Chapter 2: Introduction to Linear Polynomials

The degree of a polynomial is the highest power of the variable in the polynomial.

(i) 2x² – 5x + 3

• The terms are: 2x² (power 2), –5x (power 1), 3 (power 0)
• Highest power = 2
• Degree = 2

(ii) y³ + 2y – 1

• The terms are: y³ (power 3), 2y (power 1), –1 (power 0)
• Highest power = 3
• Degree = 3

(iii) –9

• This is a non-zero constant polynomial. It can be written as –9x⁰.
• Highest power = 0
• Degree = 0

(iv) 4z – 3

• The terms are: 4z (power 1), –3 (power 0)
• Highest power = 1
• Degree = 1

We need to write one example of a polynomial for each given degree.

• Degree 1 (Linear Polynomial):

  • Example: 3x + 5
  • The highest power of x is 1.

• Degree 2 (Quadratic Polynomial):

  • Example: x² – 4x + 4
  • The highest power of x is 2.

• Degree 3 (Cubic Polynomial):

  • Example: 2x³ + x² – x + 1
  • The highest power of x is 3.

Note: There are many possible correct answers. Any polynomial where the highest power matches the required degree is acceptable.

The given polynomial is: x⁴ – 3x³ + 6x² – 2x + 7

The coefficient of a term is the numerical factor multiplied with the variable part.

• Coefficient of x²:

  • The term containing x² is 6x².
  • Coefficient of x² = 6

• Coefficient of x³:

  • The term containing x³ is –3x³.
  • Coefficient of x³ = –3

Important: Include the sign of the coefficient. Here, the coefficient of x³ is –3 (negative), not 3.

The given polynomial is: 4z³ + 5z² – 11

We need to find the coefficient of z (i.e., the coefficient of z¹).

• Looking at the polynomial: 4z³ + 5z² + 0·z – 11
• There is no term containing z (z to the power 1) in this polynomial.
• Therefore, the coefficient of z = 0

The given polynomial is: 9x³ + 5x² – 8x – 10

The constant term is the term that does not contain any variable (i.e., the term with x⁰).

• Rewriting: 9x³ + 5x² – 8x + (–10)
• The term without any variable is –10.
• Constant term = –10

Note: The constant term is –10, not 10. Always include the sign.

We substitute each value of x into the polynomial p(x) = 5x – 3.

(i) x = 0 $p(0) = 5(0) - 3 = 0 - 3 = -3$

(ii) x = –1 $p(-1) = 5(-1) - 3 = -5 - 3 = -8$

(iii) x = 2 $p(2) = 5(2) - 3 = 10 - 3 = 7$

Summary:

• When x = 0, the value is –3
• When x = –1, the value is –8
• When x = 2, the value is 7

We substitute each value of s into the polynomial p(s) = 7s² – 4s + 6.

(i) s = 0 $p(0) = 7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$

(ii) s = –3 $p(-3) = 7(-3)^2 - 4(-3) + 6$ $= 7 \times 9 + 12 + 6$ $= 63 + 12 + 6 = 81$

(iii) s = 4 $p(4) = 7(4)^2 - 4(4) + 6$ $= 7 \times 16 - 16 + 6$ $= 112 - 16 + 6 = 102$

Summary:

• When s = 0, the value is 6
• When s = –3, the value is 81
• When s = 4, the value is 102

Let Salil's present age = x years

Then, Salil's mother's present age = 3x years

After 5 years:

• Salil's age = (x + 5) years
• Mother's age = (3x + 5) years

Setting up the equation: $(x + 5) + (3x + 5) = 70$ $4x + 10 = 70$ $4x = 70 - 10$ $4x = 60$ $x = 15$

Finding the ages:

• Salil's present age = x = 15 years
• Mother's present age = 3x = 3 × 15 = 45 years

Verification: After 5 years, Salil = 20 years, Mother = 50 years. Sum = 20 + 50 = 70 ✓

Let the two positive integers be in the ratio 2:5.

So, let the integers be 2k and 5k for some positive integer k.

