Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material
EvMexdcAaenlTtaiHn OEMptAioTnICaSl
10Book
Author
Piyush Raj Gosain
Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar
vedanta
Vedanta Publication (P) Ltd.
jb] fGt klAns;] g kf| = ln=
Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
vEMexdcAaenlTtaiHn OEMptAioTnICaSl
10Book
Author
Piyush Raj Gosain
All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.
¤ Vedanta Publication (P) Ltd.
First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B.S. 2078 (2021 A. D.)
Layout and Design
Pradeep Kandel
Printed in Nepal
Published by:
Vedanta Publication (P) Ltd.
j]bfGt klAns];g k|f= ln=
Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np
Preface
The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the
contemporary pedagogical teaching learning activities and methodologies. It is an innovative
and unique series in the sense that the contents of each textbooks of the series are written and
designed to ful ill the need of integrated teaching learning approaches.
Vedanta Excel in Opt. Mathematics has incorporated applied constructivism the latest trend of
learner centered teaching pedagogy. Every lesson of the series is written and designed in such a
manner that makes the classes automatically constructive and the learner actively participate in
the learning process to construct knowledge themselves, rather than just receiving ready made
information from their instructor. Even teachers will be able to get enough opportunities to play
the role of facilitators and guides shifting themselves from the traditional methods imposing
instructions. The idea of the presentation of every mathematical item is directly or indirectly
re lected from the writer's long experience, more than two decades, of teaching optional
mathematics.
Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out
examples, arranged in the hierarchy of the learning objectives and they are re lective to the
corresponding exercises.
Vedanta Excel in Opt. Mathematics class 10 covers the latest syllabus of CDC, the government of
Nepal, on the subject. My honest efforts have been to provide all the essential matter and practice
materials to the users. I believe that the book serves as a staircase for the students of class 10.
The book contains practice exercises in the form of simple to complex including the varieties of
problems. I have tried to establish relationship between the examples and the problems set for
practice to the maximum extent.
In the book, every chapter starts with review concepts of the same topic that the students have
studied in previous classes. Discussion questions in each topic are given to warm up the students
for the topic. Questions in each exercise are categorized into three groups - Very Short Questions,
Short Questions and Long Questions.
The project works are also given at the end of exercise as required. In my experience, the students
of class 10 require more practices on Trigonometry and Vector Geometry, the examples and the
exercise questions are given to ful ill it in the corresponding topics. The latest syllabus of the
subject speci ication grid and a model question issued by CDC are given at the end of the book.
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
colleage Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book.
I am also thankful to my respected parents and my family members for their valuable support to
bring the book out in this form. I would also like to express my hearty gratitude to all my friends,
colleagues and beloved students who always encouraged me to express my knowledge, skill and
experience in the form of books. I am highly obliged to all my known and unknown teachers who
have laid the foundation of knowledge upon me to be such a person.
Last but not the least, I am hearty thankful to Mr. Pradeep Kandel, the computer and designing
senior of icer of the publication for his skill in designing the series in such an attractive form.
Efforts have been made to clear the subject matter included in the book. I do hope that teachers
and students will best utilize the series.
Valuable suggestions and comments for its further improvement from the concerned will be
highly appreciated. Piyush Raj Gosain
CONTENT
Unit 1 Functions 5
25
Unit 2 Polynomials 44
84
Unit 3 Sequence and Series 97
113
Unit 4 Linear Programming 128
154
Unit 5 Quadratic Equations and Graphs 199
269
Unit 6 Continuity 299
347
Unit 7 Matrices
Unit 8 Co-ordinate Geometry
Unit 9 Trigonometry
Unit 10 Vectors
Unit 11 Transformations
Unit 12 Statistics
Syllabus
Specification Grid
Model Questions
vedanta Excel In Opt. Mathematics - Book 10 1
Functions
1.0 Review
Study the following relations and state which are functions and which are not functions by
giving your reasons:
R1 = {(1,2), (2,3), (3,4)}
R2 = {1,2}, (1,4), (1,6)}
R3 = {(2,4), (3,6), (4,8)}
Discuss the following types of functions with examples related to our daily life:
(a) Onto Functions (b) Into Functions
(c) One to one Functions (d) Many to one Functions
1.1 Algebraic Functions
The functions which can be generated from a real variable x by a finite number of algebraic
operations (addition, subtraction, multiplication, division, and extraction of roots) are called
algebraic functions.
Quite often algebraic functions are algebraic equations.
Some examples of algebraic functions are
f(x) = 4x + 5, y = f(x) = x2 + 3x + 5
f(x) = 5 etc.
Note: function f(x) is also denoted by y. i.e., y = f(x)
Simple Algebraic Functions
The following are some simple algebraic functions: Y
(a) Linear Function (b) Constant Function 5 y=x+3
4
(c) Identity Function (d) Quadratic Function
(e) Cubic Function 3 (0, 3)
2
Let us define each of the above functions with their graphs.
1
X
(a) Linear Function X' (-3, 0) O 1 2 3 4 5
An algebraic function of the first degree expressed Y'
in the form of y = f(x) = mx + c is called linear
function. Here, m and c are constants. The linear
function always gives a straight line when plotted
in a graph.
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vedanta Excel In Opt. Mathematics - Book 10
In the above function, put m = 1 and c = 3,
we get y = x + 3.
To draw graph of y = x + 3, we take points as (1, 4) (-3, 0), (0, 3). The graph is shown
in above figure. Y
(b) Constant Function 4 y=4
An algebraic function that is expressed in the form
of y = f(x) = c i.e. y = c, (where c is a constant) is 3
called a constant function. It represents a straight 2
line parallel to X-axis. In constant function, the X' 1 X
value of y is constant for different values of x. O 1234
For example, y = f(x) = 4,
we get f(1) = 4, f(2) = 4, f(3) = 4.
The graph of y = 4 is shown in the figure, which Y'
is parallel to x-axis.
(c) Identity Function Y
An algebraic function expressed in the form of y=x is X' 2 y=x X
called identity function in which every x is associated 1
with itself. In this function, the function has same 45°
domain and range. -3 -2 -1 O 12 3
-1
The graph of identity function is straight line passing -2
through the origin and bisecting the angle between the
axes.
Note: An identity function is one to one onto function. Y'
(d) Quadratic Function
An algebraic function expressed in the form of y = f(x) = ax2 + bx + c, where a ≠ 0, a, b,
and c are constants, is called a quadratic function.
The graph of a quadratic function is a curve called parabola. The curve is symmetric
b b 4ac - b2
about the line x = - 2a with vertex = – 2a , 4a
The standard form of equation of parabola is y = a(x – h)2 + k,
For example : y = x2 – 6x + 8
Comparing it to y = ax2 + bx + c, Y
we get, a = 1, b = –6, c = 8
where vertex (h, k) = – b , 4ac - b2 y = x2 - 6x + 8
2a 4a
= – -6 , 4.1.8 - 36 = (3, -1)
2.1 4.1
from y = x2 - 6x + 8
x0123456 X' O X
y 8 3 0 -1 0 3 8
6 Y'Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
(e) Cubic Function Y
A function expressed in the form of
y = ax3 + bx3 + cx + d, where a ≠ 0, a, b, c and d are
constants, is called a cubic function.
Examples of cubic functions are
y = x3, y = x3 + 3x, y = 4x3 + 3x2 + 4x + 5 etc.
Let's as draw a graph of cubic function. y = x3 . X
x –2 –1 0 1 2 X'
y –8 –1 0 1 8
Its graph is given as shown in the figure.
1.2 Trigonometric Functions
Y'
A function involving trigonometric ratios like sine, cosine, tangent etc. is called trigonometric function.
Trigonometric functions are related to the angles and their measurements. Trigonometric
functions of a right angled triangle are defined in terms of ratios of sides of right angled triangle.
1. Discuss the values of trigonometric ratios of angles from -2Sc to 2Sc.
2. Note the maximum and minimum values of the trigonometric ratios of sine, cosine
and tangent.
3. Do the values of the trigonometric function repeat for different values of angles ?
Discuss it.
Trigonometric functions do not satisfy the algebraic operations like addition, subtraction,
multiplication, and division. They are different from algebraic functions.
The functions which are not algebraic are called transcendental functions. Trigonometric
functions are also transcendental functions.
Trigonometric functions are periodic functions. A function f which satisfies
f(x+k) = f(x) for all x belonging to its domain and smallest positive value of k.
Examples : sin(x + 2Sc) = sinx
cos(x +2Sc) = cosx
tan(x+Sc) = tanx
We can say that sinx, cosx, and tanx are the periodic function of periods 2Sc, 2Sc, and Sc
respectively.
Graphs of Trigonometric Functions
(a) Graph of y = sinx, (-2Sc ≤ x ≤ 2Sc)
For the sine function, y = sinx, take the values of x from -2Sc to 2Sc, we get the
corresponding values of y as shown in the table.
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vedanta Excel In Opt. Mathematics - Book 10
x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc
2 2 2 2
y 0 1 0 -1 0 1 0 -1 0
What are maximum and minimum values of sinx ?
Plot the points (-2Sc,0), – 3Sc , 1 , Y
2
Sc Sc
(Sc, 0), – 2 , - 1 , (0, 0) 2 , 1 ,
(Sc, 0), 3Sc , –1 , and (-2Sc, 0) in a -1
2
X' X
graph paper and join the points. We -2cS 3Sc O Sc Sc
- 2 -Sc Sc -1 2 3Sc 2Sc
- 2 2
get a curve as shown in the figure
alongside.
(b) Grph of y = cosx, (-2Sc ≤ x≤ 2Sc) Y'
For y = cosx, take the values of x from -2Sc to 2Sc, we get corresponding values of y as
shown in the table.
x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc
2 2 1 2 2
y 1 0 -1 0 0 -1 0 -1
Note the maximum and minimum values of cosx. Plot the above points in the graph
paper and join the points and we get the curve in the figure below :
Y
-1
X' -2Sc 3Sc 0 Sc X
2 -1 2
- -Sc - 3 Sc 3Sc 2Sc
Sc 2
Y'
(c) Graph of y = tanx, (-2Sc ≤ x ≤ 2Sc)
For y = tanx, the value of y is not defined for the values x = - 32Sc, - Sc , Sc , 3Sc .
2 2 2
Image of the function for these values of x are not defined.
x -2Sc -7Sc -3Sc -5Sc -Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc Sc 5Sc 3Sc 7Sc 2Sc
4 2 4 4 2 4 4 2 4 4 2 4
y 0 1 ∞ -1 0 1 ∞ -1 0 1 ∞ -1 0 -1 ∞ -1 0
8 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Y (1.249, 3)
4
3
2
1
X' 3Sc O X
- 2
-Sc - Sc Sc Sc 3Sc 2Sc
2 2 2
-1
-2
-3
Y'
The graph of tanx is shown in the above graph.
Worked out Examples
Example 1. Name the following algebraic functions.
Solution: (a) y = 5 (b) y = ax + b,
Example 2.
(c) y = x3 + 2x2 (d) y = 2x2 + x + 5
(a) Constant Function (b) Linear Function
(c) Cubic Function (d) Quadratic Function
Name the algebraic functions of the following graphs. Y
Y
X' O X X' O X
Y' Y'
(a) The given graph represents linear function.
Solution : (b) The given graph represents quadratic function.
