O Math 10 2075

or, a× – 1 3
2
= 32

∴ a = – 256 Algebra

The given G.P. is – 256, 128, – 64, ... ... ... ... and the 15th term is

t15 = ar15 – 1 14

or, t15 = – 256 × 1
2
1
or, t15 = – 256 × 28 × 26

or, t15 = – 256 × 1 64
256 ×
1
∴ t15 = – 64

Example 6: In a geometric series, the 7th term is 16 times of the third term and the fifth term is 1 . Find
16
the 2nd term.

Solution: t7 = 16 × t3 Now, t5 = 1
Here, ar6 = 16 × ar2 16
or,
or, ar4 = 1
16

or, aarr26 = 16 or, ar × r3 = 1
16

or, r4 = 16 = 24 or, t2 × 23 = 1
∴ r = 2.  16

or, t2 = 1 = 1
16×8 128

The second term is 1128.
Example 7: The product of three numbers in G.P. is 1000. If we add 6 to its second number and 7 to its

third number, the resulting three numbers form an A.P. Find the numbers in G.P.

Solution:

Let the numbers in G.P. be ar, a and ar.

Then, a × a × ar = 1000
r

or, a3 = 1000

∴ a = 10

Hence, the numbers are 1r0, 10 and 10r.

The new numbers are 1r0, 10 + 6 and 10r + 7 and these numbers are in A.P.
10
or, 16 – r = (10r + 7) – 16

or, 16r – 10 = 10r2 – 9r

or, 10r2 – 25r + 10 = 0
or, 2r2 – 5r + 2 = 0

or, 2r2 – 4r – r + 2 = 0

Geometric Sequence and Series / 47

or, (2r – 1) (r – 2) = 0

∴ r = 2 or 1
2
Case I, when r = 2, the numbers are 120, 10, 10 × 2 i.e. 5, 10, 20

Case II, when r = 21, the numbers are 11/02, 10, 10 × 1 i.e. 20, 10, 5
2

Hence, the numbers are 5, 10, 20 or 20, 10, 5.

Exercise 1.7A

1. a. Define goemetric mean.

b. Define common ratio of a geometric sequence.

c. Write the formula to find the general term of a geometric sequence.

2. Examine whether the following sequences form geometric sequence. Also find their common ratios.

a. 25, 5, 1, 51, ... ... ... b. 118, – 217, 19, 31, ... ...

c. 4, 9, 16, 25, ... ... ... d. ab2, ab, a, ba, ... ... ...

3. From the following geometric sequences, find the common ratio and the next two terms.

a. 3, 6, 12, 24, ... ... ... b. 64, – 32, 16, – 8, ... ... ...

c. 2, 1, 1, 12, ... ... ... d. am – n, am, am + n, am + 2n, ... ...
2
4. From the following first terms and common ratios of the geometric sequences, write the first three

terms and the indicated terms.

a. First term (a) = 23, common ratio (r) = – 2 , t6 = ?
3
1
b. First term (a) = 54, common ratio (r) = 3 , t7 = ?

c. First term (a) = 5, common ratio (r) = – 3 , t5 = ?

d. First term (a) = 32, common ratio (r) = 3 , t4 = ?
2
5. a. Find the first term of a G.P. whose fifth term is 729 and the common ratio is 3.

b. If the ninth term of a G.S. whose first term is 4, is 64, find the common ratio.

c. Find the common ratio of a G.S. whose first term is 1 and the sixth term is 625.
5

6. Find the number of terms in each of the following geometric series.

a. 1 + 3 + 9 + ... ... ... ... + 243 b. 1 – 1 + 1 ... ... ... ... + 729
9 3

c. 4 + 6 + 9 + ... ... ... ... + 2483 d. 3 + 12 + 48 + ... ... .. + 192

7. a. If a, a – 2, a + 1 are in a geometric sequence, find the value of a.

b. If x + 4, x – 2 and x – 6 are in a geometric sequence, find the value of x.

48 / Optional Mathematics Class 10

c. The first three terms of a G.S. are x, x + 3, x + 9. Find the value of x and 5th term of the series. Algebra
d. Find the x, if 2x – 8, x + 2 and 3x – 2 are consecutive terms in a geometric sequence.
e. The length of the sides of a triangle form a G.P. If the perimeter of the triangle is 37cm and the

shortest side of the length 9cm, find the lengths of the other two sides.— 1.7

8. a. If the 4th term of G.P. is 8 times its first term and its fifth term is 64, find the tenth term of the

series. 49,

b. If the ratio of the third and the sixth terms of a GP is 8:27 and its 8th term is find the tenth
term of the series.

c. The 7th term of G.P. is 8 times of 4th term and 5th term is 48. Find the sequence.

9. a. Is 512 a term of the sequence 2516, 614, 116, ... ... ... ?

b. Which term of the series 1 + 1 + 1 + 3 + … … is 729 ?
9 3

Exercise 1.7B

1. a. The fourth term of a G.P. is square of its second term and the first term is – 3. Determine the
seventh term of the G.P.

b. The third term of a G.P. is square of its first term. If the second term is 8, determine the 6th term.

2. a. If the 5th and 9th terms of a G.S. are 1 and 16 respectively, find the 4th term.
3 243

b. If the 5th and 10th terms of a G.S. are 16 and 1 respectively, find the 15th term.
2

c Find the G.P. for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.

d. Find the 8th term of a geometric series whose third and sixth terms are respectively 3 and 38.

3. a. The second term of a geometric series is 27 times the 5th term and 7th term is 31. Find the series.

b. The seventh term of a geometric series is 16 times the third term and the fifth term is 116. Find
the second term.

4. a. Find the value of x for which 2x, 2x + 3 and 2x + 9 are the three terms of a geometric sequence
and calculate the fifth term.

b. If nth terms of the series 5 + 10 + 20 + ... ... ... and 1280 + 640 + 320 + ... ... are equal, find the
value of n.

5. a. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers respectively,
we obtain A.P. Find these numbers.

b. The sum of three numbers in A.P. is 15. If 8, 6, 4 be added to these respectively, the resulting
numbers are in G.P. Find the numbers.


Geometric Sequence and Series / 49

1.8 Geometric Mean

Introduction

If three numbers a, b and c are in G.P. then the middle term b is known as the geometric mean (G.M.)
between the other two. For example, 24, 12 and 6 are in G.P., so 12 is the geometric mean between 24 and 6.

If any number of terms are in G.P. then the terms between the first and the last terms are known as
the geometric means between the first and the last. For example, 1, 2, 4, 8, 16, 32 are in G.P. and 2, 4, 8, 16
are the geometric means between 1 and 32.

Geometric Mean between Two Numbers

Let G be the geometric mean between two numbers a and b then a, G, b are in G.P.

Now by definition,

Ga = b
G

or, G2 = ab

∴ G = ab
Example 1: Find the geometric mean between 2 and 8.

Solution:

Here, the first term (a) = 2

Last term (b) = 8

Geometric mean (G) = ?
Then G = ab

or, G = 2×8

∴ G = 4

Geometric Means between Two Numbers

Let x1, x2, ... ... ..., xn1 be the n1 geometric means between a and b, then a, x1, x2, x3, ... ..., xn1, b are
in G.P. If r be the common ratio, then tn = arn – 1, here n = number of terms.

As there are n1 + 2 terms in all i.e. n = n1 + 2

or, b = arn1+ 2 – 1

or, rn1+ 1 = b
a

∴ r= b1
n1 + 1

a

Now, the required geometric means are:

x1 = ar = a b 1
x2 a n1 + 1

= ar2 = a b 2
a n1 + 1

50 / Optional Mathematics Class 10

x3 = ar3 = a b 3 Algebra
a
n1 + 1

xn1 = arn1 = a b n1
n1 + 1

a

Note: The second term of the sequence is the first mean and the third term is the second mean and so on.

Example 2: Insert 5 geometric mean between 81 and 3.

Solution:

Here, the first term (a) = 81

The last term (b) = 3

Number of means (n1) = 5
Hence the common ratio,

r = b 1
a
n1 + 1

or, r= 3 5 1 1 = 1 1
81 + 27 6

∴ r= 1
3
Hence, the required geometric means are

x1 = ar =81 × 1 = 27 3
x2 3
x3 1
x4 = ar2 = 81 × 3 = 27
x5
Example 3: = ar3 = 81 × 1 = 9 3
33
Solution: 1
= ar4 = 81 × 9 = 9

= ar5 = 81 × 1 = 3 3 1
93
1
If 4, p, q, – 16 are in G.P., find p and q.

First term (a) = 4

Last term (b) = – 1
16

n1 = number of means = 2
Common
ratio, (r) = b 1
a n1 + 1

–1/16 1 –1 1 – 1 3× 1 1
4 16×4 3 3 4
or, r= 2+1 = = = –
4

Second term, t2 = p = ar = 4 × – 1 = – 1
4

Geometric Mean / 51

Third term, t3 = q = ar2 = 4 – 1 2 4 × 1 = 1
4 16 4
=

Hence, p = – 1 and q = 41.

Example 4: Two numbers are in ratio 16:1 and their G.M. is 16. Find the numbers.

Solution:

Let the two numbers be 16x and x and thier G.M. is 16.

or, (16)2 = 16x × x
or, 256 = 16x2

or, x2 = 16
or, x = ± 4

When x = 4, the numbers are 4x and x i.e. 64 and 4.

When x = – 4, the numbers are 4x and x i.e. – 64 and – 4.
Example 5: Find the number of G.M.’s inserted between 4 and 128, the ratio of the first mean and the

last mean is 1:8.
Solution:

Here, the first term (a) = 4 and the last term (b) = 128

Let the number of means = n, then the common ratio (r) = b 1
a n+1

i.e. r= 128 n 1 1= (32) n 1 1
4 + +

From question, First mean : Last mean = x1 : xn = 1:8

i.e. xx1n = 1 1 – n 5(1 – n)
+
ar = r = r1 – n = (32) n 1 = 1 – n = 1 – n = (2) n + 1
arn rn + 1 + 1
(32) n (25) n

Now, x1 = 1
xn 8

5(1 – n)

or, (2) n + 1 = 2–3

or, 5(n1+–1n) = – 3
or, 5 – 5n = – 3n – 3

or, – 5n + 3n = – 3 – 5

or, – 2n = – 8

or, n = 4

Relation between A.M. and G.M. of two Positive Numbers

Let a and b be two positive numbers and A.M. and G.M. be the arithmetic mean and geometric
mean between a and b respectively, then

A.M. = a + b and G.M. = ab
2

52 / Optional Mathematics Class 10

Now, A.M. – G.M. = a + b – ab
2
Algebra
= a+b–2 ab
2
b)2 ≥ 0
= ( a+
2

Hence, AM ≥ GM

Therefore the arithmetic mean between two positive numbers is greater than or equal to their
geometric mean.
Example 6: Find the two numbers whose A.M. is 25 and G.M. is 20.
Solution:

Let the two numbers be a and b, then

a + b = 25 and ab = 20
2 ∴ ab = 400 ... ... (ii)

∴ a + b = 50 ... ... (i) and

Now, a – b = (a + b)2 – 4ab

= (50)2 – 4×400

= 2500 – 1600

= 900

∴ a – b = 30 ... ... (iii)
Solving equations (i) and (iii), we get
a = 40 and b = 10.

Hence, the required numbers are 40 and 10.

Exercise 1.8A

1. a. What is the geometric mean between a and b?

b. Write the relation between the number of means n and the number of terms in GP?

