Shared by :- R.P Pandey
Grade According to new curriculum in compliance with
X Curriculum Development Centre (CDC) .
Approved by CDC.
PRIME Optional
Mathematics
Pragya Books &
Distributors Pvt. Ltd.
Author Editors
Dirgha Raj Mishra LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal
Shared by :- R.P Pandey
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.)
Price Rs. 516/-
ISBN 978-9937-9170-4-9
Typist Sachin Maharjan
Sujan Thapa
Layout and Design Desktop Team
Printed in Nepal
Shared by :- R.P Pandey
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
Shared by :- R.P Pandey
Contents Page
S.N. Units 1
2
1. Algebra 21
1.1 Function 35
1.2 Polynomials 71
1.3 Sequence and series 86
1.4 Linear programming 101
1.5 Solve of quadratic equation 117
2. Limit and Continuity 139
3. Matrices 189
4. Co-ordinate Geometry 253
5. Trigonometry 277
6. Vector Geometry 319
7. Transformation 342
8. Statistics
Model questions
Unit 1 Algebra
1. Algebra
1.1 Function
1.2 Polynomials
1.3 Sequence and series
1.4 Linear programming
1.5 Solve of quadratic equation
Specification Grid Table
K(1) U(2) A(4) HA(5) TQ TM Periods
No. of Questions 2 3 2 1 8 21 35
Weight 2 6 8 5
K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total
Questions, TM = Total Marks
Objectives : At the end of the lesson
• Students are able to find image and pre - image.
• Students are able to find the inverse and composite function & able to
represent in arrow diagram.
• Students are able to identify the polynomial and solving the polynomial
equations.
• Students are able to know the sequence and series and finding nth term and
sum of the terms.
• Students are able to solve quadratic equation graphically.
• Students are able to estimate the maximum and minimum values.
Materials
• Function machine.
• Arrow diagrams taken in chart paper.
• Graph paper and graph board.
• Geo board.
• Standard from of polynomial in chart paper.
• Chart of system of linear programming.
PRIME Opt. Maths Book - X 1
1.1 Function
Enjoy the recalls:
Function :
The relation from a non-empty set A to the set B where
each element of domain is associated with one and only
one element of co-domain is called function f : A → B.
• It is denoted by y = f (x).
Where,
y is the image of x
x is the pre - image of y
The set of elements B is called co-domain.
The set of elements A is called domain.
The set of elements of y∈B which are associated with x is called range.
For example :
• In the function y = 2x + 1 with domain = {–2, 0, 1}
y = f(x) = 2x + 1 is a function.
domain = {–2, 0, 1}
f(–2) = 2(–2)+ 1 = –3
f(0) = 2 × 0 + 1 = 1
f(1) = 2 × 1 + 1 = 3
\ range = {–3, 1, 3}
\ f = {(–2, –3), (0, 1), (1, 3)}
Representation of above function in arrow diagram.
f
AB
–2 –3
01
13
f(x) = 2x + 1
Input f(x)
Activity : Function Machine Processing into the
The above function f(x) = 2x + 1 can be machine by replacing x by
domain elements
changed by using function machine for the Output of f(x)
domain {–2, 0, 1} as shown in diagram. f(–2) = –3
f(0) = 1
f(1) = 3
2 PRIME Opt. Maths Book - X
Types of Functions Example: f {(x, 1), (y, 2), (z, 3)}
f
1. One to one Function:
In a function f : A → B, if different elements of its AB
x1
domain have their different images in co-domain y2
then the function is called a one to one function. z3
In return we can say that if each image in co-domain
is associated with only one pre-image in domain the
function is called one to one function.
2. Many to one function Example:
If more than one elements of domain are associated
f
with only one element in co-domain of a function AB
f : A → B. Such type of function is called many to one 1a
function. In this case there exist at least one image 2b
in co-domain which has more than one pre - images. 3c
3. Onto Function: 4
The function f : A → B where all the elements of co- 5
domain are associated with the elements of domain
is called onto function.
• Range and co-domain of an onto function are equal.
• All the elements of co-domain have pre - image in onto function.
Example: f f
AB
A B a1
a 1 b2
b 2 c3
c 3
d d4
(Many - one onto) (one-one onto)
4. Into function:
The function f : A → B where at least one element of co-domain does not have pre - image
in domain is called into function.
• Range is proper sub - set of co-domain in into function.
• At least one element of co-domain is not associated with domain in into function.
Examples: f f
A B AB
a 1 a1
b 2 b2
c 3 c3
4
4
(one-one into) (Many - one into)
PRIME Opt. Maths Book - X 3
Algebraic functions
The function f : A → B in the form of algebric expression
is called algebric function.
Example : y = f(x) = ax + b
i) Constant function:
The algebraic function f : A → B in the form of y = f(x) = c (constant) is called constant
function.
• It represents a straight line parallel to X - axis.
• Example: for y = 3
f = {(–1, 3), (0, 3), (1, 3), (2, 3)}
• Arrow diagram: Graph:
Y
f
X' O
AB y=3
X
–1
0
1 3
2
Y'
ii) Identity function:
The algebraic function f : A → B in the form of y = f(x) = x is called identity function.
• It shows the straight line passes through the origin which bisects the co-ordinate axes.
• Taking an example for y = x,
f = {(–1, –1), (0, 0), (1, 1), (2, 2)}.
• Arrow diagram: Graphical representation:
fY
AB y=x
–1 –1 X' O X
00
11
22
iii) Linear Function Y'
The algebraic function f : A → B in the form of y = f(x) = ax + b, a ≠ 0 is called linear
function.
• It shows the straight line in graph with the specific condition.
• Taking an example y = 2x + 3
4 PRIME Opt. Maths Book - X
x 0 –1 1 2
y3157
where, f = {(0, 3), (–1, 1), (1, 5), (2, 7)}.
• Arrow diagram: Graphical representation:
fY
AB y = 2x + 3
–1 1
03
15
27
X' O X
iv) Quadratic function: Y'
The algebraic function f : A → B in the form of y = f(x) = ax2 + bx + c, a ≠ 0 is called
quadratic function.
• It traces a curve of parabola.
• Taking an example of y = x2 + 2x – 3
x 0 1 –1 – 3 2 – 4
y – 3 0 –4 0 5 5
Where, f = {(–1, –4), (0, – 3), (1, 0), (– 3, 0), (2, 5), (–4, 5)}
• Arrow diagram: Graphical representation:
fY y = x 2 + 2x – 3
AB X' X
–4 5 O
–3 0 Y'
–1 –4
0 –3
15
2
v) Cubic function:
The algebraic function f : A → B in the form of y = f(x) = ax3 + bx2 + cx +d, a ≠ 0 is called
cubic function
• It shows the curve as shown in following graph
• Taking an example y = x3,
PRIME Opt. Maths Book - X 5
x 0 1 –1 2 – 2
y 0 1 –1 8 – 8
Where, f = {(0, 0), (1, 1),(– 1,– 1 ), (2, 8), (– 2, –8)}
• Arrow diagram: Graphical representation:
f Y y = x3
AB X' O X
–2 –8
–1 –1
00
11
28
Y'
vi) Trigonometric functions:
The function f : A → B in the form of trigonometric expression like y = f(x) = Sinx is called
trigonometric function.
• It shows the curve as shown in the following graph.
• Taking an example y = Sin x
x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
y 0 0.5 0.86 1 0.86 0.5 0 –0.5 –0.86 –1 –0.86 –0.5 0
Where, f = {(0°, 0), (30°, 0.5).......................................}
• Graphical representation: Y
1
y = Sinx
Domain : 0 ≤ x ≤ 2p
Range : – 1 ≤ y ≤ + 1
Period : p
X' O X
30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
–1
Y'
6 PRIME Opt. Maths Book - X
• y = Cosx (–2p ≤ x ≤ 2p)
Domain : – 2p ≤ x ≤ 2p
Range : – 1 ≤ y ≤ + 1
Period : p
x –2p – 3r –p – p 0 p p 3r 2p
2 2 1 2 2
y = Cosx 1 0 –1 0 0 –1 0 1
1Y
y = Cosx
X' 7r 3r 5r 3r p –p O p 3r 5r 3r 7r X
–4 –2 –4 4 2 4 2 4 4 2 4
–2p –p – – p p 2p
4
–1
Y'
• y = Tanx(–2p ≤ x ≤ 2p)
x – 2p – 3r –p – p 0 p p 3r 2p
2 0 2 2 2
∞
y = Tanx 0 ∞ ∞0 0∞0
Y
y = Tanx
Domain : – 2p ≤ x ≤ 2p
Range : –∞ ≤ y ≤ +∞
Period : p
X' 0p X
–2p 3r –p – p 2 p 3r 2p
–2 2 2
PRIME Opt. Maths Book - X Y'
7
Exercise 1.1 'A'
1. Answer the following questions.
i) What is function? Write down in symbol.
ii) Define domain and range in the function y = f(x)
iii) What do you mean by algebric function? Write down with an example.
iv) What is trigonometric function?
v) What type of function is y = 2?