Setting up the equation (larger – smaller = 63): $5k - 2k = 63$ $3k = 63$ $k = 21$

Finding the integers:

• Smaller integer = 2k = 2 × 21 = 42
• Larger integer = 5k = 5 × 21 = 105

Verification:

• Difference: 105 – 42 = 63 ✓
• Ratio: 42 : 105 = 2 : 5 ✓

Let the number of five-rupee coins = x

Then, the number of two-rupee coins = 3x

Total amount: $5 \times x + 2 \times 3x = 88$ $5x + 6x = 88$ $11x = 88$ $x = 8$

Finding the number of coins:

• Number of five-rupee coins = x = 8
• Number of two-rupee coins = 3x = 3 × 8 = 24

Verification: Total amount = 8 × 5 + 24 × 2 = 40 + 48 = ₹88 ✓

Let the length of the shorter piece = x feet

Then, the length of the longer piece = 4x feet

Setting up the equation (total length = 300 feet): $x + 4x = 300$ $5x = 300$ $x = 60$

Finding the lengths:

• Shorter piece = x = 60 feet
• Longer piece = 4x = 4 × 60 = 240 feet

Verification: 60 + 240 = 300 feet ✓ and 240 = 4 × 60 ✓

Let the width of the rectangle = w cm

Then, the length = (2w + 3) cm

Using the perimeter formula: $\text{Perimeter} = 2(\text{length} + \text{width})$ $24 = 2[(2w + 3) + w]$ $24 = 2[3w + 3]$ $24 = 6w + 6$ $6w = 24 - 6$ $6w = 18$ $w = 3$

Finding the dimensions:

• Width = w = 3 cm
• Length = 2w + 3 = 2(3) + 3 = 6 + 3 = 9 cm

Verification: Perimeter = 2(9 + 3) = 2 × 12 = 24 cm ✓

Given Information:

• Initial savings = ₹500
• Monthly pocket money = ₹150

Step-by-step working:

Pattern Observation:

• At the end of each month, ₹150 is added to the previous amount.
• The amount increases by a constant value of ₹150 each month → this is linear growth.

Linear Expression:

Amount at the end of the nth month: $A(n) = 500 + 150n$

where n is the month number (n = 1, 2, 3, ...).

Verification:

• n = 1: A(1) = 500 + 150(1) = ₹650 ✓
• n = 2: A(2) = 500 + 150(2) = ₹800 ✓
• n = 3: A(3) = 500 + 150(3) = ₹950 ✓

Given Information:

• Initial number of members = 120
• Members dropping out each hour = 9

Step-by-step working:

Pattern Observation:

• After each hour, the number of members decreases by a constant value of 9.
• This is an example of linear decay.

Linear Expression:

Number of members at the end of the nth hour: $M(n) = 120 - 9n$

where n is the number of hours elapsed (n = 1, 2, 3, ...).

Verification:

• n = 1: M(1) = 120 − 9(1) = 111 ✓
• n = 2: M(2) = 120 − 9(2) = 102 ✓
• n = 3: M(3) = 120 − 9(3) = 93 ✓

Given Information:

• Length of rectangle = 13 cm (fixed)
• Breadth varies: 12 cm, 10 cm, 8 cm
• Formula: Area = Length × Breadth

Calculations:

(i) Breadth = 12 cm: $\text{Area} = 13 \times 12 = 156 \text{ cm}^2$

(ii) Breadth = 10 cm: $\text{Area} = 13 \times 10 = 130 \text{ cm}^2$

(iii) Breadth = 8 cm: $\text{Area} = 13 \times 8 = 104 \text{ cm}^2$

Summary Table:

Pattern Observation:

• As the breadth decreases by 2 cm, the area decreases by 26 cm².
• The area changes by a constant amount for equal changes in breadth.

Linear Expression:

Let the breadth be b cm. Then: $A(b) = 13b$

This is a linear pattern where the area is directly proportional to the breadth, with the length (13 cm) as the constant multiplier.

Given Information:

• Length = 7 cm (fixed)
• Breadth = 11 cm (fixed)
• Height varies: 5 cm, 9 cm, 13 cm
• Formula: Volume = Length × Breadth × Height
• Base area = 7 × 11 = 77 cm² (constant)

Calculations:

(i) Height = 5 cm: $\text{Volume} = 7 \times 11 \times 5 = 77 \times 5 = 385 \text{ cm}^3$

(ii) Height = 9 cm: $\text{Volume} = 7 \times 11 \times 9 = 77 \times 9 = 693 \text{ cm}^3$

(iii) Height = 13 cm: $\text{Volume} = 7 \times 11 \times 13 = 77 \times 13 = 1001 \text{ cm}^3$

Summary Table:

Pattern Observation:

• Height increases by 4 cm each time (5 → 9 → 13).
• Volume increases by 308 cm³ each time (385 → 693 → 1001).
• The volume changes by a constant amount for equal changes in height → linear pattern.

