Vedanta Opt. Math 9 Final (2078)

vedanta Excel in Opt. Mathematics - Book 9 Polynomials

5. (a) p(x) = x2 + x, q(x) = x2 – 4x, then verify that p(x).q(x) = q(x).p(x)
(b) If f(x) = x + 4, g(x) = x + 7, verify that f(x). g(x) = g(x). f(x)

Long Questions
6. (a) If f(x) = 11x2 – 5x + 7, g(x) = 13x2 + 5x – 9 and h(x) = 3x2 – 6x + 1, then find the

value of {f(x) + g(x)} –h(x).
(b) If f(x) = x2 + 4x + 3, g(x) = x2 + 7x + 5 and h(x) = x – 7, find the value of

{f(x) + g(x)} – h(x)
7. (a) What should be subtracted from x3 – 3x2 + 6 to get x3 – 4x2 + x + 2 ?

(b) What should be subtracted from sum of y3 + 2y + 1 and y2 + 6y + 2 to get 6y + 8 ?
8. Find the product of p(x) and q(x) when

(a) p(x) = x2 – 2x + 1, q(x) = x3 – 6x2 + x + 1.
(b) p(x) = x2 + x + 1, q(x) = x2 – x + 1
9. Let p(x) = x2 + x + 2, q(x) = 2x2 + 3x + 5 and r(x) = x + 2, then verify that
{p(x) + q(x)} + r(x) = p(x) + {q(x) + r(x)}.
10. (a) Let f(x) = x – 1, g(x) = x2 + x + 1, h(x) = x2 + 2x – x3, then verify that

{f(x) . g(x)} h(x) = f(x) {g(x). h(w)}
(b) Let p(x) = x – 4, q(x) = x + 4 and r(x) = x2 + 16, then verify that

{p(x). q(x)} r(x) = p(x) {q(x). r(x)}.
11. Find the additive inverse of each of the following given polynomial:

(a) p(x) = x4 + 4x3 + 6x2 + 7x + 8
(b) q(x) = 4x4 – 6x3 + 7x2 + 22x + 2

1. (a) 3x2 + 10 (b) 7x3 – 4x2 + 4x (c) 0

2. (a) 3x4 – 2x2 + 9x – 10 (b) –x2 – 7x – 5

3. (a) x2 – y2 (b) x3 – 1 (c) x3 + y3

6. (a) 21x2 + 6x – 3 (b) 2x2 + 10x + 15 7.(a) x2 – x + 4 (b) y3 + y2 + 2y – 5

8. (a) x5 – 8x4 + 14x3 – 7x2 – x + 1 (b) x4 + x2 + 1

11. (a) – x4 – 4x3 – 6x2 – 7x – 8 (b) – 4x4 + 6x3 – 7x2 – 22x – 2

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Sequence and Series 3

3.0 Review

Group discuss the answers of the following questions :

(a) What is the degree of polynomial 4x + 5 ?

(b) If f(x) = n(n + 1) , find f(1), f(2), f(3).
2

(c) Do you find any rule in the given pattern of numbers 1, 2, 4, 8, 16, ...... ?

3.1 Introduction to Sequence

Let us observe the following patterns of numbers:

(a) 1, 2, 3, 4, ................ (b) 100, 98, 96, 94, .................

(c) 1, 5, 25, 125, .......... (d) 2, 8, 15, 19, 22, ................

Can you write next two numbers in each pattern ?
In (a), each number is increased by 1 to get the next term/ number.

In (b), each number is decreased by '2' to get the next term.

In (c), each number is multiplied by 5 to get the next term.

In each pattern of numbers in (a), (b), and (c), there is a certain rule which shows that how
the numbers are arranged. Hence, they are sequences.

But in (d), there is no certain pattern. Hence, it does not represent a sequence.

Definition : A sequence is an arrangement of numbers in a definite order according to some
special rules. The arranged numbers of the sequence are called the terms of the sequence
and they are denoted by t1,t2,t3 ..........

A sequence may have the finite or infinite number of terms. A sequence with the finite
number of terms is called a finite sequence. For Example, 1, 2, 4, 6, 8, ......, 20 form a finite
sequence.

A sequence with infinite number of terms is called an infinite sequence. In an infinite
sequence, the final term is not fixed. For example, in sequence of natural numbers, the final
term can not be written.

Generally the terms of a sequence are denoted by t1, t2, t3, etc.

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

3.2 General Term of Sequence

Let us consider a sequence of numbers 4, 6, 8, 10 ........ This sequence can be written as

46 8 10

22 2

Difference : Here, differences 6 – 4 = 2, 8 – 6 = 2, 10 – 8 = 2 This constant difference is
common difference. Each term is increased by 2 than the preceding term. Then we try to
find the nth form of the sequence.

t1 = 4 = 4 + (1 – 1) 2

t2 = 6 = 4 + (2 – 1) . 2

t3 = 8 = 4 + (3 – 1) . 2

t4 = 10 = 4 + (4 – 2 ).2

the nth term = 4 + (n – 1). 2

= 4 + 2n – 2

= 2n + 2

In above sequence, each consecutive pair of terms has a common difference.

The nth term of a sequence is called the general term of the sequence. From the formula of
general term, we can find the required term of the sequence. For example, in above sequence
of numbers, tn = 2n + 2

Put n = 100, we get, t100 = 2 × 100 + 2 = 202.

General Term of a Linear Sequence

The sequence in which each consecutive pair of terms have a common difference is called
a linear sequence. In this type of sequence, the general term is of the first degree of the form
tn = an + b.

In this type of sequence, terms are in the form of a + b, 2a + b, 3a + b, 4a + b, ....... , an + b.

Here, a+b 2a+b 3a+b 4a+b

First Difference a a a

Worked out Examples

Example 1. What are the next two terms in the following sequence of numbers ?
Solution :
(a) 1, 3, 5, 7, ..... (b) 1, 2, 4, 8, ........ (c) 100, 95, 90, .............

(a) Here, 1, 3, 5, 7, ........ the difference between two successive terms is 2,
Each term is increased by 2 than the preceding term.

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Example 2. t1 = 1, t2 = 1 + 2 = 3, t3 = 3 + 2, = 5, t4 = 5 + 2 = 7
Solution : Next two terms are, t5 = 7 + 2 = 9

Example 3. t6 = 9 + 2
Solution : (b) Here, 1, 2, 4, 8, ............., each term of the sequence is two times the

preceding term.

t1 = 1, t2 = 2t1 = 2 × 1 = 2, t3 = 2t2 = 2 × 2 = 4, t4 = 2t3 = 2 × 4 = 8
Next two terms are given by t5 and t5

t5 = 2 × t4 = 2 × 8 = 16
t6 = 2 × t5 = 2 × 16 = 32
(c) Here, 100, 95, 90, ......, the difference between two successive terms is
–5. Each term of the sequence is less than the preceding term by 5.

t1 = 100
t2 = t1 – 5 = 100 – 5 = 95
t3 = t2 – 5 = 95 – 5 = 90
Next two terms are given by t4 and t5 .
t4 = t3 – 5 = 90 – 5 = 85
t5 = t4 – 5 = 85 – 5 = 80

If f(n) = n2 + 5 is a function, where n represents a natural number. Find the
first five terms and write them in a sequence.
Here, f(n) = n2 + 5

We get the first 5 terms putting n = 1, 2, 3, 4, 5 successively.
f(1) = 12 + 5 = 1 + 5 = 6
f(2) = 22 + 5 = 4 + 5 = 9
f(3) = 32 + 5 = 9 + 5 = 14
f(4) = 42 + 5 = 16 + 5 = 21
f(5) = 52 + 5 = 25 + 5 = 30

Hence, the sequence of the numbers is written as 6, 9, 14, 21, 30,.......

Find the general term of 6, 10, 14, 18 ......

Here, given sequence is 10 14 18
6

First difference (a) = 4 44 4
Let, tn = an + b
a = the first common difference = 4
t1 = a.1 + b,

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

or, 6 = 4 + b
? b=2
Hence, tn = an + b = 4n + 2
? tn = 4n + 2
Alternative Method
In linear sequence, the general term is given by tn= a + (n – 1). d
where, a = the first term

n = the number of terms
d = the common difference
We have a sequence of of numbers 6, 10, 14, 18, ........
To find the general term of the sequence, we have the first term (a) = t1 = 6
Common difference (d) = t2 – t1 = 4
Then, tn = a + (n – 1) d = 6 + (n – 1).4 = 6 + 4n – 4 = 4n + 2
? tn = 4n + 2.

General Term of a Quadratic Sequence

The sequence of numbers in which the first difference is not constant but the second
difference constant is called quadratic sequence. General term of a quadratic sequence is
given by,

tn = an2 + bn + c.
where, a, b, and c are constants and n is the number of terms.

tn = an2 + bn + c

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

t2 = a.22 + b.2 + c = 4a + 2b + c

t3 = a.32 + b.3 + c = 9a + 3b + c

t4 = a.42 + b.4 + c = 16a + 4b + c.

Here, a+b+c 4a+2b+c 9a+3b+c 16a+4b+c

First difference 3a+b 5a+b 5a+b

Second difference 2a 2a

The second differences are constant.
Hence, the general term is given by tn = an2 + bn + c.

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Example 4. Find the general term of the sequence 1, 3, 6, 10, .........

Solution : Here, 13 6 10

First difference 2 3 4

Second difference 1 1
tn = an2 + bn + c, second difference 2a = 1

? a = 1
2

For n = 1, t1 = a.1 + b. 1 + c

or, 1 = 1 + b + c
2
1
or, b + c = 2 ........... (i)

For n = 2, t2 = a.22 + b.2 + c

or, 3 = 1 . 4 + 2b + c
2

or, 3 = 2 + 2b + c

? 2b + c = 1 ............. (ii)

Multiplying equation (i) by '2' and and (i) by '2' and subtraction it from (ii)
2b + c = 1
–2b +–2c = 1–
c=0

Put the value of c in equation (i), we get,

Hence, the required general term is given by

tn = an2 + bn + c

= 1 n2 + 1 n + 0 = 1 n (n + 1)
2 2 2

Example 5. Find the general term of the sequence 6, 11, 18, 27, .......
Solution :
Given sequence is 6, 11, 18, 27, ........

Here, 6 11 18 27

First difference 5 7 9

Second difference 2 2

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Since the second difference is constant, the sequence is quadratic sequence.

Then, the general term is given by,

tn = an2 + bn + c i.e., a = 1
The second common difference = 2a = 2,

For n = 1, t1 = 1. 12 + b.1 + c
or, 6 = 1 + b + c

? b + c = 5 ........... (i)

For, n = 2, t2 = 1. 22 + b.2 + c
or, 11 = 4 + 2b + c

? 2b + c = 7 ......... (ii)

Subtracting equation (i) from (ii), we get,
2b + c = 7

– b +–c = –5
b=2

Put the value of b in equation (i), we get,

c=3

? tn = an2 +bn + c = n2 + 2n + 3

Definition : A sequence of numbers is said to be in a geometric sequence if ratio of any term
to its preceding term is constant throughout the whole. The constant is called common ratio.
It is denoted by r.

