According to new curriculum in compliance with
Curriculum Development Centre (CDC)
PRIME Optional
Mathematics
Pragya Books &
Distributors Pvt. Ltd. Editors
LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
Author J.N. Aryal
Kadambaba Pradhan
Dirgha Raj Mishra
Dinesh Silwal
Pragya Books & Distributors Pvt. Ltd.
Lalitpur, Nepal
Tel : 5200575
email : [email protected]
© Author
Author Dirgha Raj Mishra
Editors LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal
First Edition 2076 B.S. (2019 A.D.)
Revised Edition 2077 B.S. (2020 A.D.)
Revised Edition 2078 B.S. (2021 A.D.)
Price
ISBN 978-9937-9170-6-3
Typist Sachin Maharjan
Sujan Thapa
Layout and Design Desktop Team
Printed in Nepal
Preface
Prime Optional Mathematics series is a distinctly outstanding mathematics
series designed according to new curriculum in compliance with Curriculum
Development Centre (CDC) to meet international standard in the school level
additional mathematics. The innovative, lucid and logical arrangement of the
contents make each book in their series coherent. The representation of ideas in
each volume makes the series not only unique but also a pioneer in the evaluation
of activity based mathematics teaching.
The subject is set in an easy and child-friendly pattern so that students will
discover learning mathematics is a fun thing to do even for the harder problems.
A lot of research, experimentation and careful graduation have gone into the
making of the series to ensure that the selection and presentation is systematic,
innovative, and both horizontally and vertically integrated for the students of
different levels.
Prime Optional Mathematics series is based on child-centered teaching
and learning methodologies, so that the teachers can find teaching this series
equally enjoyable.
I am optimistic that, this series shall bridge the existing inconsistencies
between the cognitive capacity of children and the subject matter.
I owe an immense dept of gratitude to the publishers (Pragya Books team)
for their creative, thoughtful and inspirational support in bringing about the
series. Similarly, I would like to acknowledge the tremendous support of editors
team, teachers, educationists and well-wishers for their contribution, assistance
and encouragement in making this series a success. I would like to express my
special thanks to Sachin Maharjan (Wonjala Desktop) for his sincere support
of designing part of the book and also Mr. Gopal Krishna Bhattarai to their
memorable support to prepare this series.
I hope this series will be another milestone in the advancement of teaching
and learning Mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that I can refine and improvise the series
in the future editions.
– Author
Algebra : Ordered Pairs Contents Page
S.N. Units 1
2
1. Algebra 11
1.1 Ordered pairs 21
1.2 Surds 28
1.3 Polynomials 37
1.4 Sequence and series 47
2. Limits 63
3. Matrices 87
4. Co-ordinate Geometry 131
5. Trigonometry 149
6. Vector Geometry 173
7. Transformation 200
8. Statistics
Model questions
4 PRIME Opt. Maths Book - VIII
Unit 1 Algebra
1. Algebra
1.1 Ordered pairs
1.2 Surds
1.3 Polynomials
1.4 Sequence and series
Specification Grid Table
K(1) U(2) A(4) HA(5) TQ TM Periods
7 16
No. of Questions 1 1 1 – 3
2 4 –
Weight 1
K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks
Objectives : At the end of the lesson
• Students are able to know the ordered pairs and Cartesian products.
• Students are able to represent Cartesian product in arrow diagram and graph.
• Students are able to know the number system including surds.
• Students are able to operate the surds.
• Students are able to know the polynomials, types and degree of polynomials.
• Students are able to identify the sequence and series.
• Student are able to know the difference between arithmetic and geometric
sequence and their nth term.
Materials Required:
• Chart paper
• Chart of number system.
• Chart of types of surds.
• Chart of types polynomials & properties of addition and multiplication.
• Chart of types of sequence and series.
PRIME Opt. Maths Book - VIII 1
Algebra : Ordered Pairs 1.1 Ordered Pair and Cartesian Product
Ordered Pair
Look at a sequence of numbers 1, 2, 3, 4, 5, 6, 7, ....., there is a certain rule
between the numbers and the next number will be 8 according to the rule.
The numbers in the above example are in order. It can be seen in another
example also which is 2, 5, 8, 11,... .
Let us consider the following examples:
(2, 3), (4, 5), (a, b), (Nepal, Kathmandu) etc.
In above pair of numbers or letters or names are kept inside a round bracket
separated by comma where the position of numbers or objects play very
important role. This type of presentation of number or letter or objects is
called an ordered pair.
Ordered pair :
The Pair of numbers or objects in the form of
(x, y) with an order is called ordered pair.
For an ordered pair (x, y)
x & y are called the elements of ordered pair, the first element is called
antecedent (x–component) and second element is called consequence
(y–component).
(x, y) is ordered pair :
x is antecedent (x – component)
y is consequence (y – component)
There is a vital role of order (position) in ordered pair, If we change the
position of elements, a new ordered pair is formed.
Examples from practical life.
• Ordered pair of country with respect to capital
(Nepal, Kathmandu), (Japan, Tokyo), (China, Beijing)
2 PRIME Opt. Maths Book - VIII
• Order pair of temples with respect to location.
(Pashupatinath, Kathmandu), (Pathiveradevi, Taplejung),
(Manakamana, Gorkha)
• Order pair of community and their dresses:
(Newari, Haku Patassi), (Sherpa, Bakhu) (Aryan, Gunyu Choli) etc.
Equal Ordered Pairs 3 6
3 3
In the ordered pair (1, 2) and ( , ), the components are equal which are
called equal ordered pairs. In this example antecedents and consequences
of both the ordered pairs are equal.
Any two ordered pairs are said to be equal ordered Algebra : Ordered Pairs
pairs if and only if the corresponding components are
equal.
• If (x, y) and (2, 3) are equal then x = 2 and y = 3.
Worked out Examples
1. Write down any five ordered pairs of currency with respect to country.
Solution:
The currencies and countries have to be written as x – component and
y – component respectively as,
(Rupiya, Nepal), (Taka, Bangladesh), (Pound, United Kingdom), (Yen,
China), (Riyal, Saudi Arebia)
2. If (x, 3) and (2, y + 2) are two equal ordered pairs, find x & y.
Solution:
Given (x, 3) = (2, y + 2).
By equating the corresponding elements, we have
x = 2 and 3 = y + 2
or 3 – 2 = y
or 1 = y
\ x = 2 and y = 1
PRIME Opt. Maths Book - VIII 3
3. If (x + y, 2x – 3) = (5, 3), find the value of x and y.
Solution :
Given, (x + y, 2x – 3) = (5, 3)
By equating the corresponding elements,
2x – 3 = 3 Consequence Consequence
or, 2x = 3 + 3 of 1st pair = of 2nd pair
or, 2x = 6
or, x = 3
Again,
x + y = 5 [Antecedent of 1st pair = Antecedent of 2nd pair]
or, 3 + y = 5 [ a x = 3]
or, y = 5 – 3
y = 2
\ x = 3
\ y = 2
Algebra : Ordered Pairs Cartesian Product
Let us consider two sets A = {1, 2} and B = {3, 4}. Let us obtain all possible
ordered pairs where antecedent is from set A and consequence from set B
and shown below.
1 3 (1, 3)
4 (1, 4)
2 3 (2, 3)
4 (2, 4)
Now, the set of these ordered pairs is {(1, 3), (1, 4), (2, 3), (2, 4)} which is
called the Cartesian product of A and B and we denote it by A × B.
The set of all possible ordered pairs (x, y) from the
non–empty set A to the non–empty set B is called
Cartesian product A × B. where x ∈ A, y ∈ B.
4 PRIME Opt. Maths Book - VIII
Thus, the Cartesian product of A and B is read as A cross B and denoted by
A × B and defined as,
A × B = {(x, y) : x ∈ A and y ∈ B}
Also, n(A × B) = number of ordered pairs of A × B is called the cardinality of
cartesian product A × B.
Example
If A = {2, 3}, B = {4, 5}
A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}
Here,
n(A × B) = 4 which is Cardinality of A × B.
Also,
B × A = {(4, 2), (4, 3), (5, 2), (5, 3)}
Here, Algebra : Ordered Pairs
n (B × A) = 4 which is Cardinality of B × A.
