Secondary 4 Additional Mathematics Preliminary Examination Paper 2

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TuitionGoWhere Practice Paper — Additional Mathematics Secondary 4

Preliminary Examination — Version 2

TuitionGoWhere Secondary School (AI)

Subject: Additional Mathematics (4049)
Level: Secondary 4
Paper: Prelim — Graphs & Coordinate Geometry
Duration: 1 hour 30 minutes
Total Marks: 80

Name: _______________________________
Class: _______________________________
Date: _______________________________

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Instructions to Candidates

1. This paper consists of 20 questions in four sections.
2. Answer all questions.
3. Write your answers in the spaces provided.
4. All working must be clearly shown. Marks are awarded for method, not only for the final answer.
5. Solutions by accurate drawing will not be accepted unless otherwise stated.
6. You are expected to use an approved scientific calculator.
7. Unless otherwise stated, give non-exact answers correct to 3 significant figures.
8. The number of marks is given in brackets [ ] at the end of each question or part question.
9. The total mark for this paper is 80.

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Formula Sheet

Quadratic Equation: For $ax^2 + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Coordinate Geometry:

• Gradient of line through $(x_1, y_1)$ and $(x_2, y_2)$: $m = \frac{y_2 - y_1}{x_2 - x_1}$
• Midpoint: $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
• Distance: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• Parallel lines: $m_1 = m_2$
• Perpendicular lines: $m_1 \cdot m_2 = -1$

Circle:

• Standard form: $(x - a)^2 + (y - b)^2 = r^2$, centre $(a, b)$, radius $r$
• General form: $x^2 + y^2 + 2gx + 2fy + c = 0$, centre $(-g, -f)$, radius $\sqrt{g^2 + f^2 - c}$

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Section A: Straight Lines and Linear Graphs (20 marks)

Answer all questions in this section.

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1. The points $A(2, 5)$ and $B(8, -3)$ lie on a straight line.

(a) Find the gradient of the line $AB$. [1]

(b) Find the equation of the line $AB$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [2]

(c) The line $AB$ meets the $x$-axis at point $C$. Find the coordinates of $C$. [1]

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2. A line $L_1$ passes through the point $P(3, -1)$ and has gradient $\frac{2}{5}$.

(a) Find the equation of $L_1$ in the form $y = mx + c$. [2]

(b) Another line $L_2$ is perpendicular to $L_1$ and passes through the point $Q(-4, 6)$. Find the equation of $L_2$. [3]

(c) Find the coordinates of the point of intersection of $L_1$ and $L_2$. [3]

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3. The points $D(-1, 4)$, $E(3, 10)$ and $F(7, 2)$ are three vertices of a triangle.

(a) Find the midpoint of $DE$. [1]

(b) Show that $DE$ is perpendicular to $DF$. [3]

(c) Hence, or otherwise, find the area of triangle $DEF$. [4]

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Section B: Quadratic Curves and Parabolas (20 marks)

Answer all questions in this section.

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4. The curve $C$ has equation $y = 2x^2 - 8x + 11$.

(a) Express $2x^2 - 8x + 11$ in the form $a(x - h)^2 + k$, where $a$, $h$ and $k$ are constants. [2]

(b) Hence, write down the coordinates of the minimum point of $C$. [1]

(c) State the equation of the line of symmetry of $C$. [1]

(d) Find the set of values of $x$ for which $y \geq 5$. [3]

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5. A parabola $P$ has equation $y = x^2 - 6x + 5$.

(a) Find the coordinates of the points where $P$ crosses the $x$-axis. [2]

(b) Find the coordinates of the point where $P$ crosses the $y$-axis. [1]

(c) The line $y = 2x - 7$ intersects $P$ at two points. By forming and solving a quadratic equation, find the coordinates of these two intersection points. [4]

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6. The quadratic function $f(x) = -x^2 + 4x + k$ has a maximum value of 9.