General Term of a Geometric Sequence

Consider the sequence : 1, 3, 9, 27, ..........

Common ratio (r) = tk + 1 , k  N.
tk
Here,
1 3 9 27

First common ratio (r) = 3 3 3 3

This type of sequence is called geometric sequence.

Hence, general term is given by a formulae:

tn = arn–1
where, r = the common ratio

a = the first term

n = the number of term.

Now, tn = arn–1 = 1. 3n–1 (from above example)
? tn = 3n–1

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Example 6. Find the general term of the sequence : 5, 25, 125, 625, ........

Solution : Here, 5 25 125 625

The first common ratio (r) 5 5 5

The first common ratio (a) = 5 is a constant .

Hence, the general term is given by,

tn = arn–1
where, a = 5, r = 5,

or, tn = 5 . 5n–1

? tn = 5n

Example 7. Find the general term of the sequence 3 , 4 , 5 , 6 , ...........
Solution 4 9 16 25
3 4 5 6
Example 8. Given sequence is 4 , 9 , 16 , 25 , ...........
Solution :
t1 = 3 = 1 + 2 t2 = 4 = 2 + 2
4 (1 + 1)2 9 (2 + 1)2
5 3+2 6 4+2
t3 = 16 = (3 + 1)2 t4 = 25 = (4 + 1)2

Hence the general term of the sequence tn = n + 2
(n + 1)2

Find the general term of the sequence : – 1 , 2 , – 3 , 4 , – 5 .........
2 3 4 5 6

The general term of the sequence 1, 2, 3, 4, 5, ....... is n. The general term

of the sequence of 2, 3, 4, 5, 6..... is (n + 1). The given sequence has signs

negative and positive alternately. It can be shown by (–1)n.

Now, t1 = – 1 = (–1)1 1
2 1+1

t2 = 2 = (–1)2 2
3 2+1

t3= – 3 = (–1)3 3
4 3+1

t4 = 4 = (–1)4 4
5 1+2

t5 = 5 = (–1)5 5
6 5+1

? tn = (–1)n n is the required general term.
n+1

Exercise 3.1

Very Short Questions :
1. (a) Define a sequence.

(b) Define a finite sequence with an example.
(c) Define an infinite sequence with an example.

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

2. (a) Define general term of a sequence with an example.

(b) Define linear sequence with an example.

(c) Define a quadratic sequence with an example.

(d) Write general term formula of linear sequence

(e) Write general term formula of quadratic sequence.

3. (a) Write next two terms of the sequences in each of given below:

(i) 1, 2, 3, ..... (ii) 2, 4, 6, ...... (iii) 100, 90, 80, .......

(b) Write a rule for each of sequence of numbers of given below :

(ii) 1, 4, 7, 10, ......... (ii) 200, 250, 300, ......

(iii) 100, 50, 25, ........ (iv) 2, 6, 18, 54

4. Find the first difference of the following sequences :

(a) 4, 8, 12, 16, 10 .................. (b) 2, 5, 8, 11, 14, ..................

(c) 18, 14, 10, 6 ......................

5. Find the ratio of consecutive pair of terms in the following sequences :

(a) 1, 2, 4, 8, 16,................. (b) 5, 10, 20, 40, .................

(c) 4 , 16 , 64 , 256
5 25 125 625

Short Questions :

6. In the following functions, find the first five terms and then write them in sequence:

(a) f(n) = 2n + 1 (b) f(n) = n2 – 1
(d) f(n) = n(n + 1)
(c) f(n) = n(n + 1) (f) f(n) = (–1)n + 1 . n2
2
n(n + 1) (2n + 1)
(e) f(n) = 6

(g) f(n) = (– 1)n . n

7. From the following general term (tn) find the first five terms:

(a) tn = 3n + 2 (b) tn = 4n – 1 (c) tn = n2 + 4n + 6

(d) tn = n2 + 5 (e) tn = n2 (–1)n 1 (f) tn = n n2 1
+n– +
n(n + 1)(n + 2) (–1)n +1
(g) tn = 3 (h) tn = n+
1

8. Find the general term (tn) of the following sequences.

(a) 1, 4, 7, 10, ............... (b) 2, 6, 10, 14, .................

(c) 5, 3, 1, –1, –3, .................. (d) 100, 80, 60, ................

(e) 1, 4, 16, 64, .................. (f) 3, 9, 27, 81, ..................

9. If a1, a2, a3 ........... are the terms of a sequence.

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

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

(b) If an+1 = 2an and a1 = 3, find the values of a2 and a3.
(c) If an+1 = 3an and a1 = 2, find the value of a2 and a3.
Long Questions :

10. Find the general term of the given sequences :

(a) 3, 6, 10, 15,...... (b) 11, 18, 27, 38, ..........

(c) 2, 6, 12, 20, 30 ....... (d) 3, 4, 7, 12, ...........

(e) 3, 10, 25, 48, ........... (f) 1 , 4 , 7 , 10 ..............
(g) –4, 9, –16, 25, ............... 3 5 7 9

(h) – (x + y) , (x + y)2 , – (x + y)3 , (x + y)4 , ...........
1 2 3 4

(i) (m – 1) , (m – 1)2 , (m – 1)3 , (m – 1)4 ..............
2 4 8 16

(j) 2 × 3, 3 × 4, 4 × 5, 4 × 5, 5 × 6, ............

(k) 1 1 2 , 2 2 3 , 3 3 4 , 4 4 5 , ..............
× × × ×

11. From the following sequence of figures, find the general term.

(a)

(b)

(c)

(d)

Project Work

12. Prepare a short note on "Sequence and Series", including the followings:

(i) Definition (ii) Linear Sequence (iii) Quadratic Sequence (iv) General term

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

2. (d) tn = an + b (e) tn = an2 + bn + c
3. (a) (i) 4, 5
(ii) 8, 10 (iii) 70, 60

(b) (i) each term is increased by 3 than the preceding term.

(ii) each term is increased by 50 than the preceding term.

(iii) each term is multiplied by 1 to get the next term.
2

(iv) each term is multiplied by 3 to get the next term.

4. (a) 4 (b) 3 (c) –4 5. (a) 2 (b) 2 (c) 4
6. (a) 3, 5, 7, 9, 11 (b) 0, 3, 8,15, 24 (c) 1, 3, 6, 10, 15 5

(d) 2, 6, 12, 20, 30

(e) 1, 5, 14, 30, 55 (f) 1, –4, 9, –16, 25 (g) –1, 2, –3, 4, –5

7. (a) 5, 8, 11, 14, 17 (b) 3, 7, 11, 15, 19 (c) 11, 18, 27, 51 (d) 6, 9, 14, 21, 30

(e) – 1, 1 , – 1 , 1 , – 1 (f) 1 , 4 , 9 , 16 , 25
5 11 19 29 2 3 4 5 6
1 1 1 1 1
(g) 2, 8, 20, 40, 70 (h) 2 , – 3 , 4 , – 5 , 6

8. (a) 3n – 2 (b) 4n – 2 (c) 7 – 2n (d) 120 – 20n

(e) 4n – 1 (f) 3n 9.(a) 3, 1 (b) 6, 12 (c) 6, 18

10. (a) (n + 1) (n + 2) (b) n2 + 4n + 6 (c) n(n + 1) (d) n2 – 2n + 4
2
3n – 2 (x + y)n
(e) 4n2 – 5n + 4 (f) 2n + 1 (g) (–1)n (n + 1)2 (h) (–1)n n

(i) m–1 n (j) (n + 1) (n + 2) (k) n
2 n(n + 1)

11. (a) 2n – 1 (b) n + 2 (c) 2n + 1 (d) 2n + 1

3.3 Introduction to series

Group discuss the answer of the following questions:

(a) From the general term, tn = n2 + n, find the sum of the first five terms.

(b) Does t1 + t2 + t3 + t4 + t5 represent the sum of the first five terms of a sequence ?

5

(c) Can you write n = 1tn for the sum t1 + t2 + t3 + t4 + t5 ?
Example : 1, 2, 3, 4, 5 is the sequence of the first five natural numbers. Its sum is written as

1 + 2 + 3 + 4 + 5.

It is called a series corresponding to the sequence 1, 2, 3, 4, 5.....

Definition : When the numbers of a sequence are connected by plus or minus signs, it is
called a series.

For Examples : (a) 1 + 2 + 3 + .......... + 10 is a finite series

(b) 1 + 2 + 3 + 4 + ............... is an infinite series.

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Definition : A sequence or series is called a progression if there exists a functional relation
between its successive terms.

Sigma Notation :

If t1, t2, t3, ......., tn is a finite sequence, the series associated with the sequence is
t1 + t2 + t3 + .......... + tn.

n
The sum of the series is denoted by Sn = t1 + t2 + t3 + ,,,,,,,, + tn = n = 1tn
n
Here, '∑' is called sigma notation or sign of summation. 1tn is read as summation of
tn, when n runs from 1 to n. n =

Partial Sum

Let t1, t2, t3, ............,, tn be a sequence. The sum of terms of this sequence is written as,

n

Sn = t1 + t2 + t3 + ............ + tn = n = 1tn.
5

Then S5 = t1 + t2 + t3 + t4 + t5 = n = 1tn is said to be the partial sum.
Also, let S4 = t1 + t2 + t3 + t4

S5 = t1 + t2 + t3 + t4 + t5 be the partial sum of the first gour and five terms of a sequence.
respectively. we get,

S5 – S4 = (t1 + t2 + t3 + t4 + t5) – (t1 + t2 + t3 + t4) = t5
Hence, tn = Sn – Sn–1

Worked Out Examples

Example 1. General term of a sequence is tn = 3n + 2. Then
Solution : (a) Find the first five terms of the sequence (b) Find the sum of the terms.

(c) Write the series of the sequence using '∑' notation.

(a) Here, tn = 3n + 2
t1 = 3.1 + 2 = 5
t2 = 3.2 + 2 = 6 + 2 = 8
t3 = 3.3 + 2 = 9 + 2 = 11
t4 = 3.4 + 2 = 12 + 2 = 14
t5 = 3.5 + 2 = 15 + 2 = 17

(b) Sum of the first five terms; so, S5 = t1 + t2 + t3 + t4 + t5
= 5 + 8 + 11 + 14 + 17
= 55

5

(c) In sigma notation, we write (3n + 2)
n=1

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

Example 2. If Sn ann(nd2+an1a)r,efitnhde sum of n terms and nth term of a sequence respectively.
Solution : Sn = S3, S4 and a4.