Then, we conclude that :
A×B≠B×A
But, n(A × B) = n(B × A)
Note : If n(A) = m, n(B) = n then
n(A × B) = n(A) × n(B) = m × n = mn
Representation of Cartesian product:
Taking the sets A = {2, 3}, B = {4, 5}
i) Listing method :
A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}
ii) Arrow diagram : B B×A A
A A×B B 42
24
3 5 5 3
PRIME Opt. Maths Book - VIII 5
iii) Tabular form :
A × B B×A
B 4 5 A 2 3
A B
2 (2, 4) (2, 5) 4 (4, 2) (4, 3)
3 (3, 4) (3, 5) 5 (5, 2) (5, 3)
Algebra : Ordered Pairs Worked out Examples
4. If A = {1, 2, 3}, B = {4, 5, 6}, find A × B and B × A. Also prove that n (A ×
B) = n(B × A)
Solution :
A = {1, 2, 3},
B = {4, 5, 6}
A × B = {(x, y) : x ∈ A and y ∈ B}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
B × A = {(x, y) : x ∈ B and y ∈ A}
= {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
∴ A × B ≠ B × A
Here,
n(A × B) = 3 × 3 = 9
n (B × A) = 3 × 3 = 9
∴ n (A × B) = n (B × A)
5. If A × B = {(a, x), (a, y), (b, x), (b, y)}, find the sets A and B. Also find A
× A and B × B.
Solution:
A × B = {(a, x), (a, y), (b, x), (b, y)}
Set A = {all the Antecedents x of A × B} = {a, b}
Set B = {all the Consequences y of A × B} = {x, y}
Again,
A × A = {a, b} × {a, b}
= {(a, a), (a, b), (b, a), (b, b)}
B × B = {x, y} × {x, y}
= {(x, x), (x, y), (y, x), (y, y)}
6 PRIME Opt. Maths Book - VIII
6. If A = {a, b, c} and A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z), (c,
....), (c, ....), (c, ....)}, find the set B, Complete A × B. Also show A × B in
arrow diagram.
Solution : A = {a, b, c}
A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z), (c, ....), (c, ....), (c, ....)}
By comparing A and A × B, we get B = {x, y, z}
Also, A × B = (a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)}
Arrow diagram:
A A×B B
ax
by
c z
7. Find x and y if (x + 2y, 1) = (3, 2x – y). Also find x – y. Algebra : Ordered Pairs
Solution :
Given, (x + 2y, 1) = (3, 2x – y)
From equality of ordered pairs, we have
x + 2y = 3 and 2x – y = 1
or, x = 3 – 2y ................. (i) or, 2(3 – 2y) – y = 1 [\ using (i)]
or, 6 – 4y – y = 1
or, –5y = – 5
or, y = 1
Putting y = 1 in (i) we have
x =3–2×1
= 3 – 2
= 1
\ x = 1
y = 1
Also,
\ x – y = 1 – 1 = 0
PRIME Opt. Maths Book - VIII 7
Exercise 1.1
1. i) Write down any five ordered pairs of districts with respect to
ii) headquarter of our country Nepal. Also show in arrow diagram.
Write down the countries and their capitals from the given arrow
diagram in ordered pair form.
India New Delhi
China Beijing
Japan Tokyo
Algebra : Ordered Pairs iii) Complete the following ordered pairs by filling the gaps with
suitable words.
(Bhanubhakta, ....), (...., Aanshukabi), (Lekhnath Poudel, .....)
iv) Taking an order ‘more than by 2’, complete the following ordered
pairs. (2, ...), (5, ...), (..., 9) (..., 12), (15, ...)
v) What is ordered pair? Write down with an example.
2. Find the value of ‘x’ and ‘y’ from the following equal ordered pairs.
i) (x, y) = (3, 4) ii) (x – 2, y + 1) = (5, 4)
iii) (2x – 1, y + 2) = (5, 6 – y) iv) (x – 1, 2x + y) = (2, 8)
v) (3x – 2, 2y + 5) = (x + 4, 1) iv) (2x – 3y, 13) = (3, 3x + 4y)
3. If A = {a, b}, B = {1, 2, 3}, find
i) A × B and show in arrow diagram.
ii) B × A and show in arrow diagram.
iii) Prove that A × B ≠ B × A
iv) Prove that n(A × B) = n(B × A)
4. If P = {2, 5}, Q = {3, 4}, find
i) P × Q and show in tabular form.
ii) Q × P and show in tabular form.
iii) Find the ordered pair of P × P and Q × Q.
iv) Prove that n(P × Q) = n(Q × P)
8 PRIME Opt. Maths Book - VIII
5. If A × B = {(a, x), (b, x), (a, y), (b, y), (a, z), (b, z)} find,
i) Set A and Set B.
ii) Prove that A × B ≠ B × A
iii) Prove that n (A × B) = n (B × A)
iv) Find the ordered pair of B × B and A × A.
6. i) From the given arrow diagram find the set A and set B. Also find
B × A & n(B × A) A×B B
A
25
47
69 Algebra : Ordered Pairs
ii) If A = {x : x ∈ 1, 2}, B = {y : y ∈ 3 < N ≤ 6}, find A × B. Also show in
tabular form.
iii) If P = {x : x ∈ N, 1 ≤ N < 4}, Q = {x : x ∈ 3, 4}, find P × Q and n(P × Q).
iv) If A × B = {(....., 3), (5, .....), (....., 4), (....., 4), (5, .....), (2, .....)} and A = {2, 5},
find B. Also find A × B and B × A and prove that A × B ≠ B × A.
v) If A = {1, 2}, B = {2, 3}, find (A ∪ B) × B.
PRIME more creative questions
7. i) If (x + y, 6) = (6, 2x – y), find the value of ‘x’ and ‘y’.
ii) If (2x, y + 3) = (y + 3, 3x – 2), find the value of ‘x’ & ‘y’.
iii) From the given table, find the sets A and B. Also prove that
n(A × B) = n(B × A)
B×A
A a b c
B (x, c)
x (x, a) (x, b)
y (y, a) (y, b) (y, c)
z (z, a) (z, b) (z, c)
iv) If A = {2, 3}, B = {3, 4}, find (A ∪ B) × A.
v) If A = {1, 2}, B = {2, 3}, C = {3, 4, 5}, find A × (B ∪ C). Also show
in tabular form.
Project work
8. Collect the cost of different vegetables from your local shop and write
down the ordered pair of them with respect to cost.
PRIME Opt. Maths Book - VIII 9
Answer
Algebra : Ordered Pairs 1. Show to your subject teacher
2. i) x = 3, y = 4 (ii) x = 7, y = 3 (iii) x = 3, y = 2
iv) x = 3, y = 2 (v) x = 3, y = –2
3. Show to your subject teacher.
4. Show to your subject teacher.
5. i) A = {a, b}, B = {x, y, z}
ii) n(A × B) = 6, n(B × A) = 6
iii) B × B = {(x, x), (x, y), (x, z), (y, x), (y, y), (y, z), (z, x), (z, y), (z, z)}
iv) Show to your subject teacher.
v) Show to your subject teacher.
6. i) A = {2, 4, 6}, B = {5, 7, 9}
B × A = {(2, 5), (2, 7), (2, 9), (4, 5), (4, 7), (4, 9), (6, 5), (6, 7), (6, 9)}
n(B × A) = 9
ii) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6)}
A×B
B 4 5 6
A
1 (1, 4) (1, 5) (1, 6)
2 (2, 4) (2, 5) (2, 6)
iii) P×Q = {(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4)}, n(P×Q) = 6
iv) Show to your subject teacher.
v) (A ∪ B) × B = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}
7. i) x = 4, y = 2 ii) x = 2, y = 1
iii) A = {a, b, c}, B = {x, y, z}, n(A × B) = 9, n(B × A) = 9
iv) (A ∪ B) × A = {(2, 2), (2, 3), (3, 2), (3, 3), (4, 2), (4, 3)}
v) A × (B ∪ C) = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)}
A × (B ∪ C)
B∪C 2 3 4 5
A
1 (1, 2) (1, 3) (1, 4) (1, 5)
2 (2, 2) (2, 3) (2, 4) (2, 5)
10 PRIME Opt. Maths Book - VIII
1.2 Surds Algebra : Surds
Number system
• The set of natural numbers is the set of all counting numbers. We denote
the set of natural numbers by N. The least natural number is 1 and the
highest is not defined.
N = {1, 2, 3, 4, 5 ....}
• The set of counting numbers including zero is called whole numbers.
The set of whole numbers is denoted by W The least whole number is
zero.
W = {0, 1, 2, 3, 4, 5, ...}
• The set of all counting numbers including zero and their negatives is
called integers. The set of integers is denoted by Z.
Z = {....., –4, –3, –2, –1, 0, 1, 2, 3, 4, ....}
In the given examples set of natural numbers
is the sub–set of whole number and set of
whole number is the sub set of integers.
