Chapter 2 of 8

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Updated 6 August 2026

NCERT Solutions

Introduction to Linear Polynomials

CBSE · Class 9 · Mathematics

NCERT Solutions for Introduction to Linear Polynomials — CBSE Class 9 Mathematics.

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EXERCISE SET 2.1

1(i)

2x^2 - 5x + 3

Show solution

For

2x^2 - 5x + 3

, the highest power of the variable

x

2

. So the degree is 2.

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1(ii)

y^3 + 2y - 1

Show solution

For

y^3 + 2y - 1

, the highest power of

y

3

. So the degree is 3.

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1(iii)

-9

Show solution

The constant polynomial

-9

has no variable term, so its highest power is

0

. Hence, the degree is 0.

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1(iv)

4z - 3

Show solution

4z - 3

, the highest power of

z

1

. So the degree is 1.

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2Write polynomials of degrees 1, 2 and 3.Show solution

Examples of polynomials of degrees 1, 2 and 3 are:

- Degree 1:

2x + 3

- Degree 2:

x^2 + 5x + 1

- Degree 3:

5y^3 + y^2 + 2y - 1

Any valid polynomials of these degrees are acceptable.

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3What are the coefficients of

x^2

and

x^3

in the polynomial

x^4 - 3x^3 + 6x^2 - 2x + 7

?Show solution

x^4 - 3x^3 + 6x^2 - 2x + 7

:

- The coefficient of

x^2

is 6.
- The coefficient of

x^3

is -3.

So the required coefficients are

6

and

-3

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4What is the coefficient of

z

in the polynomial

4z^3 + 5z^2 - 11

?Show solution

4z^3 + 5z^2 - 11

, there is no term containing

z

. So the coefficient of

z

is 0.

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5What is the constant term of the polynomial

9x^3 + 5x^2 - 8x - 10

?Show solution

9x^3 + 5x^2 - 8x - 10

, the constant term is the term without a variable, which is -10.

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EXERCISE SET 2.2

1(i)

x = 0

Show solution

Substitute

x = 0

5x - 3

5(0) - 3 = -3

.

So the value is -3.

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1(ii)

x = -1

Show solution

Substitute

x = -1

5x - 3

5(-1) - 3 = -5 - 3 = -8

.

So the value is -8.

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1(iii)

x = 2

Show solution

Substitute

x = 2

5x - 3

5(2) - 3 = 10 - 3 = 7

.

So the value is 7.

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2(i)

s = 0

Show solution

Substitute

s = 0

7s^2 - 4s + 6

7(0)^2 - 4(0) + 6 = 6

.

So the value is 6.

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2(ii)

s = -3

Show solution

Substitute

s = -3

7s^2 - 4s + 6

7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81

.

So the value is 81.

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2(iii)

s = 4

Show solution

Substitute

s = 4

7s^2 - 4s + 6

7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102

.

So the value is 102.

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3The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages.Show solution

Let Salil's present age be

x

years. Then his mother's present age is

3x

years.

After 5 years:
- Salil's age =

x + 5

- Mother's age =

3x + 5

Their sum is 70:

x + 5 + 3x + 5 = 70

4x + 10 = 70

4x = 60

x = 15

So Salil is 15 years old and his mother is 45 years old.

But this does not match the book's worked setup in the chapter excerpt? No, using the stated relation gives these ages.

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4The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.Show solution

Let the two positive integers be in the ratio

2:5

.
So, let them be

2x

and

5x

.

Their difference is 63:

5x - 2x = 63

3x = 63

x = 21

So the integers are:

2x = 42

and

5x = 105

.

Thus, the two integers are 42 and 105.

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5Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹88, how many coins does she have of each type?Show solution

Let the number of five-rupee coins be

x

.
Then the number of two-rupee coins is

3x

.

Total value:

5x + 2(3x) = 88

5x + 6x = 88

11x = 88

x = 8

So:
- five-rupee coins =

8

- two-rupee coins =

3 imes 8 = 24

Thus, she has 24 two-rupee coins and 8 five-rupee coins.

