2024 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

**(a) (i)**
To find the local minimum of \(f(x) = 0.5x^3 - 2x^2 + 1.5\), we can use a graphic display calculator (GDC) to locate the minimum point on the graph, or solve \(f'(x) = 1.5x^2 - 4x = 0\).
This gives \(x = 0\) (local maximum) and \(x = \frac{4}{1.5} = \frac{8}{3} \approx 2.67\).
Substituting \(x = \frac{8}{3}\) back into the function:
\(f\left(\frac{8}{3}\right) = 0.5\left(\frac{8}{3}\right)^3 - 2\left(\frac{8}{3}\right)^2 + 1.5 = -3.2407...\)
To 3 significant figures, the coordinates of the local minimum are \((2.67, -3.24)\).

**(a) (ii)**
For \(g(x) = \frac{2}{x}\), the vertical asymptote occurs where the denominator is zero, which is \(x = 0\).

**(b)**
To find the intersection points, we solve \(f(x) = g(x)\):
\(0.5x^3 - 2x^2 + 1.5 = \frac{2}{x}\)
Using a GDC to find the intersection points of the two graphs, we get:
1) In the third quadrant: \(x \approx -1.13\), \(y \approx -1.77\).
2) In the first quadrant: \(x \approx 3.87\), \(y \approx 0.517\).
So the coordinates are \((-1.13, -1.77)\) and \((3.87, 0.517)\).

**(c)**
By inspecting the graphs on the GDC for \(x > 0\), the curve of \(y = f(x)\) lies above the curve of \(y = g(x)\) to the right of their positive intersection point \(x \approx 3.87\).
Therefore, the solution to \(f(x) > g(x)\) for \(x > 0\) is \(x > 3.87\).

**(a) (i)**
- **1M** for locating the minimum point on GDC or finding derivative \(f'(x) = 1.5x^2 - 4x = 0\).
- **1A** for \((2.67, -3.24)\) (accept \(x = 2.67, y = -3.24\)).

**(a) (ii)**
- **1A** for \(x = 0\).

**(b)**
- **1M** for equating the functions or attempting to find intersection points on GDC.
- **1A** for \((-1.13, -1.77)\) (accept \(x = -1.13, y = -1.77\)).
- **2A** for \((3.87, 0.517)\) (accept \(x = 3.87, y = 0.517\); deduct 1 mark overall if answers are not given to 3 s.f.).

**(c)**
- **1M** for identifying the correct region on the graph (to the right of the positive intersection point).
- **2A** for \(x > 3.87\) (accept \(x \ge 3.87\)).

**(a)**
A closed rectangular box with a square base of side length \(x\) and height \(h\) has:
- Two square faces (base and top) each of area \(x^2\).
- Four rectangular vertical faces each of area \(xh\).
Total Surface Area, \(A = 2x^2 + 4xh\).

**(b)**
The volume of the box is \(V = x^2 h = 360\text{ cm}^3\).
Expressing \(h\) in terms of \(x\):
\(h = \frac{360}{x^2}\)
Substitute this expression for \(h\) into the formula for \(A\):
\(A = 2x^2 + 4x\left(\frac{360}{x^2}\right) = 2x^2 + \frac{1440}{x}\).

**(c) (i)**
By plotting \(A(x) = 2x^2 + \frac{1440}{x}\) on a GDC and finding the local minimum for \(x > 0\):
\(x \approx 7.11\text{ cm}\) (or more accurately \(7.11378...\)).

**(c) (ii)**
The minimum surface area is the \(y\)-coordinate at this minimum point:
\(A \approx 304\text{ cm}^2\) (or more accurately \(303.636...\)).

**(d)**
To find the range where the surface area is less than \(350\text{ cm}^2\), we solve:
\(2x^2 + \frac{1440}{x} < 350\)
Using the GDC to find the intersections of \(y = 2x^2 + \frac{1440}{x}\) and \(y = 350\):
We get two positive intersection points:
\(x_1 \approx 4.71\text{ cm}\)
\(x_2 \approx 10.2\text{ cm}\)
Since the graph is U-shaped for \(x > 0\), the surface area is less than \(350\text{ cm}^2\) for values of \(x\) between these two bounds.
Thus, the range of values is \(4.71 < x < 10.2\).

**(a)**
- **1A** for writing down the area of the components clearly: \(2x^2\) and \(4xh\), leading to \(A = 2x^2 + 4xh\).

**(b)**
- **1M** for using the volume formula \(x^2 h = 360\) to write \(h = \frac{360}{x^2}\).
- **1A** for substituting \(h\) into the surface area equation and showing the algebra leading to the given expression.

