Ganita Manjari ยท Grade 9 ยท Part I
Chapter 2 ยท Introduction to
Linear Polynomials
Complete step-by-step solutions to all Exercise Sets 2.1 โ 2.6 and End-of-Chapter Exercises
Exercise Set 2.1
Exercise Set 2.2
Exercise Set 2.3
Exercise Set 2.4
Exercise Set 2.5
Exercise Set 2.6
End-of-Chapter โ
Exercise Set 2.1 โ Degrees of Polynomials
Q1
Find the degrees of the following polynomials:
(i) 2xยฒ โ 5x + 3 (ii) yยณ + 2y โ 1 (iii) โ9 (iv) 4z โ 3
Solution
The degree of a polynomial is the highest power of the variable present.
(i) 2xยฒ โ 5x + 3
1
Identify powers of x: xยฒ (power 2), xยน (power 1), and constant (power 0).
2
Highest power = 2. This is a Quadratic Polynomial.
Degree = 2
(ii) yยณ + 2y โ 1
1
Powers present: yยณ (power 3), y (power 1), โ1 (power 0).
2
Highest power = 3. This is a Cubic Polynomial.
Degree = 3
(iii) โ9
1
โ9 is a constant. We can write it as โ9ยทxโฐ, so the highest power = 0.
Degree = 0 (Constant Polynomial)
(iv) 4z โ 3
1
Powers present: zยน (power 1), โ3 (power 0). Highest power = 1.
Degree = 1 (Linear Polynomial)
Q2
Write polynomials of degrees 1, 2 and 3.
Solution
Many answers are possible. Here are standard examples:
Degree 1 (Linear): 3x + 5
Degree 2 (Quadratic): xยฒ โ 4x + 7
Degree 3 (Cubic): 2xยณ + xยฒ โ x + 1
๐ Any polynomial where the highest power of the variable matches the required degree is correct.
Q3
What are the coefficients of xยฒ and xยณ in the polynomial xโด โ 3xยณ + 6xยฒ โ 2x + 7?
Solution
1
Write the polynomial term-by-term: xโด, โ3xยณ, +6xยฒ, โ2x, +7.
2
The coefficient of xยณ is the number multiplied with xยณ, which is โ3.
3
The coefficient of xยฒ is the number multiplied with xยฒ, which is 6.
Coefficient of xยฒ = 6 | Coefficient of xยณ = โ3
Q4
What is the coefficient of z in the polynomial 4zยณ + 5zยฒ โ 11?
Solution
1
Look for the term containing zยน in 4zยณ + 5zยฒ โ 11.
2
There is no z term โ it is absent (coefficient = 0).
Coefficient of z = 0
Q5
What is the constant term of the polynomial 9xยณ + 5xยฒ โ 8x โ 10?
Solution
1
The constant term is the term that has no variable (xโฐ term).
2
In 9xยณ + 5xยฒ โ 8x โ 10, the constant term is โ10.
Constant term = โ10
Exercise Set 2.2 โ Evaluating Polynomials & Linear Equations
Q1
Find the value of the linear polynomial 5x โ 3 if: (i) x = 0 (ii) x = โ1 (iii) x = 2
Solution
Substitute each value in p(x) = 5x โ 3:
(i) x = 0
p(0) = 5(0) โ 3 = 0 โ 3 = โ3
Value = โ3
(ii) x = โ1
p(โ1) = 5(โ1) โ 3 = โ5 โ 3 = โ8
Value = โ8
(iii) x = 2
p(2) = 5(2) โ 3 = 10 โ 3 = 7
Value = 7
Q2
Find the value of the quadratic polynomial 7sยฒ โ 4s + 6 if: (i) s = 0 (ii) s = โ3 (iii) s = 4
Solution
Substitute each value in q(s) = 7sยฒ โ 4s + 6:
(i) s = 0
q(0) = 7(0)ยฒ โ 4(0) + 6 = 0 โ 0 + 6 = 6
Value = 6
(ii) s = โ3
q(โ3) = 7(โ3)ยฒ โ 4(โ3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81
Value = 81
(iii) s = 4
q(4) = 7(4)ยฒ โ 4(4) + 6
= 7(16) โ 16 + 6
= 112 โ 16 + 6
= 102
Value = 102
Q3
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.
Solution
1
Let Salil's present age = x years.
Then mother's present age = 3x years.
2
After 5 years: Salil's age = x + 5, Mother's age = 3x + 5.
