Secondary 4 Additional Mathematics Preliminary Examination Paper 1

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Secondary School (AI)

Subject: Additional Mathematics (G3)
Level: Secondary 4
Paper: Preliminary Examination – Paper 1
Duration: 1 hour 30 minutes
Total Marks: 60
Version: 1 of 5

Name: _________________________
Class: _________________________
Date: _________________________

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Instructions to Candidates

1. This paper consists of 15 questions.
2. Answer ALL questions.
3. Write your answers in the spaces provided.
4. The number of marks is given in brackets [ ] at the end of each question or part question.
5. You are expected to use a scientific calculator where appropriate.
6. Unless otherwise stated, give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees.
7. Solutions by accurate drawing will not be accepted.

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Section A: Coordinate Geometry (40 marks)

Answer ALL questions in this section.

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1. The points A and B have coordinates (−2, 3) and (4, −1) respectively.

(a) Find the length of AB. [2]

(b) Find the coordinates of the midpoint of AB. [1]

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

(d) Hence find the equation of the perpendicular bisector of AB, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [3]

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2. The line $L_1$ has equation $3x - 4y + 12 = 0$. The line $L_2$ passes through the point (5, 2) and is parallel to $L_1$.

(a) Find the gradient of $L_1$. [1]

(b) Find the equation of $L_2$, giving your answer in the form $y = mx + c$. [2]

(c) Find the coordinates of the point where $L_2$ meets the $x$-axis. [2]

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3. The points P(1, 5), Q(7, 1) and R(3, −3) are three vertices of a parallelogram PQRS.

(a) Find the coordinates of S. [2]

(b) Find the area of parallelogram PQRS. [3]

(c) Determine whether PQRS is a rhombus. Justify your answer. [2]

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4. A triangle has vertices A(−1, 2), B(3, 6) and C(5, 0).

(a) Show that triangle ABC is right-angled at B. [3]

(b) Find the area of triangle ABC. [2]

(c) Find the equation of the line through C that is perpendicular to AB. Give your answer in the form $ax + by + c = 0$. [3]

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5. The points D(2, 5) and E(8, −3) 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, giving your answer in simplified surd form. [2]

(c) Write down the equation of the circle in the form $(x - a)^2 + (y - b)^2 = r^2$. [1]

(d) Determine whether the point (10, 1) lies inside, on, or outside the circle. Show your working. [2]

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

(a) Express the equation of $C_1$ in the form $(x - a)^2 + (y - b)^2 = r^2$, and hence state the coordinates of its centre and its radius. [3]

(b) Find the equation of the tangent to $C_1$ at the point (7, −1). Give your answer in the form $y = mx + c$. [4]

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7. The line $y = 2x - 3$ intersects the curve $y = x^2 - 4x + k$ at two distinct points.

(a) Form a quadratic equation in $x$ that represents the intersection of the line and the curve. [2]

(b) Using the discriminant, find the range of values of $k$ for which the line intersects the curve at two distinct points. [3]

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8. The curve $C$ has equation $y = 2x^3 - 9x^2 + 12x - 1$.

(a) Find $\frac{dy}{dx}$. [2]

(b) Find the coordinates of the stationary points of $C$. [4]

(c) Determine the nature of each stationary point. [3]

(d) Explain why the curve has no point of inflexion. [2]

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Section B: Linearisation and Applications (20 marks)

Answer ALL questions in this section.

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9. The 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 $y$.

$x$	2	4	6	8	10
$y$	5.6	22.4	50.4	89.6	140.0

(a) Using a scale of 2 cm to 0.1 units on the $\log_{10} x$ axis and 2 cm to 0.2 units on the $\log_{10} y$ axis, plot $\log_{10} y$ against $\log_{10} x$ and draw a straight line graph. [3]

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

(c) Hence find the value of $y$ when $x = 15$. [2]

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10. The variables $t$ and $P$ are related by the equation $P = kb^t$, where $k$ and $b$ are constants.

The table below shows experimental values of $t$ and $P$.

$t$	0	2	4	6	8
$P$	3.0	5.4	9.7	17.5	31.5

(a) Explain why plotting $\log_{10} P$ against $t$ will produce a straight line. [2]

(b) Using a scale of 2 cm to 1 unit on the $t$ axis and 2 cm to 0.1 units on the $\log_{10} P$ axis, plot $\log_{10} P$ against $t$ and draw a straight line graph. [3]

(c) Use your graph to estimate the values of $k$ and $b$. [4]

(d) Hence estimate the value of $P$ when $t = 10$. [2]

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END OF PAPER

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Check your work carefully. Ensure all answers are in the spaces provided.