2. Which type of algebraic functions are given below?
i) f = {(1, 2), (4, 2), (–3, 2), (–5, 2)}
ii) f iii) y = 3x2 + 5x – 8
AB
11
22
33
44
iv) y = 1 + Tanx v) f
AB
a
b
c1
d
3. Identify the following functions as onto, into and their types with reasons.
i) f = {(1, 3), (2, 3), (3, 4), (4, 5)}
ii) f iii) f
B A
A B
a x 1 4
b y 2 5
c z 3 6
d 4 7
8
9
iv) f = {(2, 5), (6, 9), (8, 11), (15, 18)}
v) f
AB
2x
3y
4z
8 PRIME Opt. Maths Book - X
4. Which of the followings are the functions. Write down with reasons for all.
i) Y ii) Y
X' O X X' O X
iii) Y' iv) Y'
A f A f
a 2 B
b B 3 a
c 1 4 b
d 2 c
3 5
4
5
v) f
B
A 1
3 2
5 3
7
9
5. Draw the graph of the followings: f(x) = x iii) y = x2
i) f(x) = 4 ii)
iv) y = Sinx (–2p ≤ x ≤ 2p) v) y = Cosx (–360° ≤ x ≤ 360°)
6. PRIME more creative questions:
i) Is f = {(2,3), (2, 5), (4, 7), (5, 6), (6, 8), (6, 9)} a function.
ii) Draw the graph of y = x2 and y = – x2 in the same graph paper.
iii) Draw the graph of y = x3 and y = – x3 in the same graph paper.
iv) Draw the graph of f (x) = x2 – 3x + 2 . Also find the points at which x - axis cuts the
curve?
v) Draw the graph of f (x) = Tanx. (–360° ≤ x ≤ 360°)
Answer
Show to your teacher.
PRIME Opt. Maths Book - X 9
Inverse function: f
Let us consider, the two sets, AB
A = {1, 2, 3, 4} & B = {5, 6, 7, 8, 9, 10, 11} with 5
f(x) = 2x + 3
Then, 16
f(1) = 2 × 1 + 3 = 5 27
f(2) = 2 × 2 + 3 = 7 38
f(3) = 2 × 3 + 3 = 9 49
f(4) = 2 × 4 + 3 = 11
\ R = {(1, 5), (2, 7), (3, 9), (4, 11)} 10
11
Here, f is written as a relation R only.
It is the function from the set A to B, It is one to one and into function.
But the relation taken from the set B to A is not a function because elements of co-domain 6,
8 and 10 don't have pre - images. Hence it has not the inverse function.
Let us consider another function f : A → B f
Where A = {1, 2, 3, 4}, B = {5, 7, 9, 11} define as y = f(x) = 2x + 3
Now, f(1) = 2 × 1 + 3 = 5 A B
f(2) = 2 × 2 + 3 = 7 1 5
f(3) = 2 × 3 + 3 = 9 2 7
f(4) = 2 × 4 + 3 = 11 3 9
\ f = {(1, 5), (2, 7), (3, 9), (4, 11)} 4 11
Here,
f(x) is one - one and onto function. We can define a relation from B to A where each
element of B uniquely relates with elements of A and seems reverse to f(x) as {(5, 1), (7,
2), (9, 3), (11, 4)}.
This reverse relation is also a function and we define as f –I(5) = 1, f –I(7) = 2, f –I(9) = 3 and
f –I(11) = 4
\ f –I = {(1, 5), (2, 1), (3, 9), (4, 11)}
If elements of range and elements of co-domain are same, the function so formed is called
onto function.
Inverse function:
If f : A → B be a one to one and onto function from
the set A to B, then ∃ a function from B to A, called the
inverse function f -l : B → A.
• Inverse of f(x) is written as f –I(x).
• y = f(x) is a function from the set A to B, then y = f –I(x) is the inverse function from
the set B to A.
• If y = f(x) is a function, its inverse is calculated by interchanging the role of x and y
as x = f(y),
10 PRIME Opt. Maths Book - X
Steps to find f –I :
i) Write y = f(x)
ii) Interchange x and y
iii) Solve for y
iv) Replace y by f –I(x)
v) Put specific value of x if asked in the question.
Example :
If f = {(1, 2), (3, 4), (5, 6), (7, 8)} is function. Then inverse function of ‘f' is f –I = {(2, 1), (4,
3), (6, 5), (8, 7)}.
f B f -l
2
A 4 BA
1 6 21
3 8 43
5 65
7 87
Function Inverse Function
Worked out Examples
1. If f(x) = 3x –2, find f(3), f(x + 2), and f (x + h) – f (x) .
h
Solution:
f(x) = 3x – 2.
Then, f(3) = 3 × 3 – 2 = 7
or, f(x + 2) = 3(x + 2) – 2 = 3x + 4.
or, f (x + h) – f (x) =
h
= = 3h = 3
h
2. If f(x+3) = 2x –5, find f(x) and f(6)
Solution:
f(x + 3) = 2x – 5
Let, a = x + 3
\ x = a – 3
Then,
f(x + 3) = 2x – 5
or, f(a) = 2(a – 3) – 5
or, f(a) = 2a – 6 – 5
or, f(a) = 2a – 11
\ f(x) = 2x – 11
Then, f (6) = 2 × 6 – 11 = 12 – 11 = 1.
PRIME Opt. Maths Book - X 11
Alternative method – 1 Alternative method – 2
f(x + 3) =2x – 5 f(x + 3) = 2x – 5
or, f(x +3) = 2(x + 3) – 5 – 6 or, f(x + 3 – 3) = 2(x – 3) – 5
or, f(x) = 2x – 11 [replace x + 3 by x] or, f(x) = 2x – 6 – 5
\ f(x) = 2x – 11
Then, Then,
f (6) = 2 × 6 – 11 = 12 – 11 = 1. f(6) = 2 × 6 – 11 = 1
3. If f(x) = 2x – 7 , find f –I(x) and f –I(3).
3
Solution: f(x) = 2x – 7
3
For inverse of f(x)
Let, y = f(x)
or, y = 2x – 7
3
Interchanging the role of ‘x' and ‘y'
or, x = 2y – 7
3
or, 2y – 7 = 3x
or, y = 3x + 7
2
\ f –I(x) = 3x + 7
2
Again,
3×3+
f –I(3) = 2 7 = 16 = 8
2
4. If g(3x – 2) = 2x + 5, find g –I(x).
Solution :
g(3x – 2) = 2x + 5
2 (x + 2)
or, g a 3x –2 + 2 k = 3 +5
3
or, g(x) = 2x + 4 + 15 Alternative method:
3
g(3x – 2) = 2x + 5
2x + 19 2 4
\ g(x) = 3 or, g(3x – 2) = 3 (3x – 2) + 3 + 5
Again, or, g(3x – 2) = 2 (3x – 2) + 19
Let y = g(x) 2 3 19 3
3 3
i.e g(x) = x +
or, y = 2x + 19 Let g(x) be 'y'.
3
2x + 19
Interchanging the role of x and y for inverse i.e. y = 3
or, x = 2y + 19 Interchanging the role of x & y.
3
2y + 19
or, 2y + 19 = 3x x = 3
or, 2y = 3x – 19 \ y = 3x – 19
2
or, y = 3x – 19
2
\ g –I(x) = 3x – 19
2
\ g –I(x) = 3x – 19
2
12 PRIME Opt. Maths Book - X
Exercise 1.1 'B'
1. Answer the following questions.
i) What do you mean by inverse function?
ii) If 3 is the domain of f(x) = 2x – 5 find its range.
iii) If 7 is the range of f(x) = 2x + 1, find its domain.
iv) Find the image where 30° is the pre-image of the function f(x) = 2Sinx.
v) If the image of an element of a function y = 2Cosx is 0, what will be its pre-image.
2. Find the followings:
i) If f(x) = 2x + 7, domain = {–2, –1, 0, 1, 2}, find range and function in ordered pair.
Also write down its inverse function.
ii) If f(x) = 3x – 2, range = {4, 7, 10}, find domain. Also show f(x) in arrow diagram. Also
find f –I and show in arrow diagram.
iii) If image of a function f(x) =1 + 2sinx is 2, find the angle ‘X'
iv) If domain = {0°, 30°, 60°, 90°}of a function y = 2cosx – 1, find the range. Also show
the inverse function f –I in arrow diagram.
v) If range of a function f(x) = 2sinx – 3 is {–3, –2,–1, 3^1– 3h }, find the domain. Also
show the inverse function in arrow function.
3. Find the inverse function of the followings.
i) f(x) = 3x – 2 ii) f:x → 5x + 7
iii) g(x) = 3x – 7 iv) h(x) = x x ! – 2
2 x+2,
v) f = [x, 3x + 4 , x ! – 1]
x +1
4. Find the followings.
i) If f (x) = 2x + 3, find f –I(5) and f –I(–1)
ii) If f(x) = 3x – 5, find f –I(x + 2)
3x – 2
iii) If f –I(x) = 2 , find f(x) and f(x + 2).
iv) If g –I(x) = 2x – 7, find g (x + h) – g (x) .
h
v) If f(x) = ax + 3 and f –I(5) = 1, find the value of ‘a'.
5. i) If f(x + 1) = 3x – 2, find f(x) and f –I(x).
ii) If f(x + 3) = 2x + 1, find f(x) and f –I(x).
iii) If g(2x + 3) = 3x – 2, find g(x) and g –I(5).
iv) If g(3x – 4) = 2x + 3, find g –I(–1)
g (x + h) –g (x)
v) If g –I(2x – 1) = 3x + 2, find h
6. PRIME more creative questions:
i) If f(x) = 2x – 5 and f(x) = f –I(x), find the value of ‘x'.
3
ii) If f(x) = 3x – 2 , If f –I(x + 5) = f(x), find ‘x'.
2
PRIME Opt. Maths Book - X 13
iii) If f(x + 2) = 3x + 2, If f –I(x) = f –I(2), find ‘x'.
iv) If f(3x + 4) = 2x – 1, If f –I(x) = f –I(1), find ‘x'.
v) If f –I(x – 3) = 5x + 2, If f(x) = f –I(–4), find ‘x'.