3
Form the equation:
(x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15
4
Salil's age = 15 years, Mother's age = 3 ร 15 = 45 years.
Salil = 15 years | Mother = 45 years
Q4
The difference between two positive integers is 63. The ratio of the two integers is 2 : 5. Find the two integers.
Solution
1
Since ratio is 2 : 5, let the integers be 2k and 5k.
2
The difference:
5k โ 2k = 63
3k = 63
k = 21
3
Smaller integer = 2 ร 21 = 42, Larger = 5 ร 21 = 105.
Verify: 105 โ 42 = 63 โ 42 : 105 = 2 : 5 โ
The two integers are 42 and 105.
Q5
Ruby has 3 times as many two-rupee coins as five-rupee coins. If she has a total โน88, how many coins of each type does she have?
Solution
1
Let number of five-rupee coins = x.
Then number of two-rupee coins = 3x.
Then number of two-rupee coins = 3x.
2
Total amount:
5(x) + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8
3
Five-rupee coins = 8, Two-rupee coins = 3 ร 8 = 24.
Verify: 8 ร 5 + 24 ร 2 = 40 + 48 = โน88 โ
Five-rupee coins: 8 | Two-rupee coins: 24
Q6
A farmer cuts a 300 feet fence into two pieces. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Solution
1
Let shorter piece = x feet, then longer piece = 4x feet.
2
x + 4x = 300
5x = 300
x = 60
3
Shorter piece = 60 ft, Longer piece = 4 ร 60 = 240 ft.
Shorter piece = 60 feet | Longer piece = 240 feet
Q7
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?
Solution
1
Let width = w cm. Then length = 2w + 3 cm.
2
Perimeter = 2(length + width):
2(2w + 3 + w) = 24
2(3w + 3) = 24
6w + 6 = 24
6w = 18
w = 3
3
Width = 3 cm, Length = 2(3) + 3 = 9 cm.
Rectangle with width = 3 cm and length = 9 cm
Width = 3 cm | Length = 9 cm
Exercise Set 2.3 โ Exploring Linear Patterns
Q1
A student has โน500 in her savings account. She gets โน150 every month as pocket money. Find a linear expression for the amount she has in the nth month.
Solution
1
Initial amount = โน500. She adds โน150 each month.
2
After n months, total amount = 500 + 150n.
Amount in n-th month = 500 + 150n (rupees)
3
| Month (n) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Amount (โน) | 650 | 800 | 950 | 1100 | 1250 |
Linear expression: A(n) = 500 + 150n
Q2
A rally starts with 120 members. Each hour, 9 members drop out. Find a linear expression for the number of members at the end of the nth hour.
Solution
1
Start = 120, decrease each hour = 9.
2
Members after n hours = 120 โ 9n
3
| Hour (n) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Members | 120 | 111 | 102 | 93 | 84 | 75 |
M(n) = 120 โ 9n (This is a linear decay pattern)
Q3
Suppose the length of a rectangle is 13 cm. Find the area if breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern for area.
Solution
1
Area = length ร breadth = 13 ร b.
(i) b = 12 cm
Area = 13 ร 12 = 156 cmยฒ
(ii) b = 10 cm
Area = 13 ร 10 = 130 cmยฒ
(iii) b = 8 cm
Area = 13 ร 8 = 104 cmยฒ
2
| Breadth b (cm) | 8 | 10 | 12 |
|---|---|---|---|
| Area (cmยฒ) | 104 | 130 | 156 |
Areas: 156, 130, 104 cmยฒ | Linear pattern: A = 13b
Q4
A rectangular box has length 7 cm, breadth 11 cm. Find the volume if height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern.
Solution
1
Volume = l ร b ร h = 7 ร 11 ร h = 77h.
(i) h = 5 cm
V = 77 ร 5 = 385 cmยณ
(ii) h = 9 cm
V = 77 ร 9 = 693 cmยณ
(iii) h = 13 cm
V = 77 ร 13 = 1001 cmยณ
2
| Height h (cm) | 5 | 9 | 13 |
|---|---|---|---|
| Volume (cmยณ) | 385 | 693 | 1001 |
Volumes: 385, 693, 1001 cmยณ | Linear pattern: V = 77h
Q5
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.
Solution
1
Pages remaining after n days = 500 โ 20n.
