Opt. Maths Book for Class-9

Vedanta

Optional Mathematics

Teacher's Guide

Grade

9

Piyush Raj Gosain
Vedanta Optional Mathematics Teacher's Guide ~ 9 1

Preface

This is a teacher's Guide of Vedanta Excel in Optional Mathematics to help the
teacher's, in teaching learning process, who are teaching Optional Mathematics in
secondary level in Grade 9 and 10.

I have tried to write this book in the form to help the teachers of Optional
Mathematics regarding what are objectives, how to teach, how to solve problems
what are required teaching materials and how to evaluate in the classroom effectively.
At the end of each chapter, there are given some questions for more practice and
evaluation of the students.

I hope that the book will be one of best friend of teachers who have been using
Vedanta Excel in Optional Mathematics for grade 9 and 10. It helps the teachers to
make lesson plan, to use required teaching materials to evaluate the students. It also
helps the teachers providing required teaching notes.

The motto of Vedanta Publication (P) Ltd is ''read, lead and succeed''. I am hopeful
that the book will also help to fulfill the objectives of the publication as well as the
objectives of curriculum of Optional Mathematics.

The idea how to write this book is coined by our respectable senior Mathematics
text book writer, educator Hukum Pd Dahal and heartfelt gratitude to him.

I am confident that the teachers will find this book as an invaluable teaching aid.
I am thankful to all the teachers who have been using Vedanta Excel in Optional
Mathematics.

My hearty thanks goes to Mr. Hukum Pd Dahal, Tara Bahadur Magar and P.L Shah,
the series editors, for their invaluable efforts in giving proper shape to the series. I am
also thankful to my colleague Mr Gyanendra Shrestha who helped me a lot during
the preparation of the book.

I would like to thank chairperson Mr Suresh Kumar Regmi, Managing Director
Mr. Jiwan Shrestha, Marketing Director Manoj Kumar Regmi for their invaluable
suggestion and support during the preparation of the series in Optional Mathematics.

Last but not the least I am thankful to Mr Daya Ram Dahal and Pradeep Kandel,
the computer and designing senior officer for their skill in designing the book in
such an attractive form.

I'm profoundly grateful to the Vedanta Publication (P) LTD to get the series
published. Valuable suggestion and comments from the concerned will be highly
appreciated in days ahead.

Piyush Raj Gosain

2 Vedanta Optional Mathematics Teacher's Guide ~ 9

CONTENT

Unit 1 Relations and Functions 4
Unit 2 Polynomials 21
Unit 3 Sequence and series 27
Unit 4
Unit 5 Limit 35
Unit 6 Matrix 51
Unit 7 Co-ordinate Geometry 65
Unit 8
Equations of Straight Lines 80
Unit 9
Trigonometry 114
Unit 10 Trigonometric Ratios & 127
Conversion of t-ratios
Unit 11 Trigonometric Ratios of 145
Some Standard Angles
Unit 12 Trigonometric Ratios of 148
Unit 13 Any Angle 171
Unit 14 Vectors 183
Transformation 195
Unit 15 201
Statistics: Partition values

Statistics: Measure of
Dispersion

Vedanta Optional Mathematics Teacher's Guide ~ 9 3

UNIT

one Relations and Functions

Estimated Teaching Hours : 15

1. Objectives

SN Level Objectives

To define ordered pair.

To define cartesian product.

i. Knowledge (K) To define a relation and a function.
To define domain and range of a relation and

function.

To define different types of relations.

To find domain and range and codomain of a
relation and a function.
To represent a relation in different ways (set of ordered pair,
tabulation method, arrow diagram method, graphic method.)
ii. Understanding (U) To find inverse of a given relation.
To check given relations are functions or not.
To be able to check given diagrams represent
function or not. (i.e. vertical line test.)
To represent different type sof functions in arrow diagrams.

To solve problems related to functions.

iii. Application (A) To draw graphs of linear constant, identity

function, linear function.

iv. Higher Ability (HA) To solve difficult problems in relations and
functions.

2. Teaching Materials
Arrow diagrams and graphs for different function in a chart paper.

Different geometrical figures to check vertical line test on chart papers.

3. Teaching Learning Strategies:
Take some examples of pair of elements used in our daily life.
Define a pair and an ordered pair with different examples.
Explain difference between a pair and an ordered pair with illustrated examples.
Define equal ordered pairs.
Solve some problems from exercise 1.1.
Taking two non-empty sets, illustrate how to form cartesian product.
Define the cartesian product and discuss how to show them in
i) set of ordered pair

4 Vedanta Optional Mathematics Teacher's Guide ~ 9

ii) set builder form
iii) tree diagram method
iv) arrow diagram method
v) tabular method
Discuss solution of some questions in exercise 1.2.
Starting from daily life relation examples define relations.
Discuss different methods of presentation of relations.
Define domain and range of a relation with at least two examples.
Define types of relations - identity relation, reflexive relation, transitive relation,

symmetric relation, equivalence relation and inverse relation with examples.
Discuss solution of questions in exercise 1.4.
Discuss functions and its types with arrow diagrams.
Explain how to find domain and range of a function.
Discuss solution of some questions from exercise 1.5(A).
Draw some figures and explain about vertical line test of a function.
Explain how to solve questions in exercise 1.5(B) (vertical line test)
Notes:
1) Let A and B be any two non empty sets. Then a relation from A to B is defined as a

sub-set of cartesian product A×B. A×B is defined by
A×B = {(x, y) : x∈A and y∈B}
Relation R from A to B is denoted by
R : A → B.
2) Domain of relation R = {x : (x, y)∈R}
Range of relation R = {y : (x, y)∈R}
3) Inverse relation of R is defined by
R–1 = {(y, x) : x∈A and y∈B}
4) Let a relation R be defined on A = {a, b, c}
a) Reflective relation, R = {(a, a)} for all a∈A.
b) Transitive relation, R = {(a, b), (b, c), (a, c)} for a, b, c∈A.
c) Symmetric relation R = {(a, b), (b, a), (a, c), (c, a), (b, c), (c, b)} for a, b, c∈A.
d) A relation R on set A = {a, b, c} is called equivalence iff
i) Is is reflexive; for every a∈R, (a, a)∈R.
ii) It is transitive; for a, b∈A, (a, b), (b, c), (a, c)∈R.
iii) It is symmetric; a, b, c∈A, (a, b), (b, a), (b,c), (c,b), (a,c), (c,a)∈A.
5) (i) A function f:A → B is called an onto if the range is equal to co-domain of f.
ii) A function f:A → B is called into if range of f is proper-subset of co-domain.
iii) A function f:A → B is called one - one if each element of A has distinct image in set B.
iv) A function f:A → B is called many to one function if two or more element of set A

have the same image in set B.
v) A function f:A → B is called one to one onto (objective) function if each element

of A has the distinct image in set B and range is equal to co-domain.

Vedanta Optional Mathematics Teacher's Guide ~ 9 5

6) A function f:A → B is called real valued function if A and B are sub-sets of real num-
ber R. It is denoted by f:R → R.

Some solved problems

1. Find the value of x and y in the following conditions:
a) (2x + y, 2) = (1, x – y)

Solution
Here, (2x + y, 2) = (1, x – y)
Equating the corresponding elements of equal ordered pair.
2x + y = 1 ... ... ... (i)
x – y = 2 ... ... ... (ii)
adding equation (i) and (ii), we get
3x = 3
 x = 1
put the value of x in equation (i), we get
2,1 + y = 1
or, y = 1 – 2
 y = –1

b) (2x + y, 2x – y) = (16, 1)

Solution
Here, (2x + y, 2x – y) = (16, 1)
Equating the corresponding elements of equal ordered pair.
2x + y = 16
or, 2x + y = 24
 x + y = 4 ... ... ... (i)
and 2x – y = 1
or, 2x – y = 20
 x – y = 0 ... ... ... (ii)
adding equations (i) and (ii), we get
2x = 4
 x = 2
put the value of x in equation (ii), we get,
 x = y = 2.

2. a) Under which condition A×B is defined ?

Solution
Here, A×B is defined if A and B both are non-empty sets.

6 Vedanta Optional Mathematics Teacher's Guide ~ 9

b) If P = {1, 2, 3}, Q = {10}, find n(P×Q).

Solution

Here, P = {1, 2, 3} and Q = {10}

n(P) = 3, n(Q) = 1

We have, n(P×Q) = n(P) × n(Q)

=3×1

= 3

Alternate Method

Here, P×Q = {1, 2, 3} × {10}

= {(1, 10), (2, 10), (3, 10)}

Now, n(P×Q) = no. of ordered pair in P×Q

= 3.

