Shubharambha OPT Mathematics 10 Final for CTP 2077

Approved by Government of Nepal, Ministry of Education,Science and Technology,
Curriculum Development Center(CDC), Sanothimi, Bhaktapur, Nepal

Infinity

Optional Mathematics

10Grade

Authors
Nil Prasad Ghimire
Shakti Prasad Acharya

Sujit Shrestha

Editors
Ramesh Subedi
Jibnath Sharma

Shubharambha Publication Pvt.Ltd.
Kathmandu, Nepal

Published by:

Shubharambha Publication Pvt. Ltd.

Kathmandu, Nepal
URL: www.shubharambhapublication.com.np
E-mail: [email protected]
www.facebook.com/shubharambhapublication

Book : Infinity Optional Mathematics-Book 10

Authors : Nil Prasad Ghimire

Shakti Prasad Acharya

Sujit Shrestha

Layout Design : Zeeta Computer Service Pvt. Ltd.

Ghantaghar, Kathmandu

Mobile No. 9841418545

Copyright © : Publisher

Edition : First: 2076 B.S.

Revised : 2077 B.S.

N©o:pPaurtbolifshtheirs book may be reproduced or transmitted by
means(electronic,

photocopying, recording or otherwise) without prior written permission from the

publisher. Any breach of this condition will entail legal action and prosecution.

Printed in Nepal

Preface

Infinity Optional Mathematics is a set of 5 books tailored to teach optional mathematics
to the students in the grades between 6 and 10. It is developed in strict compliance with
the recent school syllabus prescribed by the Curriculum Development Centre (CDC),
Ministry of Education, Science and Technology, Government of Nepal.

The text-books in the series are a result of a painstaking effort from a team of dedicated
and hard-working authors who have devised the lesson units with fitting ingenuity
and the acumen derived from years of experience as teachers.

The chapters in the books feature a fundamental theory and sufficiently-worked-out
examples right at the beginning as a ready reference for students. The classification
and the order of the chapters in each book are systematic and are put in a proper
sequence so that students can learn better and comprehend the progression of topics
and subject-matter with clarity. The authors are also confident that the books in their
present form are comprehensive, and helpful to the students in their preparation,
which is consistent with the exam requirements of the new question model (grid).

We thank the management team of Shubharambha Publication for their help in the
publication of this series. We also remain indebted to Mr Ananda Prasad Poudel,
Managing Director of Zeeta Computer Service, and his team for their expertly typing,
layout and cover design.

While every effort has been made to keep the series error-free, it is still possible for
some errors to have unwittingly crept into the work. If the beneficiaries of these books,
the reviewers and the others concerned bring to our notice any discrepancies, or areas
for potential improvement, we shall be extremely grateful to you. We would love to
receive your valuable suggestion, feedback, or queries via email at shubharambha.
[email protected].

Once again, we hope that this new edition of Infinity Optional Mathematics does
more than meet the expectations of students and teachers alike.

Authors

Contents

Unit 1 ALGEBRA 5 – 108
1.1
Inverse function and Composite Function 5
1.2 Polynomial 24
1.3 Sequence and Series 40
1.4 Linear Inequation and Graph of Quadratic Equation 85

Unit 2 CONTINUITY 109–123

Unit 3 MATRIX 124–138

3.1 Determinant 125

3.2 Solving of Simultaneous Equations by Inverse Method 131

3.3 Cramer’s Rule 133

Unit 4 CO-ORDINATE GEOMETRY 139–182

4.1 Angle Between Two Lines 139

4.2 Equation of Pair of Lines 153

4.3 Conic Section 167

4.4 Circle 170

Unit 5 TRIGONOMETRY 183–253

5.1 Multiple Angles 185
5.2
5.3 Sub-multiple Angles 199
5.4
5.5 Transformation of Trigonometric Formulae 210
5.6
Conditional Trigonometric Identities 223
Unit 6
Trigonometric Equations 232

Height and Distance 242

VECTORS 254–278

6.1 Scalar Product 254
6.2
Vector Geometry 263
Unit 7
TRANSFORMATION 279–321
7.1
7.2 Introduction of Transformation 279
7.3
7.4 Combined Transformation 282

Unit 8 Inversion Transformation and Inversion Circle 308

Transformation Using Matrix 313

STATISTICS 322–343

8.1 Classification of Measures of Dispersion 323
Answer Sheet 344

UNIT

1 ALGEBRA

1.1 Inverse Function and Composite Function
Review

Let us consider the following arrow diagrams.

(a) f (b) f (c) f
A A
a B AB –1 B
b d 0
c e 11 1 1
f 4 0
2 9

3 16

(d) f (e) f (f) f
A A
a B A B p B
b p 2 q
c q 1 3 r r
r 2 4 s
3
5

In the above arrow diagrams, figure number (a), (b), (c) and (f) represent the
function. Figure number (d), and (e) do not represent the function. For a function,
every element of set A must associate with unique element of set B. In figure no.
(d), element a has made ordered pairs with two elements p and q. In figure no. (e), 5
has not made ordered pair with any element of set B. So, they are simply relations.

Types of Function f B
A 5
6
1. Onto function 1 7
8
Let us consider an arrow diagram 2
Domain = {1, 2, 3, 4} 3 B
co-domain = {5, 6, 7, 8} 4 4
1
Range = {5, 6, 7, 8} 0
3
Here, range and co-domain are equal. So it is an onto function. 5

f 5

2. Into function A

Let us consider an arrow diagram –2
Domain = {– 2, – 1, 0, 1, 2} –1

Co-domain = {4, 1, 0, 3, 5} 0

Range = {4, 1, 0}. 1
Here, range is the proper subset of co-domain. So it is an into 2

function.

Infinity Optional Mathematics Book - 10

3. One to one function
Let us consider the following arrow diagram:

ff

AB AB

12 ae
23 f
34 b g
45 c h

di

(i) (ii)
Here, separate elements of set A have made ordered pairs with the separate

elements of set B. So the above arrow diagrams represent one to one function.
There are two types of one to one function.

(i) one to one onto function. Figure no. (i) represents one to one onto function.
(ii) one to one into function. Figure no. (ii) represents one to one into function.
4. Many to one function
Let us consider the following arrow diagrams.

ff

AB AB

1a ap
2b bq
3c cr
4 ds

(i) (ii)
Here, one element of range has more than one pre–image. So, the above arrow

diagrams represent many to one function .
There are two types of many to one function.
(i) Many to one onto function
Figure No. (i) represents many to one onto function.
(ii) Many to one into function
Figure No. (ii) represents many to one into function.

Inverse Function

Let us observe the following functions with arrow diagram.

(i) one to one onto function B A
2 1
f 3 2
4 3
A B After interchanging
1 2 domain and range
23
34

This is also a function

6 Infinity Optional Mathematics Book - 10

(ii) One to one into function.
f

A B After interchanging Be A
a e domain and range f a
b f g b
c
g hc

h

This is not a function

(iii) Many to one onto function. f
f

A B After interchanging BA
1 domain and range 1
25
3 5 2
3

4 4

This is not a function

(iv) Many to one into function.

ff

A B After interchanging B A

a p domain and range p a
b q q b
c r r c
s s

This is not a function

From the above arrow diagrams, we have found that only in one to one onto function,

when domain and range are interchanged, we get function. But in other functions,

when domain and range are interchanged, none of them form function.

Thus, a function obtained by interchanging domain and range of one to one onto

function is called inverse function of the function. The inverse function of the

function f, g and h are denoted by f–1, g–1 and h–1 respectively. The function f:A → B

and inverse function f–1: B → A are shown in the arrow diagrams.

f f–1

A BA B
13 31

24 42

35 53

46 64

Here, f = {(1. 3), (2, 4), (3, 5), (4, 6)}, f-1 = {(3, 1), (4, 2), (5, 3), (6, 4)}

Infinity Optional Mathematics Book - 10 7

WORKED OUT EXAMPLES

1. If f = {(1, 4), (2, 5), (3, 6)}, then find the inverse function of the function f.
Solution: Here,
f = {(1, 4), (2, 5), (3, 6)}
The inverse function of the function is
f-1 = {(4, 1), (5, 2), (6, 3)}

2. A function g is defined as g : x→ 5x + 2. Find the inverse of g.

Solution: Here,

g : x → 5x + 2

Let y = g(x) = 5x + 2.