2
After 15 days:
Pages left = 500 โ 20(15) = 500 โ 300 = 200 pages
3
| Day (n) | 0 | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|---|
| Pages left | 500 | 400 | 300 | 200 | 100 | 0 |
After 15 days: 200 pages left | Linear pattern: P(n) = 500 โ 20n
Exercise Set 2.4 โ Linear Growth & Linear Decay
Q1
A plant has height 1.75 feet and grows 0.5 feet each month. (i) Find height after 7 months. (ii) Table for t = 0 to 10. (iii) Expression relating h and t.
Solution
(i) Height after 7 months
h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet
Height after 7 months = 5.25 feet
(ii) Table of values (t = 0 to 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 |
(iii) Expression & why Linear Growth
h(t) = 1.75 + 0.5t
This represents linear growth because as t increases by 1 (one month), h increases by a fixed constant (0.5 feet). The slope is positive (+0.5).
Q2
A mobile phone is bought for โน10,000 and its value decreases by โน800 every year. (i) Value after 3 years. (ii) Table for t = 0 to 8. (iii) Expression for v and t.
Solution
(i) Value after 3 years
v = 10000 โ 800(3) = 10000 โ 2400 = โน7,600
Value after 3 years = โน7,600
(ii) Table of values (t = 0 to 8)
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| v (โน) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) Expression & why Linear Decay
v(t) = 10000 โ 800t
This represents linear decay because as t increases by 1 year, v decreases by a fixed amount (โน800). The slope is negative (โ800).
Q3
Initial population of a village is 750. Every year 50 people move from a city to the village. (i) Population after 6 years. (ii) Table for t = 0 to 10. (iii) Expression for P and t.
Solution
(i) Population after 6 years
P = 750 + 50(6) = 750 + 300 = 1050
Population after 6 years = 1,050
(ii) Table (t = 0 to 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 |
(iii) Expression & why Linear Growth
P(t) = 750 + 50t
This is linear growth: each year P increases by a fixed 50. Slope = +50 (positive).
Q4
A telecom company charges โน600 recharge. Balance reduces โน15 each day. (i) Equation for b(x). (ii) After how many days will balance run out? (iii) Table for x = 1 to 10.
Solution
(i) Equation for remaining balance b(x)
b(x) = 600 โ 15x
This is linear decay: balance decreases by a fixed โน15 per day (constant rate of decrease, negative slope โ15).
(ii) When will balance run out?
600 โ 15x = 0
15x = 600
x = 40 days
Balance runs out after 40 days.
(iii) Table (x = 1 to 10)
| Day x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) (โน) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Exercise Set 2.5 โ Linear Relationships
Q1
A learning platform charges a fixed fee + cost per module. 10 modules โ โน400; 14 modules โ โน500. Find a and b in y = ax + b.
Solution
1
Using y = ax + b:
When x = 10, y = 400 โ 400 = 10a + b ... (1)
When x = 14, y = 500 โ 500 = 14a + b ... (2)
2
Subtract equation (1) from (2):
500 โ 400 = (14a + b) โ (10a + b)
100 = 4a
a = 25
3
Substitute a = 25 in (1):
400 = 10(25) + b
400 = 250 + b
b = 150
a = 25 โ cost per module = โน25; b = 150 โ fixed monthly fee = โน150.
a = 25, b = 150 | y = 25x + 150
Q2
A gym charges fixed fee + cost per hour of badminton court. 10 hrs โ โน800; 15 hrs โ โน1100. Find a and b in y = ax + b.
Solution
1
When x = 10, y = 800 โ 800 = 10a + b ... (1)
When x = 15, y = 1100 โ 1100 = 15a + b ... (2)
2
Subtract (1) from (2):
300 = 5a โ a = 60
3
From (1): b = 800 โ 10(60) = 800 โ 600 = 200.
a = 60 โ โน60 per hour; b = 200 โ fixed monthly fee = โน200.
a = 60, b = 200 | y = 60x + 200
Q3
ยฐC = aยฐF + b. Ice melts: 0ยฐC = 32ยฐF. Water boils: 100ยฐC = 212ยฐF. Find a and b.
Solution
1
ยฐC = 0, ยฐF = 32 โ 0 = 32a + b ... (1)
ยฐC = 100, ยฐF = 212 โ 100 = 212a + b ... (2)
2
Subtract (1) from (2):
100 = 180a
a = 100/180 = 5/9
3
From (1):
b = โ32a = โ32 ร (5/9) = โ160/9
4
Therefore:
ยฐC = (5/9)ยฐF โ 160/9 = (5/9)(ยฐF โ 32)
This is the well-known CelsiusโFahrenheit conversion formula!
a = 5/9, b = โ160/9 | ยฐC = (5/9)(ยฐF โ 32)
Exercise Set 2.6 โ Visualising Linear Relationships
Q1
Draw the graphs of the following sets of lines and reflect on the role of 'a' and 'b' in each case.
