 17, 23, .. (ii) 18, 7, – 4, –15,
(iii) 100, 90, 75, 55, ... (iv) 1, 2, 4, 8, 16, 32,
Solutions: For the sequence (i) 5, 11, 17, 23, ... :
There are infinite number of terms. The difference between its any two consecutive terms
at first step is equal to 6. Also, its every term is increased by 6 than its previous term.
Therefore, the sequence (i) is the infinite or ascending or linear or arithmetic sequence.
For the sequence (ii) 18, 7, – 4, –15 :
There are finite number of terms. The difference between its any two consecutive terms
at first step is equal to –11 and its every term is decreased by 11 then its previous term.
Therefore, the sequence (ii) is the finite or descending or linear or arithmetic sequence.
For the sequence (iii) 100, 90, 75, 65, ....:
There are infinite number of terms. The differences between its any two consecutive terms
at first step are –10, –15, –20, ... and that of second difference is –5. Also, its every term is
decreased by 10, 15, 20 ... respectively than its previous term. Therefore, the sequence (iii)
is the infinite or descending or quadratic sequence.
For the sequence (iv) 1, 2, 4, 8, 16, 32:
There are finite number of terms. The ratio of its any two consecutive terms is equal to 2 and
its every term is formed by multiplying of 2 (or is increased by multiplication of 2) than its
previous term. Therefore, the sequence (iv) is finite or ascending or geometric sequence.
EXERCISE – 1.5 (A)
1. (a) Define arithmetic sequence and geometric sequence.
(b) Differentiate in one sentence between:
(i) finite and infinite sequence
(ii) ascending and descending sequence
(iii) linear and quadratic sequence
(c) What is common difference in arithmetic sequence?
36 Illustrated Optional Mathematics-9
2. Draw the next one shape in the following sequences of figures and represent in numeral form:
(a) (b) ALGEBRA
,, , , , , , ,.........
,......... (d) , ,.........
(c)
,, , ,,
3. Write down next three numbers in the following sequences:
(a) 7, 11, 15, 19, ..... (b) 278, 197, 161, –135, ........ (c) 1, 3, 7, 13, ....
(d) 12, 22, 36, 56, 84, 122, ... (e) 2, 8, 32, 128, ... (f) 217, 19, 1 , 1, .....
3
4. Find the common difference of the following sequences:
(a) 12, 15, 18, 21, ... (b) 22, 13, 4, –5, .... (c) 2 1 , 5, 7 12, 10, ...
2
(d) 2, 3, 5, 8, .... (e) 2, 8, 19, 37, 64, 102, .... (f) 6, 11, 20, 36, 62, 101...
5. Find the common ratio of the following sequences:
(a) 2, 4, 8, 16, ... (b) 1, 3, 9, 27, ... (c) 8, 4, 2, 1, ...
(d) 1, 110, 1010, 10100, ... (e) 0.625, 0.25, 0.1, 0.04, ... (f) 13, 313, 3613, 39193, ...
6. Which of the following sequences are linear or quadratic ? Give reason.
(a) 2, 21, 40, 59, ... (b) 2, 11, 25, 44, ... (c) 25, 17, 9, 1, ...
(d) 26, 30, 35, 41, ... (e) 5, 17, 29, 41, ... (f) – 4, 6, 21, 41, ....
7. Identify the types of sequences of the following sequences with reason.
(a) 2, 4, 6, 8, 10 (b) 21, 14, 7, 0, ... (c) 1, 2, 4, 7, 11, ..
(d) 102, 105, 110, 117 (e) –7, –5, –3, –1, ... (f) 1, 2, 4, 8, ..
(g) –3, –5, –9, –15, (i) 2.3, 3.4, 4.5, 5.6
(j) 54, 86, 181, 1104, .... (h) 0.25, 0.3, 0.35, 0.4, (l) 275, 242, 119, –162, ...
(k) 23, 46, 182, 24
16
ANSWERS
2. (a) ; 1, 5, 9, 13, .... (b) ; 1, 5, 9, 13, ... (c) ; 1, 7, 13, 19, .....
(d) ; 1, 3, 7, 13..... 3. (a) 23, 27, 31 (b) –1156, –1277, –1398 (c) 21, 31, 43
(d) 172, 236, 316 (e) 512, 2048, 8192 (f) 3, 9, 27
4. (a) 3 (b) –9 (c) 2 1 (d) 1 (e) 2 (f) 3
2
5. (a) 2 (b) 3 (c) 1 (d) 1 (e) 0.4 (f) 1 6. & 7. show to your teacher
2 10 11
Sequence and Series 37
ALGEBRA 1.5 (B) General Term of Sequence
Learning Objectives
At the end of this topic, the students will be able to:
• find the general or nth term of the linear, quadratic and geometric sequences.