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vedanta Excel In Opt. Mathematics - Book 10
Exercise 1.1
Very Short Questions
1. Classify the following functions:
(a) y = x (b) y = x2 (c) y = x3
(d) y = ax2 + bx + c (e) y = x3 - x2 (f) y = 7
(g) y = tanx
2. Define a function with an example.
3. Define an algebraic function.
4. What is a trigonometric function ?
5. What is a transcendental function ?
6. What is the minimum value of the cosine function y = cosx ?
7. Write the periods of the following functions:
(a) y = sinx (b) y = cosx (c) y = tanx
8. For what values of x, the image of tangent function y = tanx is not defined ?
9. Define each of following functions with an example:
(a) Linear Function (b) Constant Function (c) Identity Function
(d) Quadratic Function (e) Cubic Function (f) Periodic Function
10. Name the functions represented by the following graphs: Y
(a) Y (b)
4
2 3
1 2
1
45° X' X
X' X
-3 -2 -1 O 12 3 -3 -2 -1 O 12 3
Y
-1 -1
-2 -2
-3
Y' -4
(c) Y (d)
Y'
-1 -1
X' 0 X X' -2Sc 3Sc 0 Sc Sc X
2 -1 2
-2Sc - 3Sc -Sc - 3 Sc Sc 3Sc - -Sc - Sc 3Sc 2Sc
2 Sc -1 2 2 2 2
10 Y' Y'
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vedanta Excel In Opt. Mathematics - Book 10
Y
(e) Y (f)
X' 0 X
X' O X
y = -3
Y'
Short Questions
11. Draw graph for each of the following table: Y'
(a) x 2 3 4 5 6 (b) x 0 ±1 ±2 ±3
y 4 6 8 10 12 y0149
(c) x -3 -2 0 1 (d) x 1234
y -1 -2 -3 -4
y 6 4 0 -2
12. Draw graph of each of the following functions:
(a) y = 2x - 1 (b) y = x + 1 (c) y = x2
(d) y = -x (e) y = x3 (f) y = -x3
13. Draw the graphs of the following trigonometric functions:
(a) f(x) = sinx, (-Sc ≤ x ≤ Sc) (b) f(x) = cosx, (-Sc ≤ x ≤ Sc)
(c) f(S) = tanx, (-Sc ≤ x ≤ Sc)
Long Questions
14. Draw the graphs of the following trigonometric functions:
(a) f(x) = sinx, (-2Sc ≤ x ≤ 2Sc) (b) f(x) = cosx, (-2Sc ≤ x ≤ 2Sc)
(c) f(x) = tanx, (-2Sc ≤ x ≤ 2Sc)
Project Work
15. Record your body weight continuously for a week at the same time in the morning.
Taking days in X-axis and weight in Y-axis, draw the graph of it and discuss the type of
function represented by the graph drawn.
16. List the price of petrol per litre for a week. Taking days in X-axis and the price in Y-axis,
represent the information obtained in graph. Discuss the type of function represented
by the graph drawn.
17. Where are the functions f(x) = sinx and f(x) = cosx used in our daily life ? Investigate
and prepare a report and present in your class.
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vedanta Excel In Opt. Mathematics - Book 10
1. (a) Identity function (b) quadratic (c) cubic (d) quadratic
(g) trigonometric
(e) cubic (f) constant 6. –1
(c) Sc
7. (a) 2Sc (b) 2Sc (c) y = cosx 8. Sc
10. (a) identity (b) quadratic 2
(f) cubic
(e) constant (d) y = sinx
1.3 Composite Function
Let A = {1, 2, 3}, B = {2, 3, 4}, and C = {4, 6, 8} be three sets. Let us define two functions
f and g such that f : A → B and g : B → C such that
f = {(1, 2), (2, 3), (3, 4)} and g = {(2, 4), (3, 6), (4, 8)}
By representing above function in mapping diagram, we write it as below.
fg
A BC
1 24
2 36
3 48
gof
Here, f is a function defined from A to B. So write.
1 A o f(1) = 2 B
2 A o f(2) = 3 B
and 3 A o f(3) = 4 B
range of f = {2, 3, 4} which is domain for g.
Again, g is a function from B to C.
2 B o g(2) = 4 C
3 B o g (3) = 6 C
4 B o g (4) = 8 C
Now, let us define a new function from A to C. Such that,
1 A o 4 = g(2) = g(f(1)) C
2 A o 6 = g(3) = g (f(2)) C
4 B o 8 = g(4) = g (f(3)) C
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vedanta Excel In Opt. Mathematics - Book 10
The new function defined from A to C is called composite A gof
function of f and g, and it is denoted by gof. we can draw for gof
as below. 1 C
2
Writing the above composite function as a set of ordered pair, 3 4
6
we get, gof = {(1, 4), (2, 6), (3, 8)} 8
Definition
Let f : A → B and g: B → C be two functions. Then a new function defined from set A to
set C is called composite function of f and g. It is A B C
denoted by gof and defined by gof : A → C.
Then every element of A is associated with a x f(x) g(f(x))
unique element of set C.
Symbolically, we write.
If f : A → B and g: B → C, then the composite gof
function from A to C is defined by
gof : A → C, for every x A, (gof) (x) = g (f(x)) C
Note :
(i) The composite function gof is read as "f followed by g".
(ii) In general gof ≠ fog
(iii) If (fog) (x) = x and (gof) (x) = x, then, the composite function is called identity
function.
Worked Out Examples
Example 1. If f = {(1, 1), (2, 4), (3, 9)} and g = {(1, 2), (4, 8), and (9, 18)}, then, show the
function gof in arrow diagram and find it in the set of ordered pairs.
Solution: Here, f = {(1, 1), (2, 4), (3, 9)}
g = {(1,2), (4,8), (9,18)}
Arrow diagram of composite function gof is given below,
f g
BC
A
11 2
24 8
3 9 18
gof 13
Writing above composite function in the ordered pairs, we get
gof = {(1,2), (2, 8), (3, 18)}
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vedanta Excel In Opt. Mathematics - Book 10
Example 2. Let f : R → R, f(x) = 4x + 3 and g : R → R, g(x) = x2 + 4, Find the following:
Solution:
(a) (fof) (x) (b) (gog) (x)
(c) (fog) (x) (d) (got) (x)
(e) (gof) (1) (f) (fog)(2)
Also, check is (gof) (x) = (fog)(x) ?
Here, f(x) = 4x + 3, g(x) = x2 + 4
(a) (fof)(x) = f(f(x)) = f(4x + 3) = 4(4x+3) + 3
= 16x + 12 + 3
= 16x + 15
(b) (gog)(x) = g(g(x)) = g(x2 + 4) = (x2 + 4)2 + 4
= x4 + 8x2 + 16 + 4
= x4 + 8x2 + 20
(c) (fog)(x) = f(g(x)) = f(x2 + 4) = 4(x2 + 4) + 3
= 4x2 + 16 + 3
= 4x2 + 19
(d) (gof) (x) = g(f(x))= g(4x + 3) =(4x + 3)2 + 4
= 16x2 + 24x + 9 + 4
= 16x2 + 24x + 13
(e) (gof) (1) = g(f(1)) = g(4.1 + 3) Alternative Method
= g(7) we have,
= 72 + 4 (gof)(x) = 16x2 + 24x + 23
= 49 + 4 (gof)(1) = 16.12 + 24.1 + 13
= 53 = 53
(f) (fog)(2) = f(g(2)) Alternative Method
= f(22 + 4) we have,
=f(8) (fog)(x) = 4x2 + 19
=4.8 + 3 ? (fog)(2) = 4.22 + 19
= 32 + 3 = 16 + 19
= 35 = 35
From above, we get,
(fog) (x) = 4x2 + 19
(gof) (x) = 16x2 + 24x + 13
? (gof)(x) ≠ (fog)(x)
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vedanta Excel In Opt. Mathematics - Book 10
Example 3. If f: R → R, and g: R → R are defined by f(x) = 2x and g(x) = 3x + 4 and
Solution: (gof)(x) = 22, find the value of x.
Example 4. Here, f(x) = 2x and g(x) = 3x + 4
Solution:
Now, (gof)(x) = g(f(x)) = g(2x) =3(2x) + 4 = 6x + 4
Example 5.
Solution: By question, (gof)(x) = 22
Example 6. ie. 6x + 4 = 22
Solution:
or, 6x = 22 - 4
or, x= 18
6
? x = 3
If f(x) = px + 4 and g(x) = 5x and (gof)(2) =4, find the value of p.
2
px + 4
Here, f(x) = 2 and g(x) = 5x
Now, (gof)(2) = 4
or, g(f(2)) = 4
or, g p.2+4 =4
2
or, g(p + 2) = 4
or, 5(p + 2) = 4
or, p+2= 4
5
4
or, p = 5 -2
or, p = 4 - 10 =- 6
5 5
6
? p =- 5
If (fog)(x) = 7x - 1 and f(x) = 3x + 5, find g(x) where g(x) is a linear function.
3
We have, f(x) = 3x + 5
Now, (fog) (x) = 7x - 1
3
7x - 1
or, f[g(x)] = 3
or, 3g(x) + 5 = 7x- 1
3
7x - 1
or, 3g(x) = 3 – 5
g(x) = 7x - 16
9
If f: R → R ; f(x) = 8 - x, Show that (fof)(x) = x.
Here, f(x) = 8 - x
(fof)(x) = f(8 - x) = 8 - (8 - x) = 8 - 8 + x = x
? (fof)(x) = x, proved
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vedanta Excel In Opt. Mathematics - Book 10
Exercise 1.2
Very Short Questions
1. Define a composite function of two functions.
2. Write the composite function of given functions f and g as shown in the mapping
diagram.
f g C
A B
x yz
3. From the adjoining figure, write the composite function gof of function f and g in the
ordered pair form.
f g C
B
A
ad x
be y
cf z
4. Write the composite function gof of the following functions in the ordered pair form:
(a) f = {(1, 2), (3, 4), (4, 5)} and g = {(2,6), (4,12), (5,15)}
(b) f = {(1,2), (3,4), (5,6)} and g = {(2,3), (4,6), (6, 8)}
(c) f = {(3,4), (4,5), (5,6)} and g = {(4,5), (5,6), (6.7)}
5. Write the composite function fog of the following functions in the ordered pair form:
(a) f = {(3,4), (4,5), (5,6)} and g = {(2,3), (3,4), (4,5)}
(b) f = {(4,2), (8,4), (16, 8)} and g = {(2,4), (4, 8), (8, 16)}
(c) f = {(2, 8), (4, 64), (8, 256)} and g = {(1, 2), (2, 4), (4, 8)}
Short Questions
6. (a) f = {(1,3), (2,1), (3,2)} and g = {(1,2), (2,3), and (3,1)}, find the domain and range
of the composite function gof.
(b) If f = {(1,2), (3,5), (4,1)} and g = {(2,3), (5,1), and (1,3)}, find the domain and
range of the composite function gof.
7. (a) If f = (1,5), (2,1), (3,3), (5,2)} and g= {(1,3), (2,1), (3,2), (5,5)}, find gof and fog.
(b) If h = {(5,2), (6,3), k = {(2,5), (3,6)} find hok and koh.
8. (a) From the function f = {(4,1), (5,3), (6,2)} and g = {(1,2), (3,6), (2,4)}, show gof in
arrow diagram and write it in ordered pair form.