(a) Express $-x^2 + 4x + k$ in the form $-(x - p)^2 + q$, giving $p$ and $q$ in terms of $k$. [3]

(b) Hence, find the value of $k$. [2]

(c) For this value of $k$, find the range of values of $x$ for which $f(x) \leq 0$. [3]

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Section C: Circles (20 marks)

Answer all questions in this section.

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7. A circle $C_1$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

(a) Find the coordinates of the centre of $C_1$ and the radius of $C_1$. [3]

(b) The point $A(7, 1)$ lies on $C_1$. Find the equation of the tangent to $C_1$ at $A$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [4]

(c) Another circle $C_2$ has centre $B(-1, 5)$ and touches $C_1$ externally. Find the equation of $C_2$. [4]

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8. The points $P(2, 1)$ and $Q(10, 7)$ are the endpoints of a diameter of a circle.

(a) Find the coordinates of the centre of the circle. [1]

(b) Find the radius of the circle, leaving your answer in simplified surd form. [2]

(c) Hence, write down the equation of the circle in standard form. [2]

(d) Determine whether the point $R(6, 9)$ lies inside, on, or outside the circle. Show your working clearly. [3]

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9. A circle passes through the points $S(0, 0)$, $T(8, 0)$ and $U(4, 6)$.

(a) By considering the perpendicular bisector of $ST$, explain why the $x$-coordinate of the centre of the circle must be 4. [2]

(b) Let the centre of the circle be $(4, k)$. Using the fact that the circle passes through $S$, find an expression for the radius squared in terms of $k$. [2]

(c) Using the fact that the circle also passes through $U$, find the value of $k$. [3]

(d) Hence, write down the equation of the circle. [2]

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Section D: Linearisation and Applications (20 marks)

Answer all questions in this section.

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10. Two variables $x$ and $y$ are related by the equation $y = ax^n$, where $a$ and $n$ are constants. The table below shows experimental values of $x$ and $y$.

$x$	1.5	2.0	2.5	3.0	3.5
$y$	4.24	8.00	13.0	19.6	27.7

(a) Explain why plotting $\log y$ against $\log x$ will produce a straight line graph. [2]

(b) Complete the following table of values for $\log x$ and $\log y$, giving each value correct to 3 decimal places. [2]

$\log x$	0.176	0.301	0.398	0.477	0.544
$\log y$		0.903	1.114	1.292

(c) Plot the points on the grid below and draw the best-fit straight line. [2]

(Grid provided — 2 marks for correct plotting and line)

(d) Use your graph to estimate the values of $a$ and $n$. [4]

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11. A scientist models the growth of a bacterial population using the equation $P = kb^t$, where $P$ is the population after $t$ hours, and $k$ and $b$ are constants.

The following data is recorded:

$t$ (hours)	0	1	2	3	4
$P$	200	280	392	549	768

(a) By taking logarithms, show that plotting $\log P$ against $t$ will produce a straight line. State the gradient and vertical intercept of this line in terms of $k$ and $b$. [3]

(b) Using the data for $t = 0$ and $t = 4$, find the values of $k$ and $b$. [4]

(c) Hence, estimate the population when $t = 6$. [2]

(d) Find the time taken for the population to reach 2000, giving your answer correct to 1 decimal place. [3]

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12. The variables $x$ and $y$ are connected by the equation $y = \frac{p}{x} + qx$, where $p$ and $q$ are constants.

(a) Explain how the equation can be rearranged so that a graph of $xy$ against $x^2$ can be used to find the values of $p$ and $q$. State what the gradient and vertical intercept of this graph represent. [3]

(b) The table below shows values of $x$ and $y$ obtained from an experiment.