Let us consider the sequences of figures:
(i)
, , , , ............ , ............
(ii)
, , , , ............ , ............
Can you add one more figure in the above sequential patterns? Can you find the number of dots in the sixth
figure on both sequences? But can you find the number of dots in the 25th figure on the both sequences?
Can it be done easily? Are they possible?
It is so difficult. For these, at first we find the rule of the sequence and then find them easily.
The rule of the sequence is the specific term for nth term. Let t1, t2, t3, t4, .... tn, ..... be a sequence, then its
nth term is denoted by tn. Thus, the standard form of sequence is as t1, t2, t3, t4, .... tn, ..... or, a1, a2, a3, a4,
...........an, ....... or, u1, u2, u3, u3, ........... un, ...... . The nth term of a sequence is known as a general term of
the sequence. It is denoted by tn, an or un, etc. and is obtained by the rule, which describes the specific
order of the terms of the sequence.
General Term of Linear or Arithmetic Sequence
Consider a linear sequence,
2, 9, 16, 23, ............
1st Difference: 7 7 7 ...........
The linear sequence has constant first differences, that is, the differences of any two consecutive terms
are same as at the first step. So, it has the general term tn = an + b, where a, b ∈ Q and n ∈ N. It is the first
degree algebraic expression in the variable n, n ∈ N.
There are two methods of finding the general term of the linear sequence. They are
(i) Method of making pattern (ii) Method of using formula
Method-1: Method of making pattern
Consider a linear sequence,
2, 9, 16, 23, ............
1st Difference: 7 7 7 ...........
38 Illustrated Optional Mathematics-9
We make a pattern to generalize the given sequence as shown in the table below:
Terms 1st Step 2nd Step Since the first difference is 7. So, what ALGEBRA
t1 = 2 7–5 7×1–5 to do in 7, which is equal to 2 ? 7 – 5 = 2
t2 = 9 14 – 5 7×2–5 and so on.
t3 = 16 21 – 5 7×3–5
t4 = 23 28 – 5 7×4–5 We have, 7, 14, 21, 28, .... are the
........ ........ ......... multiples of 7. So, take 7 as common
tn ........ 7n – 5 in each multiple.
Hence, the general term (or nth term) of the given sequence is 7n – 5, which is also the rule
of that sequence. Also, tn = 7n – 5, in which d = 7 and (a – d) = –5. So, we formulate the
nth term for the linear sequence as tn = dn + (a – d) = dn + a – d = a + (n – 1)d.
Method-2 : Method of using formula
The given sequence is 2, 9, 16, 23, .....
Here, the first term (a) = 2
The constant or common difference (d) = 9 – 2 = 7
Now, we have
The nth term (tn) = dn + (a – d)
= 7n + (2 – 7)
= 7n – 5
Thus, the general term or the nth term of the given sequence is 7n – 5.
Alternatively
The nth term (tn) = a + (n – 1)d
= 2 + (n – 1)7
= 2 + 7n – 7 = 7n – 5. 2, 5, 10, 17, 26, ...
General Term of Quadratic Sequence
Consider a quadratic sequence 2, 5, 10, 17, ... . 1st Difference: 3 5 7 9 ...
2nd Difference: 2 2 2 ...
The above sequence has not equal difference at first, but
equal difference at second. This type of sequence is so
called the quadratic sequence.
Thus, its general term has second degree of algebraic expression as the form of an2 + bn + c, where a, b,
c ∈ Q and n ∈ N.
The finding of the general term of these types of sequences by the following methods:
Method-1: By making pattern
Consider a quadratic sequence,
2, 5, 10, 17, 26, ...
1st Difference: 3 5 7 9 ...
2nd Difference: 2 2 2 ...
Sequence and Series 39
We make a pattern to generalise the above sequence shown in the table below:
ALGEBRA Terms 1st Step 2nd Step Rough
t1 = 2 1+1 12 + 1
Method-2: t2 = 5 4+1 22 + 1 We first know,
t3 = 10 9+1 32 + 1 tn = an2 + .....
t4 = 17 16 + 1 42 + 1
t5 = 26 25 + 1 52 + 1 The second difference, 2a = 2 ∴ a = 1
∴ tn = 1 × n2 + .... = n2 + ...,
... ... ...
tn ... n2 + 1 where n∈N. So, which is added to the sequence
of natural number equals to 2 and so on ?