3. If M = {1, 2, 3} and N = {2, 4, 5}, find a relation from M to N such that: x + y ≤ 10.

Solution

Here, M = {1, 2, 3} and N = {2, 4, 5}

M×N = {1, 2, 3} × {2, 4, 5}

= {(1, 2), (1, 4), (1, 5), (2, 2), (2, 4), (2, 5), (3, 2), (3, 4), (3, 5)}

Let, R = {(x, y) : x + y ≤ 10}

= {(1, 2), (1, 4), (1, 5), (2, 2), (2, 4), (2, 5), (3, 2), (3, 4), (3, 5)}

4. Find the domain and range of the following relations. Also find the inverse relation of them:
a) R1 = {(a, x), (b, y), (c, z)}
4 (b)

R

26
4

12
5
6 10

Solution
Here, R1 = {(a, x), (b, y), (c, z)}
Domain of R1 = {a, b, c}
Range of R1 = {x, y, z}
Inverse relation of R1 is given by R–1={(x, a), (y, b), (z, a)}
b) R = {(2, 6), (6, 4), (10, 5), (12, 6)}

Solution
Here, {(2, 6), (6, 4), (10, 5), (12, 6)}

Vedanta Optional Mathematics Teacher's Guide ~ 9 7

domain of R = {2, 4, 5, 6}
range of R = {6, 12, 10}
Inverse relation of R is given by R–1 = {(2, 6), (6, 4), (10, 5), (12, 6)}

5. If A = {1, 2, 3} and a relation R is defined on A as R = {(1, 1), (1, 2), (2, 2), (3, 3), (1, 3),
(2, 3), (3, 1), (3, 2), (2, 1)}. Check whether R is reflexive, symmetric or transitive relation.

Solution

Here, R = {(1, 1), (1, 2), (2, 2), (3, 3), (1, 3), (2, 3), (3, 1), (3, 2), (1, 2), (2, 1)}

i. Now (1, 1), (2, 2), (3,3)∈R

Hence, a is a reflexive relation.

ii. (1, 2)∈R and (2, 1)∈R,

(1, 3)∈R and (3, 1)∈R

(2, 3)∈R and (3, 2)∈R

Hence, relation R is a symmetric relation

iii. Since, (1, 2), (2, 1)∈R and (1, 1)∈R

(1, 2), (3, 1)∈R and (1, 1)∈R

(2, 3), (3, 1)∈R and (2, 1)∈R

Hence R is also a transitive relation.

 R is reflexive, symmetric and transitive. Hence R is an equivalence relation.

6. Which of the following relations are functions ?

a) f = {(a, x), (b, y), (c, z)}

Solution
Here, each pre-image has distinct image.
It is a function i.e. every element of the first component has a unique image.

b) k = {(1, 2), (2, 3), (4, 5), (2, 8)}

Solution
Here, an element '2' has two images.
Hence, the relation k does not represent a function.

7. State whether the following mapping diagrams represent the functions or not. Give
reason also.

a) A fB

14

28

3 12

8 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution
Here, each element of the first set A has distinct image in the second set B.
So f is a function.
b) X f Y

21

42

63

4

Solution
Here, each element of the first set x has distinct image in the second set.
Hence, f is a function.

c) P f Q

ax

b
y

c

Solution
Here, each element of the first set P has unique element in the second set.
Hence f is a function.

d) h

0
12
48
5 10

Solution
Here, an element '0' does not have image in the second set.
Hence h is not a function.
8. Let f:A → B be a function with A = {1, 2, 3, 4} and B = 21, 1, 32, 2 . Find the function

defined 'is double of.' Then represent the function in the following ways:
a) Arrow diagram

Vedanta Optional Mathematics Teacher's Guide ~ 9 9

Solution Af B
Here,
1 1
2 2
3 1
4 3
2
1

b) Set builder form

Solution
Here, f = {(x, y): x=2y, x∈A and y∈B}

c) Formula

Solution

Here, f = {(x, y): x = 2y}

i.e. f= (x, y): y = x
2

d) Set of ordered pair

Solution

Here,

f = {(x, y): x is double of y}

=  1, 1 , (2, 1), 3, 3 , (4, 2)
 2 2

9. By using vertical line test, find which of the following represent function:

a) yM

O x
x'

N
y'
10 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution
Here, the vertical line cuts the given line only at a single point.
Hence it represents a function.
b) y

x' O x

y'

Solution
Here, the vertical line cuts the given curve (circle) at two points.
Hence, the curve does not represent a function.
c) y

x' O x

y'
Solution
Here, the vertical line cuts the given curve at a single point.
Hence the curve represents a function.

10. Draw graphs of the following functions and also state the type of function with
reasons.

a) f = {(1, 2), (2, 3), (3, 4)}

Solution
Here, f = {(1, 2), (2, 3), (3, 4)}

Vedanta Optional Mathematics Teacher's Guide ~ 9 11

y (3,4)
4 (2,3)
3 (1,2)

2
1

x' –2 –1 O 1 2 3 4x
–1

y'
The plotting of points on graph, we get a straight line as shown in the figure. It is one - one

function because each element of first component has distinct image in second set (set of
y-component). Moreover it represents a one - one onto function. It is also a linear function.

b) k = {(1, 2), (2, 4), (3, 9), (4, 16)}

Solution
Here, k = {(1, 2), (2, 4), (3, 9), (4, 16)}

y

16 (4,16)

14

12

10
(3,9)

8

6

x' 4 (2,4) 6
–4 2
(1,1)
–2 o 24

y'
The given points are plotted on a graph paper. We get a curve as shown in the figure. It

represents a quatratic function because it is in the form of y = x2.

12 Vedanta Optional Mathematics Teacher's Guide ~ 9

11. If f(x) = x2 – 3x + 1, find f(1) and f(2).
x+ 1

Solution

Here, f(x) = x2 – 3x + 1
x+ 1

For, x = 1, we get f(1) = 12 – 3.1 + 1
1+1

= 2 – 3 = –12
2

For x = 2, we get f(2) = 22 – 3.2 + 1
2+1

= 4 – 6+ 1 = – 1
3 3

12. If f  1  = 1, f  13  = 23, f  23  = 4 then write the function f in set ordered pair and hence
 2    3

find the domain and range of the function.

Solution

Here, f  1  = 1, f  1  = 23, f  2 = 4
2 3 3 3

f=  21, 1, 13, 32, 23, 4 in set of ordered pair.
3

Domain of f = 12, 31, 2
3

Range of f = 1, 32, 4
3

13. a) If f(x) = 3x + 4, find the image of 4.

Solution
Here, for x = 4, y = f(4) = 3.4 + 4 = 16
 16 s the image of 4.
b) If g(x) = x2+ 5 defined –6 ≤ x ≤ 7, find g(2), g(7) and g(8).

Solution

Here, g(x) = x2 + 5, –6 ≤ x ≤ 7

Here, the values of x are from –6 to 7.

Hence, g(8) is not defined i.e. we cannot find the value of g(8).

Now, for x = 2, g(2) = 22 + 5 = 9

g(7) = 72+ 5 = 54

14. a) A function f:N → N is defined by f(x) = x + 3, is f one - one and onto function ?

Solution
Here, f(x) = x + 3, x∈N
Domain and range are both set of natural numbers. For simplicity we wirte,

Vedanta Optional Mathematics Teacher's Guide ~ 9 13

y = x + 3

For natural numbers 1 and 2, there are no pre-images for them.
It means that the given function is not onto. Again for every value of x,
there is unique value of y.
Hence f is a one - one function.

b) If f:N → N is defined by f(x) = x2 + 1. What type of the function f is ? Write with examples.

Solution
Here, f:N → N and f(x) = x2 + 1
For x = 1, 2, 3, 4 ... ... ... we get
f(1) = 2, f(2) = 5, f(3) = 10, f(4) = 17
Domain and range both are set of natural numbers.
Some elements of the second set (N) do not have their pre-images.
For simplicity, we write y = x2 + 1
or, x2 = y – 1
 x = y – 1
For all values of y, x is not a natural number.
eg. For y = 3, x = 2∉N
Hence, f is not one - one onto function.
But for every value of x, y is a natural number.
Hence the function is one - one function
In conclusion we say that f is one - one into function.
In arrow diagram, we write

MN

1 1
2
3
24
5
6
3 7

8
9
4 10

17
5 26

This shows that some elements of second set do not have their pre-images.