2. Find the geometric mean between the following two numbers.

a. 3 and 27 b. 9 and 1
27

c. 6 and 24 d. (a + b)2 and (a – b)2

3. a. If 23, x, y, – 1 are in geometric series, find the values of x and y.
96
b. If 81, a, b, c, 16 are in G.P., find the values of a, b and c.

4. a. If the geometric mean between k – 2 and k + 4 is 4, find the value of k.

b. If the geometric mean between 4 and 4k is k, find the value of k.
1
c. If the geometric mean between 9 and m is 2, find the value of m.

d. If 16 is the geometric mean between m and m3, then find the value of m.

5. a. If x, – 3, 9, y are in geometric sequence, find the value of x and y.

Geometric Mean / 53

b. If 2, x, 18, y are in G.P., find the values of x and y.
c. If a, 3, b, 12, c are in G.P., find the values of a, b and c.

Exercise 1.8B

1. a. Insert 3 geometric means between 5 and 80.

b. Insert 3 geometric means between 1 and 9.
9
c. Insert 4 geometric means between 27 and 392.

d. Insert 5 geometric means between 35 and 2240.

2. a. There are some geometric means between 5 and 80. If the second mean be 20. Find the number

of means between the two numbers. Also find the last mean.

b. If n geometric means are inserted between 4 and 1 such that second mean is 1, find the value of n.
32

c. If n geometric means are inserted between 7 and 448 such that third mean is 56, find the value of n.

3. a. There are 7 geometric means between x and 64. If the fourth mean is 4, find the value of x.

b. There are 4 geometric means between 4 and x. If the third mean is 32, find the value of x.

c. There are 4 geometric means between 5 and x. If the third mean is 40, then find x.

4. a. Find the number of geometric means inserted between 3 and 729 when the ratio of the first

mean and the last mean is 1:27. 16 1
3 24
b. If n geometric means are inserted between and such that (n – 1)th mean: 2nd mean = 1:8,

find the value of n.

c. There are n geometric means between 9 and 8 such that 27(nth mean) = 8(2nd mean). Find the
8 81
value of n.

5. a. If two numbers are in ratio 1:4 and their geometric mean is 8, then find the numbers .

b. If two numbers are in ratio 1:9 and their goemetric mean is 27, then find the numbers.
c. If two numbers are in ratio 1:16 and their goemetric mean is 14, then find the numbers.

6. a. Find two numbers whose arithmetic mean is 5 and geometric mean 4.

b. The A.M. between two numbers is 15 and G.M. 9. Find the numbers.

c. The A.M. between two numbers is 45 and G.M. is 27. Find the numbers.

d. If the arithmetic mean of two numbers is 50 and G.M. is 80% of A.M., find the numbers.

7. a. Find two numbers whose difference is 12 and whose A.M. exceeds their G.M. by 2.

b. The sum of the four consecutive terms of a G.S. is 30. If the arithmetice mean of the first two terms

is 3, find the terms.

8. a. If an arithmetic mean m and two geometric means x and y are inserted between two number a

and b, prove that: x2 + y2 = 2m.
y x

b. If a, b and c are in A.P. and x, y and z are in G.P., prove that xb – c xc – a za – b­ = 1.



54 / Optional Mathematics Class 10

1.9 Sum of Geometric Series

The series associated with a geometric sequence is known as a geometric series. For example, Algebra

9 + 3 + 1 + 1 + 1 is geometric series associated with the geometric sequence 9, 3, 1, 13, 91.
3 9

Let a be the first term, r be the common ratio, n be the number of terms,  be the last term and Sn be
the sum to n terms of G.S., then

Sn = a + ar + ar2 + ... ... + arn – 2 + arn – 1 ... ... (i)
Multiplying both terms by r,

r Sn = ar + ar2 + ar3 + ... ... ... + arn – 1 + arn
–Sn =–a +– ar +– ar2 –+ ... ... ... –+ arn – 2 –+ arn – 1
(r – 1)Sn = – a + arn

∴ Sn = a(rn – 1)
r–1

Again, Sn = a(rn – 1)
r–1

or, Sn = arn – a = arn – 1.r – a
r–1 r –1

∴ Sn = lr – a [  = arn – 1]
r–1

Example 1: Find the sum of the geometric series 1 + 2 + 4 + ... ... 7 terms.

Solution:

First term (a) = 1 2
1
Common ratio (r) = = 2

Number of terms (n) = 7

Sum of the terms (Sn) = ?

By using formula,

Sn = a(rn – 1)
r–1

or, Sn = a(27 – 1)
r–1

or, Sn = 1(128 – 1)
2– 1

∴ Sn = 127
Example 2: In a G.P., if the first term is 7, the last term is 448 and the sum is 889, find the common

ratio.

Solution:

First term (a) = 7
Last term () = 448

Sum of Geometric Series / 55

Sum of terms (Sn) = 889
Common ratio (r) = ?
By using formula,

lr – a
Sn = r – 1

or, 889 = 448r – 7
r– 1

or, 889r – 889 = 448r – 7

or, 441r = 882

∴ r=2
Example 3:
5
Solution:
Find the sum of ∑ 3n + 1
n=1

5

Here, ∑ 3n + 1
n=1

= 31+1 + 32+1 + 33+1 + 34+1 + 35+1

= 32 + 33 + 34 + 35 + 36

= 9 + 27 + 81 + 243 + 729

= 1089
Example 4: How many terms of the series 32 + 48 + 72 + ... ... will add up to 665 ?
Solution:

First term (a) = 32

Common ratio (r) = 48 = 3
32 2

Sum of terms (Sn) = 665

Number of terms (n) = ?

By using formula,

Sn = a(rn – 1)
r–1

3 n
2
32 – 1
665 =
or, 3
2
– 1

3 n
2
or, 32 – 1
665 =
1

2

3 n
2
or, 665 = 64 – 1

or, 23 n – 1 = 665
64

56 / Optional Mathematics Class 10

or, 32 n = 665 + 1
64
Algebra
or, 32 n = 665 + 64
64

or, 23 n = 729
64

or, 23 n = 3 6
2

∴ n = 6

Number of terms = 6

Example 5: If S3 and S6 of a G.P. are 7 and 63 respectively, find the common ratio.
Solution:

Here, S3 = 7
or,
And, 7 = a(r3 – 1) .. ... .. (1)
r–1
or,
S6 = 63

63 = a(r6 – 1) ... ... ... (2)
r–1

Now solving equations (1) and (2), we have

or, SS36 = a(r6 – 1)

r–1
a(r3 – 1)

r–1

or, 673 = a(r3 + 1) (r3 – 1)× r–1
r– 1 a(r3 – 1)

or, 9 = r3 + 1

or, r3 = 9 – 1 = 8

or, r3 = 23

or, r = 2

Common ratio = 2

Example 6: The second and fifth term of G.S. are 3 and 81 respectively. Find the sum of the first five
terms.

Solution:

The second term of G.S. is
t2 = 3
or, ar = 3
Again, the fifth term of G.S. is
t5 = 81
or, ar4 = 81

Sum of Geometric Series / 57

or, ar . r3 = 81

or, 3 . r3 = 81 [As ar = 3]

or, r3 = 81
3

or, r3 = 27 = 33

∴ r = 3

and ar = 3

or, a × 3 = 3
∴ a = 1

Now, S5 = a(r5 – 1) = 1(35 – 1) = 1(243 – 1) = 242 = 121.
r–1 3–1 2 2

Example 7: Find the G.P. for which the sum of the first two terms is – 4 and the fifth term is 4 times the

third term.

Solution:

Let S2 = t1 + t2 = – 4
or, a + ar = – 4

or, a(1 + r) = – 4 ... ... ... (1)

and t5 = 4 × t3

or, ar4 = 4 × ar2

or, r2 = 4
or, r = ± 2

Taking r = 2, from equation (1), we have

a(1 + 2) = – 4

or, a = – 4
3

The GP is a, ar, ar2, ...

i.e. –3 4, –3 4 × 2, –3 4 × 22, ...
i.e. –3 4, –3 8, – 316, .. ..

When r = – 2, from equation (1), we get

Taking r = – 2, from equation (1), we have

a(1 – 2) = – 4

or, a = 4

The GP is a, ar, ar2, ...

i.e. 4, 4 × – 2, 4 × (– 2)2

i.e. 4, – 8, 16, ...

58 / Optional Mathematics Class 10

Example 8: If the sum of the first three terms of a G.P. is 1 and the sum of the first six terms is 28, find Algebra
the sum of the first 9 terms of the series.

Solution:

Let a be the first term and r be the common ratio of the series, then

Sn = a(rn – 1)
r–1

or, S3 = a(r3 – 1)
r–1

or, 1 = a(r3 – 1) ... ... (i)
r–1

Again, S6 = a(r6 – 1)
r–1

or, 28 = a(rr6––11) ... ... (ii)

Dividing (ii) by (i), Now, the sum of the first 9 terms,

or, 28 = r6 – 1 Sn = a(r3 – 1)
r3 – 1 r–1

or, 28 = r3 + 1 113(39 – 1)
or, S9 = 3 – 1
∴ r=3

Substituting r = 3 in (i), 1
26
1 = a(33 – 1) or, S9 = (19683 – 1)
3–1
19682
or, a(27 – 1) = 2 or, S9 = 26

∴ a = 1 ∴ S9 = 757
13

Example 9: The sum of the first three terms of a G.P. is 7 and the sum of their squares is 21. Determine

the first five terms of a G.P.

Solution:

Let the G.P. be a, ar, ar2, ... ...

By given conditions,

a + ar + ar2 = 7

or, a(1 + r + r2) = 7 ... ... (1)

and a2 + (ar)2 + (ar2)2 = 21

or, a2(1 + r2 + r4) = 21 ... ... (2)

From equations (1) and (2), we have

aa22((11 + r2 + r4) = 21
+ r + r2)2 72

Sum of Geometric Series / 59

or, (1 +(1r2+)2r–+2rr22)2+ r2 = 3
7

or, (1 + r2 + – r)(1 + r2 + r2) = 3
(1 + r + r2)2 7

or, 7(1 + r2 – r) = 3(1 + r + r2)

or, 7r2 – 3r2 – 7r – 3r + 7 – 3 = 0

or, 4r2 – 10r + 4 = 0

or, 2r2 – 5r + 2 = 0

or, 2r2 – 4r – r + 2 = 0

or, (2r – 1)(r – 2) = 0

∴ r = 2 or 1
2

Case I, when r = 2,

a(1 + 2 + 4) = 7

or, a = 1

∴ G.P. is 1, 1 × 2, 1 × 22, ... ... i.e. 1, 2, 4, ... ...

The first five terms are 1, 2, 4, 8, 16.

Case II, when r = 1 ,
2

a 1+ 1 + 1 =7
2 4

or, a = 4
∴ G.P. is 4, 4 × 21, 4 × 41, ... ... i.e. 4, 2, 1, ... ...
The first five terms are 4, 2, 1, 12, 14.

Exercise 1.9A

1. Find the sum of the following series.
a. 2 + 6 + 18 + ... ... ... ... ... 6 terms b. 2 + 10 + 50 + ... ... ... ... ... 5 terms

c. 4 – 8 + 16 – ... ... ... ... ... 5 terms d. 27 + 18 + 12 + ... ... ... ... ... 6 terms

2. Find the sum of the following series.

a. 7 + 14 + 28 + ... ... ... ... ... + 1792 b. 3 – 1 + 1 – ... ... ... ... ... – 1
3 81
c. 6225 2 2
+ 125 + 25 + ... ... ... ... ... + 50 d. 2 – 2 + 2 2 – ... ... ... ... ... + 64 2

3. a. Find the first term of a G.P. whose common ratio is 3, the last term is 189 and the sum of the

series 280.

b. Find the first term of a G.P. whose common ratio is 2 and the sum of the first 9 terms 1533.