Since 1, 4, 9, 16, 25 ... are the square of natural
numbers. So, square each of them.
Hence, the general term (or nth term) of the given sequence is n2 + 1, which is also the rule
of that sequence.
By using formula:
The given sequence is
2, 5, 10, 17, 26, ...
1st Difference: 3 5 7 9 ...
2nd Difference: 2 2 2 ...
Now, we have
The general term of the given quadratic sequence is tn = an2 + bn + c ............... (i)
Second difference, 2a = 2
or, a = 1
Substituting the value of a in eqn (i), we get tn = n2 + bn + c ............................ (ii)
If n = 1, then t1 = 12 + b × 1 + c
or, 2 = 1 + b + c or, b + c = 1 ................................ (iii)
If n = 2, then, t2 = 22 + b × 2 + c
or, 5 = 4 + 2b + c or, 2b + c = 1 .............................. (iv)
Subtracting the eqn (iii) from eqn (iv), we have
2b + c = 1
–b +–c = –1
b = 0
Putting the value of b in eqn (iii), we get,
0+c=1 or, c = 1.
Also, putting the values of b and c in eqn (ii), we get
tn = n2 + 0 × n + 1 = n2 + 1, which is the required general term (or nth term).
General Term of a Geometric Sequence
Consider a sequence, 2, 6, 18, 54, ............
1st Ratio: 3 3 3 ...........
40 Illustrated Optional Mathematics-9
Here, the above sequence has the same backward ratio (r) at first step. This constant backward ratio (r) ALGEBRA
is commonly known as common ratio. Thus, its general term has the (n – 1) degree on r of algebraic
expression as the form of a . rn–1, where a = first term, r = common ratio and n ∈ N.
Find the general term of these types of sequence by the following methods:
Method-1 : By making pattern
Consider a geometric sequence
2, 6, 18, 54, ............
1st Ratio: 3 3 3 ...........
Making a pattern to generalise the above sequence shown in the given table.
Terms 1st Step 2nd Step 3rd Step
t1 = 2 2×1 2 × 30 2 × 31–1
t2 = 6 2×3 2 × 31 2 × 32–1
t3 = 18 2×9 2 × 32 2 × 33–1
t4 = 54 2 × 27 2 × 33 2 × 34–1
........ ........ ......... .........
tn ........ ......... 2 × 3n – 1
Hence, the general term (or nth term) of the given sequence is 2 × 3n–1, which is the rule of
the geometric sequence.
Method-2 : By using formula:
The given sequence is 2, 6, 18, 54, ....
Here, the first term (a) = 2
The common ratio (r) = 6 = 18 = 54 =3
2 6 18
∴ The general (nth) term (tn) = arn–1 = 2 × 3n–1.
Example 1 : Find the nth term of the sequence 25, 12, –1, –14, ..... Also evaluate the values of 25th
term and 50th term of the same sequence.
Solution: Given,
25, 12, –1, 14, ............
1st Difference: –13 –13 –13 ...........
Making a table for arranging the pattern: What to do in –13, it
comes 25 ?
Terms 1st Step 2nd Step
t1 = 25 38 – 13 38 – 13 × 1 Sequence and Series 41
t2 = 12 38 – 26 38 – 13 × 2
t3 = –1 38 – 39 38 – 13 × 3
t4 = –14 38 – 52 38 – 13 × 4
........ ........ .........
tn
38 – 13n
Again, the 25th term (t25) = 38 – 13 × 25 = – 287 and the
ALGEBRA 50th term (t50) = 38 – 13 × 50 = – 612.
Alternatively
Given,
Here, the first term (a) = 25
The common difference (d) = –13
The nth term (tn) = dn + (a – d)
= –13n + (25 – (–13))
= –13n + 38
= 38 – 13n
∴ The 25th term (t25) = 38 – 13 × 25 = –287 and
The 50th term (t50) = 38 – 13 × 50 = – 612.
Example 2 : Find the 24th term and 36th term of the sequence 1, 3, 9, 19, 33, ...
Solution : Given,
1, 3, 9, 19, 33,............
1st Difference: 2 6 10 14 .......