Rational numbers:
Let us consider the elements of integers taken in different mathematical
operations as follows,
4 – 6 = –2 (Integer) 4 + 6 = 10 (Integer)
4 × 6 = 24 (Integer) 4÷6= 2 (Not Integer)
3
But 4 ÷ 2 = 2 (Integer)
Thus, the division of any two integers may not be always an integer.
To define such numbers a new number system is introduced which is
called the rational number.
The set of numbers in the form of p/q where p & q are the
integers and q ≠ 0 is called the set of rational numbers.
i.e. Q = {...., –2, –3/2, –1, 0, 1/2, 1, 3/2, 3, ............} is the
set of rational numbers.
PRIME Opt. Maths Book - VIII 11
Note :
i. The set of rational numbers is denoted by Q.
ii. All integers can be expressed as a rational number.
p
e.g. 2 = 2 = 4 etc which is in the form q , q ≠ 0 and p, q ∉ z.
1 2
So, all integers are rational numbers. But all rational numbers are not
integers.
So, Z ⊂ Q. Also N ⊂ W ⊂ Z ⊂ Q.
Let us consider the rational number 19 .
4
We have
4 19 4.75
16
30
28
20
20
Algebra : Surds ×
Here, the process of division is terminated (finished) with the decimal
quotient 4.75. This shows that a rational number some time possesses
a terminating decimal number.
Again consider the rational number 19
3
We have
3 19 6.33
18
10
9
10
9
1
Here, the process of division doesn’t terminate and a number is
occurred again and again after decimal point. Similarly, we can see
sometimes a block of numbers repeats after decimal point.
Thus a rational number possesses an integer or a terminating decimal
number or a non terminating recurring decimal number.
12 PRIME Opt. Maths Book - VIII
Try to Tfihnedrethaerereicnufirnriitnegnduemcibmeralonf uramtiboenraflonrum272bers between any two
Note :
rational numbers.
Irrational numbers:
Some Examples :
2 = 1.414213562....................
3 = 1.732...............
5 = 2.236067977..
4 =2
9 =3
Here, 2 , 3 , 5 have the non terminating and non recurring
decimal parts while 4 & 9 have the integer numbers. The numbers
with roots 2 , 3 , 5 are called the irrational numbers.
Note: Algebra : Surds
i. All the roots which are not rational are irrational
ii. The non terminating and non recurring decimal numbers are irrational.
iii. p= length of circumference of a circle is an irrational number.
diameter
iv. Length of a body is either a rational number or an irrational number.
Introduction of Surds:
The numbers 2 , 3 , 3 2 etc are the roots of rational numbers where
the result is irrational are called surds.
2 , 3 are unlike surds of same order (same orders but different
radicands)
2 , 3 2 are unlike surds of different order (different orders but same
radicands)
3 , 2 3 are the like surds (same orders and same radicands)
3 5 , 3 10 are unlike surds
Any number in the from n a which can not be expressed in the form
p
of q , where q ≠ 0, p, q ∉ z, is called a surd. n is called the order, ‘a’ is
called the radicand and called the sign of radical.
PRIME Opt. Maths Book - VIII 13
Like surds are the surds having same
orders and same radicands.
12 is the pure surd (having only irrational factors)
2 3 is the mixed surd (having both irrational factors and rational
factors rather than 1.)
• For the operation of the surds, pure surds should be changed into
mixed surds.
• Only like surds can be operated using plus and minus sign.
Worked out Examples
1. Insert any three rational numbers between 2 and 3 .
Solution : 3 4
Algebra : Surds The given rational numbers are 2 and 3
3 4
The other three rational numbers between 2 and 3 are as follows.
3 4
2 + 3
3 4
1st rational no. =
2
= 1 ` 17 j = 17
2 12 24
2nd rational no. = 1 a 2 + 17 k
2 3 24
= 1 ` 33 j = 11
2 24 16
3rd rational no. = 1 ` 17 + 3 j
2 24 4
= 1 a 35 k = 35
2 24 48
So on any other can be found out.
2. Express the surd 50 into mixed surd and 2 3 5 into pure surd.
Solution :
50 = 52 × 2 = 5 2 (mixed surd)
2 3 5 = 3 23 × 5 = 3 40 (pure surd)
14 PRIME Opt. Maths Book - VIII
3. Express the surds 2, 3 5, 4 6 into the surds of same order and
arranged in ascending order.
Solution :
The given surds are 2, 3 5, 4 6
L.C.M. of 2, 3, & 4 is 12.
Where,
2 = 2× 6 26 = 12 26 = 12 64
3 5 = 3× 4 54 = 12 54 = 12 625
4 6 = 4 × 3 63 = 12 63 = 12 216
The surds in ascending order
12 64 , 12 216 , 12 625
i.e. 2 , 4 6 , 3 5
Note : If n a and n b are two surds and a > b then n a > n b
4. Simplify : 5 18 + 72 – 3 32 5. Multiply the surds 3 and 3 2 . Algebra : Surds
Solution : Solution
5 18 + 72 – 3 32 3 ×3 4
= 5 32 × 2 + 62 × 2 – 3 42 × 2 LCM of 2 and 3 is 6.
= 15 2 + 6 2 – 12 2 = 6 33 × 6 42
= (15 + 6 – 12) 2 = 6 27 × 16
=9 2 = 6 432
6. Rationalise the denominate of 1 2.
Solution : 3+
1 2= 1 2× 3– 2 [ a Multiply both numerator
3+ 3+ 3– 2 and denominator by conjugate
3– 2 3– 2]
3h2 – ^ 2h2
= ^
= 3– 2
3–2
= 3– 2
1
= 3– 2
PRIME Opt. Maths Book - VIII 15
7. Simplify : 4 2– 3 3– 1
6– 6+ 3+ 2
Solution :
= 4 2× 6+ 2 – 3 3× 6+ 3 – 1 2× 3+ 2
6– 6+ 2 6+ 6+ 3 3+ 3+ 2
= 4( 6 + 2) – 3( 6 – 3) – ( 3 + 2)
( 6)2 – ( 2)2 ( 6)2 – ( 3)2 3)2 +( 2)2
= 4( 6+ 2) – 3( 6– 3) – 3+ 2
4 3 1
= 6+ 2 – 6+ 3 – 3– 2
=0
Algebra : Surds 8. Solve for ‘x’ of : x – 2 = 3
Solution:
" x –2=3
or x = 3 + 2
or x = 5
Squaring on both sides ^ xh2 = (5)2
∴ x = 25
Checking for x = 25 on x – 2 = 3
or, 25 – 2 = 3
or, 5 – 2 = 3
or, 3 = 3 (true)
∴ x = 25
9. Solve : x + 9 – x = 1
Solution :
x+9 – x =1
or, x + 9 = x + 1
Squaring on both side
or, ( x + 9 )2 = ( x + 1)2
or, x + 9 = ( x )2 + 2 x – 1 + (1)2
or, x + 9 = x +2 x + 1
or, 8 = 2 x
or, 4 = x
16 PRIME Opt. Maths Book - VIII
Again,
Squaring on both sides,
or, (4)2 = ( x )2
or, 16 = x
∴ x = 16
Note : If a + b is a surd then a – b is called the conjugate surd of a + b .
10. Find the square root of 4 – 2 3 .
Solution:
Square root of 4 – 2 3 is,
4 – 2 3 = ^ 3h2 – 2. 3 .1 + 12
= ( 3 – 1)2
= 3 –1
Exercise 1.2 Algebra : Surds
1. i) What is natural number? Write down the set of natural numbers.
ii) How can you say that set of natural numbers is the sub–set of
whole number?
iii) What do you mean by irrational numbers?
iv) What is the set of rational numbers?
v) Differentiate between like surds and unlike surds with
examples.
2. Pick out the like surds from the following with reason.
i) 3 2 , 2 , 5 2 ii) 2 3 5 , 5 3 2 , 3 3
iii) 3 24 , 3 3 3 , 3 81 iv) 210, 75, 4 144
v) 20 , 2 45 , 3 250
3. Convert the followings as indicated in brackets.
i) 98 (into mixed surd) ii) 3 500 (into mixed surd)
iii) 2 5 (into pure surd) iv) 33 2 (into pure surd)
v) 25 2 (into pure surd)
PRIME Opt. Maths Book - VIII 17
4. Express the followings as indicated in brackets.
i) 3 and 3 2 (into same order)
ii) 3 3, 4 4 (which is greater)
iii) 3, 3 4, 6 20 (in ascending order)
iv) 3 3, 2, 4 5 (in descending order)
v) 3 4, 4 5, 12 50 (in ascending order)
5. Simplify:
i) 18 + 50 – 3 8 ii) 2 108 – 6 75 + 5 48
iii) 3 40 + 2 3 320 – 3 3 135 iv) 3 700 – 2800 – 2 63
v) 4 3 432 – 5 3 128 + 7 3 2000
6. Multiply the surds. ii) 3 4 × 6 8
i) 2 × 3 3 iv) 2 5 × 3 3 4
iii) 3 × 3 5 × 6 4
v) 3 2 × 2 4 8 × 5 6 4
Algebra : Surds 7. Rationalise the denominator.