4. Find the sum of the following. 5 8

66 c. ∑ (3m + 2) d. ∑ 2(– 2)r
m=1 r=3
a. ∑ 3k b. ∑ 2 × 3k
k=2 k=1

60 / Optional Mathematics Class 10

Exercise 1.9B Algebra

1. a. How many terms of the series 81 + 54 + 36 + ....... will add up to 211?
b. How many terms of the series 64 + 32 + 16 + … … must be taken so that the sum may be 12712?
2. a. If S2 and S4 of a geometric series are 3 and 15 respectively, find the common ratio.
b. If S3 and S6 of a geometric series are 13 and 1066 respectively, find the common ratio.
3. a. The second and the 5th term of a geometric series are – 4 and 108 respectively. Find the sum of

the first 7 terms of the series.

b. If the second and the sixth terms of a GS are 1 and 16 respectively, find the sum of the first 10
terms.

c. If the third and the fifth terms of a GS are 1 and 9 respectively, find the sum of the first 8 terms.

4. a. The sum of first four terms is 40 and the sum of first two terms is 4 of a geometric series whose

common ratio is positive. Find the sum of first eight terms.

b. In a G.P., the sum of the 4 terms is 5152 and that of the first 2 terms is 343. Find the sum of the
first 6 terms.

c. If the sum of the first four terms and the eight terms of a G.P. are 15 and 255 respectively, find
2 32
the sum of the first 11 terms.

5. a. The 6th term of a G.P. is 16 times its second term and the sum of the first 7 terms is 1427. Find
the first term.

b. If the 6th term of a G.P. is 8 times the third term and the sum of the first 5 terms of the series is
4621, find the sum of the first 7 terms of the series.
39
6. a. The sum of three numbers in G.P. is 10 and their product is 1. Find these numbers.

b. Find three numbers in G.P. whose sum is 38 and product is 1728.

c. Find the three numbers in G.P. whose sum is 19 and the sum of their squares is 133.

7. a. A machine is depreciated at the rate of 10% yearly and the ultimate final value was Rs.6561.
Find the effective life of the machine. The price of the machine is Rs.10,000.

b. A machine depreciates in the value of the rate of 30% every year on the reducing balance. If
the original cost be Rs.20000 and ultimate final value be Rs.4802, find the effective life of the
machine.



Sum of Geometric Series / 61

1.10 Linear Programming

Inequalities

Two numbers a and b are said to be equal if a – b = 0. But they are unequal if a – b ≠ 0. If a and b are
unequal, a may be greater than or less than b.

a > b if a – b is positive.

a < b if a – b is negative.

a = b if a – b is zero.

Sometimes, equality and inequality signs may simultaneously occur. Let us study the following
mathematical statements.

a ≥ b a = b or a > b and a ≤ b a = b or a < b
A mathematical statement containing the sign of type >, <, ≥ or ≤ is said to be an inequality or
inequation.

For a set of real numbers, the statement 'x denotes any x>4

real number greater than 4' is written symbolically as x > 4. -1 0 1 2 3 4 5 6 7 8
Similarly, the symbol x < 4 represents the statement 'x is a real
x<4

number less than 4.' The graphs of inequalities x > 4 and x < 4 -1 0 1 2 3 4 5 6 7 8
are as shown the figure alongside.

The symbol x ≥ 4 represents the statement 'x is greater x≥4
than or equal to 4' and the symbol x ≤ 4 rerpesents 'x is less
than or equal to 4'. The graphs of x ≥ 4 and x ≤ 4 are shown in -1 0 1 2 3 4 5 6 7 8

the figure alongside. x≤4

-1 0 1 2 3 4 5 6 7 8

The graphical solution of linear inequalities of one variable in co-ordinate plane as shwon below.

Y YY Y

x=0 x>0 x≤0 x=0 y≥0 y=0

O O y=0 X' O X

X' O X X' X y<0
X'
X

Y' Y' Y' Y'

Y YYY

x>1 x≤2 x=2

X' O y = –2 X X' O y ≥ –3 X
X
X' O X X' O y < –2 y = –3
x=1

Y' Y' Y' Y'

62 / Optional Mathematics Class 10

Graphs of Inequalities in Two Variables

Any one of the relation represented by ax + by + c > 0, ax + by + c < 0, Algebra
ax + by + c ≥ 0 or ax + by + c ≤ 0 is known as the linear inequality in two variables x and y. The associated
equation of all these linear inequalities in two variable in x and y is ax + by + c = 0. The equation is also the
boundary line of each of the inequalities. For graphing of any linear inequality in two variables, we have to
follow the following steps.

Step 1: Draw the graph of the corresponding equation i.e. ax + by + c = 0 by taking at least two points
on the boundary line. If the inequality posses the symbol ≥ or ≤, which indicate the boundary

line is a part of the solution, draw a solid line, otherwise a broken (dotted) line.

Step 2: Take any point as a test point. The test point should not lie on the boundary line.

Step 3: Substitute the test point in the inequality and check whether the test point gives true or false
result for the given inequality.

Step 4: If the reslut is true, the half plane of of the inequality contains the test point. So shade the half

plane towards the test point as solution. If the result is false, the half plane of the inequality does

not contain the test point. So, shade the half plane opposite to the test point as the solultion.

Example 1: Draw the graph of x – y ≥ 4.

Solution: Y
The given inequality is x – y ≥ 4 and its corresponding equation

is x – y = 4 which is the boundary line of the given inequality. For the

boundary line, X' O A(4,0)
when x = 0, then y = – 4 and when y = 0, then x = 4
X

So, the boundary line passes through the points A(4, 0) and B(0, – 4). B(0,– 4)
Y'
Draw the boundary line passing through A and B as shown. Take
(0, 0) as test point, substituting in given inequality
0 – 0 ≥ 4, i.e. 0 ≥ 4 which is false.

Hence, the graph of x – y ≥ 4 does not contain the origin. Y
Example 2: Find the inequality that represents the shaded region in

the graph.

Solution: (0,2)
(3,0)
From the graph, the boundary line passes through (3, 0) and X'
OX
(0, 2). So, the equation of the boundary line is
Y'
y – y1 = y2 – y1 (x – x1)
x2 – x1
or,
or, y – 0 = 2 – 0 (x – 3)
0 – 3
–2
y = 3 (x – 3)

Linear Programming / 63

or, 3y = – 2x + 6
or, 2x + 3y = 6
Take a test point (0, 0), then 2×0 + 3×0 = 0 < 6.

Since the boundary line is solid, the required inequality is 2x + 3y ≤ 6.

System of Linear Inequalities

A combination of two or more linear inequalities is called a system of linear inequalities. If the

common region of a system of inequalities is called a solution region or feasible region which is either

closed or opened. The closed solution region (feasible region) is called convex polygonal region.

Example 3: Draw the graph of 2x + y ≤ 6 and y – x < 4.

Solution: AY C
Consider the inequality 2x + y ≤ 6 whose boundary line

is 2x + y = 6. (0,6)

x03 (0,4)
y60

The boundary line passes through (0, 6) and (3, 0). Plot

these points and join them by a solid line AB. (-4,0) (3,0)

On substituting the test point (0, 0) in 2x + y ≤ 6, 0 + 0 ≤ 6 X' OX
i.e. 0 ≤ 6 which is true. Hence, the graph of 2x + y ≤ 6 contains
D Y' B

the origin. Again consider y – x < 4 whose boundary line is y – x

= 4. x 0 –4

y40

The boundary line passes through (0, 4) and (– 4, 0) and joining them by dotted line. On substituting
the test point (0, 0) in y – x < 4, we get 0 – 0 < 4 i.e. 0 < 4 which is true, the graph of y – x < 4 contains the
origin. Hence, the graphs of the given inequalities are as shown. The double shaded region is the solution
region (or feasible region) of the given system of inequalities.

Exercise 1.10A

1. Draw the graph of the following inequalities.

a. x ≤ 0 b. x > 0 c. x ≤ 0 d. x < 0
h. y ≥ 0
e. y > 0 f. y < 0 g. y ≤ 0
d. x < – 2
2. Draw the graph of the following inequalities. h. y ≥ 3

a. x ≥ 2 b. x > – 3 c. x ≤ 3 d. x ≥ y
h. x < – y
e. y < 2 f. y ≤ – 1 g. y > 4

3. Draw the graph of the inequalities given below.

a. x < y b. x > – y c. y ≤ – x

e. y ≥ – x f. y < x g. x ≤ – y

64 / Optional Mathematics Class 10

4. Draw the graph of the given inequalities. c. 2x – 3y ≥ 0 d. x + 2y ≤ 4 Algebra
a. x + y ≥ 2 b. x – y ≤ 4 g. x – 2y > 2 h. 4x + 3y ≤ 12
e. 3x + y ≥ 9 f. 2x + 3y < 6

5. Find the enequality that represents the shaded region on the given graph.

a. Y b. Y c. Y

X' O X X' O X X' O X

Y' Y' Y'
d. Y e. Y f. Y

X' O X X' O X X' O X

Y' Y' Y'
g. Y h. Y i. Y

X' O X X' O X X' O X

Y' Y' Y'

6. Draw the inequalities on graph paper and shade the region that represents its solution region.

a. x ≥ 0 and y ≥ 0 b. x ≤ 0 and y < 0

c. x ≤ 2 and y ≥ – 1 d. x ≥ y and y ≤ 2
e. x ≤ – 1 and x + 3y ≥ 6 f. y < 3 and 2x + 3y ≤ 6
g. x + y ≤ 2 and x – 2y ≥ 4 h. 3x + 4y ≤ 12 and 3x – 4y ≥ 12

7. Draw the graphs of the following inequalities and shade the common solution region.

a. x – 2y ≥ 4 b. x + y ≥ 1 a. 3x – 2y ≥ 6 d. x + 2y ≤ 12,
2x + y ≤ 8 3x + 2y ≤ 24,
y ≥ – 1 3x + y ≤ 3 x + 4y ≤ 8 x ≥ 0 and y ≥ 0

x ≥ 0, y ≥ 0 y ≥ – 3, x < 0



Linear Programming / 65

Linear Programming

The objective of every business activity is to make maximum profit with minimum investment,
maximum production with limited resources. Such types of problems are tackled by the mathematical
technique known as Linear Programming.

The linear function whose value is to be maximized or minimized is said to be an Objective Function.
For example, x and y are the number of two types of articles produced from a certain industry and Rs.3 and
Rs.5 are the profits per unit of the first and the second type of articles respectively, then maximum profit P
= 3x + 5y is the objective function.

The conditions to be satisfied by the variables of the objective function are said to be constraints.

The closed plane region bounded by the intersection of boundary lines is called the Feasible Region or

convex polygonal region. The values of the variables x and y involved in the objective function satisfying

all the given conditions are known as Feasible Solution. Y
Example 1: Find the maximum and minimum values of the

objective function F = 3x + y for the given convex

polygonal region. S(3,4)
Solution: R(6,3)

The values of objective function on the vertices of the P(3,1) Q(8,0) X
convex polygonal region can be calculated as follows: X' O

At point P(3, 1) At point Q(8, 0) At point R(6, 3) At point S(3, 4) Y'
F = 3x + y F = 3x + y F = 3x + y F = 3x + y

= 3×3 + 1 = 3×8 + 0 = 3×6 + 3 = 3×3 + 4

=9+1 = 24 = 18 + 3 = 13

= 10 = 21

The maximum value is 24 at the point Q(8, 0) and the minimum vlaue is 10 at point P(3, 1).
Example 5: Maximize P = 3x + y under the following constraints:

2y ≥ x – 1, x + y ≤ 4, x ≥ 0, y ≥ 0 Y
Solution: C(0,4)

The boundary lines of given inequalities:

2y = x – 1 ... ... ... (1)

x + y = 4 ... ... ... (2) B(3,1) X
x = 0 ... ... ... (3) X' O A(1,0)
y = 0 ... ... ... (4)

In equation (1), when x = 1 then y = 0, when x = 3 then y = 1

∴ The line (1) passes through the points (1, 0) and (3, 1). Y'
For (0, 0) the inequality 2y ≥ x – 1 is true, so it contains the

origin.