Linear Expression:

Let the height be h cm. Then: $V(h) = 77h$

This is a linear pattern where the volume is directly proportional to the height, with 77 (= length × breadth) as the constant multiplier.

Given Information:

• Total pages in the book = 500
• Pages read per day = 20

Step-by-step working:

Pages left after d days: $P(d) = 500 - 20d$

Pages left after 15 days: $P(15) = 500 - 20 \times 15 = 500 - 300 = 200 \text{ pages}$

Verification Table:

Pattern Observation:

• After each day, the number of pages remaining decreases by a constant 20 pages.
• This is an example of linear decay.

Linear Expression: $P(d) = 500 - 20d$

where d is the number of days.

Answer: After 15 days, Sarita will have 200 pages left to read.

Note: The book will be completely read after $\frac{500}{20} = 25$ days.

Solution

Part (i): Height after 7 months

• Initial height = 1.75 feet
• Growth per month = 0.5 feet
• Height after 7 months = Initial height + (growth per month × number of months)

$h = 1.75 + 0.5 \times 7 = 1.75 + 3.5 = \mathbf{5.25 \text{ feet}}$

Part (ii): Table of values (t = 0 to 10 months)

The height increases by a constant 0.5 feet every month.

Part (iii): Expression relating h and t

$\boxed{h = 0.5t + 1.75}$

Why it represents linear growth:

• This is of the form h = at + b, where a = 0.5 (rate of growth) and b = 1.75 (initial height).
• The height increases by an equal amount (0.5 feet) for every unit increase in time.
• Since the rate of change is constant, the relationship is linear.
• The graph of h vs t will be a straight line with positive slope, confirming linear growth.

Solution

Part (i): Value after 3 years

• Initial value = ₹10,000
• Depreciation per year = ₹800
• Value after 3 years = Initial value − (depreciation per year × number of years)

$v = 10000 - 800 \times 3 = 10000 - 2400 = \mathbf{₹7600}$

Part (ii): Table of values (t = 0 to 8 years)

The value decreases by a constant ₹800 every year.

Part (iii): Expression relating v and t

$\boxed{v = 10000 - 800t}$

This can also be written as: v = −800t + 10000

Why it represents linear decay:

• This is of the form v = at + b, where a = −800 (rate of depreciation) and b = 10,000 (initial value).
• The value decreases by an equal amount (₹800) for every unit increase in time.
• Since the rate of change is constant and negative, the relationship is linear decay.
• The graph of v vs t will be a straight line with negative slope, confirming linear decay.

Solution

Part (i): Population after 6 years

• Initial population = 750
• Increase per year = 50 people
• Population after 6 years = Initial population + (increase per year × number of years)

$P = 750 + 50 \times 6 = 750 + 300 = \mathbf{1050}$

Part (ii): Table of values (t = 0 to 10 years)

The population increases by a constant 50 people every year.

Part (iii): Expression relating P and t

$\boxed{P = 50t + 750}$

Why it represents linear growth:

• This is of the form P = at + b, where a = 50 (rate of population increase) and b = 750 (initial population).
• The population increases by an equal amount (50 people) every year.
• Since the rate of change is constant and positive, the relationship is linear growth.
• The graph of P vs t is a straight line with positive slope, confirming linear growth.

Solution

Part (i): Equation for remaining balance b(x)

• Initial balance = ₹600
• Reduction per day = ₹15

$\boxed{b(x) = 600 - 15x}$

Why it represents linear decay:

• This is of the form b(x) = −15x + 600, which matches y = ax + b with a = −15 and b = 600.
• The balance decreases by an equal amount (₹15) for every day that passes.
• Since the rate of change is constant and negative, the relationship is linear decay.
• The graph of b(x) vs x is a straight line with negative slope.

Part (ii): Number of days for balance to run out

The balance runs out when b(x) = 0:

$600 - 15x = 0$ $15x = 600$ $x = \frac{600}{15} = \mathbf{40 \text{ days}}$

The balance will run out after 40 days.

Part (iii): Table of values (x = 1 to 10 days)

The balance decreases by a constant ₹15 every day.