1 1
i) 3– 2 ii) 5+ 3
iii) 4 iv) x– a
3 + 2 –1 x+ a
v) x+2 – x–2
x+2 + x–2
8. Simplify the following.
i) 2 3+ 1
5+ 3– 2
ii) 1 2– 3 3+ 4
3+ 6– 6– 2
iii) 3 2+ 2 5+ 5
5+ 7+ 2– 7
iv) x+ a + x– a
x– a x+ a
18 PRIME Opt. Maths Book - VIII
v) x+2 – x–2 + x+2 + x–2
x+2 + x–2 x+2 – x–2
9. Solve the followings. ii) x – 1 – 2 = 3
i) x + 3 = 5 iv) x + 6 – 2 = 1
iii) x – 2 = 3
v) x2 – 5 + 3 = 5
10. Solve the followings. ii) x + 3 – x = 1
iv) x + 7 = x + 1
i) x + 5 + x = 5
iii) x = x + 11 – 1
v) x2 + 9 – x = 3
11. Find the square root of: ii) 42 + 32 + 22 + 12 – 5
i) 12 + 22 + 22 × 5 iv) 5 + 2. 5 .1 + 1
iii) ( 3 )2 + 2. 3 .2 + 22
v) 2 + 2. 2 . 3 + 3 Algebra : Surds
12. PRIME more creative questions:
i) Simplify : 3 200 – 2 1250 + 5 4 1024
12
ii) Rationalise : 5+ 3+ 2
iii) Simplify : 5 10 – 32 6 + 43
15 + 12 – 18 – 6
iv) Solve : x–3 – x–2 =2
x+3 + x–3
v) Find the square root of 4 + 2 3
vi) Find the square root of 7 – 4 3 .
vii) What is the order of the surds n xm ?
viii) What is the order of the single surds of 3 × 3 2 ?
ix) Insert the irrational numbers between 2 and 11.
x) Insert 3 rational numbers between 1 and 1 .
24
PRIME Opt. Maths Book - VIII 19
Answer
1. Show to your subject teacher.
2. Show to your subject teacher.
3. Show to your subject teacher.
4. i) 6 27 & 6 4 ii) 3 3 is greater.
iii) 3 4 < 6 20 < 3 iv) 4 5 > 3 3 > 2
v) 12 50 < 4 5 < 3 4
5. i) 2 2 ii) 2 3 iii) 3 5
iv) 4 7 v) 74 3 2
6. i) 6 72 ii) 2 6 2 iii) 6 4050
iv) 6 6 2000 v) 20 12 32
Algebra : Surds 7. i) ^ 3 + 2h ii) 12 ^ 5 – 3h iii) 6 + 2 – 2
x+a–2 ax v) 12 ^x – x2 – 4h
iv) x – a
8. i) 5 + 2 ii) 0 iii) –2 2
iv) 2^xx–+aah v) x
9. i) 4 ii) 26 iii) 25
iv) 3 v) ±3
10. i) 4 ii) 1 iii) 25
iv) 9 v) 4
11. i) 5 ii) 5 iii) 2 + 3
iv) 5 + 1 v) 3 + 2
12. i) 0 ii) – 30 + 3 2 + 2 3 iii) 0
iv) x = 3 v) 3 + 1 vi) 2 – 3
2 ix) 3 , 5 ,
7
vii) n viii) 6
x) 3 , 5 , 7
8 16 16
20 PRIME Opt. Maths Book - VIII
1.3 Polynomials
• An algebraic term is the product or quotient of number(s) with
variable(s).
Examples : 3x2, 2xy, 3 x3, 2x2 etc.
2 y2
Let us consider an example of algebraic term 2ax3 in x. Algebra : Polynomials
Here,
2 is called numeral coefficient.
x is called base.
3 is called power (index) or exponent
a is called literal coefficient of x.
• The combination of the algebraic terms with (+) or (–) sign is called
algebraic expression.
Examples : 2x2 – 3x + 5, 2x + 3, 2x etc.
• The algebraic expression is called a polynomial under positive power
and real coefficient.
The algebraic expression having non–negative
exponents (power) of the variables and having
coefficient a real number is called polynomial.
Standard form of polynomial in x :
A polynomial written in descending or ascending order of the exponent
(power) of the variable x is called the polynomial in standard form. The
highest value of exponent of the variable is called the degree of the variable.
For example :
p(x) = a0xn + a1xn-1 + a2xn–2 + ..... + anx0, a0 ≠ 0
• Where, n is non–negative integer and
a0, a1, a2, ..... an are the real numbers (coefficient) & a0 ≠ 0.
• n is called the degree of the polynomial p(x)
a0xn, a1xn-1 ... are the terms of the polynomial.
PRIME Opt. Maths Book - VIII 21
Types of Polynomials
i) According to degree
• The polynomial having degree ‘1’ is called a linear polynomial.
p(x) = ax + b, a ≠ 0 (First degree polynomial)
• The polynomial having degree ‘2’ is called a quadratic polynomial.
p(x) = ax2 + bx + c, a ≠ 0 (Second degree polynomial)
• The polynomial having degree ‘3’ is called cubic polynomial.
p(x) = ax3 + bx2 + cx + d, a ≠ 0 (Third degree polynomial)
• The polynomial having degree ‘4’ is called biquadratic polynomial.
p(x) = ax4 + bx3 + cx2 + dx + e, a ≠ 0 (Fourth degree polynomial)
Algebra : Polynomials ii) According to number of terms, polynomials are classified as the
following.
i) Monomial → Having only one term in the expression.
p(x) = 2x
ii) Binomial → Having two terms in the expression.
p(x) = 2x2 + 3
iii) Trinomial → Having three terms in the expression.
p(x) = 2x3 + 3x + 2
iv) Multinomial → Having more terms in the expression.
or Polynomial p(x) = 2x4 + 3x3 + 5x2 – 3x + 7
Other information on polynomials
• The polynomial may have more than one variable also.
P(xy) = ax2y + bxy + cxy2
p(xyz) = ax2yz + bxy2z + cxyz2.
• The polynomials having degree and coefficient of the corresponding
terms are same are called equal polynomials.
p(x) = 2x3 + 5x2 – 3x + 2
q (x) = 4 x3 + 5x2 – 6x + 8
2 2 4
= 2x3 + 5x2 – 3x + 2
Here, p(x) = q(x)
Note : All polynomial expressions are algebraic expression but all algebraic
expressions are not polynomials.
22 PRIME Opt. Maths Book - VIII
Worked out Examples
1. If p(x) = 12x3 + 5x2 – 10x + 7 and q(x) = (2m + 2)x3 + 5x2 – (3n + 1)x +
7 are equal polynomials, find the value of ‘m’ and ‘n’.
Solution :
p(x) = 12x3 + 5x2 – 10x + 7
q(x) = (2m + 2)x3 + 5x2 – (3n + 1)x + 7
Since, the polynomial are equal,
Corresponding coefficients are also equal,
i.e. 2m + 2 = 12 and –(3n + 1) = –10
or, 2m = 10 or, 3n = 9
∴ m = 5 \ n = 3
2. If f(x) = x3 + 2x2 + 3x – 2 and g(x) 3x3 + 5x2 – 7x – 1, find f(x) + g(x). Algebra : Polynomials
Solution :
f(x) = x3 + 2x2 + 3x – 2
g(x) = 3x3 + 5x2 – 7x – 1
Then,
f(x) + g(x) = (x3 + 2x2 + 3x – 2) + (3x3 + 5x2 – 7x – 1)
= x3 + 2x2 + 3x – 2 + 3x3 + 5x2 – 7x – 1
= (x3 + 3x3) + (2x2 + 5x2) + (3x – 7x) + (– 2 – 1)
= 4x3 + 7x2 – 4x – 3
3. What must be subtracted from the polynomial 5x3 – 3x2 + 2x + 5 to get
the polynomial 2x3 – x2 + 3x – 2.
Solution :
Let, the subtracted polynomial be ‘K’.