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6A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?Show solution

Let the shorter piece be

x

ft. Then the longer piece is

4x

ft.

Total length is 300 ft:

x + 4x = 300

5x = 300

x = 60

So the two pieces are:
- shorter piece = 60 ft
- longer piece = 240 ft

So the lengths are 60 ft and 240 ft.

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7If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?Show solution

Let the width be

w

cm.
Then length =

2w + 3

cm.

Perimeter of rectangle = 24 cm:

2( ext{length} + ext{width}) = 24

2((2w + 3) + w) = 24

2(3w + 3) = 24

6w + 6 = 24

6w = 18

w = 3

Then length =

2(3) + 3 = 9

cm.

So the dimensions are 3 cm by 9 cm.

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1Find the value of the linear polynomial

5x - 3

if:Show solution

For

5x - 3

:
- If

x = 0

, then

5(0) - 3 = -3

.
- If

x = -1

, then

5(-1) - 3 = -8

.
- If

x = 2

, then

5(2) - 3 = 7

.

So the values are -3, -8, 7.

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2Find the value of the quadratic polynomial

7s^2 - 4s + 6

if:Show solution

For

7s^2 - 4s + 6

:
- If

s = 0

, value =

6

.
- If

s = -3

, value =

81

.
- If

s = 4

, value =

7(16) - 16 + 6 = 102

.

So the values are 6, 81, 102.

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EXERCISE SET 2.3

1A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the

n^{th}

month.Show solution

Let the amount at the end of

n

months be

A

.
She starts with ₹500 and gets ₹150 each month, so:

A = 500 + 150n

Thus:
- 1st month: ₹650
- 2nd month: ₹800
- 3rd month: ₹950
- 4th month: ₹1100
- and so on.

The required linear expression is **

500 + 150n

**.

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2A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, ... hours? Find a linear expression to represent the number of members at the end of the nth hour.Show solution

The rally starts with 120 members and loses 9 members each hour.
So after

n

hours, the number of members is

N = 120 - 9n

The first few values are:
- after 1 hour:

120 - 9 = 111

- after 2 hours:

120 - 18 = 102

- after 3 hours:

120 - 27 = 93

So the linear expression is **

120 - 9n

**.

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3Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.Show solution

Area of rectangle = length × breadth =

13 imes b

.

- If breadth = 12 cm, area =

13 \times 12 = 156

cm²
- If breadth = 10 cm, area =

13 \times 10 = 130

cm²
- If breadth = 8 cm, area =

13 \times 8 = 104

cm²

So the linear pattern is:

156, 130, 104, \dots

and the expression for area is **

13b

**.

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4Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.Show solution

Volume of the box = length × breadth × height =

7 \times 11 \times h = 77h

.

- If

h = 5

cm, volume =

77 \times 5 = 385

cm³
- If

h = 9

cm, volume =

77 \times 9 = 693

cm³
- If

h = 13

cm, volume =

77 \times 13 = 1001

cm³

So the linear pattern is:

385, 693, 1001, \dots

and the expression is **

77h

**.

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5Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.Show solution

Sarita reads

20

pages each day, so in

15

days she reads

20 \times 15 = 300

pages.

Pages left after 15 days:

500 - 300 = 200

So the linear pattern for pages left is

500, 480, 460, 440, \dots

or, after

n

days,

500 - 20n

Thus after 15 days, 200 pages will be left.

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3Suppose the length of a rectangle is 13 cm. Find the area if the breadth isShow solution

The book source for this exercise gives the rectangle example as: length

13

cm and breadth values

12

cm,

10

cm, and

8

cm.

Area of a rectangle = length × breadth.

- If breadth =

12

cm, area =

13 \times 12 = 156\text{ cm}^2

- If breadth =

10

cm, area =

13 \times 10 = 130\text{ cm}^2

- If breadth =

8

cm, area =

13 \times 8 = 104\text{ cm}^2

So the linear pattern for the area is:

156, 130, 104, \dots

and in general, if breadth is

b

, then area is

A = 13b

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4Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height isShow solution

The book source for this exercise gives the rectangular box example as: length

7

cm, breadth

11

cm, and height values

5

cm,

9

cm, and

13

cm.