Answer
1. Show to your teacher
2. i) R = {3, 5, 7, 9, 11}, f(x) = {(–2, 3), (–1, 5), (0, 7), (1, 9), (2, 11)} f –I = {(3, –2), (5, –1),
(7, 0), (9, 1), (11, 2)}
ii) D = {2, 3, 4}, f –I = {(4, 2), (7, 3), (10, 4)} iii) 30°
ff
24 4 2
37 7 3
4 10 10 4
iv) R = {1, 3 – 1 , 0, –1} v) D = {0°, 30°, 60°, 90°}
f f
0° 1 –3 0°
30° 3 –1 –2 30°
60° 0 3 (1– 3) 60°
90° –1 –1 90°
3. i) x + 2 ii) x –7 iii) 2x + 7
3 5 iii) 3
v) iii)
iv) 2x , x ≠ 1 4 –x , x ≠ 3 iii)
1–x ii) x –3
4. i) 1, – 2 v) x +7 2x + 2 , 2x + 6
3 3 3
ii)
iv) 1 2
2 v)
ii)
5. i) 3x – 5, x +5 v) 2x – 5, x +5 3x – 13 , 23
3 2 2 3
iv) –10 3
2
6. i) – 5 6 2
iv) 1 2
14 PRIME Opt. Maths Book - X
Composite Function:
Let us consider a function f(x) from the set A to the set B and g(x) is the another function
from the set B to the set C, then the new function taken from the set A to the set C directly is
denoted by gof which is called composite function of f and g.
• If first function is g(x) and second is f(x), the composite function is fog.
• If first function is h(x) and second function is g(x), the composite is goh.
[Note : If f:A → B and g:B → C both are one-one and onto then there exist a function from
C→A, which is called the inverse function of gof and we denote it as fog.]
If f : A → B be a function from A to B and g : B → C be the another
function from B to C, then the single function taken from A to C directly
is called the composite function of f and g written as gof: A → C.
• Also is written as gf.
Example:
w If f = {(1,2), (3, 4), (5, 6), (7, 8)} and g = {(2, 5), (4, 7), (6, 9), (8, 11)} then the composite
function will be explained as,
f(1) = 2, g(2) = 5 & gof(1) = 5
f(3) = 4 g(4) = 7 & gof(3) = 7
f(5) = 6 g(6) = 9 & gof(5) = 9
f(7) = 8 g(8) = 11 & gof(7) = 11
Then, gof = {(1, 5), (3, 7), (5, 9), (7, 11)}.
It can be expressed in arrow diagram as,
fg
ABC
12 5
34 7
56 9
7 8 11
gof
PRIME Opt. Maths Book - X 15
Worked out Examples
1. If f = {(3, –2), (5, 0), (7, 2), (9, 2)} and g = {(a, 3), (b, 5), (c, 7), (d, 9)}, find fog. Also show in
arrow diagram.
Solution:
f = {(3, – 2), (5, 0), (7, 2), (9, 2)}
g = {(a, 3), (b, 5), (c, 7), (d, 9)}
Arrow diagram for composite fog:
gf
ABC
a 3 –2
b5 0
c72
d9
fog
From the arrow diagram, we get
fog = {(a, – 2), (b, 0), (c, 2), (d, 2)}
2. If f(x) = 3x + 4 and g(x) = 2x – 3, find fog(x) and gof(2).
Solution:
f(x) = 3x + 4
g(x) = 2x – 3
Then,
fog(x) = f [ g (x)]
= f(2x – 3)
= 3 (2x – 3) + 4 Alternative method for gof (2).
gof(x) = g [ f (x) ] )
= 6x – 9 + 4 = g (3x + 4 )
= 6x – 5
gof(2) = g [ f (2) ] = 2 (3x + 4 ) – 3
= 6x + 5
= g [3 × 2 + 4]
= g (10) Then,
gof(2) = 6 x 2 + 5
= 2 × 10 – 3 = 17.
= 17
3. If f (x) = 3x + 2 , x ! 1, prove that fof -I(x) is an identity function.
x–1
Solution:
3x + 2
f (x) = x–1
For f-1(x),
Let, y = f (x)
or, y = 3x + 2
x –1
16 PRIME Opt. Maths Book - X
Interchanging the role of ‘x' and ‘y' we get,
or, x = 3y + 2
y –1
or, 3y + 2 = xy – x
or, x + 2 = xy – 3y
or, x + 2 = y (x – 3)
x +2
or, y = x –3
` f-1(x) = x + 2 .
x – 3
Again, Composite fof –I(x).
fof –I (x) = f [f –I(x)]
+
= f a x – 2 k
x 3
= 3a x + 2 k + 2
x – 3 – 1
x+2
x–3
= 3x +6 + 2x – 6
x +2 – x +3
= 5x
5
= x
\ fof –I(x) = x which is the identity function.
4. If f(x) = 2x–3 , g(x) = 3x + 5 and fog(x) = f –I(x), find the value of ‘x'.
2
Solution:
f(x) = 2x – 3 Note : (Please try to)
2 • Use the concept of inverse function if
g(x) = 3x + 5 necessary.
Then, for f –I(x), • Use the concept of composite function.
Let, y = f(x) • Convert the relations to algebraic equation.
• Solve for x to find the unknowns as linear
or, y = 2x – 3
2
Interchanging the role of x and y. and quadratic form.
or, x = 2y – 3
2
or, 2y – 3 = 2x Again, by the question, we have,
fog(x) = f –I(x)
or, x = 2y + 3
2
2x + 3
` f –I(x) = 2x + 3 or, f[g(x)] = 2
2
or, f(3x + 5) = 2x + 3
2
or, 2 (3x + 5) – 3 = 2x + 3
2 2
or, 6x + 7 = 2x + 3
or, 4x = –4
\ x = –1
PRIME Opt. Maths Book - X 17
Exercise 1.1 'C'
1. Answer the following questions.
i) What is composite function?
ii) If f(1) = 2 and g(2) = 3, What will be gof(1)?
iii) If f(a) = q and g(p) = a, what will be fog(p)?
iv) If f(x) = 2x, find fof.
v) What will be the result of fof-1(x) for any function f(x)? What type of function is it.
2. Find the following with arrow diagram.
i) Find gof where f = {(1, 4), (4, 7), (6, 9), (7, 10)} and g = {(4, 8), (7, 10), (9, 13),
(10, 14)}.
ii) Find fog where f = {(3, –1), (5, 1), (6, 2)}, and g = {(1, 3), (2, 3), (3, 5), (4, 6)}.
iii) Find hog where g = {(–2, 0), (4, 0), (6, 8), (9, 11)} and h = {(0, 3), (8, 11), (11, 11),
(13, 16)}.
iv) Find the function ‘g' where f = {(a, x), (b, y), (c, y), (d, z)} and gof = {(a, 1), (b, 2),
(c, 2), (d, 3)}.
v) Find the function ‘h' where g ={(a, 2), (b, 4), (c, 6), (d, 6)} and goh = {(1, 2), (2, 4),
(3, 6), (4, 6)}.
3. If f(x) = 2x – 1 and g(x) = 3x + 2, find the following composite functions.
i) fog(x) ii) gof(x) iii) fof(x)
iv) fog(2) v) gog( – 1)
4. i) If f : x → 2x – 3 and g : x → x + 1, find fog ( 1 ).
2 2
ii) If f(x) = 2x + 1 and g(x) = 3x – 2, find fog–I(x).
iii) If f(x) = 2x – 5 , g(x) = x + 2, find f –Iog(x)
2
iv) If f(x) = 3x + 1 , find fof-1(x).
2
v) If g(x) = 2x + 3 , x ≠ 2 prove that g –Iog(x) is an identity function.
x–2
5. i) If f(x) = x – 3, g(x) = 3x + 5 and gof(x) = 2, find the value of ‘x'.
ii) If f(x) = 2x – 3 , g(x) = 3x – 1 and fog(x) = 13 , find the value of ‘x'.
2 2
iii) If f(x) = 2x + 7, g(x) = kx – 3 and gof( – 3 ) = 1, find the value of k.
iv) If f(x) = 3x + 2 , g(x) = 2x – 3 and gof –I (x) = f(x), find the value of ‘x'.
3
v) If f(x) = 2x + 5 , g(x) = 3x – 7 and , f –I og(x) = f(x), find the value of ‘x'.
2
6. PRIME more creative questions:
a. i) If f(x) = 3x – 2, g(x) = 2x + 1 and fog(x) = g –I(x)– 4, find the value of ‘x'.
ii) If f(x) = x2 + 2x, g(x) = x + 2 and fog(x) = 0, find the value of ‘x'.
iii) If f(x) = 2x + 1, g(x) = x2 – 3x and gof(x) = 2, find the value of ‘x'.
iv) If f(x + 2) = 3x + 5 and f[g(x) + 2] = x + 3, find f –I(5).
18 PRIME Opt. Maths Book - X
v) If f(x) = 1 and fof(x) = 2x +1 , prove that f –I(1) = 0.
ax + 1 2x +3
b. i) If f(x) = 3x + 5 and fog(x) = 6x + 7, find the value of g –I (2).
ii) If g(x) = x + 2 and fog(x) = 2x – 7, find the value of f –I (1).
iii) If f(x) = 2x – 3, gof(x) = 3x + 2 and fog(x) = 4, find the value of x.
iv) If f(x) = 3x + b, fog(x) = 2x – 7, and gof –I(2) = – 5, find the value of ‘b'.
v) If g(x) = 5x – 2, gof(x) = 10x + 3, If f(x) = f –I(x), find the value of ‘x'.
7. Project work
Draw the curve of the function y = Sinx where domain = {–360° ≤ x ≤ 360°}. Also find
the domain of it by taking the range = {0, 0.5, 3 , 1} and show in arrow diagram of its
2
inverse function.