Then, by the question,
(5x3 – 3x2 + 2x + 5) – K = 2x3 – x2 + 3x – 2
or, (5x3 – 3x2 + 2x + 5) – (2x3 – x2 + 3x – 2) = K
or, 5x3 – 3x2 + 2x + 5 – 2x3 + x2 – 3x + 2 = K
∴ K = 3x3 – 2x2 – x + 7
∴ The subtracted polynomial is, 3x3 – 2x2 – x + 7
Remember : The required polynomial is
= (Subtracted from the polynomial) – (to get the polynomial)
= (5x3 – 3x2 + 2x + 5) – (2x3 – x2 + 3x – 2)
PRIME Opt. Maths Book - VIII 23
4. What should be added with 3x3 + 5x2 – 1 – 7x to get x3 + 2x2 + 3x – 2?
Solution :
Let the required polynomial be K.
According to question
(3x3 + 5x2 – 1 – 7x) + k = x3 + 2x2 + 3x – 2
or, K = (x3 + 2x2 + 3x – 2) – (3x3 + 5x2 – 1 – 7x)
= x3 + 2x2 + 3x – 2 – 3x3 – 5x2 + 1 + 7x
= x3 – 3x3 + 2x2 – 5x2 + 3x + 7x – 2 + 1
= – 2x3 – 3x2 + 10x – 1
Remember : The required polynomial is
= (To get the polynomial) – (added with the polynomial)
= (3x3 + 5x2 – 1 – 7x) – (x3 + 2x2 + 3x – 2)
Algebra : Polynomials 5. If P(x) = 3x3 – 4 and Q(x) = 3 – 4x + 7x2 – 5x3, find P(x).Q(x)
Solution :
We have P(x).Q(x) = (3x3 – 4).(3 – 4x + 7x2 – 5x3)
= 3x3(3 – 4x + 7x2 – 5x3) – 4(3 – 4x + 7x2 – 5x3)
= 9x3 – 12x4 + 21x5 – 15x6 – 12 + 16x – 28x2 + 20x3
= –15x6 + 21x5 – 12x4 + 9x3 + 20x3 – 28x2 + 16x – 12
= –15x6 + 21x5 – 12x4 + 29x3 – 28x2 + 16x – 12
24 PRIME Opt. Maths Book - VIII
Exercise 1.3
1. Answer the following questions.
i) What is polynomial? Write down its types according to degree of
polynomial.
ii) What is algebraic expression? Write down the types of polynomial
according to number of terms of the polynomials.
iii) What do you mean by equal polynomials? Explain with an
example.
iv) Write down numerial coefficient, literal coefficient, base and
power of the polynomial
(a) 3x3y2. in y. (b) 3x3y2 in x . (c) 3ax3y2 in ‘a’.
v) Write down the degree of the polynomial 3x4 – 2x3 + 5x2 – 2x + 7.
Also write down its type according to degree of the polynomial.
2. Which of the following algebraic expressions are the polynomials ? Algebra : Polynomials
Write down with reasons.
i) 4x3 + 5x2 – 3x + 2 ii) 3 x3 + 4x2 + 2x
3
iii) x12 – x + 4 +x2 iv) x4 ( 1 + 2 + 3 )+ 10
x3 x2 x
v) 73 x3 + 3x2 – 5 x–7+ 5
2 x
3. Write down the following polynomials in standard form and find their
degrees.
i) 3x – 5 – 3x2 + x4 – 2x3
ii) 7 – 2x2 – 3x + 5x3
iii) 3 – 5x2 – x3 + 2x4 – 3x + x5
iv) 3x – 2 + 5x2 – x5 + 2x3 – 3x4
v) 3x4 + 2x5 – 7 – 3x – 2x2 + x3
4. Write down the types of the polynomials according to degree and
number of terms.
i) 3x – 2x2 – 5 + 2x4 – x3 ii) 5 + 3x2 – 2x + 3x3
iii) –3 + 2x + x2 iv) 7 – 3x
v) 2
PRIME Opt. Maths Book - VIII 25
Algebra : Polynomials 5. If the following polynomials are equal, find the value of ‘a’ and ‘b’.
i) p(x) = (2a + 1)x3 – 7x2 + 3x + 2 and
q(x) = 7x3 + (2b – 3)x2 + 3x + 2
ii) f(x) = 10x4 – 3x3 + 5x2 – 12x + 5 and
g(x) = 10x4 – (3a – 3)x3 + 5x2 + 3bx + 5.
iii) p(x) = 5x2a – 1 + (3b + 2)x2 – 7x + 3.
q(x) = 5x3 + 8x2 – 7x + 3.
iv) f(x) = 2ax4 – 3x3 + bx2 – 2
g(x) = 4x4 – 3x3 – 2
v) p(x) = (3a – 2)x4 – 2x2b – 4 – 3x2 + 2x + 5
q(x) = 7x4 – 3x2 + 2x + 5
6. Find the following polynomials.
i) Add the polynomials: p(x) = x3 – 2x2 – x + 5 and q(x) = 3x3 + x2 – 2x – 1
ii) Add the polynomials: f(x) = 2 – 3x + x2 + 2x3 and g(x) = 2x – x2 +3x3 – 3
iii) Find the sum of the polynomials p(x) = 3x2 – 2x + 3x3 – 3 and
q(x) = –2x3 – x2 – x + 15.
iv) Subtract the polynomials p(x) = 2x3 – x2 + 2x + 3 and
q(x) = x3 + x2 – x + 5.
v) Subtract the polynomials x3 + 3x2 – 2x – 7 from 4x3 – 5 +2x – x2.
7. Find the following.
i) If p(x) = 3x3 + 2x2 – 5x + 2 and q(x) = 2x3 – 3x2 – 2x + 3, find p(x) +
q(x).
ii) If p(x) + q(x) = 5x3 – 3x2 + 2x – 5 and p(x) = 3x3 – x2 + 3x – 2, find
q(x).
iii) What must be subtracted from the polynomial 4x4 – 3x3 + 2x2 – 5x + 1
to get 3x4 – x3 – 5x2 + 2x + 3 ?
iv) What must be added to the polynomial x3 – 3x2 – 2x + 3 to get
x4 + 3x3 – x2 + 3x – 2?
v) What must be subtracted from the sum of x4 + 2x3 – 3x2 + 2x – 5
and 2x4 + x3 – x2 – 3x + 2 to get x4 – x3 + 2x2 – x – 2?
8. PRIME more creative questions
i) If f(x) = 3x3 + 2x2 – 3x + 2 and g(x) = x4 – x3 + x2 + x – 5, find
f(x) + g(x). Also write down the types of polynomial of the result
according to degree and number of terms.
26 PRIME Opt. Maths Book - VIII
ii) A polynomial x3 + x2 – 3x – 1 is subtracted from p( ) results x4 – x3
+ 5x2 – x + 2, find the polynomial p(x).
iii) If p(x) = (x2 + 2x – 3) and q( ) = (2x2 – x + 2). Find the value of
p(x) × q(x). Also write down the type of polynomial according to
degree.
iv) Multiply the polynomial x2 + 2x – 3 and 3x3 – 2x2 + 3x – 5.
v) If f(x) = (x2 + 2), g(x) = 2x2 – x + 3, what must be subtracted from
the product of f(x) and g(x) to get x4 – x3 + 2x2 – x – 2?
Answer Algebra : Polynomials
1. Consult to your subject teacher.
2. Consult to your subject teacher.
3. Consult to your subject teacher.
4. Consult to your subject teacher.
5. i) a = 3, b = –2 ii) a = 2, b = –4 iii) a = 2, b = 2
iv) a = 2, b = 0 v) a = 3, b = 2
6. i) 4x3 – x2 – 3x + 4
ii) 5x3 – x – 1
iii) x3 + 2x2 – 3x + 12
iv) x3 – 2x2 + 3x – 2
v) 3x3 – 4x2 + 4x + 2
7. i) 5x3 – x2 – 7x + 5
ii) 2x3 – 2x2 – x – 3
iii) x4 – 2x3 + 7x2 – 7x – 2
iv) x4 + 2x3 + 2x2 + 5x – 5
v) 2x4 + 4x3 – 6x2 – 1
8. i) x4 + 2x3 + 3x2 – 2x – 3; Biquadratic, multinomial
ii) x4 – 2x3 + 4x2 + 2x + 3
iii) 2x4 + 3x3 – 6x2 + 7x – 6 ; Biquadratic.
iv) 3x5 + 4x4 – 10x3 + 7x2 – 19x + 15
v) x4 + 5x2 – x + 8
PRIME Opt. Maths Book - VIII 27
Algebra : Sequence and Series 1.4 Sequence and series
Let us take examples of the set of numbers taken in order.