Example 3. Here, Sn = n(n + 1)
Solution : 2
Example 4. 3(3 + 1) 3.4
Solution : S3 = 2 = 2 = 6
Example 5.
S4 = 4(4 + 1) = 2. 5 = 10
2

Now, a4 = S4 – S3 = 10 – 6 = 4

4

Write in the expanded form and find the sum of (4n + 3)
n=1
4

Here, (4n + 3) = (4.1+3) + (4.2+3) + (4.3+3) + (4.4 + 3)
n=1
= 7 + 11 + 15 + 19 = 52

How many terms are there in the series given in the form of sigma notation
20
3n2 + n
n+1 ? It represents a series.

n=4
The number of terms in the series is (20 – 3) = 17

Eleventh term is t11 = 3.112 + 11 = 363 + 11 = 374 = 187 = 3116
11 + 1 12 12 6

Study the given patterns of the figures and answer the following questions:

Solution : Then,
(a) Add two more figures in the same pattern.
(b) Find the general term of the sequence
(c) Find the 10th term of the sequence.
(a) Two more figures in the same patterns are given below:

(b) The sequence of number of dots in above figures.

14 7 10

First difference 3 3 3

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

The above sequence is a linear sequence.

Its general term is tn = an + b
a = the first constant difference

i.e., a = 3 ( t1 = 1) tn = 3n + b
For, n = 1 t1 = 3.1 + b
or, 1 = 3 + b ?
? b = – 2 tn = 3n – 2
Alternative Method,

Here, 1, 4, 7, 10, ..... are in linear sequence.

Its common difference d = 4 – 1 = 7 – 4 = 10 = 3

First term (a) = 1

General term (tn) = a + (n – 1) . d = 1 + (n – 1) . 3
? tn = 3n – 2
(c) The tenth term is t10 = 3.10 – 2 = 30 – 2 = 28.

Exercise 3.2

Very Short Questions :

1. (a) Define a series.

(b) What is sigma notation used for ?

(c) Write the formula to calculate the nth term when sums of the first n term and (n –1)
terms are given.

2. State which of the following are sequence or series.

(a) 1, 2, 3, 4, ................... 10 (b) 4 + 8 + 12 + 16.

14 (d) {2n + 5}
(f) 1 – 3 + 9 – 27 + 81
(c) (2n + 5)
n=2

(e) 95 + 90 + 85 + 80 + 75 + 70

(g) {(2, 4), (3, 9), (4, 16), (5, 25)} (h) 1 + 5 + 25 + 125 + 625

Short Questions :

3. Write the series corresponding to given sequence using sigma notation '∑'.

(a) 1, 2, 3, 4, ............... 40

(b) – 1, 2, – 3, 4, –5, 6, –7, 8, –9, 10

(c) (x – 1), (x – 2)2, (x – 3)3, (x – 4)4, (x – 5)5

(d) 3 , 4 , 5 , 6 , 7
5 6 7 8 9

(e) a, ar, ar2, ar3, ar4

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

4. Expand and find the sum of the given series

5 8 8

(a) 4n (b) (3n + 1) (c) (n2 + 1)
n=1 n=1 n=4

(d) 5 2n – 1 (e) 5 n(n + 1) 6
2n + 1 2
n=1 n=1 (f) (–1)n (2n + 3)
n=1

5. If Sn and an are the sum of n terms and in terns of a sequence. respectively. Then,

(a) If Sn = 2n + 1, find S3, S4 and a4

(b) If Sn = n(n + 1) , find S5, S6 and a6
2
n(n + 1)(n + 2)
(c) If Sn = 6 , find S4 , S5 and a5

(d) If, Sn = n(n2 + 6n + 11) , find S2, S3 ans a3
3

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

6

(a) (2x + 3), 5th term
n=1

6 n(n + 1),
2
(b) 3rd term

n=2

20

(c) (–1)n+1 (n2 + 3), 10th term
n=2

25 n(n + 1)(2n + 1)
6
(d) , 20th term

n=1

20 4k + 2
k+2
(e) (– 1)k. . 15th term

k=5

Long Questions

7. Study the patterns given below and answer the following questions.

(a)

(b)

(c) 65

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vedanta Excel in Opt. Mathematics - Book 9 Sequence and Series

(d)

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

2. (a) Sequence (b) Series (c) Series (d) Sequence
(e) Series (f) Series (g) Sequence
(l) Series
5
40 10 (d) 5 n+2
(c) (x – n)n n+4
3. (a) n (b) (–1)n n n=1
n=1 n=1
55 (b) 116 n=1

(e) arn – 1 or a rn – 1 4.(a) 60 (c) 195
1n =121 39 n=1
(d) 3465 (f) 6
(e) 35

5. (a) S3 = 7, S4 = 9, a4 = 2 (b) S5 = 15, S6 = 21, a4 = 6

(c) S4 = 20, S5 = 35, a4 = 15 (d) S2 = 18, S3 = 38 26
(c) 19, 24 (d) 25, 2870 7
6. (a) 6, 13 (b) 5, 10 (e) 16, –

7. (a) (ii) 2n + 4 (iii) 24 (b) (ii) 4n – 3 (iii) 37
(c) (ii) 3n (iii) 30
(d) (ii) 5n – 4 (iii) 46

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vedanta Excel in Opt. Mathematics - Book 9 Limit

Limit 4

4.0 Review

Group discuss the answers of the following questions.

(a) If f(x) = x2 + 4x + 3, find f(3).

(b) Define a function. Give an example.

(c) Is there any difference between a function and a relation ?

(d) Write the domain of the function f given by f = {(1, 2), (2, 3), (4, 5), (6, 7)}.

(e) Give an example of sequence of each of the following:

(i) a sequence of numbers which are in increasing order.

(ii) a sequence of numbers which are in decreasing order.

(f) What is the value of 1 as n increases ?
2n

(g) What are the nearest whole numbers of each number of the sequence 6.1, 6.001, 6.0001

by rounding off ?

Limit of Sequence of Numbers.

Let us take a line segment AB of length 16cm.

Find the mid-point C of it. Again, find the mid-point D of AC. In the similar way, find the
mid point of DC as E.

A D E FGC B

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Find the mid-point F of EC. Again, find the mid-point G of FC. Continue this process of
bisection. Discuss it.

Does the length of line segments AB, AC, AD, DE, EF, EG, ........ 16, 8, 4, 2, 1, 21, ...... represent
1 1 1 1
a sequence? If we continue above process of bisection, can we write 16, 8, 4, 1, 2 , 4 , 8 , 16

, ....... as lengths of the line segments?

(a) What is the distance between C and the final mid point ?

We finally come at the conclusion that the distance between the final mid-point and C
tends to zero.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

(b) What is the general term of the sequence 16, 8, 4, 2, 1 , 1 , ....... ?
2 4

Its general term will be 25–n .

Here '0' is called the limit of sequence 16, 8, 4, 2, 1, 12, 1 , ......... The final length of the
2
segment on bisection is nearly equal to 0 cm but not exactly 0 cm.

Definition : The limit of sequence of numbers is the finite value that the terms of the sequence
tends to that value. If the limit of a sequence exists, the sequence is called convergent. If the
sequence is not convergent, it is called divergent.

In a real number line, left number tends to –∞ and right number tends to + ∞ .

–∞ +∞

–2 –1 –0 1 2

For example, left tn = 1 , if n is infinitely large then tn tends to zero (but not exactly zero).
2n

Worked Out Examples

Example 1.

S.N Sequence of numbers Limit of the sequence
1 10.1, 10.01. 10.001, 10.0001,.... 10
10
2 9.9, 9.99, 9.999, 9.9999,.......... 2
2
3 2.1, 2.01, 2.001, 2.0001,........ 6
6
4 1.9. 1.99, 1.999, 1.9999, ......... +∞
–∞
5 6.5, 6.05, 6.005, 6.0005, ......... 0
0
6 5.9, 5.99, 5.999, 5.9999, .........

7 1, 2, 3, 4, 5, ...........

8 10, 9, 8, 7, 6, .............

9 1 , 1 , 1 , 1 , 1 , 1 , ......
10 2 4 8 16 32 64

0.5, 0.05, 0.005, 0.00005, ...........

Example 2. Write down the sequence of numbers given by the general term. tn = 31n–1.
Solution : Also, find the limit of the sequence.

Here, tn = 1
3n–1
1 1
t1 = 31–1 = 30 = 1

t2 = 1 = 1
32–1 3
1 1 1
t3 = 33–1 = 32 = 9

t4 = 1 = 1 = 1
34–1 33 27

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Hence the required sequence is 1, 1 , 1 , 1 , ...........
3 9 27

The values of each succeeding terms decreases; hence, the limit of the

sequence is zero.

Example 3. Find the tenth term of the sequence 1 , 1 , 1 , ....... Convert the
10 100 1000
sequence in decimal form and find the limit of the sequence.

Solution : Writing the sequence in decimal numbers,

we get, 0.1, 0.01, 0.001, ..... Tenth term is = 1 = 0.0000000001.
10000000000

The values of each succeeding terms decrease; hence, the limit of the

sequence is zero.

Exercise 4.1

Very Short Question
1. (a) Define the limit sequence of numbers.

(b) Define the convergent sequence with an example.
(c) Define the divergent sequence with an example.
Short Questions
2. A sequence is given as 4.1, 4.01, 4.001, 4.0001, .......
(a) Find the eighth, ninth, and tenth terms of the sequence.
(b) Find the difference between the tenth and ninth terms with the first term.
(c) What is the value of the last term if the number of term tends to infinity ?
(d) What is the limit of the above sequence ?
3. Write the limit of the given sequence of numbers.

S.N Sequence of numbers limit of the sequence
(a) 4.9, 4.99, 4.999, 4.9999, ........

(b) 5.1, 5.01. 5.001, 5.0001, ..............

(c) 1, 4, 16, 64, ..........

(d) 1 , 1 , 1 , 1 , .......
9 81 729 6561

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

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

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

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

5. Find out the whole number nearer to each of the following sequences.

(a) 1.1, 1.01, 1.001, 1.0001, ..........

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vedanta Excel in Opt. Mathematics - Book 9 Limit

(b) 2.9, 2.99, 2.999, 2.9999, ..........

(c) 3.1, 3.01, 3.001, 3.0001, ..........

(d) 1.9, 1.99, 1.999, 1.9999, .........

Long Questions

6. Write down the sequences of terms whose general terms are given by,

(a) tn = 1 , (b) tn = 1 –1
4n 10n
1 1
(c) tn = n + 2 (d) tn = n3

Also write the limit of each of the above sequence.

7. (a) Draw a line segment PQ of length 16cm. Bisect it 10 times and write the sequence
of lengths obtained after bisecting. Write the limiting value of the sequence.