Now, interchanging domain (x) and range (y), we get

x = 5y + 2

or, x – 2 = 5y
o∴r , yg-1=(xx)–5=2x
– 2
5

3. If f: R → R be defined by f(x) = 2x + 5 , x ≠ –2, x ∈ R, find f-1 (x) and
x+2
f-1 (5).

Solution: 2x + 5
x+2
Here, f(x) =

Let y = f(x) = 2x + 5
x+2

Now, interchanging domain (x) and range (y), we get

x = 2y + 5
y+2

or, xy + 2x = 2y + 5

or, xy – 2y = – 2x + 5

or, y(x – 2) = –2x + 5

or, y = 5 – 2x
x–2

∴ f– 1(x) = 5 – 2x
x–2

Now, f-1 (5) = 5 – 2 × 5 = 5 – 10 = –35.
5 – 2 3

8 Infinity Optional Mathematics Book - 10

4. If a function f(x + 3) = 5x + 4, then find f-1 (x).
Solution:
Here, f(x + 3) = 5x + 4
Let, x + 3 = a then
x = a – 3
Now, f(a) = 5 (a – 3) + 4
f(a) = 5a – 15 + 4
f(a) = 5a – 11
∴ f(x) = 5x – 11
Let, y = f(x) = 5x – 11

Now, interchanging domain (x) and range (y), we get

or, x = 5y – 11

or, y = x + 11
5

∴ f-1(x) = x + 11
5

Exercise 1.1

Section 'A'

1. Which of the following arrow diagram represent function?

(a) 1 3 (b) 1 9 (c) a e
–1 b f
2 4 2 4 c g
3
5 –2 8d N
A
BP QM

(d) 4 1 (e) p a (f) 1 a
B
5 2q b2
6 3r c
7 4 d3
C DE FA

(g) a (h) 1 5 (i) 1 e
f
3 b2 62 g
c3 73
M N
4 4
NA
BM

Infinity Optional Mathematics Book - 10 9

2. Write down the type of function represented by the following arrow
diagram.

(a) a d (b) 1 3 (c) –1 1
0
b e 2 4 –2 4
c 3 0 9
f 5 1 B
A
62

BA BA

(d) a (e) 16 4 (f) a 3
4
b e9 3b 5
c 4 2c 6
d 1d
A 1 0e B
BA
BA

3. Given below are the mapping diagram of the functions from set A to
set B. Which of the function has its inverse? Explain with reason:



(a) f (b) g (c) f

1a ps p
2b qt q4
3c ru r
4
AB AB AB

(d) f (e) h (f) g

42 –2 21
93 –1 1 3
16 4 3 4
04 4 5
AB 10 5 6
2
MN
AB

4. Show the inverse function of each of the functions in mapping
diagrams.



(a) f (b) g

12 ae
23 bf
34 cg
45 dh
AB MN

10 Infinity Optional Mathematics Book - 10

(c) f (d) f

11 7 49
24 6 36
39 5 25
4 16 4 16
PN AB

5. Find the inverse function of the following functions.

(a) f = {(1, 1), (2, 4), (3, 9), (4, 16)}

(b) g = {(1, 3), (2, 3), (3, 5), (4, 6), (5, 7)}

(c) h = {(a, 3), (b, 5), (c, 7), (d, 8), (e, 9), (f, 2)}

(d) f = {(–4, 4), (–5, 5), (–6, 6), (–7, 7)}

(e) g= –3, –2 , –4, –3 , –5, –4
5 5 5
Section 'B'

6. Find the inverse of each of the following functions:

(a) f(x) = x + 3 (b) g(x) = x – 5

(c) h(x) = 5x – 2 (d) t(x) = 5 – 7x

(e) s(x) = 5x – 3 (f) f(x) = 3x
4

(g) g(x) = 3x + 1 (h) h(x) = –5x – 7
4 4

(i) t(x) = x 5 4 x ∈ R, x ≠ – 4 (j) f(x) = 5 3
+ – 2x

(k) f(x) = 2x 6 (l) h(x) = 3x – 1 x ∈ R, x ≠ –2
5x + x+2

(m) f(x) = 5x + 21 (n) g(x) = 4x + 3
2x – 3x + 4

(o) h(x) = x – 11, x ≠ 1, x R (p) k(x) = {(x, y): y = 1x, x R}
x +

(q) t(x) = {(x, y): y = x, x ∈R} (r) f(x) = {(x, x + 5) : x R}
3

(s) h(x) = {(x, 5x – 3) : x ∈ R} (t) g(x) = {(x : 3xx+–11): x∈R}
7. If f(x) = 3x – 5, find:

(a) f-­ 1(x) (b) f-1(2) (c) f-1 (–3)

(d) f-1(2x) (e) f-1 3
2

Infinity Optional Mathematics Book - 10 11

8. If f(x) = 2x + 3, find:
4
3
(a) f-1(x) (b) f-1(3x) (c) f-1 5
(e) f-1(2x – 5).
(d) f-1(x + 1)

9. If f(x) = 3x + k and f-1 (5) = 2, find the value of k.

10. If g(x) = 5x – 3 and f(x) = 7x – 5, then find the value of x when g-1 (x) = f-1(3).
2 3
11. Find the value of f-1(x) from the following.

(a) f(x + 3) = 2x + 5 (b) f(x + 1) = x + 5
x
3x – 1
(c) f(x – 5) = x+2

12. If f-1 (x) = 3x + 5, find f(x) and f(3).

Composite Function

Let us consider the following arrow diagrams of the function.
fg

AB BC

14 4 16

25 5 25

3 6 6 36

Here, f is a function defined from A to B and g is another function defined from set

B to C. Combining these two arrow diagrams, we get
fg

A BC

1 4 16

2 5 25

3 6 36

When we apply both function f and g at a time, then a new function can be

defined from A to C. f g

A BC

1 4 16

2 5 25

3 6 36

gof
Here, the new function is defined from A to C and is called the composite

12 Infinity Optional Mathematics Book - 10

function of f and g. From the arrow diagram .

Composite function of f and g = {(1, 16), (2, 25), (3, 36)}
Thus if f: A → B be a function from set A to set B and g: B → C be another

function from set B to set C, then the function from set A to set C is known as
the composite function of f and g and denoted by gof or gf.
Symbolically, the composite function of f and g is written as
gof: A → C such that for every x∈A.
gof(x) = g(f(x))∈C.

Note: For two functions f and g.
1. The composite function of f and g is written as gof or gf and read as f
followed by g.
2. The composite function of g and f is written as fog or fg and read as g
followed by f.
3. The meaning of f2 (x) means fof(x).
4. The composite function of f(x) and its inverse is x.

i.e. fof-1 (x) = f-1of (x) = x.

WORKED OUT EXAMPLES

1. If f = {(1, 2,), (2, 3), (3, 4), (4, 5)} and g = {(2, 4), (3, 9), (4, 16), (5, 25)},

then find the composite function of f and g in ordered pairs form by

representing in arrow diagram.

Solution: Here, f = {(1, 2), (2, 3), (3, 4), (4, 5)}

g = {(2, 4), (3, 9), (4, 16), (5, 25)}

In arrow diagram, fg

ABC
1 24
2 39
3 4 16
4 5 25

gof
Infinity Optional Mathematics Book - 10 13

From arrow diagram.
gof = {(1, 4), (2, 9), (3, 16), (4, 25)}

2. If f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} are two functions
then find gof and fog in ordered pairs form by representing in mapping
diagram.

Solution:
Here, f = {(1, 2), (3, 5), (4, 1)}
g = {(2, 3), (5, 1), (1, 3)}
For gof:

fg From the arrow diagram of gof,

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

1 2 3 g(2) = 3, g(5) = 1, g(1) = 3
3 5 1 Now, gof (1) = g(f(1)) = g(2) = 3
4 1
gof(3) = g(f(3)) = g(5) = 1

gof (4) = g(f(4)) = g(1) = 3

g gof C ∴ gof = {(1, 3), (3, 1), (4, 3)}

A f 2 From the arrow diagram of fog,
2 B 5 g(2) = 3, g(5) = 1g(1) = 3
5 1 f(3) = 5, f(1) = 2, f(4) = 1
1 3 Now, fog(2) = f(g(2) = f(3) = 5
1 fog (5) = f(g(5)) = f(1) = 2
4

fog (1) = f(g(1)) = f(3) = 5

∴ fog = {(2, 5), (5, 2), (1, 5)}
fog

3. Let f : R → R be defined by f(x) = 3x + 5 and g: R → R be defined by
g(x) = x – 1, find the

(i) composite function of f and g (i.e. find gof)

(ii) composite function of g and f (i.e. find fog)
Solution:

Here, f(x) = 3x + 5

g(x) = x – 1

(i) Composite function of f and g (ii) Composite function g and f

= gof (x) = fog (x)

= g(f(x)) = f(g(x))

14 Infinity Optional Mathematics Book - 10

= g(3x + 5) = f(x – 1)
= 3x + 5 – 1 = 3(x – 1) + 5
= 3x + 4 = 3x – 3 + 5
= 3x + 2.