$x$	1.0	1.5	2.0	2.5	3.0
$y$	5.0	3.8	3.5	3.4	3.5

Complete the following table, giving values correct to 2 decimal places where appropriate. [2]

$x^2$	1.00	2.25	4.00	6.25	9.00
$xy$

(c) Plot $xy$ against $x^2$ on the grid below and draw the best-fit straight line. [2]

(Grid provided — 2 marks for correct plotting and line)

(d) Use your graph to estimate the values of $p$ and $q$. [3]

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— End of Paper —

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TuitionGoWhere Practice Paper — Additional Mathematics Secondary 4

Preliminary Examination — Version 2: ANSWER KEY AND MARKING SCHEME

TuitionGoWhere Secondary School (AI)

Subject: Additional Mathematics (4049)
Level: Secondary 4
Paper: Prelim — Graphs & Coordinate Geometry
Total Marks: 80

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Section A: Straight Lines and Linear Graphs (20 marks)

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Question 1

(a) Gradient of $AB$ [1 mark]

$m = \frac{-3 - 5}{8 - 2} = \frac{-8}{6} = -\frac{4}{3}$

Answer: $-\frac{4}{3}$ ✓ [A1]

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(b) Equation of $AB$ [2 marks]

Using point $A(2, 5)$ and $m = -\frac{4}{3}$: $y - 5 = -\frac{4}{3}(x - 2)$ [M1 — correct substitution into point-gradient form]

$3(y - 5) = -4(x - 2)$ $3y - 15 = -4x + 8$ $4x + 3y - 23 = 0$ [A1 — correct simplified integer form]

Answer: $4x + 3y - 23 = 0$

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(c) Coordinates of $C$ (where $AB$ meets $x$-axis) [1 mark]

At $x$-axis, $y = 0$: $4x + 3(0) - 23 = 0$ $4x = 23$ $x = \frac{23}{4} = 5.75$ [A1]

Answer: $C(5.75, 0)$ or $C\left(\frac{23}{4}, 0\right)$

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Question 2

(a) Equation of $L_1$ [2 marks]

Using $P(3, -1)$ and $m = \frac{2}{5}$: $y - (-1) = \frac{2}{5}(x - 3)$ [M1] $y + 1 = \frac{2}{5}x - \frac{6}{5}$ $y = \frac{2}{5}x - \frac{11}{5}$ [A1]

Answer: $y = \frac{2}{5}x - \frac{11}{5}$

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(b) Equation of $L_2$ [3 marks]

Gradient of $L_2$: $m_2 = -\frac{1}{m_1} = -\frac{5}{2}$ [M1 — correct perpendicular gradient]

Using $Q(-4, 6)$: $y - 6 = -\frac{5}{2}(x - (-4))$ [M1 — correct substitution] $y - 6 = -\frac{5}{2}(x + 4)$ $y - 6 = -\frac{5}{2}x - 10$ $y = -\frac{5}{2}x - 4$ [A1]

Answer: $y = -\frac{5}{2}x - 4$

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(c) Intersection of $L_1$ and $L_2$ [3 marks]

Solve simultaneously: $\frac{2}{5}x - \frac{11}{5} = -\frac{5}{2}x - 4$ [M1 — equating $y$ values]

Multiply by 10: $4x - 22 = -25x - 40$ $29x = -18$ [M1 — correct algebra] $x = -\frac{18}{29}$

Substitute into $L_1$: $y = \frac{2}{5}\left(-\frac{18}{29}\right) - \frac{11}{5} = -\frac{36}{145} - \frac{319}{145} = -\frac{355}{145} = -\frac{71}{29}$ [A1]

Answer: $\left(-\frac{18}{29}, -\frac{71}{29}\right)$

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Question 3

(a) Midpoint of $DE$ [1 mark]

$M = \left(\frac{-1 + 3}{2}, \frac{4 + 10}{2}\right) = (1, 7)$ [A1]

Answer: $(1, 7)$

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(b) Show $DE \perp DF$ [3 marks]

Gradient of $DE$: $m_{DE} = \frac{10 - 4}{3 - (-1)} = \frac{6}{4} = \frac{3}{2}$ [M1]

Gradient of $DF$: $m_{DF} = \frac{2 - 4}{7 - (-1)} = \frac{-2}{8} = -\frac{1}{4}$ [M1]

$m_{DE} \times m_{DF} = \frac{3}{2} \times \left(-\frac{1}{4}\right) = -\frac{3}{8} \neq -1$

Correction: Let me recalculate. The question states $F(7, 2)$.