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

2. (a) 4.00000001, 4.000000001, 4.0000000001 (b) 0.0999999999, 0.099999999

(c) 4 (d) 4 3.(a) 5 (b) 5

(c) f (d) 0 4. 70.4, 70.37 (c) 70

5. (a) 1 (b) 3 (c) 3 (d) 2

6. (a) 1 , 1 , 1 , 1 , ....... , limit 0 (b) 1, 1 , 1 , 1 , ..... , limit 0
4 16 64 256 10 100 10000
1 1 1 1 1 1 116, 1
(c) 3 , 4 , 5 , 6 , 7 , ......... , limit 0 (d) 1, 8 , 64 , ........ , limit 0

7. (a) 16, 8, 4, 2, 1, 1 , ..... , limit 0 (b) 10, 5, 2.5, 1.25, ...... , limit zero.
2

4.2 Concept of Limit from sequence of Figures

Let us observe the following figures and discuss the answers of the following questions.

n=3 n=4 n=5 n=6
(i) (ii) (iii) (iv)

(a) In which figure, the area of shaded region is least ?

(b) Compare the area of shaded region in figure (ii) and (iii) (use greater or smaller term).

(c) If the number of sides of regular polygon is increased, what is the difference in area of
circle and area of polygon ?

In each of the above inscribed regular polygon, the circumference of the circle is constant.
When the number of sides of the polygon increases, the difference of the area of circle and

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polygon (i.e. area of shaded region) decreases.

When the inscribed regular polygon has n sides, where n is infinitely large in number, the
difference between the area of the polygon and the circle is nearer to zero (but not equal to
zero) Hence, the area of the polygon when the number of sides increases, the limiting value
of area of the inscribed polygon is equal to the area of the circle.

Worked Out Examples

Example 1. Draw a square ABCD and inscribe another square by joining the middle
points of each side successively. Repeat the process six times. If the process
is continued for sufficiently large in number of times, what would be the
perimeter and area of the squares so formed ?

A EB

IM J
QU

F NV XO

R ws
L Pk

Solution : DG C

As the number of squares increases, the area and perimeter of squares
decrease. If the process continues repeatedly and sufficiently, both the area
and the perimeter tend to zero. This means that the limit of area and the
perimeter are zeros.

Example 2. A circle is drawn with centre at O. A second concentric circle is drawn

radius equal to half of the first circle. Again a third

concentric circle with radius half of the third

circle. The process is continued indefinitely many

times. What is the limiting area of the sequence of

circles so drawn ? O

Solution : As the concentric circles are drawn with the radii
half of the previous circles, the area of the circles
becomes smaller and smaller. Finally, the circle is of
a point whose area is zero. Hence, the limiting area
of the sequence of circles so drawn is zero.

Exercise 4.2

1. ∆PQR is an equilateral triangle. A second triangle is drawn P

joining the middle points of the sides of the ∆PQR. In the similar

manner, a third triangle is drawn in the second triangle. The

process of drawing the triangles is continued for sufficiently

large number of times, what is the limiting values of the area

and perimeter of the sequence of triangles so formed ? Q R

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2. A circle has the centre O. A number of circles are drawn within the circle. Such that the

radius of the succeeding circle is one third of the radius of the

preceding circles. The process of drawing triangles is continued for

sufficiently large number of times. What is the limiting value of O

perimeter of the circles so drawn ?

3. PQRS is a square consisting of 9 squares of equal are formed by 4
vertical and 4 horizontal lines. Diagonals parallel to PR are drawn by
joining the opposite vertices of all small squares. Then,

(a) How many small right angled triangles of equal area are formed in square PQRS ?

(b) If two more vertical lines next to QR are drawn and the S R
diagonals are produced forming all right angled triangles
of equal area, find the number of small right angled
triangles.

(c) If three more horizontal lines next to SR are drawn and

the diagonals are produced forming all small right angled

triangles of equal area, how many small right angled P Q

triangles are formed ?

(d) If two more vertical lines next to RQ and three more horizontal lines above SR are
drawn, what is the number of small right angled triangles of equal area formed ?

(e) If the process of drawing vertical and horizontal lines is continued as for as possible,
what is the limiting area of the sequence of small right angled triangle?

4. In the figure, three types of parallel lines are intersected. As the number of parallel
sides increases, what is the limit of the number of triangles formed ?

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Project Work

Draw a rectangle of length 24 cm and breadth 8 cm. Bisect the rectangle along the length 4
times with drawing figures. Write down the sequence of rectangle formed by the sides on the
basis of perimeters and areas of the rectangles so formed. What is the limit of each sequence
if the process is continued infinitely?

1. zero 2. zero 3.(a) 18 (b) 36
(c) 48 (d) 72 (e) f 4. f

4.3 Concept of Limit from Sum of Infinite Series

I. Let us consider the following infinite series :

1 + 1 + 1 +213 +214 + ............
2 22

In the example, every succeeding term is half of the preceding term. So, they are based

on the continuous bisection of the straight line segment: one part is left and other part

is again bisected. The process of bisection is continues.

Q
P

R S TU

1

1
2

1
4

1
8

Then, the series formed is in the form of 1 + 1 + 1 + 1 + ..........
2 4 8

In the figure, PQ is a line segment of length 1 unit. On bisecting PQ, we get,

PR = RQ = 1 unit
2
1
PQ is again bisected RS = SQ = 4

SQ is again bisected, ST = TQ = 1
8
1
TQ is again bisected, TU = UQ = 16

Now, PQ = 1, PS = PR + RS = 1 + 1
2 4
1 1
PT = PR + RS + ST = 2 + 4 + ..........

PU = PR + RS + ST + TU = 1 + 1 + 1 + .........
2 4 8

As the line segment is bisected as possible as many times, we finally come closer to

point Q.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

Sum of the first piece (S1) = PR = 1 = 0.5 unit.
2
1 1
Sum of the two pieces (S2) = PS = 2 + 4 + ... = 0.75

Sum of the three pieces (S3) = PT = 1 + 1 + 1 = 0.875 units.
2 4 8
1 1 1 1
Sum of the four pieces (S4) = PU = 2 + 4 + 8 + 16 = 0.9375

In the similar manner, we can compute s5, s6, ...... Sn (finite bisection). Then, Sn is
approximately equal to 1. It can be concluded that-

∞ 1 =1
2n
Sn =

Here,n1= i1s called the limit of the sequence.

In the above example, the absolute value of common ratio (r) is less than unity.

Common ratio (r) = t2 = t3 = t4 = 1 < 1.
t1 t2 t3 2

II. Let us take the series : 1 + 2 + 4 + 8 + ..........

Sum of the first term (S1) = 1

Sum of the first two terms = 1 + 2 = 3

Sum of the first three terms = 1 + 2 + 4 = 7

Sum of the first four terms = 1 + 2 + 4 + 8 = 15

As the number of terms increases, the sum of the series increases. If we find the sum of

infinite terms of the series, the sum is very large. So, it is not possible to find the limit of the

sequence. In this series, absolute value of common ratio (r) is 2 which is greater than 1. i.e. r

=whtte12 n=thtt32e=abtt34so=lu2te. It is concluded that the limiting value of the infinite sequence is possible
value of common ratio is less than unity.

Now, let us take some examples of infinite series.

1 1 1 t2 1 1
2 4 8 t1 2 2
(i) 1 + + + + ........., common ratio (r) = = 1 =

1 1 1 1 1 1
3 9 27 81 3 3
(ii) 1 + + + + + .........., common ratio (r) = 1 =

1 1 1 t2 1 1
5 25 125 t1 5 5
(iii) 1 + + + + ..........., common ratio (r) = = 1 =

Above series is called infinite geometric series.

Definition of Geometric Progression (G.P) : A sequence of numbers is said to be in a
geometric progression if the ratio of any term to its preceding term is constant throughout
the whole sequence.

Common ratio (r) = tk + 1 , k  N
tk

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For an example :

1, 1 , 1 , 1 , ........ is an example infinite of geometric series.
10 100 1000

Sum of Infinite Geometric Series,

Let us take an infinite geometric series.

∞ 1 1 1 1 1 1
2n 2 4 8 16 32
= + + + + + .........

n=1 of the first term (S1) = 1
2
Sum

Sum of the first two terms (S2) = 1 + 1 = 3
2 4 4
1 1 1 7
Sum of the first three terms (S3) = 2 + 4 + 8 = 8

Sum of the first four terms (S4) = 1 + 1 + 1 + 1 = 15
2 4 8 16 16
1 1 1 1 1 31
Sum of the first five terms (S5) = 2 + 4 + 8 + 16 + 32 = 32

As the number of terms increases in above given infinite geometric series, the sum is

approximately equal to a certain real number (i.e. 1). 1 is the limit of the sequence.

The sum of infinite geometric series can also be computed by using formulae,

S∞ = a r
1–

Where, a = the first term, r = common ratio, |r| < 1.

If (r) > 1, the sum does not exist.

e.g. 1 + 1 + 1 + ...........
5 25
1
t2 5 1
Here, common ratio (r) = t1 = 1 = 5

First term (a) = 1

S∞ = 1 r = 1 1 1 1 1 1 = 5 1 1 = 5 = 1.25
1– – 5 – 5 – 4
5

Worked Out Examples

Example 1. Find the limit of the following infinite series :
Solution:
(a) 1 + 1 + 1 + 1 + 1 + .............
2 4 8 16 32
1 1 1 1
(b) 1 – 2 + 4 – 8 + 16 – 1 ....... (c) 0. 3

(a) Here, 1 + 1 + 1 + 1 + 1 + ..........
2 4 8 16 32

= 0.5 + 0.25 + 0.125 + 0.0025 + 0.03125 + ........

The value of each term decreases and the absolute value of common
1
ratio (r) = 2 which less than 1. Hence the last term tends to zero.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

? Limit of the sequence is 0.

(b) 1 – 1 + 1 – 1 + 1 + ......
2 4 8 16
1
Common ratio (r) = t2 = – 2 = – 1
t1 2
1

| |Absolute common ratio |r| = – 1 = 1 < 1
2 2

1

The value of each term decreases and the absolute value common ratio
1
is 2 , less then 1. Hence, the limit of the sequence is zero.

(c) 0. 3 = 0.03 + 0.003 + 0.0003 + .......

= 3 + 3 + 3 + 3 + .......
10 100 1000 10000
3
100 1
The absolute value common ratio (r) = 3 = 10

Which is less than 1, 10

When the number of terms increases continuously, the last term
approaches zero.

? Limit of the sequence is zero.

Example 2. Find the limiting value of the sum of the infinite series:
Solution :
1 + 1 + 1 + 1 + ........
76 5 25 125
1 1 1
Here, 1 + 5 + 25 + 125 + ........

Now, S1 = 1

S2 = 1 + 1 = 1 + 0.2 = 1.2
5
1 1
S3 = 1 + 5 + 25 = 1.24

S4 = 1 + 1 + 3 + 1 = 1.248
5 25 125
1 3 1 1
S5 = 1 + 5 + 25 + 125 + 625 +........ =1.2496

S6 = 1 + 1 + 3 + 1 + 1 + 1 + ........... = 1.24992
5 25 125 625 3125

Similarly, we can find S7, S8, S9, .............