2nd Difference: 4 4 4
Method – 1: Making a table for arranging the pattern
Terms 1st Step 2nd Step 3rd Step Rough
t1 = 1 2 × 12 – 1 2 × 12 – (4 – 3) 2 × 12 – (4 × 1 – 3) 2nd Difference:
t2 = 3 2 × 22 – 5 2 × 22 – (8 – 3) 2 × 22 – (4 × 2 – 3) 2a = 4 , or, a = 2
t3 = 9 2 × 32 – 9 2 × 32 – (12 – 3) 2 × 32 – (4 × 3 – 3) ∴ tn = 2n2 + .....
= 2 × 12 + ...
t4 = 19 2 × 42 – 13 2 × 42 – (16 – 3) 2 × 42 – (4 × 4 – 3)
2 × 22 + ......
t5 = 33 2 × 52 – 17 2 × 52 – (20 – 3) 2 × 52 – (4 × 5 – 3) 2 × 32 + ......
........ ........ ........ ........ For 1, 5, 9, 13, ....
d = 4 ∴tn = 4n – 3.
tn 2n2 – (4n – 3)
= 2n2 – 4n + 3
Method-2: By using formula:
Let the nth term of the given quadratic sequence be tn = an2 + bn + c ......... (i)
Then, the second difference, 2a = 4 or, a = 2
Put a = 2 in eqn (i), we obtain
tn = 2n2 + bn + c .......... (ii)
If n = 1, then t1 = 2 × 12 + b × 1 + c
or, 1 = 2 + b + c
or, b + c = –1 ......... (iii)
42 Illustrated Optional Mathematics-9
If n = 2, then t2 = 2 × 22 + b × 2 + c ALGEBRA
or, 3 = 8 + 2b + c
or, 2b + c = –5 ........ (iv)
Subtracting the eqn (iii) from eqn (iv), we have,
2b + c = –5
b + c = –1
–– +
b = – 4
Put b = –4 in eqn (iii), we get
– 4 + c = –1 ∴ c = 3
Also, put b = – 4 and c = 3 in eqn (ii), we have
tn = 2n2 + (– 4)n + 3 = 2n2 – 4n + 3
Thus, the 24th term (t24) = 2 × 242 – 4 × 24 + 3 = 1059 and
The 36th term (t36) = 2 × 362 – 4 × 36 + 3 = 2451
Example 3 : Write down the first five terms and 30th term of the sequence whose nth term is (–1)n (3n – 2).
4n – 1
(–1)n (3n – 2)
Solution: Given, tn = 4n – 1
Put n = 1, the first term (t1) = (–1)1 (3 × 1 – 2) = – 1
4 ×1–1 3
Put n = 2, the second term (t2) = (–1)2 (3 × 2 – 2) = 4
4 ×2–1 7
Put n = 3, the third term (t3) = (–1)3 (3 × 3 – 2) = – 7
4 ×3–1 11
Put n = 4, the fourth term (t4) = (–1)4 (3 × 4 – 2) = 10
4 ×4–1 15
Put n = 5, the fifth term (t5) = (–1)5 (3 × 5 – 2) = – 13
4 ×5–1 19
Thus, the required first five terms are – 31, 74, – 171, 10 and – 1193.
15
Again, the 30th term (t30) = (–1)30 (3 × 30 – 2) = 88 .
4 × 30 –1 119
EXERCISE – 1.5 (B)
1. (a) What is the general term of a sequence?
(b) Write the formula to compute the fourth term of an arithmetic sequence.
(c) What is the standard form of quadratic sequence?
2. Write down the first five terms of the sequences whose nth term are given below:
(a) tn = 2n + 5 (b) tn = 2 – 7n (c) an = (n + 1) (2n – 1)
(d) un = (3n – 1)2 (e) tn = 2n2 + 4 (f) an = 1 – 2n
3n + 1
(–1)n (3n + 1)
(g) un = (–1)n + 1 (2n2 – 1) (h) tn = 2n
Sequence and Series 43
3. Find the general term of the following sequences:
ALGEBRA (a) 8, 11, 14, 17 .... (b) 27, 29, 31, 33, .... (c) 50, 45, 40, 35, ....
4. What is the rule which forms the following sequences?
(a) 4, 7, 12, 19, .... (b) 1, 7, 17, 31, .... (c) 2, 1, –2, –7, ...
5. Find the nth term of the following sequences:
(a) 1, 3, 9, 27, ... (b) 2, 6, 18, 54, ... (c) 8, 16, 32, 64, ...