15. If A = {x: x ≤ 4, x∈N}, B = {x: x ≤ 3, x∈N} and C = = {x: 3 ≤ x ≤ 5, x∈N}, verify the following:
a) A×(B∪C) = (A×B)∪(A×C)

Solution
Here, given sets can be written as follows
A = {1, 2, 3, 4}, B = {1, 2, 3}, C = {3, 4, 5}

14 Vedanta Optional Mathematics Teacher's Guide ~ 9

Now, B∪C = {1, 2, 3, 4, 5}

LHS = A×(B∪C)

= {1, 2, 3, 4} × {1, 2, 3, 4, 5}

= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1),

(3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5)}

Again, A×B = {1, 2, 3, 4} × {1, 2, 3}

= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)}

A×C = {1, 2, 3, 4} × {3, 4, 5}

= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5)}

Now, (A×B)∪(A×C) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4),

(4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}

 A×(B∪C) = (A×B)∪(A×C)

b) (A – B)×C = (A×C) – (B×C)

Solution

Here, A – B = {1, 2, 3, 4} – {1, 2, 3}

= {4}

LHS = (A × B)×C = {4} × {3, 4, 5}

= {(4, 3), (4, 4), (4, 5)}

A×C = {1, 2, 3, 4} × {3, 4, 5}

= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5)}

B×C = {1, 2, 3} × {3, 4, 5}

= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}

RHS = (A×C) – (B×C)

= {(4, 3), (4, 4), (4, 5)}

 (A – B)×C = (A×C) – (B×C) proved

16. Let, R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, Show that R is
reflexive, symmetric and transitive relation and hence equivalence relation.

Solution

Here, R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

Then we show that R is reflexive, symmetric and transitive, then it is an equivalence relation.

i) Reflexive: since, (1, 1), (2, 2), (3, 3)∈R, R is a reflexive relation.

ii) Symmetric: since, (1, 2), (2, 1), (2, 3), (3, 2)∈R, R is a symmetric relation.

iii) Transitive: since (1, 2), (2, 3)∈R and (1, 3)∈R

(1, 2), (2, 1)∈R and (1, 1)∈R

(1, 3), (3, 1)∈R and (1, 1)∈R

(2, 3), (3, 2)∈R and (2, 2)∈R

 R is also a transitive relation.

Since R is reflexive, symmetric and transitive relation, it is an equivalence relation.

Vedanta Optional Mathematics Teacher's Guide ~ 9 15

17. Let, A = {1, 2, 3}, find symmetric, reflexive, transitive and equivalence relation.

Solution
To show a relation R is an equivalence, we show that it is symmetric, reflexive and

transitive.
Let, R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
be a relation defined on given set A.
Also, R1 = {(1, 1), (2, 2), (3, 3)}, a reflexive relation.
R2 = {(1, 2), (2, 1), (1, 3), (3, 1)}, a symmetric relation.
R3 = {(1, 2), (2, 3), (1, 3)}, a transitive relation
Above R has the properties of reflexive, symmetric and transitive, hence an equivalence relation.

18. Let, R = {(1, 1), (2, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

Solution
Hints: show it is symmetric, reflexive and transitive, then is will be an equivalence relation.

19. a) If f(x + 5) = f(x) + f(5), x∈R, prove that: (i) f(o) = 0 (ii) f(–5) = –f(5)

Solution
i) f(o) = 0
Here, f(x + 5) = f(x) + f(5), x∈R
put x = 0 we get
f(0 + 5) = f(0) + f(5)
or, f(5) = f(0) + f(5)
 f(0) = 0 ... ... ... (i)
ii) f(–5) = –f(5)
again, put x = –5, we get
f(–5 + 5) = f(–5) + f(5)
or, f(0) = f(–5) + f(5)
by using (i)
f(–5) + f(5) = 0
 f(–5) = f(–5) proved.

b) If f(x + 3) = f(x) + f(9) prove that: (i) f(6) = 0 (ii) f(3) = –f(9)

Solution
i) f(6) = 0
Here, f(x + 3) = f(x) + f(9)
put x = 6 we get
f(6 + 3) = f(6) + f(9)
 f(6) = 0 ... ... ... (i)
ii) f(–5) = –f(5)
again, put x = 3, we get
f(3 + 3) = f(3) + f(9)

16 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, f(6) = f(3) + f(9)
 0 = f(3 + f(9)
by using (i)
f(3) = –f(9) proved.

20. a) If f(x) = ax + b, f(5) = 26 and f(2) = 14, find the values of a and b.

Solution
Here, f(x) = ax + b, f(5) = 26, f(2) = 14
put x = 5, f(5) = a.5 + b
or, 26 = 5a + b
 5a + b = 26 ... ... ... (i)
Again, for x = 2, f(2) = a.2 + b
or, f(2) = 2a + b
or, 14 = 2a + b
 2a + b = 14 ... ... ... (ii)
Subtracting equation (ii) from (i), we get
3a = 12
 a = 4
put the value of 'a' in (i), we get
5.4 + b = 26
or, b = 26 – 20
 b = 6
Hence, f(x) = 4x + 6.

b) If f(x) = mx + c, f(4) = 11 and f(5) = 13, find the values of m and c.

Solution
Here, f(x) = mx + c
we have f(4) = 11
i.e. m4 + c = 11
 4m + c = 11 ... ... ... (i)
and f(5) = 13
5m + c = 13 ... ... ... (ii)
Subtracting (ii) from (i), we get
–m = –2
 m = 2
put the value of m in equation (i)
4×2 + c = 11
or, c = 11 – 8
 c = 3
 f(x) = 2x + 3.

21. a) If f(x + 2) = 4x + 3, find f(x) and f(2).

Vedanta Optional Mathematics Teacher's Guide ~ 9 17

Solution
Here, f(x + 2) = 4x + 3
Replacing x by x – 2, we get
f(x – 2 + 2) = 4(x – 2) + 3
or, f(x) = 4x – 5
 f(x) = 4x – 5
When x = 2, f(2) = 4×2 – 5 = 3.

Alternative method
Here, f(x + 2) = 4x + 3
or, f(x + 2) = 4(x + 2) – 5
Replacing x + 2 by x, we get
f(x) = 4x – 5
and f(2) = 4.2 – 5
= 3.

b) If f(3x + 2) = 6x + 7, find f(x) and f(3).

Solution

Here, f(3x + 2) = 6x + 7

x is replaced by 3x, we get

f 3 . x + 2  = 6 . x + 7
3 3

or, f(x + 2) = 2x + 7

Again x is replaced by x – 2, we get

f(x) = 2(x – 22) + 7

 f(x) = 2x + 3

For x = 3, f(3) = 2.3 + 3

= 6 + 3 = 9

 f(x) = 2x + 3 and f(3) = 9

Alternative method

Here, f(3x + 2) = 6x + 7

or, f(3x + 2) = 2(3x + 2) + 3

Replacing (3x + 2) by x, we get

f(x) = 2x + 3

and f(3) = 2.3 + 3

= 6 + 3 = 9.

 f(x) = 2x + 3 and f(3) = 9.

22. If f(x) = 4x2 – 2x + 5 and g(x) = 7x2 – x + 3, and f(x) = g(x), find the values of x.

Solution
Here, f(x) = 4x2 – 2x + 5 and g(x) = 7x2 – x + 3
Now, we have f(x) = g(x)
4x2 – 2x + 5 = 7x2 – x + 3

18 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, 3x2 + x – 2 = 0

or, 3x2 + 3x – 2x – 2 = 0

or, 3x(x + 1) –2(x + 1) = 0

or, (x + 1) (3x – 2) = 0

Either x + 1 = 0 ⇒ x = –1

or, 3x – 2 = 0 ⇒ x = 2
3

23. If f(x) = 9x, then prove that: f(m + n + p) = f(m) . f(n) . f(p)

Solution

Here, f(x) = 9x

f(m + n + p) = 9m + n + p

= 9m . 9n . 9p

= f(m) . f(n) . f(p)

 f(m + n + p) = f(m) . f(n) . f(p) proved

24. For what value of domain has its image 2 under the function. f(x) = x2 + 6x + 7

Solution

Here, y = f(x) = x2 + 6x + 7
Image is 2 means y = 2
Now, y = x2 + 6x + 7
or, 2 = x2 + 6x + 7
or, x2 + 6x + 5 = 0
or, x2 + 5x + x + 5 = 0
or, x(x + 5) + 1(x + 5) = 0
 (x + 5) (x + 1) = 0
Either x + 5 = 0 ⇒ x = –5
x + 1 = 0 ⇒ x = –1
 x = –5, –1
Hence, –5 and –1 has the image 2.

23. A function g:R → R is defined by

x – 1 for 0 ≤ x < 4
g(x) = 2x + 1 for 4 ≤ x < 6

3x + 10 for 6 ≤ x < 8
Find the value of f(2), f(5) and f(7).