(b) From the functions f = {(a,b), (b,c), (c,d)} and g = {(b,e), (c,f), (d,h)}, find gof in
ordered pair form and also show in mapping diagram.
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vedanta Excel In Opt. Mathematics - Book 10
9, If (fog) = {(2,1), (4,2), (6,3), and (8,4)} and g = {(2,4), (4,8), (6,12), and (8,16)}, find f in
ordered pair form. Also show fog in arrow diagram.
10. (a) If f(x+2) = 4x + 5, find f(x) and (fof) (x)
(b) If g(x+5) = x + 20, find g(x) and (gog) (x).
Long Questions
11. (a) Let f: R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x3. Then (i) find
gof and fog. (ii) Is (gof) (x) = (fog)(x)?
(b) Let f: R → R and g: R→R be defined by f(x) = x3 - 1 and g(x) = x2
(i) Find (gof) (x) and (fog) (x) (ii) Is (gof)(x) = (fog)(x)
12. (a) If f: R → R and g: R → R are defined by f(x) = 2x + 3 and g(x) = 2x - 1,
find (gof) (3) and (gof)(-2).
(b) If f: R → R and g: R → R be defined by f(x) = x + 5 and g(x) = 5x - 5. Find (gof)(-2)
and (gog)(2).
13. (a) Find the value of x if f(x) = 4x + 5 and g(x) = 6x + 3, (gof)(x) = 75
(b) Find the value of x if f(x) = 3x + 4 and g(x) = 2x + 5, (fog)(x) = 25.
14. (a) If f(x) = px + 8 and g(x) = 3x + 5, (gof)(3) = 60, find the value of p.
(b) If f(x) = 7x + q and g(x) = x + 8, (fog) (4) = 24, find the value of q.
(c) If g(x) = 62 and h(x)= ax2 - 1 and (goh)(3)= 17, find the value of a.
x-2
(d) If f(x)= 3x + a and g(x) = 4x + 5 and (gof)(5) = 20, find the value of a.
(e) If f(x) = 3x , g(x)= Ex - 4, and (gof)(1) = 10, find the value of E.
x+2
15. (a) If f(x) = 4x - 5 and g(x) is a linear function and (gof) (x) = 5x + 1, find g(x).
(b) If g(x) = 5x + 3 and (gof)(x) = 2x + 5 and f(x) is a linear function, find the value
of f(x).
(c) If g(x) = 2x and (fog)(x) = 6x-2, find f(x), where f(x) is a linear function.
(d) If f(x) = 2x + 3 and (fog) (x) = 10x + 1, find the value of g(x).
(e) If f(x) = 2x + 8, (fog)(x) = 3x+4, find the function g(x).
(f) If f(x)= 2x -1, g(x) is a quadratic function and (fog)(x) = 6x2+4x + 1, find g(x).
16. (a) If g(x) = 10 - x, show that (gog)(x) = x.
(b) If h(x) = 20 -x, show that (hoh)(x) = x
17. (a) If p(x) = 4x+5 and q(x)= 9x-5 prove that (poq)(x) is an identity function.
9 4
3x+2
(b) If p(x) = 3 and q(x) = 3x-2 , prove that (poq) (x) is an identity function.
(a) If g(x3) = {(3,1), (5,1), (7,1)} prove that (gof) is constant
18. f = {(1,3), (4,5), (6,7)} and
function.
(b) If f = {(2,4), (4,5), (7,10)} and g = {(4,2), (5,2), (10,2)}. Show that gof is a constant
function.
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Project Work
19. Number of bacteria in food kept in a refrigerator is expressed by N(T)=20T2 - 80T +
500, (2 ≤ T ≤ 14), where T denotes temperature and
T(t) = 4t + 2, (0 ≤ t ≤ 3), where t represents time in hour.
(a) Find (NoT) (t)
(b) What is the number of bacteria in 2 hours?
(c) In how many hours does the number of bacteria in the food reach 3300 ?
2. gof 3. gof = {(a, x), (b, y), (c, z)}
4. (a) {(1, 6), (3, 12), (4, 15)} (b) {(1, 3), (3, 6), (5, 8)}
(c) {(3, 5), (4, 6), (5, 7)} 5. (a) {(2, 4), (3, 5), (4, 6)}
(b) {(2, 2), (4, 4), (8, 8) (c) {(1, 8), (2, 64), (4, 256)}
6. (a) D = {1, 2, 3), R = {1, 2, 3} (b) D = {1, 3, 4}, R = {3, 1}
7. (a) gof = {(1, 5), (2, 3), (3, 2), (5, 1)} fog = {(1, 3), (2, 5), (3, 1), (5, 2)}
(b) hok = {(2, 2), (3, 3)} koh = {(5, 5), (6, 6)}
8. (a) gof = {(4, 2), (5, 6), (6, 4)} (b) fog = {(a, e), (b, f), (c, h)}
9. F = {(4, 1), (8, 2), (12, 3), (16, 4)} 10.(a) 4x – 3, 16x – 15 (b) x + 15, x + 30
11. (a) gof = x3 + 3x2 + 3x + 1, fog = x3 + 1, No. (b) gof = x6 – 2x3 + 1, fog = x6 – 1, No.
12. (a) 17, –3 (b) 10, 20 13. (a) 7 (b) 1
4
31 19 45
14. (a) 9 (b) –36 (c) 51 (d) – 4 (e) 14
15. (a) 5x + 29 (b) 2(x + 1) (c) 3x – 2 (d) 5x – 1
4 5
3x
(e) 2 – 2 (f) 3x2 + 2x + 1 19. (a) 320t2 + 420 (b)1700 (c) 3 hours
1.4 Inverse Function
Let A = {1, 2, 3} and B = {1, 8, 27} be two sets and the function f is defined from A to B by
f = {(1,1), (2,8), (3,27)}
A fB
11
28
3 27
Here, domain of f = {1,2,3}
range of f = {1, 8, 27}
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co-domain of f = {1, 8, 27}
since range of f = co-domain of f, f is one to one onto function.
(or bijective function)
Now, let us define a new function g from B to A so that the domain and range of g are
obtained by interchanging the domain and range of f respectively.
Then g = {(1,1), (8,2), (27,3)}
g
11
82
27 3
domain of g = {1, 8, 27}
range of g = {1, 2, 3}
Here, g is also one onto function. g is called inverse of function of f.
This new function defined from B to A is called inverse of f. It is denoted by f -1.
Therefore the function must be one to one onto, to have its inverse function.
Let us take another function g as shown in the mapping diagram.
g
1x
2y
3z
Here, g is not a onto function. It is many to one into function.
Now, let us interchange the domain and range of g. Is it again a function ?
g -1
x1
y2
z3
This g-1 does not represent a function.
Hence, to have inverse of a function, the function must be one to one onto, i.e., bijective.
A function f is called bijective if it is one to and onto. Only the bijective functions have
their inverse functions.
Remember, f-1 ≠ 1 and f-1 (x) ≠ 1
f f(x)
Definition:
Let f: A → B be a one to one onto function. Then, the function f-1 : B → A is called inverse of
f for every x B there corresponds f-1 (x) = y A
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In functions f and f-1, the roles of x and y are interchanged.
Steps for finding inverse of an algebraic functions:
(i) Put y in place of f(x) i.e. y = f(x)
(ii) Interchange the roles of x and y.
(iii) Get new y in terms of x solving equation obtained in (ii).
Note :
(i) If the two functions f and g are inverse to each other, f-1 = g or g-1 = f and
(gof)(x) = x, (fog)(x) = x.
(ii) If f is a bijective function, (f-1)-1 = f.
(iii) If f–1 exists, then, domain of f is range of f–1 and range of f is domain of f–1.
Worked Out Examples
Example 1. Let f: A → B be a bijective function defined by f = {(1,2), (2,3), (3,4),(4,5)}.
Solution: Find f-1 in ordered pair form.
Example 2.
Solution : Here, f = {(1,2), (2,3), (3,4), (4,5)} Interchanging the elements of ordered
pairs, we get f-1 = {(2,1), (3,2), (4,3), (5,4)}
Let f: A → B be one to one and onto function, defined by f = {(2,4), (3,6), (4,8),
(5,10)}. Find f-1 and show f and f-1 in mapping diagrams.
Here, f = {(2,4), (3,6), (4,8), (5,10)}. Interchanging the components in each
ordered pair, we get f-1 = {(4,2), (6,3), (8,4), (10,5)}
In mapping diagrams, we show that
Af B B f A
2 44 2
3 66 3
4 88 4
5 10 10 5
Example 3. (a) If f(x) = 7x - 2, find f-1 (x)
Solution:
(b) If g : 5x - 3 , x ≠ - 1 , find g-1 (x) and g-1 (2)
2x + 1 2
(a) Here, f(x) = 7x - 2
To find f-1, y = 7x - 2
Interchanging the roles of x and y, we get
x = 7y - 2
or, x + 2 = 7y
y= x+2
7
x+2
? f-1 (x) = 7
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(b) Here, g (x) = 5x - 3
2x+ 1
To find g-1 (x), interchanging the roles of x and y, we get
x = 5y-3
2y+1
or, 2xy + x = 5y - 3
or, 2xy - 5y = - x - 3
or, y(2x - 5) = - (x + 3)
or, y = - (x+3)
2x - 5
x+3 5
y = 5 - 2x , x ≠ 2
g-1 (x) = x+3
5 - 2x
2+3 5
Also, g-1 (2) = 5-2×2 = 5 - 4 =5
? g–1 (2) = 5
Example 5. If f = {(x,y) : y = 7x - 4} find f-1 (x) and f-1 (4).
Solution:
Here, y = 7x - 4, to find f-1(x), interchanging the roles of x and y, we get
Example 6.
Solution: x = 7y - 4
or, 7y = x + 4
y = x+4
7
x+4
? f-1 (x) = 7
Now, f-1 (4) = 4 + 4 = 8
7 7
8
? f–1(4) = 7
If f = {(x, y) : y = 3x - k} and f-1(5) = 4, find the value of k.
Here, f = {(x,y) : y = 3x - k}
y = 3x - k
i.e. f(x) = 3x - k
To find f-1(x), interchanging the roles of x and y, we get,
x = 3y - k
or, y = x + k
3
x+k
f-1(x) = 3
and f-1 (5) = 5+k
3
But, f-1(5) = 4
or, 5+k =4
3
or, 5 + k = 12
? k=7
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Example 7. Given a function f(x+3) = 3x + 5, find f-1(x)
Solution:
Here, f(x+3) = 3x + 5
Example 8.
Solution: replacing x by x - 3, we get,
f(x - 3 + 3) = 3(x - 3) + 5
or, f(x) = 3x - 4 or, y = 3x - 4
To find f-1 (x) ,
Interchanging the roles of x and y, we get,
x = 3y - 4
or, y = x+4
3
x+4
? f-1(x) = 3
Alternate Method.
Here, f(x + 3) = 3x + 5
or, f(x + 3) = 3 (x + 3) - 4
f(D) = 3D - 4
where, D = x + 3
Since D is a dummy variable, it can be replaced by any other variable.
f(x) = 3x - 4 or, y = 3x - 4
To find f-1 (x),
interchanging the role of x and y, we get,
x = 3y - 4
or, y = x+4
3
x + 4
? f-1(x) = 3
If f(2x + 3) = 6x + 7, show that (fof-1) (x) is an identity function.