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Preliminary Examination – Paper 1 (Version 1) — ANSWER KEY

Subject: Additional Mathematics (G3)
Level: Secondary 4
Total Marks: 60

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Section A: Coordinate Geometry (40 marks)

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1. A(−2, 3), B(4, −1)

(a) Length of AB [2 marks]

$AB = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}$ $= \sqrt{6^2 + (-4)^2}$ $= \sqrt{36 + 16}$ $= \sqrt{52}$ $= 2\sqrt{13} \text{ units}$

Marking: M1 for correct substitution into distance formula, A1 for correct simplified surd.

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(b) Midpoint of AB [1 mark]

$\text{Midpoint} = \left(\frac{-2 + 4}{2}, \frac{3 + (-1)}{2}\right) = (1, 1)$

Marking: A1 for correct coordinates.

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(c) Gradient of AB [1 mark]

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

Marking: A1 for correct gradient.

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(d) Equation of perpendicular bisector [3 marks]

Gradient of perpendicular bisector: $m_{\perp} = \frac{3}{2}$ (since $m_{AB} \cdot m_{\perp} = -1$)

Passes through midpoint (1, 1):

$y - 1 = \frac{3}{2}(x - 1)$ $2y - 2 = 3x - 3$ $3x - 2y - 1 = 0$

Marking: M1 for perpendicular gradient, M1 for using midpoint, A1 for correct equation in required form.

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2. $L_1: 3x - 4y + 12 = 0$

(a) Gradient of $L_1$ [1 mark]

$3x - 4y + 12 = 0$ $4y = 3x + 12$ $y = \frac{3}{4}x + 3$ $m_1 = \frac{3}{4}$

Marking: A1 for correct gradient.

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(b) Equation of $L_2$ [2 marks]

$L_2 \parallel L_1$, so $m_2 = \frac{3}{4}$.

Passes through (5, 2):

$y - 2 = \frac{3}{4}(x - 5)$ $y = \frac{3}{4}x - \frac{15}{4} + 2$ $y = \frac{3}{4}x - \frac{7}{4}$

Marking: M1 for using parallel gradient, A1 for correct equation.

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(c) $L_2$ meets $x$-axis [2 marks]

At $x$-axis, $y = 0$:

$0 = \frac{3}{4}x - \frac{7}{4}$ $\frac{3}{4}x = \frac{7}{4}$ $x = \frac{7}{3}$

Coordinates: $\left(\frac{7}{3}, 0\right)$

Marking: M1 for setting $y = 0$, A1 for correct coordinates.

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3. P(1, 5), Q(7, 1), R(3, −3)

(a) Coordinates of S [2 marks]

In parallelogram PQRS, $\vec{PQ} = \vec{SR}$.

$\vec{PQ} = (7-1, 1-5) = (6, -4)$

$S = R + \vec{PQ} = (3+6, -3+(-4)) = (9, -7)$

Alternatively: $\vec{PS} = \vec{QR} = (3-7, -3-1) = (-4, -4)$, so $S = (1-4, 5-4) = (-3, 1)$

Using midpoint of diagonals: Midpoint of PR = (2, 1), so S = (4-7, 2-1) = (-3, 1)

Correct answer: S = (−3, 1)

Marking: M1 for correct vector approach, A1 for correct coordinates.

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(b) Area of parallelogram PQRS [3 marks]

Area = $|\vec{PQ} \times \vec{PS}|$ (magnitude of cross product in 2D)

$\vec{PQ} = (6, -4)$ $\vec{PS} = (-3-1, 1-5) = (-4, -4)$

Area = $|6(-4) - (-4)(-4)| = |-24 - 16| = |-40| = 40$ square units

Marking: M1 for finding vectors, M1 for correct determinant calculation, A1 for correct area.

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(c) Is PQRS a rhombus? [2 marks]

For a rhombus, all sides must be equal.

$PQ = \sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}$ $PS = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$

Since $PQ \neq PS$, PQRS is not a rhombus.

Marking: M1 for calculating at least two adjacent side lengths, A1 for correct conclusion with justification.

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4. A(−1, 2), B(3, 6), C(5, 0)

(a) Show triangle ABC is right-angled at B [3 marks]

$\vec{BA} = (-1-3, 2-6) = (-4, -4)$ $\vec{BC} = (5-3, 0-6) = (2, -6)$

$\vec{BA} \cdot \vec{BC} = (-4)(2) + (-4)(-6) = -8 + 24 = 16 \neq 0$

Alternative approach using gradients: $m_{AB} = \frac{6-2}{3-(-1)} = \frac{4}{4} = 1$ $m_{BC} = \frac{0-6}{5-3} = \frac{-6}{2} = -3$

$m_{AB} \cdot m_{BC} = 1 \cdot (-3) = -3 \neq -1$

Check right angle at B using Pythagoras: $AB^2 = (3-(-1))^2 + (6-2)^2 = 16 + 16 = 32$ $BC^2 = (5-3)^2 + (0-6)^2 = 4 + 36 = 40$ $AC^2 = (5-(-1))^2 + (0-2)^2 = 36 + 4 = 40$

$AB^2 + BC^2 = 32 + 40 = 72 \neq 40 = AC^2$

Correction: The triangle is not right-angled at B. Recheck: right angle at A?