$m_{DF} = \frac{2 - 4}{7 - (-1)} = \frac{-2}{8} = -\frac{1}{4}$

$m_{DE} \times m_{DF} = \frac{3}{2} \times \left(-\frac{1}{4}\right) = -\frac{3}{8}$

This does NOT equal $-1$. Let me re-examine the coordinates.

Given $D(-1, 4)$, $E(3, 10)$, $F(7, 2)$:

$m_{DE} = \frac{10-4}{3-(-1)} = \frac{6}{4} = \frac{3}{2}$

$m_{DF} = \frac{2-4}{7-(-1)} = \frac{-2}{8} = -\frac{1}{4}$

Product $= \frac{3}{2} \times (-\frac{1}{4}) = -\frac{3}{8} \neq -1$

The points as given do not form a right angle at $D$. However, for the purpose of this answer key, let me verify if perhaps $DE \perp EF$:

$m_{EF} = \frac{2-10}{7-3} = \frac{-8}{4} = -2$

$m_{DE} \times m_{EF} = \frac{3}{2} \times (-2) = -3 \neq -1$

$m_{DF} \times m_{EF} = (-\frac{1}{4}) \times (-2) = \frac{1}{2} \neq -1$

Note to examiner: The given coordinates do not produce perpendicular lines. The question should be adjusted. For marking purposes, accept the working method:

Expected method:

• Find $m_{DE} = \frac{3}{2}$ [M1]
• Find $m_{DF} = -\frac{1}{4}$ [M1]
• Show product $= -\frac{3}{8}$ [A1 — but this doesn't equal $-1$]

Revised coordinates suggestion: Use $F(7, -2)$ instead of $F(7, 2)$.

Then $m_{DF} = \frac{-2-4}{7-(-1)} = \frac{-6}{8} = -\frac{3}{4}$

$m_{DE} \times m_{DF} = \frac{3}{2} \times (-\frac{3}{4}) = -\frac{9}{8} \neq -1$

Alternative: Use $D(-1, 4)$, $E(5, 8)$, $F(1, 12)$.

$m_{DE} = \frac{8-4}{5-(-1)} = \frac{4}{6} = \frac{2}{3}$

$m_{DF} = \frac{12-4}{1-(-1)} = \frac{8}{2} = 4$

Product $= \frac{2}{3} \times 4 = \frac{8}{3} \neq -1$

Better alternative: $D(0, 0)$, $E(6, 0)$, $F(0, 8)$ — right angle at $D$.

$m_{DE} = 0$, $m_{DF}$ undefined — perpendicular.

For the given coordinates, accept the method marks and note the error.

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(c) Area of triangle $DEF$ [4 marks]

Using the shoelace formula with $D(-1, 4)$, $E(3, 10)$, $F(7, 2)$:

$\text{Area} = \frac{1}{2}\left|(-1)(10) + (3)(2) + (7)(4) - (4)(3) - (10)(7) - (2)(-1)\right|$ [M1 — correct setup]

$= \frac{1}{2}\left|-10 + 6 + 28 - 12 - 70 + 2\right|$ [M1 — correct multiplication]

$= \frac{1}{2}\left|-56\right|$ [M1 — correct simplification]

$= 28 \text{ square units}$ [A1]

Answer: 28 square units

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Section B: Quadratic Curves and Parabolas (20 marks)

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Question 4

(a) Express in completed square form [2 marks]

$y = 2x^2 - 8x + 11$ $= 2(x^2 - 4x) + 11$ [M1 — factor out 2] $= 2(x^2 - 4x + 4 - 4) + 11$ $= 2[(x - 2)^2 - 4] + 11$ $= 2(x - 2)^2 - 8 + 11$ $= 2(x - 2)^2 + 3$ [A1]

Answer: $2(x - 2)^2 + 3$

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(b) Minimum point [1 mark]

From $y = 2(x - 2)^2 + 3$: Minimum at $(2, 3)$ [A1]

Answer: $(2, 3)$

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(c) Line of symmetry [1 mark]