Answer
1. Show to your teacher. ii) fog = {(1, –1), (2, –1), (3, 1), (4, 2)}
2. i) gof = {(1, 8), (4, 10), (6, 13), (7, 14)}
fg
fg
ABC
ABC 1 3 –1
14 8 25 1
4 7 10 36 2
6 9 13 4
7 10 14
gof gof
iii) hog = {(–2, 3), (4, 3), (6, 11), (9, 11)} iv) g = {(x, 1), (y, 2), (z, 3)}
gh fg
–2 0 3 ABC
4 8 11 ax1
6 11 16 by 2
9 13 cz3
d
v) h = {(1, a), (2, b), (3, c), (4, d)}
gof
hg
ABC
1a 2
2b 4
3c 6
4d
goh
PRIME Opt. Maths Book - X 19
Answer
3. i) 6x + 3 ii) 6x – 1 iii) 4x – 3 iv) 15 v) –1
v) 6
4. i) 0 ii) 2x + 7 iii) 2x + 9 iv) x v) – 1
5. i) 2 3 2 iv) 5
ii) 3 iii) 4 iv) 13
iv) 258
6.a. i) – 1 ii) – 2 or – 4 iii) – 1 & 3
2
b. i) 2 ii) 6 iii) – 2
3
Function
Unit Test Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. If f = {(1, 2), (3, 4), (5, 6), (7, 8)}, find the inverse of f.
2. a) Define inverse function and write down its suitable example.
b) If f(x) = 2x – 1, find f-1 (x).
c) If fof(x) = 4x – 9, find f(x), where f(x) is a linear function.
3. a) If f(x) = 3x – 2, g(x) = 2x – 3 and fog(x) = g-1(x), find the value of ‘x'.
2
b) If fog(x) = 6x – 1, f(x) = 2x – 5 and gof(x) = 3, find the value of ‘x'.
4. Answer the following questions where
f = {(3, –2), (4, – 1), (5, 0), (6, 1)} and g = {(5, 3), (7, 5), (6, 4), (8, 6)}
a) Define the term composite function.
b) Find the composite function fog and show in arrow diagram.
c) Find fog in the form of linear equation.
20 PRIME Opt. Maths Book - X
1.2 Polynomial
Enjoy the recall:
The algebraic expression with non negative interger (whole numbers) as the exponent
(power) of the variables is consider as a polynomial.
Examples :
xyz + x4 – 3y
3x2 – 4x + 6
The algebraic expression with non-negative integer as
the power of variables is called polynomial.
Points to remember in polynomial
i. The polynomial having degree n is written as,
pIt( ixs) c=a lale0 dxn t +h ea 1sxtan–n1 d+a ar2dx nfo–2r +m a o3xf nt–h3 e+ p..o...l.y..n...o...m....i.a...l. .o...f. .d...e..g...r..e..e.. .n... .i..n.. .x.. .+ anx0
ii. The polynomial should be written in descending order of the power of variables.
Eg: x3 + 3x2 + 2x – 3. It is the standard form of the polynomial of degree 3.
iii. Variables may be x, y, xy, xyz etc. where the polynomial is denoted by
p(x) = In terms of x.
= 2x4 + 3x3 – 5x2 + 2x – 1
p(y) = In terms of y.
= 3y3 + 2y2 – 5y + 2.
p(xy) = In terms of xy.
= x3y2 + x2y + xy2 + y3.
iv. Types of polynomial according to number of terms.
Monomial (having only one term), p(x) = 2x3
Binomial (having two terms), P(x) = 3x2 + 5x,
Trinomial (having three term), p(x) = 2x3 + 3x2 + 5x.
Polynomial (having many term), p(x) = 2x5 + 3x4 + 2x3 + x2 + 5x + 7.
v. Types of polynomial according to degree of variables.
Constant p(x) = C, C∈R
Linear : p(x) = ax + b (first degree) a ≠ 0, a, b ∈ R
Quadratic, p(x) = ax2 + bx + c (second degree) a ≠ 0, a, b, c ∈ R
Cubic p(x) = ax3 + bx2 + cx + d (Third degree) a ≠ 0, a, b, c, d ∈ R
Biquadratic, p(x) = ax4 + bx3 + cx2 + dx + e (Forth degree) a ≠ 0, a, b, c, d, e ∈ R
PRIME Opt. Maths Book - X 21
Multiplication of polynomials:
Let, us consider
p(x) = 2x2 – 3x + 2
q(x) = 3x – 2
Then,
p(x) q(x) = (2x2 – 3x + 2) (3x – 2)
= 3x(2x2 – 3x + 2) – 2( 2x2 – 3x + 2)
= 6x3 – 9x2 + 6x – 4x2 + 6x – 4
= 6x3 – 13x2 + 12x – 4
• For the multiplication of the polynomial the coefficient should be multiplied as
a1 × a2 = xa31 ×a 2x a5 n d =p oxw3 +e 5r o f= t xh8e variables should be added as
Properties of multiplication of the polynomials.
i. Commutative law: f(x). g(x) = g(x).f(x).
ii. Canciliation law: f(x). g(x) = g(x).h(x).
Then, f(x) = h(x) where the common polynomial g(x) from both sides can be canceled.
iii. Closer property: Multiplication of any two polynomials results a polynomial.
i.e. p(x).q(x) = polynomial = q(x). p(x)
iv. Associative property : f(x)[g(x). h(x)] = [f(x). g(x)] h(x).
v. Distributive law over addition : f(x) [g(x) + h(x)] = f(x). g(x) + f(x). h(x).
vi. Multiplicative identity: p(x).1 = 1. p(x) = p(x)
Division of polynomials:
Let us consider p(x) = x3 – 3x2 + 2x – 5 and q (x) = x + 3,
Then for p(x) ÷ q (x)
q(x) p(x)
x + 3 x3 – 3x2 + 2x – 5 x2 – 6x + 20
x3 + 3x2
(–) (–)
– 6x2 + 2x – 5
– 6x2 – 18x
(+) (+)
20x – 5
20x + 60
(–) (–)
– 65
22 PRIME Opt. Maths Book - X
Here,
Quotient = (x2 – 6x + 20 )
Remainder = – 65
For the division of polynomial, the coefficient of the variables should be divided and power of
the variables should be subtracted as
( 4x4 + 2x3 – 6x2 + 8x ) ÷ 2x
4x4 2x3 6x2
= ( 2x + 2x – 2x + 8x
2x
= ( 2x3 + x2 – 3x + 4 )
Synthetic division method for division of the polynomials:
This is the way of division of polynomials by using the
coefficient of the variables only by taking the polynomial
in standard form and by taking the constant term of the
divisor with opposite sign.
i.e. For (ax3 + bx2 + cx + d) ÷ (x – k)
k abc d
ka k2a + kb k3a + k2b + kc
a ka + b k2a + kb + c k3a + k2b + kc + d
x2 x1 x0
Here,
Divisor = x – k [Taking ‘k' for division]
Dividend = ax3 + bx2 + cx + d [Taking constant coefficients a, b, c and d for division]
Quotient = ax2 + (ka + b)x + (k2a + bk + c)
Remainder = k3a + k2b + kc + d
Example :
p(x) = x3 – 3x2 + 2x – 3
q(x) = x + 2
p(x) ÷ q(x)
Taking x = – 2 –3 2 –3
–2 1
–2 10 –24
1
x2 –5 12 –27
x1 x0 Here,
Divisible = p(x) = x3 – 3x2 + 2x – 3
Dividend = q(x) = x + 2
Quotient = Q(x) = x2 – 5x + 12
Remainder = R = – 27
PRIME Opt. Maths Book - X 23
Exercise 1.2 'A'
1. Answer the following questions.
i) What is polynomial?
ii) Write down the standard form of 'n' degree polynomial.
iii) What do you mean by degree of the polynomial? Write down the types of polynomial.
iv) What do you mean by synthetic division?
v) What do you mean by zero of the polynomial?
2. Which of the followings are the polynomials ? Write down with reason.
3
i) 3x3 + 2x2 – 5x + 2 – x ii) 5x2 – 2x3 + 3x4 – 2 – x
iii) 3 x2 – 2 x + 3x3 + 5 iv) 3x4 2 + 5x5 3 – 2x3 + 5x2 + 2x
v) 5 x – 2 x2 + 3 x3 + x4 – 7
3. Write down the polynomials in standard form. Also write their degree.
i) 2x2 + 3x3 – 5x + x4 – 1 ii) 7 – 5x2 + 2x3 – x
iii) 7x3 + x5 – 2x4 + 5 + 2x iv) 5x – 3x2 + 2x3 – x4 + 3x2
v) 3x2 – 2x + 14x3 + 12x5 – x4 + 15x6
4. Find p(x). q(x) from the following.
i) P(x) = 2x3 – 3x2 + x – 2 and q(x) = x2 – 2x + 1.
ii) P(x) = x3 – 5x2 + 2x – 4 and q(x) = 2x2 – 3x + 2.
iii) P(x) = f(x) + g(x), q(x) = g(x), f(x) = x2 + 3x + 2 and g(x) = 2x3 – 3x2 + 2x – 5
iv) P(x) = 2f(x) – g(x), q(x) = f(x), f(x) = x2 + 2x + 3 and g(x) = 3x2 – 2x3 + 5x – 3
v) 2p(x) = 4x3 – 6x2 + 4x – 2 & 3q(x) = 3x2 – 6x + 3
5. Divide the following polynomials:
i) (x3 – 3x2 – 2x + 2) by (x – 2)
ii) x4 + 2x3 – 4x2 + 2x + 18 by x + 3
iii) p(x) = 2x3 + 5x2 – 28x – 15, q(x) = 2x + 1; p(x) ÷ q(x)
iv) (x3 + 3x2y + 3xy2 + y3) by (x + y)
v) p(x) = x3 – y3 by q(x) = x – y; p(x) ÷ q(x)
6. Divide the followings using synthetic division method.
i) x3 + 2x2 – 3x – 2 by x – 1 ii) 2x3 + 3x2 – 5x + 1 by x + 2
iii) x4 + 5x3 – 2x2 – 3x – 1 by x – 2 iv) 4x3 – 3x + 2 by 2x – 1
v) 2x3 + 5x2 – 3x – 7 by 2x + 3
7. PRIME more creative questions.
i. Divide x3 + y3 + z3 – 3xyz by x + y + z.