1, 4, 7, 10, ..., ..., ..., ..., ..., ..., ..., ...
1, 3, 9, 27, ..., ..., ..., ..., ..., ..., ..., ...
40, 30, 20, 10, 0, ..., ..., ..., ..., ..., ..., ..., ...
Here, the set of numbers are written in ascending and descending order
under the rule of increased by (+3) in first, by (×3) in second and decreased
by 10 in third example respectively.
Such type of arrangement of the numbers as the above examples taken
in ascending or in descending order under a certain rule with a constant
number is called sequence.
Sequence : The set of the numbers taken in order
under a certain rule is called sequence. Eg.: 2, 6,
10, 14, 18, ..., ..., ..., ..., ..., ..., ..., ...
• A sequence having finite number of terms is called finite sequence.
For example : 2, 5, 8, 11, 14, 17, 20.
• A sequence having infinite number of terms is called infinite
sequence. For example : 3, 7, 11, 15, 19, ..., ..., ..., ..., ..., ..., ..., ...
• Set of numbers of a sequence is also called the progression.
The sum of the terms of a sequence is called the
series.
• If the no. of terms in the sequence is finite, the
corresponding series is called the finite series.
• If the number of terms is infinite, the series is
called infinite series.
• Finite series is 2 + 5 + 8 + 11 + 14 + 17
• Infinite series is 3 + 7 + 11 + 15 + 19 + ..., ..., ..., ..., ..., ..., ..., ...
28 PRIME Opt. Maths Book - VIII
Types of sequence Algebra : Sequence and Series
1. Arithemetic Sequence : A sequence is said to be an Arithmetic Sequence
(AS) or Arithmetic Progression (AP), if the difference between any two
consecutive terms remains constant. The constant difference is called
the common difference (d) and the next consecutive term is obtained
by adding d with the term.
Arithmetic Sequence : The set of number of a sequence
which is written under a certain rule of addition with a
constant number is called arithmetic sequence.
Example : 2, 7, 12, 17, 22, 27.
Here, the numbers taken in order are increased by (+5)
• The constant difference is called common difference.
d = t2 – t1
• t1, t2, t3, t4, t5, ..., ..., ..., ..., ..., ..., ..., ... etc are taken symbolic form of the
terms of sequence or series.
• Symbolically the constant addition number of the terms is ‘d’
called common difference where,
t1 = a, = a + 0 . d = a + (1 – 1)d
t2 = a + d = a + 1.d = a + (2 – 1)d
t3 = a + 2d = a + (3 – 1)d
t4 = a + 3d = a + (4 – 1)d and so on.
nth term of the sequence can be written as tn = a + (n – 1)d
Here, d = t2 – t1 = t3 – t2 = t4 – t3 ...................................... etc.
\ tn = a + (n – 1)d
• More examples of arithmetic sequence.
2, 7, 12, 17, 22, ..., ..., ..., ..., ..., ..., ..., ...
40, 30, 20, 10, 0, –10, –20, ..., ..., ..., ..., ..., ..., ..., ...
a, a + d, a + 2d, a + 3d, a + 4d, ..., ..., ..., ..., ..., ..., ..., ...
a, a – d, a – 2d, a – 3d, a – 4d, ..., ..., ..., ..., ..., ..., ..., ...
PRIME Opt. Maths Book - VIII 29
• Above examples in the form of series.
2 + 7 + 12 + 17 + 22 + ..., ..., ..., ..., ..., ..., ..., ...
40 + 30 + 20 + 10 + 0 + (–10) + (–20) + ..., ..., ..., ..., ..., ..., ..., ...
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + ..., ..., ..., ..., ..., ..., ...
a + (a – d) + (a – 2d) + (a – 3d) + (a – 4d) + ..., ..., ..., ..., ..., ..., ..., ...
2. Geometric Sequence : A sequence is said to be a Geometric sequence
(GS) or Geometric progression (GP) if the ratio between any two
consecutive terms remains constant. The constant ratio is called
the common ratio (r) and the next consecutive term to any term is
obtained by multiplying the term with common ratio ‘r’.
Algebra : Sequence and Series Geometric Sequence : The set of number of a sequence
which is written under a certain rule of multiplication
with a constant number is called geometric sequence.
Example : 2, 6, 18, 54, ..............
Here, the numbers are written in ascending order under
the multiplication by a constant number by (× 3).
• The constant ratio is called common ratio.
t2
r= t1
• t1, t2, t3, t4, t5, ..., ..., ..., ..., ..., ..., ..., ... etc. are the symbolic form of the
terms of the sequence or series.
• The constant multiplicative number is symbolised by ‘r’ which is
called common ratio, where
t1 = a = ar0 = ar1 – 1
t2 = ar = ar1 = ar2 – 1
t3 = ar2 = ar3 – 1
t4 = ar3 = ar4 – 1
Similarly as above by inspection,
nth term of the s=eqttu32en=cetti34s,=tn.=.....a..r..n..–.1....
t2
Where, r = t1 etc.
\ tn = arn–1
30 PRIME Opt. Maths Book - VIII
• More example of geometric sequence.
3, 6, 12, 24, 48, ..., ..., ..., ..., ..., ..., ..., ...
80, 40, 20, 10, 5, 5 , 5 , ..., ..., ..., ..., ..., ..., ..., ...
2 4
a, ar, ar2, ar3, ar4, ..., ..., ..., ..., ..., ..., ..., ...
a a a a a
a, r , r2 , r3 , r4 , r5
• Geometric series for the above examples.
3 + 6 + 12 + 24 + 48 + ... + ... + ... + ... + ... + ... + ...
80 + 40 + 20 + 10 + 5 + 5 + 5 + ... + ... + ... + ... + ... + ... + ...
2 4
a + ar + ar2 + ar3 + ar4 + ... + ... + ... + ... + ... + ... + ...
a a a a a
a+ r + r2 + r3 + r4 + r5
Worked out Examples Algebra : Sequence and Series
1. What type of sequence is given below? Also find the 10th term of 2, 6,
10, 14, 18, ..., ..., ..., ..., ..., ...
Solution:
The given sequence is, 2, 6, 10, 14, 18, .....................
Here,
First term (a) = 2
Common difference (d) = t2 – t1
= 6 – 2
= 4
Also,
d = t3 – t2 = 10 – 6 = 4
d = t4 – t3 = 14 – 10 = 4
Hence, It is arithmetic sequence.
We have,
nth term of the sequence is,
tn = a + (n – 1)d
t10 = 2 + (10 – 1)4
= 2 + 9 × 4
= 38
PRIME Opt. Maths Book - VIII 31
2. Which sequence represents the given pattern? Also find the no. of dots
found in 6th diagram.
Algebra : Sequence and Series
Solution:
The sequence formed according to the no. of dots present in the given
diagrams is 3, 5, 7, ..., ..., ..., ..., ...
Here, t2 – t1 = 5 – 3 = 2
t3 – t2 = 7 – 5 = 2
Hence, it is arithmetic sequence
Where,
First term (a) = 3
Common difference (d) = t2 – t1
= 5 – 3 = 2
\ No of dots found in 6th diagram is
tn = a + (n – 1)d
t6 = 3 + (6 – 1)2
= 3 + 5 × 2
= 13
3. Find the 8th term of the sequence 512, 256, 128, 64, ..., ..., ..., ..., ...
Solution :
The given sequence is, 512, 256, 128, 64, ..., ..., ..., ..., ..., ..., ..., ...
t2
Here, t1 = 256 = 1
512 2
t3 = 128 = 1
t2 256 2
Hence, It is geometric sequence.
Then, First term (a) = 512
t2
Common ratio (r) = t1 = 256 = 1
512 2
8th term (t8) = ?
We have,
or, tn = arn–1 ` 1 8–1
t8 = (512) 2
j
32 PRIME Opt. Maths Book - VIII
= 512 × ` 1 7
2
j
= 512 × 1
128
\ t8 = 4
4. If first term is 3 of an arithmetic sequence having 9th term 35, find the Algebra : Sequence and Series
common difference.
Solution:
First term (a) = 3
9th term (t9) = 39
common difference (d) = ?
We have,
tn = a + (n – 1)d
or, t9 = a + (9 – 1)d
or, 35 = 3 + 8d
or, 32 = 8d
\d =4
Exercise 1.4
1. What types of sequences are given below? Write down with the reason
after inspection.
i) 3, 8, 13, 18, 23, ..., ..., ..., ..., ..., ..., ..., ...
ii) 5, 10, 20, 40, ..., ..., ..., ..., ..., ..., ..., ...
iii) 2, 6, 18, 54, ..., ..., ..., ..., ..., ..., ..., ...
iv) a, a + d, a + 2d, a + 3d, ..., ..., ..., ..., ..., ..., ..., ...
v) a, ar, ar2, ar3, ar4, ..., ..., ..., ..., ..., ..., ..., ...