6. Find the nth term of the following sequences. Also, evaluate the 30th term and 45th term.
(a) 1, 2, 3, 4, .... (b) 8, 3, – 2, –7, .... (c) 5, 11, 19, 29, ...
(d) 3 , 4 , 6 , 9 , ... (e) 0.28, 0.35, 0.42, 0.49, ... (f) 3 , 3 , 3 , 3 , ...
5 7 9 11 2048 1024 512 256
7. (a) Add the next one figure on the following sequence , ,, , ....
and find the number of triangles in the 25th figure:
(b) Draw the next one shape on the following sequence , ,, , ....
of figures and find the number of glasses in the 20th
shape:
(c) Construct the next one shape on the following , , , , ....
sequence of figures and also find the number of
vertices in the 30th shape:
ANSWERS
2. (a) 7, 9, 11, 13, 15 (b) –5, –12, –19, –26, –33 (c) 2, 9, 20, 35, 54
(f) – 1 , – 3 , – 5 , – 7 , – 9
(d) 4, 25, 64, 121, 196 (e) 6, 12, 22, 36, 54
4 7 10 13 16
(g) 1, –7, 17, –31, 49 (h) –4 , 7 , –10 , 13 , –16
2 4 6 8 10 (c) 55 – 5n
3. (a) 3n + 5 (c) 1 + 2n – n2
4. (a) n2 + 3 (b) 2n + 25 (c) 2n + 2
5. (a) 3n – 1
(b) 2n2 – 1
(b) 2 × 3n – 1
6. (a) tn = n, t30 = 30, t45 = 45 (b) tn = 13 – 5n, t30 = –137 t45 = –212
(c) tn = n2 + 3n + 1, t30 = 991, t45 = 2161
1 n2 – 1 n + 3 438, 993
(d) tn = 2 2 63 93
3 + 2n , t30 = t45 =
(e) tn = 0.07n + 0.21, t30 = 2.31, t45 = 3.36 (f) 3 × 2n–12, t30 = 786432, t45 = 25769803776
7. (a) ; 73 (b) ; 210 (c) ; 536870912
44 Illustrated Optional Mathematics-9
1.5 (C) Introduction and Types of Series ALGEBRA
Learning Objectives
At the end of this topic, the students will be able to:
• define series and its types. • solve the problems related to partial sum.
Series
Aarisha saves the money everyday adding Rs. 5 more per next day. She starts to save Rs. 10 from the first
day for the 7th day of a month.
How much does money she save on the seventh day?
How much does money she save all together?
To find the answer of the first question, we complete the saving amount per day in the given table below:
Day 1st 2nd 3rd 4th 5th 6th 7th
Amount 10 10 + 5 15 + 5 20 + 5 25 + 5 30 + 5 35 + 5
Sequence 10 15 20 25 30 35 40
Thus, she saves Rs. 40 on the seventh day. But the second question, we need to add all of these amounts
of seven days and arranged as a sequence 10, 15, 20, 25, 30, 35, 40.
i.e. the sum = 10 + 15 + 20 + 25 + 30 + 35 + 40 = 175.
Thus, she saves Rs. 175 all together for these seven days.
The representation of the terms of a sequence in the sum or addition is called a series associated with
that sequence. For example, the series of the sequence 2, 6, 14, 26, 42, .... is 2 + 6 + 14 + 26 + 42 + .......
In general, t1, t2, t3, t4, ... , tn, .... is a sequence, then its associated series is represented by Sn and is written
as Sn = t1 + t2 + t3 + t4 + .... + tn + ..........
Types of Series
There are four types of series as in the case of sequence.
Finite and Infinite Series
A series having finite number of terms is called a finite series. The general form of the finite series is t1 +
t2 + t3 + t4 +... + tn. For example;
(i) 4 + 7 + 10 + 13 + ... to 30 terms, (ii) 1 + 5 + 9 + 13 + 17 + 21 + 25
A series having infinite number of terms is called an infinite series. The general form of the infinite series
is t1 + t2 + t3 + t4 + .... + tn + ..... For examples;
(i) 2 + 5 + 8 + 11 + ............... (ii) – 2 + 6 – 11 + 17 – ............