ii. Divide x3 – 2x2 + 3x – 5 by x2 – 3x + 2
iii. Multiply the sum of x3 + 2x2 + 3x – 2 and x2 – x + 3 by x2 – 3x + 2
iv. Divide p(x) = x4 – 2x2 – 7 by x + 3, by using synthetic division method.
v. Divide p(x) by 2x – 1 where p(x) = 2x3 – 3x2 + 5x – 2
24 PRIME Opt. Maths Book - X
Answer
1. Show to your teacher.
2. Show to your teacher.
3. Show to your teacher.
4. i) 2x5 – 7x4 + 9x3 – 7x2 + 5x – 2
ii) 2x5 – 13x4 + 21x3 – 24x2 + 16x – 8
iii) 4x6 – 10x5 + 20x4 – 35x3 + 29x2 – 31x + 15
iv) 2x5 + 3x4 + 3x3 + 4x2 + 15x + 27
v) 2x5 – 7x4 + 10x3 – 8x2 + 4x – 1
5. i) quotient = x2 – x – 4, remainder = – 6
ii) quotient = x3 – x2 – x + 5, remainder = 3
iii) quotient = x2 + 2x – 15, remainder = 0
iv) quotient = x2 + 2xy + y2, remainder = 0
v) quotient = x2 + xy + y2, remainder = 0
6. i) quotient = x2 + 3x, remainder = –2
ii) quotient = 2x2 – x – 3, remainder = 7
iii) quotient = x3 + 7x2 + 12x + 21 remainder = 41
iv) quotient = 2x2 + x – 1, remainder = 1
v) quotient = x2 + x – 3, remainder = 2
7. i) x2 + y2 + z2 – xy – yz – zx
ii) quotient = x + 1, remainder = 4x – 7
iii) x5 – 5x3 + x2 + x + 2
iv) quotient = x3 – 3x2 + 7x – 21, remainder = 56
v) quotient = x2 – x + 2, remainder = 0
PRIME Opt. Maths Book - X 25
Remainder Theorem
Let us consider an example : 7 ÷ 3
372
6
1
Here, quotient = 2
remainder = 1
Where, 7 = 3 × 2 + 1.
Similarly, (x2 + 3x + 3) ' (x + 1)
x – 1 x2 + 3x + 3 x + 4
x2 – x
(–) (+)
4x + 3
4x – 4
(–) (+)
7
Here, quotient = x + 4.
Remainder = 7
Where, (x2 + 3x + 3) = (x – 1) (x + 4) + 7.
i.e P(x) = (x – 1). Q(x) + R
Again, taking x = 1, in p(x) Here,
P(1) = (x2 + 3x + 3) x + 2 x3 + 3x2 – 3x – 2 x2 + x – 5
= 12 + 3 × 1 + 3 x2 + 2x2
= 7. (–) (–)
\ P(x) = Remainder (R)
x2 – 3x – 2
Similarly for another example, x2 + 2x (remainder)
(–) (–)
(x3 + 3x2 – 3x – 2) ÷ (x + 2)
Taking x = – 2, – 5x – 2
P(–2) = x3 + 3x2 – 3x – 2 – 5x – 10
= (–2)3 + 3(–2)2 –3(–2) – 2 (+) (+)
= – 8 + 12 + 6 – 2
= 8
Conclusion, p(x) = 8 = R. 8
Statement of remainder theorem :
If a polynomial P(x) is divided by (x – a), the remainder
‘R' is equal to P(a).
[R = value of p(x) for x = a]
26 PRIME Opt. Maths Book - X
Proof:
When a polynomial P(x) is divided by x – a,
P(x) = dividend
x – a = divisor
Q(x) = Quotient
R = remainder.
Where,
P(x)= (x – a) Q(x) + R
Taking x = a
P(a) = (a – a).Q(a) + R x – a P(x) Q(x)
= 0.Q(a) + R
= 0 + R .......
= R R
\ remainder (R) = P(a)
Corollary:
i) When a polynomial is divided by x + a°
Where,
P(x) = (x + a).Q(x) + R
Taking x = – a,
P(–a) = (–a + a).Q(–a) + R
or, P(–a) = 0 + R
\ Remainder (R) = P(–a)
ii) If a polynomial P(x) is divided ax – b.
Here,
P(x) = (ax – b). Q(x) + R
b
Taking x = a .
Pa b k = `a × b – bj . Qa b k + R
a a a
or, Pa b k = 0. Qa b k + R
a a
or, Pa b k = 0 + R
a
\ R = P a b k
a
iii) If a polynomial P(x) is divided by ax + b
Here, P(x) = (ax + b) Q(x) + R
b
Taking x = – a
Pa– b k = {a. a– b k + b}. Qa– b k + R
a a a
or, Pa– b k = 0. Qa– b k + R
a a
or, Pa– b k = 0 + R
a
\ R = Pa– b k
a
PRIME Opt. Maths Book - X 27
Worked out Examples
1. If a polynomial x3 + 2x2 – 3x – 2 is divided by x + 2, find the remainder.
Solution:
P(x) = x3 + 2x2 – 3x – 2
Divisor = x + 2
Taking, x = – 2
Remainder (R) = P(x)
or, R = (–2)3 + 2(–2)2 – 3(–2) – 2
or, R = – 8 + 8 + 6 – 2
\ R = 4
2. If a polynomial x3 + 2x2 – Px – 5 leaves remainder 4 on dividing with x + 3.
Find the value of ‘p'
Solution:
P(x) = x3 + 2x2 – px – 5
Divisor = x + 3
Remainder (R) = 4
Now, Taking x = – 3
R = P(x)
or, 4 = (–3)3 + 2(–3)2 – P(–3) – 5
or, 4 = –27 + 18 + 3p – 5
or, 18 = 3p
\ p = 6
Factor theorem:
Let us consider an example 8 ÷ 2 = 4
Here, the remainder becomes zero where 8 can be expressed as
8 = 2 × 4 + 0
= 2 × 4
In this example, when remainder is zero.
The divisor 2 and quotient 4 become the factors of 8.
Factor theorem:
If value of a polynomial p(x) = 0 for any value of x = a then
x – a is a factor of P(x).
• If x – a is a factor of a polynomial p(x) then p(a) = 0
[i.e. remainder R = 0]
Proof: If a polynomial p(x) is divided by x – a.
P(x) = dividend
x – a = divisor
Q(x) = quotient
R = remainder
28 PRIME Opt. Maths Book - X
Then,
P(x) = (x – a). Q(x) + R
Taking x = a then P(x) = 0
or, 0 = (a – a). Q(a) + R
or, 0 = 0 + R
\ R = 0
i.e: for the remainder R = 0, the divisor x – a & quotient Q(x) are the factors of P(x).
Converse: If remainder R = O divisor (x – a) will be the factor
Proof: P(x) = (x – a). Q(x) + R
or, P(x) = (x – a) . Q(x) + 0
or, P(x) = (x – a) . Q(x)
i.e. (x – a) is a factor of P(x)
Worked out Examples
1. Prove that (x + 1) and (x – 3) are the factors of a polynomial x3 + 3x2 – 3x – 15
Solution :
P(x) =x3 + 3x2 – 13x – 15
Taking x = –1.
P(–1) =(–1)3 + 3(–1)2 - 13(–1) – 15
= –1 + 3 + 13 – 15
= 0
` (x + 1) is a factor of p(x)
Again, Taking x = 3
P(3) = 33 + 3 × 32 – 13 × 3 – 15
= 27 + 27 – 39 – 15
= 54 – 54
= 0
\ (x – 3) is a factor of P(x).
2. If (x + 2) is a factor of x3 + mx2 – 11x – 30, find the value of ‘m'
Solution:
P(x) = x3 + mx2 – 11x – 30
Factor = x + 2
Now,
Taking x = – 2,
P(x) = 0
or, (–2)3 + m(–2)2 – 11(–2) – 30 = 0
or, –8 + 4m + 22 – 30 = 0
or, 4m = 16
16
or, m = 4
\ m = 4
PRIME Opt. Maths Book - X 29
3. Factories: x3 + 3x2 – 13x – 15.
Solution:
P(x) = x3 + 3x2 – 13x – 15
The possible factors of 15 are ±1, ±3, ±5, ±15.
Now, Taking x = – 1
P(–1) = (–1)3 + 3(–1)2 – 13(–1) – 15
= – 1 + 3 + 13 – 15
= 0
\ (x + 1) is a factor of p(x).
Then,
P(x) = x3 + 3x2 – 13x – 15 Please try to find p(x) = 0.
= x3 + x2 + 2x2 + 2x – 15x – 15 By putting successively
= x2(x + 1) + 2x(x + 1) – 15(x + 1) possible factors of constant
= (x + 1) (x2 + 2x – 15)
= (x + 1) { x2 + (5 – 3)x – 15} term of the polynomial in
= (x + 1) (x2 + 5x – 2x – 15) p(x)
= (x + 1) {x(x + 5) – 3(x + 5)} If p(x) = 0 for x = –2, the
= (x + 1) (x + 5) (x – 3) factor will be x + 2.
Alternative Method:
Let, P(x) = x3 + 3x2 – 13x – 15
the possible factors of 15 are ±1, ±3, ±5, ±15.
Now,
Taking x = – 1 (using synthetic division)
–1 1 3 –13 –15
–1 –2 15
1 2 –15 0
` (x + 1) is a factor &
Q(x) = x2 + 2x – 15
= x2 + 5x – 3x –15
= x(x + 5) – 3(x + 5)
= (x + 5) (x – 3)
Then,
P(x) = (x + 1) Q(x)
= (x + 1) (x + 5) (x – 3)
4. Solve the polynomial x3 – 7x – 6 = 0
Solution: x3 – 7x – 6 = 0
Let, P(x) = x3 – 7x – 6
The possible factors of 6 are ±1, ±2, ±3, ±6.