4. If f(x) = 3x – 1 and g(x) = 7 – 2x, find (a) f-1og(x), (b) g-1of(x)

(c)­ f-1og-1(x) (d) f-1 og-1 (3).
Solution:

Here, f(x) = 3x – 1 and g( x) = 7 – 2x

Taking f(x) = 3x – 1

Let, y = f(x) = 3x – 1

Now, interchanging domain (x) and range (y), we get

x = 3y – 1

or, x + 1 = y
3

∴ f-1(x) = x + 1
3

Taking g(x) = 7 – 2x

Let y = g(x) = 7 – 2x

Now interchaning domain (x) and range (y), we get

x = 7 – 2y

or, 2y = 7 – x

or, y= 7–x
2
7–x
\ g–1 (x) = 2

(a) Now, f-1g(x) = f-1(g(x)) = f-1(7 – 2x)= 7 – 2x + 1 = 8 –32x
3

(b) g-1of(x) = g-1(f(x)) = g-1(3x – 1) = 7 – (3x – 1) = 7 – 3x + 1 = 8 – 3x
7 +1 = 2 2
–2x
(c) f-1og-1(x) = f-1(g-1(x))= f-1 7–x = 7 – x + 2 = 9 – x
2 2 6 6
3

(d) f-1og-1(3) = 9 – 3 = 6 = 1
6 6
Infinity Optional Mathematics Book - 10 15

5. If f(x) = 3x + 5 and fog (x) = 7x – 1, find g(x).
3

Solution: Alternative Method:

Here, f(x) = 3x + 5 Here, f(x) = 3x + 5

fog(x) = 7x – 1 fog(x) = 7x – 1
3 3

Let g(x) = k Let g(x) = ax + b

Now, fog(x) = 7x – 1 Now, fog(x) = 7x – 1
3 3

or, f(g(x)) = 7x – 1 or, f(ax + b) = 7x – 1
3 or, 3(ax + b) +
7x – 1 5 3 7x – 1
3 = 3
or, f(k) =

or, 3k + 5 = 7x – 1 or, 9ax + 9b + 15 = 7x – 1
3
Now, equating the corresponding coefficient,

or, 9k + 15 = 7x – 1 9a = 7 9b + 15 = – 1

or, k = 7x – 16 ∴ a = 7 ∴ b = – 16
9 9 9
7x – 16
∴ g(x) = 9 Hence, g(x) = ax + b

= 7x + – 16
9 9

= 7x – 16
9

6. If fog(x) = 19 – 7x and g(x) = 2 – x, find f(x).

Solution:
Here, fog(x) = 19 – 7x
g(x) = 2 – x
Let, f(x) = ax + b
Now, fog(x) = 19 – 7x
or, f(2 – x) = 19 – 7x
or, a(2 – x) + b = 19 – 7x
or, 2a – ax + b = 19 – 7x
or, 2a + b – ax = 19 – 7x
Now, comparing the corresponding coefficients, we get
2a + b = 19
b = 19 – 2a ...............(1)

16 Infinity Optional Mathematics Book - 10

and –a = – 7
∴ a = 7 ................ (2)
Also,
b = 19 – 2a
= 19 – 2 × 7
= 19 – 14
∴ b = 5

Hence, f(x) = ax + b = 7x + 5

7. Given that f(x) = 3x + 5, g(x) = x + 1 and h(x) = 2x – 3. Find f(gh), gf(h)

and gf(h) (2).

Solution:

Here, f(x) = 3x + 5 g(x) = x + 1 and h(x) = 2x – 3

(i) f(gh) = fg[h(x)] (ii) gf(h) = gf[h(x)]

= fg(2x – 3) = gf(2x – 3)

= f[g(2x – 3)] = g[f(2x – 3)]

= f(2x – 3 + 1) = g(3(2x – 3) + 5)

= f(2x – 2) = g(6x – 4)

= 3(2x – 2) + 5 = 6x – 4 + 1

= 6x – 6 + 5 = 6x –1 = 6x – 3

(iii) gf(h( 2) = 6 × 2 – 3

= 9

8. Given that f(x) = 3x – 1 and g(x) = 4x – 1 , find the value of x when
2

ff(x) = g-1(x).

Solution:

Here, f(x) = 3x – 1

Now, ff(x) = f(f(x)) = f(3x – 1) = 3(3x – 1) – 1 = 9x – 4

Taking g(x) = 4x – 1 Let, y = g(x) = 4x – 1
2 2

Interchanging domain (x) and range (y), we get

x = 4y – 1
2

or, 2x + 1 = 4y

or, yg-1=(x2)x=4+21x4+ 1
∴

Infinity Optional Mathematics Book - 10 17

By question

ff(x) = g-1(x)
2x + 1
or, 9x – 4 = 4

or, 36x – 16 = 2x + 1

or, x= 17
34

∴ x = 1
2

Exercise 1.2

Section 'A'

1. Find the composite function of the two function f and g (i.e. (gof) from
the given arrow diagram.



(a) f g (b) f g

1 39 a pd
2 4 16 b qe
3 5 25 c rf
BC P
A (d) f QR
g
(c) f 1 g
40 2
-2 82 3 2 3
-1 0 -1 A 5 1
0 6 1
1 BC
2 BC
A

2. From the following arrow diagrams. Find the elements of set B and
set C. Also find the composite function of the two functions f and g
(i.e. gof) in ordered pair form.

(a) f g (b) f g

1 ... ... f(x)=x2 4 ... ...
2 ... ... g(x)=2x 5 ... ... f(x)=2x
3 ... ... 6 ... ... g(x)=2x–1

A BC A BC

18 Infinity Optional Mathematics Book - 10

(c) f g (d) f g

1 ... ... f(x)=3 x -1 ... ...
8 ... g(x)=x+2 -8 ... ... f(x)=x+3
27 ... ... -3 ... ... g(x)=x2+1
...
BC
A BC A

Section 'B'
3. From the following functions f and g, find the ordered pairs for gof

(composite function of f and g) by representing in mapping diagram.

(a) f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}

(b) f = {(1, x), (2, y), (3, z)} and g = {(x, 4), (y, 5), (z, 6)}

(c) f = {(–2, 4) (–1, 1) (0, 0), (1, 1) (2, 4)} and g = {(4, 5) (1, 2) (0, 1)}

4. (a) If f = {(1, 3), (0, 0), (–1, –3)} and gof = {(1, 5), (0, 2), (–1, – 1)}, find the

function g in terms of ordered pairs.

(b) If f = {(1, 2), (2, 3), (3, 4) and g = {(2, 3), (3, 1), (4, 2)}, calculate

gof (1), gof(3), fog(2), fog(3) and fog(4).

(c) If h = {(1, 1) (8, 2), (27, 3)} and g = {(1, 3) (2, 4), (3, 5)}, calculate

goh (1) and goh (8).

5. (a) If f(x) = 3x and g(x) = 7x find fog (x) and gof(x).

(b) If f(x) = x + 3 and g(x) = x – 5, find fog (x) and gof(x).

(c) If f(x) = 2x – 1 and g(x) = 5x + 3, find fog (x) and gof(x).

(d) If f(x) = 7x + 5 and g(x) = 2 – x, find gof(x) and fog(x).

(e) If f(x) = 5x – 1 and g(x) = 3x – 3, find fog(x) and gof(x).
3
5x – 1 7x + 1,
(f) If f(x) = 7 and g(x) = 5 find fog(x) and gof(x).

(g) If h(x) = 5 – 2x and g(x) = 5 – 4x , find (hog) (x) and (goh) (x).
4 2

(h) If f( x) = 3x2 + 1 and g(x) = x + 5, find fog(x) and gof(x).

(i) If f(x) = x2 – 1 and g(x) = x2 + 1, find fog(x) and gof(x).

(j) If f( x) = 2x + 1 and g(x) = x + 7, find (fog)x and (gof) (x).
2x
6. (a) If f(x) = 3x + 2 and g(x) = 2x – 1, find fog(x), fog(2), gof(x) and gof (–3).

(b) If h(x) = 5 – 4x and g(x) = x + 1, find goh(x), goh(–3), hog(x) and hog 3 .
2
7x – 2
(c) If f(x) = 3 and g(x) = 3 – 2x, find fog (1) and gof(–3).

Infinity Optional Mathematics Book - 10 19

(d) If f(x) = 3–x and g(x) = 2x + 7, find fog(–2) and gof(5).
5x – 1 3

(e) If f(x) = x3 – 1 and g(x) = x2, find fog(–3) and gof(2).

Section 'C'

7. (a) If f(x) = 3x – 4 and g(x) = x + 5, find f-1 og(x) and f-1og(3).

(b) If h(x) = 7x + 2 and g(x) = 3x + 1, find goh-1 (x) and goh-1(–5).
3
3x – 1
(c) If f(x) = 2 and g(x) = 3 – 2x, find f-1og-1(x), g-1of-1(x), f-1og-1(–2) and

g-1of-1(–1).

(d) If g(x) = 7 – 5x and h(x) = 3 + x, find (gh)-1(x), (hg)-1(x), (gh)-1(3) and (hg)-1(–2)

(e) If f(x) = 5x – 2, find f-1(x), fof-1(x), f-1f(x) and fof-1(3).

(f) If g(x) = x + x7, find gog-1(x), g-1og(x), g-1og(3) and gog-1(–3).
1 –

8. (a) Let f(x) = 3x – 2, g(x) = x + 2 and h(x) = 1 – 2x, find

(i) (hog).f (ii) (f.g).h (iii) (gof)h (iv) f(gh) (–3)

(b) Let f(x) = –2x + 1, g(x) = 3x – 4 and h(x) = x + 3, find

(i) f(gh) (ii) (fg)h (iii) (gh)f (iv) g(hf) (3)

9. (a) If f(x) = x2 and g(x) = 1x, show that fog = gof.

(b) If f(x) = 3x + 4 and g(x) = 2(x + 1), show that fog = gof.

(c) If f(x) = x 6 2 and g(x) = mx2 – 1, find the value of m when gf(5) = 7.
–
(d) It is given that h(x) = x2 – 2x and g(x) = 2x + 3. If hg-1(x) = 3, calculate

the value of x.

(e) If f(x) = x 4 (x ≠ 4) and f(x) = f-1(x), find the value of x.
(f) If f(x) = x– – the value of
4x 17 and g(x) = 2x + 8, then find x when ff(x) =
5
g-1(x).

(g) If f(x) = 2x – 7, g(x) = x + 2 and fog (x) = g-1(x), find the value of x.
3

(h) It is given that the function f(x) = 4x + 7 and g(x) = 3x – 5, find the value

of x when fog-1(x) = 15

(i) If f(x) = 3x + 5 and fog(x) = 5x – 2, find g(x) and g-1(x).