Setting Up the Equations

We are given that y = ax + b, where:

• y = monthly bill (in ₹)
• x = number of modules accessed
• a = cost per module
• b = fixed monthly fee

Using the two given conditions:

Condition 1: When x = 10, y = 400 $10a + b = 400 \quad \text{...(i)}$

Condition 2: When x = 14, y = 500 $14a + b = 500 \quad \text{...(ii)}$

Solving the System of Equations

Subtract equation (i) from equation (ii): $14a + b - (10a + b) = 500 - 400$ $14a + b - 10a - b = 100$ $4a = 100$ $\boxed{a = 25}$

Substitute a = 25 back into equation (i): $10(25) + b = 400$ $250 + b = 400$ $b = 400 - 250$ $\boxed{b = 150}$

Verification

• When x = 10: y = 25(10) + 150 = 250 + 150 = ₹400 ✓
• When x = 14: y = 25(14) + 150 = 350 + 150 = ₹500 ✓

Conclusion

a = 25 (cost per module = ₹25)

b = 150 (fixed monthly fee = ₹150)

The linear relationship is: y = 25x + 150

Setting Up the Equations

We are given that y = ax + b, where:

• y = monthly bill (in ₹)
• x = hours of badminton court usage
• a = cost per hour
• b = fixed monthly fee

Using the two given conditions:

Condition 1: When x = 10, y = 800 $10a + b = 800 \quad \text{...(i)}$

Condition 2: When x = 15, y = 1100 $15a + b = 1100 \quad \text{...(ii)}$

Solving the System of Equations

Subtract equation (i) from equation (ii): $15a + b - (10a + b) = 1100 - 800$ $15a + b - 10a - b = 300$ $5a = 300$ $\boxed{a = 60}$

Substitute a = 60 back into equation (i): $10(60) + b = 800$ $600 + b = 800$ $b = 800 - 600$ $\boxed{b = 200}$

Verification

• When x = 10: y = 60(10) + 200 = 600 + 200 = ₹800 ✓
• When x = 15: y = 60(15) + 200 = 900 + 200 = ₹1100 ✓

Conclusion

a = 60 (cost per hour of badminton court = ₹60)

b = 200 (fixed monthly fee = ₹200)

The linear relationship is: y = 60x + 200

Setting Up the Equations

We are given that °C = a·°F + b

Using the two given conditions:

Condition 1: Ice melts at 0°C and 32°F $0 = a(32) + b$ $32a + b = 0 \quad \text{...(i)}$

Condition 2: Water boils at 100°C and 212°F $100 = a(212) + b$ $212a + b = 100 \quad \text{...(ii)}$

Solving the System of Equations

Subtract equation (i) from equation (ii): $212a + b - (32a + b) = 100 - 0$ $212a + b - 32a - b = 100$ $180a = 100$ $a = \frac{100}{180} = \frac{5}{9}$ $\boxed{a = \frac{5}{9}}$

Substitute a = 5/9 back into equation (i): $32 \times \frac{5}{9} + b = 0$ $\frac{160}{9} + b = 0$ $b = -\frac{160}{9}$ $\boxed{b = -\frac{160}{9}}$

Verification

• When °F = 32: $\text{°C} = \frac{5}{9}(32) - \frac{160}{9} = \frac{160}{9} - \frac{160}{9} = \mathbf{0°C}$ ✓

• When °F = 212: $\text{°C} = \frac{5}{9}(212) - \frac{160}{9} = \frac{1060}{9} - \frac{160}{9} = \frac{900}{9} = \mathbf{100°C}$ ✓

Conclusion

$a = \frac{5}{9}, \quad b = -\frac{160}{9}$

The complete relationship between Celsius and Fahrenheit is: $°C = \frac{5}{9} \times °F - \frac{160}{9}$

This can also be written as the well-known formula: $°C = \frac{5}{9}(°F - 32)$

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

Step 1: Prepare a table of values for each line.

All three lines pass through the origin (0, 0) since they are of the form y = ax.

Step 2: Plot the points and draw the lines.

• y = x: Plot (0,0), (1,1), (2,2). This line makes a 45° angle with the x-axis.
• y = 2x: Plot (0,0), (1,2), (2,4). This line is steeper than y = x.
• y = 4x: Plot (0,0), (1,4), (2,8). This line is the steepest among the three.

Step 3: Reflect on the role of 'a'.