Solution
(i) y = 4x, y = 2x, y = x
Fig: y = x, y = 2x, y = 4x all pass through origin. Steeper as slope increases.
All lines y = ax pass through the origin (0,0). As a increases, the line becomes steeper. Since b = 0, the y-intercept is 0.
(ii) y = โ6x, y = โ3x, y = โx
Fig: y = โx, y = โ3x, y = โ6x โ all pass through origin, sloping downward (negative slope).
All lines with negative slope fall from left to right. Steepness increases as |a| increases.
(iii) y = 5x, y = โ5x
Fig: y = 5x rises steeply; y = โ5x is its mirror image, falling steeply. Both pass through origin.
y = 5x and y = โ5x are reflections of each other in the x-axis (or y-axis). Equal magnitude, opposite direction of slope.
(iv) y = 3x โ 1, y = 3x, y = 3x + 1
Fig: Parallel lines y = 3xโ1, y = 3x, y = 3x+1 โ same slope (3), different y-intercepts.
Key insight: When the slope 'a' is fixed but 'b' varies, the lines are parallel. They shift up/down by the value of b without changing direction.
(v) y = โ2x โ 3, y = โ2x, y = 2x + 3
Fig: y=โ2xโ3 and y=โ2x are parallel (same slope โ2). y=2x+3 has opposite slope.
y=โ2x and y=โ2xโ3 are parallel (both have slope โ2). y=2x+3 has slope +2, opposite direction, so it's not parallel to them.
End-of-Chapter Exercises
Q1
Write a polynomial of degree 3 in variable x, where the coefficient of xยฒ is โ7.
Solution
1
A degree-3 polynomial has highest power xยณ. The coefficient of xยฒ must be โ7.
p(x) = xยณ โ 7xยฒ + 2x + 1
๐ Many answers are possible; any polynomial with degree 3 and coefficient of xยฒ equal to โ7 is correct.
Example: xยณ โ 7xยฒ + 2x + 1
Q2
Find the values of: (i) 5xยฒ โ 3x + 7 if x = 1 (ii) 4tยณ โ tยฒ + 6 if t = a
Solution
(i) 5xยฒ โ 3x + 7 at x = 1
= 5(1)ยฒ โ 3(1) + 7
= 5 โ 3 + 7
= 9
Value = 9
(ii) 4tยณ โ tยฒ + 6 at t = a
= 4(a)ยณ โ (a)ยฒ + 6
= 4aยณ โ aยฒ + 6
Value = 4aยณ โ aยฒ + 6
Q3
If we multiply a number by 5/2 and add 2/3 to the product, we get โ7/12. Find the number.
Solution
1
Let the number = n. Form the equation:
(5/2)n + 2/3 = โ7/12
2
Isolate the term with n:
(5/2)n = โ7/12 โ 2/3
= โ7/12 โ 8/12 [converting 2/3 = 8/12]
= โ15/12
= โ5/4
3
n = (โ5/4) รท (5/2)
= (โ5/4) ร (2/5)
= โ10/20
= โ1/2
Verify: (5/2)(โ1/2) + 2/3 = โ5/4 + 2/3 = โ15/12 + 8/12 = โ7/12 โ
The number = โ1/2
Q4
A positive number is 5 times another number. If 21 is added to both, one of the new numbers becomes twice the other. Find the numbers.
Solution
1
Let smaller number = n. Then larger = 5n.
2
After adding 21: n + 21 and 5n + 21.
3
The larger becomes twice the smaller:
5n + 21 = 2(n + 21)
5n + 21 = 2n + 42
3n = 21
n = 7
4
Numbers: 7 and 5 ร 7 = 35.
Verify: 35 + 21 = 56 = 2 ร (7 + 21) = 2 ร 28 = 56 โ
The two numbers are 7 and 35.
Q5
If you have โน800 and save โน250 every month, find the amount after (i) 6 months and (ii) 2 years. Express as a linear pattern.
Solution
1
Linear pattern: A(n) = 800 + 250n, where n = number of months.