(j) If h(x) = 5x – 3 and hog(x) = 7x – 3, find g(x) and g(3).
x+1

(k) If g(x) = x – 5 and fog(x) = x – 2, find f(x).

(l) If f(x) = 2x – 1 and gof(x) = 6x – 1, find g(x).
20 Infinity Optional Mathematics Book - 10

A. Simple Algebraic Function Y
5
Constant function: A B
O X
A function of the form y = f(x) = c, Where c is a Y'
Y X
constant term is known as constant function. It X'
always represents a straight line parallel to X–axis. O

For the constant function y = f(x) = 5, the nature of Y'

graph is shown alongside. Y

x 0 2 –5

y55 5

(x, y) (0, 5) (2, 5) (–5, 5)

Linear function

An algebraic function of the first degree is
called linear function. Liner function is in X'

the form of y = f(x) = ax + b where a and b

are real constant. It represents a straight

line.

For example, y = f(x) = – 3x + 2, y = g(x) =

3x – 5 etc.

For the liner function y = f(x) = 2x – 1, the

nature of the graph is shown in the adjoining diagram.



x012

y –1 1 3

(x, y) (0, –1) (1, 1) (2, 3)

Identity function y=x
A function f defined by y = f(x) = x is X' OX
known as identify function. It maps every
element into itself. This means domain Y'
and range of identify function are same.
It represents a straight line bisecting the
angle between the axes of co-ordinates. For

Infinity Optional Mathematics Book - 10 21

the identify function y = f(x) = x, the nature of the graph is shown in the adjoining

diagram.
x –2 0 3
y –2 0 3

(x, y) (–2, –2) (0, 0) (3, 3)

Y

Quadratic function y = x2 X
An algebraic function of second degree in O
the form y = f(x) = ax2 + bx + c where a, b
and c are real constants and a ≠ 0 known
as quadratic function. The nature of the X'
graph of the quadratic function is called
parabola.

Some of the examples of quadratic

function are y = f(x) = x2 + 5x + 6, y = Y'
g(x) = 3x2 – 4, y = h(x) = x2 etc. For the

quadratic function y = f(x) = x2, the nature of the graph is shown below.

x 0 1 –1 2 –2
y 0 1 1 4 4
(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)

Cubic function
An algebraic function of third degree in the form of y = f(x) = ax3 + bx2 + cx + d, a ≠

0 where a, b, c and d are real constant is called a cubic function.
For example: y = f(x) = 3x3, y = g(x) = x3 – 3x + 5 etc.
For the cubic function y = f(x) = x3, the nature of the graph is shown below.

x 0 1 –1 2 –2
y 0 1 –1 8 –8
(x, y) (0, 0) (1, 1) (–1, –1) (2, 8) (–2, –8)

22 Infinity Optional Mathematics Book - 10

Y
y = x2

X' O X

Y'

B. Trigonometric Function

A function f is said to be trigonometric function if it involves trigonometric
ratios like sine, cosine, tangent etc. For example: y = f(x) = sinx, y = g(x) = cosx
etc. For the trigonometric function y = f(x) = sinx, the nature of the graph is

shown below.

x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360°

y 0 0.5 0.7 0.85 1 0.86 0.7 0.5 0 -0.5 -0.7 -.86 -1 -.86 -0.7 -0.5 0

Y

0.1
0.8

0.6
0.4

0.2

X' O 30º 60º 90º 120º 150º 180º 210º 240º 270º 300º 330º 360º X
-0.2

-0.4

-0.6

-0.8 y = sinx
-1
Y' Infinity Optional Mathematics Book - 10

23

Exercise 1.3

Section 'A'

1. Identify whether the function given below are constant, linear,

quadratic, cubic or trigonometric.

(a) f(x) = 3x + 5 (b) f(x) = 3x2 + 7x – 2 (c) f(x) = 4

(d) f(x) = x. (e) f(x) = 3x3 – 7 (f) f(θ) = cosθ

(g) g(x) = – 3x – 5 (h) g(θ) = sinθ + cosθ

Section 'B'

2. Draw the graph of the following function

(a) f(x) = – 5x + 3 (b) g(x) = – x (c) f(θ) = cosθ (0≤q ≤90°)
(f) f(x) = – x3
(d) h(x) = – 3 (e) f(x) = 1 x2
2

1.2 Polynomial

Review

Consider the following algebraic functions.
3x
(i) f(x) = 2 + 5 52 (ii) g(x) = 5x2 + 6x + 1
(iii) (iv) f(y) = 5y4 – 3y2 – 7y + 7
h(x) = 5x3 – 4x2 – 2x +

In the above algebraic functions, the power of the variables are positive real number.

They are the polynomials.

Hence, the algebraic functions whose variable consist of only whole numbers in
exponents are called the polynomials.

The highest power of the variable involved in the polynomial is known as the degree

of the polynomial.

In the polynomials (i) f(x) = 3x3 + 5x2 – 2x + 1, the degree is 3.

(ii) g(x) = 7x5 + 2x2 – 5x – 1, the degree is 5.
3 1
(iii) h(y) = 8y6 – 5y5 – 4 y3 – 2 y + 2, the degree is 6.

The above polynomials contain only one variable.

But if the polynomial contains two or more variables, then the degree of the

polynomial is the sum of the exponents of the variables involved in that polynomial.

24 Infinity Optional Mathematics Book - 10

In the polynomials (i) f(x, y) = 8x3y5 + 4x3y2 – 7x2y – 5 the degree is 8.

(ii) g(x, y, z) = 7x3y5z4 + 7x3y2z – x2yz, the degree is 12.

Generally, the terms of the polynomials are arranged in ascending or descending

power of the variable.

Multiplication of Polynomials

Consider the polynomials
g(x) = 5x2 + 3x – 2 and h(x) = 3x3 – 5x2 + 7x – 2

Now, g(x) × h(x)
= (5x2 + 3x – 2) (3x3 – 5x2 + 7x – 2)
= 5x2 (3x3 – 5x2 + 7x – 2) + 3x( 3x3 – 5x2 + 7x – 2) – 2(3x3 – 5x2 + 7x – 2)
= 15x5 – 25x4 + 35x3 – 10x2 + 9x4 – 15x3 + 21x2 – 6x – 6x3 + 10x2 – 14x + 4
= 15x5 – 16x4 + 14x3 + 21x2 – 20x + 4

In the above example, while multiplying two polynomials, each term of the

polynomial g(x) multiples every term of the polynomials h(x) and combine the like

terms. Here the degree of the product g(x) × h(x) is the sum of the degree of g(x) and

h(x)

Properties of multiplication of polynomials.
1. Closure property
If f(x) and g(x) be two polynomials of any degree then their product f(x). g(x)

or g(x). f(x) is also a polynomial whose degree is the sum of the degree of f(x)

and g(x).
2. Commutative property
For two polynomial f(x) and g(x) of any degree,
f(x). g(x) = g(x). f(x)
3. Associative property
For three polynomials f(x), g(x) and h(x) of any degree then
f(x). {g(x). h(x)} = {f(x) . g(x)} . h(x)
4. Multiplicative identity
For a polynomial f(x), there exists the unit polynomial I(x) = 1 such that f(x).

I (x) = f(x) Here, I is said to be the multiplicative identity.
5. Distributive law
If f(x), g(x) and h(x) be three polynomials, then

f(x) × {g(x) + h(x)} = f(x). g(x) + f(x). h(x).

Infinity Optional Mathematics Book - 10 25

Division of Polynomial

Consider the two polynomials
f(x) = x3 + 5x2 – 3x – 4 and g(x) = x2 + 3x + 1

Now, f(x) ÷ g(x) is shown below.

x2 + 3x +1) x3 + 5x2 – 3x – 4(x + 2
x3 + 3x2 + x

–– –

2x2 – 4x – 4

2x2 + 6x + 2

–– –

– 10x – 6

Here, g(x) = x2 + 3x + 1 is a divisor.

f(x) = x3 + 5x2 – 3x – 4 is a dividend.

q(x) = x + 2 is a quotient.

R = – 10x – 6 is a remainder.

This shows that if the polynomial f(x) is divided by g(x) then we can write it as

f(x) = g(x). Q(x) + R

i.e. Divident = Divisor × quotient + Remainder
Here, degree of R is always less than g(x) and the division process is continued

till the degree of the remainder is less than the degree of the divisor.
When remainder R is zero, then f(x) is exactly divisible by g(x) and g(x) is

called a factor of f(x).

Exercise 1.4

1. Find the product of : (b) 5x3 – 3x + 7 and 2x – 5
(a) 3x3 – 2x2 – 5x – 7 and x + 2 (d) 5x5 – 3x2 + 7x – 9 and 2x + 3
(c) 7x5 – 2x and x2 + 3x + 2

2. Divide the polynomial f(x) by h(x) in the following:

(a) f(x) = 3x3 + 5x2 – 4x + 3 h(x) = x + 3

(b) f(x) = 5x5 – 3x3 + 5x + 4 h(x) = x – 2

(c) f(x) = 4x3 – 5x2 – 3x h(x) = 2x + 1

(d) f(x) = x6 – 8 h(x) = x – 3

3. Find Q(x) and R when 3x3 – 4x2 – 9x + 2 = (3x + 1) Q(x) + R

26 Infinity Optional Mathematics Book - 10

Synthestic Division Method

Synthetic division is the process which helps us to find the quotient and remainder
in a short time way when a polynomial f(x) is divided by a linear polynomial (x – a).