Volume of a rectangular box = length × breadth × height.

First find the base area:

7 \times 11 = 77

Now multiply by the height:

- If height =

5

cm, volume =

77 \times 5 = 385\text{ cm}^3

- If height =

9

cm, volume =

77 \times 9 = 693\text{ cm}^3

- If height =

13

cm, volume =

77 \times 13 = 1001\text{ cm}^3

So the linear pattern for the volume is:

385, 693, 1001, \dots

and in general, if height is

h

, then

V = 77h

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EXERCISE SET 2.4

1(i)Find the height after 7 months.Show solution

The plant’s initial height is

1.75

feet and it grows by

0.5

feet each month.

After 7 months, height =

1.75 + 7 \times 0.5

= 1.75 + 3.5

= 5.25

But the question asks for the source exercise answer, which in the chapter is based on a plant growth example with the same pattern. For this specific item, the height after 7 months is 5.25 feet.

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1(ii)Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.Show solution

The height starts at

1.75

feet and increases by

0.5

feet every month.

So for

t

months,

h = 1.75 + 0.5t

The table of values from

t=0

10

is:

|

t

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
|

h

(feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |

Each month the height increases by a constant amount of 0.5 feet, so this is a linear growth pattern.

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1(iii)Find an expression that relates h and t, and explain why it represents linear growth.Show solution

The expression relating height

h

and time

t

h = 1.75 + 0.5t

Here,

1.75

is the initial height and

0.5

is the fixed amount added each month.

This represents linear growth because the height increases by the same amount,

0.5

feet, over each equal interval of one month. Since the rate of increase is constant, the relation is linear.

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2(i)Find the value of the phone after 3 years.Show solution

The phone’s initial value is ₹10,000 and it decreases by ₹800 each year.

After 3 years:

10000 - 3 \times 800 = 10000 - 2400 = 7600

So the value of the phone after 3 years is ₹7600.

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2(ii)Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.Show solution

The phone is worth ₹10,000 at

t=0

and loses ₹800 every year.

So the values from

t=0

8

years are:

|

t

(years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
|

v

(₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |

The value decreases by a constant amount of ₹800 each year, so it shows linear decay.

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2(iii)Find an expression that relates v and t, and explain why it represents linear decay.Show solution

The relation between value

v

and time

t

v = 10000 - 800t

This represents linear decay because the value decreases by a fixed amount, ₹800, in each equal time interval of one year. The decrease is constant, so the graph is linear with negative slope.

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3(i)Find the population of the village after 6 years.Show solution

The initial population is

750

and every year

50

people move to the village.

After 6 years:

750 + 6 \times 50 = 750 + 300 = 1050

So the population after 6 years is 1050.

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3(ii)Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.Show solution

For each year, the village starts with 750 people and gains 50 people every year.

So the population after

t

years is:

P = 750 + 50t

Now make the table for

t = 0

10

:

|

t

(years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
|

P

| 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |

This shows that the population increases by 50 every year.

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3(iii)Find an expression that relates P and t, and explain why it represents linear growth.Show solution

The village has an initial population of 750, so when

t=0

P=750

.

Each year, 50 people move to the village, so the population increases by 50 every year.

Therefore, the expression relating population

P

and time

t

is:

P = 750 + 50t

This represents linear growth because the population increases by a constant amount of 50 for each equal interval of 1 year.

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4(i)Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.Show solution

The scheme starts with a fixed balance of ₹600.
Each day, the balance is reduced by ₹15.

So after

x

days, the remaining balance is:

b(x) = 600 - 15x

This represents linear decay because the balance decreases by the same constant amount of ₹15 every day.

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4(ii)After how many days will the balance run out?Show solution

The balance runs out when

b(x)=0

.

Given

b(x)=600-15x

Set it equal to zero:

600 - 15x = 0

15x = 600

x = \frac{600}{15} = 40

So, the balance will run out after 40 days.

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4(iii)Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.Show solution

The balance decreases by ₹15 per day from an initial ₹600.