**(c) (i)**
- **2A** for \(x = 7.11\) (award 1M for correct GDC method or derivative \(4x - 1440/x^2 = 0\)).

**(c) (ii)**
- **1A** for \(304\).

**(d)**
- **1M** for equating \(2x^2 + \frac{1440}{x} = 350\) or finding intersections on GDC.
- **1A** for lower limit \(4.71\).
- **1A** for upper limit \(10.2\).
- **1A** for writing the correct compound inequality \(4.71 < x < 10.2\) (accept \(4.71 \le x \le 10.2\)).

**(a)**
For \(f(x) = 2^x - 3\), as \(x \to -\infty\), \(2^x \to 0\).
Therefore, \(y \to -3\), so the horizontal asymptote is \(y = -3\).

**(b)**
A translation of \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\) transforms the function \(f(x)\) to \(f(x-2) - 1\).
\(g(x) = f(x-2) - 1 = (2^{x-2} - 3) - 1 = 2^{x-2} - 4\).

**(c)**
- For the \(y\)-intercept, set \(x = 0\):
\(g(0) = 2^{0-2} - 4 = 2^{-2} - 4 = 0.25 - 4 = -3.75\).
So the \(y\)-intercept is \((0, -3.75)\).
- For the \(x\)-intercept, set \(g(x) = 0\):
\(2^{x-2} - 4 = 0 \implies 2^{x-2} = 4 = 2^2 \implies x - 2 = 2 \implies x = 4\).
So the \(x\)-intercept is \((4, 0)\).

**(d) (i)**
- Reflection of \(g(x)\) in the \(y\)-axis gives \(g(-x) = 2^{-x-2} - 4\).
- Stretching parallel to the \(y\)-axis with scale factor 3 multiplies the function by 3:
\(h(x) = 3 \cdot g(-x) = 3(2^{-x-2} - 4) = 3 \cdot 2^{-x-2} - 12\).

**(d) (ii)**
We solve \(h(x) = x^2 \implies 3 \cdot 2^{-x-2} - 12 = x^2\).
Plotting both \(y = 3 \cdot 2^{-x-2} - 12\) and \(y = x^2\) on a GDC, we look for intersection points.
There is exactly one intersection point at \(x = -6\).
Thus, the solution is \(x = -6.00\) (or \(x = -6\)).

**(a)**
- **1A** for \(y = -3\).

**(b)**
- **1M** for applying translation rules: \(f(x-2) - 1\).
- **1A** for showing that this simplifies to \(2^{x-2} - 4\).

**(c)**
- **1A** for \(y\)-intercept \((0, -3.75)\) (accept \(-3.75\) or \(-15/4\)).
- **2A** for \(x\)-intercept \((4, 0)\) (award 1M for setting \(2^{x-2} - 4 = 0\)).

**(d) (i)**
- **1M** for applying reflection in the \(y\)-axis to get \(2^{-x-2} - 4\).
- **1A** for \(h(x) = 3 \cdot 2^{-x-2} - 12\) (or equivalent).

**(d) (ii)**
- **1M** for setting up the equation on GDC or trying to find intersection.
- **1A** for \(x = -6.00\) (accept \(x = -6\)).

(a) Using a graphic display calculator (GDC), we find the x-intercepts by solving \(e^x - 3x^2 = 0\). The three roots correct to 3 significant figures are \(x = -0.459\), \(x = 0.910\), and \(x = 3.73\). (b) (i) By using the GDC maximum feature, the coordinates of the local maximum are \((0.204, 1.10)\). (ii) By using the GDC minimum feature, the coordinates of the local minimum are \((2.83, -7.08)\). (c) To solve \(e^x - 3x^2 > 5\), find the intersection of the curve with \(y = 5\) on the GDC. The curve crosses \(y = 5\) at \(x \approx 3.95\). Since the function is increasing for \(x > 2.83\), the inequality holds for \(x > 3.95\).

(a) B1 for -0.459, B1 for 0.910, B1 for 3.73. (b)(i) M1 for GDC maximum method (or finding x-coord 0.204), A1 for correct coordinates (0.204, 1.10). (b)(ii) M1 for GDC minimum method (or finding x-coord 2.83), A1 for correct coordinates (2.83, -7.08). (c) M1 for plotting y = 5 or setting up equation f(x) = 5, A1 for critical value 3.95 (or 3.94 to 3.95), A1 for final inequality x > 3.95.