Here, f(2x + 3) = 6x + 7
x is replaced by x , we get,
2
( )f2. x =6× x 7
2 +3 2 +
or, f(x + 3) = 3x + 7
Again, x is replaced by x - 3, we get,
f(x) = 3(x - 3) + 7
or, f(x) = 3x - 2 or, y = 3x - 2
To find the f-1 (x), interchanging the role of x and y, we get,
x = 3y – 2
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y= x+2
3
( )Now, (fof-1) (x) = f (f-1 (x)) = f x
x+2 = 3. + 2 –2
3 3
= x + 2 - 2 = x.
Since (fof-1) (x) = x
Therefore, (fof-1)(x) is an identity function.
Exercise 1.3
Very Short Questions
1. Define one-to-one function with an example.
2. What is the required condition for existence of inverse of a function ?
3. Do all the functions have their inverses ?
4. Find the inverse functions of the following functions.
(a) f = {(a,x), (b,y), (c,z)}
(b) g = {(2,6), (3,9), (4, 12), (5, 15)}
(c) h = {(1,2), (2,3), (3,4), (4,5), (5,6)}
5. If f = {(2, 1), (4, 1), (5, 1)} Does f-1 exist ?
6. Let f : R →R be defined by f(x) = 2x - 3, find the formula that defines the inverse
function f-1.
7. Let f : R → R, then, find the inverse function in each of the following cases:
(a) f = {(x,y): y = 4x + 5}
{ }(b) g = 3x-1 3 ,
(x,y) : y = 3-2x , x≠ 2
{ }(c) h =
(x,y) : y = 4x+3 , x ≠ 1 ,
x-1
8. If f = {(a,x), (b,y), (c,z), (d,w)}, then, show f and f-1 in a mapping diagram.
Short Questions
9. (a) If g(x) = 3x - 5, the find g-1 (5), where g is a one to one onto function.
(b) If h(x) = 2x-3 , find h-1 (4), where h(x) is one to one.
4
2x+8
10. (a) If g(x) = 5 , find g-1(x) and g-1(10).
(b) If k(x) = 5x+7 , x ≠ 2, find k-1 (x) and k-1(4).
x-2
11. (a) If f(x) = 2x + k, f-1(4) = 20, find the value of k.
(b) If f(x) = 7x + k, f-1(3) = 12, find the value of k.
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12. (a) If f(x+2) = 4x + 5, find f-1(x)
(b) If f(2x + 5) = 10x + 7, find f-1(x)
(c) If f(4x + 5) = 12x + 20, find f-1 (x)
13. (a) If f(x + 2) = 7x + 13, find f-1(2)
(b) If f(x + 5) = 10x + 11, find f-1(4)
(c) If f(4x + 5) = 20x + 24, find f-1(x).
Long Questions
14. (a) If f and g are two one-to-one and onto functions defined by f(x) = 2x – 3 and
g(x) = 4x + 5, (fof)(x) = g-1 (x), find the value of x.
(b) If f(x) = 4x - 7 and g(x) = 3x - 5 are two one to one onto functions, and (fog-1)
(x) = 15, find the value of x.
(c) If f(x) = x -2 and g(x) = 1 and f-1(x) = (gof)(x), find the value of x.
2x +1 x
2x + 8
(d) If f(x) 4x-17 and g(x)= 5 and (fof)(x)= g-1(x), find the value of x.
15. (a) If f(x) = 2x - 4, then prove that, (fof-1)(x) is an identify function.
(b) If f(x) = 2x + 5 , x ≠ - 2, then, prove that (fof-1)(x) is an identity function.
x+2
2x + 1,
(c) If f(x) = 4 prove that (f-1of)(x) is an identify function.
16. If f(x) = x + 1 and g(x) = 3 - x , x ≠ 0, are two one-to-one onto function, then prove
that (f-1og-1)(2) = 0 x
17. If f = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}, g = {(4, 9), (5, 11), (6, 13), (7, 15), (8, 17)} and
3
(fog)–1(x) = x + 2, find the positive value of x.
Project Work
18. State the formula of conversion of temperature from degree Celsius to degree Fahrenheit
as a function. Write the inverse of the function. Convert 37°c into °F and vice versa.
2. One one and onto function 3. No.
4. (a) {(x, a), (y, b), (z, c)} (b) {(6, 2), (9, 3), (12, 4), (15, 5)}
(c) {(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)} 5. does not exist
6. x+3 7.(a) x–5 (b) 3x + 1 , x ≠ – 3
2 4 2x + 3 2
x+ 3 10 19
(c) x– 4 , x ≠ 4 9.(a) 3 (b) 2
10. (a) 21 (b) –15 11.(a) –36 (b) –81
(d) 6
12. (a) x+3 (b) x + 18 (c) x–5
4 5 3
3 43 x+1
13. (a) 7 (b) 10 (c) 5
14. (a) 31 (b) 23 (c) ± 1 17. 4
15 2
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vedanta Excel In Opt. Mathematics - Book 10 2
Polynomials
2.0 Review
Discuss the following :
(a) What is the difference between a polynomial and an algebraic expression ?
(b) Find the sum and difference of f(x) and g(x), when f(x) = 4x2 + 3x - 5 and
g(x) = 2x2 + 5x + 3.
(c) Are the sum f (x) + g(x) and difference f(x) - g(x) again a polynomials ?
2.1 Division of polynomials
Let f(x) = 7x3 + 8x2 - 5x + 10 be a polynomial. It is divided by x - 1. We have the process of
divisions as follows.
) (x-1 7x3 + 8x2 - 5x + 11 7x2 + 15x + 10
7x3 - 7x2
-+
15x2 - 5x
15x2 - 15x
-+
10x + 11
10x - 10
-+
21
Now, by division algorithm,
Dividend = Divisor × Quotient + Remainder
i.e. f(x) = (x - 1). Q(x) + R
Where, Q(x) = Quotient
R = Remainder
f(x) = Dividend.
If R = 0,
f(x) = (x - 1) × Q(x)
Then, (x - 1) is called factor of f(x).
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Worked out Examples
Example 1. If f(x) = x4 + 7x3 + x2 +2x + 7 is divided by d(x) = x2 + 4, find the quotient
Solution: Q(x), and Remainder R(x).
Example 2. Here, f(x) = x4 + 7x3 + 6x2 + 3x + 5
Solution: d(x) (x2 + 4)
) (Now, x2 + 4 x4 + 7x3 + 6x2 + 3x + 5 x2 + 7x + 2
2.1 x4 + 4x2
--
7x3 + 2x2 + 3x
7x3 + 28x
--
2x2 - 25x + 5
2x2 + 8
--
- 25x - 3
quotient = x2 + 7x + 2
remainder R (x) = - 25x - 3
If quotient Q(x) = x3 + 4x2 + 5 and remainder R(x) = x - 4 and divisor
d(x) = x2 + 3, find the polynomial.
Here, quotient, Q(x) = x3 + 4x2 + 5
divisor, d(x) = x2 + 3
remainder, R(x) = x - 4
Then, by division algorithm, we have,
dividend = divisor × quotient + remainder
i.e. f(x) = (x2 + 3) × (x3 + 4x2 + 5) + (x - 4)
= x5 + 4x4 + 5x2 + 3x3 + 12x2 + 15 + x - 4
= x5 + 4x4 + 3x3 + 17x2 + x + 11
Synthetic Division
Synthetic division is the process of division which is used to find the quotient and remainder
when a polynomial f(x) is divided by a linear polynomial x - a.
For an example, let f(x) = 4x3 + 3x2 - 7x + 5 be a polynomial and it is to be divided by x - 2
by division method, we have,
) (x - 2 4x3 + 3x2 - 7x + 5 4x2 + 11x + 15
-4x3 +- 8x2
11x2 - 7x
-1+11x52x+- 22x
+5
- 15x +- 30
35
? Divisor = d(x) = x - 2
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Quotient = Q(x) = 4x2 + 11x + 15
Remainder = R = 35
Again, polynomial f(x) can be divided by (x - 2) by writing the coefficients only.
We write the coefficients in descending order of variable x.
24 3 -7 5
30
8 22
4 11 15 35
coefficient of x2 remainder
coefficient of x
coefficient of x°
? Quotient = Q(x) = 4x2 + 11x + 15
Remainder R = 35
This process of division of a polynomial by a linear polynomial is called synthetic
division.
How do we operate synthetic division? Let us note the followings:
Following Steps are performed for synthetic division.
(i) Change the sign of the constant term in the linear divisor. For above example, x - 2 is
compared with x - a, we get a = 2. (in divisor),
(ii) Write down the coefficients of dividend f(x) with their signs. In above example. We
write 4, 3, -7, 5 as coefficients of x3, x2, x constant term respectively.
(iii) Bring down the leading coefficient (i.e., 4 in above example)
(iv) Multiply it (4) by the constant term of the divisor with sign changed in above example
4 is multiplied by 2, we get 8.
(v) Write the result under next coefficient (i.e., 3 in above example) and add them (we get
11 in above example)
(vi) Again, multiply the result obtained (i.e., 11 is multiplied by 2 to get 22).
(vii) Repeat the process until we get the last term.
The last number so obtained is the remainder. The number just before the last term is
constant term for quotient.
Note :
That the degree of quotient is 1 degree less than the dividend polynomial f(x), when f(x) is
divided by a linear factor. In above example, the degree of f(x) is 3 and that of divisor is 1.
(a) If the divisor is x+2, we write x+2 = x - (-2). Comparing it to x - a,
we get a = -3, i.e., the constant term of the divisor is -3 with sign changed.
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{ ( )}(b) 2
If the divisor is 3x + 2, we write, 3x + 2 = 3 x – 3 . Then divide the polynomial
( )first by x =–2 , the quotient obtained is divided by 3 to get actual quotient.
3
Worked out Examples
Example 1. What is degree of quotient when the polynomial of degree 4 is divided by a
Solution: linear divisor?
Example 2. Degree of polynomial = 4
Solution: Degree of linear polynomial = 1
Degree of quotient is less than degree of polynomial by 1.
Degree of quotient = 4 - 1 = 3.
Divide f(x) = 4x4 + 7x3 + 5x2 + 3x + 5 by linear polynomial d(x) = x - 2
Comparing x - 2 with x - a, we get a = 2,
Now, by using synthetic division method, we get,
2 47 53 5
8 30 70 146
4 15 35 73 151
quotient = Q(x) = 4x3 + 15x2 + 35x + 73
remainder = R = 151
Example 3. Find the quotient and remainder when f(x) = 2x3 + 7x2 + x - 15 is divided by
Solution: x + 2, by using synthetic division method.
Here, f(x) = 2x3 + 7x2 + x - 15
divisor x + 2 = x -(-2), comparing it with x - a, we get a = -2
Writing the coefficients in order,
-2 2 7 1 -15
-4 -6 10
23 -5 -5
Quotient = Q(x) = 2x2 + 3x - 5,
Remainder = R = -5
Example 4. Find the quotient and remainder when 2x3 - 9x2 + 5x - 5 is divided by 2x - 3.
Solution:
Here, dividend, f(x) = 2x3 - 9x2 + 5x - 5
28
( )divisor, d(x)= 2x-3=2 x - 3
x - a, we get a = 2
Comparing it with 3 .