2. Identify the following as sequence and series. Also find the common
constant term associated with them.
i) 5 + 12 + 19 + 26 + ... + ... + ... + ... + ... + ... + ...
ii) 100, 90, 80, 70, 60, ..., ..., ..., ..., ..., ..., ..., ...
iii) 3 + 6 + 12 + 24 + 48 + 96 + 192 + ... + ... + ... + ... + ... + ... + ...
iv) a + 10d, a + 9d, a + 8d, a + 7d + ..., ..., ..., ..., ..., ..., ..., ...
a a a a
v) a, r , r2 , r3 , r4
PRIME Opt. Maths Book - VIII 33
3. Identify the sequences or series given below as finite and infinite.
i) 5 + 12 + 19 + 26 + ... + ... + ... + ... + ... + ... + ...
ii) 100, 90, 80, 70, 60, ..., ..., ..., ..., ..., ..., ..., ...
iii) 3 + 6 + 12 + 24 + 48 + 96 + 192 + ... + ... + ... + ... + ... + ... + ...
iv) a + 10d, a + 9d, a + 8d, a + 7d + ..., ..., ..., ..., ..., ..., ..., ...
a a a a
v) a, r , r2 , r3 , r4
Algebra : Sequence and Series 4. i) Find the 12th term of the sequence : 2, 6, 10, 14, 18 ..., ..., ..., ..., ..., ...
ii) Find the 8th term of the series 40 + 34 + 28 + 22 + 16 ..., ..., ..., ..., ...
iii) If first term and common difference of an arithmetic sequence
are 6 and 8 respectively. Find the 15th term.
iv) If first term is 50 and common difference is –6, find the 11th term
of the arithmetic sequence.
v) If 10th term of the arithmetic sequence having common difference
4 is 50, find the first term.
5. i) Find the 8th term of the sequence 3, 6, 12, 24, ..., ..., ..., ..., ..., ..., ..., ...
ii) Find the 6th term of the series 128 + 64 + 32 + ..., ..., ..., ..., ..., ..., ..., ...
iii) Find the 10th term of the sequence 5, 10, 20 40, ..., ..., ..., ..., ..., ..., ...
iv) If first term and common ratio of a geometric sequence are 10
and 2 respectively, find the 5th term of the sequence.
v) If first term and common ratio of a geometric sequence are 80
1
and 2 respectively, find the 6th term of the sequence.
6. Find the followings:
i) If first term and 6th term of an AS are 7 and 27 respectively, find
the common difference of the sequence.
ii) If common difference and 9th term of an AS are 3 and 44
respectively, find the first term of the sequence.
iii) If first term and common ratio of a GS are 2 and 3 respectively,
find the fifth term of the sequence.
iv) If first term and 4th term of a GP are 5 and 135 respectively, find
the common ratio of the sequence.
v) If common ratio and 6th term of a GS are 4 and 5120, find the first
term of the sequence.
34 PRIME Opt. Maths Book - VIII
7. PRIME more creative questions
i) Find the number of dots present in 8th figure in the given pattern
of triangles.
ii) Add two more diagrams for the followings. Also find the number
of dots found in 10th figure in the given pattern.
Algebra : Sequence and Series
iii) Monthly salary of a person in the month of Baishakh is Rs. 4000
where monthly increment is given for first 6 months by Rs. 500.
What will be his salary in the month of Aswin?
iv) If first term of an arithmetic sequence having 8th term 25 is 4,
find the common difference of the sequence.
v) Find the 7th term of the series 320 + 160 + 80 + ... + ... + ... + ...
Answer
1. Show to your teacher.
2. Show to your teacher.
3. Show to your teacher.
4. i) 46 ii) –2 iii) 118 iv) –10 v) 14
v) 2.5
5. i) 384 ii) 4 iii) 2560 iv) 160 iv) 3
v) 5
6. i) 4 and 7,11,15,... ii) 20 iii) 162 v) 20
7. i) 24 ii) 50 iii) Rs. 6,500 iv) 3
PRIME Opt. Maths Book - VIII 35
Alzebra
Unit Test - 1
Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Algebra : Ordered Pairs Attempt all the questions:
1. Name the x-component and y-component of a ordered pair (x, y).
2. a. If (3x – 2, 4) = (x + 4, 3y + 16) are two equal ordered pairs, find ‘x’
any ‘y’.
b. Which is greater between 3 or 3 5 ?
c. Write down the polynomial 5x3 – 2x2 + x4 – 3x + 2 in standard form
and mention its type based on degree.
3. a. Simplify:
3 – 4 – 1
6– 3 6– 2 3– 2
b. If A = {1, 2, 3} and B = {4, 5}, find A × B and show in arrow diagram.
Also find the cardinality of A × B.
4. Find 8th term of the sequence 3, 7, 11, 15, ..............................
Unit Test - 2
Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. Write down one difference between sequence and series.
2. a. What is ordered pair? Write down any three ordered pairs of
district with respect to headquarter.
b. Simplify: 18 + 3 50 – 98
c. Find the value of p(x) + q(x) where p(x) = 3x2 – 5x + 2 and q(x) =
x3 – 2x2 + x – 3.
3. a. If A × B = {(a, p), (a, q), (a, r), (b, p), (b, q), (b, r)} find the sets A and
B. Also find B × A and n(B × A).
b. Write down in ascending order of: 2, 3 4 and 4 5 .
4. Find the 6th term of the series. 3 + 6 + 12 + ..........................
36 PRIME Opt. Maths Book - VIII
Unit 2 Limits
Specification Grid Table
K(1) U(2) A(4) HA(5) TQ TM Periods
No. of Questions – – 1 – 1 4 4
Weight – – 4 –
K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks
Objectives : At the end of the lesson
• Students are able to know sequence and series in descending and
ascending order.
• Students are able to round up the numbers in decimal and whole number.
• Students are able to know the limiting value of the sequence of numbers.
• Students are able find the limiting value of algebraic expressions.
Materials Required:
• Chart paper.
• Chart of numbers taken in order.
• chart of examples of limit values.
• Some of the sequences in diagrams.
PRIME Opt. Maths Book - VIII 37
Limit
Let us take example of sequences of infinite terms. The infinite term will be
nearly equal to a definite number which is called absolute value.
• 0.1, 0.01, 0.001, 0.0001, 0.00001, ..., ..., ..., ..., ...
• 0.9, 0.99, 0.999, 0.9999, 0.99999, ..., ..., ..., ..., ...
• 2.1, 2.01, 2.001, 2.0001, 2.00001, ..., ..., ..., ..., ...
• 2.9, 2.99, 2.999, 2.9999, 2.99999, ..., ..., ..., ..., ...
The above sequence can be defined to their infinite terms where the term
will be nearly equal to zero (0), 1, 2, 3 respectively. They are also called the
absolute values. The absolute value of such sequences are also called the
limiting value of the sequence which is nearest whole number.
Diagrammatic representation of sequence to get absolute value.
i) Let us consider a line segment AB having length 20cm. It is divided
by a point at mid-point and so on as given in diagram.
10cm 20cm
5cm
Limits 2.5cm
A SR Q P B
The line segment AB is divided by half and continuously to half again.
Here,
AB = 20cm, AP = 10cm,
AQ = 5cm, AR = 2.5cm,
AS = 1.25cm
The sequence of the length of the line segments so formed will be,
20, 10, 5, 5 , 5 , .................... .
2 4
As the value of the partition ‘n’ increases and for very large ‘n’ the
length of the intersected line becomes very small which is about to
zero i.e. as n → ∞, l → 0 and we write this information as
n lim l = 0
"3
38 PRIME Opt. Maths Book - VIII
Let us consider the another situation,
Let add the intersected half parts successively as shown in the fig
which is denoted by S (say).
i.e. Sn = BP + PQ + QR + RS + ....................
= 10 + 5 + 2.5 + 1.25 + .................
The sum of the all parts AB will the equal to 20(nearly).
As n → an (i.e. infinitely intersected)
S → 20 (but not equal to 20)
We write this information in limit as
n lim Sn = 20
"3
The sequence to denote whole line segment AB
from its parts as,
• 10, 5, 5 , 5 , 5 , .................... ≈ 0
2 4 8
• 10 + 5 + 5 + 5 + 5 + ..................... ≈ 0 Limits
2 1 8
ii) When a basketball is dropped on the ground, it is going in
dropping on decreasing height and finally comes at rest (nearly)
as shown in the diagram after infinite drops.