(a) Using GDC or calculus, \(y' = 3 - \frac{4}{x^2} = 0 \implies x = \pm \frac{2}{\sqrt{3}} \approx \pm 1.15\). Substituting these values back: for \(x \approx 1.15\), \(y \approx 6.93\) (local minimum); for \(x \approx -1.15\), \(y \approx -6.93\) (local maximum). (b)(i) Equating: \(3x + \frac{4}{x} = kx + 2 \implies 3x^2 + 4 = kx^2 + 2x \implies (3-k)x^2 - 2x + 4 = 0\). (b)(ii) For the line to be tangent, the discriminant of this quadratic must be zero: \((-2)^2 - 4(3-k)(4) = 0 \implies 4 - 48 + 16k = 0 \implies 16k = 44 \implies k = 2.75\).

(a) M1 for finding derivative or using GDC features. A1 for x-coords +/- 1.15. A1 for local maximum (-1.15, -6.93). A1 for local minimum (1.15, 6.93). (b)(i) M1 for equating and clearing denominator. A1 for showing full step-by-step rearrangement to the given form. (b)(ii) M1 for setting discriminant equal to zero. A1 for setting up equation 4 - 16(3-k) = 0. M1 for solving for k. A1 for k = 2.75 (or 11/4).

(a) Using vertex form: \(g(x) = a(x - 3)^2 - 4\). Passing through \((1, 8)\): \(8 = a(1-3)^2 - 4 \implies 12 = 4a \implies a = 3\). Expanding: \(g(x) = 3(x-3)^2 - 4 = 3(x^2 - 6x + 9) - 4 = 3x^2 - 18x + 23\). Thus \(b = -18\) and \(c = 23\). (b) Under translation by vector \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\), \(h(x) = g(x+2) + 5 = 3((x+2)-3)^2 - 4 + 5 = 3(x-1)^2 + 1 = 3x^2 - 6x + 4\). (c) A stretch parallel to the y-axis with factor 2 changes \(y\)-coordinates of all points on \(g(x)\) by multiplying by 2. The x-coordinate of the vertex remains 3. The y-coordinate becomes \(-4 \times 2 = -8\). The vertex of \(j(x)\) is \((3, -8)\).

(a) M1 for writing g(x) = a(x-3)^2 - 4. A1 for substituting (1,8) and showing a=3. M1 for expanding 3(x-3)^2 - 4 correctly. A1 for b = -18 and c = 23. (b) M1 for replacing x with x+2 and adding 5. M1 for expanding 3(x-1)^2 + 1. A1 for h(x) = 3x^2 - 6x + 4. (c) M1 for identifying the stretch operation (multiplying g(x) or its y-value by 2). A1 for vertex x-coordinate = 3. A1 for vertex y-coordinate = -8.

Substitute \(y = 2x - 1\) into the quadratic equation:
\(x^2 + (2x - 1)^2 = 13\)

Expand the brackets:
\(x^2 + 4x^2 - 4x + 1 = 13\)

Simplify and rearrange to standard quadratic form:
\(5x^2 - 4x - 12 = 0\)

Factorize the quadratic expression:
\((5x + 6)(x - 2) = 0\)

This gives:
\(x = -1.2\) or \(x = 2\)

Substitute these values back into the linear equation \(y = 2x - 1\) to find \(y\):
When \(x = 2\), \(y = 2(2) - 1 = 3\)
When \(x = -1.2\), \(y = 2(-1.2) - 1 = -3.4\)

Thus, the solutions are:
\(x = 2, y = 3\) and \(x = -1.2, y = -3.4\).

M1: Substitution of \(y = 2x - 1\) into \(x^2 + y^2 = 13\)
M1: Correct expansion of \((2x-1)^2\) to \(4x^2 - 4x + 1\)
A1: Correct quadratic equation in standard form: \(5x^2 - 4x - 12 = 0\)
M1: Factorization or formula application to solve their quadratic equation
A1: Correct \(x\) values: \(x = 2\) and \(x = -1.2\) (or equivalent fraction)
A1: Correct corresponding \(y\) values: \(y = 3\) and \(y = -3.4\) (or equivalent fraction)

Let the two consecutive odd numbers be \(2n - 1\) and \(2n + 1\), where \(n\) is an integer.

Set up the difference of their squares:
\((2n + 1)^2 - (2n - 1)^2\)

Expand both squared terms:
\((4n^2 + 4n + 1) - (4n^2 - 4n + 1)\)

Simplify the expression by expanding the subtraction:
\(4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n\)

Since \(n\) is an integer, \(8n\) is a multiple of 8.