$AB^2 = 32$, $AC^2 = 40$, $BC^2 = 40$

$AB^2 + AC^2 = 32 + 40 = 72 \neq BC^2$

Right angle at C? $AC^2 + BC^2 = 40 + 40 = 80 \neq AB^2 = 32$

The triangle is isosceles (AC = BC) but not right-angled.

Note: The question as written contains an error. If intended as a right-angled triangle, the coordinates would need adjustment. For marking purposes, accept correct working showing it is NOT right-angled at B, or adjust coordinates.

Revised marking: M1 for finding two vectors or gradients, M1 for dot product or gradient product, A1 for correct conclusion (not right-angled at B).

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(b) Area of triangle ABC [2 marks]

Using coordinates formula: Area = $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

= $\frac{1}{2}|(-1)(6 - 0) + 3(0 - 2) + 5(2 - 6)|$ = $\frac{1}{2}|(-1)(6) + 3(-2) + 5(-4)|$ = $\frac{1}{2}|-6 - 6 - 20|$ = $\frac{1}{2}|-32|$ = 16 square units

Marking: M1 for correct substitution into area formula, A1 for correct area.

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(c) Equation of line through C perpendicular to AB [3 marks]

$m_{AB} = 1$ (from part a) $m_{\perp} = -1$

Line through C(5, 0): $y - 0 = -1(x - 5)$ $y = -x + 5$ $x + y - 5 = 0$

Marking: M1 for perpendicular gradient, M1 for using point C, A1 for correct equation in required form.

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5. D(2, 5), E(8, −3)

(a) Centre of circle [1 mark]

Centre = midpoint of DE = $\left(\frac{2+8}{2}, \frac{5+(-3)}{2}\right) = (5, 1)$

Marking: A1 for correct coordinates.

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(b) Radius of circle [2 marks]

$DE = \sqrt{(8-2)^2 + (-3-5)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Radius = $\frac{1}{2} \times DE = 5$ units

Marking: M1 for finding diameter length, A1 for correct radius.

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(c) Equation of circle [1 mark]

$(x - 5)^2 + (y - 1)^2 = 25$

Marking: A1 for correct equation.

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(d) Position of (10, 1) relative to circle [2 marks]

Distance from centre (5, 1) to (10, 1): $d = \sqrt{(10-5)^2 + (1-1)^2} = \sqrt{25} = 5$

Since $d = 5 = r$, the point (10, 1) lies on the circle.

Marking: M1 for calculating distance from centre, A1 for correct conclusion.

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6. $C_1: x^2 + y^2 - 6x + 4y - 12 = 0$

(a) Centre and radius [3 marks]

$x^2 - 6x + y^2 + 4y = 12$ $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$ $(x - 3)^2 + (y + 2)^2 = 25$

Centre: (3, −2), Radius: 5 units

Marking: M1 for completing square for $x$, M1 for completing square for $y$, A1 for correct centre and radius.

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(b) Tangent at (7, −1) [4 marks]

Centre C = (3, −2), point P = (7, −1)

Gradient of radius CP: $m_{CP} = \frac{-1 - (-2)}{7 - 3} = \frac{1}{4}$

Gradient of tangent: $m_{\text{tangent}} = -4$ (perpendicular to radius)

Equation of tangent through (7, −1): $y - (-1) = -4(x - 7)$ $y + 1 = -4x + 28$ $y = -4x + 27$

Marking: M1 for finding gradient of radius, M1 for perpendicular gradient, M1 for using point (7, −1), A1 for correct equation.

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7. Line: $y = 2x - 3$, Curve: $y = x^2 - 4x + k$

(a) Quadratic equation [2 marks]

$2x - 3 = x^2 - 4x + k$ $0 = x^2 - 6x + k + 3$ $x^2 - 6x + (k + 3) = 0$

Marking: M1 for equating and rearranging, A1 for correct quadratic.

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(b) Range of $k$ for two distinct points [3 marks]

For two distinct intersection points, discriminant > 0.

$a = 1$, $b = -6$, $c = k + 3$

$b^2 - 4ac > 0$ $(-6)^2 - 4(1)(k + 3) > 0$ $36 - 4k - 12 > 0$ $24 - 4k > 0$ $4k < 24$ $k < 6$

Marking: M1 for stating discriminant > 0, M1 for correct substitution, A1 for correct inequality.

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8. $y = 2x^3 - 9x^2 + 12x - 1$

(a) $\frac{dy}{dx}$ [2 marks]

$\frac{dy}{dx} = 6x^2 - 18x + 12$

Marking: M1 for differentiating each term correctly, A1 for correct expression.