Solution ( 0 ≤ x < 4)
( 4 ≤ x < 6)
Here, For x = 2, we take
f(x) = x – 1,
or, f(2) = 2 – 1
 f(2) = 1
For x = 5, we take
f(x) = 2x + 1,
or, f(5) = 2.5 + 1

 f(5) = 11

For x = 7, we take

Vedanta Optional Mathematics Teacher's Guide ~ 9 19

f(x) = 3x + 10, ( 6 ≤ x < 8)
or, f(7) = 3.7 + 10
 f(7) = 31

26. If f(x) = 2x2 + 5, find

a) f(a + h) b) f(a + h) – f(a)
Solution h

Here, f(x) = 2x2 + 5

or, f(a + h) = 2(a + h)2 + 5

= 2(a2 + 2ah + h2) + 5

= 2a2 + 4ah + 2h2 + + 5

and f(a) = 2a2 + 5

Now, f(a + h) – f(a)
h
= 2a2 + 4ah + 2h2 + 5 – 2a2– 5
h
= 4ah + 2h2
h
= 2h(2ah+ 2h)
= 2(2a + h)

Questions for practice

1. If (3x, 4y­) = (27, 64) find the values of x and y.

2. If A = {x:x ≤ 4, x∈N} and B = {x:x2 – 9 = 0}, then find A×B and B×A.

3. Let R be a relation defined on A = {a, b, c}, then find the following relation on A.

a) Reflexive

b) Symmetric

c) Transitive

4. Let R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} then prove that R is

an equivalence relation.

5. If f(x) = 4x2 + 5x + 3, find the values of f(2), f(–2), f(3), f(–3).

6. If f(x) = 5x – 4, what is the pre-image of 16 ?

7. If f(x) = 3x2 + 5, then find:

a) f(a + h)

b) f(a + h) – f(a)
h
8. Let A = {1, 2} and B = {5, 10}, then

a) How many one to one function can be defined from A to B ?

b) How many one to many into function can be defined ?

c) How many many to one onto function can be defined ?

9. If f(x) = 5x, then prove that f(m + n + p) = f(m) . f(n) . f(p).

10. If f(x + 7) = 7x + 64, find f(x) and f(8).

20 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

two Polynomials

Estimated Teaching Hours : 4

1. Objectives

SN Level Objectives

i. Knowledge (K) To define a polynomial.
To define degree of polynomial, standard polynomial

To define different types of polynomials with examples.

To write degree of given polynomials.

ii. Understanding (U) To write given polynomials in standard form.

To add or subtract simple polynomials.

To state properties addition and multiplication of polynomials.

To use properties addition of polynomials.

iii. Application (A) To use properties of multiplication of polynomials.

To find additive inverse of given polynomials.

iv. Higher Ability (HA) To verify properties of addition of polynomials (with three
polynomials)

2. Teaching Materials
Chart papers with definition of polynomials and the types of polynomials.

Chart paper with properties of addition and multiplication of polynomials.

3. Teaching Learning Strategies:
Discuss difference between algebraic expressions and polynomials with examples (use

chart peper)
Discuss about standard form of polynomials.
Discuss standard polynomials, degree of polynomials, equal polynomials, with

appropriate examples.
Discuss numerical and literal coefficient of polynomials.
Taking two or three polynomials discuss how to add them.
State and prove the properties of addition of polynomials with examples.
Discuss about multiplication and properties of multiplication of polynomials.
Notes :
1. Algebraic expressions having non-negative integer as the power of variables in each

term is called a polynomial.
2. The polynomials in which the terms are arranged in ascending or descending powers of

variable are called standard polynomials.

Vedanta Optional Mathematics Teacher's Guide ~ 9 21

3. i) The highest power of the variables involved in the polynomial with one variable is
called the degree of the polynomial in one variable.

ii) The degree of polynomial in two or more variables is the highest sum of the powers of
the variables involved in the polynomial.

4. The product of two polynomial is a polynomial.
5. Two polynomials with same degree and the same coefficients of the corresponding

terms are called equal polynomials.
6. A number or alphabet which is the multiple of a variable in an algebraic / polynomial

term is known as coefficients. Coefficients are of two types: Numerical coefficient and
literal coefficient.

Some solved problems

1. Which of the following algebraic expressions are polynomials. Give your reason.
a) 4x2 + 3x + 6

Solution
Here, 4x2 + 3x + 6
It is a polynomial because exponent of variables of each term is a positive integer.

b) 37x4 + 3x2 + 3 x
Solution

Here, 7x4 + 3x2 + 3x
3

= 73x4 + 3x2 + x1/3

It is not a polynomial because the exponent or power of variable x in third term is 1 which
3
is not an positive integer.

c) 4x2 + x + 1
x
Solution

Here, 4x2 + x + 1
x

= 4x2 + x + x–1

Since the exponent of variable x in the third term is (–1), it is not a polynomial.

2. Find the numerical and literal coefficient of the following polynomials.
a) –0.9x

Solution
Here, –0.9x
Numerical coefficient = –0.9
Literal coefficient = x

22 Vedanta Optional Mathematics Teacher's Guide ~ 9

b) 5x + 9
13
Solution

Here, 5x + 9
13

= 153x + 9
13

Numerical coefficient of x in first term 153x = 5
13

Literal coefficient of x in the first term 153x = x

 9 is called constant term.
 13

c) x in 7xy

Solution
Here, numerical coefficient = 7
literal coefficient of x is = y

3. Find the degree of polynomials.
a) 6 x + y5

Solution
Here, 6 x + y5
Highest exponent = 5
⸫ Degree of polynomial = 5

b) 5xyz + 3y2 + x2

Solution
Here, 5xyz + 3y2 + x2
Sum of degree of the variables in the first term is 1 + 1 + 1 = 3,
which is the highest degree.
 degree of polynomial = 3

4. Which of the following polynomials are equal.

f(x) = 43x3 – 5x2 + 0.25x and g(x) = 0.75x3 + 1 – 5x2.
4
Solution

Here, f(x) = 34x3 – 5x2 + 1
4

= 0.75x3 – 5x2 + 0.25
and g(x) = 0.75x3 – 5x2 + 0.25
Here, f(x) and g(x) are equal polynomials because the coefficients of their corresponding
terms are equal.

Vedanta Optional Mathematics Teacher's Guide ~ 9 23

5. Write the given polynomials in standard form

a) ascending order
b) descending order
3x2 + 4x6 + 9x3 + 3x + 7

Solution
a) writing the given polynomials in ascending order
7 + 3x + 3x2 + 9x3 + 4x6
b) writing the given polynomials in descending order
4x6 + 9x3 + 3x2 + 3x + 7

6. Find the sum of p(x) and q(x) if p(x) = 3x3 – 8x2 + 9x + 8, q(x) = 4x3 + 4x2 – 5x – 8

Solution

Here, p(x) = 3x3 – 8x2 + 9x + 8

q(x) = 4x3 + 4x2 – 5x – 8

Sum of p(x) and q(x) is given by

p(x) + q(x) = (3x3 – 8x2 + 9x + 8) + (4x3 + 4x2 – 5x – 8)

= (3 + 4)x3 + (–8 + 4)x2 + (9 – 5)x + (8 – 8)

= 7x3 – 4x2 + 4x.

7. Subtract 4x2 + 5x + 7 from 3x2 – 2x + 2

Solution
Here, to subtract 4x2 + 5x + 7 from 3x2 – 2x + 2
we have, (3x2 – 2x + 2) – (4x2 + 5x + 7)
= (3 – 4)x2 + (–2 – 5)x + (2 – 7)
= –x2 – 7x – 5.

8. If p(x) = x2 + x, q(x) = x2 – 4x, then verify that: p(x) . q(x) = q(x) . p(x)

Solution

Here, p(x) = x2 + x, q(x) = x2 – 4x

LHS = p(x) . q(x)

= (x2 + x) (x2 – 4x)

= x2(x2 – 4x) + x(x2 – 4x)

= x4 – 4x3 + x3 – 4x2

= x4 – 3x3 – 4x2

Again, RHS = q(x) . p(x)

= (x2 – 4x) (x2 + x)

= x2(x2 + x) – 4x(x2 + x)

= x4 + x3 – 4x3 – 4x2

= x4 – 3x3 – 4x2

 LHS = RHS proved.