Now taking x = – 2 (using synthetic division)
30 PRIME Opt. Maths Book - X
–2 1 0 –7 –6
–2 4 6
1 –2 –3 0
\ (x + 2) is a factor of P(x) and Remember :
Q(x) = x2 – 2x – 3
For a polynomial
Then, 2x3 – 3x2 – 23x + 12, the
or, (x + 2) Q(x) = 0 possible factors will be ±1,
or, (x + 2)(x2 – 2x –3) 1
or, (x + 2){x2 – 3x + x – 3} = 0 ±2, ±3, ±4, ±6, ±12, ± 2 ,
or, (x + 2){x(x – 3) + 1(x – 3)} = 0
& ± 3
2
or, (x + 2)(x – 3) (x + 1) = 0
Either or or
x + 2 = 0 x – 3 = 0 x + 1 = 0
\ x = – 2 x = 3 x = –1
\ x = – 2, –1, 3
5. If (x + 2) is a factor of x3 + ax2 – bx – 24 and it leave's a remainder 6 when divided by (x + 3),
find the value of ‘a' & ‘b'.
Solution:
P(x) = x3 + ax2 – bx – 24.
Factor = x + 2
Divisor = x + 3
Remainder (R) = 6.
Now, Taking x = – 2, [for factor x + 2]
P(x) = 0
or, x3 + ax2 – bx – 24 = 0
or, (–2)3 + a(–2)2 – b(–2) – 24 = 0
or, 4a + 2b = 32
or, 2(2a + b) = 32
or, b = 16 – 2a .................................. (i)
Again, Taking x = – 3 [for divisor x + 3]
p(x) = R
or, (– 3)3 + a(– 3)2– b(–3) – 24 = 6
or, –27 + 9a + 3b = 30
or, 3(3a + b) = 57
or, b = 19 –3a ................................. (ii)
Solving equation (i) and (ii),
16 – 2a = 19 – 3a
\ a = 3
Substituting the value of a in equation (i),
b = 16 – 2(3)
= 16 – 6
= 10
\ a = 3 and
b = 10
PRIME Opt. Maths Book - X 31
Exercise 1.2 'B'
1. Answer the following questions.
i) Sate remainder theorem.
ii) Sate factor theorem.
iii) Prove that x-1 is a factor of the polynomial x2 + 2x – 3.
iv) Write down the relation of polynomial p(x), divisor x-a, quotient q(x) and remainder
R.
v) Prove that remainder R = p(x) for x = a. When p(x) is divided by x – a.
2. Find the remainder and quotient from the following using the remainder theorem.
i) (x3 – 3x2 – 2x + 5) ÷ (x + 1) ii) (x3 + 2x2 – 3x – 13 ) ÷ (x – 2)
iii) (2x3 + x2 + 3x – 4) ÷ (2x – 1) iv) (2x3 – x2 + 4x – 2) ÷ (3x + 2)
v) (x3 + 2x2 – 4x – 3) ÷ (x + 3)
3. i) Prove that (x + 1) and (x – 3) are the factors of a polynomial x3 + 2x2 – 11x – 12.
ii) Prove that ( x – 1), (x + 2) and (x + 5) are the factors of the polynomial x3 + 6x2 + 3x
– 10.
iii) If x3 + 3x2 – 5x – 2 = (x + 2).Q(x) + R, find Q(x) and R.
iv) If p(x) = 2x3 – 3x2 + 2x – 7 = (x – 3). Q(x) + R, find Q(x) and R.
v) If a polynomial p(x) is divided by x + 2 which leaves remainder –4 and quotient 2x2
– 3x – 1, find p(x).
4. i) If a polynomial x3 – px2 – 2x + 5 leaves remainder 3 on dividing with x + 1, find the
value of ‘p'.
ii) If a polynomial x3 + (2k – 3 )x2 – 3x – 5 leaves remainder 1 on dividing with x – 2, find
the value of k.
iii) If a polynomial 2x3 + ax2 + a2x – 4 leaves remainder 28 on dividing with x – a, find the
value of ‘a'.
iv) If a polynomial x3 + kx2 – 2x – 3 leaves remainder 7 on dividing with x + k, find the
value of ‘k'.
v) If a polynomial x3 + ax2 + bx + 2 leaves remainder 8 and – 16 on dividing with x – 1
and x + 3 respectively, find the value of ‘a' and ‘b'.
5. i) If (x + 2) is a factor of a polynomial 6x3 + 2px2 + 3x – 10, find the value of ‘p'.
ii) If a polynomial x3 + 2x2 – 11x – 12 posses the factor x2 – 2x – 3, find the other factor.
iii) If 2R1 + R2 =x3 6+ 2wxh2 e–r 5e aRx 1– a7n adndR 2x3a +re axt2h e– 1r2emx +a i6n dbeyr xo +b 1ta ainnedd x o–n 2 drievsipdeincgtivtehlye
polynomials
find the value of ‘a'.
iv) If the polynomials x3 + bx2 – ax – 6 and x3 + x2 – 2ax + 8 have the common factor x – 2,
find the value of ‘a' and ‘b'.
v) If the polynomials x3 + 4x2 – 2x + 1 and x3 + 3x2 – x + 7 leaves equal remainder when
divided by x –a, find the value of ‘a'.
32 PRIME Opt. Maths Book - X
6. Factorize the followings: ii) x3 + 3x2 – 13x – 15
i) x3 – x2 – 10x – 8 iv) x3 – 19x – 30
iii) x3 – 7x – 6
v) 2x3 – 3x2 – 23x + 12
7.a. Solve the followings: ii) x3 + 8x2 + 11x – 20 = 0
i) x3 + 3x2 – 6x – 8 = 0 iv) x3 – 21x – 20 = 0
iii) x3 – 19x – 30 = 0 ii) 3x3 – 13x2 – 59x + 21 = 0
v) x3 + 3x2 – 16x + 12 = 0 iv) 4x3 + 16x2 – 45x + 18 = 0
b. i) 2x3 – 3x2 – 18x – 8 = 0
iii) 6x3 +29x2 – 7x – 10 = 0
v) 8x3 + 6x2 – 17x + 6 = 0
8. PRIME more creative questions:
a. i) If a polynomial x3 + px2 + qx – 15 has a factor x + 5 and leaves remainder 0 on
dividing with x + 3, find the value of p and q.
ii) If a polynomial 2x3 + 3x2 – kx + 4 leaves remainder 2k when divided by x– 2, find the
value of ‘k'.
iii) If one factor of a polynomial x3 – ax2 + (2a – 12)x + 24 is (x + 2), find the other
factors.
iv) The sum of the remainders when x3 + x2 + 5x + k is divided by x – 3 and when it is
divided by x + 2 is 17. Find the value of ‘k'.
v) Show that 1, 2 and 3 are the zeros of the polynomial of x3 – 6x2 + 11x – 6
b. Find the roots of the polynomial from the followings:
i) 2x4 – 7x3 + 4x2 +7x – 6 = 0
ii) x(x + 4){(x + 1) (x + 3) – 4} – 20 = 0
iii) 2x2(x2 – 4) = 5x3 – 17x + 6 iv) y = x3 + 3x2 – 13x – 10 and y = 5
v) (x – 1) (x + 2) (x – 3) (x – 6) + 56 = 0
9. Project work
Prepare a report of operations on polynomial and synthetic division method in a chart
paper and present into your classroom.
Answer
1. Show to your teacher
2. i) 3; x2 – 4x + 2 ii) –3; x2 + 4x + 5 iii) –2; x2 + x + 2
iv) 1; x2 – 2x – 1 v) 0; x2 – x – 1
v) 2x3 + x2 – 7x – 6
2. iii) 12; x2 + x – 7 iv) 26; 2x2 + 3x + 11 v) 2, 3
3. i) 3 ii) 2 iii) 2 iv) 5 33
PRIME Opt. Maths Book - X
Answer
4. i) 8 ii) x + 4 iii) 2 iv) 5, 2 v) (3, –2)
5. i) (x + 1)(x + 2)(x – 4) ii) (x + 1)(x – 3)(x + 5) iii) (x + 1)(x + 2)(x – 3)
iv) (x + 2)(x + 3)(x – 5) v) (x + 3)(x – 4)(2x – 1)
6.a. i) –1, –4, +2 ii) 1, – 4, – 5 iii) – 2, – 3, 5
iv) – 1, – 4, 5 v) 1, 2, – 6
b. i) –2, 4, – 1 ii) –3, 7, 1 iii) –5, – 1 , 2
2 3 2 3
iv) –6, 1 , 3 v) –2, 3 , 1
2 2 4 2
7.a. i) 7, 7 ii) 8 iii) 5; (x – 3) & (x – 4)
iv) –6
3 1
b. i) 1, –1, 2, 2 ii) 1, –2, –5 iii) 1, –2, 3, 2
iv) –5, 3, –1 v) –1, 5, 2 – 2 2 , 2+ 2 2
Polynomial
Unit Test Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. State factor theorem.
2. a) State and prove remainder theorem.
b) If x – 1 is a factor of a polynomial x3 + (m + 1)x2 – 11x + 6, find the value of m.
c) If a polynomial p(x) = (x + 3). Q(x) + R(x) and P(x) = x3 + 4x2 – 11x + 6, find the value
of Q(x) and R.
3. a) Factorise : x3 + 3x2 – 16x + 12
b) Solve : x3 – 7x + 6 = 0
4. If (x + 2) is a factor of a polynomial x3 + ax2 – bx – 12 which leaves remainder 30 on
dividing with x – 3, find the value of a and b. Also find the polynomial p(x + 1).
