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NCERT Class 9 Maths Chapter 2 Introduction to Linear Polynomials
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Introduction to Linear Polynomials
Chapter – 2
| Exercise Set 2.1 |
1. Find the degrees of the following polynomials.
(i) 2×2 – 5x + 3
Ans: 2x2 – 5x + 3
Highest power of x = 2
∴ Degree = 2
(ii) y3 + 2y – 1
Ans: y3 + 2y – 1
Highest power of y = 3
∴ Degree = 3
(iii) -9
Ans: -9
This is a constant polynomial (no variable)
∴ Degree = 0
(iv) 4z – 3
Ans: 4z – 3
Highest power of z = 1
∴ Degree = 1
2. Write the polynomials of degree 1, 2 and 3.
Ans: Degree 1 (Linear): 4x + 2
Degree 2 (Quadratic): 3x2 – 4x +7
Degree 3 (Cubic): x3 + 2x2 – x + 10
3. What are the coefficients of x2 and x3 in the polynomial x4 – 3x3 + 6x2 – 2x + 7?
Ans: Given polynomial: x4 – 3x3 + 6x2 – 2x + 7
Coefficient of x3 = -3
Coefficient of x2 = 6
4. What is the coefficient of z in the polynomial 4z3 + 5z2 – 11?
Ans: Since we can rewrite 4z3 + 5z2 – 11 as 4z3 + 5z2 + 0z – 11,
∴ Coefficient of z in the given polynomial is 0.
5. What is the constant term of the polynomial 9x3 + 5x2 – 8x – 10?
Ans: The constant term of the given polynomial is -10.
| Exercise Set 2.2 |
1. Find the value of the linear polynomial 5x – 3 if:
(i) x = 0
Ans: x = 0
= 5(0) – 3 = -3
(ii) x = -1
Ans: x = -1
= 5(-1) – 3 = -5 – 3 = -8
(iii) x = 2
Ans: x = 2
= 5(2) – 3 = 10 – 3 = 7
2. Find the value of the quadratic polynomial 7s2 -4s+ 6 if:
(i) s = 0
Ans: s = 0; the quadratic polynomial
7s2 – 4s + 6 = 7(0)2 – 4(0) + 6 = 0 – 0 + 6 = 6
(ii) s = -3
Ans: s = -3; the quadratic polynomial
7s2 – 4s + 6 = 7(-3)2 – 4(-3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81
(iii) s = 4
Ans: s = 4; the quadratic polynomial
7s2 – 4s + 6
= 7(4)2 – 4(4) + 6
= 7(16) – 16 + 6
= 112 – 16 + 6
= 102
3. The 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.
Ans: Let Salil’s present age = x years
Mother’s present age = 3x years
After 5 years:
Salil’s age = x + 5
Mother’s age = 3x + 5
According to question: (x + 5) + (3x + 5) = 70
⇒ 4x + 10 = 70
⇒ 4x = 60
⇒ x = 15
Therefore, Salil’s age = 15 years
Mother’s age = 45 years
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Ans: Let the two integers be 2x and 5x.
Their difference: 5x – 2x = 63
⇒ 3x = 63
⇒ x = 21
∴ The first integer = 2 × 21 = 42
And the second integer = 5 × 21 = 105
Thus, the two integers are 42 and 105.
5. Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total of ₹88, how many coins does she have of each type?
Ans: Let the number of five-rupee coins = x
Number of two-rupee coins = 3x
Total value: 5x + 2(3x) = 88
⇒ 5x + 6x = 88
⇒ 11x = 88
⇒ x = 8
Hence:
Five-rupee coins = 8
Two-rupee coins = 24
6. A 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?
Ans: Let the length of the shorter piece be x.
Then, the length of the longer piece is 4x.
Total length of the fence = x + 4x = 300
⇒ 5x = 300
⇒ x = 60
Thus, the shorter piece is 60 feet long, and the longer piece is 60 × 4 = 240 feet long.