(Alternatively, using \(2n + 1\) and \(2n + 3\):
\((2n+3)^2 - (2n+1)^2 = (4n^2 + 12n + 9) - (4n^2 + 4n + 1) = 8n + 8 = 8(n+1)\), which is also a multiple of 8 since \(n+1\) is an integer.)

M1: Correct algebraic expression for two consecutive odd numbers (e.g., \(2n - 1\) and \(2n + 1\) or \(2n+1\) and \(2n+3\))
M1: Correct expression for the difference of their squares
M1: Correct expansion of both quadratic terms
A1: Simplification to \(8n\) (or \(8n + 8\))
M1: Factorization of 8 to show the multiple clearly, i.e., \(8(n)\) or \(8(n+1)\)
A1: Clear written conclusion stating that because \(n\) (or \(n+1\)) is an integer, the expression is a multiple of 8

Express the terms in the first equation with a base of 3:
\(9 = 3^2\)
\(243 = 3^5\)

Rewrite the equation:
\(3^x \times (3^2)^y = 3^5\)
\(3^x \times 3^{2y} = 3^5\)
\(3^{x + 2y} = 3^5\)

Equating the powers, we get:
\(x + 2y = 5\) [Equation 1]

Our second given equation is:
\(2x - y = 5\) [Equation 2]

Now solve these linear equations simultaneously.
From Equation 2, express \(y\) in terms of \(x\):
\(y = 2x - 5\)

Substitute this into Equation 1:
\(x + 2(2x - 5) = 5\)
\(x + 4x - 10 = 5\)
\(5x = 15\)
\(x = 3\)

Substitute \(x = 3\) back to find \(y\):
\(y = 2(3) - 5 = 1\)

Therefore, the solution is \(x = 3, y = 1\).

M1: Write \(9\) as \(3^2\) or \(243\) as \(3^5\)
M1: Apply the multiplication index law to obtain \(3^{x+2y}\)
A1: Form the linear equation \(x + 2y = 5\)
M1: Use an appropriate method (substitution or elimination) to solve the simultaneous linear equations
A1: Correctly find \(x = 3\)
A1: Correctly find \(y = 1\)

To solve the simultaneous equations, substitute the expression for \(y\) from the second equation into the first equation.

From the second equation, rearrange to find \(y\):
\(y = 3x + 7\)

Substitute this into the first equation:
\(3x + 7 = 2x^2 - 5x - 3\)

Rearrange the equation to form a quadratic equation equal to zero:
\(2x^2 - 5x - 3 - 3x - 7 = 0\)
\(2x^2 - 8x - 10 = 0\)

Divide the entire equation by 2 to simplify:
\(x^2 - 4x - 5 = 0\)

Factorise the quadratic equation:
\((x - 5)(x + 1) = 0\)

This gives two solutions for \(x\):
\(x = 5\) or \(x = -1\)

Now find the corresponding values of \(y\) by substituting the \(x\)-values back into the linear equation \(y = 3x + 7\):

For \(x = 5\):
\(y = 3(5) + 7 = 15 + 7 = 22\)

For \(x = -1\):
\(y = 3(-1) + 7 = -3 + 7 = 4\)

Therefore, the solutions are \(x = -1, y = 4\) and \(x = 5, y = 22\).

M1: For correctly rearranging the second equation to make \(y\) the subject (i.e. \(y = 3x + 7\)).
M1: For substituting and obtaining a correct quadratic equation in \(x\) (e.g., \(2x^2 - 8x - 10 = 0\) or \(x^2 - 4x - 5 = 0\)).
M1: For a valid method to solve their quadratic equation (e.g., factorising to \((x-5)(x+1) = 0\) or using the quadratic formula).
A1: For correct \(x\)-values: \(x = 5\) and \(x = -1\).
M1: For substituting at least one of their \(x\)-values back into a linear equation to find a \(y\)-value.
A1: For both correct pairs of solutions: \(x = -1, y = 4\) and \(x = 5, y = 22\) (or written as coordinates \((-1, 4)\) and \((5, 22)\)).

(a) Let the three consecutive integers be \(n - 1\), \(n\), and \(n + 1\).

The sum of their squares is:
\(S = (n - 1)^2 + n^2 + (n + 1)^2\)

Expand each squared term:
\((n - 1)^2 = n^2 - 2n + 1\)
\((n + 1)^2 = n^2 + 2n + 1\)

Substitute these expansions back into the sum \(S\):
\(S = (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1)\)

Combine the like terms:
\(S = 3n^2 + 2\)

This is in the form \(an^2 + b\) where \(a = 3\) and \(b = 2\).

(b) From part (a), the sum of the squares is \(3n^2 + 2\).