Ascending and Descending Series
A series in which its every term is increased one after another is called an ascending series. For examples;
(i) 2 + 5 + 10 + 17 + ......... (ii) 10 + 15 + 20 + 25 + 30 + 25
A series in which its every term is decreased one after another is called a descending series. For examples;
(i) 10 + 7 + 4 + 1........ (ii) 1001 + 992 + 983 + 974 + ....
Sequence and Series 45
ALGEBRA Linear and Quadratic Series
A series in which the first backward differences of every two consecutive terms are constant is called a
linear series. For examples;
(i) 2 + 5 + 8 + 11 + ... (ii) 100 + 95 + 90 + 85 + ...
1st Difference: 3 3 3 ... 1st Difference: – 5 – 5 – 5 ...
A series in which the first backward differences of every two consecutive terms are not equal and the
second backward differences of every two successive terms are constant, is called a quadratic series.
For examples;
(i) 5 + 10 + 17 + 26 + ...........
1st Difference: 5 7 9 ....... Linear Sequence
2nd Difference: 2 2
(ii) 25 + 23 + 20 + 10 + ...........
1st Difference: – 2 – 3 – 4 ....... Linear Sequence
2nd Difference: –1 –1
Arithmetic and Geometric Series
A series having the constant first backward difference of any two successive terms is called an arithmetic
series (AS). It is a linear series. For examples;
(i) 5 + 8 + 11 + 14 + .... .. (ii) 15 +5 – 5 – 10 – ......
1st Difference: 3 3 3 ...... 1st Difference: –10 –10 –10 ......
A series having the constant first backward ratio of any two consecutive terms is called a geometric series.
For example;
(i) 2 +4 +8 +16 + ...... (ii) 1 + 1 + 1 + 1 + ......
10 100 1000 10000
1st Ratio: 2 2 2 ...... 1st Ratio: 1 1 1 ......
10 10 10
Progression
A sequence (or series) of numbers in which there is a constant relation between any two consecutive terms, is
called a progression. Each term of progression is obtained by operating on the previous term under specific
rule. The most common progressions are arithmetic progression (AP) and geometric progression (GP).
Note: (i) The AP is same as arithmetic sequence or series. (ii) The GP is same as geometric sequence or series.
46 Illustrated Optional Mathematics-9
Sigma Notation (Using of Summation) ALGEBRA
Consider a finite series 15 + 21 + 27 + 33 + ... to 10 terms.
Now, making pattern 1st Step 2nd Step We have, the first difference
Terms 6+9 6×1+9 is 6. So, 6 + 9 = 15 and so
t1 = 15 12 + 9 6×2+9 on.
t2 = 21 18 + 9 6×3+9
t3 = 27 24 + 9 6×4+9
t4 = 33
........ .........
........
........ 6×n+9
tn
Thus, the sum of the first 10 terms is represented as:
10
∑S10 = 15 + 21 + 27 + 33 + .... to 10 terms = (6n + 9)
n=1
If t1 , t2, t3, t4, t5, ..........., tn is a finite sequence, then the series associated with that sequence is t1 + t2 + t3
+ t4 + t5 + ........ + tn . The sum of the series is denoted by Sn = t1 + t2 + t3 + t4 + t5 + ........ + tn. and its sigma
i
∑(summation) notation is symbolized as [tn] , i ∈ N. where, i = specific term.
n=1
i
∑Here, ∑ is the summation sign. [tn] is read as "summation of tn 's when n runs from 1 to i".
n=1
Partial Sum
Consider a finite sequence 1, 5, 9, 13, 17, 21, 25, then
The sum of the first term, S1 = 1
The sum of the first two terms, S2 = 1 + 5 = 6
The sum of the first three terms, S3 = 1 + 5 + 9 = 15
The sum of the first four terms, S4 = 1 + 5 + 9 + 13 = 28
The sum of the first five terms, S5 = 1 + 5 + 9 + 13 + 17 = 45
The sum of the first six terms, S6 = 1 + 5 + 9 + 13 + 17 + 21 = 66
The sum of the first seven terms, S7 = 1 + 5 + 9 + 13 + 17 + 21 + 25 = 91
Now, taking backward difference every two successive terms,
S2 – S1 = 6 – 1 = 5 = t2 (Second term) S3 – S2 = 15 – 6 = 9 = t3 (Third term)
S4 – S3 = 28 – 15 = 13 = t4 (Fourth term) ............. S7 – S6 = 91 – 66 = 25 = t7 (Seventh term)
Similarly, we generalise for the finite sequence t1, t2, t3, ....... tn–1, tn.