$x = 2$ [A1]

Answer: $x = 2$

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(d) Values of $x$ for $y \geq 5$ [3 marks]

$2(x - 2)^2 + 3 \geq 5$ [M1 — set up inequality] $2(x - 2)^2 \geq 2$ $(x - 2)^2 \geq 1$ [M1 — simplify] $x - 2 \leq -1 \quad \text{or} \quad x - 2 \geq 1$ $x \leq 1 \quad \text{or} \quad x \geq 3$ [A1]

Answer: $x \leq 1$ or $x \geq 3$

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Question 5

(a) $x$-intercepts of $P$ [2 marks]

$x^2 - 6x + 5 = 0$ [M1] $(x - 1)(x - 5) = 0$ $x = 1 \text{ or } x = 5$ [A1]

Answer: $(1, 0)$ and $(5, 0)$

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(b) $y$-intercept [1 mark]

When $x = 0$: $y = 0^2 - 6(0) + 5 = 5$ [A1]

Answer: $(0, 5)$

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(c) Intersection of $P$ and $y = 2x - 7$ [4 marks]

$x^2 - 6x + 5 = 2x - 7$ [M1 — equate] $x^2 - 8x + 12 = 0$ [M1 — rearrange] $(x - 2)(x - 6) = 0$ [M1 — factorise] $x = 2 \text{ or } x = 6$

When $x = 2$: $y = 2(2) - 7 = -3$ → $(2, -3)$ When $x = 6$: $y = 2(6) - 7 = 5$ → $(6, 5)$ [A1 — both points]

Answer: $(2, -3)$ and $(6, 5)$

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Question 6

(a) Express in completed square form [3 marks]

$f(x) = -x^2 + 4x + k$ $= -(x^2 - 4x) + k$ [M1 — factor out $-1$] $= -(x^2 - 4x + 4 - 4) + k$ $= -[(x - 2)^2 - 4] + k$ [M1 — complete square] $= -(x - 2)^2 + 4 + k$ [A1]

Answer: $p = 2$, $q = 4 + k$

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(b) Find $k$ [2 marks]

Maximum value is 9, so $q = 9$: $4 + k = 9$ [M1] $k = 5$ [A1]

Answer: $k = 5$

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(c) Range of $x$ for $f(x) \leq 0$ [3 marks]

$f(x) = -(x - 2)^2 + 9 \leq 0$ $-(x - 2)^2 \leq -9$ [M1] $(x - 2)^2 \geq 9$ [M1] $x - 2 \leq -3 \quad \text{or} \quad x - 2 \geq 3$ $x \leq -1 \quad \text{or} \quad x \geq 5$ [A1]

Answer: $x \leq -1$ or $x \geq 5$

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Section C: Circles (20 marks)

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Question 7

(a) Centre and radius of $C_1$ [3 marks]

$x^2 + y^2 - 6x + 4y - 12 = 0$

Complete the square: $(x^2 - 6x) + (y^2 + 4y) = 12$ [M1] $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$ [M1] $(x - 3)^2 + (y + 2)^2 = 25$

Centre: $(3, -2)$ Radius: $\sqrt{25} = 5$ [A1 — both centre and radius]

Answer: Centre $(3, -2)$, radius 5

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(b) Tangent at $A(7, 1)$ [4 marks]

Gradient of radius $CA$: $m_{CA} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}$ [M1]

Gradient of tangent: $m_{\text{tangent}} = -\frac{1}{m_{CA}} = -\frac{4}{3}$ [M1 — perpendicular to radius]

Equation of tangent through $A(7, 1)$: $y - 1 = -\frac{4}{3}(x - 7)$ [M1] $3(y - 1) = -4(x - 7)$ $3y - 3 = -4x + 28$ $4x + 3y - 31 = 0$ [A1]

Answer: $4x + 3y - 31 = 0$

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(c) Equation of $C_2$ [4 marks]

$C_1$: centre $(3, -2)$, radius $r_1 = 5$ $C_2$: centre $B(-1, 5)$, radius $r_2$ (unknown)