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(b) Stationary points [4 marks]

$\frac{dy}{dx} = 0$: $6x^2 - 18x + 12 = 0$ $x^2 - 3x + 2 = 0$ $(x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$

When $x = 1$: $y = 2(1)^3 - 9(1)^2 + 12(1) - 1 = 2 - 9 + 12 - 1 = 4$ When $x = 2$: $y = 2(8) - 9(4) + 12(2) - 1 = 16 - 36 + 24 - 1 = 3$

Stationary points: (1, 4) and (2, 3)

Marking: M1 for setting derivative to zero, M1 for solving quadratic, A1 for each correct coordinate pair.

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(c) Nature of stationary points [3 marks]

$\frac{d^2y}{dx^2} = 12x - 18$

At $x = 1$: $\frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0$ → maximum point (1, 4)

At $x = 2$: $\frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0$ → minimum point (2, 3)

Marking: M1 for second derivative, A1 for each correct nature determination.

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(d) Why no point of inflexion? [2 marks]

A point of inflexion occurs when $\frac{d^2y}{dx^2} = 0$ and changes sign.

$\frac{d^2y}{dx^2} = 12x - 18 = 0$ $x = \frac{3}{2}$

At $x = \frac{3}{2}$: $\frac{dy}{dx} = 6\left(\frac{3}{2}\right)^2 - 18\left(\frac{3}{2}\right) + 12 = 6\left(\frac{9}{4}\right) - 27 + 12 = \frac{27}{2} - 15 = -\frac{3}{2} \neq 0$

Since $\frac{dy}{dx} \neq 0$ at $x = \frac{3}{2}$, this is not a stationary point, and therefore not a point of inflexion (which must be a stationary point in this context).

Alternative: The curve has only two stationary points (both turning points), and no other point where concavity changes while the gradient is zero.

Marking: M1 for finding where second derivative is zero, A1 for correct reasoning and conclusion.

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Section B: Linearisation and Applications (20 marks)

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9. $y = ax^n$

(a) Graph plot [3 marks]

$x$	$\log_{10} x$	$y$	$\log_{10} y$
2	0.301	5.6	0.748
4	0.602	22.4	1.350
6	0.778	50.4	1.702
8	0.903	89.6	1.952
10	1.000	140.0	2.146

Marking: M1 for correct computation of log values, M1 for correct plotting, A1 for reasonable straight line.

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(b) Values of $a$ and $n$ [4 marks]

$\log_{10} y = \log_{10} a + n \log_{10} x$

From graph: gradient $n = \frac{2.146 - 0.748}{1.000 - 0.301} = \frac{1.398}{0.699} \approx 2.0$

Vertical intercept $\log_{10} a \approx 0.146$

$a = 10^{0.146} \approx 1.40$

Marking: M1 for identifying gradient as $n$, M1 for identifying intercept as $\log_{10} a$, A1 for $n$, A1 for $a$.

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(c) $y$ when $x = 15$ [2 marks]

$y = 1.40 \times 15^2 = 1.40 \times 225 = 315$

Marking: M1 for substitution, A1 for correct value.

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10. $P = kb^t$

(a) Why $\log_{10} P$ vs $t$ is linear [2 marks]

$\log_{10} P = \log_{10}(kb^t) = \log_{10} k + t \log_{10} b$

This is of the form $Y = c + mt$, where $Y = \log_{10} P$, $c = \log_{10} k$, and $m = \log_{10} b$.

Since it is a linear equation in $t$, plotting $\log_{10} P$ against $t$ produces a straight line.

Marking: M1 for taking logarithms correctly, A1 for explaining linear relationship.

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(b) Graph plot [3 marks]

$t$	$P$	$\log_{10} P$
0	3.0	0.477
2	5.4	0.732
4	9.7	0.987
6	17.5	1.243
8	31.5	1.498

Marking: M1 for correct log values, M1 for correct plotting, A1 for reasonable straight line.

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(c) Values of $k$ and $b$ [4 marks]

From graph: gradient $= \log_{10} b$

Using points (0, 0.477) and (8, 1.498): $\log_{10} b = \frac{1.498 - 0.477}{8 - 0} = \frac{1.021}{8} \approx 0.1276$

$b = 10^{0.1276} \approx 1.34$

Vertical intercept $= \log_{10} k \approx 0.477$

$k = 10^{0.477} \approx 3.00$

Marking: M1 for finding gradient, M1 for identifying as $\log_{10} b$, A1 for $b$, A1 for $k$.

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(d) $P$ when $t = 10$ [2 marks]

$P = 3.00 \times (1.34)^{10} \approx 3.00 \times 18.7 \approx 56.1$

Marking: M1 for substitution, A1 for correct value.

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END OF ANSWER KEY

Total: 60 marks