• All three lines pass through the origin (0, 0).
• The value of 'a' is the slope of the line — it tells us how steep the line is.
• Larger the value of 'a', the steeper the line (it rises more quickly as x increases).
• As 'a' increases from 1 → 2 → 4, the line rotates closer to the y-axis (becomes more vertical).
• Conclusion: In y = ax, 'a' controls the steepness (slope) of the line. All lines with different values of 'a' but no constant term pass through the origin.

Graphing y = –6x, y = –3x, y = –x

Step 1: Prepare a table of values for each line.

Step 2: Plot the points and draw the lines.

• y = –x: Plot (0,0), (1,–1), (–1,1). Line goes from top-left to bottom-right.
• y = –3x: Plot (0,0), (1,–3), (–1,3). Steeper than y = –x.
• y = –6x: Plot (0,0), (1,–6), (–1,6). The steepest of the three.

Step 3: Reflect on the role of 'a'.

• All three lines pass through the origin (0, 0).
• The negative sign of 'a' means the lines have a negative slope — they go downward from left to right.
• As the magnitude of 'a' increases (1 → 3 → 6), the line becomes steeper in the negative direction.
• Conclusion: When 'a' is negative in y = ax, the line slopes downward (falls as x increases). The larger the magnitude of 'a', the steeper the downward slope.

Graphing y = 5x and y = –5x

Step 1: Prepare a table of values.

Step 2: Plot the points and draw the lines.

• y = 5x: Plot (0,0), (1,5), (–1,–5). Line rises steeply from left to right.
• y = –5x: Plot (0,0), (1,–5), (–1,5). Line falls steeply from left to right.

Step 3: Reflect on the role of 'a'.

• Both lines pass through the origin (0, 0) and have the same steepness (magnitude of slope = 5).
• y = 5x has a positive slope → rises from left to right.
• y = –5x has a negative slope → falls from left to right.
• The two lines are mirror images of each other with respect to the x-axis (or equivalently, they are reflections across the y-axis).
• Conclusion: The sign of 'a' determines the direction of the line. Positive 'a' → upward slope; Negative 'a' → downward slope. Lines y = ax and y = –ax are symmetric about the x-axis.

Graphing y = 3x – 1, y = 3x, y = 3x + 1

Step 1: Prepare a table of values for each line.

Step 2: Plot the points and draw the lines.

• y = 3x – 1: Crosses y-axis at (0, –1). Slopes upward with slope 3.
• y = 3x: Passes through origin (0, 0). Slopes upward with slope 3.
• y = 3x + 1: Crosses y-axis at (0, 1). Slopes upward with slope 3.

Step 3: Reflect on the role of 'b'.

• All three lines have the same slope (3), so they are parallel to each other.
• The value of 'b' is the y-intercept — it tells us where the line crosses the y-axis.
  • b = –1 → line crosses y-axis below the origin.
  • b = 0 → line passes through the origin.
  • b = 1 → line crosses y-axis above the origin.
• Changing 'b' shifts the line up or down without changing its direction or steepness.
• Conclusion: In y = ax + b, 'b' controls the vertical position of the line (y-intercept). All lines with the same 'a' but different 'b' are parallel.

Graphing y = –2x – 3, y = –2x, y = –2x + 3

Step 1: Prepare a table of values for each line.

Step 2: Plot the points and draw the lines.

• y = –2x – 3: Crosses y-axis at (0, –3). Slopes downward with slope –2.
• y = –2x: Passes through origin (0, 0). Slopes downward with slope –2.
• y = –2x + 3: Crosses y-axis at (0, 3). Slopes downward with slope –2.

Step 3: Reflect on the role of 'b'.

• All three lines have the same slope (–2), so they are parallel to each other.
• All three lines slope downward from left to right (negative slope).
• The value of 'b' is the y-intercept:
  • b = –3 → line is shifted 3 units downward from y = –2x.
  • b = 0 → line passes through the origin.
  • b = 3 → line is shifted 3 units upward from y = –2x.
• Conclusion: In y = ax + b, 'b' shifts the line vertically without affecting its slope. Positive 'b' shifts the line up; negative 'b' shifts it down. Lines with the same 'a' are always parallel, regardless of 'b'.

A polynomial of degree 3 in x has the general form:

$p(x) = ax^3 + bx^2 + cx + d$

where $a \neq 0$.

Given condition: The coefficient of $x^2$ must be –7, so $b = -7$.