7. If 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?
Ans: Let width = x cm
Length = 2x + 3 cm
Perimeter = 2(length + width)
⇒ 2[(2x + 3) + x] = 24
⇒ 2(3x + 3) = 24
⇒ 6x + 6 = 24
⇒ 6x = 18
⇒ x = 3
Therefore:
Width = 3 cm
Length = 2(3) + 3 = 9 cm
| Exercise Set 2.3 |
Solve the following:
1. A 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 nth month.
Ans: Initial amount = ₹500
Pocket money per month = ₹150
Money she will have at the end of every month:
| Month | 2 | 3 | 4 | 5 | … |
| Money (₹) | 500 + 150 × 2 = 800 | 500 + 150 × 3 = 950 | 500 + 150 × 4 = 1100 | 500 + 150 × 5 = 1250 | … |
The total amount, A(n), at the end of the nth month can be expressed as:
A(n) = ₹(500 + 150n), where n is the number of months.
2. A 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.
Ans: Initial members = 120
Members leaving each hour = 9
After:
1 hour = 120 – 9 = 111
2 hours = 120 – 18 = 102
3 hours = 120 – 27 = 93
Let number of members after n hours = Mn
Mn = 120 – 9n
Thus, required linear expression is Mn = 120 – 9n.
3. Suppose 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.
Ans: Length = 13 cm
Area = Length × Breadth
(i) When the breadth is 12 cm:
Area = 13 × 12 = 156 cm2
(ii) When the breadth is 10 cm:
Area = 13 × 10 = 130 cm2
(iii) When the breadth is 8 cm:
Area = 13 × 8 = 104 cm2
The linear expression representing the area A(b) in terms of the breadth b is:
A (b) = 13 b, where b is the breadth of the rectangle.
4. Suppose 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.
Ans: Length = 7 cm, Breadth = 11 cm
Volume = Length × Breadth × Height
= 7 × 11 × h = 77h
Height = 5 cm
Volume = 77 × 5 = 385 cm3
Height = 9 cm
Volume = 77 × 9 = 693 cm3
Height = 13 cm
Volume = 77 × 13 = 1001 cm3 Let height = h cm
Volume = 77h
Thus, linear pattern is V = 77h.
5. Sarita 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.
Ans: Total pages in the book = 500
The number of pages read per day = 20
The number of pages left after n days, L(n), can be expressed as:
L(n) = 500 – 20n,
For n = 15 days:
L(15) = 500 – 20(15)
= 500 – 300 = 200
So, after 15 days, 200 pages of the book will be left to read.
| Exercise Set 2.4 |
1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
Ans: Initial height = 1.75 feet
Growth per month = 0.5 feet
Height after 7 months = Initial height + (Growth per month × 7)
∴ h= 1.75+ (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
Ans:
| Time (months) | Height (h feet) |
| 0 | 1.75 |
| 1 | 2.25 |
| 2 | 2.75 |
| 3 | 3.25 |
| 4 | 3.75 |
| 5 | 4.25 |
| 6 | 4.75 |
| 7 | 5.25 |
| 8 | 5.75 |
| 9 | 6.25 |
| 10 | 6.75 |
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Ans: Let height after t months = h
h = 1.75 + 0.5t
This represents linear growth because height increases by a constant amount (0.5 feet) every month.
2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value of the phone after 3 years.
Ans: Initial value = ₹10,000
Decrease per year = ₹800
Decrease in value after 1 year = ₹800
So, Value after 3 years:
v = 10000 – (800 × 3)
= 10000 – 2400
= ₹7600
(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.
Ans:
| Time (t years) | Value (v in ₹) |
| 0 | 10,000 |
| 1 | 9,200 |
| 2 | 8,400 |
| 3 | 7,600 |
| 4 | 6,800 |
| 5 | 6,000 |
| 6 | 5,200 |
| 7 | 4,400 |
| 8 | 3,600 |
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Ans: Expression:
Let value after t years = v
v = 10000 – 800t
This represents linear decay because the value decreases by a constant amount (₹800) every year.