So,

b(x)=600-15x

Now find values for

x=1

10

:

|

x

(days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
|

b(x)

(₹) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |

This table shows that the balance reduces by 15 each day.

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1Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.Show solution

The plant’s initial height is 1.75 feet and it grows by 0.5 feet each month.

So after

t

months, its height is:

h = 1.75 + 0.5t

For

t=7

h = 1.75 + 0.5 \times 7 = 1.75 + 3.5 = 5.25

feet

Table of values for

t=0

10

:

|

t

(months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
|

h

(feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |

The expression is a linear growth pattern because the height increases by a constant amount of 0.5 feet every month.

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2A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.Show solution

The phone’s initial value is ₹10,000 and it decreases by ₹800 every year.

So after

t

years, the value is:

v = 10000 - 800t

For the required values:
- After 3 years:

v = 10000 - 800\times 3 = 10000 - 2400 = ₹7600

Table for

t=0

8

:

|

t

(years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
|

v

(₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |

This is linear decay because the value decreases by a constant amount of ₹800 every year.

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3The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.Show solution

The village starts with 750 people and gains 50 people every year.

So after

t

years, the population is:

P = 750 + 50t

For the required year:
- After 6 years:

P = 750 + 50\times 6 = 750 + 300 = 1050

Table for

t=0

10

:

|

t

(years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
|

P

| 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |

This is linear growth because the population increases by the same amount, 50, every year.

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EXERCISE SET 2.5

1A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill

y

depends on the number of modules accessed,

x

, according to the relation

y = ax + b

, find the values of

a

and

b

.Show solution

We use the relation

y = ax + b

.

Given points:
- When

x=10

y=400

- When

x=14

y=500

So,

400 = 10a + b

500 = 14a + b

Subtract the first equation from the second:

500 - 400 = (14a + b) - (10a + b)

100 = 4a

a = 25

Now substitute in

400 = 10a + b

400 = 10(25) + b

400 = 250 + b

b = 150

So, **

a=25

and

b=150

**.

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2A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill

y

depends on the hours of the use of the badminton court,

x

, according to the relation

y = ax + b

, find the values of

a

and

b

.Show solution

We use

y = ax + b

.

Given:
- When

x=10

y=800

- When

x=15

y=1100

So,

800 = 10a + b

1100 = 15a + b

Subtract the first equation from the second:

1100 - 800 = (15a + b) - (10a + b)

300 = 5a

a = 60

Now substitute into

800 = 10a + b

800 = 10(60) + b

800 = 600 + b

b = 200

So, **

a=60

and

b=200

**.

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3Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C =

a

°F +

b

. Find

a

and

b

, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.Show solution

Let the relation be

\,^{\circ}C = a\,^{\circ}F + b.

Using the two given points:

- When

^{\circ}C = 0

^{\circ}F = 32

0 = 32a + b

- When

^{\circ}C = 100

^{\circ}F = 212

100 = 212a + b

Subtract the first equation from the second:

100 = (212-32)a = 180a

So,

a = \frac{100}{180} = \frac{5}{9}

Now substitute in

0 = 32a + b

0 = 32\cdot\frac{5}{9} + b

b = -\frac{160}{9}

So the linear relation is

^{\circ}C = \frac{5}{9}\,^{\circ}F - \frac{160}{9}.

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3Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C =

a

°F +

b

. Find

a

and

b

, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find

a

and

b

, and thus, the linear relationship between °C and °F.)Show solution

We use the given relation

^{\circ}C = a\,^{\circ}F + b.

From the data:

- Ice melts:

^{\circ}C = 0

^{\circ}F = 32

gives

0 = 32a + b

- Water boils:

^{\circ}C = 100

^{\circ}F = 212

gives

100 = 212a + b

Subtracting the first equation from the second:

100 = (212-32)a = 180a

Hence,

a = \frac{100}{180} = \frac{5}{9}.

Now substitute into

0 = 32a + b

0 = 32\cdot\frac{5}{9} + b

b = -\frac{160}{9}.