Sn – Sn–1 = (t1 + t2 + t3 + ... + tn – 1 + tn) – (t1 + t2 + t3 + ... + tn – 1) = tn.
∴ tn = Sn – Sn – 1 where, n∈N
Sequence and Series 47
ALGEBRA Example 1 : Find the sum of the sequence 6, 9, 13, 18, 24, ...... to n terms.
Solution: Here, the given sequence is
6, 9, 13, 18, 24, ....... to n terms
1st Difference: 3 4 5 6 .......
2nd Difference: 1 1 1
Making a pattern to find the nth term; Rough
Terms 1st Step 2nd Step We have,
t1 = 6
t2 = 9 1 × 12 + 5 1 1 × 12 + 121 × 1 + 4 1st difference 2a = 1 ∴ a = 1
t3 = 13 2 2 2 2
t4 = 18 1
t5 = 24 1 1 112 × 2 + 4 ∴ tn = 2 n2 + ...
........ 2 2
× 22 + 7 × 22 + 1 11
2 2
1 1 1 112 × 3 + 4 ∴ t1 = × 1 +
2 2 2
× 32 + 8 × 32 + 1 1
2 2
1 1 121 × 4 + 4 = × 12 + 5 and so on.
2 2
× 42 + 10 × 42 + Again, 521, 7, 821, 10, ... is the linear
sequence. So, d = 112 and a = 5.
1 × 52 + 11 1 1 × 52 + 121 × 5 + 4
2 2 2
........ ........ ∴ tn = 112 n + 521 – 112 = 112 n + 4
tn 1 × n2 + 121 × n + 4
2
Thus, the sum of first n terms of the given sequence
Sn = 6 + 9 + 13 + 18 + 24 + ..... + 1 n2 + 112 n + 4
2
7
∑Example 2 : Write down the expanded form and find the value of (4n – 25).
7 n=3
Solution:
∑Here, (4n – 25) = (4 × 3 – 25) + (4 × 4 – 25) + (4 × 5 – 25) + (4 × 6 – 25) + (4 × 7 – 25)
n=3
= – 13 – 9 – 5 – 1 + 3
= – 25. 30
∑Example 3 : 3n2 – 21
How many terms are there in the series as in the form of sigma notation 2n – 21 ?
Find its 15th term. n=7
30 3n2 – 21
2n – 21
Solution: ∑Here, the given series as in the form of sigma notation is
n=7
There are 30 – 6 = 24 terms in that series and for the 15th term, n = 15 + 6 = 21
So, the required 15th term (t15) = 3 × 212 – 21 = 1302 = 62.
2 × 21 – 21 21
48 Illustrated Optional Mathematics-9
Example 4 : Find the 15th term of the sequence whose sum of the first n terms is 4n2 – 2n + 1.
Solution: Here, the sum of the first n terms of the sequence, Sn = 4n2 – 2n + 1 ALGEBRA
∴ Sn–1 = 4(n – 1)2 – 2(n – 1) + 1 = 4(n2 – 2n + 1) – 2n + 2 + 1
= 4n2 – 8n + 4 – 2n + 3 = 4n2 – 10n + 7
Now, we have, the nth term, "Alternatively"
tn = Sn – Sn – 1 Given,
= 4n2 – 2n + 1 – (4n2 – 10 n + 7) Sn = 4n2 – 2n + 1
= 4n2 – 2n + 1 – 4n2 + 10 n – 7 ∴ S15 = 4 × 152 – 2 × 15 + 1 = 871 and
= 8n – 6 S14 = 4 × 142 – 2 × 14 + 1 = 575.
∴ The 15th term, t15 = 8 × 15 – 6 = 114. ∴ t15 = S15 – S14 = 871 – 757 = 114.
Example 5 : Write the sigma notation of the series: 1 + 3 + 7 + 13 + 21 + 31 + 43
Solution: Here, the given series is
1 + 3 + 7 + 13 + 21 + 31 + 43
1st Difference: 2 4 6 8 10 12
2nd Difference: 2 2 2 2 .....
Now, making pattern
Terms 1st Step 2nd Step Rough
t1 = 1 12 – 0 12 – (1 – 1) The second difference, 2a = 2
t2 = 3 22 – 1 22 – (2 – 1)
t3 = 7 32 – 2 32 – (3 – 1) a = 2 ∴ tn = n2 + ...
t4 = 13 42 – 3 42 – (4 – 1)
........ ........ .........
t7 = 43 72 – 6 72 – (7 – 1)
........ ........ .........
tn = n2 – (n – 1) = n2 – n + 1
7
∑ Thus, the required sigma notation is (n2 – n + 1).
n=1
EXERCISE – 1.5 (C)
1. (a) Define series.