WORKED OUT EXAMPLES

1. Find the remainder and quotient by using synthetic division method
when f(x) = 3x3 – 5x2 + 7 is divided by g(x) = x + 2

Solution:

Following steps are performed to find the remainder and quotient by synthetic

division method.

1. Write the dividend in the standard form
i.e. 3x3 – 5x2 + 0.x + 7. (Here the term x is missing, so we have to put zero as

coefficient for x)

2. Compare the divisor x + 2 with x – a, then a = – 2

3. Write down the coefficients of x and constant term of f(x) including sign.
Here, 3, – 5, 0, are the coefficient of x3, x2 and x respctively and 7 is the constant

term.

4. Bring down the first coefficient 3.

5. Multiply 3 by the value of a.

6. Write down the product under the coefficient – 5 and add.

7. Multiply the result by – 2 and write down the result under next coefficient 0

and add

8. Continue the same process till the constant term.

9. The last result is the remainder, the second last number is the constant term

of quotient and the third last number is the coefficient of x and so on

Here, Dividend f(x) = 3x3 – 5x2 + 7 and divisor g(x) = x + 2.

Now, by synthetic division method,

x2 x Constant
x3 7
3 –5 0
–44
–2 –6 22
3 –37
–11 22
x2 x Constant Remainder

Hence, remainder (R) = 37
Quotient Q(x) = 3x2 – 11x + 22

Infinity Optional Mathematics Book - 10 27

2. Find the remainder and quotient by synthetic division method when
x3 – 5x2 + 2x – 5 is divided by x – 3

Solution:

Here, dividend f(x) = x3 – 5x2 + 2x – 5

divisor = (x – 3) (Comparing with (x – a), a = 3)

Now, by synthetic division method,

x3 x2 x Constant

1 –5 2 –5

3 3 –6 –12
1 –2 –4 –17

x2 x Constant Remainder

Hence, remainder (R) = – 17

Quotient Q(x) = x2 – 2x – 4

3. Find the remainder and quotient when 2x4 – 8x2 + 7x – 2 is divided by
3x + 2 by synthetic division method.

Solution:
Here, dividend f(x) = 2x4 – 8x2 + 7x – 2

Divisor = 3x + 2

= 3 x + 2 Comparing with (x – a)‚ a = –2
3 3

Now, by synthetic division method

x2 x3 x2 x Constant

–2 2 0 –8 7 –2
3↓
– 4 8 128 – 634
2 3 9 27 81
–64 317
x3 – 4 27 –796
3 9 constant
x 81

x2 remainder

∴ Remainder (R) = –78916, Quotient Q(x) = 1 2x3 – 4 x2 – 64x + 317
3 3 9 27

Remainder theorem :

Statement: If any polynomial f(x) is divided by a linear polynomial (x – a), then the
remainder is f(a)

Proof : When a polynomial f(x) is divided by a linear polynomial x – a, then, we

28 Infinity Optional Mathematics Book - 10

get quotient Q(x) and remainder R.
Then f(x) = (x – a) Q(x) + R
Putting x = a, we get
f(a) = (a – a) Q(a) + R
or, f(a) = R
∴ R = f(a)
Hence remainder (R) = f(a)
This proves the remainder theorem.
Hence, if the polynomial f(x) is divided by linear polynomial (x – a), then the

remainder is f(a).

Note: 1. If a polynomial f(x) is divided by (x + a), then the remainder = f ( – a).

2. If a polynomial f(x) is divided by ax + b, then the remainder = f –b .
a

3. If a polynomial f(x) is divided by (ax – b), then the remainder = f b
a

WORKED OUT EXAMPLES

1. If f(x) = 4x3 – 5x2 – 2x + 10 is divided by x – 1, find the remainder by
using remainder theorem.

Solution:

Here, f(x) = 4x3 – 5x2 – 2x + 10

Comparing (x – 1) with (x – a), we get a = 1

Now, by remainder theorem,

remainder = f(a) = f(1)

= 4(1)3 – 5(1)2 – 2(1) + 10

= 4 – 5 – 2 + 10 = 7

2. If f(x) = 3x3 – 4x2 + 5 is divided by 3x – 2, use remainder theorem to find
the remainder.

Solution:
Here, f(x) = 3x3 – 4x2 + 5

Infinity Optional Mathematics Book - 10 29

Comparing 3x – 2 with ax – b we get b = 2
a 3

Now, by remainder theorem

Remainder (R) =f b =f 2
a 3

=3 2 3 2 2
3 3
–4 +5

= 3 × 8 – 4 × 4 + 5
27 9

= 8 – 16 + 5
9 9

= 8 – 16 + 45
9

= 37
9

Factor theorem :

Statement : If a polynomial f(x) is divided by a linear polynomial (x – a), then
(x – a) is said to be a factor of f(x), if the remainder f(a) is 0.

Proof : When a polynomial f(x) is divided by a linear polynomial (x – a), then
we get quotient Q(x) and Remainder R.
Then, f(x) = (x – a) Q(x) + R
Putting x = a, we get
f(a) = (a – a) Q(a) + R
or, f(a) = R
When f(a) = 0, then R = 0, we get
f(a) = (x – a) Q (x)
So, (x – a) is a factor of f(x).

Converse of the factor theorem:
If (x – a) is a factor of the polynomial f(x), then remainder f(a) = 0
Proof: When (x – a) is a factor of f(x), then
f(x) = (x – a) Q(x)
Putting x = a, we get
f(a) = (a – a). Q(a)
∴ f(a) = 0 (= R)
Hence if (x – a) is a factor of f(x), then remainder f(a) = 0

30 Infinity Optional Mathematics Book - 10

Note: 1. If f(x) is a polynomial and a is a real number, then (x + a) is a factor of

f(x) when f(- a) = 0

2. (ax + b) is a factor of polynomial f(x), if f –b = 0, where a and b are
a
real numbers.

3. (ax – b) is a factor of a polynomial f(x), if f b = 0, where a and b are real
a
numbers.

WORKED OUT EXAMPLES

1. Show that (x – 3) is a factor of x3 – 4x2 + x + 6.

Solution:

Let f(x) = x3 – 4x2 + x + 6

Comparing (x – 3) with (x – a), then a = 3

Now, remainder = f(a) = f(3)

= 33 – 4(3)2 + 3 + 6

= 27 – 36 + 9

=0

Since remainder f(3) = 0

Hence, (x – 3) is a factor of f(x).

2. Show that (3x + 2) is a factor of 6x3 + 13x2 – 4.
Solution:
Let f(x) = 6x3 + 13x2 – 4

Comparing (3x + 2) with (ax + b), then –b = – 2
a 3

Now, remainder = f –b =f –2
a 3

=6× –2 3 –2 2
3 3
+ 13 –4

= – 6 × 8 + 13 × 4 – 4
27 9

= – 16 + 52 – 4
9 9

Infinity Optional Mathematics Book - 10 31

= –16 + 52 – 36 = 0
Since, remainder f 9
2
– 3 =0

∴ (3x + 2) is a factor of f(x).

3. If (x + 1) is a factor of the polynomial x3 – 5x2 – mx + 7, find the value
of m.

Solution:
Let, f(x) = x³ – 5x2 – mx + 7
Here, (x + 1) is a factor of f(x).
Then remainder f( – 1) = 0
or, ( – 1)3 – 5( – 1)2 – m( – 1) + 7 = 0
or, – 1 – 5 + m + 7 = 0
∴ m = – 1

4. The polynomial 3x3 + 2x2 – bx + a is exactly divisible by (x – 1) but
leaves a remainder of 10 when divided by (x + 4). Find the values of a
and b.

Solution:
Let, f(x) = 3x3 + 2x2 – bx + a

When, f(x) is exactly divisible by (x – 1), then
remainder f(1) = 0
or, 3(1)3 + 2(1)2 – b(1) + a = 0
or, 3 + 2 – b + a = 0
or, a = b – 5 …………… (1)

Again, when f(x) is divided by (x + 4); then remainder f( – 4) = 10
or, 3 ( – 4)3 + 2( – 4)2 – b( – 4) + a = 10
or, 3 × – 64 + 32 + 4b + a = 10
or, a + 4b = 170 …………… (2)

Putting the value of a from equation (1) in equation (2)

b – 5 + 4b = 170

or, b= 175
5

32 Infinity Optional Mathematics Book - 10

∴ b = 35
And, a = b – 5 (from equation (1))
a = 35 – 5
∴ a = 30
Hence, a = 30 and b = 35.

Zero of a Polynomial

Let f(x) be a polynomial in x. Then the value of x which when substituted in the given
polynomial, the value of the polynomial is zero called the zero of the polynomial.
For the polynomial f(x), if f(x) = 0 for x = a, then a is called zero of the polynomial.
The number of zeros of the polynomial is equal to the degree of that polynomial.

WORKED OUT EXAMPLES

1. Show that 3 is a zero of the polynomial f(x) = 2x – 6
Solution:
Here, f(x) = 2x – 6
When x = 3, then
f(3) = 2 ×3 – 6 = 6 – 6 = 0
∴ 3 is a zero of the given polynomial.
2. Show that 1 and – 3 are the zeros of the polynomial f(x) = x2 + 2x – 3
Solution:
Here, f(x) = x2 + 2x – 3
When x = 1, then
f(1) = 12 + 2 ×1 – 3 = 1 + 2 – 3 = 0
When x = – 3, then
f( – 3) = ( – 3)2 + 2( – 3) – 3 = 9 – 6 – 3 = 0
∴ 1 and – 3 are the zeros of the given polynomial.