66 / Optional Mathematics Class 10

In equation (2), when x = 0 then y = 4, when x = 4 then y = 0 Algebra
∴ The line (2) passes through the points (0, 4) and (4, 0). For (0, 0) the inequality x + y ≤ 4 is true, so
it contains the origin.
Similarly, x ≥ 0 is half plane right side the Y-axis and y ≥ 0 is half plane above the X-axis.
∴ In this situation, the inequalities and their solutions are expressed as in the figure given above.

The shaded area OABC is required solution of inequality where O(0, 0), A(1, 0), B(3, 1), C(0, 4).

SN. Vertices (x, y) P = 3x + y

1. O(0, 0) 0+0=0
2. A(1, 0) 3+0=3
3. B(3, 1) 3 × 3 + 1 = 10 (max.)
4. C(0, 4) 0+4=4

Maximum value is in the point B(3, 1), which is 10. Y
Example 2: Write down the system of inequalities that
X' B A X
describe the shaded region in the graph and find the O C
minimum value of F = 3x + 7y.

Solution:

Here, the co-ordinate of the vertices of the shaded region
∆ABC are A(4, 4), B(2, 2) and C(4, 0).

Producing the side AB of triangle ABC, it passes through
(0, 0) and (4, 4) as shown. As it is the line making 45° with X-ax-
is its equation is y = x or x – y = 0.

From figure, point (3, 2) is a point on the solution region. YY'
The inequality having boundary line AB must be satisfied by
point (3, 2). A
B
Hence, the required inequality is x – y ≥ 0 ... ... (1) OC

Similarly, the side BC produced intersect Y-axis at (0, 4)
as shown. Hence, the line passing through (4, 0) and (0, 4) is

x + y = 1
a b

i.e. x + 4y =1 or, x + y = 4 X' X
4

As the inequality with boundary line BC does not contain Y'
the point (0, 0), the required inequality is x + y ≥ 4 ... ... (2)

The line AC is parallel to Y-axis and at distance 4 units from Y-axis its equation is x = 4. As it does

not contain (0, 0), the required inequality is x ≥ 4 ... ... (3)

The value of the function F = 3x + 7y at the vertices of the triangle ABC can be calculated in

following table. SN. Vertices (x, y) F = 3x + 7y

1. A(4, 4) 3×4 + 7×4 = 40

2. B(2, 2) 3×2 + 7×2 = 20

3. C(4, 0) 3×4 + 7×0 = 12 (min.)

Minimum value is 12 at the point C(4, 0).

Linear Programming / 67

Example 7: Two tailors A and B earn Rs.150 and Rs.200 per day respectively. A can stitch 6 shirts and
4 pants per day while B can stitch 10 shirts and 4 pants per day. How many days shall each work
if they are supposed to stitch at least 60 shirts and 32 pants at a minimum labour cost ?

Solution:

Let tailor A work for x days and tailor B for y days. Y

Total labour cost, C = 150x + 200y.

Given constraints,

Number of shirts is at least 60 and the number of A(0,8)

pants is at least 32. Here A stitches 6x shirts in x days and D(0,6)

B stitches 10y shirts in y days. B(5,3)
Hence, 6x +10y ≥ 60
Similarly, 4x + 4y ≥ 32, x ≥ 0, y ≥ 0

The boundary lines of the given constraints are X' O E(8,0) A(10,0) X

6x + 10y = 60

or, 3x + 5y = 30 ... ... ... (1) Y'

4x + 4y = 32

or, x + y = 8 ... ... ... (2)

x = 0 ... ... ... (3)

y = 0 ... ... ... (4)

As the boundary line (1) passes through the points (0, 6) and (10, 0), drawing the boundary line
and as 0 + 0 ≥ 30 or 0 ≥ 30 which is false, so the graph of the first inequality does not contain the origin.

Similarly, the second boundary line passes through the points (0, 8) and (8, 0) and the graph of the second
constraint does not contain origin as 0 ≥ 8 is false.

Plotting the graphs of all four constraints we get the above figure, the feasible region is the intersection
of the portions above these lines as shown in the graph. The feasible region has the boarder vertices A(0, 8),
B(5, 3) and C(10, 0).

SN. Vertices (x, y) Objective Function C = 150x + 200y
1. A(0, 8) 150×0 + 200×8 = 1600

2. B(5, 3) 150×5 + 200×3 = 1350 (min.)

3. C(10, 0) 150×10 + 200×0 = 1500

Therefore, A should work 5 days and B should work 3 days to minimize the labour cost.

68 / Optional Mathematics Class 10

Exercise 1.10B Algebra

1. Determine the maximum and minimum values of the function F = 2x + 7y for each of the following

feasible regions:

a. Y b. Y c. Y
B(5,6) C(0,5) B(2,5)

C(2,4) C(-2,3)

X' O A(7,1) X' O A(3,0) B(7,0) X X' O A(3,1)
X X

Y' Y' Y'

2. Determine the maximum and minimum values of the function P = 3x + 2y for each of the following

feasible regions:

a. Y b. Y C(5,7) c. Y
B(3,6) C(3,6) B(6,5)

C(0,4) D(3,5) B(6,5)

X' O (0,0) A(5,0) X X' O A(4,1) X X' O(0,0) A(5,-1)X

Y' Y' Y'

3. a. Maximize the objective function P = 3x + 5y under the following constraints:

x + y ≤ 6, x – y ≥ 2, x ≥ 0 and y ≥ 0.

b. Calculate the maximum value of P = 9x + 7y under the following constraints:

x + 2y ≤ 8, x – y ≤ 4, x ≥ 0 and y ≥ 0.

c. Find the maximum value of P = 14x + 16y under the following constraints:

3x + 2y ≤ 12, 7x + 5y ≤ 28, x ≥ 0 and y ≥ 0.

4. a. Minimize the objective function Z under the following constraints:

x + y ≥ 6, x – y ≥ 4, x ≤ 6 and y ≥ 0: Z = 2x + y

b. Minimize the objective function F under the following constraints:
x + y ≤ 8, 4x + y ≥ 8 and y ≥ 3: F = 2x + 3y

5. a. Find the maximum and minimum values of objective function P under the constraints:

2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2: P(x, y) = 4x – y

b. Find the maximum and minimum values of the objective function P under the constraints:
x + y ≤ 6, x – 2y ≤ 2, x ≥ 0 and y ≥ 0 : P = 2x + 5y

6. Find the maximum values of the following objective functions with the given constraints.

a. F = 3x + 5y b. F = 5x + 6y a. P = 10x + 12y
x + y ≤ 6, x + 2y ≤ 6, x + 2y ≤ 12,
x – y ≥ 2, 3x + 2y ≤ 12, 3x + 2y ≤ 24,
x ≥ 0 and y ≥ 0 x ≥ 0 and y ≥ 0 x ≥ 0 and y ≥ 0

Linear Programming / 69

7. Find the minimum value of the following objective functions with the given constraints.

a. Q = 5x + 4y b. F = 3x + 5y a. Q = 10x + 12y

x + 2y ≤ 4, 2x + y ≤ 6, 4x + 3y ≤ 24,

3x + y ≤ 3, x + y ≥ 3, x + 2y ≥ 4,

x ≥ 0, y ≥ 0 2y – 3x ≤ 6 x ≥ 0, y ≥ 0

8. a. In the given diagram, the co-ordinates of A, B and C are B 6Y

(3, 2), (– 3, 6) and (– 3, – 4) respectively. The shaded region 4
inside the ∆ABC is represented by three inequalities. Write
2A

down the equations of these three inequalities and also X' O 2 4 6 X
calculate the minimum value of 4x – 5y from the values

which satisfy all three in equalities. C
b. In the given diagram, the co-ordinates of B and C are
Y'
Y
C
8

(– 2, – 1) and (– 2, 8) respectively. The shaded region inside 6

the triangle ABC is represented by three inequalities, one 4
of these is x + y ≤ 6. Write down the co-ordinates of A and
2A

other two inequalities. Also calculate the maximum value of X' B O 2 4 6 X
x + 2y from the values which satisfy all three inequalities. Y'

c. In the adjoining graph the co-ordinates of A, B and C are Y

(1, 0), (3, 1) and (0, 4) respectively. If one of the inequalities C
represented by shaded region is x + y ≤ 4, find the remaining
B
inequalities. Calculate the maximum value of P = 2x + 5y.
X' O A
X

9. a. Two foods A and B are available at the prices of Rs.10 and Rs.12 per unit.YC'ertain amounts

of nutrients N1, N2 and N3 are essential for an individual. The table gives these minimum
requirements and the nutrients available from one unit of each food A and B.

Nurient Minimum requirement One unit contains
AB
N1 50 5 10
N2 32 84
N3 24 44
Price per unit –
10 12

Find an optimum combination of foods A and B to minimize the total cost.

b. The parking area in a garage is 600 sq.m. The average area required for a car is 6 sq. m and for
a bus 30 sq.m. How many of each should the garage arrange to park for maximum income if a
car pays Rs.25 and a bus pays Rs.75 and not more than 60 vehicles can be taken owing to the
time required to handle them.



70 / Optional Mathematics Class 10

1.11 Equations and Graphs

Graphs of Linear Functions Algebra

A linear function is the function in which the power of variable is 1. The equations x + y = 3,
x + 2 = 0, y = x etc. are the examples of linear functions. The general form of the linear function is
y = ax + b where a and b are constants.

The graph of linear function y = ax + b gives a straight line in which a indicates the slope a and b

indicates the y-intercept of the straight line.
Example 1: Draw the graphs of linear functions y = 2x – 3 and y = – x – 1.

Solution: Y (4,5)
For y = 2x – 3, make a table for the value of x and y.
(-5,4)
x –1 1 2 4

y –5 –1 1 5 y = –x y = 2x – 3
Plot these points on the graph and join them by a straight
line as shown in the graph alongside. X' (2,1) X
For y = –  x – 1, make a table for the value of x and y.
(-1,0) O
x 0 3 –1 –5
(0,-1) (1,-1)

y –1 –4 0 4 (3,-4)
Plot these points on the graph and join them by a straight
(-1,-5)

line as shown in the graph alongside. Y'

The straight line y = 2x – 3 passes through the points

(– 1, – 5), (1, – 1), (2, 1) and (4, 5). Similarly, the straight line y = – x – 1 passes through the points (0, – 1),

(3, – 4), (– 1, 0) and (– 5, 4). Plotting these points on the graph and drawing the corresponding lines.