This shows that the partial sum of the series increases slightly as the number
of terms increases but the sum is approximately equal to 1.25. Hence, the
limit of given infinite series has sum 1.25.

Alternatively :

Here, 1 + 1 + 1 + 1 + 1 + 1 + .......
5 25 125 625 3125
1
t2 5 1
Common ratio of the series (r) = t1 = 1 = 5 < 1.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

First term (a) =1

Sum of the series S∞ = 1 a r = 1 1 1 = 5 = 1.25
– – 5 4

Hence, the limiting value of sum of series is 1.25.

Exercise 4.3

1. Which of the following series have a limit as a fixed real number ? Give your reason.
Also write the limit if exists.

(a) 1 + 1 + 1 + 1 + ............ (b) 1 + 3 + 9 + 27 + 81 + ........
3 9 27 2 4 8 16

(c) 32 – 16 + 8 – 4 + 2 ............. (d) 16 – 8 + 4 – 2 + .........

(e) 16 + 8 + 4 + 2 + 1 + ........ (f) 1.6+ 8 + 40 + 200

(g) 0 . 6 (h) 0. 78

(i) 0. 45 (j) 0. 15

2. Find the limiting values of the sum of the following infinite series if exists.

(a) 8 – 4 + 2 – 1 + ........ (b) 2 + 1 + 1 + 1 + .........
2 4
1 1 1 1 1
(c) 25 + 5 + 1 + 5 + 25 + ....... (d) 1 + 3 + 9 + 27 + ......

(e) –16 + 8 – 4 + 2 – ........ (f) 1 + 3 + 9 + 27 + .......
2 4 8

(g) 0. 45 (h) 0. 6

(i) 1 + 2 + 4 + 8 + 16 +....... (j) 1 + 5 + 25 + 125 + .........

3. Find the first five terms of the following sequences. Also find the limit of the nth term if exists.

(a) tn = 1 n–1 (b) tn = 1n (c) tn = 2n
2 3 3

(d) tn = 5 –1 n (e) tn = 5n + 3 (f) tn = n + 2
2

(g) tn = 4 – 1 n
3

4. Write the sequence of numbers represented by the following diagrams. If Sn represents
the area of shaded portion of the diagram, find the limit of sum of sequence.

(a) Fill in the given table. ABCD is a square of length 1cm.

n Sn A DD CD C DC

1 1 cm2
2

2 ....... B 1 cm C A 1 cm B A 1 cm B A 1 cm B
Fig (i) Fig (ii) Fig (iii) Fig (iv)
3 .......

4 ....... Draw two more figures in the same pattern.
5 .......

6 .......

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(b) Given circle is of radius 14cm. Fig (iv)
n Sn
1 .......

2 ....... Fig (i) Fig (ii) Fig (iii)
3 .......
4 ....... Draw two more figures in the same pattern.
5 .......
6 .......

1. (a) 0 (b) 0 (c) 0 (d) 0

(e) 0 (f) It does not have a limit 1 (g) 0
3
(h) 0 (j) 0 2.(a) 5 (b) 4

(c) 31 1 (d) 1.5 (e) – 10 2 (f) does not exist
4 3
5 2
(g) 11 (h) 3 (i) sum does not exists

(j) sum does not exist
1 1 1 1
3. (a) 1, 2 , 4 , 8 , 16 , limiting value of sum = 2

(b) 1 , 1 , 1 , 1 , 1 , limiting value of sum = 1
3 9 27 81 243 2
2 4 8 16 32
(c) 3 , 9 , 27 , 81 , 243 , limiting value of sum = 2

(d) – 5 , 5 , – 5 , 5 , – 5 , limiting value of sum = – 5
2 4 8 16 32 3

(e) 8, 13, 18, 23, 28, limiting value of sum does not exists.

(f) 3, 4, 5, 6, 7, limiting value does not exists.

(g) – 4 , 4 , – 4 , 4 , limiting value = – 1
3 9 27 81
1 3 7 15
4. (a) 2 cm2, 4 cm2, 8 cm2, 16 cm2, limit 1 cm2

(b) 308 cm2, 462 cm2, 539 cm2, 577.5 cm2, 596.7 cm2, 606.38 cm2, limit 616 cm2

4.4 Limit of Function

Function :

Let x and y be two variables which are so related that every value of x. There, is a unique
value of y. Then, y is said to be a function of x. It is written as y = f(x).

Example : y = f(x) = x + 2

Value of the function :

Let y = f(x) be a function of x. Then the value of the function f(x) at x = a is denoted by f(a).
f(a) is called the functional value of f(x) at x = a. Let us take y = f(x) = x + 2.

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Now, let us fill the table given below :

x f(x) (x, f(x) or (x, y)

–2 0 (–2, 0)

–1 1 (1, 0)

0 ............. ...........

1 ............. ...........

2 ............. ...........

3 ............. ...........

4 ............. ...........

5 ............. ...........
6 ............. ...........

Draw a graph of f(x) = x + 2, taking the points from above table. Is there unique value of y
for every value of x ? Discuss from above table.

Does f(x) = x2 – 9 exists for x = 3 ?
x – 3

If f(x) is not defined for x = 3, find (4).

Is f(x) = x2 – 9 = (x + 3) for x = 4, 5 ? Discuss it.
x – 3

Limit of Function :

I. Let y = f(x) = x + 1. Draw a graph of it. Find the value of f(x) at x = 1, 2, 3, 3.5, 3.9,
3.999, .....

x 1 2 3 3.5 3.9 3.99 3.999 3.9999

y=x+1 2 3 4 4.5 4.9 4.99 4.999 4.9999

In the above table, the values of x are taken from left side of x = 4. As these values come
closer and closer to x = 3, the value of f(x) becomes closer to 5 or approximately equal
to 5. Hence, 5 is called limiting value of given function f(x) which is called left hand
limit (LHL) of the function at x = 4.

9 1234567
8
7
6
5
4
3
2
1
-5 -4 -3 -2 -1-O1
-2
-3

Similarly, find the values of f(x) at x = 7, 6, 5, 4.5, 4.1

x7 65 4.5 4.1 4.01 4.001 3.9999
y=x+1 8 76 5.5 5.1 5.01 5.001 4.9999

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vedanta Excel in Opt. Mathematics - Book 9 Limit

This table shows that as the value of x come closure and closer to 4 from right side of
x = 4, the values of f(x) also come closer and closure to 5. This is called right hand limit
of the given function, at x = 4.

From above two tables we conclude that as the values of x are closure and closure to 4, the
values of f(x) also come closer and closer t0 5. Here, 5 is called the limit of the function at
x = 4. In symbol it is written as,

II. Let f(x) = x2 – 4
x – 2
The function is not defined for x = 2.

Leaving the value of x = 2, we take the values of x slightly greater or less than 2. Let as
tabulate the values of x.

For x < 2 For x > 2

x f(x) x2 – 4 x f(x) = x2 – 4
x – 2 x – 2
2.1
1.9 1.92 – 4 = 3.9 2.12 – 4 = 4 .1
1.9 – 2 2.01 2.1 – 2
1.99 2.001
1.999 3.99 2.001 4.01
2.0001
3.999 2.00001 4.001

1.9999 3.999 4.001
1.9999 3.9999 4.0001

1.99999 3.99999 4.00001

From above table, it is clear that as the values of x become closure and closure to 2 from
either LHS and RHS, the values of f(x) become closure and closure to 4. The number 4 is
called the limit of f(x) as x tends or approaches 2. In notation, we write,

As x o 2, x2 – 4 o 4
x – 2

lim f(x) = lim x2 – 4 = lim (x + 2) (x – 2)
xo2 xo2 x – 2 xo2 (x – 2)

= lim (x + 2)
xo2

=2+2=4

Definition : Let f(x) be a function of x. If f(x) takes the values closer and closer to f(a)
whenever x takes the value closer and closer to the value 'a', then f(x) is said to have limit
f(a) as x tends to 'a'.

In notation we write

lim f(x) = f(a) xoa

Example : lim f(x) = lim x2 – 16
Solution: x – 4
lim lim x2 – 16
Here, lim f(x) = xo4 f(x) = xo4 x – 4

When x = 4, the function is not defined

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vedanta Excel in Opt. Mathematics - Book 9 Limit

lim (x + 4) (x – 4) = lim (x + 4) = 4 + 4 = 8
xo4 (x – y) xo4

Hence, limit of f(x) at x = 4 is 8.

Indeterminate Forms

Let y = f(x) = x2 – 1 and x = 1
x – 1
0
When x = 1, y = f(1) = 0 .

This does not give any real number. It does not have any meaning. Such a form is called
∞
indeterminate form. Other indeterminate forms are ∞ , ∞ – ∞, ∞ + ∞, 1∞ and 0∞ etc.

Note :

1. ∞ + ∞ = ∞ 2. ∞ + finite value = ∞ 3. ∞ – finite value = ∞

4. finite × ∞ = ∞ 5. ∞ × ∞ = ∞ 6. finite value = 0
8. ∞ – ∞ = ∞ ∞
finite value
7. 0 =∞

Worked out Examples

Example 1. If f(x) = 3x – 1 , find f(4) and f(3.99) and f(4.01).
Solution : 2
3x – 1
Example 2. Here, f(x) = 2
Solution :
Put x = 4 we get,

f(4) = 3 ×4 – 1 = 12 – 1 = 11 = 5.5
2 2 2

Put x = 3.99

f(3.99) = 3 × 3.99 – 1 = 10.97 = 5.485
2 2
3 × 4.01 – 1 12.03 – 1 11.03
Put x = 4.01, f(4.01) = 2 = 2 = 2 = 5.515

Evaluate : f(x) = x2 – 25 , x ≠ 5 for x = 4, 4.9, 5.1, 5.01. Write the conclusion.
x – 5
x2 – 25
Here, f(x) = x–5 , x ≠ 5

Now, f(4) = 42 – 25 = 16 – 25 = 9
4 – 5 –1
4.92 – 25
f(4.9) = 4.9 – 5 = 9.9

f(5.1) = 5.12 – 25 = 10.1
5.1 – 5
5.012 – 25
f(5. 01) = 5.01 – 5 = 10.01

As the values of x are closer and closer to 5, the values of f(x) are closer to 10.

We conclude that the limit of the f(x) is 10 as x tends to 5.

i.e. lim f(x)=xloim5 x2 – 255=xloim5 (x + 5) (x – 5) =xloim5 (x+5) =5+5 = 10
xo5 x – x–5

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vedanta Excel in Opt. Mathematics - Book 9 Limit

Exercise 4.4

Very Short Questions

1. (a) Define functional value of f(x) at x = a.

(b) Define limit of a function.

(c) Define indeterminate forms.

2. (a) Find the value of f(x) = 4x2 – 5 at x = 2

(b) Find the value of g(x) = 2x + 5 at x = 2.