(b) Write the types of series based on difference of successive terms.
(c) What is sigma notation?
(d) Write the formula to compute nth term when the sum of first n terms is given.
2. Represent the following sequences into their associated series:
(a) 1, 6, 11, 16, .... to n terms (b) 15, 6, –3, –12, .... to n terms
(c) 15, 20, 30, 45, .... to n terms (d) 45, 39, 31, 21, .... to n terms
(e) 2, 4, 6, 8, .... to n terms (e) 5, 10, 20, 40, .... to n terms
Sequence and Series 49
3. Find the sum of the first five terms (S5) of the following sequences whose nth term are given below:
ALGEBRA (a) tn = 2n2 – n + 3 (b) an = 2n + 3 (c) un = (–1)n – 1 3n (d) tn = (–1)n 3n
2n – 1
4. Which of the following sequences are the arithmetic progression (AP) or geometric progression
(GP)?
(a) 5, 11, 17, 23, .... (b) 1, 5, 25, 125, ... (c) 28, 21, 14, 7, ...
(d) 210, 110, 1 , 2 , .... (e) 6, 3, 32, 43, ... (f) –25, –11, 3, 17, ...
5 5
5. Write down the expanded form and find the value of the following in sigma notations:
4 6 7 2n2 + 3
(2n + 1) (3n – 1) n–2
∑ (a) (2n2 – 2n + 3) ∑ ∑(b) (c)
n=1 n=2 n=4
6 8 7 (– 1)n (2n + 1)
∑ (d) ∑(f) 3n – 1
(–1)n (2n + 1) ∑(e) (–1)n+1 (n2 + 2)
n=2 n=3 n=2
6. Find the number of terms in the series represented as the form of the following sigma notation
and find their 12th term.
16 25 20 4k – 1
k+1
∑ (a) xn ∑(b) (2n2 + 3) ∑(c) (–1)k
n=3
n=2 k=5
7. Write each of the following by using sigma notation: 10 14 18
2 6 9 11 13 22
(a) 4 + 7 + 10 + 13 + 16 + 19 + 22 (b) 5 + 7 + + + + 15
(c) – 3 + 5 – 7 + 9 – 2111 (d) 1 + 5 + 25 + 125 + 625 + 3125
5 9 13 17
(e) (r – 1) + (r – 1)2 + (r – 1)3 + (r – 1)4 + (r – 1)5 + (r – 1)6
2 4 8 16 32 64
(f) a + ar + ar2 + ar3 + ar4 + ar5 + ar6
8. Find the first four terms and 25th term of the following series whose sum of the first n terms
are given below:
(a) Sn = 2n + 1 (b) Sn = 3n2 – 1 (c) Sn = (2n – 1)2 (d) Sn = 2n2 + 4n
ANSWERS
2. (a) 1 + 6 + 11 + 16 + .. + (5n – 4) (b) 15 + 6 – 3 – 12 – ... – (24 – 9n)
(c) 15 + 20 + 30 + 45 + .. + 5 n2 – 5 n + 15 (d) 45 + 39 + 31 + 21 + ... + (– n2 – 3n + 49)
22
(e) 2 + 4 + 8 + 16 + .... + (2n) (f) 5 + 10 + 20 + 40 + .... + (5 × 2n–1)
3. (a) 110 47 (d) –183
(b) 12 315 (c) 183 (d) geometric (e) geometric (f) arithmetic
4. (a) arithmetic (b) geometric (c) arithmetic
5. (a) 52 (b) 555 (c) 4447 (d) 9 (e) –33 (f) 1803
60 10472
(c) 16, 63
6. (a) 15, x13 (b) 23, 395 17
7. 7 6 4n – 2 (c) 5 (–1)n (2n + 1) (d) 6 (e) 6 (r – 1)n (f) 7
(a) (3n + 1) (b) n=1 2n + 3 n=1 4n + 1 5n–1 n=1 2n arn–1
n=1 n=1 n=1
8. (a) 2, 2, 2, 2; 2 (b) 3, 9, 15, 21;147 (c) 0, 8, 16, 24; 192 (d) 6, 10, 14, 18;102
50 Illustrated Optional Mathematics-9
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