Infinity Optional Mathematics Book - 10 33

Note: 1. If the sum of all the coefficients and constant term of the given polynomial
is zero then x = 1 is one of the zero of the polynomial and (x – 1) becomes
a factor.

2. If the sum of the coefficients of odd power is equal to the sum of the
coefficients of even power and constant term of the given polynomial
then x = – 1 is one of the zero of the polynomial and (x + 1) becomes a
factor.

3. Factorise: x3 – 4x2 – 7x + 10

Solution:

Here, x3 – 4x2 – 7x + 10

Here, the sum of all coefficients of the variables and constant term

= 1 – 4 – 7 + 10

= 0

∴ x = 1 is one of the zero of the given polynomial .

So, (x – 1) is a factor.

Now, splitting the second and third term so that the terms can be grouped to

have a factor (x – 1), we get

x3 – 4x2 – 7x + 10 = x3 – x2 – 3x2 + 3x – 10x + 10

= x2 (x – 1) – 3x(x – 1) – 10 (x – 1)

= (x – 1) (x2 – 3x – 10)

= (x – 1) {x2 – (5 – 2)x – 10}

= (x – 1) (x2 – 5x + 2x – 10)

= (x – 1) {x(x – 5) + 2(x – 5)}

= (x – 1) (x + 2) (x – 5)

Alternative method:

After finding (x – 1) as one of the factor, we can find other factor by using

synthetic division method.

1 x3 x2 x Constant
1 –4 –7 10
–10
↓1 –3 0

1 –3 –10 Remainder

x2 x Constant

34 Infinity Optional Mathematics Book - 10

∴ Remainder = 0

Quotient = x2 – 3x – 10

Now, x3 – 4x2 – 7x + 10

= (x – 1) (x2 – 3x – 10)

= (x – 1) (x2 – 5x + 2x – 10)

= (x – 1) {x(x – 5) + 2(x – 5)}

= (x – 1) (x + 2) (x – 5)

4. Factorise: 2x3 + 13x2 – 36

Solution:

Let, f(x) = 2x3 + 13x2 – 36

Here, constant term of the polynomial f(x) is 36 and the possible factors of 36

are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.

When x = 1, then

f (1) = 2 (1)3 + 13 (1)2 – 36

= 2 + 13 – 36

= – 21 ≠ 0.

So (x – 1) is not a factor of f(x).

When x = – 2, then

f ( – 2) = 2 ( – 2)3 + 13 ( – 2)2 – 36

= – 16 + 52 – 36

=0

∴ (x + 2) is a factor of f(x).

For other factors, using synthetic division method.


x3 x2 x Constant
–2 13 0 –36

2

↓ –4 –18 36

2 9 –18 0
x2 x Constant Remainder

∴ Remainder = 0
Quotient = 2x2 + 9x – 18
Now, 2x3 + 13x2 – 36
= (x + 2) (2x2 + 9x – 18)

Infinity Optional Mathematics Book - 10 35

= (x + 2) (2x2 + 12x – 3x – 18)
= (x + 2) {2x (x + 6) – 3(x + 6)}
= (x + 2) (x + 6) (2x – 3)

Polynomial Equation

Let f(x) be a polynomial in x. Then f(x) = 0 is called polynomial equation in x.
☻ 3x + 5 = 0 is the linear equation or first degree equation.
☻ 3x2 + 5x + 2 = 0 is the quadratic equation or second degree equation.
☻ x3 – 4x2 + x + 6 = 0 is the cubic equation or third degree equation.
If θ is a real number such that f(θ) = 0, then θ is known as a root of the polynomial
equation f(x) = 0.
Consider a polynomial equation x3 – 4x2 + x + 6 = 0.
When x = 2, then
(2)3 – 4 (2)2 + 2 + 6 = 8 – 16 + 8 = 0
∴ x = 2 is a root of a polynomial equation.
x3 – 4x2 + x + 6 = 0.
The number of the roots of the polynomial equation is equal to the degree of that
polynomial.

WORKED OUT EXAMPLES

1. Solve: 2x3 + 3x2 – 11x – 6 = 0
Solution:
Here, 2x3 + 3x2 – 11x – 6 = 0
Here, the possible factors of 12 are ±1, ±2, ±3, ± 6, ± 12.
when x = 1, then 2(1)3 + 3(1)2 – 11(1) – 6 = 0
or, 2 + 3 – 11 – 6 = 0
or, – 12 = 0 (false)
∴ (x – 1) is not a factor
When x = 2, then 2(2)3 + 3(2)2 – 11(2) – 6 = 0
or, 16 + 12 – 22 – 6 = 0
or, 28 – 28 = 0
or, 0 = 0
∴ (x – 2) is a factor of the given equation.
For other factor, using synthetic division method.

36 Infinity Optional Mathematics Book - 10

x3 x2 x Constant
2 2 3 –11 –6
↓ 4 14 6
73 0
2 x Constant
x2 Remainder

∴ Remainder = 0

Quotient = 2x2 + 7x + 3

Now, 2x3 + 3x2 – 11x – 6 = 0

or, (x – 2) (2x2 + 7x + 3) = 0

or, (x – 2) {2x2 + (6 + 1)x + 3} = 0

or, (x – 2) (2x2 + 6x + x + 3} = 0

or, (x – 2) {2x (x + 3) + 1(x + 3)} = 0

or, (x – 2) (x + 3) (2x + 1) = 0

Either x – 2 = 0

∴ x = 2

or, x + 3 = 0

∴ x = – 3

or, 2x + 1 = 0

∴ x = –1
2
–1
Hence x = 2, – 3, 2

Exercise 1.5

Section 'A'
1. Define reminder theorem. Use this theorem and find the remainder when

x3 + 5x2 – 3x + 4 is divided by x + 1.
2. What do you mean by factor theorem. Using this theorem, show that (x – 1) is

a factor of 5x3 + 3x2 – 2x – 6.

3. Use remainder theorem to find the remainder when
(a) x3 – x2 + 5x + 1 is divided by (x – 1).
(b) x4 – 3x2 + 5x + 7 is divided by (x – 2).

Infinity Optional Mathematics Book - 10 37

(c) 8x3 – 4x2 + 6x – 7 is divided by (2x – 1).
(d) 4x3 – 6x2 + 15x – 2 is divided by (x + 3).
(e) x5 – x3 + 20 is divided by (2x – 1).
(f) x4 – 3x3 – 5x + 7 is divided by (x – 2).

4. Using remainder theorem, find the value of k in the following cases.
(a) kx3 + 5x2 + 4x – 7 leaves a remainder 4 when divided by x + 3.
(b) x3 + 3x2 – kx + 4 leaves a remainder k when divided by x – 2.
(c) 2x3 – 5x2 + 7x – k leaves a remainder 15 when divided by x + 1.
(d) g(x) = 4x3 – 3x2 + 3x – k and g(2) = 12
(e) h(y) = 5y3 – 7y2 – ky + 3 and h (2) = k + 3
(f) f(p) = 3p3 – 7p2 – 7kp + 3 and f( – 1) = 5

5. Use factor theorem and show that:
(a) (x – 2) is a factor of x3 + x2 + x – 14
(b) (y + 3) is a factor of y3 + 4y2 + 5y + 6
(c) (3x + 2) is a factor of 6x3 + 13x2 – 4
(d) (2x – 1) is a factor of 8x3 + 16x – 9
(e) (3p – 6) is a factor of 3p3 – 15p + 6
(f) (x – 5) is a factor of x(x – 1) (x – 2)(x – 3) – 120
(g) (x + 1) is a factor of (x + 2)(x – 3)(x – 6)(x – 1) + 56

6. Decide whether or not
(a) (x – 3) is a factor of 2x3 – 5x2 – 6x + 8.
(b) (x + 2) is a factor of x3 – 3x2 – 4x + 12.
(c) (2y + 3) is a factor of 4y3 + 5y2 + 9y + 7.
(d) (x – 2) is a factor of x3 – 5x + 3 2.
(e) (2x – 1) is a factor of 5x3 – 3x – 5.

7. Find the value of k in each of the following cases, when
(a) (x – 1) is a factor of x3 – 5x2 – kx + 5.
(b) (x + 3) is a factor of 5x3 + 7x2 + x – k.
(c) (x + 1) is a factor of x3 + k2x2 – 6.
(d) (x + 3) is a factor of x3 + (k + 2)x2 + 5x + 3.
(e) (2x + 1) is a factor of 2x3 + 5kx2 + 5x + 2.

38 Infinity Optional Mathematics Book - 10

(f) (2x – 3) is a factor of kx3 + 5x – 7.

(g) (x – 1) is a factor of k2x3 + 5kx – 6.