Graphs of Quadratic Functions

A quadratic function is a function in which the greatest power of the variable is 2. The general
form of quadratic function is y = ax2 + bx + c where a, b and c are constants and a ≠ 0. The equations
y = x2, y = x2 – 3, y = 2x2 – 3, y = x2 + 4x + 4 etc. are the examples of quadratic equations. The quadratic

equation always gives two values of the variable when it is solved. The values of the variable are also called

the roots of the quadratic equation. If the quadratic equation is represented in the graph, it forms an opened

mouth curve, called a parabola. The graph of parabola has either upward or downward opened mouth. The

turning point of the parabola is called its vertex. Y

Let us draw the graph of y = ax2 at different values of a as given below.

a. When a = 1, then y = x2 y = 2x2

x 0 1 –1 2 –2 3 –3 y = x2
y0114499

b. When a = 2, then y = 2x2 1
2
x 0 1 –1 2 –2 y = x2
y02288

c. When a = 21, then y = 1 x2
2

x 0 2 –2 4 –4 X' O X
Y'
y02288

Equations and Graphs / 71

d. When a = – 1, then y = – x2 X' Y X
x 0 1 –1 2 –2 3 –3 O X
y = – 1 x2
y 0 –1 –1 –4 –4 –9 –9 2 Y'

e. When a = – 2, then y = – 2x2 y = – x2 Y

x 0 1 –1 2 –2 y = – 2x2 O
y 0 –2 –2 –8 –8 Y'

f. When a = – 12, then y = – 21 x2
x 0 2 –2 4 –4

y 0 –2 –2 –8 –8

Conclusion: a. If a > 0, the parabola is upward.

b. If a < 0, the parabola is downward.

c. If |a| is greater, the parabola becomes narrower.

d. If |a| is smaller, the parabola becomes wider.

Now, let us draw the graph of y = x2 + c at different values of c.

a. When c = 1, then y = x2 + 1 y = x2 + 3

x 0 1 –1 2 –2 3 –3 y = x2 + 1
y 1 2 2 5 5 10 10 y = x2 – 1

b. When c = 3, then y = x2 + 3

x 0 1 –1 2 –2 y = x2 – 2

y34477

c. When c = – 1, then y = x2 – 1 X'

x 0 1 –1 2 –2 3 –3
y –1 0 0 3 3 8 8

b. When c = – 2, then y = x2 – 2

x 0 1 –1 2 –2 3 –3
y –2 –1 –1 2 2 7 7

Conclusion: a. If c > 0, the vertex lies c units above the origin.

b. If c < 0, the vertex lies c units below the origin.

c. The constant c indicates Y-intercept.

72 / Optional Mathematics Class 10

Turning Point of Graph Algebra

To draw the graph of quadratic equation of the form y = ax2 + bx + c, a ≠ 0, we have to find the
turning point of the curve or the vertex. Plotting the points on the either side of the vertex, we can have the
idea about the shape of its graph. The given equation is y = ax2 + bx + c.

or, y = a x2 + b x + c
a a

x2 + 2.x.2ba + b 2 b 2 c
2a 2a a
or, y=a – +

b 2 b2 – 4ac The vertex or the turning point of parabola
2a 4a2
or, y=a x + –

2 y = ax2 + bx + c (a ≠ 0) is – 2ba, 4ac – b2 and the
4a
or, y= a x + b + 4ac – b2
2a 4a equation of line of symmetry is x = – 2ba.

∴ y= a x– – b 2 4ac – b2
2a 4a
+

Example 1: Draw the graph of the following parabolas and also find the vertex and the line of symmetry.

a. y = x2 + x – 2 b. y = – 3x2 + 6x + 10
Solution:

a. The given equation is y = x2 + x – 2.

Comparing it with y = ax2 + bx + c, we have a = 1, b = 1 and c = – 2.

The turning point or vertex of the given parabola is – 2ba, 4ac – b2
4a
Y
i.e. – 2×11, 4×1×(–2) – 12 = – 21, – 8– 1 = – 21, –9
4×1 4 4

Now making the table for the values of x and y, we get y = x2 + x – 2

x – 1 0 1 2 –1 –2 3
2

y – 9 –2 0 4 – 2 0 10
4

The line of symmetry is x = – 2ba i.e. x = – 21.

b. The given equation is y = – 3x2 + 6x + 10.

Comparing the given equation with parabola y = ax2 + bx + c,
we get a = – 3, b = 6 and c = 10

The turning point of the parabola is – 2ba, 4ac – b2
4a

i.e. – 2×(6–3), 4×(–3)×10 – 62 = – –66, – 120 – 36 O X
4×(–3) – 12
X'

= 1, – 156 = 1, 13 y = – 3x2 + 6x + 10
– 12

Now making the table for the values of x and y, we get Y'

x 1 0 –1 2 3 The line of symmetry is x = – b i.e. x = 1.
2a
y 13 10 1 10 1

Equations and Graphs / 73

Example 2: Find the equation of the parabola given in the graph. Y
Solution:
(-1,6)
Let the equation of the given parabola is y = ax2 + bx + c.

From figure, the parabola passes through the points (3,3)
(0, 1), (– 1, 6) and (2, 3).

For (0, 1) : 1 = a × 02 + b × 0 + c ⇒c=1

For (– 1, 6) : 6 = a × (– 1)2 + b × (– 1) + c ⇒ 6 = a – b + 1 (0,1)

or, a – b = 5 ... (1) X' O X
⇒ 3 = 4a + 2b + 1
For (2, 3) : 3 = a × 22 + b × 2 + c

or, 4a + 2b = 2 ⇒ 2a + b = 1 ... (2) Y'

Solving (1) and (2), we get
a–b=5

2a + b = 1

3a = 6
\ a = 2 and from equation (1), b = a – 5 = 2 – 5 = – 3

Substituting the values of a, b and c, the equation of parabola is y = 2x2 – 3x + 1.

Graphs of Cubic Functions

A cubic function is a function in which the greatest power of the variable is 3. The general form

of a cubic function is y = ax3 + bx2 + cx + d where a, b, c and d are constants and a ≠ 0. The equations

y = x3, y = x3 – 3x, y = (x – 2)3 etc. are the examples of the cubic functions. Y
Example 3: Draw the graphs of y = x3 and y = – x3.

Solution:

For table for the values of x and y in the equation y = x3 y = – x3 y = x3
x 0 1 2 –1 –2

y 0 1 8 –1 –8

Plot the above points on a graph and draw the
freehand curve for its graph (curve in red colour).

X' O X

For table for the values of x and y in the equation y = – x3

x 0 1 2 –1 –2
y 0 –1 –8 1 8

Plot the above points on graph and draw the
freehand curve for its graph (curve in black colour).

Y'

74 / Optional Mathematics Class 10

Exercise 1.11A Y c. Y Algebra

1. Find the vertex of the following parabola.
a. Y b.

X' O X X' O X X' O X

Y' Y' Y'
2. Find the vertex of the following parabola.

a. y = x2 b. y = – x2 c. y = 2x2
f. y = 2x2 – 1
d. y = x2 – 2 e. y = x2 + 4

3. Draw the graphs of the following linear equations.

a. y = – 2x + 3 b. y = 4x – 5 c. 2y = 2x + 5

4. Find the vertex of the following parabola.

a. y = x2 + 2x + 1 b. y = x2 – 2x – 5 c. y = x2 – x – 6

d. y = 2x2 + 4x + 7 e. y = x2 – 2x + 3 f. y = 3x2 – 6x + 5

5. Draw the graph of the following parabola.

a. y = 2x2 b. y = 3x2 c. y = – 1 x2
3
d. y = x2 + 5 e. y = x2 – 4 f. y = 2x2 – 3

6. Draw the graph of the following parabola and find its vertex and the line of symmetry.

a. y = x2 + x + 3 b. y = x2 – 2x – 8 c. y = x2 – 2x – 10

d. y = x2 – 3x + 5 e. y = 2x2 – 3x + 7 f. y = 3x2 – 6x + 10

7. Find the equation of the following parabola.

a. Y b. Y c. Y

(-2,8) (0,9) (2,2)

(5,4) X' (-1,-1) O X

(1,2)

X' O (0,0) X X' O (3,0) X Y' (4,–6)

Y' Y'

8. Draw the graphs of the following cubic equations.

a. y = x3 + 3 b. y = – x3 – 1 c. y = (x – 2)3


Equations and Graphs / 75

Solving a Quadratic and a Linear Equation

We can solve a quadratic and a linear equation by substitution method and graphical method also.
Study the following example.
Example 1: Solve the equation y = x2 – x – 3 and y = x.
Solution:

The given two equations are

y = x2 – x – 3 ... ... (i) and

y = x ... ... (ii)

Substituting y = x in equation (i), we get

x = x2 – x – 3

or, x2 – 2x – 3 = 0

or, x2 – 3x + x – 3 = 0

or, (x – 3) (x + 1) = 0

Either, x – 3 = 0 or, x + 1 = 0

\ x = 3 \ x = – 1

Hence, (3, 3) and (– 1, – 1) are the required solutions.

Graphical Method,

For y = x2 – x – 3

Comparing it with y = ax2 + bx + c, a = 1, b = – 1 and c = – 3.

The x co-ordinate of the turning point is – 2ba, 4ac – b2
4a
y = x2 – x – 3 Y
i.e. – 2–×11, 4×1×(–3) – (–1)2 = 21, – 12 – 1 y=x
4×1 4 (-2,3) (3,3)

i.e. 12, – 13
4

Make a table for values of x and y X' (-1,-1) O (2,-1) X
x 1/2 0 1 – 1 3 3 – 2
(0,-3) (1,-3)
y –13/4 – 3 – 3 – 1 – 1 3 3 (0.5,-3.25)

Plot these points on graph and draw a freehand curve.

and for y = x. Y'

x 0 2 4 –2 –5
y 0 2 4 –2 –5

Drawing this line on the graph, the curve and the line intersect at the point (– 1, – 1) and (3, 3) which
are the solutions of given two equations.

76 / Optional Mathematics Class 10

Solving Quadratic Equation Using Graph Algebra

Quadratic equations can be solved by using graph. Study the following examples.

Example 2: Solve the equation x2 + 2x – 3 = 0 using graph. Y
Solution:
(3,9)

The given equation is x2 + 2x – 3 = 0 or x2 = – 2x + 3. (-2,7)

Let x2 = – 2x + 3 = y then,

y = x2 ... ... ... (i) (-2,4) (2,4)
(0,3)

and y = – 2x + 3 ... ... ... (ii) (-1,1)

Make a table for the values of x and y for y = x2, X' O (2,-1) X

x 0 1 2 3 –1 –2 –3 (3,-3)
y0149149
Y'
Plot these points and draw a freehand curve which gives
the graph of y = x2.

Make a table for the values of x and y for the equation y = – 2x + 3,

x 0 3 –2 2
y 3 –3 7 –1

Drawing the graphs of y = x2 and y = – 2x + 3, we have the graph alongside. The point of intersections
of the parabola y = x2 and the line y = – 2x + 3 are (1, 1) and (– 3, 9). As the equation given is a quadratic in
x only, the required solutions are x = 1, – 3.
Alternately,

The given equation is x2 + 2x – 3 = 0. (-4,5) Y (2,5)
Let y = x2 + 2x – 3 (= 0) and comparing it with y = ax2 + bx +

c, its turning point of the parabola is

– 2ba, 4ac – b2 = – 2×21, 4×1×(– 3) – 22
4a 4×1
X' (-3,0) O (1,0) X
= – 22, – 12 – 4 = (– 1, – 4)
4

Making a table for the value of x and y for the given equation. (-1,-3) (0,-3)

x –1 0 1 2 –2 –3 –4 (-1,-4)
y –4 –3 0 5 –3 0 5
Y'

As, y = x2 + 2x – 3 intersect X-axis (y = 0) at (1, 0) and (– 3, 0), hence the required solution of the given

equation is x = 1, – 3.