3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
Ans: Initial population of village = 750
Population increase per year = 50
∴ Population after 6 years = Initial population + (Increase per year × 6)
⇒ Population (P) = 750 + (50 × 6)
= 750 + 300
= 1050
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
Ans:
| Time (year) | Population (P) |
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1,000 |
| 6 | 1,050 |
| 7 | 1,100 |
| 8 | 1,150 |
| 9 | 1,200 |
| 10 | 1,250 |
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
Ans: Expression:
Let population after t years = P
P = 750 + 50t
This represents linear growth because the population increases by a constant number (50 people) every year.
4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
Ans: Initial balance = ₹600
Reduction per day = ₹15
∴ Remaining balance after x days can be expressed as: b(x) = 600 – 15x
This expression represents linear decay because the balance decreases by a constant amount (₹15) each day.
(ii) After how many days will the balance run out?
Ans: When balance runs out:
b(x) = 0
600 – 15x = 0
15x = 600
x = 40
So, the balance will run out after 40 days.
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
Ans:
| Days (x) | Remaining Balance b(x) (₹) |
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |
| Exercise Set 2.5 |
1. A 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.
Ans: Given:
When x = 10, y = 400
When x = 14, y = 500
Using y = ax + b
For x = 10: 400 = 10a + b …(1)
For x = 14: 500 = 14a + b …(2)
Subtracting (1) from (2), we get:
500 – 400 = 14a – 10a
⇒ 100 = 4a
⇒ a = 25
Substituting a = 25 in (1), we get:
400 = 10(25) + b
⇒ 400 = 250 + b
⇒ b = 150
Therefore, a = 25 and b = 150.
2. A 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.
Ans: Given the linear relation y = ax + b, where y is the monthly bill and x is the number of hours:
When x = 10, y = 800, and when x = 15, y = 1100
We now have two equations:
800 = 10a + b … (i)
1100 = 15a + b …(ii)
Subtract the first equation from the second one:
1100 – 800 = (15a + 6) – (10a + 6)
⇒ 300 = 5a
⇒ a = 60
Substitute a = 60 in first equation:
800 = 10(60) + b
⇒ 800 = 600 + 6
⇒ b = 200
Thus, the equation for the gym’s monthly bill is
y = 60x + 200, and a = 60 and b = 200.
3. Consider 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.)
Ans: Given relation: °C = a°F + b
Using the point (°F, °C) = (32, 0), we have:
0 = 32a + b …(1)
Using the point (°F, °C) = (212, 100), we have:
100 = 212a + b …(2)
Subtracting (1) from (2), we get:
100 – 0 = 212a – 32a
⇒ 100 = 180a
⇒ a = 100/180
⇒ a = 5/9
Substitute a = 5/9 in (1), we get:
0 = 32(5/9) + b
⇒ 0 = 160/9 + b
⇒ b = -160/9
Therefore, a = 5/9 and b = -160/9.
Hence, the linear relationship is °C = (5/9)°F – 160/9.
It can also be written as °C = (5/9)(°F – 32).
| Exercise Set 2.6 |
1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
(i) y = 4x, y = 2x, y = x
Ans: y = 4x, y = 2x, y = x
All lines are of the form y = ax (b = 0).
Observation:
All lines pass through the origin (0, 0).
The value of ‘a’ (slope) determines steepness.
Larger ‘a’ ⇒ steeper line.
(ii) y = – 6x, y = – 3x, y = – x
Ans: y = – 6x, y = -3x, y =-x
All lines are of the form y = -ax (b = 0).
Observation:
All lines pass through the origin.
Negative ‘a’ means lines slope downward.
Larger magnitude of‘a’ ⇒ steeper downward slope.
(iii) y = 5x, y = -5x
Ans: Observation:
Both Lines pass through the origin.
y = 5x slopes upward, y = -5x slopes downward.