Section 'B'

8. If (x – 2) is a factor of x2 + kx + 6, then show that (x – 3) is also its another
factor.

9. If (x – 1) is one of the factor of x3 – kx2 + 11x – 6, then show that (x – 2) is also
its another factor.

10. Apply synthetic division method and find the remainder and quotient
in each of the following cases:

(a) (5x3 + 3x2 – 5x + 7) ÷ (x – 1) (b) (3x3 + 7x2 – 10) ÷ (x + 2)

(c) (2x3 – 5x + 7) ÷ (x + 3) (d) (x3 – 5x + 2) ÷ (x – 2)

(e) (x4 – 3x2 – 7) ÷ (x – 3) (f) (x6 – 6) ÷ (x – 1)

(g) (4x3 – 3x2 + 5) ÷ (2x + 1) (h) (4y3 + 2y2 – 4y + 3) ÷ (2y + 3)

(i) (3p3 – 7p – 2) ÷ (2p – 1)

11. Find the remainder R and polynomial Q(x) in the following cases:

(a) 2x3 – 7x2 + x + 10 = (x + 1) Q(x) + R.

(b) x3 – 4x2 + 3x – 5 = (x – 2) Q(x) + R.

(c) 3x3 – 5x + 2 = (2x – 1) Q(x) + R.

(d) 4x3 – 3x – 5 = (2x + 3) Q(x) + R.

12. Find the value of k when one root of the polynomial 2y3 + 4y2 + ky + 6 is – 2.
13. If 2 is one of the zero of the polynomial x3 – kx – 2, then show that – 1 is also

a zero of the given polynomial.
Section 'C'

14. Find the values of k and m in each of the following cases:
(a) (x – 2) and (x – 3) are the factors of x3 – kx2 + x + m.
(b) (x – 5) and (x + 2) are the factors of x3 – mx2 – kx + 10.

(c) (x + 1) and (x – 2) are the factors of mx3 – kx – 2.

(d) 2y3 + ky2 + my – 2 has a factor (y + 2) and leaves a remainder 7 when
divided by 2y – 3.

(e) kp3 + 3p2 + mp – 3 has a factor (2p + 3) and leaves a reminder – 3 when
divided by (p + 2).

(f) kx2 + mx + 3 has a remainder 10 and 0 when divided by (x – 1) and
(2x + 3) respectively.

(g) mx3 – 8x2 + kx + 6 has a factor x2 – 2x – 3.

Infinity Optional Mathematics Book - 10 39

15. Factorize the followings: (b) 2x3 + 3x2 – 3x – 2
(a) x3 + 6x2 + 11x + 6 (d) z3 – 5z – 4
(b) 4x3 + 12x2 + 5x – 6 (f) y3 – 19y – 30
(e) 2x3 + 3x2 – 1
(g) (p – 1) (2p2 + 15p + 15) – 21

16. Solve the following:

(a) 2x3 + 3x2 – 3x – 2 = 0 (b) 4x3 + 12x2 + 5x – 6 = 0

(c) x3 + 6x2 + 11x = – 6 (d) 8y3 – 2y2 – 5y – 1 = 0

(e) y3 – 3y – 2 = 0 (f) 6x3 + 7x2 – x – 2 = 0

(g) k3 – 13k – 12 = 0 (h) (x + 1)(x2 – 5x + 10) – 12 = 0

(i) (p – 1)(2p2 + 15p + 15) – 21 = 0 (j) 6x3 + 11x2 – 26x – 15 = 0

1.3 Sequence and Series

Review

The arrangement of numbers arranged in a definite order determined by a rule is
called a sequence.

For example

(a) 1, 4, 7, 10…

(b) 1, 4, 9, 16…

(c) 1 , 41, 21, 1, 2,… etc.
8

When the terms of a sequence are connected by signs '–' or '+', then it is called a

series corresponding to the sequence. For example, 3, 6, 9, 12…is a sequence and 3

+ 6 + 9 + 12+ …is a series corresponding to the sequence.

Progression
A sequence or series is said to be progression, if the functional relation between its
any two successive terms is constant.
In the sequence 3, 6, 9, 12, 15…, the difference between any two successive terms
is 3. In the series 2 + 4 + 8 + 16 + 32 + …, the ratio of two successive term is 2. So
both of them are the progression.
There are two types of progressions:
(i) Arithmetic progression (A.P)
(ii) Geometric progression (G.P)

40 Infinity Optional Mathematics Book - 10

Arithmetic Progression (A.P)

Let us consider the two examples

(a) 3 + 7 + 11 + 15 + 19…

Here, the numbers are increased by a constant difference 4.

(b) 80 + 70 + 60 + 50 + 40 + …

Here, the numbers are decreased by a constant difference 10.

In the above two examples, the numbers are increased or decreased by a constant

difference. They are arithmetic progression. An arithmetic sequence or series (A.S)

is also known as arithmetic progression (A.P).

A sequence or series of numbers that increases or decreases with a constant

difference is called an arithmetic sequence or series.

The constant difference is called common difference and is denoted by d.

Common difference = succeeding term-preceding term = t2 – t1 = t3 – t2…
For example

(i) 7 + 12 + 17 + 22 + 27 + … (ii) – 5, – 10, – 15, – 20, – 25 – …

(iii) – 3, – 1, 1, 3, 5, … (iv) 90, 85, 80, 75, 70,…etc.

General term or nth term of an A.P. (or A.S)

In an A.P., the following notations are used.

First term = t1 = a.
common difference = d

number of terms = n.
last term or nth term = tn = l
sum of first n terms = sn
Let us consider an A.P

3 + 7 + 11 + 15 + 19 + …

common difference (d) = t2 – t1 = 7 – 3 = 4

First term (t1) = 3 = 3 + (1 – 1) × 4 = a + (1 – 1) d

2nd term (t2) = 7 = 3 + 4 = a + d = a + ( 2 – 1)d

3rd term (t3) = 11 = 3 + 8 = 3 + 2 × 4 = a + 2d = a + (3 – 1)d

4th term (t4) = 15 = 3 + 12 = 3 + 3 × 4 = a + 3d = a + (4 – 1)d

5th term (t5) = 19 = 3 + 16 = 3 + 4 × 4 = a + 4d = a + (5 – 1) d.

nth term (tn) = l = a + (n – 1)d.

Hence the nth term or the general term of an A.P. is tn = a + (n – 1)d.

Note: for first term, n = 1

∴ t1 = a + (1 – 1) d = a

Infinity Optional Mathematics Book - 10 41

For second term, n = 2

∴ t2 = a + (2 – 1) d = a + d
For third term, n = 3

t3 = a + (3 – 1) d = a + 2d

Similarly.

8th term (t8) = a + 7d
10th term (t10) = a+ 9d
15th term (t15) = a + 14d and so on.

WORKED OUT EXAMPLES

1. In the given arithmetic sequence 3, 8, 13, 18, ...., find i) common

difference (d) (ii) the general term (tn) and (iii) next three terms.
Solution:

Here, the given arithmetic sequence is 3, 8, 13, 18

(i) Common difference (d) = t2 – t1 = 8 – 3 = 5

(ii) The general term (tn) = a + (n – 1) d = 3 + (n – 1) 5

= 3 + 5n – 5 = 5n – 2

(iii) 5th term (t5) = a + 4d = 3 + 4 × 5 = 23
6th term (t6) = a + 5d = 3 + 5 × 5 = 28
7th term (t7) = a + 6d = 3 + 6 × 5 = 33.

2. Find the 10th term of an A.P. whose first term is 3 and common

difference is –2.

Solution:

Here, first term (a) = 3

common difference (d) = – 2

Now, 10th term (t10) = a + 9d
= 3 + 9 × (– 2) = – 15

3. Find the 50th term of the sequence 2, 6, 10…

Solution:

Here,the given sequence is 2, 6, 10,…

first term (a) = 2

common difference (d) = t2 – t1 = 6 – 2 = 4

Now, 50th term (t50) = a + 49d

= 2 + 49 × 4 = 198.

42 Infinity Optional Mathematics Book - 10

4. Find the number of terms of the give series.

3 + 6 + 9 + …+ 231
Solution :
Here, the given series is 3 + 6 + 9 + …+ 231
Here, first term (a) = 3
common difference (d) = t2 – t1 = 6 – 3 = 3
last term (tn) = 231
Now, we have
tn = a + (n – 1)d
or, 231 = 3 + (n – 1) 3
or, 231 = 3n
∴ n = 77

5. If 4th term and 9th term of an A.P. are 18 and 38 respectively, find the
series.

Solution:
Here, 4th term (t4) = 18
or , a + 3d = 18
or, a = 18 – 3d……………(i)
9th term (t9) = 38
or, a + 8d = 38
or, a = 38 – 8d………………(ii)

From equations (i) and (ii),
18 – 3d = 38 – 8d
or, 5d = 20
∴ d=4
putting the value of d in equation (i), we get
a = 18 – 3d = 18 – 3 × 4 = 6

Now, the required A.P. is
a, a + d, a + 2d,…
= 6, 6 + 4, 6 + 2 × 4,…
= 6, 10, 14,…

6. Is 350 a term of an A.S 10, 14, 18,…?
Solution:
Here, first term (a) = 10
common difference (d) = t2 – t1 = 14 – 10 = 4

Infinity Optional Mathematics Book - 10 43

If tn = 350, then
tn = a + (n – 1) d

or, 350 = 10 + (n – 1) 4

or, 340 + 1 = n
4

∴ n = 86

Since, the value of n is a whole number,

∴ 350 is a term of the given A.S.

7. If 3rd term and 11th term of an A.P. are 18 and 58 respectively. Find

the 25th term.