Equations and Graphs / 77

Example 3: Solve graphically x2 + y2 = 9 and x + y = 3.
Solution:

The given equations are x2 + y2 = 9 ... ... (1) and x + y = 3 ... ... (2) Y

For x2 + y2 = 9, make a table for the values of x and y.

x 0 3 –3 0 (0,3)
y 3 0 0 –3

Plot these points and draw a freehand curve for the graph X' (-3,0) (2,1) (3,0) X
of the given equation. The equation (1) is the circle with centre (4,-1)
at origin and radius 3 units. O

For, x + y = 3, make a table for the values of x and y. (0,-3)

x0324 Y'
y 3 0 1 –1

Plot these points and draw a line through these points. From graph, it is clear that the line x + y = 3
intersects the circle at two different points (3, 0) and (0, 3).

Hence, the required solutions are (3, 0) and (0, 3).

Exercise 1.11B

1. Solve the following system of equations by graphical method and the method of substitution.

a. y = x2 and y = 2 – x b. y = x2 – 5 and y = 2x + 3

b. y = x2 – 4x + 9 and y = 2x + 1 d. y = x2 + x – 9 and y = 2x + 3

2. Solve the following equations by graphical method.

a. x2 – x – 2 = 0 b. x2 + 2x – 3 = 0

c. x2 – 2x – 15 = 0 d. 2x2 + x – 6 = 0

e. 2x2 + x – 3 = 0 f. 2x2 – 11x + 5 = 0

3. Solve graphically the following circles and straight lines.

a. x2 + y2 = 16 and x + y – 4 = 0 b. x2 + y2 = 25 and x + y – 1 = 0

4. a. Find the points of intersection of the parabola y = x2 – 3 and the line passing through the points
(0, 0) and (– 2, 4).

b. Find the points of intersection of the parabola y = – x2 + 6x – 4 and the line passing through the
points (4, 2) and (1, 5).



78 / Optional Mathematics Class 10

2. LIMITS AND CONTINUITY

Curriculum Contents Included Periods

Simple Concept of Continuity 10 Continuity

– Testing continuity in the set of numbers (Natural numbers, Whole number,
Rational number)

– Testing continuity in the set of real numbers

– Testing continuity and discontinuity in the set of various numbers (number
line and construcion)

– Testing continuity of function by graph and symbolic representation

Objectives

On completion of this unit, students will be able to:
state the meaning of continuity and discontinuity.
taste the continuity of the numbers in set of different numbers.
taste the continuity of the given function using graph.

Teaching Materials

The meaning of continuity is being continuous. A function is said to be continuous if it has a
‘smooth’ or unbroken graph at every point at the interval. The graph of a discontinuous function is
‘broken’ at one or more points. In fig.(a) given below, the function is continuous at every point at the
interval (a, b) as its graph does not have any break during this interval and in fig.(b) the funcion is
discontinuous at point x = c at the interval (a, b).

Y Y
5
5
4 4
3
2 3
1
ab 2

X' -5 -4 -3 -2 --11 1 2 3 4 5 X 1 cb X
a
-2 X' -5 -4 -3 -2 --11 1 2 3 4 5
-3
-4 -2
-5
-3
Y' Fig. (a)
-4

-5

Y' Fig. (b)

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Continuity Each of 1 mark Each of 5 marks Quesitons Marks
Each of 2 marks Each of 4 marks
1 – 2 5
–1

/ 79

Limits and Continuity

2 LIMITS AND CONTINUITY

Limits Review

We have discussed the limit of a function at the given point in the previous class. The limit of the
function f(x) at x = a is said to exist if the left hand limit and the right hand limit of the function at x = a are

equal. i.e. lim f(x) = lim f(x) [= L]. It is written as xl→imaf(x) = L.
x→a– x→a+

Study the following examples for the revision of limit.

Example 1: Use the left hand limit and the right hand limit of f(x) = x + 1, x ≤ 1 to estimate the
limit of the function at x = 1. Also draw its graph. 2x, x > 1

Solution:

Here, the given function is f(x) = x + 1, x≤1
2x, x>1

To estimate lim f(x), we calculate the left hand limit and the right Y
x→1
5
hand limit of f(x) at x = 1. 4
3
When x tends to 1 from the left of 1, it is denoted by x→1–. We have 2
1

the left hand limit of f(x) as lim f(x) = lim (x + 1) = 1 + 1 = 2 X' X-5 -4 -3 -2 -1- 1 1 2 3 4 5
x→1– x→1–
-2

When x tends to 1 from the right side of 1, it is denoted by x→1+ . -3
-4
We have the right hand limit of f(x) as lim f(x) = lim 2x = 2 × 1 = 2 -5
x→1+ x→1+
Y'
lim lim
Now, x→1– f(x) = x→1+ f(x)

Hence, the limit of f(x) at x = 1 exists and written as lim f(x) = 2 Y
x→1
6
The graph of the function is shown in the figure alongside. 5

Example 2: Find the limit of f(x) = 3x, x≤2 at x = 2 4
6 – x, x>2 3
Solution: 2
1

3x, x≤2 X' -3 -2 -1- 1 1 2 3 4 5 6 7 X
6 – x, x>2
Here, the given function is f(x) = -2
-3
lim -4
x→2
We have to estimate the limit of f(x) at x = 2, i.e. f(x) Y'

Now, when x tends to 2 from the left side of 2, it is denoted by x→2–. We have the left hand limit of

f(x) as lim f(x) = lim 3x = 3 × 2 = 6
x→2– x→2–

When x tends to 2 from the right side of 2, it is denoted by x→2+. We have the right hand limit of

f(x) as lim f(x) = lim (6 – x) = 6 – 2 = 4
x→2+ x→2+

80 / Optional Mathematics Class 10

Now, lim f(x) ≠ lim f(x)
x→2– x→2+

Therefore, the limit of f(x) at x = 2 does not exist.

The graph of the function is shown in the figure alongside.

Techniques of Evaluating Limits

In the above examples, we calculate the limit of a function at the given point by finding the numerical Continuity

values from the both sides of the point.

Let us consider f(x) = x2 – 4 be a given function. At x = 2, the value of the function is f(2) = 22 – 4
x–2 2– 2
4 – 4 00,
= 2 – 2 = which is indeterminate form. Hence, the function is not defined at x = 2. When we calulate

the limit of the function at x = 2, i.e. lim f(x), it is the value of the function f(x) when x→2. The function
x→2

is same on either side of 2. Hence, instead of calculating the left hand limit and the right hand limit of the

function at x = 2, we can calculate the limiting value of f(x) at x = 2 in the following way.

Here, lim f(x) = lim x2 – 4 = lim (x + 2)(x – 2)
x→2 x→2 x – 2 x→2 (x – 2)

As, x→2 means x – 2 = 0, so we can divide the numerator and the dinominator by (x – 2).

= lim (x + 2) = 2 + 2 = 4
x→2

Hence, lim f(x) = 4.
x→2

In this example, we calculate the limit of the given function directly instead of calculating the left
hand limit and the right hand limit of the function at the given point separately.

Example 3: Evaluate : a. lim 2x3 + 7x b. lim √x – √3
x→3 x2 + 9 x→3 x2 – 9

Solution:

a. Here, lim 2x3 + 7x = 2(3)3 + 7 × 3 2 × 27 + 21
x→3 x2 + 9 (3)2 + 9 = 9+9

= 54 + 21 = 75 = 25 = 4 1
18 18 6 6

b. Here, lim √x – √93 =xl→im3 √x – √3 × √x + √3
x→3 x2 – x2 – 32 √x + √3

= lim (x + (√x)2 – (√3)2 √3)
x→3 3)(x – 3) (√x +

= lim (x – 3)
x→3 (x + 3)(x – 3)(√x + √3)

= lim (x + 1 + √3)
x→3 3)(√x

= (3 + 1 + √3) = 6 1 = 1
3)(√3 × 2√3 12√3

Limits and Continuity / 81

Continuity

The meaning of continuity is being continuous. The water from the mountain continuously flows
to reach the sea. A car is continuously driven to reach from one place to another place. When water is
transfered from one tank to another using a pipe, the water flown inside the pipe is continuous. Here, we
study the continuity in the set of different numbers.

a. A = {x : 3 ≤ x < 10, x∈N} = {3, 4, 5, 6, 7, 8, 9}

2 3 4 5 6 7 8 9 10

A is a set of natural numbers from 3 to 9. If we plot the numbers on a number line, there is some gap
between 3 and 4, 4 and 5 and so on. Hence, the graph of numbers in set A is not continuous.

b. B = {x : – 3 < x ≤ 4, x∈Z} = {– 2, – 1, 0, 1, 2, 3, 4}

-3 -2 -1 0 1 2 3 4 5

B is a set of integers greater than 3 and less than or equal to 4. If we plot the numbers on a number
line, there is some gap between – 2 and – 1, – 1 and 0 and so on. Hence, the graph of numbers in set B is not
continuous.

c. C = {x : – 4 ≤ x < 3 x∈R}

-5 -4 -3 -2 -1 0 1 2 3 4

C is a set of real numbers greater than or equal to – 4 and less than 3. If we plot the numbers on a
number line, there is no any gap between – 4 and 3. Hence, the graph of numbers in set C is continuous.

Example 4: Draw the graph of the following equations.

a. y = x, where x, y∈N b. x + y = 4 where x, y∈Z

c. 2x + y = 4 where x, y∈R

Solution: Y

a. Here, given equation is y = x, x, y∈N. As x and y both 7
are natural numbers, we can draw graph of the equation 6
taking some sample points. 5
4
x123456 3
y123456 2
1

X' -3 -2 -1- 1 1 2 3 4 5 6 7 X

-2

-3

Y'

82 / Optional Mathematics Class 10

b. Here, given equation is x + y = 4, x, y∈Z. As x and Y
y both are integers, we can draw graph of the equation
taking some sample points. 8
7
x –2 –1 0 1 2 3 6
y654321 5
4
3
2
1

X' Y'-5 -4 -3 -2 -1- 10 1 2 3 4 5 X Continuity
Y
c. Here, given equation is 2x + y = 4, x, y ∈ R. As x
and y both are real numbers, we can draw its graph as 5 X
shown with some sample points. As there are infinite real 4
numbers between – 2 and – 1, the graph is continuous. It 3
is different from the above examples in (a) and (b). 2
1
x – 2 – 1 0 1 2 3
y654321 X' -5 -4 -3 -2 -1 0 1 2 3 4 5
-1
Continuity and Discontinuity of a Function -2
-3
-4
Y'

When a graph of any function is ‘broken’at one or more points, the function is said to be discontinuous,

otherwise it is continuous. In fig.(a) given below, the function is continuous at every point in the interval

(a, b) as it has no break in the interval (a, b). In fig.(b) the funcion is discontinuous at point x = c in the

interval (a, b). The limit of the function from both the sides exist and equal, but the function is not defined

at x = c. In fig.(c), the function is discontinuous at x = s in the interval [p, q]. At x = s, the limit from both

the sides exist and are not equal to each other.

YYY
555

444

333

222

1 b 1 cb p1 s q
a a

X' X X' X X' X-5 -4 -3 -2 --11 1 2 3 4 5
-5 -4 -3 -2 --11 1 2 3 4 5 -5 -4 -3 -2 --11 1 2 3 4 5

-2 -2 -2

-3 -3 -3

-4 -4 -4

-5 -5 -5

Fig.(a) Y' Fig.(b) Y' Fig.(c) Y'

Let’s study the graph of some more functions.