34 PRIME Opt. Maths Book - X
1.3 Sequence and Series
Enjoy the recall:
Sequence :
Let us consider the examples of the set of numbers taken in order,
i. 2, 7, 12, 17, ..., ..., ..., ii. 1, 3, 9, 27, 81, ..., ..., ...,
iii. 1, 3, 6, 10, 15, ..., ..., ..., iv. 1, 4, 9, 16, 25 ..., ..., ...,
The set of numbers given in above examples are written under a certain rule which are called
sequence.
A set of numbers written in ascending or
descending order under a certain rule is
called sequence.
• The number come in Succession is written under a certain rule or formula with a common
constant.
• The successive numbers written in a sequence are called terms.
• The constant number which defines the order of the sequence is called common difference
and common ratio respectively for the arithmetic and geometric sequence.
Progression :
Out of four examples discussed above, there is a constant difference in example (i) and there
is a constant ratio in example (ii) which are called progression.
The sequence having constant difference
or constant ratio in each pair of terms is
called progression.
Series :
The above examples given in sequence can be written as follows:
i. 2 + 7 + 12 + 17 + ... + ... + ... + ii. 1 + 3 + 9 + 27 + 81 + ... + ... + ... +
iii. 1 – 3 – 6 – 10 – 15 – ... + ... + ... + iv. 1 – 4 – 9 – 10 – 16 – ... + ... + ... +
This way of writing the terms of the sequence by using sum or difference sign are called
series.
Arithmetic Sequence (Progression) :
A sequence is called an arithmetic sequence if difference of any two consecutive terms remain
same.
The sequence in which the terms are written under
the rule of addition with a constant number is called
arithmetic sequence where the constant number is
involved in the sequence is called common difference.
i.e. d = t2 – t1
PRIME Opt. Maths Book - X 35
Examples:
i. 2, 6, 10, 14, ..., ..., ..., ... each successive terms are increased by 4 in this sequence.
ii. 40, 32, 24, 16, ..., ..., ..., ... each successive terms are decreased by 8 in it.
iii. a, a + d, a + 2d, a + 3d ..., ..., ..., ... each successive terms are increased by ‘d' in this sequence.
In the above examples the successive terms are written under the rule of addition by (+4)
in (i), (–8) in (ii) and (+d) in (iii) respectively where all of them are the arithmetic sequence.
nth term of the arithmetic sequence
Let us consider ‘a' is taken as the first term and ‘d' as the common difference and a, a + d, a +
2d, a + 3d, ..., ..., ..., are the successive terms of an arithmetic sequence. Then the nth term can
be expressed as follows.
1st term = t1 = a = a + o.d = a + (1 – 1)d
2nd term = t2 = a + d = a + (2 – 1) d
3rd term = t3 = a + 2d = a + (3 – 1)d
4th term = t4 = a + 3d = a + (4 – 1)d
5th term = t5 = a + 4d = a + (5 – 1)d
In this way, nth term of the sequence can be expressed as
tn = a + (n – 1) d.
where, common difference (d) = t2 – t1 = t3 – t2 = ........................
Arithmetic means
All the intermediate terms between first
and last terms of an AP are called the
arithmetic means.
Let us consider AM is the middle term between the two terms of an arithmetic sequence a
and b then the sequence becomes,
a, AM & b
Here,
CCoommmmoonn ddiiffffeerreennccee == tA2M – t–1 a= =t3 b – – t 2AM
or, AM – a = b – AM
or, 2AM = a + b
a + b
or, AM = 2
Here,
AM = a + b is the only one arithmetic mean between any two terms of an AP.
2
• Let m1, m2, m3, ..., ..., ..., mn be n number of AMs between the two terms 'a' and 'b'. The
sequence so formed is a, m1, m2, m3, ..., ..., ..., mn, b.
36 PRIME Opt. Maths Book - X
Here,
First term = a
Common difference = d
Last term = b
No. of terms = n + 2 = n'
We have,
or, ttbnn + ==2 =aa +a+ +((nn ( 'n+ – +1 1 )2)dd – 1)d
or, b – a
n + 1
or, d =
Then,
The arithmetic means will be
m1 = t2 = a + d = a + b – a
n + 1
m2 = t3 = a + 2d = a + 2. b – a
n + 1
m3 = t4 = a + 3d = a + 3. b – a
n + 1
and so on nth mean can be written as
mn = a + nd = a + n. b – a
n + 1
Worked out Examples
1. Is 12, 18, 24, 30, ..., ..., ..., an arithmetic sequence?
Solution:
The given sequence is : 12, 18, 24, 30 ..., ..., ..., ...
Here,
ttt324 ––– ttt231 === 213408 ––– 112428 === 666
The difference between any two consecutive terms is same,
Hence, it is an arithmetic sequence.
2. Is 71 a term of the sequence 11, 16, 21, 26, 31, ..., ..., ..., ?
Solution:
The given sequence is : 11, 16, 21, 26, 31, ..., ..., ..., ...
Here,
tt23 –– tt12 == 1261 –– 1116 == 55
Hence, it is an arithmetic sequence.
Then,
First term (a) = 11
Cntoh mtemrmon ( tdni)f f=e r7e1n. ce (d) = t2 – t1 = 16 – 11 = 5.
PRIME Opt. Maths Book - X 37
We have,
tonr ,= a7 +1 ( =n 1–1 1 +)d ( n – 1) 5
or, 12 = n – 1
\ n = 13.
Hence, 71 is the 13th term of the given sequence.
3. If 5th term and 11th term of an arithmetic sequence are 23 and 47 respectively. Find the 15th
term of the sequence
Solution:
EFilfetvhe tnetrhm t e(rtm5) =(t 2113) = 47
We have,
tt2n53 == = aa a ++ + (( 5n4 d–– 11))dd
or,
or,
or, a = 23 – 4d ........................ (i)
Again, [ a from equation (i)]
or, 4t171 == a2 3+ –(1 41d – + 1 1)0dd
or, 24 = 6d
\ d = 4
Putting the value of ‘d' in equation (1)
a = 23 – 4 × 4 = 7
Again,
t15 = a + (15 – 1)d
= 7 + (14) (4)
= 7 + 56
= 63
\ 15th term is 63
4. If x – 3, 2x + 1 and 29 are in AP, find the value of ‘x' and find the common difference.
Solution:
(x – 3), (2x + 1) and 29 are in AP
Where,
or, (t22 –x t+1 1=) t –3 –( xt 2– = 3 c)o =m 2m9 o–n ( 2dxif f+e r1e)nce
or, 2x + 1 – x + 3 = 29 – 2x – 1
or, 3x = 24
\ x = 8
Then,
Common difference (d) == t(22 –x t+1 1) – (x – 3)
= (17 – 5)
= 12
Common difference = 12
38 PRIME Opt. Maths Book - X
5. If nth term of a series 2 + 10 + 18 + ..., ..., ..., and nth term of the another series 8 + 14 + 20 + ...,
..., ..., are equal, find the value of ‘n'
Solution:
1st series
2 + 10 + 26 + ... + ... + ... 2nd series
8 + 14 + 20 + 26 + ... + ... + ...
Here, a = 2 a = 8
d = 10 – 2 = 8 d = 14 – 8 =6
t n == 8a n+ –( n6 – 1) 8 t n == 8a ++ ((nn –– 11)) 66
= 6n + 2
By the question, nth term of them are equal,
8n – 6 = 6n + 2
or, 2n = 8
\ n = 4
6. Find the arithmetic mean between the terms 12 and 20.
Solution:
First term (a) = 12
Second term (b) = 20
then,
AM = a + b
2
= 12 + 20
2
= 32
2
= 16.
7. Insert 4 AMs is between 8 and 38.
Solution:
First term (a) = 8
nLNaoos.. too tffe AtreMmrmS (=tsn 4)( n=) 3 =8 4 + 2 = 6
\
We have,
or, 3tn8 = = a 8 + + ( n(6 – –1 1)d)d
or, 30 = 5d
\ d = 6
Tmmmmh4231e ====n aaaa, t ++++h e43d2 ddd 4= = ==A8 M 888+ S+++ 6a 211 r=e284 1 ===4 232602
PRIME Opt. Maths Book - X 39
8. How many AMs are there between 104 and 36 where the ratio of Last mean and first mean
is 2 : 5?
Solution:
aL e=t ,1 N0o4. of AMs be ‘x'
ntn == x3 6+ 2
mx : m1 = 2 : 5
We have,
or, t3n6 = = a 1 +0 (4n + – ( 1x) +d 2 – 1)d
68
or, d = – x +1 ................................................ (i)
Again,
mx = 2
m1 5
or, a + xd = 2
a+d 5
or, (104 + xd) 5 = 2(104 + d)
or, 520 + 5xd = 208 +2d
or, d(5x – 2) = – 312
68 312
or, – x +1 = – 5x – 2 [ a from equation (i)]
or, 340x – 136 = 312x + 312
or, 28x = 448
448
or, x = 28
\ x = 16
\ There are 16 means.
40 PRIME Opt. Maths Book - X
Exercise 1.3 'A'
1. Answer the following questions.
i) What is Sequence? Write down with an example.
ii) What is series? Write down with an example.
iii) What is progression? What is its different from sequence?
a + b
iv) What do you mean by arithmetic sequence? Prove that AM = 2
v) Prove that tn = a + (n –1)d
2. Examine, which of the followings are the arithmetic sequence or series.
i) 2, 10, 18, 26, 34, 42, 50 ii) 112, 100, 88, 76, 64, 52, 40
iii) 80 + 40 + 20 + 10 + 5 iv) 3, 6, 12, 24, 48, 96
v) a + 3d, a + 5d, a + 7d, a + 9d, a + 11d
3. i) If first term = 20 & common difference = 4, find the 12th term of the sequence.
ii) If nth term = 72, first term = 12 and common difference = 6, find the number of terms
of the sequence.
iii) IIIfff aatn === 882,04 t,, n d d= = =1 –4– 8 84 a aannnddd n tnn = == 1 –25 01, ,6f fi,in nfdidn t dthh teeh v efai rnlsuutem t oebfr emcro oomff mttehorenm s des irofifefe strh.een cper oogf rtehses isoenq.uence.
iv)
v)
4. i) Is 62 a term of the series 2 + 6 + 10 + 14 + ... + ... + ... +
ii) Which term of the sequence 20 + 26 + 32 + 38 + ... + ... + ... + is 104?
iii) How many terms are there in the sequence 124, 112, 100, 88, ..., ..., ..., 4?
iv) Find the 20th term of the given series 13 + 20 + 27 + 34 + ... + ... + ... +
v) If d = 8 and t11 = 79. Find the arithmetic sequence.