Example answer:

$p(x) = 2x^3 - 7x^2 + 3x + 1$

• Degree = 3 ✓
• Coefficient of $x^2$ = –7 ✓

Note: Many correct answers are possible. Any polynomial of the form $ax^3 - 7x^2 + cx + d$ (with $a \neq 0$) is valid.

Let $p(x) = ax + b$.

(i) Finding p(x):

Since the graph passes through (1, 5): $a(1) + b = 5 \implies a + b = 5 \quad \text{...(1)}$

Since the graph passes through (3, 11): $a(3) + b = 11 \implies 3a + b = 11 \quad \text{...(2)}$

Subtract (1) from (2): $2a = 6 \implies a = 3$

Substitute in (1): $3 + b = 5 \implies b = 2$

$\boxed{p(x) = 3x + 2}$

(ii) Where does the graph cut the axes?

Cuts y-axis (set x = 0): $p(0) = 3(0) + 2 = 2$ → y-axis at (0, 2)

Cuts x-axis (set p(x) = 0): $3x + 2 = 0 \implies x = -\frac{2}{3}$ → x-axis at $\left(-\dfrac{2}{3}, 0\right)$

(iii) Verification by graph:

Plot points: (0, 2), (1, 5), (3, 11), (–2/3, 0).

Draw the straight line through these points.

• The line passes through (1, 5): $3(1) + 2 = 5$ ✓
• The line passes through (3, 11): $3(3) + 2 = 11$ ✓

Given:

• $p(x) = ax + b$, $q(x) = cx + d$

Step 1: Use condition (i) — p(0) = 5:

$p(0) = a(0) + b = b = 5$

So $b = 5$, and $p(x) = ax + 5$.

Step 2: Use condition (iii) — p(x) + q(x) = 6x + 4:

$(ax + 5) + (cx + d) = 6x + 4$ $(a + c)x + (5 + d) = 6x + 4$

Comparing coefficients:

• $a + c = 6 \quad \text{...(1)}$
• $5 + d = 4 \implies d = -1 \quad \text{...(2)}$

Step 3: Use condition (ii) — p(x) – q(x) cuts x-axis at (3, 0):

$p(x) - q(x) = (ax + 5) - (cx - 1) = (a - c)x + 6$

At x = 3, this equals 0: $(a - c)(3) + 6 = 0$ $3(a - c) = -6$ $a - c = -2 \quad \text{...(3)}$

Step 4: Solve equations (1) and (3):

Adding (1) and (3): $2a = 4 \implies a = 2$

From (1): $c = 6 - 2 = 4$

Result:

$\boxed{p(x) = 2x + 5}$ $\boxed{q(x) = 4x - 1}$

Verification:

• $p(0) = 5$ ✓
• $p(x) + q(x) = 6x + 4$ ✓
• $p(x) - q(x) = -2x + 6$; at $x = 3$: $-6 + 6 = 0$ ✓

Understanding the pattern:

• Stage 1: 1 hexagon = 6 matchsticks
• Stage 2: 2 hexagons sharing one side → 6 + 5 = 11 matchsticks
• Stage 3: 3 hexagons → 11 + 5 = 16 matchsticks

Each new hexagon adds 5 matchsticks (since 1 side is shared with the previous hexagon).

(i) Stages 4 and 5:

• Stage 4: 16 + 5 = 21 matchsticks
• Stage 5: 21 + 5 = 26 matchsticks

(Draw 4 and 5 hexagons in a row, each sharing one side with the next.)

(ii) Complete Table:

(iii) Rule for nth stage:

The number of matchsticks at stage $n$:

• Stage 1: 6 = 5(1) + 1
• Stage 2: 11 = 5(2) + 1
• Stage 3: 16 = 5(3) + 1

$\boxed{M(n) = 5n + 1}$

This is a linear polynomial in $n$.

(iv) Matchsticks at Stage 15:

$M(15) = 5(15) + 1 = 75 + 1 = \boxed{76 \text{ matchsticks}}$

(v) Can 200 matchsticks form a stage?

We need to check if $5n + 1 = 200$ has a positive integer solution:

$5n = 199$ $n = \frac{199}{5} = 39.8$

Since $n$ is not a whole number, 200 matchsticks cannot form a stage in this pattern.

Justification: Every stage requires $5n + 1$ matchsticks. Since 200 – 1 = 199 is not divisible by 5, no stage corresponds to 200 matchsticks.