3. (a) Find the value of f(x) = x2 – 1 at x = 2
x – 1
x2 – 36
(b) Find the value of f(x) = x – 6 at x = 5.

Short Questions

4. (a) If f(x) = 3x + 2, find f(1. 9), f(1.99), f(2).

(b) If f(x) = 2x, then find f(0.9), f(0.99), f(0.999)

(c) Given that f(x) = 8x + 1, find f(8.9), f(8.99) f(9.01).

Long Questions

5. (a) Given that f(x) = 2x + 1, find the nearest whole number of f(2.99), f(2.999) f(3.01),
f(3.001). What conclusion can you draw from these results ?

(b) Let f(x) = 2x – 1, find f(1.9), f(1.99), and f(2.1), f(2.01), find the nearest whole
number nearer to there values, what conclusion can you draw from there results ?

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

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

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

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

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

7. Let f(x) = x2 – 47y9ou, x ≠ 7, evaluate f(x) for x = 7.1, 7.01, and x = 6.9, 6.99. What
conclusion cxan– draw from these values of x?

2. (a) 11 (b) 9 3.(a) 3 (b) 11

4. (a) 7.7, 7.97, 7.997 (b) 1.8, 1.98, 1.998 (c) 72.2, 72.92, 73.08

5. (a) 6.98, 6.998, 7.02, lim f(x) = 7 (b) 2.8, 2.98, 3.2, 3.02, 3, lim (f)x = 3
xo3 xo2

6. (a) f(x) is not defined for x = 4. (b) 7.9, 7.99, 7.999, 7.9999

(c) 8.1, 8.02, 8.002 (d) 8 (e) limit of f(x) at x = 4 si 8.

7. 14.1, 14.2, 13.9, 13.99, limit of f(x) at x = 7 is 14.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

4.5 Notational Representation of Limit

I. Discuss the meaning of the following.
(a) The value of x is approximately equal to 5. What does it mean?

(b) Due to over speed of car, Hary tends to meet with an accident.

(c) What notation do you use as the value of x is approximately equal to a and the
value of f(x) is nearly equal to f(a) ?

II. Let y = f(x) = x + 2, then complete the table given below :

x 1 1.5 1.9 1.99 1.999 2.1 2.01 2.001 2.0001
y = x + 2 ....... ....... ....... ....... ....... ....... ....... ....... .......

Then answer the following questions from above table :
(a) For x = 1 1.99, to what whole number the value of f(x) = x + 2 can be rounded off ?

(b) For x = 2.001, to what whole number the value of f(x) = x + 2 can be rounded off ?

(c) For 1.99 and 2.001, what is the nearest whole number approximately equal to them ?

(d) For x = 2, what is the value of f(x) ?

From above activities. Can we draw conclusion that as the values of x become nearer and
nearer 2, the values of f(x) became nearer and nearer 4 ?

When the values of x approaches to 2 from both of LHS and RHS, the value of f(x) approaches
to 4. It is symbolically represented as if x o 2, then f(x) o 4.

In the above example, the value of f(x) approaches to 4 as x approaches to 2. 4 is the limit of
f(x) as x approaches to 2.

Symbolically, we write,

lim f(x) = lim (x + 2) = 2 + 2 = 4
xo2 xo2

The limit of a function f(x) as x approaches (or tends) to 'a is f(a).
Symbolically, we write;

lim f(x) = f(a)
xoa

Exercise 4.5

Very Short Questions :

1. Write the notations for the following mathematical statements:

(a) x approaches to a. (b) x tends to 4.

(c) t is closer to to 2. (d) x approximately equals to 5.

(e) As limit x tends to a f(a) is equal to f(a). (f) The limit of f(x) is l as x tends to a.

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vedanta Excel in Opt. Mathematics - Book 9 Limit

(g) x approaches to 'a' from the right. (h) x approaches to 'a' from the left.

2. Write the following notations of limits into mathematical statements.

(a) x o b (b) x o 4+ (c) x o 4–

(d) x o 2+ (e) lim (4x + 5) (f) lim f(x) = f(a)
xo0 xoa

(g) lim x2 – 16 = 8 (h) lim x2 – 25 = 10
xo4 x +4 xo3 x – 5

Short Questions

3. Evaluate :

(a) lim (4x + 1) (b) lim (3x – 1) (c) lim x2 – 4
xo2 xo2 xo2 x–2

(d) lim x–5 (e) lim 39– x (f) lim x2 – 25
xo9 xo3 xo5 x–5

Long Questions

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

(a) f(x) = x + 2 x 2.0 2.5 2.9 2.99 3.1 3.01 3.001 xo
....... ....... ....... f(x) o
f(x)=x+2 ....... ....... ....... .......

(b) f(x)= x2 – 4 x 2.5 2.9 2.99 3.1 3.01 3.001 3.0001 xo
x– 2 ....... f(x) o
f(x)= x2 – 4 ....... ....... ....... ....... ....... .......
x – 2

(c) f(x)= x + 4 x 7.5 7.9 7.99 7.99 8.2 8.01 8.001 xo
x + 1 ....... ....... ....... ....... ....... ....... ....... f(x) o
f(x)= x + 4
x + 1

Project Work
5. Discuss use of limit in our daily life with any five examples.

1. (a) x o a (b) x o 4 (c) t o 2 (d) x o 5
(h) x o a–
(e) lim f(x) = f(a) (f) lim f(x) = l (g) x o a+
xoa xoa

2. (a) x tends to b (b) x tends to 4 from right

(c) x tends to 4 from left (d) x tends to 2 from right

(e) limit of (4x + 5) is 5 as x tends to 0 (f) limit of f(x) is f(a) as x tends to a

(g) limit of x2 – 16 is 8 as x tends to 4 (h) limit of x2 – 25 is 10 as x tends to 5
x – 4 x – 5

3. (a) 9 (b) 5 (c) 4 (d) 2 (e) 6 (f) 10

4. (a) lim f(x) = 5 (b) lim f(x) = 5 (c) lim f(x) = 4
xo3 xo3 xo8 3

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Matrix 5

5.0 Review

Observe the adjoint calender of 2020 June and answer the following questions
(a) Name the day of 20th June.
(b) How many Saturdays are there in June month ?
(c) How many days are there in June month ?

5.1 Introduction of Matrix

Observe the marks of three students in the first terminal exam for two subjects; Maths and
Science.

Subject Jeny Jack Jaya
Math 88 75 90
Science 60 63 70

(a) What marks did Jeny obtained in Maths ?

(b) What does 63 represent ?

Above table can be written as,

Jeny Jack Jaya

Maths 88 75 90
Science 60 63 70

This representation of marks is called a matrix. In a matrix, the numbers are arranged in rows
and columns. The numbers in the horizontal lines are called rows and that the numbers in
vertical lines are called columns. The numbers 88, 75, and 90 are in the first row (R1) and
that of 60, 63 and 70 are in the second row. The numbers 88 and 60 are in the first column
(C1) and so on. C1 C2 C3

88 75 90 R1
60 63 70 R2

Definition : A matrix is defined as a rectangular arrangement of numbers (elements) into
rows and columns enclosed by a pair of round or square brackets. Each number of a matrix
is called an element or component or entry.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Notation : Generally matrices are denoted by capital letters like A, B, C, D,........ X, Y, Z. The
elements are denoted by the corresponding small letters along two suffices. The first suffix
indicates the number of row and the latter one indicates the number of column in which the
element appears. For examples,

(i) aij represents the element of matrix A in the ith row and jth column.

(ii) a13 represents the element of matrix A in the first row and the third column.

Order of a Matrix

Let us consider a matrix A = 1 2 3 .
4 5 6
The matrix A has 2 rows and 3 columns. This matrix is of order 2 × 3 (read as 2 by 3). A

matrix of order 2 × 3 is also denoted by A2 × 3.

Definition : The number of "rows × columns" of a matrix is known as the order of the matrix.

Example : A = 123 is an order of 2 × 3.
321
Let A be a matrix of order 3 × 3, then we write in double suffix notation as below.

a11 a12 a13
A = a21 a22 a23

a31 a32 a33

Worked out Examples

Example 1. Write down the orders of the following matrices :
Solution :
(a) A = [1 2 3] 1
Example 2. (b) B = 7
Solution :
3

(c) P= 123 1 40
478 (d) Q = 2 3 4

6 07

(a) The matrix A has one row and three columns. Hence, the order of
matrix A is 1 × 3

(b) The matrix B has three rows and one column. Hence, it is of order 3 × 1.

(c) The matrix P has two rows and three columns. Hence, matrix P is of
order 2 × 3.

(d) The matrix Q has three rows and three columns. Hence, the order of
matrix Q is 3 × 3.

2 45
If A = 6 7 2 , do the following:

0 41
(a) Write the number of elements of matrix A.

(b) Write down the order of the matrix A.

(a) Here the matrix A has 9 elements.

(b) The order of matrix A is 3 × 3

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Example 3. Let A = 2 4 5 . Write the matrix using double suffix notation like a11, a12,
Solution : 6 7 8

...... a23. Then, write the values of A11m a12,........

Here, A = 245
678
In double suffix notation, we can write matrix A as,

A= a11 a12 a13
a21 a22 a23
Equating the corresponding elements, we get,

a11 = 2, a12 = 4, a13 = 5, a21 = 6, a22 = 7, a23 = 8

Exercise 5.1

Very short Questions :

1. (a) Define a matrix with an example.

(b) What is order of matrix ? Give an example of it.

(c) What is the meaning of a23, an element of matrix A ?

(d) If R = 2 3 4 , what is the number of elements in the matrix R ?
2 4 5
2. Write down the order of the following matrices:

123 246
321
(a) A = 6 4 2 (b) B=

452 m
(e) E = n
(c) C= c11 c12 c13 (d) D = [1 2 3]
c21 c22 c23 p

124

3. if A = 6 7 8 , write down the order of the matrix A. How is it written to show the

234
order of matrix A ? Write it in notation.

Short Questions :

–4 6 2
4. (a) If A = 3 2 1 , write the matrix A using double suffix notation, like

4 21
a11, a12, ........ a33. Write the values of a11, a12, ........., a33.

123 b11 b12 b13

(b) If 4 6 7 = b21 b22 b23 , write the values of b11, b12, ......., b33.
542 b31 b32 b33

Which letter is appropriate to denote the above matrix?

1. (d) 6 2. (a) 3 × 3 (b) 2 × 3 (d) 1 × 3

(e) 3 × 1 3. 3 × 3, A3 × 3

4. (a) a11 = –4, a12 = 6, a13 = 2, a21 = 3, a22 = 2, a23 = 1, a31 = 4, a32 = 2, a33 = 1, A

(b) b11 = 1, b12 = 2, b13 = 3, b21 = 4, b22 = 6, b23 = 7, b31 = 5, b32 = 4, b33 = 2

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

5.2 Types of Matrices

On the basis of structures of matrices, they are of different types. Some special types of
matrices are defined below :

(a) Row Matrix :

A matrix having only one row is called a row matrix.