1. y = f(x) = x Y 2. y = f(x) = x2 Y

5 5
4 4
3 3
2 2
1 1

X' -5 -4 -3 -2 --11 1 2 3 4 5 X X' -5 -4 -3 -2 --11 1 2 3 4 5 X

-2 -2
-3 -3
-4 -4
-5 -5

Y' Y'
Continuous
Continuous

Limits and Continuity / 83

3. y = f(x) = 1/x 1, x≤1
2, x>2
Y 4. y = f(x) =

5 Y
4
3 5
2 4
1 3
2
X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X 1

-2 X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X
-3
-4 -2
-5 -3
-4
Y' -5

Y'

Discontinuous at x = 0 Discontinuous at x = 0

x2, x < 1 6. y = f(x) = x + 1, x<2
5. y = f(x) = 2 – x, x ≥ 1 x, x≥2

Y Y

5 5
4
3 4
2 3
1
2

1

X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X

-2 -2
-3 -3
-4 -4
-5 -5

Y' Y'

Continuous Discontinuous at x = 2

7. y = f(x) = 2x, x≠1 x2 – 1
1, x=1 8. y = f(x) = x – 1

Y Y

5 5
4 4
3 3
2 2
1 1

X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X X' -5 -4 -3 -2 -1- 1 1 2 3 4 5 X

-2 -2
-3 -3
-4 -4
-5 -5

Y' Y'

Discontinuous at x = 1 Discontinuous at x = 1

84 / Optional Mathematics Class 10

Here, we will focus on the continuity of the given function Y
at a fixed point. The function will be continuous at a fixed point,
if its left hand limit, right hand limit and the value of the function 5
at the given point are equal.
From the given figure of the graph of the function 4

3 f(c)
xlimc+f(x)
2 lim
x c-
lim 1 f(x)
x→c–
y = f(x). It is cearly seen that the left hand limit f(x), X' -2 -1 0 1 2 3 4 5 6 7 8 X

the right hand limit lim f(x) and the value of the function -1 c– c c+ Continuity
x→c+ -2

at c [f(c)] are equal and the function is continuous at x = c. -3

-4

lim lim -5
x→c– x→c+
i.e. f(x) = f(c) = f(x) Y'

Therefore, y = f(x) is continuous at x = c.

Definition of Continuity

Let c be a number in the interval (a, b) and f(x) be a function whose Y
5

domain contains the interval (a, b). The function f(x) is continuous at the 4
point c if the following conditions are true. 3
2

a. f(c) is defined. 1a c b X

b. lim f(x) exits i.e. lim f(x) = lim f(x) X' -2 -1 0 1 2 3 4 5 6 7 8
x→c x→c– x→c+ -1

-2

-3

-4

c. lim f(x) = f(c) -5
x→c
Y'

In other words, we can say that a function is continuous on an interval if its graph on the interval can
be traced using a pencil and paper without lifting the pencil from the paper.

Example 5: Discuss the continuity of each of the following function by calculating the left hand limit,

right and limit and the value of the given function at specified point.

a. f(x) = 2, x ≤ 2 at x = 2 b. f(x) = 2x, x≤1 at x = 1
1, x>2 3 – x, x>1

x2, x≠0 2x – 3, x < 2
1, x=0
c. f(x) = at x = 0 d. f(x) = 1, x = 2 at x = 2
3x – 5, x > 2

e. f(x) = x2 – 9 at x = 3
Solution: x–3

2, x≤2
a. The given function is f(x) = 1, x > 2 at x = 2

To discuss the continuity of the function at x = 2, the left hand limit of f(x) at x = 2 is lim f(x) = 2.
x→2–

Limits and Continuity / 85

The right hand limit of f(x) at x = 2 is lim f(x) = 1. The value of the function at x = 2 is f(2) = 2.
x→2+

Here, lim f(x) = f(2) = 2, lim f(x) = 1 Y
x→2– x→2+ 5
4
Therefore, the function f(x) is discontinuous at x = 2.
y = f(x) = 23
y = f(x) = 1
2
1

The graph of the function is as shown in the figure. X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X
-1
2x, x ≤ 1 -2
b. The given function is f(x) = 3 – x, x > 1 at x = 1 -3
-4
-5
Y'

To discuss the continuity of the function at x = 1, the left hand Y
5
limit of f(x) at x = 1 is lim f(x) = lim 2x = 2 × 1 = 2. The right 4
x→1– x→1– 3
2
lim lim 1 y = f(x) = 3–x
x→1+ x→1+
hand limit of f(x) at x = 1 is f(x) = (3 – x) = (3 – 1) = 2. X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X
-1

The value of the function at x = 1 is f(1) = 2 × 1 = 2. y = f(x) = 2x -2
-3

-4

lim lim -5
x→1– x→1+
Now, f(x) = f(1) = f(x) = 2. Y'

Therefore, the function f(x) is continuous at x = 1. The graph of the function is as shown in the figure.

x2, x≠0 Y y = x2
c. The given function is f(x) = 1, x = 0 at x = 0 5
4

To examine the continuity of the function at x = 0, the left hand limit of 3
2

1

f(x) at x = 0 is lim f(x) = lim x2 = 02 = 0. X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X
x→0– x→0– -1
-2
The right hand limit of f(x) at x = 0 is lim f(x) = lim x2 = 02 = 0. -3
x→0+ x→0+ -4
-5

The value of the function at x = 0 is f(0) = 1. Y'

Now, lim f(x) = lim f(x) ≠ f(0).
x→0– x→0+

Therefore, the function f(x) is discontinuous at x = 0. The graph of the function is shown in the figure.

x – 1, x < 2 Y
d. Here, the given function is f(x) = 1, x = 2 . 7
6
3x – 5, x > 2 5
4
To discuss the continuity of f(x) at x = 2, we see the following 3
2
results. The left hand limit of f(x) at x = 2 is 1

lim f(x) = lim (x – 1) = 2 – 1 = 1. X' -4 -3 -2 -1 0 1 2 3 4 5 6 X
x→2– x→2– -1

The right hand limit of f(x) at x = 2 is -2

-3

Y'

86 / Optional Mathematics Class 10

lim f(x) = lim (3x – 5) = 3 × 2 – 5 = 6 – 5 = 1.
x→2+ x→2+

The value of the function at x = 2 is f(2) = 1.

Here, lim f(x) = f(2) = lim f(x) = 1.
x→2– x→2+

Therefore, the function f(x) is continuous at x = 2. The graph of the function is shown in the figure.

e. Here, the given function is f(x) = x2 – 9 . Continuity
x–3

To discuss the continuity of f(x) at x = 3, we see the following results. The left hand limit of f(x) at

x = 3 is lim f(x) = lim x2 –9 = lim (x + 3)(x – 3) Y
x→3– x→3– x –3 x→3– (x – 3) 7
6

= lim (x + 3) = 3 + 3 = 6. 5
x→3– 4
3
The right hand limit of f(x) at x = 3 is 2

lim f(x) = lim x2 – 9 = lim (x + 3)(x – 3) 1 X
x→3+ x→3+ x – 3 x→3+ (x – 3)
X' -4 -3 -2 -1 0 1 2 3 4 5 6
-1
-2

= lim (x + 3) = 3 + 3 = 6. -3
x→3+
Y'

The value of the function at x = 3 is f(3) = (3)2 – 9 = 9 – 9 = 0 (undefined)
3–3 3 – 3 0

Here, lim f(x) = lim f(x) and f(3) is not defined.
x→3– x→3+

Hence, the function is discontinuous at x = 3. The graph of the function is shown in figure above.

Example 6: Discuss the continuity of the function f(x) = 2x2 – 2 , x ≠ 1 at x = 1.
x–1

Solution: 4, x = 1

The given function is f(x) = 2x2 – 2 , x≠1
x–1 . To test the continuity of f(x) at x = 1, we calcu-

4, x = 1

late the lim f(x) and value of the function at x = 1.
x→1 lim 2x2 – 2
lim x→1 x – 1
Now, x→1 f(x) =

= lim 2(x2 – 1)
x→1 x–1

= lim 2(x + 1)(x – 1)
x→1 (x – 1)

= lim 2(x + 1)
x→1

= 2(1 + 1) = 2 × 2 = 4

The value of the function at x = 1 is f(1) = 4.

Hence, lim f(x) = f(1) = 4, so, the function is continuous at x = 1.
x→1

Limits and Continuity / 87

Example 7: A function f(x) is defined as f(x) = 3x2 – 12 , x ≠ 2 . Find the vlaue of k so that f(x)
x – 2
is continuous at x = 2.
Solution: k + 10, x=2

Here, the limiting value of f(x) at x = 2 is As the function f(x) is continuous at x = 2.

lim f(x) = lim 3x2 – 12 lim f(x) = f(2)
x→2 x→2 x–2 x→2

= lim 3(x2 – 4) i.e. 12 = k + 10
x→2 x–2
or, k = 12 – 10
lim 3(x + 2)(x – 2)
= x→2 (x – 2) \ k = 2

= lim 3(x + 2)
x→2

= 3(2 + 2) = 3 × 4 = 12

Removable Discontinuity

There are various types of discontinuity such as : Removable discontinuity, Infinite discontinuity,
Jump discontinuity etc. A function f(x) is said to have a removable discontinuity at x = c in its domain if

lim f(x) ≠ f(c) i.e. lim f(x) = lim f(x) ≠ f(c).
x→c x→c– x→c+
x2 – 24.
For example, let us take a function f(x) = x – To check the continuity of f(x) at x = 2, we have

lim f(x) = lim x2 – 4 = lim (x + 2)(x – 2) = lim (x + 2) = 2 + 2 = 4.
x→2 x→2 x – 2 x→2 (x – 2) x→2
Y
The value of f(x) at x = 2 is f(2) = 22 – 4 = 4 – 4 = 0 = undefined. 6
2–2 2 – 2 0 5
4
So, lim f(x) ≠ f(2) and x = 2 is the removable discontinuity. This 3
x→2 2
1

discontinuity can be removed by redefining the function f(x) at x = 2 such that X' -3 -2 -1 0 1 2 3 4 5 X
-1

lim f(x) = f(2) = 4. -2
x→2
Y'

x2 – 4 , x ≠ 2 is continuous at x = 2.
Hence, f(x) = x – 2
4 , x = 2

Similarly, for the function f(x) = (x – 1)2, x ≠ 3. To check the continuity of f(x) at x = 3, we have
2 , x = 3
Y
6
lim f(x) = lim (x – 1)2 = (3 – 1)2 = 4. The value of f(x) at x = 3 is f(3) = 2. 5
x→3 x→3 4

So, lim f(x) ≠ f(3) and x = 3 is the removable discontinuity of this function. 3
x→3 2

This discontinuity can be removed by redefining the function f(x) at x = 3 such 1

that lim f(x) = f(3) = 4. X' X-3 -2 -1 0 1 2 3 4 5
x→3 -1

Hence, f(x) = (x – 1)2, x ≠ 3 is continuous at x = 3. -2

Y'

4 , x = 3

88 / Optional Mathematics Class 10

Exercise 2

1. Draw the graph of the following equations.

a. 2x = 3y, x, y ∈ N b. 2x – y = 5, x, y ∈ Z

c. 3x + 2y = 6, x, y ∈ R d. 4x – 2y = 6, x, y ∈ R

2. Use left hand limit and right hand limit of f(x) at given point to estimate the limit of the function at the Continuity
point. Aslo draw its graph.

x – 2, x ≤ 3 3x, x < 1
a. f(x) = 4 – x, x > 3 at x = 3 b. f(x) = x + 2, x ≥ 1 at x = 1

x – 5, x ≤ 5 x2, x < 2
c. f(x) = 5 – x, x > 5 at x = 5 d. f(x) = 10 – 3x, x ≥ 2 at x = 2

5 – 2x, x ≤ 2
3. a. Find the limit of f(x) = x – 1, x > 2 at x = 2. Also draw its graph.