Finding p(x):

Let $p(x) = ax + b$.

Using point (2, 3): $2a + b = 3 \quad \text{...(1)}$

Using point (6, 11): $6a + b = 11 \quad \text{...(2)}$

Subtract (1) from (2): $4a = 8 \implies a = 2$

From (1): $b = 3 - 4 = -1$

$\boxed{p(x) = 2x - 1}$

Finding q(x):

Since $q(x)$ is parallel to $p(x)$, they have the same slope: $c = 2 \implies q(x) = 2x + d$

Using point (4, –1): $2(4) + d = -1$ $8 + d = -1$ $d = -9$

$\boxed{q(x) = 2x - 9}$

Where do the lines meet the x-axis?

For p(x): Set $p(x) = 0$: $2x - 1 = 0 \implies x = \frac{1}{2}$ → p(x) meets x-axis at $\left(\dfrac{1}{2}, 0\right)$

For q(x): Set $q(x) = 0$: $2x - 9 = 0 \implies x = \frac{9}{2}$ → q(x) meets x-axis at $\left(\dfrac{9}{2}, 0\right)$

Verification:

• $p(2) = 4 - 1 = 3$ ✓
• $p(6) = 12 - 1 = 11$ ✓
• $q(4) = 8 - 9 = -1$ ✓
• Both have slope 2, so lines are parallel ✓

Analysing f(x) = ax + a:

We can factor this as:

$f(x) = a(x + 1)$

Finding the x-intercept (set f(x) = 0):

$a(x + 1) = 0$

Since $a > 0$, $a \neq 0$, so:

$x + 1 = 0 \implies x = -1$

The x-intercept is always (-1, 0), regardless of the value of $a$.

Key Properties:

Conclusion:

All linear functions of the form $f(x) = ax + a$ (with $a > 0$) pass through the common point (–1, 0).

In other words, every such line intersects the x-axis at x = –1, regardless of the value of the positive constant $a$.

Geometrically: All these lines form a family of lines passing through the fixed point (–1, 0), with different positive slopes depending on the value of $a$.

(i) p(x) = 5x² – 3x + 7 at x = 1

Substitute $x = 1$:

$p(1) = 5(1)^2 - 3(1) + 7$ $= 5(1) - 3 + 7$ $= 5 - 3 + 7$ $= \boxed{9}$

(ii) p(t) = 4t³ – t² + 6 at t = a

Substitute $t = a$:

$p(a) = 4(a)^3 - (a)^2 + 6$ $= \boxed{4a^3 - a^2 + 6}$

Let the unknown number be x.

Setting up the equation:

$\frac{5}{2} \cdot x + \frac{2}{3} = -\frac{7}{12}$

Solving for x:

Subtract $\dfrac{2}{3}$ from both sides:

$\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3}$

Convert $\dfrac{2}{3}$ to twelfths: $\dfrac{2}{3} = \dfrac{8}{12}$

$\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4}$

Divide both sides by $\dfrac{5}{2}$ (i.e., multiply by $\dfrac{2}{5}$):

$x = -\frac{5}{4} \times \frac{2}{5} = -\frac{10}{20} = -\frac{1}{2}$

Verification: $\frac{5}{2} \times \left(-\frac{1}{2}\right) + \frac{2}{3} = -\frac{5}{4} + \frac{2}{3} = -\frac{15}{12} + \frac{8}{12} = -\frac{7}{12} \checkmark$

$\boxed{x = -\dfrac{1}{2}}$

Let the smaller number be $x$.

Then the larger number = $5x$.

After adding 21 to both:

• Smaller number becomes: $x + 21$
• Larger number becomes: $5x + 21$

Condition: One of the new numbers is twice the other.

Since the larger number was originally 5 times the smaller, after adding 21 the larger should be twice the smaller:

$5x + 21 = 2(x + 21)$

$5x + 21 = 2x + 42$

$5x - 2x = 42 - 21$

$3x = 21$

$x = 7$

So the smaller number = 7 and the larger number = 5 × 7 = 35.

Verification:

• After adding 21: smaller = 7 + 21 = 28, larger = 35 + 21 = 56
• Is 56 = 2 × 28? → 56 = 56 ✓

$\boxed{\text{The numbers are } 7 \text{ and } 35.}$

Linear Pattern:

Let $n$ = number of months.