Distance between centres: $d = \sqrt{(-1 - 3)^2 + (5 - (-2))^2} = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$ [M1]

For external tangency: $d = r_1 + r_2$ $\sqrt{65} = 5 + r_2$ [M1] $r_2 = \sqrt{65} - 5$ [M1]

Equation of $C_2$: $(x + 1)^2 + (y - 5)^2 = (\sqrt{65} - 5)^2$ [A1]

Answer: $(x + 1)^2 + (y - 5)^2 = (\sqrt{65} - 5)^2$

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Question 8

(a) Centre of circle [1 mark]

Midpoint of $P(2, 1)$ and $Q(10, 7)$: $\left(\frac{2 + 10}{2}, \frac{1 + 7}{2}\right) = (6, 4)$ [A1]

Answer: $(6, 4)$

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(b) Radius [2 marks]

$r = \frac{1}{2} \times PQ = \frac{1}{2}\sqrt{(10 - 2)^2 + (7 - 1)^2}$ [M1] $= \frac{1}{2}\sqrt{64 + 36} = \frac{1}{2}\sqrt{100} = 5$ [A1]

Answer: 5

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(c) Equation of circle [2 marks]

$(x - 6)^2 + (y - 4)^2 = 25$ [A2 — correct centre and radius]

Answer: $(x - 6)^2 + (y - 4)^2 = 25$

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(d) Position of $R(6, 9)$ [3 marks]

Distance from centre $(6, 4)$ to $R(6, 9)$: $d = \sqrt{(6 - 6)^2 + (9 - 4)^2} = \sqrt{0 + 25} = 5$ [M1]

Since $d = r = 5$, the point lies on the circle. [M1 — comparison with radius] [A1 — correct conclusion]

Answer: $R$ lies on the circle.

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Question 9

(a) $x$-coordinate of centre [2 marks]

$S(0, 0)$ and $T(8, 0)$ have midpoint $(4, 0)$. [M1]

The perpendicular bisector of $ST$ is the vertical line $x = 4$. Since the centre lies on the perpendicular bisector of any chord, the $x$-coordinate of the centre is 4. [A1 — explanation]

Answer: The perpendicular bisector of $ST$ is $x = 4$, so the centre has $x$-coordinate 4.

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(b) Radius squared in terms of $k$ [2 marks]

Centre $(4, k)$, passes through $S(0, 0)$: $r^2 = (0 - 4)^2 + (0 - k)^2 = 16 + k^2$ [M1, A1]

Answer: $r^2 = 16 + k^2$

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(c) Find $k$ [3 marks]

Circle also passes through $U(4, 6)$: $(4 - 4)^2 + (6 - k)^2 = 16 + k^2$ [M1] $(6 - k)^2 = 16 + k^2$ $36 - 12k + k^2 = 16 + k^2$ [M1] $36 - 12k = 16$ $-12k = -20$ $k = \frac{5}{3}$ [A1]

Answer: $k = \frac{5}{3}$

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(d) Equation of circle [2 marks]

Centre $\left(4, \frac{5}{3}\right)$, $r^2 = 16 + \left(\frac{5}{3}\right)^2 = 16 + \frac{25}{9} = \frac{144 + 25}{9} = \frac{169}{9}$ [M1]

$(x - 4)^2 + \left(y - \frac{5}{3}\right)^2 = \frac{169}{9}$ [A1]

Answer: $(x - 4)^2 + \left(y - \frac{5}{3}\right)^2 = \frac{169}{9}$

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Section D: Linearisation and Applications (20 marks)

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Question 10

(a) Explanation [2 marks]

$y = ax^n$ $\log y = \log(ax^n) = \log a + \log(x^n) = \log a + n\log x$ [M1]

This is of the form $Y = nX + \log a$, where $Y = \log y$ and $X = \log x$, which is a linear equation. Hence, plotting $\log y$ against $\log x$ produces a straight line. [A1]

Answer: $\log y = n\log x + \log a$ is a linear relationship.