24 Vedanta Optional Mathematics Teacher's Guide ~ 9

9. What should be subtracted from sum of y3 + 2y + 1 and y2 + 6y + 2 to get 6y + 8 ?

Solution
Here, sum of y3 + 2y + 1 and y2 + 6y + 2 is given by
(y3 + 2y + 1) + (y2 + 6y + 2)
= y3 + y2 + 8y + 3
Let p be subtracted from y3 + y2 + 8y + 3
Then, y3 + y2 + 8y + 3 – p = 6y + 8
or, y3 + y2 + 8y + 3 – 6y – 8 = p
 p = y3 + y2 + 2y – 5

10. Let p(x) = x – 4, q(x) = x + 4 and r(x) = x2 + 16, then verify that:
{p(x) . q(x)} r(x) = p(x){q(x) . r(x)}

Solution
Here, p(x) = x – 4, q(x) = x + 4, r(x) = x2 + 16
Now, p(x) . q(x) = (x – 4) (x + 4) = x2 – 16
LHS = {p(x) . q(x)} r(x)
= (x2 – 16) (x2 + 16)
= x4 – 256
Again, q(x) . r(x)
= (x + 4) (x2 + 16)
= x(x2 + 16) + 4(x2 + 16)
= x3 + 16x + 4x2 + 64
= x3 + 4x2 + 16x + 64
RHS = p(x){q(x) . r(x)}
= (x – 4) (x3 + 4x2 + 16x + 64)
= x4 + 4x3 + 16x2 +64x – 4x3 – 16x2 – 64x – 256
= x4 – 256
 {p(x) . q(x)} r(x) = p(x){q(x) . r(x)}. Proved

11. Find the additive inverse of q(x) = 4x4 – 6x2 + 7x2 + 22x + 2

Solution

Here, 4x4 – 6x2 + 7x2 + 22x + 2

By definition additive inverse of q(x) is –q(x)

–q(x) = (4x4 – 6x2 + 7x2 + 22x + 2)

= –4x4 + 6x2 – 7x2 – 22x – 2

Alternative method

Let p(x) be the additive inverse of q(x).

Then by definition,

q(x) + p(x) = 0

i.e. p(x) = 0 – q(x)

= –4x4 + 6x2 – 7x2 – 22x – 2.

Vedanta Optional Mathematics Teacher's Guide ~ 9 25

Questions for practice

1. Find the numerical coefficient of 15 6 x.
2. Find the literal coefficient of y in x7y.
3. Write the following polynomials in descending order of their degree.
a) x2 + 7x5 – 8x3 + 8x – 9
b) 5 x + 14x2 + 8x3 + 9x5
4. Add p(x) and q(x) where
p(x) = 4x3 – 8x2 + 7x and q(x) = 6x3 + 4x2 + 8x + 9
5. Subtract q(x) from p(x) where,
p(x) = 4x4 + 6x2 + 8x + 9 and q(x) = 6x4 – 8x2 + 5x3 + 6x – 27
6. If p(x) = 2x2 + 4x – 6, g(x) = x2 + x + 2, h(x) = x2 + 4x + 5 then verify that
{p(x) + g(x)} + h(x) = p(x) + {g(x) + h(x)}
7. If p(x) = x + 5, q(x) = x – 5 and r(x) = x2 + 25 then verify that
{p(x) . g(x)} r(x) = p(x) {q(x) . r(x)}
8. If p(x) = x2 – 3x + 1, q(x) = 3x2 + 5x + 3, r(x) = –4x2 – 2x – 4, show that
p(x) + q(x) + r(x) is a zero polynomial.

26 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

three Sequence and Series

Estimated Teaching Hours : 5

1. Objectives

SN Level Objectives

i Knowledge (K) To define sequence and series.
To define general term of a sequence.

To identify linear and quadratic sequence.

ii Understanding (U) To find stated terms of linear and quadratic
sequence when its general term is given.

To use sigma notation for given series.

iii Application (A) To find the nth term of linear and quadratic sequence.
To expand and find the sum of series given in sigma notation.

iv Higher Ability (HA) To find general terms of given sequence in diagrams with dots.

2. Teaching Materials
Chart papers with definition of sequence and series.
Chart paper with diagram with dots, patterns.
3. Teaching Learning Strategies:
Stating examples define sequence, series and general term of sequence and series.
Discuss about sigma notation.
Define a linear and a quadratic sequence with examples.
With an example, illustrate how to find the general term of a linear sequence.

(tn = an + b). (eg. 1, 5, 9, 13, ... ... ...)
With an example, illustrate how to find the general term of a quadratic sequence.

(tn = an2 + bn + c). (example: 6, 9, 14, 21, 30)
From given pattern of figures with dots, discuss to find the general term, examples:

Notes:
1. A sequence of numbers in which each consecutive pair of terms have a common

difference is called a linear sequence. General term of a linear s given by tn = an + b,
a = the first difference.
2. A sequence of numbers in which the first difference is not constant but the second

Vedanta Optional Mathematics Teacher's Guide ~ 9 27

difference is constant is called a quadratic sequence. General term of a quadratic
sequence is given by tn = an2 + bn + c, 2a = second difference.
3. A series is formed by adding or subtracting the successive terms.
4. tn = Sn – Sn – 1

Some solved problems

1. Find the first difference of the following sequences
a) 1, 4, 7, 10

Solution 1 4 7 10
Here, 1, 4, 7, 10

First difference 4–1=3 7–4=3 10–7=3

It means each term is increased by 3.

b) 18, 14, 10, 6

Solution 6
Here, 18, 14, 10, 6

18 14 10

14–18=–4 10–14=–4 6–10=–4
It means each term is decreased by 4.

2. Find the first five terms from the given general term: tn = (–1)n +1
n+
Solution 1

Here, tn = (–1)n +1
n+
1

t1 = (–1)1 +1 = 1
1+ 2
1

t2 = (–1)2 +1 = –13
2+
1

t3 = (–1)3 +1 = 1
3+ 4
1

t4 = (–1)4 +1 = –15
4+
1

t5 = (–1)5 +1 = 1
5+ 6
1

 The required five terms are 12, –31, 41, –51, 16.

3. a) If an = aann – 1 and a1 = 1, a2 = 3, (n > 2) find the values of a3 and a4.
– 2

28 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution aann

Here, an = – 1 and a1 = 1, a2 = 3
– 2
aa33 = aa12
For n = 3, a3 = – 1 = 3 =3
– 2 1 =1

For n = 4, a4 = a4 – 1 = a3 = 3
 a3 = 3, a4 = 1. a4 – 2 a2 3

b) an + 1 = 2an and a1 = 3, find the values of a2 and a3.

Solution
Here, an + 1 = 2an, a1 = 3
put n = 1, a2 = 2.a1 = 2.3 = 6
put n = 2, a3 = 2.a2 = 2.6 = 12
 a2 = 6 and a3 = 12.

4. Find the general term (tn) of the following sequences.
a) 1, 4, 7, 10, ... ... ...

Solution 14 7 10
Here, 1, 4, 7, 10, ... ... ...

First difference 33 3

Since the first difference is constant, it is 'a' linear sequence.

 tn = an + b, where a = the first difference
⇒ t1 = a.1 + b
⇒ a = 3
 a = 3

For n = 1, t1 = a.1 + b
or, 1 = 3 + b

 b = –2

 tn = 3n – 2

b) 3, 9, 27, 81

Solution
Here, 3, 9, 27, 81

9 = 3, 27 = 3, 81 = 3
3 9 27
Ratio of any two consecutive term is constant.

t1 = 3 = 31
t2 = 9 = 32
t3 = 27 = 33
... ... ... ... ...

tn = 3n

Vedanta Optional Mathematics Teacher's Guide ~ 9 29

OR
Here, 3, 9, 27, 81

a = the first term = 3

ratio is constant, r = tt13 – tt32 = tt34

i.e. r = 9 = 27 = 81 =3
3 9 27

Its general term is given by

tn = arn – 1
= 3.3n – 1
= 31 + n – 1
= 3n

5. Find the general term of the following sequences.

a) 3, 6, 10, 15

Solution 3 6 10 15
Here, 3, 6, 10, 15

First difference 3 4 5

Second difference bu1t the seco1nd difference is constant.
First difference is not constant

So given sequence is a quadratic sequence.

Let tn = an2 + bn + c
where, 2a = second difference

⇒ 2a = 1

 a = 1
2

For n = 1,

t1 = 1 . 12 + b . 1 + c
2

or, 3 = 1 + b + c
2

 b + c = 5 ... ... ... (i)
2

2b + 2c = 5

For n = 2,

t2 = 1 . 22 + b . 2 + c
2

30 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, 6 = 1 + 2 + 2b + c
2

 2b + c = 4 ... ... ... (ii)

Solving equations (i) and (ii), we get

c = 1, b = 3
2

Now, tn = an2 + bn + c
= 12n2 + 23n + 1

= n2 + 3n + 2
2

= n2 + 2n + n + 2
2

= (n + 1) (n + 2) 11 18 27 38
2

b) 11, 18, 27, 38, ... ... ...

Solution 79 11
Here, 11, 18, 27, 38, ... ... ...

First difference

Second difference 22

First difference is not constant but the second difference is constant.

So given sequence is a quadratic sequence.