(i) After 6 months
A(6) = 800 + 250(6) = 800 + 1500 = โน2,300
โน2,300
(ii) After 2 years = 24 months
A(24) = 800 + 250(24) = 800 + 6000 = โน6,800
โน6,800
Linear pattern: A(n) = 800 + 250n (rupees after n months)
โ
Q6
The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original, we get 143. Find both numbers.
Solution
1
Let tens digit = a, units digit = b, where a โ b = 3 (so a = b + 3).
2
Original number = 10a + b. Reversed = 10b + a.
3
(10a + b) + (10b + a) = 143
11a + 11b = 143
a + b = 13
4
Solve the system: a + b = 13 and a โ b = 3:
2a = 16 โ a = 8
b = 13 โ 8 = 5
5
Original number = 85, reversed = 58.
Verify: 85 + 58 = 143 โ | Digits differ by 8 โ 5 = 3 โ
The two numbers are 85 and 58.
โ
Q7
Draw graphs of (i) y = โ3x+4 (ii) 2y = 4x+7 (iii) 5y = 6xโ10 (iv) 3y = 6xโ11. Identify slopes and y-intercepts. Are any lines parallel?
Solution
1
Rewrite all equations in slope-intercept form y = ax + b:
| Equation | Slope-Intercept Form | Slope (a) | y-intercept (b) | y-axis point |
|---|---|---|---|---|
| (i) y = โ3x + 4 | y = โ3x + 4 | โ3 | 4 | (0, 4) |
| (ii) 2y = 4x + 7 | y = 2x + 3.5 | 2 | 3.5 | (0, 3.5) |
| (iii) 5y = 6x โ 10 | y = (6/5)x โ 2 | 6/5 | โ2 | (0, โ2) |
| (iv) 3y = 6x โ 11 | y = 2x โ 11/3 | 2 | โ11/3 | (0, โ11/3) |
2
Lines (ii) and (iv) both have slope = 2 but different y-intercepts (3.5 and โ11/3). Therefore they are parallel to each other.
Fig: Lines (ii) and (iv) shown as parallel (dashed). Both have slope = 2.
Lines (ii) y = 2x + 3.5 and (iv) y = 2x โ 11/3 are parallel (same slope = 2).
โ
Q8
y = (9/5)(x โ 273) + 32 relates Kelvin (x) to Fahrenheit (y). (i) Find ยฐF when x = 313 K. (ii) Find K when y = 158ยฐF.
Solution
(i) x = 313 K โ y = ?
y = (9/5)(313 โ 273) + 32
= (9/5)(40) + 32
= 72 + 32
= 104 ยฐF
Temperature = 104ยฐF
(ii) y = 158ยฐF โ x = ?
158 = (9/5)(x โ 273) + 32
158 โ 32 = (9/5)(x โ 273)
126 = (9/5)(x โ 273)
x โ 273 = 126 ร (5/9)
x โ 273 = 70
x = 343 K
Temperature = 343 K
โ
Q9
Work = force ร distance. Express in linear form with constant force = 3 units. What is work done when distance = 2 units?
Solution
1
Work w = F ร d. With F = 3:
w = 3d
2
When d = 2:
w = 3 ร 2 = 6 units
Fig: w = 3d passes through origin. Point (2, 6) is verified on the graph.
w = 3d | Work done when d = 2 is 6 units.
โ
Q10
Graph of linear polynomial p(x) passes through (1, 5) and (3, 11). (i) Find p(x). (ii) Find where it cuts the axes. (iii) Draw graph and verify.
Solution
1
Let p(x) = ax + b. Use the two given points:
At (1, 5): a(1) + b = 5 โ a + b = 5 ...(1)
At (3, 11): a(3) + b = 11 โ 3a + b = 11 ...(2)
2
Subtract (1) from (2):
2a = 6 โ a = 3
From (1): b = 5 โ 3 = 2
3
p(x) = 3x + 2
4
Where it cuts axes:
x-axis (y=0): 3x + 2 = 0 โ x = โ2/3 โ Point (โ2/3, 0)
y-axis (x=0): p(0) = 2 โ Point (0, 2)
Fig: p(x) = 3x + 2. Cuts x-axis at (โ2/3, 0) and y-axis at (0, 2).
p(x) = 3x + 2 | Cuts x-axis at (โ2/3, 0) | Cuts y-axis at (0, 2)
โ
Q11
p(x) = ax+b and q(x) = cx+d such that: (i) p(0)=5 (ii) p(x)โq(x) cuts x-axis at (3,0) (iii) p(x)+q(x) = 6x+4 for all x. Find p(x) and q(x).