So, the values are

a = \frac{5}{9}

and

b = -\frac{160}{9}

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Think and Reflect

1Can you identify the terms, variables and coefficients of this algebraic expression?Show solution

In the algebraic expression from Example 2,

200l + 160w + 50lw,

the terms are:
-

200l

160w

50lw

The variables are:
-

l

w

The coefficients are:
- of

l

, the coefficient is

200

- of

w

, the coefficient is

160

- of

lw

, the coefficient is

50

So the expression has three terms, two variables, and coefficients 200, 160, and 50.

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2How is it different from the algebraic expression in Example 1?

3Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?

4Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?

5If a player paid ₹750, how many matches did he play?

6We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle,

10x - x^2

, is a function of

x

. Can you interpret this as an input-output process? What value does the expression take when

x = 6

cm?

7Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.

8Using the expression

2n - 1

, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?

9What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?

10For how many km will the fare be ₹130?

11What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?

12What will be the height of the water at the end of 5 months?

13Can you guess what the numbers 20 and 150 in the equation

y = 20x + 150

represent?

14Identify other points on the line by completing the following table.

15Differentiate between the graphs of the equations

y = 3x + 1

, and

y = -3x + 1

16Does this help you to conclude anything about the linear equation

y = ax + b

when

a

is fixed but

b

varies?

EXERCISE SET 2.6

1(i)

y = 4x, y = 2x, y = x

1(ii)

y = -6x, y = -3x, y = -x

1(iii)

y = 5x, y = -5x

1(iv)

y = 3x - 1, y = 3x, y = 3x + 1

1(v)

y = -2x - 3, y = -2x, y = 2x + 3

END-OF-CHAPTER EXERCISES

1Write a polynomial of degree 3 in the variable

x

, in which the coefficient of the

x^2

term is

-7

2(i)

5x^2 - 3x + 7

x = 1

2(ii)

4t^3 - t^2 + 6

t = a

3If we multiply a number by

\frac{5}{2}

and add

\frac{2}{3}

to the product, we get

\frac{-7}{12}

. Find the number.

4A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

5If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

6The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

7(i)

y = -3x + 4

7(ii)

2y = 4x + 7

7(iii)

5y = 6x - 10

7(iv)

3y = 6x - 11

8(i)Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

8(ii)If the temperature is 158 °F, then find the temperature in Kelvin.

9The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work

w

and distance

d

), and draw its graph by taking the constant force as 3 units. What is

10(i)Find the polynomial

p(x)

10(ii)Find the coordinates where the graph of

p(x)

cuts the axes.

10(iii)Draw the graph of

p(x)

and verify your answers.

11Let

p(x) = ax + b

and

q(x) = cx + d

be two linear polynomials such that:

- (i)

p(0) = 5

.
- (ii) The polynomial

p(x) - q(x)

cuts the x-axis at (3, 0).
- (iii) The sum

p(x) + q(x)

is equal to

6x + 4

for all real

x

.

Find the polynomials

p(x)

and

q(x)

12(i)Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

12(ii)Complete the following table.

12(iii)Find a rule to determine the number of matchsticks required for the

n^{th}

stage.

12(iv)How many matchsticks will be required for the 15th stage of the pattern?

12(v)Can 200 matchsticks form a stage in this pattern? Justify your answer.

13Let

p(x) = ax + b

and

q(x) = cx + d

be two linear polynomials such that:

(i) The graph of

p(x)

passes through the points (2, 3) and (6, 11).
(ii) The graph of

q(x)

passes through the point (4, -1).
(iii) The graph of

q(x)

is parallel to the graph of

p(x)

.

Find the polynomials

p(x)

and

q(x)

. Also, find the coordinates of the point where these lines meet the x-axis.

14What do all linear functions of the form

f(x) = ax + a

a > 0

, have in common?

11Let

p(x) = ax + b

and

q(x) = cx + d

be two linear polynomials such that:

5If you have ₹800 and you save ₹250 every month, find the amount you have after

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EXERCISE SET 2.1

EXERCISE SET 2.2

EXERCISE SET 2.3

EXERCISE SET 2.4

EXERCISE SET 2.5

Think and Reflect

EXERCISE SET 2.6

END-OF-CHAPTER EXERCISES

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