Let, tn = an2 + bn + c
where, 2a = second difference = 2

⇒ a = 1

For n = 1,

t1 = 1 . 12 + b . 1 + c or, 11 =
 b + c = 10 ... ... ... (i)

For n = 2,

t2 = 1 . 22 + b . 2 + c
or, 18 = 4 + 2b + c

 2b + c = 14 ... ... ... (ii)

using (i)

⇒ b + b + c = 14

or, 10 + b = 14

 b = 4

put the value of b in (i), we get

c=6

 tn = n2 + 4n + 6.

c) 3, 10, 25, 48

Vedanta Optional Mathematics Teacher's Guide ~ 9 31

Solution 3 10 25 48
Here, 3, 10, 25, 48

First difference 10–3=7 25–10=15 48–25=23

Second difference

Since the first difference is not constant but the second differen8ce is cons8tant,
the given sequence is quadratic.
Let tn = an2 + bn + c
where, 2a = second difference
⇒ 2a = 8 ⇒ a = 4
For n = 1,
t1 = 4 . 12 + b . 1 + c
or, 3 = 4 + b + c
 b + c = –1 ... ... ... (i)
For n = 2,
t2 = 4 . 22 + b . 2 + c
or, 10 = 16 + 2b + c
or, 2b + c = –6
 b + b + c = –6
bu using (i), we get
b – 1 = –6
 b = –5
put the value of b in (i), we get
c=4
 tn = 4n2 – 5n + 4.

d) 31, 54, 77, 10
Solution 9

Here, 13, 45, 77, 10
9

Hints: sequence of numbers in numerator 1, 4, 7, 10, ... ... ...

sequence of numbers in denominator 3, 5, 7, 9, ... ... ...

Both are linear sequences use tn = an + b.

6. If Sn = n(n + 1) (n + 2) S4, S5 and a5
6
Solution

Here, Sn = n(n + 1) (n + 2)
6

32 Vedanta Optional Mathematics Teacher's Guide ~ 9

Now, Sn = 4(4 + 1) (4 + 2)
6

= 4 . 5 . 6
6
= 20

S5 = 5 . 6 . 7
6
= 35

a5 = S5 – S4  tn = Sn – Sn – 1
= 35 – 20

= 15

7. Find the number of terms in the following series and find the stated term.

25 n(n + 1) (2n + 1), 12th term
6
∑

n =12

Solution

Here, 25 n(n + 1) (2n + 1)
6
∑

n =12

Number of terms = 25 – 11 = 14

For 12th term, we take n = 23

23(23 + 1) (2×23 + 1)
6

= 23 . 24 . 47 = 4324
6

8. Study the patterns given below and answer the following questions:

i) Add two patterns in each of the sequence.
ii) Find the general term of each of the sequence.
iii) Find the 10th term of the sequence.

Solution
i)

Vedanta Optional Mathematics Teacher's Guide ~ 9 33

ii) From the above pattern of figure, we get

Sequence of numbers as 1, 6, 111, 16, 21 6 11 16 21
Now,

First difference

5555
Since the first difference is constant, the given sequence is linear
Let, tn = an + b
where a = the first difference
a = 5
For n = 1, t1 = 5.1 + b
or, 1 = 5 + b
 b = 1 – 5 = –4
 tn = 5n – 4

Questions for practice

1. Find the formula for nth term of given sequence or series.
a) 14, 49, 196, 2165, 3256, ... ... ...
b) 2.4 + 3.5 + 4.6 + 5.7 + ... ... ...
c) –1, 2, –3, 4, – 5
2. Evaluate:

5

a) ∑ (3n + 2)
n =2

4

b) ∑ (–1)n (2n + 3)
n =1

6

c) ∑ (–1)k . (4k + 1)
k=1

3. Find S6, S7 and a7 from given formula of sum of n terms of a sequence

Sn = n(n + 1) (n + 2)
3
4. Find the nth term of the sequence 2, 6, 12, 20, 30, .........

5. Study the following pattern of figures. Then

i) Add two more patterns in the sequence.

ii) Find the formula of general term.

iii) Find the 8th term of the sequence.

34 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

four Limit

1. Objectives Estimated Teaching Hours : 10
SN Level Objectives

To define limit of sequence of numbers.

i Knowledge (K) To define finite and infinite series.
To define limit of a function.
lim
To write meaning of x→a f(x).

ii Understanding (U) To write limit of sequence of numbers observing pattern of numbers.
To use notation of limits.

To calculate limit from given pattern.

iii Application (A) To complete sequence of numbers in a table.

To evaluate limit of a function.

iv Higher Ability (HA) To write the sequence of numbers by using given sequences of
diagrams and to calculate their limits.

2.Teaching Materials
→ Tables in chart paper to find limits.

→ Sequence of diagrams in chart papers to find their limits.

3. Teaching Learning Strategies:

Draw a number line of a finite length. Bisect it infinitely many times and give concept of limit.

Example: (text book page 67)

A 16cm B

16, 8, 4, 2, 1, 12, 41, 18, 1
16

Explain how the limit of above sequences of numbers is 0.

Use a table to explain limit of sequence of numbers (page 69, Q.N. 3, text book)

Give some illustrated examples, of limits solving questions from text book (Q.N. 4, page 69).

Give concept of limit from given sequence of diagrams (page 71, Example 1).

Discuss to give concept of limit from sum of infinite series (page 77, Q.N. 2).

Give concept of lim f(x), lim f(x) and lim f(x) (page 82, Q.N. 5)
x→a+ x→a– x→a

Discuss how to find limit of a function at a given point with examples. (page 83, II table

for example)

Vedanta Optional Mathematics Teacher's Guide ~ 9 35

Notes:
1. The limit of sequence of numbers is the finite value that the terms of the sequence

tends to that value.
2. Sum of a series of n terms is the sum of the first n terms. It is also called nth partial sum.
3. If the sum of the series tends to a real number, then it is called limit of sum of the series.
4. A series can have a sum only if the individual terms tend to zero. This type of series is

called convergent series.
5. If the ratio between two successive terms is less than 1, then the limit of sum of given

series is possible. It can be calculated by using formula

S = 1 a r
∞ –

Where a = the first term of the series

r = common ratio

Some solved problems

1. Write limit of given sequence of numbers.

SN Sequence of numbers Limit of sequence

i) 4.9, 4.99, 4.999, ... ... ...

ii) 5.1, 5.01, 5.001, 5.0001, ... ... ...

iii) 1, 4, 16, 64, ... ... ...
iv) 19, 811, 7219, 65161, ... ... ...

v) 4, 2, 0, –2, –4, –8, ... ... ...

Solution Sequence of numbers Limit of sequence
SN

i) 4.9, 4.99, 4.999, ... ... ... 5

ii) 5.1, 5.01, 5.001, 5.0001, ... ... ... 5

iii) 1, 4, 16, 64, ... ... ... ∞
iv) 91, 811, 7129, 65161, ... ... ... 0
v) 4, 2, 0, –2, –4, –8, ... ... ... –∞

2. Round off 70.3682 to the tenth, hundredth and whole number.

a) What is the limiting value when round is made to the tenth digit ?
b) What is the limiting value when round off is made to the hundredth digit ?

36 Vedanta Optional Mathematics Teacher's Guide ~ 9

c) What is the limiting value when the round off is the whole number ?

Solution
a) 70.4
b) 70.37
c) 70

3. Write down the sequence of terms given whose general term is by

a) tn = 1
4n
Solution

Here, tn = 1
4n

Sequence terms are:

41, 116, 614, 2156, ... ... ...

As n tends to ∞, 1 tend to 0.
4n

Hence the limit of sequence of numbers is zero.

b) tn = 1
10n – 1
Solution

Here, tn = 1 – 1
10n

the sequence of numbers is given by

1 , 1 , 1 , 1 , 1 , ... ... ...
101 – 1 102 – 1 103 – 1 104 – 1 105 – 1

i.e. 1, 110, 1010, 10100, 101000, ... ... ...

 1, 0.1, 0.01, 0.001, 0.0001, ... ... ...

As n tends to ∞, then 1 – 1 tends to 0.
10n

Hence the limit of the sequence of terms is zero.

4. Draw a line segment MN of length 10cm. Bisect it 8 times and show the bisection obtained
in the number line. Write the limit of the sequence. Thus obtained.

Solution
Draw a line segment AB of 10cm.
Divide it in two equal parts. Let C divide it two equal parts. AC = CB = 5cm
Again, divide CB in two equal parts CD = DB = 2.5cm

Vedanta Optional Mathematics Teacher's Guide ~ 9 37

A

1 2 3 4 C5 6 7 D8 9 B10
E

The process of bisection is continued, we get sequence of lengths as

10, 5, 2.5, 1.25, 0.625, 0.3125, 0.15625, ... ... ...

If the process of bisection is continued infinitely, the final segment will be length of

nearly 0cm.

Hence the sequence of lengths is zero. P
5. (text book, page 71, Q.N. 1)

Solution

Here, ∆PQR is an equilateral triangle.