2
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Writing the coefficients in order, we get
3 2 -9 5 -5
2 3 -9 -6
2 -6 -4 -11
quotient = Q(x) = 1 (2x2 - 6x - 4) = x2 - 3x - 2
2
remainder = R = -11
Example 5. Find the quotient and remainder when 4x3 + 2x2 - 4x + 3 is divided by 2x + 3.
Solution:
dividend = 4x3 + 2x2 - 4x + 3
Here, divisor = 2x + 3 = 2 {x – (– 3 )}, 3
2 2
The constant term in the divisor with sign changed is - .
Writing the coefficients in order, we get
-4 3
- 3 4 2 6 -3
2 -6
4 -4 20
Quotient = Q(x) = 1 (4x2 – 4x + 2) = 2x2 – 2x + 1
2
Remainder R = 0
Exercise 2.1
Very Short Questions
1. (a) If dividend = f(x), quotient = Q(x), divisior = d(x), and remainder = R, write their
relation.
(b) If x3 - 8 is divided by x - 2, find the quotient.
(c) Is division of a polynomial by a polynomial again a polynomial ? Divide x3 + 27
by x2 and justify your answer.
(d) What is the degree of quotient when polynomial of degree n is divided by a linear
factor?
2. (a) What should be the degree of divisor in synthetic division method?
(b) What should be the difference of degree of dividend and quotient in synthetic
division ?
3. (a) When f(x) = x4 + 3x3 + 7x2 + 4x + 7 is divided by d(x) = x - 2, write the degree
of quotient.
(b) If f(x) = (x-2) × (x2 + 4x + 2) + 3, write the degree of f(x).
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Short Questions
4. Divide f(x) by d(x) in each of the following cases. Also find remainder if exists. (Use
actual division method)
(a) f(x) = x3 + 3x2 + 5x + 6, d(x) = x + 3
(b) f(x) = x2 + 5x + 6, and f(x) = x + 3
(c) f(x) = 3x4 - 9x2 - 4 and d(x) = x - 3
(d) f(x) = 2x3 + 3x2 - 5x + 6, and d(x) = 2x - 3.
5. (a) If 2x3 - 11x2 + 20x - 15 = (x - 6). Q(x) + R(x). Find Q(x) and R(x).
(b) If 4x3 - 10x2 + 25x - 20 = (4x - 3). Q(x) + R(x). Find the value of Q(x) and R(x).
6. By using synthetic division, find the quotient and remainder when f(x) is divided by
d(x) in each of the following by synthetic division:
(a) f(x) = 3x3 - 2x2 + 13x + 10, d(x) = x - 2
(b) f(x) = 4x3 - x2 + 10x + 2, d(x) = x - 4
7. Find the quotient and remainder when f(x) is divided by d(x) in each of the following
by synthetic division:
(a) f(x) = x4 + 3x3 + 7x2 + 3x + 8, d(x) = x + 3
(b) f(x) = 2x3 + 4x2 + 3x + 7, d(x) = x + 2
8. Find the quotient and remainder in each of the following cases when f(x) divided by
d(x) (use synthetic division method):
(a) f(x) = 2x3 - 9x2 + 5x - 5, d(x) = 2x - 3.
(b) f(x) = 8x3 + 4x2 + 6x - 7, d(x) = 2x - 1.
(c) f(x)=3x3 - 3x2 + 2x + 5, d(x) = 2x – 1
(d) f(x)=4x3 -7x2 + 6x - 2, d(x) = 3x + 2
(e) f(x) = 4x3 - 3x2 + 7x + 8, d(x) = 2x + 3
(f) f(x) = 4x3 + 2x2 - 4x + 3, d(x) = 2x + 3
Long Questions
9. Let p(x) = x4 + x2 + 1, q(x) = x2 - x + 1, and r(x) = x2 + 2x + 5, find the value of
{p(x) ÷ q(x)} + r(x).
10. If p(x) = x8 + x4 + 1, q(x) = x4 + x2 + 1, and r(x) = x4 + x2 + 6, then, find the value of
{p(x) ÷ q(x)} + r (x).
11. (a) When the polynomial 2x3 + 9x2 - 7x + k, is exactly divisible by 2x + 3, find the
value of k by using synthetic division.
(b) When the polynomial 8x3 + 4x2 + 6x + p is exactly divisible by 2x - 1, find the
value of p by using synthetic division.
12. Divide f(x) = x3 + 2x2 – 5x – 6 by (x + 1) and (x + 3) by using synthetic division. On
the basis of remainder on the division process. Draw your conclusion.
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1. (a) f(x) = d(x) . Q(x) + R (b) x2 + 2x + 4 (c) No (d) n – 1
2. (a) 1 (b) 1 3.(a) 3 (b) 3
4. (a) x2 + 5, – 9 (b) x + 2, 0
(c) 3x3 + 9x2 + 18x + 54, 158 (d) x2 + 3x + 2, 12
5. (a) 2x2 + x + 26, 141 (b) x2 – 7x + 79 , – 83
4 16 16
6. (a) 3x2 + 4x + 21, 52 (b) 4x2 + 15x + 70, 282
7. (a) x2 + 7x – 18, 62 (b) 2x2 + 3, 1
8. (a) x2 – 3x – 2, –11 (b) 4x2 + 4x + 5, – 2
(c) 3x2 – 3x + 5 , 45 (d) 4x2 – 29x + 112 , – 278
2 4 8 8 3 9 27 27
9x 41 91
(e) 2x2 – 2 + 4 , – 4 (f) 2x2 – 2x + 1, 0
9. 2x2 + 3x + 6 10. 2x4 + 7
11. (a) – 24 (b) –5
12. Both are factors of polynomial f(x).
2.3 Remainder Theorem
Let f(x) = x2 + 4x + 5 be divided by x-2. Find the remainder and the value of f(x) at x = 2.
Are they equal?
Obviously, the answer is yes.
Hence, a remainder can be calculated when a polynomial f(x) is divided by linear polynomial
without actual division or synthetic division.
Remainder Theorem
Statement : If a polynomial f(x) is divided by x-a, then the remainder is R = f(a).
Proof : Let Q(x) be the quotient when the polynomial f(x) is divided by (x-a), R be the
remainder independent of x.
Then by division algorithm, we get,
f(x) = (x-a) × Q(x) + R
This is an identity and is true for all values of x. So, put x = a, we get
f(a) = (a -a) × Q(a) + R
f(a) = R
Hence, the remainder is R = f(a). Proved.
Note :
If a polynomial f(x) is divided by (x+a), then, the remainder is R = f(-a).
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Extension of the Remainder Theorem.
( )Statement : If a polynomial f(x) is divided by (ax-b), then, the remainder is R = f ba .
( )Proof : We have ax - b = a x -
b .
a
Let Q(x) be the quotient when f(x) is divided by (ax-b) and the remainder R.
Then, by division algorithm,
f(x) = (ax - b) Q(x) + R
( )or, x- b
f(x) = a a Q(x) + R
Putting, x = b , we get,
a
f( b ) = a ( b - b ). ( )b +R
a a a
a
( )R = f b .
a
( )Hence, the remainder is R = f b = value of f(x) at x = b .
a a
Note :
( )If a polynomial f(x) is divided by (ax+b), the remainder is R = f - b .
a
Remainder Theorem Table
Dividend Divisor Remainder
f(x) x-a f(a)
f(x) x+a f(-a)
f(x) ax-b f( b )
a
f(x) ax+b f(- b )
a
Worked out Examples
Example 1. What is the remainder when polynomial f(x) is divided by x-5? Use the
Solution: remainder theorem.
Dividend = f(x)
Divisor = x - 5
Comparing x - 5 with x - a, we get a = 5 by the remainder theorem, R = f(5).
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Example 2. By using the remainder theorem, find the remainder when 4x3 + 6x2 - 5x + 2
Solution: is divided by x - 2.
Example 3. Here, f(x) = 4x3 + 6x2 - 5x + 2 = dividend, (x - 2) = divisor, Comparing
Solution:
it with x-a,
Example 4.
Solution: we get a = 2
By using the remainder theorem,
Example 5.
Solution: R = f(a) = 4.23 + 6.22 - 5.2 + 2
= 32 + 24 - 10 + 2 = 58 - 10 = 48
By using the remainder theorem, find the remainder when 4x4 + x3 + 2x - 50
is divided by 2x + 1.
Let, f(x) = dividend = 4x4 + x3 + 2x - 50
divisor = 2x + 1 = 2{x – (– 1 )}
2
By the remainder theorem,
( )Remainder – 1
=R=f 2
= 4(– 1 )4 + (– 1 )3 +2 (– )1 - 50
2 2
1 1 2
4 8
= – – 51
= 2- 1 - 408 =- 407
8 8
Find the value of k when x3 + 3x2 - kx + 4 is divided by x + 2 if remainder
is k(use the remainder theorem).
Let, f(x) = x3 + 3x2 - kx + 4, dividend
d(x) = x + 2 = x - (-2), divisor.
By remainder theorem,
R = f(-2) = (-2)3 + 3(-2)2 - k(-2) + 4
= -8 + 12 + 2k + 4 = 8 + 2k
By question,
R = 8 + 2k
or, k = 8 + 2k
? k = -8
If f(x) = 4x4 - 8x2 + 3x - 20 is divided by g(x) = x + 2, prove that the remainder
is 6.
Here, f(x) = 4x4 - 8x2 + 3x - 20, dividend
g(x) = x + 2, divisor
= x - (-2), comparing it with x - a,
we get a = -2
Now, by using remainder theorem,
R = f(-2) = 4(-2)4 - 8(-2)2 + 3(-2) - 20
= 64 - 32 - 6 - 20 = 64 - 58 = 6
R = f(-2) = 6 Proved.
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Exercise 2.2
Very Short Questions
1. (a) State the remainder theorem.
(b) What is the remainder when f(x) is divided by x - b (use remainder theorem)?
(c) What is remainder when polynomial f(x) is divided by ax + b?
(d) What is the remainder when the polynomial p(x) is divided by mx + n2?
2. Find the remainder in each of the following when f(x) is divided by d(x), by using
remainder theorem:
(a) f(x) = x3 + 3x2 + 9x - 7 d(x) = x - 2
(b) f(x) = x3 + 4x2 - 7x - 20 d(x) = x - 4
(c) f(x) = 2x3 + 3x2 - 8x + 50 d(x) = x + 3
(d) f(x) = x3 + 4x2 – 20x + 20 d(x) = x + 2
(e) f(x) = 2x3 + 4x2 - 10x + 20 d(x) = 2x - 1
(f) f(x) = 16x3 + 12x2 + 24x + 5 d(x) = 2x + 3
3. Find the value of p in each of the following when f(x) is divided by d(x), with the help
of remainder theorem:
(a) f(x) = px3 + 4x - 10 d(x) = x + 3 R = 5
(b) f(x) = x3 + 3x2 - px + 4 d(x) = x - 2 R = 8
(c) f(x) =2x3 - 4x2 + 6x - p d(x) = x-2 R = 18
(d) f(x) = 4x3 - 3x2 + 3x - p d(x) = x - 2 R = 12
(e) f(x) = 2x3 - 3x2 + px - 8 R = f(3) = 10
(f) f(x) = x3 + 5x2 - px + 6 R = 2p d(x) = x - 1
(g) f(x) = x4 + x3 + =px2 + x + 20 f(2) = 20
Long Questions
4. (a) If 4x2 - 6x + p and x3 - 2x2 + px + 7 are polynomials, they are divided by x + 2
and remainders are equal. What is the value of p ?