Solution
1
From condition (i): p(0) = b = 5.
2
From condition (iii): p(x) + q(x) = (a+c)x + (b+d) = 6x + 4.
a + c = 6 ...(A)
b + d = 4 โ 5 + d = 4 โ d = โ1
3
From condition (ii): p(x)โq(x) = (aโc)x + (bโd) passes through (3, 0):
(aโc)(3) + (bโd) = 0
3(aโc) + (5โ(โ1)) = 0
3(aโc) + 6 = 0
a โ c = โ2 ...(B)
4
Solve (A) and (B):
From (A): a + c = 6
From (B): a โ c = โ2
Add: 2a = 4 โ a = 2
Then c = 6 โ 2 = 4
5
So p(x) = 2x + 5 and q(x) = 4x โ 1.
Verify: p(0)=5 โ | p(x)+q(x)=6x+4 โ | p(x)โq(x)=โ2x+6; at x=3: โ6+6=0 โ
p(x) = 2x + 5 | q(x) = 4x โ 1
โ
Q12
Growing pattern of hexagons made with matchsticks. A new hexagon sharing a side is added each stage. Find the rule for the nth stage.
Solution
Fig: Hexagon matchstick pattern โ each new hexagon shares one side (saves 1 matchstick).
1
Stage 1 (1 hexagon): 6 matchsticks.
Each new hexagon added shares 1 side, so requires 5 new matchsticks (not 6).
Each new hexagon added shares 1 side, so requires 5 new matchsticks (not 6).
2
| Stage (n) | 1 | 2 | 3 | 4 | 5 | n |
|---|---|---|---|---|---|---|
| Matchsticks | 6 | 11 | 16 | 21 | 26 | 5n + 1 |
3
Rule: Matchsticks at stage n = 5n + 1.
M(n) = 5n + 1
4
(iv) Stage 15:
M(15) = 5(15) + 1 = 75 + 1 = 76 matchsticks
5
(v) Can 200 matchsticks form a stage?
5n + 1 = 200
5n = 199
n = 39.8 (not a whole number)
Since n must be a positive integer, 200 matchsticks cannot form any stage of this pattern.
Rule: M(n) = 5n + 1 | Stage 15 โ 76 matchsticks | 200 matchsticks โ Not possible.
โ
Q13
p(x) passes through (2,3) and (6,11). q(x) passes through (4,โ1) and is parallel to p(x). Find p(x), q(x), and where they cut the x-axis.
Solution
1
Find p(x) = ax + b using (2,3) and (6,11):
3 = 2a + b ...(1)
11 = 6a + b ...(2)
Subtract: 8 = 4a โ a = 2
Then b = 3 โ 2(2) = โ1
2
p(x) = 2x โ 1
3
q(x) is parallel to p(x) โ same slope: q(x) = 2x + d.
Using point (4, โ1):
Using point (4, โ1):
โ1 = 2(4) + d
โ1 = 8 + d
d = โ9
4
q(x) = 2x โ 9
5
x-axis intercepts:
p(x) = 0: 2x โ 1 = 0 โ x = 1/2 โ Point (1/2, 0)
q(x) = 0: 2x โ 9 = 0 โ x = 9/2 โ Point (9/2, 0)
p(x) = 2xโ1 (cuts x-axis at (1/2, 0)) | q(x) = 2xโ9 (cuts x-axis at (9/2, 0))
โ
Q14
What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution
1
Rewrite: f(x) = ax + a = a(x + 1).
2
Find the x-intercept (where f(x) = 0):
a(x + 1) = 0
Since a โ 0 (a > 0), we get x + 1 = 0 โ x = โ1
3
All such lines pass through the point (โ1, 0), regardless of the value of a.
Fig: All lines f(x) = ax + a (for different values of a > 0) pass through the common point (โ1, 0).
All functions of the form f(x) = ax + a pass through the common point (โ1, 0).
Chapter Summary
POLYNOMIAL TYPES
Degree 0 โ Constant | Degree 1 โ Linear
Degree 2 โ Quadratic | Degree 3 โ Cubic
KEY FORMULAS
y = ax + b (slope a, y-intercept b)
Parallel lines: same slope a, different b
LINEAR GROWTH / DECAY
Growth: +constant per step โ positive slope
Decay: โconstant per step โ negative slope
GRAPH RULES
y = ax passes through origin (0,0)
y = ax + b cuts y-axis at (0, b)