Again triangle XYZ is formed joining the mid points of x uz
the sides of the triangle PQR.

∆XYZ is also equilateral triangle.

Similarly ∆UVW is formed by joining the mid points of vw
sides of XYZ.

We know that perimeter of equilateral triangle p = 3l Q y R
(where l = length of side of a triangle)

As we continue the process, the equilateral triangles will be smaller and smaller and

consequently the perimeters also will be smaller and smaller.

Illustration Let ∆PQR be an equilateral triangle with length of a side 10cm, then perimeter

p1 = 3 × 10cm = 30cm
Again, for ∆XYZ, length of side = 5cm

perimeter (p2) = 15cm
Similarly, perimeter of third equilateral triangles

perimeter (p3) = 3 × 2.5 = 7.5cm
Sequence of perimeters is

30cm, 15cm, 7.5cm

If the process is continued infinitely many times, the perimeter of triangle will be zero.

Hence, the sequence of perimeter of triangle is 0.

6. (text book, page 72, Q.N. 3)

Solution S RN
a) There are 18 small right angled triangles of P QM

equal area.
b) If two more vertical lines next to QR are drawn

and the diagonals are produced forming all
right angled triangles of equal area. It means in
rectangle PMNS as shown in the figure 30 right
angled triangles are formed.

38 Vedanta Optional Mathematics Teacher's Guide ~ 9

c) If the next three lines are drawn next to SR, and diagonals are drawn as in care (a) threre
will be 36 small right angled triangles of same area.

UV UV

SR S RN

PQ P QM

d) If three horizontal and two vertical lines are added simultaneously in given figure and
diagonals are drawn as in (a), the number of small right angled triangled of equal area is 48.

e) If the process is continued as for as possible, there will infinite, number of small right
angled triangles.

7. Which of the following series have a limit as a fixed real number ? Give your reason.

a) 1 + 1 + 1 + 1 + ... ... ...
3 9 27
Solution

Here, 1 + 1 + 1 + 1 + ... ... ...
3 9 27

common ratio (r) = tt12 1

=3 = 1
1 3

|r| = 1
3

Hence absolute value of common ratio is 1 which is less than unity. So the given infinite
3
geometric series has a fixed sum

sum (S∞) = 1 a r =1 = 3 = 1.5
– 1– 1 2

3

When the number of terms increases continuously, the last term approaches zero.
 Limit of the sequence is zero.

Vedanta Optional Mathematics Teacher's Guide ~ 9 39

b) 32 – 16 + 8 – 4 + 2 ... ... ...

Solution
Here, 32 – 16 + 8 – 4 + 2 ... ... ...
absolute value of common ratio is given by

|r| = –16 =1
32
2

The value of each term decreases and final term is nearly 0 and absolute value of common

ratio is 1 which is less than 1.
2
Hence limit of the sequence is zero .

c) 1.6 + 8 + 40 + 200 + ... ... ...

Solution

Here, 1.6 + 8 + 40 + 200 + ... ... ...

common ratio (r) = 8 = 5
1.6

Here absolute value of common ratio is greater than unity and each term is increasing.
Hence the limit of sequence does not exist.

d) 0.78

Solution
Here, 0.78

0.78 = 0.78 + 0.0078 + 0.000078 + ... ... ...

common ratio (r) = 0.0078 = 1 < 1
0.78 100

Common ratio is less than unity and as value of each term decreases and the value of the last

term approaches zero.

Hence the limit of sequence is zero.

8. Find the limiting values of the sum of the following series if exists.
a) 8 – 4 + 2 – 1 + ... ... ...

Solution

Here, 8 – 4 + 2 – 1 + ... ... ...

S1 = 8
S2 = 8 – 4 = 4
S3 = 8 – 4 + 2 = 6

S4 = 8 – 4 + 2 – 1 = 5
1
S5 = 8 – 4 + 2 – 1 + 2 = 5.5

S6 = 8 – 4 + 2 – 1 + 1 – 1 = 5.25
2 4

40 Vedanta Optional Mathematics Teacher's Guide ~ 9

S7 = 8 – 4 + 2 – 1 + 1 – 1 + 1 = 5.375
2 4 8

S8 = 8 – 4 + 2 – 1 + 1 – 1 + 1 – 1 = 5.3125
2 4 8 16

As the number of terms increases, the sum approaches a fixed number 5.3.

Hence the limiting value of given series 5.3.

 S = 5.3
∞

Alternative Method

Here, S = 8 – 4 + 2 – 1 + 1 – 1 + 1 – 1 + ... ... ...
∞ 2 4 8 16

Given series is an infinite geometric series.

common ratio (r) = –48 = –21 < 1
Hence the sum of given series exists.

Now a = the first term (a) = 8, r = –12

S = 1 a r
∞ –

= 8 = 16 = 531 = 5.3
1+ 1 3

b) 0.45 2

Solution
Here, 0.45

0.45 = 0.45 + 0.0045 + 0.000045 + ... ... ...

common ratio (r) = 0.0045 = 1 < 1
0.45 100

Given series is an infinite geometric series

Hence it has a finite sum.

S = 1 a r
∞ –

= 0.45 = 45 = 5 = 0.45
1– 1 99 11

100

OR
S1 = 0.45
S2 = 0.45 + 0.0045 = 0.4545
S3 = 0.45 + 0.45 + 0.45 = 0.454545
It shows that as the number of terms increases, the sum of above series tends to a fixed real
number 0.45
 S = 0.45

∞

Vedanta Optional Mathematics Teacher's Guide ~ 9 41

c) 1 + 5 + 5 + 125 + ... ... ...

Solution
Here, 1 + 5 + 5 + 125 + ... ... ...

Let S = 1 + 5 + 5 + 125 + ... ... ...
∞

S1 = 1
S2 = 1 + 5 = 6
S3 = 1 + 5 + 25 = 31

common ratio (r) = 5 = 5
1

As the number of terms increases the sum of the series also increases.

This is an infinite geometric series with common ratio 5 > 1.

Thus sum of the given series does not exist.

9. Find the first five terms of the following from the given general term. Also find the limiting
value of sum of the sequence (page 77, Q.N.3).

a) tn =  1 n – 1
2
Solution

Here, tn =  1 n – 1
2 

t1 =  1 1 – 1 =  1 0 =1
 2   2 

t2 =  1 2 – 1 =  1 1 = 1
2  2  2

t3 =  1 3 – 1 =  1 2 = 1
2  2  4

t4 =  1 4 – 1 =  1 3 = 1
 2   2  8

t5 =  1 5 – 1 =  1 4 = 1
2  2  16

The sequence of number is 1, 21, 41, 81, 116, ... ... ...
1

common ratio (r) = = 2 = 1
1 2

Since common ratio is less than unity. The sum of the sequence exists.

The limiting value of sum of above sequence is given by

S = 1 a r, a = 1, r = 1
∞ – 2

= 1
1–1
2

= 2

42 Vedanta Optional Mathematics Teacher's Guide ~ 9

b) tn = 1 +  –n1 n
n 

Solution

Here, tn = 1 +  –n1 n
n 

t1 = 1 +  –11 1 =0
1  

t2 = 1 +  –21 2 = 1 + 1 = 3
2   2 4 4

t3 = 1 + –31 3 = 1 – 1 = 8
3  3 27 27

t4 = 1 +  –14 4 = 1 + 1 = 65
4   4 256 256

t5 = 1 +  –15 5 = 1 – 1 = 624
5   5 3125 3125

Here sequence of numbers 0, 43, 287, 26556, 3612245, ... ... ...

Here, common ratio is not constant as the number of terms increases, the sum of terms
increases slightly

S1 = 0

S2 = 0 + 3 = 0.75
4

S3 = 0.75 + 8 = 1.046
27

S4 = 1.046 + 65 = 1.2999
256

S5 = 1.2999 + t5 = 1.2999 + 624 = 1.4996
3125

S6 = 1.4996 + t6 = 1.4996 + 1 +  – 1 6 = 1.6663
6  6 

S7 = 1.663 + 1 +  – 1 7 = 1.8092
7 7 

S8 = 1.8092 + 1 +  –18 8 = 1.9342
8  

S9 = 1.9342 + 1 +  –91 9 = 2.0453
9  

S10 = 2.0453 + 1 +  –110 10 = 2.0454
10 

It shows that sum approaches to a fixed number 2.

Hence the limiting value of the sum of the series is 2.

c) tn= 4–31 n


Vedanta Optional Mathematics Teacher's Guide ~ 9 43

Solution

Here, tn= 4 – 1 n
3 

t1 = 4 –13 1 = –43


t2 = 4 –13 2 = 4 . 1 = 4
 9 9

t3 = 4 –31 3 = –247


t4 = 4 –13 4 = 4
 81

t5 = 4 –13 5 = –2443


 Sequence of terms is –34, 49, –247, 841, –2443
t2
common ratio (r) = t1

4

= 9 = 4 × –3 = –13
9 4
– 4
3

Since common ratio is constant and less then unity.