(b) If x3 - px2 + 10x + 11 and 2x3 - 2px2 + 7px + 23 are divided by x - 1, and each
gives equal remainder, find the value of p.
(c) If x3p2 + x2p - 8 is divided by x - 1, then, remainder is -2, find the values of p.
(d) Find the value of p in polynomial 2x2p2 - 5px + 3 when it is divided by x-1 and
remainder is zero.
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1. (b) f(b) (c) f – b (d) f – n2
a m
2. (a) 31 (b) 80 (c) 47 (d) 68
(e) 65 (f) –58 3.(a) –1 (b) 8
4
13
(c) –6 (d) 14 (e) –3 (f) 4 (g) – 2
4. (a) – 37 (b) – 1 (c) 2, –3 (d) 1, 3
3 2 2
2.4 Factor Theorem
(a) If f(x) = x2 - 25, then find the values of f(x) for x = 5 and x = -5 i.e. f(5) and f(-5). Discuss
the following:
(b) Is each of f(5) and f(-5) zero?
(c) What are the factors of f(x)?
We note that if f(5) = 0, then, (x-5) is a factor of f(x), and f(-5) = 0, (x+5) is also a factor of
f(x). We say that (x+5) and (x-5) are the factors of f(x). This is called factor theorem.
Factor Theorem
Statement : If a polynomial f(x) in x is divided by (x-a) and the remainder
Proof : R = f(a) = 0, then, (x-a) is a factor of f(x).
Let Q(x) be the quotient when f(x) is divided by (x-a) and R is the
Corollary (I) : remainder. Then, division algorithm, we get
Corollary (II) : f(x) = (x-a) Q(x) + R
If remainder R = f(a) = 0, then, (x-a) is a factor of f(x)
If a polynomial f(x) in x is divided by (x+a) and the remainder
R = f(-a) = 0, then, (x + a) is a factor of f(x)
If a polynomial f(x) in x is divided by ax + b and the remainder
–( )R = f b = 0, then, ax + b is a factor of f(x).
a
Converse of Factor Theorem
If (x-a) is a factor of f(x), then, R = f(a) = 0
Proof : Let Q(x) be the quotient when a polynomial f(x) is divided by (x-a).
Since (x-a) is a factor of f(x), we write by division algorithm,
f(x) = (x-a).Q(x)
or, f(x) = (x-a). Q(x) + 0
Also, f(x) = (x - a). Q(x) + R
Comparing these two, we get R = 0. Proved.
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Worked out Examples
Example 1. Examine that (x-3) is a factor of f(x) = x3 - 27.
Solution:
Here, f(x) = x3 - 27
Example 2. comparing (x-3) with (x-a),
Solution:
Example 3. we get a = 3,
Solution: Now, f(3) = 33 - 27
Example 4. R = f(x) = 0
Solution: By factor theorem, (x-3) is a factor of f(x).
36 Show that (x-3) is a factor of x3 - 6x2 + 11x - 6.
Let f(x) = x3 - 6x2 + 11x - 6
comparing x - 3 with x-a,
we get a = 3
f(3) = 33 - 6.32 + 11.3 - 6 = 27 - 54 + 33 - 6 = 60 - 60 = 0
Since R = f(3) = 0, (x-3) is 6 factor of f(x)
What must be added to f(x) = x3 - 6x2 + 10x + 8, so that (x-3) is a factor of
f(x)?
Here, f(x) = x3 - 6x2 + 10x + 8.
Let k be added, f(x) = x3 - 6x2 + 10x + 8 + k.
comparing x - 3 with x - a, we get x = 3
f(3) = 33 - 6.32 + 10.3 + 8 + k = 27 - 54 + 30 + 8
= 65 - 54 + k = 11 + k
By factor theorem,
f(3) = 0
or, 11 + k = 0
k = –11
Therefore, -11 must be added.
What must be subtracted from 3x3 + 5x2 - 6x + 7 so that x + 2 is a factor of
it?
Let f(x) = 3x3 + 5x2 - 6x + 7
Let k be subtracted from f(x).
Then f(x) = 3x3 + 5x2 - 6x + 7 - k
comparing (x+2) with x - a, we get a = -2
Now, f(-2) = 3(-2)3 + 5(-2)2 - 6(-2) + 7 - k
= -24 + 20 + 12 + 7 - k = 15 - k
Since, (x + 2) is a factor of f(x), by factor theorem, we get,
R = f(-2) = 0
or, 15 - k = 0
? k = 15 ? 15 should be subtracted.
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Application of Factor Theorem and Remainder Theorem
Factor theorem is used to factorize polynomials. Remainder theorem is used to find remainder
when a polynomial is divided by linear polynomial without actual division. If remainder
R = f(a) = 0, for real value of a, then, (x-a) is a factor of f(x).
Example 5. Factorize f(x) = x3 - 3x2 - 6x + 8.
Solution:
Here, f(x) = x3 - 3x2 - 6x + 8
The possible factors of constant (8) are ± 1, ±2, ±4, ±8
For x = 1, f(1) = 13 - 3.12 - 6.1 + 8
=9-9=0
Hence, (x-1) is a factor of f(x).
Now, f(x) = x3 - 3x2 - 6x + 8 = x3 - x2 - 2x2 + 2x – 8x + 8
= x2(x-1) - 2x(x-1) - 8(x-1) = (x – 1) (x2 – 2x – 8)
= (x-1) (x2 - 4x + 2x - 8) = (x-1) {x(x-4) + 2(x-4)}
? f(x) = (x-1) (x-4) (x+2)
Alternate Method :
Here, f(x) = x3 - 3x2 - 6x + 8
f(1) = 13 - 3.12 - 6.1 + 8
=9-9=0
Hence, (x-1) is a factor of (x)
Now, by using synthetic division.
Writing the coefficients of x in order.
1 1 -3 -6 8
1 -2 -8
1 -2 -8 0
Example 6. Quotient, Q(x) = x2 - 2x - 8 = x2 - 4x + 2x - 8
Solution: = x(x-4) + 2(x-4) =(x-4) (x+2)
? f(x) = (x - 1) (x + 2) (x - 4)
Factorize : 2x3 + 3x2 - 11x - 6
Let, f(x) = 2x3 + 3x2 - 11x - 6
The possible factors of constant term (-6), are ± 1, ±2, ±3, ±6
For x = 2,
f(2) = 2.23 + 3.22 - 11.2 - 6 = 16 + 12 - 22 - 6
= 28 - 28 = 0
Hence (x - 2) is a factor of f(x)
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Now, by using the synthetic division, we get
22 3 -11 -6
4 14 6
27 30
Quotient, Q(x) = 2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x+3) + 1(x + 3) = (x + 3) (2x + 1)
? f(x) = (x-2) (x+3) (2x + 1).
Example 7. If f(x) = x3 + px2 - 6x + 8 has a factor (x-1), find the value of p. Then,
Solution: factorize f(x) completely.
Example 8. Since (x-1) is a factor of f(x),
Solution: f(1) = 0
i.e. 12 + p.12 - 6.1 + 8 = 0
or, 1 + p - 6 + 8 = 0
or, p + 3 = 0
? p = -3
Now, f(x) = x3 - 3x2 - 6x + 8 = x3 - x2 - 2x2 + 2x - 8x + 8
= x2 (x - 1) - 2x(x-1) -8(x-1) = (x-1) (x2 - 2x - 8)
= (x-1){x(x-4) + 2(x-4)} = (x-1) (x+2) (x-4)
Factorize (x + 1) (x + 3) (x + 5) (x + 7) + 16
Here, (x + 1) (x + 3) (x + 5)(x + 7) + 16
= {(x + 1)(x + 7)} {(x +3)(x + 5)} + 16
= (x2 + 8x + 7)(x2 + 8x + 15) + 16
= (D + 7) (D + 15) + 16, where D = x2 + 8x
= D2 + 22x + 105 + 16
= D2 + 2.D.11 + (11)2
= (D + 11)2
= (x2 + 8x + 11)2 (replacing D = x2 + 8)
= (x2 + 8x + 11) (x2 + 8x + 11)
2.5 Polynomial Equation
Let f(x) = anxn + an-1xn-1 + .... + a0, an ≠ 0 be a polynomial of degree n of x. Then, f(x) = 0 is
called the polynomial equation of degree n in x.
In above polynomial an, an-1, .... a0 R.
For examples,
ax + b = 0, linear equation
ax2 + bx + c = 0, quadratic equation.
ax3 + bx2 + cx + d = 0, cubic equation.
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If k be a real number such that f(k) = 0, k is called a root of the polynomial equation
f(x) = 0, k is also called zero of polynomial.
Remember that the degree of a polynomial, the highest exponent, dictates the maximum
number of roots it can have.
Example: f(x) = 2x3 + 13x2 - 36 = 0
Here, f(x) is of degree 3. Hence, it may have maximum three roots. It has roots -2, -6, 3
2
Example 9. Solve x3 - 13x - 12 = 0
Solution: Let, f(x) = x3 - 13x - 12 = 0
for x = - 1,
f(-1) = (-1)3 - 13(-1) -12) = -1 + 13 - 12 = 13 - 13 = 0
Since f(-1) = 0, by factor theorem, (x + 1) is a factor of f(x).
Now, f(x) = x3 + x2 - x2 - x - 12x - 12
= x2 (x + 1) - x(x + 1) - 12(x + 1)
= (x +1)(x2 - x - 12) = (x + 1)(x2 - 4x + 3x - 12)
=(x +1) {x(x-4) + 3(x-4)} = (x + 1) (x - 4) (x + 3)
To solve, f(x) = 0 Alternative method
or, (x + 1) (x - 4) (x + 3) = 0 Here, f(–1) = 0, x + 1 is a factor of f(x)
by factor theorem.
Either x + 1= 0....... (i)
By using synthetic division method.
or, x - 4 = 0..... (ii)
-1 1 0 -13 -12
or, x + 3 = 0 .......(iii)
From equation (i), x = -1 -1 1 12
From equation (ii), x = 4 1 -1 -12 0
From equation (iii), x =-3 Now, f(x) = 0
? x = -3, -1, 4 or, (x + 1) (x2 – x – 12) = 0
Example 10. Solve: x3 - 9x2 + 24x - 20 = 0 or, (x + 1) (x2 – 4x + 3x – 12) = 0
Solution: or, (x + 1) (x – 4) (x + 3) = 0
Let f(x) = x3 - 9x2 + 24x - 20 =0 ? x = –3, –1, 4
For x =2,
f(2) = 23 - 9.22 + 24.2 - 20 = 8 - 36 + 48 - 20
= 56 - 56 = 0
Hence, by factor theorem, (x-2) is a factor of f(x).