• As the dropping times increases, the height travelled by the ball
decreases.
• Finally the height travelled by the ball nearly equal to zero.
PRIME Opt. Maths Book - VIII 39
iii) Let a full cup of water of height 16cm is made half in each attempt
as shown in the figure.
16cm
8cm 4cm
2cm 1cm
The sequence so formed is 16, 8, 4, 2, 1 ,...................................
Discuss the absolute value of the above examples for their sequences
mentioned as above for their infinite term.
Limits
A function or expression f(x) is said to be have a finite
limit ‘l’ at x = a, if the value of f(x) approaches towards
the constant ‘l’ as the value of x approaches towards
the number ‘a’.
i.e. As x → a implies f(x) → l. We denote it as lim f(x) =l
x"a
i. In a sequence discussed above as 0.1, 0.01, 0.001, 0.0001, ..., ..., ..., the
infinite term will be nearly zero which is called its limiting value. It can
be symbolized as, x lim f(x) = 0
"3
ii. Lets us take an example given in table.
x 0.9 0.99 0.999 0.9999 0.99999 ....... ≈ 1
f(x) =3x + 2 4.7 4.97 4.997 4.9997 4.99997 ....... ≈ 5
It can be symbolized in limit as,
lim f(x) = lim (3x + 2) = 5
x "1 x "1
40 PRIME Opt. Maths Book - VIII
Note :
1. For a function y = f(x), f(a) is called the functional value at x = a.
[Put x = a in the expression]
2. For a polynomial function, the functional value is the limiting value of
any point at x = a
i.e. lim f(x) = f(a)
x"a
3. For a rational function, the functional value is the limiting value at
0033orfor33m.,
x = a if it does not take the form
If on putting x = a it takes 0
4. 0 or factorize the denominator
or numerator and cancel the common factors. Then put x = a to get the
limit at x = a.
Worked out Examples
1. Write down the 10th term of the sequence 1.1, 1.01, 1.001,.................... .
Also write down its limiting value.
Solution: Limits
The given sequence is,
1.1, 1.01, 1.001,.................... .
It’s 10th term can be defined by inspection as,
t10 = 1.0000000001.
Here,
The infinite term of the sequence is very close to 1 due to which its
limit value is 1.
i.e. lim (for expression on ‘x’) = 1.
x"3
2. Area of a triangle is 32cm2 which is going to be divided by taking
the median of the triangle continuously in a sequence. Show the
information in diagram and examine the limiting value of area.
Solution:
Taking a triangle ABC having area = 32cm2 and dividing as the given
informations by taking median as.
41
PRIME Opt. Maths Book - VIII
The sequence of the area of triangle so formed is 32, 16, 8, 4, 2, 1,
.................... . The limit value of the area so formed is very close to zero
for infinite diagram.
i.e. lim (Area) =0
x"3
3. Complete the given table and examine the limit value.
x 0.1 0.01 0.001 0.0001 0.00001 0.000001
f(x) =2x + 3 ......... ......... ......... ......... ......... .........
Solution:
Sequence of variable ‘x’ is given 0.1, 0.01, 0.001, 0.0001, 0.00001,
0.000001. The algebraic expression is f(x) = 2x + 3
Limits Where,
Taking x = 0.1
f(x) = 2x + 3 = 2(0.1) + 3 = 3.2
Taking x = 0.01,
f(x) = 2x + 3 = 2(0.01) + 3 = 3.02. Then the table is completed as,
x 0.1 0.01 0.001 0.0001 0.00001 0.000001
2x + 3 3.2 3.02 3.002 3.0002 3.00002 3.000002
It’s absolute value for the infinite term is very close to 3 which is taken
as the limit.
i.e. limit = lim (2x + 3) = 3
x"0
4. Evaluate : lim (x2 + 2x – 3)
x"2
Solution:
lim (x2 + 2x – 3)
x"2
Taking x = 2 (very close to 2)
x2 + 2x – 3 = 22 + 2 × 2 – 3
= 4 + 4 – 3
= 5.
\ lim (x2 + 2x – 3) = 5
x"2
42 PRIME Opt. Maths Book - VIII
5. Evaluate: lim x2 – 9
x"3 x–3
Solution:
lim x2 – 9 [when x = 3 then function becomes 0 form]
x"3 x–3 0
= lim (x + 3)(x–3)
x"3 (x–3)
=3+3
=6
\ lim x2 – 9 =6
x"3 x–3
Exercise 2.1
1. Find the next two terms and limit of the given sequences.
i. 1.9, 1.99, 1.999, 1.9999, ..............................
ii. 3.1, 3.01, 3.001, 3.0001, ..............................
iii. 5, 5 , 5 , 5 , 5 , .............................. Limits
2 4 8 16
iv. 0.19, 0.199, 0.1999, 0.19999, ..............................
v. 4.101, 4.1001, 4.10001, 4.100001, , ..............................
2. Complete the following table and diagrams to compute the limit value.
i. x 10 100 1000 10000 100000
........... ........... ........... ........... ...........
f(x) = 1
x
ii. x 0.1 0.01 0.001 0.0001 0.00001 0.000001
f(x) =3x + 4 ......... ........... ........... ........... ........... ...........
iii. A BA BA B
P PQ
D CD C D C
upto 6 terms. Where area of rectangle ABCD = 80cm2 and P & Q
are taken as the mid-point of sides.
PRIME Opt. Maths Book - VIII 43
iv.
upto 6 terms where area of circle is 200cm2 & going on dividing
v. by half.
Draw a line segment of length 24cm and going on dividing by half
of the line segment continuously.
3. Evaluate the following limits:
i. xli"m0 (3x + 5) ii. lim (5x – 2)
x"1
iii. lim (2x2 – x + 3) iv. lim (x2 + 2x – 3)
x"2 x"3
v. xli"ma (x2 + 2ax + a2)
4. PRIME more creative questions:
Limits i. From the given diagram write down the sequence of area of
shaded region where area of DPQR = 100cm2 and taking the mid-
point of sides as shown in diagrams. P
PPP
Q RQ RQ RQ R
ii. Complete the given table and find the absolute value.
x 10 100 1000 10000 100000 1000000
1 ........... ...........
x2 ........... ........... ........... ...........
iii. A sphere of radius 14cm is going on cutting down to hemi-sphere,
half of hemisphere and so on up to 5 times. Draw the diagram
and write down the sequence for volume. Also find its limit value.
iv. Evaluate: lim x2 – 4
x"2 x–2
v. Evaluate : lim x2 – a2
44 x"a x–a
PRIME Opt. Maths Book - VIII
Answer
1. i. 1.99999, 1.999999; limit value = lim (expression on x) = 2
x"3
ii. 3.00001, 3.000001; limit value = lim (expression on x) = 3
x"3
iii. 5 , 5 ; limit value = lim (expression on x) = 0
32 64 x"3
iv. 0.199999, 0.1999999; limit value = lim (expression on x) = 0.2
x"3
v. 4.1000001, 4.10000001, limit value = lim (expression on x) = 4.1
x"3
2. i. Limit value = lim (expression on x) = 0
x"3
ii. Limit value = lim (expression on x) = 4
x"3
iii. Limit value = lim (area) = 80cm2 Limits
x"3
iv. Limit value = lim (area) = 0
x"3
v. Limit value = lim (length) = 0
x"3
3. i. 5 ii. 3 iii. 9 iv. 12 v. 4a2
4. i. Limit value = lim (area) = 100cm2
x"3
ii. Limit value = lim (expression on x) = 0.
x"3
iii. Limit value = lim = (volume) =0
x"3
iv. 4 v. 2a.
PRIME Opt. Maths Book - VIII 45
Limits
Unit Test
Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. Write down next two terms of the sequence 2.01, 2.001, 2.0001, 2.00001,
..., ..., ..., ..., ..., ...
2. a. Write down the limit value of the sequence by finding next 2 terms
of, 4.9, 4.99, 4.999, 4.9999, ..., ..., ..., ..., ..., ...
b. What is limit? Write down a suitable example.
c. Evaluate: lim (3x + 5)
x"1
3. Divide a line segment of length 16cm making half continuously 5 times
for a part. Show it in diagram and write down the limit value of the
sequence of its length.
4. Write down the sequence of area of triangle having 80cm2 for the given
diagrams by adding two more diagram. Also write down its limit value.
Limits
46 PRIME Opt. Maths Book - VIII
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