Sum of sequence is possible limiting value can be calculated as follows

S1 = –34

S2 = –34 + 4 = –89
9

S3 = –98 + –247 = –2278

S4 = – 28 + 4 = – 84
27 81 81

S5 = – 84 + – 4  = – 244
81 243  243

Each of above sum is approximately equal to –1.
Hence the limiting value of the sum of the sequence is –1.
Alternative Method
Sequence of numbers is given by
t1, t2, t3, t4, t5, ... ... ...
(from above)

i.e. – 43, 94, – 94, 841, – 4
243

common ratio (r) = tt12

44 Vedanta Optional Mathematics Teacher's Guide ~ 9

4

=9 4 = 4 × –3 = – 1
3 9 4 3
–

Since r = –31 < 1, the sum of the sequence of numbers exists.
a 1
Now, S = 1 – r, where a = – 3
∞

= – 4 = – 4 × 3 = –1
3 3 4

1 + 1
3

Hence the limiting value of the sum of the sequence is –1.

10.

Solution

(i) (ii) (iii) (iv)
Here, Area of shaded parts (Sn)

Fig No.

i) S1 = 1 cm2
2

ii) S2 = 1 + 1 = 3 cm2
2 4 4

iii) S3 = 3 + 1 = 7 cm2
4 8 8

iv) S4 = 7 + 1 = 15 cm2
8 16 16

v) S5 = 15 + 1 = 31 cm2
16 32 32

vi) S6 = 31 + 1 = 63 cm2
32 64 64

Vedanta Optional Mathematics Teacher's Guide ~ 9 45

As the shaded part is increased, then area of the portion increases and approaches to 1cm2.

Sequence of area are 1. 3, 7, 15, 31 cm2, ... ... ...
2 4 8 16 32

Hence the limiting value of the sequence tends to 1cm2.

11. Let f(x) = 2x – 1, find f(1.9), f(2.1), f(2.01), find the nearest number nearer to these values,
what conclusion can you draw from these results ?

Solution
Here, f(x) = 2x – 1
f(1.9) = 2 × 1.9 – 1 = 2.8
f(2.1) = 2 × 2.1 – 1 = 3.1
f(2.01) = 2 × 2.01 – 1 = 3.01
From above values we observed that they are nearer to a constant 3 as that value of x nearer to 2.
This constant value 3 is called limit of the function as the value of x nearer to 2.
In notation of limit we write

lim f(x) = lim 2x – 1
x→2 x→2

=2×2–1

= 3

12. (a) When does f(x) = x2 – 16 give certain value? Does f(4) give a finite value.
x – 4

Solution

a) we have f(x) = x2 – 16
x – 4

For x = 4, f(4) = 42 – 16 = 0
4 – 4 0

Which is an indeterminate form and it does not have any meaning.

i.e. f(4) does not give certain value. So f(4) does not exist.

b) Find the value of f(3.9), f(3.99), f(3.999), f(3.9999).

f(3.9) = 3.92 – 16 = 7.9
3.9 – 4

f(3.99) = (3.99)2 – 16 = 7.99
3.99 – 4

f(3.999) = (3.999)2 – 16 = 7.999
3.999 – 4

f(3.9999) = (3.9999)2 – 16 = 7.999
3.9999 – 4

It means that as x approaches to 4, the value of f(x) approaches to a finite number 8.

c) Find the values of f(4.1), f(4.01), f(4.02), f(4.002).

46 Vedanta Optional Mathematics Teacher's Guide ~ 9

f(4.1) = 4.12 – 16 = 8.1
4.1 – 4

f(4.01) = (4.01)2 – 16 = 8.01
4.01 – 4

f(4.02) = (4.02)2 – 16 = 8.02
4.02 – 4

f(4.002) = (4.002)2 – 16 = 8.002
4.002 – 4

d) Round off these values of f(x) to their nearest whole number.

From the values in (b) and (c), we observe that the value of f(x) is nearly equal 8 as the

value of x is nearly equal to 4.

In notation of limit we write

xl→im4 f(x) = lim x2 – 16 =8
x→4 x – 4

e) Find the limiting value of f(x) from the sequence in (b) and (c)

The limiting value of the function f(x) at x = 4 is 8.

xl→im4 f(x) = lim x2 – 16 =8
x→4 x – 4

Note: In f(x) = x2 – 16
x – 4

For the value of x other then 4, we can write

f(x) = (x + 4) (x – 4) =x+4
x–4

Then f(3.9) = 3.9 + 4 = 7.9

f(4.01) = 4.01 + 4 = 8.01 etc.

13. Let f(x) = x2 – 49, x ≠ 7, evaluate f(x) for x = 7.1, 7.01, and x = 6.9, 6.99. What conclusion
x–7

can you draw from these values of x ?

Solution

Here, f(x) = x2 – 49, x≠7
x–7

For x ≠ 7 we can write

f(x) = (x + 7) (x – 7) =x+7
x–7

Now, f(7.1) = 7.1 + 7 = 14.1

f(7.01) = 7.01 + 7 = 14.01

Also f(6.9) = 6.9 + 7 = 13.9

f(6.99) = 6.99 + 7 = 13.99

Conclusion:

As the value of x approaches to 7, the value of function f(x) approaches to 14.

Vedanta Optional Mathematics Teacher's Guide ~ 9 47

We say that the limit of the function f(x) is 14 at x = 7.
In notation of limit we write,

xl→im7 f(x) = lim x2 – 49 = 14
x→7 x – 7

14. Evaluate the following:

a) lim x2 – 4
x→2 x – 2

Solution

Here, lim x2 – 4 (form)
x→2 x – 2

(Factorize the numerator)

= xl→im2 (x + 2) (x – 2)
x–2

= xl→im2 (x + 2)

= 2 + 2 = 4

b) lim x2 – 25
x→5 x – 5
Solution

Here, lim x2 – 255 (÷ form)
x→5 x –

(Factorize the numerator)

= xl→im5 (x + 5) (x – 5)
x–5

= xl→im5 (x + 5)

= 5 + 5 = 10

15. Fill in the following tables and write the limiting value of the function in symbolic form

f(x) = x2 – 4
x – 2
Solution

Here, f(x) = x2 – 4
x – 2

For x = 2, the function is not defined.

So we can write it as f(x) = x2 – 24, x≠2
x–

= (x + 2) (x – 2) =x+2
x–2

48 Vedanta Optional Mathematics Teacher's Guide ~ 9

x f(x) = x2 – 4 = (x + 2) for x ≠ 2
x – 2

2.5 2.5 + 2 = 4.5

2.9 2.9 + 2 = 4.9

2.99 2.99 + 2 = 4.99
3.1 3.1 + 2 = 5.1
3.01
3.001 3.01 + 2 = 5.02
3.0001 3.001 + 2 = 5.001
x→3
In symbolic notation, we write f(x) → 5

lim f(x) = lim x2 – 4 =5
x→3 x→3 x– 2

The limit of the function f(x) is 5 at x = 3.

Questions for practice

1. Find the fifth term of the sequence 4.1, 4.01, 4.001. P 24cm D S
What will be the limit of the sequence of numbers. A B C
2. Each side of a square is 24cm. A second square is R

inscribed by joining the mid points of the sides
successively. The process is continued upto huge
number of steps.
i) Find the length of 5th and 6th squares.
ii) Find the limit of the squences of sides.
3. The general term of a numerical sequence is defined

by tn = 4n – 1, find the first 5 terms of the sequence.
n
Q
Find the limit value of the sequence.

4. a) Evaluate: lim x2 – 215
x→15 x – 15

b) Evaluate: lim x2 – 6x – 16
x→8 x–8

5. Let f(x) = x2 – 684, does f(8) exists?
x –

Evaluate: lim x2 – 64
x→∞ x – 8

6. Fill in the table and answer the given questions:

Vedanta Optional Mathematics Teacher's Guide ~ 9 49

x 6.1 6.01 6.001 6.00 5.9 5.99 5.999
f(x) = 2x + 5

a) Find the limit of the function at x = 6.
b) Write the meaning of limit from above table.
c) Write above function in notation of limit.
7. Fill in the table and answer the given questions.

x f(x) = x2 – 64
x – 8
7.9
7.99
8.1
8.01
8.001
8.0001

a) What is the value of function at x = 8 ?

b) What is the nearest integer of the value of x ?

c) What is the limit of the above function at x = 8.

d) Write the meaning of lim f(x) = lim x2 – 684.
x→8 x→8 x –

50 Vedanta Optional Mathematics Teacher's Guide ~ 9