Now, f(x) = x3 - 9x2 + 24x - 20 = x3 - 2x2 - 7x2 + 14x + 10x - 20
= x2(x -2) - 7x (x -2)+ 10(x-2) = (x-2) (x2 - 7x + 10)
= (x-2) (x2 -5x - 2x + 10) = (x-2) {x(x-5) -2(x-5)}
= (x-2)(x-5)(x-2)
To solve, f(x) = (x-2)2 (x-5) = 0
Either (x - 2)2 = 0 .......(i)
x-5 = 0 ........ (ii)
From equation (i),
x=2
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From equation (ii),
x=5
? x = 2, 5
Example 11. If (x-1) is a factor of x3 + ax2 - x + b and leaves remainder 15 when divided
by x - 2. Find the values of a and b.
Solution: Let f(x) = x3 + ax2 - x + b
Since (x-1) is a factor of f(x),
f(1) = 0
i.e., 13 + a.12 - 1 + b = 0
or, a + b = 0 ...............(i)
Again, when f(x) is divided by x - 2, remainder (R) = 2
i.e., f(2) = 15
or, 23 + a.22 - 2 + b = 15
or, 4a + b = 9 .......... (ii)
Solving equation (i) and (ii), we get,
a+b=0
Subtracting -4a +- b =-9
- 3a = -9
? a=3
Putting the value of a in equation (i),
3+b=0
b = -3
? a = 3, b = -3
Exercise 2.3
Very Short Questions
1. (a) State factor theorem.
(b) Is there any relation between the factor theorem and remainder theorem when a
polynomial is factorized ?
(c) If f(x) = (x - a) × Q(x) + R and x - a is a factor of f(x), what is the value of R?
(d) If (x-a) is a factor of f(x), what is the value of R, f(a)?
( )(e) If f – b = 0 and f(x) is a polynomial, find one of its factor.
a
( )(f) – 2 = 0. What is one of factor of f(x) ?
If f(x) is a polynomial and f 3
2. (a) If (x-k) is a factor of f(x) = x2 - 9 What is the value of f(k) ?
(b) If (x - a ) is a factor of f(x) = x2 - 121, what is the value of a?
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(c) If (x - a) is a factor of f(x) = x3 - 27, find the value of a.
3. (a) If (x + 2) is a factor of f(x) = x3 - 8, find f(-2).
(b) If (x - 3) is a factor of f(x) = x3 - 27, find the value of f(3).
(c) If f(x) = 2x3 + 3x2 + - 11x - 6 has a factor (x-2), find f(2).
4. (a) By using factor theorem check, that (x - 5) is a factor of 2x2 - 11x + 5 or not.
(b) By using factor theorem check, that (x-3) is a factor of x3 - 4x2 + x + 6 or not.
(c) Show that (x + 1) is a factor of x3 - 4x2 + x + 6.
5. (a) What is meaning of roots of a polynomial equation f(x) = 0?
(b) What is zero of polynomial ?
(c) What is a polynomial equation ?
(d) What is the difference between a zero polynomial and the zero of a polynomial ?
(e) Give an example of each of zero polynomial and zero of polynomial.
(f) What are the maximum number of roots of polynomial equation f(x) = 0, whose
degree is 3 ?
Short Questions
6. (a) What must be added to f(x) = 2x3 + 6x2 + 4x + 8 so that (x + 3) is a factor of it ?
(b) What must be added to f(x) = x3 - 6x2 + 11x + 8 so that (x - 3) is a factor of it ?
(c) What must be subtracted from f(x) = x3 + 8x2 + 4x + 10 so that (x+2) is a factor of it ?
7. (a) If 4x2 + px + 8 has a factor x - 2, find the value of p.
(b) If 2x3 + 4x2 + px + 7 has a factor of (x + 2), find the value of p.
(c) If f(x) = x3 + 4x2 + kx + 7 has a factor (2x - 1), find the value of k.
8
(d) If x3 - bx2 - 2x + b + 5 has a factor (x - b), find the value of b.
(e) If f(x) = 3x3 + 3x + k and f(-1) = 0, find the value of k.
Long Questions
8. (a) By using the factor theorem, show that (x + 1), (x-2), and (x-3) are the factors of
f(x) = x3 - 4x2 + x + 6.
(b) By using the factor theorem, show that (x + 2), (x + 6), and (2x - 3) are the factors
of f(x) = 2x3 + 13x2 - 36.
(c) If f(x) = 6x3 + 11x2 -26x - 15, determine which are the factors of f(x) given below:
(i) x + 1 (ii) (x + 3) (iii) 2x + 1
(iv) 3x-5 (v) 2x - 1
9. Factorize :
(a) x2 + 3x + 2 (b) 3x2 + 8x + 4 (c) x2 - 5x + 6
(d) x2 - 4x + 3 (e) x2 - 6x + 9 (f) x2 + 4x - 12
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10. Factorize by using the factor theorem :
(a) x3 - 6x2 + 11x - 6 (b) x3 - 4x2 + x + 6 (c) x3 - 11x2 + 31x - 21
(d) x3 - 3x2 + 4x - 4 (e) x3 + 2x2 - 5x - 6 (f) 3x3 - 6x2 + 4x - 8
(g) 6x3 + 7x2 - x - 2 (h) 2x3 + 3x2 - 11x - 6 (i) x3 + 5x2 - 2x - 24
(j) x3 - 9x2 + 26x - 24 (k) x3 - 9x2 + 23x - 15 (l) x3 - 19x - 30
(m) x3 - 5x + 4 (n) 2x3 + 3x2 - 11x - 6
11. Factorize :
(a) (x + 3) (x + 5) (x + 7) (x + 9) + 16 (b) (x + 2)(x - 3) (x - 1) (x - 6) + 56
(c) (x - 1) (2x2 + 5x + 5) + 4 (d) (x - 1) (2x2 + 15x + 15) - 21
(e) (x - 3) (x2 - 5x + 8) - 4x + 12
12. Solve the following polynomial equations :
(a) x3 - 3x2 - 4x + 12 = 0 (b) x3 - 9x2 + 24x - 20 = 0
(c) x3 - 7x2 + 7x + 15 = 0 (d) x3 - 4x2 + x + 6 = 0
(e) 3x3 - 19x2 + 32x - 16 = 0 (f) 6x3 + 7x2 - x - 2 = 0
(g) x3 - 6x2 + 11x - 6 = 0 (h) 5x3 + 2x2 - 20x - 8 = 0
(i) y3 - 13y - 12 = 0 (j) y3 - 3y - 2 = 0
(k) 4y3 - 3y - 1 = 0 (l) y3 - 19y - 30 = 0
(m) x4 - x3 - 19x2 + 49x - 30 = 0 (n) 2x4 - 13x3 + 28x2 - 23x + 6 = 0.
13. Find the values of a and b in each of the following cases :
(a) (x - 1) and (x - 2) are the factors of x3 - ax2 + bx - 8.
(b) (x + 3) and (2x - 7) are factors of ax2 - bx - 21.
(c) 2x3 + ax2 + bx - 2 has a factor (x + 2) and leaves remainder 7 when divided by
2x - 3.
(d) 2x3 + ax2 - 11x + b leaves a remainder 0 and 42 when divided by (x-2) and (x-3)
respectively.
Project Work
14. Write two functions f(x) and g(x) of degree 4 both of which have two common factors.
How many factors are there in each of the functions f(x) and g(x) ?
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1. (c) 0 (d) 0 (e) ax + b (f) 3x + 2
2. (a) ± 3 (b) ± 11 (c) 3 3.(a) 0
(b) 0 (c) 0 4.(a) yes (b) yes
5. (f) 3 6.(a) 4 (b) –14 (c) 26
7. (a) –12 (b) 7 (c) –4 (d) 5 (e) 6
8. (c) (ii), (iii), (iv) 2
9. (a) (x + 1) (x + 2) (b) (x + 2) (3x + 2) (c) (x - 2) (x - 3)
(d) (x - 3) (x - 1) (e) (x - 3) (x - 3) (f) (x + 6) (x - 2)
10. (a) (x - 1) (x - 2) (x - 3) (b) (x - 2) (x - 3) (x + 1)
(c) (x - 1) (x - 3) (x - 7) (d) (x - 2) (x2 - x + 2)
(e) (x + 1) (x - 2) (x + 3) (f) (x - 2) (3x2 + 4)
(g) (x + 1) (2x - 1) (3x + 2) (h) (x - 2) (x + 3) (2x + 1)
(i) (x - 2) (x + 3) (x + 4) (j) (x - 2) (x - 3) (x - 4)
(k) (x - 1) (x - 3) (x - 5) (l) (x + 2) (x + 3) (x - 5)
(m) (x - 1) (x2 + x - 4) (n) (x - 2) (x + 3) (2x + 1)
11. (a) (x2 + 12x + 31)2 (b) (x - 5) (x + 1) (x2 - 4x - 4)
(c) (x + 1)2 (2x - 1) (d) (x + 2) (x + 6) (2x - 3)
(e) (x - 1) (x - 3) (x - 4)
12. (a) ±2, 3 (b) 2, 5 (c) -1, 3, 5 (d) -1, 2, 3
(e) 1, 4, 4 (f) -1, - 23, 1 (g) 1, 2, 3 (h) ±2, - 2
3 2 5
1
(i) -1, 4, -3 (j) -1, 2 (k) 1, - 2 (l) -2, -3, 5
(m) 1, 2, 3, -5 (n) 1, 2, 3, 1
13. (a) 7, 14 2
(b) 2, 1 (c) 3, -3 (d) 3, -6
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vedanta Excel In Opt. Mathematics - Book 10 3
Sequence and Series
3.0 Review
Study the following pattern of numbers.
(i) 1, 2, 3,4, 5, .................
(ii) 5, 10, 15, 20, 25 ............................, 100.
(iii) 2, 4, 8, 16, 32,.........................
(iv) 1, 2,4, 7, 11, 16
Now, answer the following questions:
(a) Does the each of above pattern of number follow a certain rule?
(b) What are the rules followed by each pattern of numbers?
(c) Can we find next three terms in each pattern of numbers?
(d) Can we find the nth term in each of these cases?
(e) Among them, which are finite and which are infinite?
(f) What is the series of each of above pattern of numbers that follows a certain rule?
Discuss answers of above questions and arrive at the conclusions.
Definition: A set of numbers presented in a definite order and formed according to a certain
rule is known a sequence.
Each number of a sequence is called term of the sequence. The terms of a sequence are
denoted by t1, t2, t3, ......, tn in case of finite sequence. The nth term of the sequence is called
the general term. The series of a sequence t1, t2, t ..... . tn is written as t1 + t2 +..... + tn.
3
Progression : A sequence (or series) is called a progression if the functional relations
exist between its two successive terms is constant.
Examples : (i) 3 + 6 + 9 + 12,........ series
(ii) 7, 9, 11, 13, ...... sequence
(iii) 2, 4, 8, 16, …, sequence.
All of these three are examples of progression, why?
3.1 Arithmetic Sequence and Series.
Let us consider the following sequence of number.
(i) 1, 5, 9, 13, 17, ......
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(ii) 100, 95, 90, 85, 80,.....
Each of above sequence of numbers are in a certain rule.
In sequence (i), each term is obtained by adding 4 on the preceding term. Difference between
a term and its preceding term is constant throughout the whole sequence.
i.e., 5 - 1 = 4, 9 - 5 = 4, 13 - 9 = 4, 17 - 13 = 4,........, etc.
The difference is called common difference and it is denoted by d. This type of sequence
is called arithmetic sequence and the corresponding series is called arithmetic series(A.S).
i.e., 1 + 5 + 9 +13 + 17 ...... is an arithmetic series.