Same magnitude of ‘a’ → same steepness but opposite direction.
(iv) y = 3x – 1, y = 3x, y = 3x + 1
Ans: y = 3x – 1, y = 3x, y = 3x + 1
All lines have same slope (a = 3).
Observation: Lines are parallel as slopes are the same (3).
Different values of ‘b’ shift the line. b = -1 line below origin, b = 0 passes through origin b = 1 line above origin.
(v) y = -2x – 3, y = -2x, y = 2x + 3
Ans: y = -2x -3 ,y = -2x, y = 2x + 3
Observation:
y = -2x – 3 and y = -2x have the same slope (-2), so the lines are parallel lines. y = 2x + 3 has positive slope and represents different direction.
‘b’ changes vertical position of the line.
Conclusion:
1. a (coefficient of x) controls slope (steepness and direction).
2. b (constant term) controls vertical shift (y-intercept).
| END-OF-CHAPTER EXERCISES |
1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is -7.
Ans: A polynomial of degree 3 has the general form: ax3 + bx2 + cx + d
Given that coefficient of x2 is -7, so b = -7
One such polynomial is x3 – 7x2 + 2x + 1.
(Any polynomial of degree 3 with -7 as the coefficient of x2 is correct)
2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x2 – 3x + 7 if x = 1
Ans: For 5x2 – 3x + 7 if x = 1
⇒ 5 × 12 – 3 × 1 + 7 = 5 – 3 + 7 = 9
(ii) 4t3 – t2 + 6 if t = a
Ans: For 4t3 – t2 + 6 if t = a
⇒ 4 × a3 – a2 + 6
⇒ 4a3 – a2 + 6
3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.
Ans: Let the number be x.
According to the question:
(5/2)x + ⅔ = -7/12
Subtract 2/3 from both sides, we get:
(5/2)x = -7/12 – 2/3
⇒ (5/2)x = -7/12 – 8/12
⇒ (5/2)x = -15/12
⇒ (5/2)x = -5/4
Now multiply both sides by 2/5, we get:
x = (-5/4) × (2/5)
⇒ x = -1/2
Therefore, the number is -1/2
4. A 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?
Ans: Let the two numbers be x and y, where x = 5y.
ATQ, x + 21 = 2(y + 21)
Substituting x = 5y:
⇒ 5y + 21 = 2(y + 21)
⇒ 5y + 21 = 2y + 42
⇒ 3y = 21
⇒ y = 7
So, x = 5y = 35.
The numbers are 35 and 7.
5. If 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.
Ans: Initial amount = ₹800
Saving every month = ₹250
Linear pattern: Amount after n months = 800 + 250n
Amount after 6 months:
= 800 + 250(6)
= 800 + 1500
= ₹2300
Amount after 2 years: 2 years = 24 months
Amount after 24 months:
= 800 + 250(24)
= 800 + 6000
= ₹6800
Therefore:
Amount after 6 months = ₹2300
Amount after 2 years = ₹6800
Linear pattern: A = 800 + 250n, where n is the number of months.
6. The 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.
Ans: Let the two-digit number be represented as 10a + b, where a and b are the tens and ones digits, respectively.
The digits differ by 3: |a – b| = 3
When the digits are interchanged, the number becomes 10b + a.
The sum of the original and the interchanged numbers is 143:
⇒ (10a + 6) + (106 + a) = 143
⇒ 11a + 11 b = 143
⇒ a + b = 13
Solving the system of equations: a – b = 3 and a + b = 13
Adding these equations: 2a = 16
⇒ a = 8
Substituting a = 8 in a + b = 13:
8 + 6 = 13 ⇒ 6 = 5
The original number = 85 and the interchanged number = 58
Therefore, the numbers are 85 and 58.
7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Ans:
(i) y = -3x + 4
Slope: a = -3
y-intercept: b = 4
Coordinates where the line cuts the y-axis: (0, 4)
Explanation: When x = 0, substitute it into the equation: y = -3(0) + 4 = 4.