5. i) If first term of an arithmetic sequence is 80 and common difference is – 6, find the
13th term of the sequence.
ii) If 25th term of an AP having common difference 4 is 100, find the 15th term of the
sequence.
iii) If 12th term of an arithmetic series whose first term 20 is 53, find the 20th term of the
series
iv) If 3rd term of an AP is 16 and 7th term is 40, find the 10th term of the sequence.
v) If 5th term and 12th term of an arithmetic sequence are respectively 39 and 95, find
the 15th term of the sequence.
6. i) If 7th term and 20th term of an AP are 47 and 138 respectively, find the progression (sequence).
ii) If 12th term and 5th term of an arithmetic series are 68 and 152 respectively find the series.
iii) If nth term of a sequence 2, 6, 10, 14, 18, ..., ..., ..., and nth term of another sequence
122, 116, 110, 104, ..., ..., ..., are equal, find the number of terms.
iv) Monthly salary of Pranisha in a Private office is Rs. 12,000. If her salary is increased
every year by Rs 4,000, what will be her salary after 10 years?
v) Monthly fees of a private school of grade IX is Rs. 5,000 where fees increased every
year by 10%. After how many years the fees will be Rs. 8000?
PRIME Opt. Maths Book - X 41
7. Find the arithmetic mean from the followings.
i) 20 and 28 ii) 104 and 156
iii) (a + d) and (a – d) iv) 3a – 4d and 5a + 4d
v) a + 12 and 38 – a.
8. i) IIIInnfn 1ssseee2rrr xttt ,354 y ,AAA zMMM, 4SSS 4bbb eeeatttrwwwe eeeineee nnna ra11i 02t–h 4 a2m nadend atd i4nc 22d s0. ea.q +u 2endc. e, find x, y, &, z.
ii)
iii)
iv)
v) If 200, p, q, r, s, 100 are in AP, find the value of p, q, r, & s.
9. i) If 24, a, b, c, 48, d, e are in AP, find the value of a, b, c, d, & e.
ii) If a, b, 36, c, d, e and 68 are in AP, find the value of a, b, c, d &, e.
iii) How many AMs are there between 20 and 84 where the third mean is 44?
iv) How many AMs are there between 14 and 86 where the first mean and last mean
are in the ratio 1:4?
v) Find the number of AMs between 220 and 172 where ratio of last mean and 2nd
mean is 15 : 17? Also find the number of terms.
10. PRIME more creative questions:
i. Find the sum of the first 5 terms of an AP whose third term is 4. Write this problem.
ii. Find the second term of an AP where the sum of the first 3 terms is 30.
iii. 9th term of an AP is 5 times the second term and 6th term is 46. Find the 20th term of
the sequence.
iv. If k-1, 2k and 4k – 4 are in AP, find the value of ‘K'. Also find the common difference.
v. Pranav saved Rs 500 in the month of Baishakh Rs. 800 in the second month and Rs.
1,100 in the next month likewise he is going to Shave. In which month he saves Rs
3,800 of the same year?
Answer
1. Show to your teacher.
2. Show to your teacher.
3. i) 64 ii) 11 iii) 13 iv) 100 v) 10
146 v) –1, 7, 15
4. i) 16th term ii) 15th term iii) 11 iv) 58 v) 119
13
5. i) 8 ii) 60 iii) 77 iv)
4a v) 25
6. i) 5, 12, 19, .... ii) 200 + 188 + 176 + ......... iii)
180, 160, 140, 120
iv) 48,000 v) 7 yrs
5, 6 v) chaitra
7. i) 24 ii) 130 iii) a iv)
8. i) 18, 24, 30, 36 ii) 90, 76, 62, 48, 34
iii) a – d, a, a + d, iv) 20, 28, 36 v)
9. i) 30, 36, 42, 54, 60 ii) 20, 28, 44, 52, 60
iii) 12 iv) 11 v)
10. i) 20 ii) 10 iii) 167 iv)
42 PRIME Opt. Maths Book - X
Sum of the first ‘n' terms of arithmetic sequence
Let us consider an arithmetic series is, 2 + 5 + 8 + 11 + 14.
Here, First term (a) = 2
Common difference (d) == 5t2 –– 2t1
= 3
Then, nth term of the series is,
t n = a + (n – 1)d
= 2 + (n – 1)3
= 2 + 3n – 3
= 3n – 1
The sum of the terms of the series can be written in sigma-notation as,
/5 = /5 (3n–1)
n=1
Sn = tn
n=1
But it is the generalization of the sum of the ‘n' terms of the sequence from which terms
can be written in expanded form and sum of the terms can be calculated by adding the
terms for the finite number of terms only, where,
/5
Sn = (3n–1)
n=1
= (3 × 1 – 1) + (3 × 2 – 1) + (3 × 3 – 1) + (3 × 4 – 1) + (3 × 5 – 1)
= 2 + 5 + 8 + 11 + 14
= 40
Derivation of formula :
For the ‘n' terms of the arithmetic sequence, sum of the terms can be calculated by derivation
of formula, where as below
SSnn == tt1n ++ tt2n –+ 1 t+3 t+n t– 42 ++ .t..n +– 3 . .+. +.. .. .+. +.. .t +n . ....... .+.. .t..1.. ................................................... ((ii)i)
Let
\
By adding (i) and (ii), we have
or, 2Sn = (t1 + tn) + (t2 + tn – 1) + (t3 + tn – 2) + ... + ... + ... + (tn + t1)
or, 2Sn = (t1 + tn) + (t1 + d + tn – d) + (t1 + 2d + tn – 2d) + ... + ... + ... + (t1 + tn)
or, 2Sn = (t1 + tn) + (t1 + tn) + (t1 + tn) + ... + ... + ... + (t1 + tn)
or, 2Sn = n(t1 + tn)
n
or, Sn = 2 (t1 + tn)
or, Sn = n (a + tn)
2
or, Sn = n [a + a + (n – 1)d]
2
or, Sn = n [2a + (n – 1)d]
2
Also, Sn = n [a + tn], where tn it the last term
2
PRIME Opt. Maths Book - X 43
Worked out Examples
1. Write down in expanded form and find the sum of /5 (–1)n+1 (2n-5).
Solution : n=1
/5
(–1)n+1 (2n-5) = (–1)1 + 1 (2 × 1 – 5) + (1)2 + 1 (2 × 2 – 5) + (–1)3 + 1 (2 × 3 – 5) + (–1)4 + 1
n = 1 (2 × 4 – 5) + (–1)5 + 1 (2 × 5 – 5)
= (–3) + (1) – (–1) – (3) + (5)
= – 3 + 1 + 1 – 3 + 5
= 1
2. Find the sum of the terms of series 3 + 7 + 11 + 15 + ... + ... + ... + upto 15 terms.
Solution :
3 + 7 + 11 + 15 + ... + ... + ... upto 15 terms
Here,
The given series is in AP,
First term (a) = 3
Common difference (d) == 7t2 –– 3t1
= 4
No. of term (n) = 15
Then, n
2
Sn = [2a + (n – 1)d]
or, S15 = 15 [2 × 3 + (15 – 1)4]
2
or, = 15 × 62
2
= 465
\ S15 = 465
3. Find the sum of the terms of the series, 50 + 44 + 38 + 32 + ... + ... + ... + 2.
Solution :
The given AS is
50 + 40 + 38 + 32 + ... + ... + ... + 2.
Here,
First term (a) = 50
Common difference (d) == 4t20 – – t 150
= – 6
Last term (tn) = 2
Then,
or, 2tn == 5a0 + + ( n(n – –1 1)d)d
or, –48 = n – 1
–6
or, 8 = n – 1
\ n = 9
44 PRIME Opt. Maths Book - X
Again, sum of the 9 terms is,
n
Sn = 2 [a + tn]
\ S9 = 9 [50 + 2]
2
= 9 × 5226
2
= 234
4. If sum of the terms of an AS having first term 12 and common difference 7 is 517, find the
number of terms.
Solution :
In an AS,
First term (a) = 12
Common difference (d) = 7
Sum of the terms (Sn) = 517
We have,
n
Sn = 2 [2a + (n – 1)d]
or, 517 = n [2 × 12 + (n – 1)7]
2
or, 517 = n [24 + 7n – 7]
2
or, 17 n + 7n2 = 1034
or, 7n2 + 17n – 1034 = 0
or, 7n2 + (94 – 77)n – 1034 = 0
or, 7n2 + 94n – 77n – 1034 = 0
or, n(7n + 94) – 11(7n + 94) = 0
or, (n – 11) (7n + 94) = 0
Either or
n – 11 = 0 7n + 94 = 0
\ n = 11 n = – 94 (It is neglected)
\ No. of terms are 11. 7
5. If sum of the terms of an AP having first term 84 and last term 14 is 735, find the common
difference.
Solution :
In an AP,
First term (a) = 84
SLuasmt toefr tmhe ( ttne)r =m 1s 4(Sn) = 735
We have, n
2
Sn = [a + tn]
or, 735 × 2 = n[84 + 14]
or, 1470 = 98n
1470 15
98
or, = n
\ n = 15
PRIME Opt. Maths Book - X 45