In sequence (ii) each term is obtained by subtracting 5 from the preceding term. Common
difference is -5. Arithmetic series of above sequence is given by
100 + 95 + 90 + 85 + 80 + ...................
Definition : A sequence of numbers with a constant difference between two consecutive
terms is called an arithmetic sequence. The constant difference is called common difference
and it is denoted by d. If ‘a’ is the first term and common difference d, a, a + d, a + 2d, a +
3d ...... are the terms of an arithmetic sequence.
Common difference (d) = tk+1 - tk , k ≥ 1, k N, for k = 1, d = t2 - t1
General Term of Arithmetic Sequence
Let us consider an arithmetic sequence 5, 8, 11, 14, 17,.....
In this sequence,
Common difference (d) = 8-5 = 3
The first term (t1) = a = 5
Now, the first term (t1) = 5 + (1-1) 3 = a + (1 - 1)d
the second term (t2) = 8 = 5 + (2-1)3 = 5 + (2-1)d
the third term (t3) = 11 = 5 + (3-1)3 = a + (3-1)d
the nth term (tn) = a + (n-1)d = 5 + (n - 1)3 = 3n + 2
The general term (tn) = a + (n - 1)d, where n is the number of terms.
If ‘l’ is the last term of an arithmetic sequence, we write tn = l = a + (n - 1)d.
Worked Out Examples
Example 1. Show that each of the sequence is in arithmetic:
Solution:
(a) 2, 4, 6, 8, 10, 12..... (b) 120, 100, 80, 60,.....
(a) In 2, 4, 6, 8, 10,12
Here, t2- t1= 4-2=2,t3-t2=6-4= 2, t4-t3=8-6=2t5, - t4=10-8 = 2,
t6 - t5=12 – 10 =2
Therefore, the difference between any two successive terms is 2 throughout
the whole sequence.
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Given sequence is arithmetic,
Here, 120, 100, 80, 60,......
(b) t2 - t1 = 100 - 120 = -20
t3 - t2 = 80 - 100 = -20
t4 - t3 = 60 - 80 = -20
Therefore, the difference between any two successive terms is -20 throughout
the whole sequence. Given sequence is arithmetic.
Example 2. Find the general term of the given arithmetic sequences:
Solution:
(a) 7, 14, 21, 28,....... (b) 10, 25, 40, ......
Example 3.
Solution: (i) Here, the first term (a) = 7
Example 4. The common difference (d) = t2 - t1 = 14 - 7 = 7
Solution: The general term (tn) = a + (n - 1)7 = 7 + 7n - 7 = 7n
(ii) Here, the first term (a) = 10
The common difference (d) = t2 - t1 = 25 - 10 = 15
Now, the general term (tn) = a + (n - 1) d = 10 + (n - 1) 15
= 10 + 15n - 15 = 15n - 5
? tn = 5(3n - 1)
Find the next two terms of given arithmetic sequence.
1, 4, 7, 10, 13,........
In given arithmetic sequence,
the first term (a) = 1
the common difference (d) = 4 - 1 =3
By using formula,
tn = a + (n - 1) d
or, t6 = 1 + (6 - 1).3 = 1 + 15 = 16
and t7 = 1 + (7 - 1) 3 = 19
Find the common difference of an arithmetic sequence whose first term is 9
and fifth term is 17.
Here, the first term (t1) = 9
the fifth term (t5) = 17
the common difference (d) = ?
By using formula,
tn = a + (n - 1) d
or, t5 = 9 +(5 - 1) .d
or, 17 = 9 + 4d
or, 4d = 8
? d=2
Therefore, the common difference (d) is 2
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Example 5. Find the number of terms in the series 13 + 16 + 19 + ........... + 79
Solution:
Here, the first term (a) = 13
Example 6.
Solution: the common difference (d) = 16 - 13 = 3
Example 7. the last term (tn) = 79
Solution: By using formula,
tn = a + (n-1)d
or, 79 = 13 + (n - 1) 3
or, 66 + 3 = 3n
or, 3n = 69
or, n= 69 =23.
3
Therefore, number of terms (n) is 23.
If the 3rd and 8th terms of an A.P. are 14 and 34 respectively, find the first
term, the common difference, and the 12th term of the sequence.
In given A.P., ( tn = a + (n – 1)d)
the third term (t3) = 14
or, a + 2d = 14 .......... (i)
and the 8th term (t8) = 34
or, a + 7d = 34.......... (ii)
Subtracting (i) from (ii) we get,
a + 7d = 34
–a +– 2d =–14
5d = 20
?d = 2v50alu=e 4 d in equation (i), we get,
Putting the of
a + 2 × 4 = 14
or, a + 8 = 14 or, a = 14 - 8
⸫ a=6
Again, t12 = a + (12 - 1)d = 6 + 11 (4) = 6 + 44 = 50
If 5th term of an AP is 28 and the 9th term is 48. Are 63 and 95 the terms of the
AP?
In given AP,
the fifth term (t5) = 28
and the ninth term (t9) = 48
Then, by using formula,
tn = a + (n - 1)d
or, t5 = a + 4d.
28 = a + 4d ...........(i)
and t9 = a + 8d
or, 48 = a + 8d ....... (ii)
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Subtracting (i) from (ii), we get,
48 = a + 8d
–28 =–a +– 4d
20 = 4d
? d=5
Putting the value of ‘d’ in (i), we get,
28 = a + 4 × 5
or, 28 = a + 20 or, a =28 - 20 = 8
Let tn = 63 if possible, then find n.
or, a + (n - 1) d = 63
or, 8 + (n - 1).5 = 63 or, (n - 1) . 5 = 55
or. n -1 = 11 ? n = 12
Therefore, 63 is the 12th term of the A.P.
Again, let tm = 95. If possible, then find the value of m.
tm = a + (m - 1) d.
or, 95 = 8 + (m-1).5
or, 95 - 8 = (m - 1) 5
or, m857==85m7 -1 = 87 + 5 = 92 = 18 2 .
or. +1 5 5 5
The number of terms cannot be in fraction.
Therefore, 95 is not the term of the given A.P.
Example 8. The nth term of the series 9 + 7 + 5 + ....... is equal to the nth term of the series
Solution: 15 + 12 + 9 + ........, find the value of n.
For the first series, 9 + 7 + 5 + .....
the first term (a) = 9
the common difference (d) = 7 - 9 = - 2
tn = a + (n - 1) d = 9 + (n - 1) (-2)
= 9 - 2n + 2 = 11 - 2n
for the second series 15 + 12 + 9 + .......
the first term (a) = 15
the common difference (d) = 12 - 15 = - 3
tn = a +(n - 1) d = 15 + (n - 1) (-3)
= 15 - 3n + 3 = 18 - 3n
By question,
11 - 2n = 18 - 3n
or, -2n + 3n = 18 - 11
? n=7
Therefore, the 7th term of both series is equal.
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Example 9. The starting salary of a man is Rs. 7200 per month. If he gets an increment
of Rs. 200 every year, what is his salary in the 10th year ?
Solution: Starting salary = the first term (a) = Rs. 7200
yearly salary increment = d = Rs. 200.
Salary in the 10th year is the 10th term of the arithmetic sequence.
By using formula,
tn = a + (n - 1) d
or, t10 = 7200 + (10 - 1). 200 = 7200 + 1800 = 9000.
Therefore, his salary in the 10th year is Rs. 9000.
Exercise 3.1
Very Short Questions
1. (a) Define sequence and series.
(b) Write a difference between a sequence and series with an example of each.
(c) Define a progression with an example.
(d) What is a general term of a sequence ?
2. (a) Define an arithmetic sequence.
(b) Define common difference of an A.P.
(c) If a = the first term, d = the common difference n = the number of terms of an A.P.
Write formula for common difference in an A.P.
3. Find common difference in each of given A.P.
(a) 2, 4, 6, 8,...... (b) 10, 15, 20, 25,....... (c) 20, 18, 16, 14, ..........
4. Which of the following sequences are in A.P. ?
(a) 2, 7, 12, 17, ........... (b) 12, 8, 6, 4, ..............
(c) 10, 20, 30, 40, ........ (d) 200, 180, 160, 140, .............
5. (a) If the first term of an A.P. is 10 and the common difference 5. Find the 4th term of
the progression.
(b) If the 10th term of an A.P. is 100 and common difference is 10, find the first term.
6. A man deposits sequence of payments Rs. 200, Rs. 400, Rs. 600, Rs. 800, and
Rs. 1000 for six days. What type of sequence is it ?
Short Questions
7. Find the general terms for each of the following arithmetic sequence.
(a) 3,4,5,6,7,....... (b) 7, 10, 13, 16, .....
(c) 6, 9, 12, 15,..... (d) 51, 45, 39, .....
8. Compute the first term and the 10th term in each of the following sequence.
(a) d = 3, t4 = 12 (b) d = -4, t6 = 24
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(c) d = 7, t8 = 63 (d) d = - 8, t5 = 80
9. Determine the number of terms in each of the following A.P.
(a) the first term = 2, the common difference = 4, l = 102
(b) the first term = 3, the common difference 6, l = 57.
10. Find the number of terms in each of the following A.P.:
(a) 2 + 4 + 6 + ....... + 100
(b) 3 + 6 + 9 + ......... + 99
11. (a) Which term of the series 2 + 4 + 6 + ..... + 512 is 256 ?
(b) Which term of the series is 5 + 10 + 15 + ..... + 200 ?
12. (a) Is 75 a term of the series 9 + 11 + 13 + .....?
(b) Is -100 a term of series - 10 - 20 -30 - ...... ?
Long Questions
13. Find the common difference and the first term of arithmetic progression in each of the
following cases:
(a) t4 = 13, t6 = 7 (b) t5 = 15, t7 = 9
(c) t4 = 75, t10 = 117 (d) t3 = -40, t13 = 0
(e) t5 = 15, t7 =9
Also, find the 10th term in each of above progression.
14. (a) In an A.P. the 3rd and the 13th terms are 40 and 0 respectively, which term will be
28 ?
(b) The 6th and the 9th terms of an A.P. 23 and 35. Which term is 67 ?
15. (a) If the nth term of the A.S. 7, 12, 17, 22,...... is equal to the nth term of another A.S.
27, 30, 33, 36,...... find the value of n.
(b) If the nth term of an A.P. 23, 26, 29, 32, ...... is equal to the nth term of another A.P.
59, 58, 57, 56,....... , find the number of terms.
16. (a) In an A.P., the 7th term is four times the second terms and the 10th term is 29, find
the progression.
(b) In an A.P. the 10 times of 10th term is fifteen times the fifteen term. If the first term
is 48, find the progression.
17. (a) If x + 6, 3x, and 2x + 9 are in A.P., find the value of x and the next three terms the
progression.
(b) If 4k + 10, 6k + 2, and 10k are in A.P., find the value of k and terms t7, t8, and t9.
18. (a) If the pth term of an AS is q and qth term is p. Show that mth term is p+q-m.
(b) If m times the mth term of an A.P. equal to n times its nth term, show that (m+n)th
term of the A.P. is zero.
19. If the pth, qth, and rth terms of an A.P. are a,b, and c respectively.
prove that : p(b-c) + q (c-a) + r(a-b) = 0
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