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(b) Complete table [2 marks]

For $x = 1.5$, $y = 4.24$: $\log y = \log 4.24 = 0.627$ (3 d.p.) For $x = 3.5$, $y = 27.7$: $\log y = \log 27.7 = 1.442$ (3 d.p.)

$\log x$	0.176	0.301	0.398	0.477	0.544
$\log y$	0.627	0.903	1.114	1.292	1.442

[A2 — all four values correct; A1 if 2-3 correct]

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(c) Plot and line [2 marks]

[A2 — correct plotting of all 5 points and reasonable best-fit line]

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(d) Estimate $a$ and $n$ [4 marks]

From $\log y = n\log x + \log a$:

Gradient $n$: Using two points on the best-fit line (not necessarily data points): $n = \frac{1.442 - 0.627}{0.544 - 0.176} = \frac{0.815}{0.368} \approx 2.21$ [M1, A1 — accept values consistent with drawn line]

Vertical intercept $\log a$: Extend line to $\log x = 0$: $\log a \approx 0.24$ [M1] $a = 10^{0.24} \approx 1.74$ [A1]

Answer: $n \approx 2.2$, $a \approx 1.7$ (accept values consistent with student's graph)

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Question 11

(a) Linearisation [3 marks]

$P = kb^t$ $\log P = \log(kb^t) = \log k + \log(b^t) = \log k + t\log b$ [M1]

This is of the form $Y = (\log b)t + \log k$, where $Y = \log P$. [A1]

Gradient $= \log b$ Vertical intercept $= \log k$ [A1]

Answer: Gradient $= \log b$, vertical intercept $= \log k$

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(b) Find $k$ and $b$ [4 marks]

When $t = 0$, $P = 200$: $200 = kb^0 = k$ $k = 200$ [M1, A1]

When $t = 4$, $P = 768$: $768 = 200b^4$ [M1] $b^4 = \frac{768}{200} = 3.84$ $b = \sqrt[4]{3.84} \approx 1.40$ [A1]

Answer: $k = 200$, $b \approx 1.40$

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(c) Population at $t = 6$ [2 marks]

$P = 200(1.40)^6$ [M1] $P = 200 \times 7.53 \approx 1510$ [A1]

Answer: Approximately 1510

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(d) Time to reach 2000 [3 marks]

$2000 = 200(1.40)^t$ [M1] $(1.40)^t = 10$ $t \log 1.40 = \log 10$ [M1] $t = \frac{\log 10}{\log 1.40} = \frac{1}{0.1461} \approx 6.8$ [A1]

Answer: 6.8 hours (1 d.p.)

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Question 12

(a) Rearrangement [3 marks]

$y = \frac{p}{x} + qx$ Multiply both sides by $x$: $xy = p + qx^2$ [M1]

This is of the form $Y = qX + p$, where $Y = xy$ and $X = x^2$. [A1]

Gradient $= q$ Vertical intercept $= p$ [A1]

Answer: Plot $xy$ against $x^2$; gradient $= q$, vertical intercept $= p$

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(b) Complete table [2 marks]

$x$	1.0	1.5	2.0	2.5	3.0
$y$	5.0	3.8	3.5	3.4	3.5
$xy$	5.0	5.7	7.0	8.5	10.5

$x^2$	1.00	2.25	4.00	6.25	9.00
$xy$	5.00	5.70	7.00	8.50	10.50

[A2 — all values correct; A1 if 3-4 correct]

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(c) Plot and line [2 marks]

[A2 — correct plotting and best-fit line]

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(d) Estimate $p$ and $q$ [3 marks]

From $xy = qx^2 + p$:

Gradient $q$: Using two points on the line: $q = \frac{10.50 - 5.00}{9.00 - 1.00} = \frac{5.50}{8.00} = 0.6875 \approx 0.688$ [M1, A1]

Vertical intercept $p$: Extend line to $x^2 = 0$: $p \approx 4.3$ [A1]

Answer: $p \approx 4.3$, $q \approx 0.69$ (accept values consistent with student's graph)

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— End of Marking Scheme —