Solution:

Here, 3rd term (t3) = 18
or, a + 2d = 18

or, a = 18 – 2d …(i)
and 11th term (t11) = 58
or, a + 10d = 58

or, a = 58 – 10d … (ii)

From the equations (i) and (ii),

18 – 2d = 58 – 10d

or, 8d = 40

∴ d = 5

putting the value of d in equation (i), we get

a = 18 – 2 × 5 = 8

Again, 25th term (t25) = a + 24d

= 8 + 24 × 5 = 8 + 120 = 128

Exercise 1.6

Section 'A'

1. Determine with reason, which of the following are in arithmetic
progression:

(a) 4, 7, 10, 13 … (b) 4, 7, 2, –3 …
(c) 5 + 12 + 15 + 20 …
(e) 2, 4, 8, 16 … (d) –3 , –2, –25 , – 3 …
2

44 Infinity Optional Mathematics Book - 10

2. From the following arithmetic progressions given below, find: (i)
the common difference (ii) the general term (tn) (iii) the next three
terms.

(a) 2, 6, 10 … (b) –5 –8 –11 …

(c) 90+70+50+ … (d) 5, 1, – 3, …

3. Write down the first 4 terms of the following arithmetic sequences:

(a) First term(a) =7, common difference (d) =4.

(b) First term (a)= – 50, common difference (d) =10.

(c) First term (a) = 125, common difference (d) = –15.

(d) First term (a) = 5, common difference (d) = 43.
Section 'B'

4. Determine the 5th term and 15th term of the given arithmetic sequences:

(a) 5, 9, 13…

(b) –7, –2, 3, …

(c) First term = 5 and common difference = 7.

(d) First term= 100 and common difference = –8.

(e) First term = 80 and second term = 85.

5. Find the first term from the following arithmetic progressions.
(a) Common difference =5 and 8th term = 35.
(b) Common difference = -8 and 10th term = 240.

6. Find the common difference of the following series:
(a) First term = 2 and 9th term = 34.
(b) First term = 150 and 12th term = 40.

7. Find the number of terms in each of the following arithmetic
progressions:

(a) First term = 4, common difference =3, last term = 25

(b) First term= – 8, common difference =5, last term= 87

(c) First term = 15, second term = 25, last term = 115.

(d) 5 + 12 + 19 + …+ 54

8. Which term of the series:

(a) 1 + 3 + 5 + 7 + … is 23 ? (b) 15, 10, 5 …is – 200 ?

Infinity Optional Mathematics Book - 10 45

9. (a) Is 73 a term of the series 7 + 9 + 11 + … ?
(b) Is –150 a term of the series 150 + 142 + 134 + … ?
10. Find the first term and common difference for the following arithmetic

sequences:
(a) 5 th term = 10 and 9th term = 18
(b) 8th term = - 31 and 21st term = – 83
11. Find the terms indicated as follows:
(a) 5th term = 15 and 11th term = 33, 15th term=?
(b) 17th term= –54 and 14th term = –42, 25th term =?
12. (a) If 5th term and 8th term of an A.P. are 19 and 31 respectively, which term

is 67?
(b) The 11th term and 15th term of an A.P. are 23 and 35 respectively. Is 49

a term of the series ?
(c) An A.P. contains 60 terms. If the 1st term and the last term are 7 and 125

respectively, find the 34th term.
(d) The 12th term, 85th term and nth term of an A.P. are 38, 257 and 395

respectively. Find the no. of term in nth term.
(e) If the nth term of an A.P. 23, 26, 29, 32, …and nth term of A.P. 59, 58, 57,

56, …are equal, then find the value of n.
13. (a) The fourth term of an A.P. is three times the first term and the seventh

term exceeds twice the third term by 1. Find the 10th term.
[Hint: t4 = 3t1 & t7 = 2t3 + 1]
(b) If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, show

that the 18th term is zero.
14. Find the three numbers in A.P., whose common difference is 1, such that the

product of the second and the third exceeds that of the first and second by 12.

46 Infinity Optional Mathematics Book - 10

Arithmetic Mean (A.M.)

Consider the following arithmetic sequences

(i) 2, 5, 8 (ii) 15, 19, 23, 27 (iii) 100, 90, 80, 70, 60, 50, 40

A.M A.Ms' A.Ms'

In the above A.P. the terms except the first term and last term are the arithmetic
means.

Thus, the terms to be placed between first term and last term of an arithmetic
progression are called arithmetic means.

Let us consider the three terms a, m and b of an A.P.

Then, m is the A.M. of a and b. When a, m, b are in A.P, their common difference
are equal.

i.e. m – a = b – m [ t2 – t1 = t3 – t2]

or, m + m = a + b

∴ m= a+b
2

Hence, A.M. between a and b = a+b
2

The arithmetic means between first term and last term are denoted by m1, m2, m3,
m4, m5 …etc.

where m1 = t2, m2 = t3, m3 = t4, m4 = t5 and so on.

WORKED OUT EXAMPLES

1. Find the arithmetic mean between 10 and 16.

Solution:

Here, the two given number are 10 and 16.

Now, A.M. between 10 and 16

= 10 + 16 = 13 [ A.M. = a + b ]
2 2

2. Find the 10th term of A.P., whose 9th term = – 50 and 11th term = 100.

Solution:

Here, 10th term is the arithmetic mean of 9th term and 11th term.
9th term + 11th term
10th term = 2

Infinity Optional Mathematics Book - 10 47

= 50 + 100
2

= 25

3. Find the value of x and y when 5, x, 15, y are in A.P.

Solution:

In the given A.P. 5, x, 15, y,

x is the A.M. of 5 and 15.

So, x= 5 + 15 = 10
2

Again,15 is the A.M. of x and y.

So, 15 = x+y
2

or, 30 = 10 + y

∴ y = 20

Hence, x = 10 and y = 20.

4. Insert 4 arithmetic means between 3 and 23.

Solution:

Here, first term (a) = 3

Last term (tn) = 23
Number of terms (n) = 4 + 2 = 6

Now, tn = a + (n – 1) d

or, 23 = 3 + (6 – 1) d
250 = d

∴ d=4

Now, first mean (m1) = t2 = a + d = 3 + 4 = 7
second mean (m2) = t3 = a + 2d = 3 + 2 × 4 = 11
third mean (m3) = t4 = a + 3d = 3 + 3 × 4 = 15
fourth mean (m4) = t5 = a + 4d = 3 + 4 × 4 = 19
Hence, the four A.MS' are 7, 11, 15, 19.

5. If 8, p, q, r, 24 are in A.S. find the value of p, q, and r.

Solution:

Here,the given A.S. is 8, p, q, r, 24

Here,first term (a) = 8

Last term (tn) = 24

48 Infinity Optional Mathematics Book - 10

No. of terms (n) = 5
Now, tn = a + (n – 1)d
or, 24 = 8 + (5 – 1) d
or 24 – 84 = d
∴ d = 4
Hence, first mean = (p) = t2 = a + d = 8 + 4 = 12
second mean = (q) = t3 = a + 2d = 8 + 2 × 4 = 16
third mean = (r) = t4 = a + 3d = 8 + 3 × 4 = 20
6. There are 6 A.M. between x and y. If the 2nd mean and 5th mean are

11 and 23 respectively, then find the value of x and y.
Solution:
Here, 2nd mean (m2) = t3 = 11
or, a + 2d = 11
or, a = 11 – 2d … (i)
5th mean (m5) = t6 = 23
or, a + 5d = 23
or, a = 23 – 5d … (ii)
From equations (i) and (ii),
11 – 2d = 23 – 5d
or, 3d = 12
∴ d = 4
Putting the value of d in equation (i), we get
a = 11 – 2d = 11 – 2× 4 = 3
∴ First term (a) = x = 3
No. of terms (n) = 6 + 2 = 8
Now, last term (tn) = y = a + (n – 1)d
i.e. y = 3 + (8 – 1)4
= 3 + 28 = 31
Hence, x = 3 and y = 31

Infinity Optional Mathematics Book - 10 49

7. Some A.M. are inserted between 7 and 43. If the 5th mean is 27, find
the number of A.M.

Solution:

Here, First term (a) = 7

Last term (tn) = 43
5th mean (m5) = t6 = 27
or, a + 5d = 27

or, 7 + 5d = 27

or, d = 20
5

∴ d = 4

Again, tn = a + (n – 1)d

or, 43 = 7 + (n – 1)4

or, 346 = n – 1
∴ n = 9 + 1 = 10

Hence, no. of A.M. = n – 2 = 10 – 2 = 8

Note: In an A.P.

Term t1 t2 t3 t4 t5 t6 t7 t8 ...... tN-3 tN - 2 tN - 1 tN

Mean m1 m2 m3 m4 m5 m6 m7 ...... mn-2 mn-1 mn

a a+b a+2d a+3d a+4d a+5d a+6d a+7d ...... tN–3d tN–2d tN – d

8. There are n arithmetic means between 4 and 44. The ratio of 3rd mean
to the 2nd last mean (i.e. (n – 1) mean) is 4:9. Find the value of n.

Solution:

Here, First term (a) = 4

Last term (tN) = 44
No. of terms (N) = n + 2

By question

3rd mean: 2nd last mean = 4:9

i.e. 3rd mean = 4
2nd last mean 9

or, 3rd4lthasttertmerm = 4
9

or, 4 + 3d = 4
44 – 2d 9

50 Infinity Optional Mathematics Book - 10