The line cuts the y-axis at (0, 4).
(ii) 2y = 4x + 7
Convert to slope-intercept form: y = 2x + 3.5
Slope: a = 2
y-intercept: b = 3.5
Coordinates where the line cuts the y-axis: (0, 3.5)
Explanation: When x = 0, substitute it into the equation: y = 2(0) + 3.5 = 3.5.
The line cuts the y-axis at (0, 3.5).
(iii) 5y = 6x – 10
Dividing both sides by 5: y = (6/5)x – 2
Slope a = 6/5
y-intercept b = -2
So, the point, where the line cuts the y-axis is (0, -2).
(iv) 3y = 6x – 11
Dividing both sides by 3: y = 2x – 11/3
Slope a = 2
y-intercept b = -11/3
So, the point, where the line cuts the y-axis is (0, -11/3).
8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5 (x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
Ans: When x = 313 K, We have to find y:
Using equation y = (9/5)(x – 273) + 32, we have
y = (9/5)(313 – 273) + 32
⇒ y = (9/5)(40) + 32
⇒ y = 72 + 32
⇒ y = 104
Therefore, the temperature is 104 °F.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Ans: If the temperature is 158 °F, then 158 = 9/5 (x – 273) + 32
⇒ 126 = 9/5 (x – 273)
⇒ x = 273 + 40 = 343
Hence, the temperature of liquid in Kelvin is 343 K.
9. The 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 the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Ans: The work done is given by the equation.
Work done = Force × Distance
ATQ, w = F × d, where F = 3.
⇒ w = 3d, is the required linear equation in two variables.
For d= 2: w = 3 × 2 = 6 units of work.
By taking w on y-axis and d on x-axis, we can plot the graph as follows:
10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
Ans: Let p(x) = ax + b
Since the graph passes through (1, 5),
a(1) + b = 5
⇒ a + b = 5 …(1)
Since the graph passes through (3, 11),
a(3) + b = 11
⇒ 3a + b = 11 …(2)
Subtracting (1) from (2), we get:
3a + b – (a + b) = 11 – 5
⇒ 2a = 6
⇒ a = 3
Substituting a = 3 in (1), we have:
3 + b = 5
⇒ b = 2
Therefore, p(x) = 3x + 2.
(ii) Find the coordinates where the graph of p(x) cuts the axes.
Ans: Coordinates where the graph cuts the axes:
To find the y-axis intercept: Put x = 0
⇒ y = p(0) = 3(0) + 2 = 2
So, the graph cuts the y-axis at (0, 2).
To find the x-axis intercept: Put y = 0
⇒ 3x + 2 = 0
⇒ 3x = -2
⇒ x = −2/3.
So, the graph cuts the x-axis at (-2/3, 0)
(iii) Draw the graph of p(x) and verify your answers.
Ans: Graph:
11. Let 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).
Ans: Let p(x) = ax + b and q(x) = cx + d
Using the condition (i), we have p(0) = 5
⇒ a(0) + b = 5
⇒ b = 5
So, p(x) = ax + 5
Now, using condition (iii), we have
p(x) + q(x) = 6x + 4
⇒ (ax + 5) + (cx + d) = 6x + 4
⇒ (a + c)x + (5 + d) = 6x + 4
Comparing coefficients, we get a + c = 6 …(1)
5 + d = 4
⇒ d = -1
So, q(x) = cx – 1
Using condition (ii), p(x) – q(x) cuts x-axis at (3, 0)
⇒ p(3) – q(3) = 0
Here:
p(3) = 3a + 5
q(3) = 3c – 1
So, (3a + 5) – (3c – 1) = 0
⇒ 3a + 5 – 3c + 1 = 0
⇒ 3a – 3c + 6 = 0
⇒ a – c = -2 …(2)
Solving equations
From (1): a + c = 6
From (2): a – c = -2
Adding both: 2a = 4 ⇒ a = 2
Then: 2 + c = 6 ⇒ c = 4
Hence:
p(x) = 2x + 5
q(x) = 4x – 1.