For examples : (i) A = [a b c] is a row matrix of order 1 × 3.

(ii) P = [1 4] is a row matrix of order 1 × 2.

(b) Column Matrix :

A matrix having only one column is called a column matrix.

a

For examples : (i) A = b is a column matrix of order 3 × 1.

c

(ii) P = p is a column matrix of order 2 × 1
q
(c) Zero Matrix :

A matrix of any order whose all elements are zeros is called a zero matrix or null
matrix.

For examples : (i) O = [0 0 0] is a zero matrix of order 1 × 3.

000

(ii) O = 0 0 0 is a zero matrix of order 3 × 3.

(d) Rectangular Matrix : 000

A matrix having unequal number of rows and columns is called a rectangular matrix.

For examples : Let a b d
m p q
Here, the number of rows = 2

The number of columns = 3

? 2≠3

Hence, A is a rectangular matrix of order 2 × 3.

(e) Square Matrix :

A matrix whose number of rows and columns are equal is called a square matrix.

123
For example : P = 4 5 6

789
Here, the number of rows = 3

The number of columns = 3

Hence, the given matrix P is a square matrix of order 3 × 3.

Similarly, R = 1 4 is a square matrix of order 2 × 2.
3 9

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Leading Diagonal :

In a square matrix the diagonal that runs from the top left to right bottom is called
leading diagonal or principal diagonal,

123
Let P = 2 6 7

324

Then the diagonal that runs from top left to right bottom with elements 1, 7, and 4 is a

leading, principal/ major diagonal.

(f) Diagonal Matrix :

A square matrix in which the elements except the leading (or principal) diagonal are
zero is called a diagonal matrix.

For examples : Let A = 2 0 ,B= 1 0 0
0 3 0 2 0
0 0 3

Here A and B are diagonal matrices of order 2 × 2 and 3 × 3 respectively.

(g) Scalar Matrix :

A diagonal matrix in which all the elements in the leading (or principal) diagonal are
equal is called a scalar matrix.

For examples : Let A = 2 0 ,B= a 0 k00
0 2 0 a and C = 0 k 0

Here, A, B and C are scalar matrices. 00k

(h) Unit/ Identity Matrix :

A scalar matrix in which all the leading diagonal elements are unity and others zeros is
called a unit matrix. It is denoted by I.

For examples : I = 1 0 100
0 1 I1 = 0 1 0

001

Here, I and I1 are unit or identity matrices of orders 2 × 2 and 3 × 3 respectively.

(i) Triangular Matrix :

A square matrix whose elements either below or above the leading (or principal)
diagonal are each of zero is called a triangular matrix. Triangular matrices are of two
types.

Upper Triangular Matrix

A square matrix whose elements below the leading diagonal each of zero is called the
upper triangular matrix.

Let P = 1 2 ,Q= a b c 1 4 6
0 3 0 d e ,T= 0 3 2
0 o 0 0 8
g

Here, P, Q and T are upper triangular matrices.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Lower Triangular Matrix

A square matrix whose elements above the leading diagonal each of zero is called the
lower triangular matrix.

For examples : A = 2 0 ,B= 1 0 0 are lower triangular matrices.
Equal Matrices : 3 4 2 3 0
5 6 7
(j)

Two matrices of the same orders are said to be equal if their corresponding elements are
equal. If P and Q are equal matrices, then we write P = Q.

For examples : (i) P= 1 2 and Q = 1 2 are equal matrices.
3 4 3 4

(ii) If p q = 2 7 , then their corresponding elements are equal.
r s 8 9

? p =2, q = 7, r = 8, s = 9 .

(k) Symmetric Matrix :

A square matrix is said to be symmetric matrix if the interchange of rows or columns
makes no change in the given matrix.

In other words a square matrix A = (aij) is called a symmetric if aij = aji for all i and j.

For Examples : (i) A= 2 4 , a12 = a21 = 4
4 3

123
(ii) A = 2 4 5

358
Here, a12 = a21 = 2, a13 = a31 = 3, a23 =a32 = 5,

? A is a symmetric matrix of order 3 × 3. For a symmetric matrix A, AT = A.

(l) Skew – symmetric Matrix :

A square matrix A = (aij) is called skew–symmetric matrix if aij = – aji for all i, j all
diagonal elements are zero.

0 37

For examples, let A = –3 0 5 is a skew – symmetric matrix of order 3 × 3.

–7 –5 0

Here, a12 = – a21 = 3,

a13 = a31, a23, = – a32 = 5.

For a skew - symmetric matrix, AT = – A

(m) Sub – matrix :

A new matrix obtained by omitting some rows or columns from a given matrix is called
a sub–matrix of the given matrix.

Let A = 1 2 3 1 2 ,C= 1 2 3 are sub matrices of matrix A.
3 4 5 ,B= 3 4 3 4 5
7 8
9

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Worked out Examples

Examples 1. State the types of the following matrices with their orders :

1

(a) A = [a b c ] (b) B = 2

3

(c) C= 2 0 100
2 3 (d) D = 0 1 0

100 001

(d) D = 0 1 0 (e) E= 1 2
2 4
001

123 124
(f) A = 2 5 7 (g) M = 0 5 6

379 007

Solution : (a) A = [a b c]. It is a row matrix of order 1 × 3

Example 2. 1
Solution :
(b) B = 2 . It is a column matrix of order 3 × 1.

3

(c) C= 2 0 . It is a lower triangular matrix of order 2 × 2.
2 4

100

(d) D = 0 1 0 . It is an identity matrix of order 3 × 3.

001

(e) E= 1 2 , It is a symmetric matrix of order. 2 × 2.
2 4

123
(f) A = 2 5 7 . Here, a12 = a21 = 2, a13 = a31, a23 = a32 = 7

379
? A is a symmetric matrix of order 3 × 3.

129
(g) M = 0 5 6 , M is an upper triangular matrix of order 3 × 3.

007
If matrix A = [aij] is of order 3 × 3 and aij = 2i – j, construct the matrix A.

Here, A = [aij] , aij = 2i – j, A is matrix of order 3 × 3.

a11 a12 a13
Let A = a21 a22 a23

a31 a32 a33
Now, a11 = 2.1 – 1 = 1, a12 = 2.1 – 2 = 0, a13 = 2.1 – 3 = –1.

a21 = 2.2 – 1 = 3, a22 = 2.2 – 2 = 2, a23 = 2.2 – 3 = 1
a31 = 2.3 – 1 = 5, a32 = 2.3 – 2 = 4, a33 = 2.3 – 3 = 3

1 0 –1
? $ 3 2 1

54 3

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Example 3. If x–4 2y – 3 = 1 1 , find the values of x, y, p and q.
Solution : p–2 8 – 3q 2 –1

Example 4. Here, x–4 2y – 3 = 1 1
Solution : p–2 8 – 3q 2 –1

Example 5. Equating the corresponding elements of equal matrices, we get
Solution :
x–4=1 or, x = 5

2y – 3 = 1 or, 2y = 4 or y = 2

p–2=2 or, p = 2 + 2 = 4

8 – 3q = – 1 or, –3q = –9 or, q = 3

? p = 4, p = – 3, x = 5 and y = 2

If the matrix a–1 c+1 is an identity matrix, find the values of a, b, c
and d. b–1 d+1

Here, a–1 c+1 is an identity matrix.
b–1 d+1

So, we can write a–1 c+1 = 1 0
b–1 d+1 0 1

Equating the corresponding elements of equal matrices,

we get, a – 1 = 1 or, a = 2

c+1=0 or, c = – 1

b–1=0 or, b = 1

and d + 1 = 1 or, d = 0

? a = 2, b = 1, c = –1, d = 0

If x – 2y x – 2 = 3 1 , find the values of a, b, x and y.
a + 2b 3a – b 5 1

Here, x – 2y x – 2 = 31
a + 2b 3a – b 51

Equating the corresponding elements of equal matrices,

we get, x–2=1

or x = 3

? x=3

x – 2y =3

or 3 – 2y = 3

or, y = 0

a + 2b = 5 .............. (i)

3a – b = 1 ............... (ii)

From equation (ii), b = 3a – 1

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put the value of b in equation (i)
a + 2 (3a – 1) = 5
or, a + 6a – 2 = 5

or, 7a = 7
? a = 1.
Put the value of a in equation (ii), we get,

b = 3.1 – 1 = 2
? a = 1, b = 2, x = 3, y = 0

Exercise 5.2

Very Short Question

1. Give on example of each of the following matrices:

(a) Scalar Matrix (b) Diagonal Matrix

(c) Scalar Matrix (d) Square matrix

(e) Row matrix (f) Column Matrix

2. State the types of the following matrices with their orders.

(a) 14 (b) 00 (c) 20
56 00 02

123 100 123
(e) 3 4 5 (e) 2 1 0 (f) 0 4 5

789 432 007

78 9 (h) 08
(g) 8 4 3 –8 0

9 3 10

Short Questions

3. Write down an example of the each of the matrices of the following order :

(a) Column Matrix of order 3 × 1 (b) Row matrix of order 1 × 3

(c) Unit Matrix of order 3 × 3 (d) Diagonal Matrix of order 3 × 3

(e) Scalar Matrix of order 3 × 3. (f) Null matrix of order 3 × 3

(g) Square matrix of order 3 × 3

4. Construct a 2 × 2 matrix whose elements aij are given by :

(a) aij = i + j (b) aij = 2i + j (c) aij = (i) j

(d) aij = 4i – 3j (e) aij = i.j

5. Find the values of x and y when,

(a) (x 2y) = (5 6) (b) 3x –5 = 6 –5
1 y 1 1

(c) x+y = 5 (d) 2x + y = 5
x–y 1 3x – 2y 4

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6. If a matrix has 6 elements in total, what are the possible orders it can have? Given
reasons.

7. (a) If x–1 p is an identity matrix, find the values of x, y, p, and q.
q y–1

(b) If the matrix 4x – 7 2y – 6 is an identity matrix, find the values of x and y.
y–3 2x – 3

(c) For what values of x and y, the matrix 2 3x – 8 is a scalar matrix.
y–4 2

(d) For what values of x and y, the matrix 8 x–y is a scalar matrix.
x+y 8

Long Questions :

8. Construct a matrix A whose order is 3 × 3 and elements are given by:

(a) aij = i + j (b) aij = (i)j

(c) aij = 2i + 3j (d) aij = 2i – j

(e) aij = (–1)i + j

9. (a) Find the values of x, y and z if x+y z–x = 3 2
y + 2z x 8 1

(b) If p+1 5 = 6 y–3 , find the values of x, y, p and q.
–3 q x –2

Project Work

10. List the trigonometric ratios of standard angles from 0° to 90° in a matrix form.

2. (a) Square, 2 × 2 (b) Null, 2 × 2 (c) Scalar, 2 × 2 (d) Square, 3 × 3

(e) Lower triangular, 3 × 3 (f) Upper triangular, 3 × 3 (g) Symmetric, 3 × 3
(h) Skew-symmetric, 2 × 2

4. (a) 23 (b) 34 (c) 11 (d) 1 –2
34 56 24 52

(e) 12 5.(a) x = 5, y = 3 (b) x = 2, y = 1 (c) x = 3, y = 2
24
(d) x = 2, y = 1 6. 1 × 6, 6 × 1, 2 × 3, 3 × 2 7.(a) x = 2, p = 0, q = 0, y = 2
(b) x = 2, y = 3 (c) x = 38, y = 4
(d) x = 0, y = 0
234
11 1 5 8 11

8. (a) 3 4 5 (b) 2 4 8 (c) 7 10 13

456 3 9 27 9 12 15

1 0 –1 1 –1 1

(d) 3 2 1 (e) –1 1 –1

54 3 1 –1 1 (b) x = –3, y = 8, p = 5, q = –2
9. (a) x = 1, y = 2, z = 3

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5.3 Operations on Matrices

Addition of Matrices.