2x, x < 3
b. Find the limit of f(x) = 9 – x, x ≥ 3 at x = 3. Also draw the graph of the function f(x).

2x2, x ≤ 1
c. Find the limit of f(x) = 3x – 1, x > 1 at x = 1. Also draw its graph.

4. a. If f(x) = 2x + 3, x>1 , find lim f(x). Draw the graph of the function f(x).
5, x≥1 x→1

b. If f(x) = 8x – 3, x<2 , find lim f(x). Sketch the graph of the function f(x).
13 x>2 x→2
7 – 3x,

c. If f(x) = x2 + 3, x < – 3 , find lim f(x). Also sketch the graph of the function.
3 – 2x, x > – 3 x→ – 3

5. Evaluate the limit of the following.

a. xl→im1 (x + 3)(x – 3) b. xl→im1 x2x–2 6–x5+x
(x – 2)
2

c. xl→im1 x2 + 3x – 4 d. xl→im1 x2 x+–x1– 2
x –1

e. xl→im3 x2 – 96 f. xl→im2 xx22––74xx++140
2x –

g. ul→im5 u+4– 3 h. xl→im1 xx22+–31– 2
u–5

Limits and Continuity / 89

6. Following are the functions and their corresponding graphs. State whether the function is continuous
or discontinuous at the specified point.

1, x≥0 2, x≤2
a. y = f(x) = – 3, x<0 b. y = f(x) = x, x>2

Y Y
5
5 4
4 3
3 2
2 1
1
X' -5 -4 -3 -2 -1 0 1 2 3 4 5
X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X -1 X
-1 -2
-2 -3
-3 -4
-4 -5
-5 Y'
Y' at point x = 2

at point x = 0

c. y = f(x) = x2 – 4 d. y = f(x) = x2
x – 2 2

Y X Y
5 7
4 6
3 5
2 4
1 3
2
X' -5 -4 -3 -2 -1 0 1 2 3 4 5 1
-1
-2 X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X
-3 -1
-4
-5 -2
Y'
at point x = 2 -3

Y'
at point x = 2

e. y = f(x) = (x – 2)2 1
f. y = f(x) = x – 2

Y Y
5
7 4
6 3
5 2
4 1
3
2 X' -3 -2 -1 0 1 2 3 4 5 6 7 X
1 -1
X -2
X' -3 -2 -1 0 1 2 3 4 5 6 7 -3
-1 -4
-2 -5
-3 Y'
Y' at point x = 2
at point x = 1

90 / Optional Mathematics Class 10

x2, x<0 2x2, x < 1
g. y = f(x) = 2x, x≥0 h. y = f(x) = 2x, x ≥ 1

Y Y
7 5
6 4
5 3
4 2
3 1
2
1 X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X Continuity
-1
X' -5 -4 -3 -2 -1 0 1 2 3 4 5 X -2
-1 -3
-4
-2 -5
Y'
-3 at point x = 1

Y'
at point x = 0

i. y = f(x) = x2 + 3x – 4 x2, x≠1
x –1 j. y = f(x) = 2, x=1

Y Y X
7 5
6 X 4
5 3
4 2
3 1
2
1 X' -5 -4 -3 -2 -1 0 1 2 3 4 5
-1
X' -5 -4 -3 -2 -1 0 1 2 3 4 5 -2
-1 -3
-2 -4
-3 -5
Y' Y'
at point x = 1 at point x = 1

7. Identify which of the functions given in question number 6 above have removable discontinuity at

specified point? Also make them continuous at the points.

8. Discuss the continuity of each of the following functions by calculating left hand limit, right hand

hand limit and the value of the function at the specified point. Also draw the graph of the function.

4, x>3 3x, x ≤ 1
a. y = f(x) = 2, x < 3 at x = 3 b. y = f(x) = x + 2, x > 1 at x = 1

x + 5, x < 2 2x + 1, x < 1
c. y = f(x) = 8 – x, x ≥ 2 at x = 2 d. y = f(x) = 4 – x, x ≥ 1 at x = 1

x2 + 1, x ≥ 2 (x + 1)2, x ≤ 2
f. y = f(x) = 2x, x > 2 at x = 2
e. y = f(x) = 52x, x < 2 at x = 2

g. y = f(x) = x2 – 1 at x = 1 h. y = f(x) = x2 – 3x + 2 at x = 1
x – 1 x –1

4x – 1, x < 1 x2 + 2, x < 5

i. y = f(x) = 2x, x = 1 at x = 1 j. y = f(x) = 6x – 3, x = 5 at x = 5

3x x > 1  3x + 12 x > 5

Limits and Continuity / 91

3. MATRICES

Curriculum Contents Included Periods
17
a. Determinant of a 2×2 matrix

b. Inverse of a 2×2 matrix

b. Solution of simultaneous linear equations by matrix method

d. Cramer's rule and its use (2 × 2 matrix only)

Objectives

On completion of this unit, students will be able to:

find the product of two matrices and multiply the given matrix by a scalar.
state the properties of matrix multiplication.
find the determinant and inverse of the given square matrix.
solve the simultaneous equations of two variables using matrix method.

Teaching Materials

For two given equations,

a1x + b1y = c1 ... ... (1) and

a2x + b2y = c2 ... ... (2).

The values of x and y can be expressed in the form of determinant as follows.

c1 b1 a1 c1
a2 c2
i.e. x= c2 b2 = DDx and y = = Dy
a1 b1 a1 b1 D
a2 b2
a2 b2

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Matrices Each of 1 mark Each of 2 marks Each of 4 marks Each of 5 marks Quesitons Marks

1 2 1 –4 9

92 / Optional Mathematics Class 10

3 MATRICES

3.1 Determinant

Matrix and Its Determinant

If A = a b is a square matrix of order 2; then the number (ad – bc) associated with the matrix A
c d

is said to be the determinant of the matrix A. The determinant of matrix A is denoted by |A| or det. A.

For matrix A = a b , det. A = a b Matrices
c d c d = ad – bc.

Similarly, if B = 1 –3 , then |B| 1 –3
5 –2 = 5 –2 = 1×(– 2) – 5×(– 3) = – 2 + 15 = 13.

The members a, b, c and d are the elements of the determinant A. Since there are two rows and two
columns, the order of the determinant is 2×2. ad – bc is also known as the expansion of determinant A.

Again, if A = ( a ) is a square matrix of order 1, then the determinant of A is simply the number 'a'.

Example 1: Find the determinant of B = 3 5 .
Solution: 4 2

Given matrix is B = 3 5 and its determinant
4 2

3 5
det. B = 4 2 = 3 × 2 – 4 × 5 = 6 – 20 = – 14.

Some Properties of Determinant

a. The value of a determinant remains same if its rows and columns are interchanged.
a b

Let ∆ = c d be a given deterninant, then ∆ = ad – bc.
Interchanging the rows and columns of this determinant, we get

a c
∆1 = b d and ∆1 = ad – bc.
Hence, ∆ = ∆1.

Determinant / 93

b. If the rows or columns of a determinant are interchanged, then the sign of the determinant is changed.

a b
Let ∆ = c d , then ∆ = ad – bc.
Interchanging the rows of this deternimant, we get

c d
∆1 = a b , then ∆1 = bc – ad = – (ad – bc).

Hence, ∆1 = – ∆.
c. If two rows or columns of a determinant are identical, then its determinant is zero.

p p
The first column and the second column of q q are identical.

p p
Here, q q = pq – pq = 0.
d. If each element of a row or a column of a determinant is multiplied by a constant k, then its

determinant is k times the given determinant.

a b
Let ∆ = c d , then ∆ = ad – bc.

Multiplying each element of the second row by k, we get

a b
∆1 = kc kd = kad – kbc = k(ad – bc).

Hence, ∆1 = k∆.

Example 2: a + b a–b
Prove that a – b a + b = 4ab
Solution:

a + b a – b
Here, a – b a + b

= (a + b)(a + b) – (a – b)(a – b)
= (a + b)2 – (a – b)2
= a2 + 2ab + b2 – (a2 – 2ab + b2)
= a2 + 2ab + b2 – a2 + 2ab – b2
= 4ab proved.

Singular and Non-Singular Matrix

A square matrix A with its determinant |A| = 0 is known as a singular matrix. A square matrix B
with its determinant |B| ≠ 0 is known as a non-singular matrix.

Let A = –4 –6 – 4 –6
2 3 , then its determinant is |A| = 2 3 = – 12 – (– 12) = 0, hence A is a singu-

94 / Optional Mathematics Class 10

lar matrix. And matrix B = 2 3 2 3
3 –4 , its determinant is |B| = 3 –4 = – 8 – 9 = – 17 ≠ 0, hence B is

a non singular matrix.

Example 3: If A = 2 a is a singular matrix, find the value of a.
4 6

Solution:
As A is a singular matrix, its determinant is zero, i.e. |A| = 0.

Here, 2 a
|A| = 4 6

or, 0 = 2 × 6 – 4 × a = 12 – 4a

or, 0 = 12 – 4a Matrices

or, 12 – 4a = 0

or, 4a = 12

∴ a = 3

Example 4: 2 7 7 3 – k
If –5 2k = 17 3 , find the value of k.
Solution:

Here, 2 7 7 3 – k
–5 2k = 17 3

or, 2 × 2k – (– 5) × 7 = 7 × 3 – 17(3 – k)
or, 4k + 35 = 21 – 51 + 17k
or, 4k – 17k = – 30 – 35

or, – 13k = – 65

or, k = – 65
– 13

\ k = 5.

Example 5: If A = 1 –2 and B = –3 0 , find the determinant of 5A – 2B + 3I.
3 4 1 –2
Solution:

Here given matrices are

A= 1 –2 , B = –3 0 and I = 1 0
3 4 1 –2 0 1

Now, 5A – 2B + 3I

= 5 1 –2 – 2 –3 0 + 3 1 0
3 4 1 –2 0 1

Determinant / 95

= 5 –10 – –6 0 + 3 0
15 20 2 –4 0 3

= 5 + 6 + 3 –10 – 0 + 0
15 – 2 + 0 20 + 4 + 3

= 14 –10
13 27

Determinant of 5A – 2B + 3I = |5A – 2B + 3I|

14 –10
= 13 27 = 14 × 27 + 13 × 10

= 378 + 130 = 508

Inverse of a Matrix

If A be a non-singular matrix i.e. |A| ≠ 0 and there exists another square matrix B such that AB = BA
= I where I is a unit matrix of same order as that of A and B. The matrix B is said to be inverse of A and
vice-versa. The inverse of A is denoted by A–1.

∴ B = A–1.

Example 6: Find the inverse of the matrix A = 5 2 .
–8 –3
Solution:

Let A–1 = w x be the inverse of matrix A = 5 2 then AA–1 = I
y z –8 –3

or, 5 2 w x = 1 0
–8 –3 y z 0 1

or, 5w + 2y 5x + 2z = 1 0
–8w – 3y –8x – 3z 0 1

i.e. 5w + 2y = 1 ... ... ... (i)

– 8w – 3y = 0 ... ... ... (ii)

5x + 2z = 0 ... ... ... (iii)

– 8x – 3z = 1 ... ... ... (iv)

Solving (i) and (ii), w = – 3, y = 8

Solving (iii) and (iv), x = – 2, z = 5

Hence, A–1 = w x = –3 –2
y z 8 5

Method of Finding an Inverse Matrix

Let A = a b be a given square matrix, then its determinant |A| = ad – bc ≠ 0 (suppose) and A–1
c d

exists.

96 / Optional Mathematics Class 10