12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. a new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
| Stage Number | 1 | 2 | 3 | 4 | 5 | … | n |
| Number of matchsticks |
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Ans: A single hexagon needs 6 matchsticks.
Since each new hexagon shares one side with the previous hexagon, only 5 new matchsticks are added at every new stage.
So the pattern is:
Stage 1: 6
Stage 2: 6 + 5 = 11
Stage 3: 11 + 5= 16
(i) Next two stages:
Stage 4: Number of matchsticks = 16 + 5 = 21
Stage 5: Number of matchsticks = 21 + 5 = 26
(ii)
| Stage Number | 1 | 2 | 3 | 4 | 5 | … | n |
| Number of matchsticks | 6 | 11 | 16 | 21 | 26 | 5n + 1 |
(iii) Rule for the nth stage:
The number of matchsticks forms a pattern: 6, 11, 16, 21, 26,…
Where, the first term = 6,
difference of two consecutive terms = 5
∴ Number of matchsticks in the nth stage
= 6 + (n – 1) × 5 = 6 + 5n – 5 = 5n + 1
Hence, the rule to determine the number of matchsticks required for the nth stage is given by Sn = 5n + 1
(iv) Matchsticks required for the 15th stage
S15= 5(15) + 1 = 75 + 1 = 76
Therefore, 76 matchsticks will be required for the 15th stage.
(v) Let for some stage n,
5n + 1 = 200
n = 199/5 = 39.8
Since, 39.8 is not a whole number. Therefore, 200 matchsticks cannot form a stage in this pattern.
13. Let 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.
Ans: Given:
p(x) = ax + b
q(x) = cx + d
First, to find p(x).
Since p(x) passes through (2, 3) and (6, 11), its slope is m = (11 – 3)/(6 – 2)
= 8/4 = 2
So, p(x) = 2x + b
Using the point (2, 3), we have:
3 = 2(2) + b
⇒ 3 = 4 + b
⇒ b = -1
Therefore, p(x) = 2x – 1.
Now to find q(x).
Since q(x) is parallel to p(x), it has the same slope.
So, slope of q(x) = 2
Hence, q(x) = 2x + d
Since q(x) passes through (4, -1), so -1 = 2(4) + d
⇒ -1 = 8 + d
⇒ d = -9
Therefore, q(x) = 2x – 9
Now we have to find where these lines meet the x-axis.
A line meets the x-axis where y = 0.
For p(x) = 2x – 1: 0 = 2x – 1
⇒ 2x = 1
⇒ x = 1/2
So, p(x) meets the x-axis at (1/2, 0).
For q(x) = 2x – 9: 0 = 2x – 9
⇒ 2x = 9
⇒ x = 9/2
So, q(x) meets the x-axis at (9/2, 0).
Hence, p(x) = 2x – 1 and q(x) = 2x – 9.
The x-axis intercepts are:
For p(x): (1/2, 0)
For q(x): (9/2, 0)
14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Ans: Given: f(x) = ax + a, where a > 0
Common properties of all such linear functions are:
1. Slope:
The slope is a, and since a > 0 all the lines have positive slope. So, all these lines rise from left to right.
2. y-intercept:
Putting x = 0
f(0) = a So, the y-intercept is (0, a).
Since a > 0 all the lines cut they-axis above the origin.
3. x-intercept:
ax + a = 0
⇒ a(x + 1) = 0
To find where the line cuts the x-axis, put f(x) = 0, Since a > 0, – a is not zero.
So, x + 1 = 0
⇒ x = -1
Thus, every line cuts the x-axis at the same point (-1,0).
Therefore, all linear functions of the form f(x) = ax + a, a > 0 have the following common characteristic:
1. All have positive slope.
2. All cut the y-axis above the origin.
3. All pass through the fixed point (-1, 0).
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