Let us observe the sales of three types of news papers in two days in two shops M and N:

Shop M Shop N

News papers News papers

Day A B C Day A B C

Sunday 200 150 400 Sunday 150 300 250

Monday 240 160 250 Monday 120 250 260

Now answer the following questions
(a) Find the total sales of each type of newspapers on the two days ?

(b) Which type of newspaper is sold maximum in number ?

To find the sum of newspapers on the two days. Let the sales in M and N be denoted by
matrices P and Q.

P = Sunday 200 150 400 Q = Sunday 150 300 250
Monday 240 160 250 Monday 120 250 260

Matrices P ans Q can added as :

P+Q = 200 150 400 + 150 300 250
Definition 240 160 250 120 250 260

= 200 + 150 150 + 300 400 + 250
240 + 120 160 + 250 250 + 260

= 350 450 650
360 410 510

Let A and B be two matrices of same orders. Then, A and B are said to be conformable for
sum and the sum is denoted by A + B. It is obtained by adding the corresponding elements
of A and B.

For Example : Let A = 1 2 3 and B = 2 3 2
4 6 7 1 2 1
Here, each of A and B are of orders 2 × 3.

Then, A+B = 1 2 3 + 2 32
4 6 7 1 21

= 1+2 2+3 3+2
4+1 6+2 7+1

= 355
588

Properties of Matrix Addition

Addition of matrices satisfies the following properties.

Let A, B, and C be of same order matrices.

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(a) Closure Property :

If A and B are two matrices of same orders, their sum A + B is also of same order as that
of A and B.

Let A = 1 2 and B = 2 6 be two matrices of order 2 × 2.
3 4 3 5

Then A + B = 1 2 + 2 6 = 1+2 2+6
3 4 3 5 3+3 4+5

= 3 8 , which is of order 2 × 2.
6 9

Hence, matrix addition satisfies closure property.

(b) Commutative Property :

If A and B are two matrices of same orders, A+B = B+A.

Let A= 1 2 and B = 1 4
6 7 3 6

Then, A+B= 1 2 + 1 4 = 2 6
6 7 3 6 9 13

and B+A = 14 + 12
36 67

= 2 6 26
9 13 9 13

? A+B=B+A

Hence, the addition of two matrices satisfies commutative properly.

(c) Associative Properly :

If A, B, and C are three matrices of same orders,

then (A + B) + C= A + (B + C).

Let A= 1 2 ,B= 2 1 and C = 2 4
Now, 3 4 5 6 6 2

Again, A+B= 1 2 + 21 = 3 3
3 4 56 8 10
?
(A + B) + C = 3 3 + 2 4 = 5 7
8 10 6 2 14 12

B+C= 2 1 + 24 = 4 5
5 6 62 11 8

A + (B + C) = 1 2 + 4 5 = 5 7
3 4 11 8 14 12

(A + B) + C = A + (B + C)

Hence, matrix addition satisfies associative properly.

(d) Existence of Identity Element :

If A is any matrix and O be a null matrix of same order as that of A, then

A+O=O+A=A

Let, A= 1 6 and O = 0 0
3 7 0 0

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Now, A+O= 1 6 + 0 0 = 1 6
and 3 7 0 0 3 7
?
O+A = 0 0 + 1 6 = 1 6
0 0 3 7 3 7

A+O=O+A=A

Here, O is called additive identity of matrix A. Hence, null matrix is additive identity .

(e) Existence of Additive Inverse : If A is a matrix of any order, then there exists another
matrix (–A) of same order as that of A. Then A + (–A) = O

Where O is a null matrix of same order as that of A .

Let A= 2 5 and (–A) = –2 –5 = 2–2 5–5 = 0 0 =O
7 8 –7 –8 7–7 8–8 0 0
Here, (–A) is called additive inverse of matrix A.

Subtraction of Matrices

Let A and B be two matrices of same order. Then subtraction of matrix B from A is denoted
by A – B and it is obtained by subtracting the elements of B from the corresponding elements
of A. The order of A – B is same order of A or B.

Let A = a11 a12 – b11 b12 = a11 – b11 a12 – b12
a21 a22 b21 b22 a21 – b21 a22 – b22

For example : Let A = 6 7 and B = 4 6
5 2 7 8

Then, A – B = 6 7 – 4 6 = 6–4 7–6 = 2 1
5 2 7 8 5–7 2–8 –2 –6

Worked Out Examples

Example 1. Let A = a11 a12 and P = p11 p12 , find A + P .
Solution : a21 a22 p21 p22

Example 2. Here, A = a11 a12 and P = p11 p12
Solution : a21 a22 p21 p22

Example 3. Now, A + P = a11 a12 + p11 p12 = a11 + p11 a12 + p12
Solution : a21 a22 p21 p22 a21 + p21 a22 + p22

If P = 4 56 and Q = 2 6 7 , then find P + Q.
5 21 4 2 5

Here, P + Q= 4 5 6 + 2 6 7
5 2 1 4 2 5

= 4+2 5+6 6+7 = 6 11 13
5+4 2+2 1+5 94 6

Let P = 2 6 7 and Q = 1 23
2 3 4 2 56

Here, P = 2 67 and Q = 1 2 3 ,
2 34 2 5 6

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Now, P – Q = 2 6 7 – 1 2 3
2 3 4 2 5 6

= 2–1 6–2 7–3 = 14 4
2–2 3–5 4–6 0 –2 –2

Example 4. Find the additive inverse of P = 6 5 .
Solution : 2 3
6 5 a b
Example 5. Here, P = 2 3 and let Q = c d be additive inverse of P.
Solution :
By definition of additive inverse, we get

P + Q = O, where O is a null matrix of order 2 × 2

or, 6 5 + a b = 0 0
2 3 c d 0 0

or, 6+a 5+b = 0 0
2+c 3+d 0 0
Equating the corresponding elements, we gets

6+a=0 or, a = –6

5+b=0 or, b= –5

2+c=0 or, c = –2 ?Q= –6 –5
3+d=0 or, d = –3 –2 –3

214
Let P = 6 2 3 , find the matrix Q such that P + Q = I, where 1 is a unit

542
matrix of order 3 × 3.

214 100

Here, P = 6 2 3 , I = 0 1 0

542 001
Now, P + Q = I

i.e., Q = I – P

1 0 0 2 1 4 –1 –1 –4
0 1 0 – 6 2 3 = –6 –1 –3

0 0 1 5 4 2 –5 –4 –1

Exercise 5.3

Very Short Questions
1. (a) State the required condition for addition of two matrices with an example.

(b) State the required condition for subtraction of two matrices with an example.
(c) Define additive inverse of a matrix with an example
(d) Define additive identity of a matrix with an example.
(e) State the closure property of matrix addition with an example.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

2. Which of the following matrices can be added and subtracted? Give your reasons.

A= 2 4 ,B= 7 8 , C = [2 4 6]
6 7 2 3

D = [2 1 7], E = 1 and F = [2 7 2]
2

3. (a) If P = 2 3 4 and Q = 1 2 6 , then find (i) P + Q (ii) P – Q.
2 4 7 7 3 2

(b) If M = 1 2 and N = 3 2 , find (i) M + N (ii) M – N
2 5 5 7

Short Questions

4. Let P = 12 and Q = 2 4 , and R = 21 verify the following. Also state the
34 6 7 46
property of each of them.

(a) P + Q = Q + P (b) Q + R = R + Q

(c) P + (–P) = O, where O is a null matrix of order 2 × 2.

(d) (P + Q) + R = P + (Q + R)

5. (a) If A = p q , find additive inverse of A
r s

(b) If P = 6 7 , then find the additive inverse of P.
5 8

6. (a) If P = p q and Q = 0 0 , then verify that P + Q = Q + P = P. What property
r s 0 0

does it state? What is Q called ?

(b) If M = 2 7 , find the additive identity of matrix M.
8 9

7. (a) If 6 5 + 4 y = z 6 , then find the values of p, x, y and z.
2 x 2 1 p 5

(b) If a 2 + 4 x = 6 3 , find the values of a, b, x and y.
b 5 8 y 10 5

(c) If 4p – 1 8 = p+5 s–4 , find he value of p, q, r and s.
q+3 4 5 r+4

8. (a) Let P = 2 3 6 ,Q= –4 6 7 and R = –4 –8 –4 , verify that
2 4 1 82 1 3 2 1

(i) ( P + Q) + R = P + (Q + R)

(b) Find the matrix X of P = [2 4 6], Q = [7 –4 8] and R = [–10 8 7]
and P + Q = R + X.

9. (a) If P + Q = 5 2 and P – Q = 3 6 , find the matrices P and Q.
0 9 –3 2

(b) If A – B = 6 7 and A + B = 10 4 , find the matrices A and B.
8 2 6 7

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Project Work

10. List the number of boy and girl students in your school in each of the class 8, 9, and 10.
Write them in matrix form. Add them to find the total number of boy and girl students.

2. A and B matrices can be added, subtracted, equal orders.

C, D and F matrices can be added, C – D, C – F, D – C, F – D, F – C, D – F

can be performed, equal orders.

3. (a) (i) 3 5 10 (ii) 1 1 –2 (b) (i) 4 4 (ii) –2 0
9 7 9 –5 1 5 7 12 –3 –2
4. (a) Commutative (b) Commutative

5. (a) –p –q (b) –6 –7 (c) Additive inverse (d) Associative
–r –s –5 –8
00
6. (a) Existence of additive identity, additive identity (b) 00

7. (a) x = 4, y = 1, z = 10, p = 4 (b) a = 2, b = 2, x = 1, y = 0

(c) p = 2, q = 2, r = 0, s = 12

8. (b) P = (19 –8 7) (c) (19 –8 7)

44 1 –2 8 11 2 – 3
–1 2
9. (a) P = 3 11 , Q = 3 7 (b) A= 2 ,B=
– 2 22 2 9 5
7
2 2

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