The amount after $n$ months is:

$A(n) = 800 + 250n$

This is a linear polynomial in $n$ with:

• Initial amount (y-intercept) = ₹800
• Rate of saving (slope) = ₹250 per month

(i) After 6 months ($n = 6$):

$A(6) = 800 + 250 \times 6 = 800 + 1500 = \boxed{₹2300}$

(ii) After 2 years ($n = 24$ months):

$A(24) = 800 + 250 \times 24 = 800 + 6000 = \boxed{₹6800}$

Linear Pattern Table:

Let the tens digit be $x$ and the units digit be $y$.

Original number = $10x + y$

Interchanged number = $10y + x$

Condition 1: Digits differ by 3: $x - y = 3 \quad \text{...(1)}$

Condition 2: Sum of original and interchanged number = 143: $(10x + y) + (10y + x) = 143$ $11x + 11y = 143$ $x + y = 13 \quad \text{...(2)}$

Solving equations (1) and (2):

Adding (1) and (2): $2x = 16 \implies x = 8$

Substituting in (2): $8 + y = 13 \implies y = 5$

The original number = $10(8) + 5 = \mathbf{85}$

The interchanged number = $10(5) + 8 = \mathbf{58}$

Verification:

• Digits differ by: 8 – 5 = 3 ✓
• Sum: 85 + 58 = 143 ✓

$\boxed{\text{The two numbers are } 85 \text{ and } 58.}$

Convert each equation to the standard form y = mx + c (slope-intercept form):

(i) y = –3x + 4

Already in standard form.

• Slope (m) = –3
• y-intercept (c) = 4
• Cuts y-axis at: (0, 4)

Plot points: At x = 0, y = 4; At x = 1, y = 1; At x = 2, y = –2.

(ii) 2y = 4x + 7

$y = 2x + \frac{7}{2} = 2x + 3.5$

• Slope (m) = 2
• y-intercept (c) = 3.5
• Cuts y-axis at: (0, 3.5)

Plot points: At x = 0, y = 3.5; At x = 1, y = 5.5; At x = –1, y = 1.5.

(iii) 5y = 6x – 10

$y = \frac{6}{5}x - 2 = 1.2x - 2$

• Slope (m) = 6/5
• y-intercept (c) = –2
• Cuts y-axis at: (0, –2)

Plot points: At x = 0, y = –2; At x = 5, y = 4; At x = –5, y = –8.

(iv) 3y = 6x – 11

$y = 2x - \frac{11}{3}$

• Slope (m) = 2
• y-intercept (c) = –11/3 ≈ –3.67
• Cuts y-axis at: (0, –11/3)

Plot points: At x = 0, y = –11/3; At x = 2, y = 4 – 11/3 = 1/3; At x = –1, y = –2 – 11/3 = –17/3.

Summary Table:

Parallel Lines:

Lines (ii) and (iv) both have slope = 2 but different y-intercepts (7/2 and –11/3), so they are parallel to each other.

For graphing: Plot the y-intercept on the y-axis, then use the slope to find a second point, and draw the line through both points.

Given equation: $y = \dfrac{9}{5}(x - 273) + 32$

(i) Find y when x = 313 K:

$y = \frac{9}{5}(313 - 273) + 32$

$= \frac{9}{5}(40) + 32$

$= \frac{360}{5} + 32$

$= 72 + 32$

$= \boxed{104 °F}$

(ii) Find x when y = 158 °F:

$158 = \frac{9}{5}(x - 273) + 32$

Subtract 32 from both sides:

$126 = \frac{9}{5}(x - 273)$

Multiply both sides by $\dfrac{5}{9}$:

$x - 273 = 126 \times \frac{5}{9} = \frac{630}{9} = 70$

$x = 273 + 70 = \boxed{343 K}$

Concept: Work = Force × Distance

Linear Equation (with constant force = 3 units):

$w = 3d$

This is a linear equation in two variables $w$ and $d$.

• Slope = 3
• y-intercept = 0 (line passes through the origin)

Table of values for graphing:

Work done when d = 2 units:

$w = 3 \times 2 = \boxed{6 \text{ units}}$

Verification from graph: Plot the point (2, 6) on the graph of $w = 3d$. It lies on the straight line, confirming the answer.

Graph: Plot the points (0, 0), (1, 3), (2, 6), (3, 9) on the d-w plane and draw a straight line through them. The point (2, 6) lies on this line.