Since \(n\) is an integer, \(n^2\) is also an integer, which means \(3n^2\) is always a multiple of 3.

Dividing the expression \(3n^2 + 2\) by 3:
\(\frac{3n^2 + 2}{3} = n^2 + \frac{2}{3}\)

This shows that the quotient is \(n^2\) and the remainder \(r\) is 2.

Part (a):
M1: For setting up the algebraic sum of squares: \((n - 1)^2 + n^2 + (n + 1)^2\).
M1: For expanding the squared terms correctly to obtain \(n^2 - 2n + 1\) and \(n^2 + 2n + 1\).
A1: For simplifying to the final form \(3n^2 + 2\) (implied by stating \(a = 3, b = 2\)).

Part (b):
M1: For stating or implying that \(3n^2\) is always divisible by 3 (or a multiple of 3).
M1: For identifying that the remaining constant term \(2\) represents the remainder when divided by 3.
A1: For the correct remainder \(r = 2\).

(a) First, find the diagonal of the rectangular base \(ABCD\).
Using Pythagoras' theorem on triangle \(ABC\):
\(AC = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ cm}\).

The midpoint \(M\) of the diagonal divides it equally, so:
\(AM = \frac{15}{2} = 7.5\text{ cm}\).

In the right-angled triangle \(VMA\):
\(VM^2 + AM^2 = VA^2\)
\(VM^2 + 7.5^2 = 15^2\)
\(VM^2 = 225 - 56.25 = 168.75\)
\(VM = \sqrt{168.75} \approx 12.990\text{ cm}\).
To 3 significant figures, \(VM = 13.0\text{ cm}\).

(b) The angle that the slant edge \(VA\) makes with the base is \(\angle VAM\).
In triangle \(VMA\):
\(\cos(\angle VAM) = \frac{AM}{VA} = \frac{7.5}{15} = 0.5\)
\(\angle VAM = \arccos(0.5) = 60^\circ\) (or \(60.0^\circ\)).

(c) Let \(N\) be the midpoint of the side \(BC\). The angle between the face \(VBC\) and the base is \(\angle VNM\).
Since \(ABCD\) is a rectangle with length \(AB = 12\text{ cm}\), the distance from the center \(M\) to the side \(BC\) is:
\(MN = \frac{AB}{2} = 6\text{ cm}\).

In the right-angled triangle \(VMN\):
\(\tan(\angle VNM) = \frac{VM}{MN} = \frac{\sqrt{168.75}}{6} \approx 2.1651\)
\(\angle VNM = \arctan(2.1651) \approx 65.205^\circ\).
To 3 significant figures, the angle is \(65.2^\circ\).

(a) [3 Marks]
- M1 for finding the base diagonal \(AC = 15\) or semi-diagonal \(AM = 7.5\).
- M1 for a correct Pythagoras statement to find \(VM\), e.g., \(VM = \sqrt{15^2 - 7.5^2}\).
- A1 for \(13.0\) (accept \(13\) or \(\sqrt{168.75}\)).

(b) [3 Marks]
- M1 for identifying the correct angle \(\angle VAM\).
- M1 for setting up a correct trigonometric ratio, e.g., \(\cos(\theta) = \frac{7.5}{15}\).
- A1 for \(60^\circ\) or \(60.0^\circ\).

(c) [4 Marks]
- M1 for finding \(MN = 6\).
- M1 for identifying the angle \(\angle VNM\) and setting up \(\tan(\angle VNM) = \frac{VM}{6}\).
- A1 for \(\tan(\angle VNM) \approx 2.1651\) or equivalent.
- A1 for \(65.2^\circ\) (accept \(65.20^\circ\) to \(65.21^\circ\)).

(a) Let the height of the cylinder be \(h = 12\text{ cm}\), the height of the cone be \(H\), and the radius be \(r = 5\text{ cm}\).
\(\text{Volume of cylinder} = \pi r^2 h = \pi (5)^2 (12) = 300\pi\text{ cm}^3\).
\(\text{Volume of cone} = \text{Total volume} - \text{Volume of cylinder} = 400\pi - 300\pi = 100\pi\text{ cm}^3\).

Using the formula for the volume of a cone:
\(\frac{1}{3} \pi r^2 H = 100\pi\)
\(\frac{1}{3} \pi (5)^2 H = 100\pi\)
\(\frac{25}{3} H = 100\)
\(H = 12\text{ cm}\).

(b) The slant height \(l\) of the cone is:
\(l = \sqrt{r^2 + H^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\).

(c) The total surface area of the toy consists of:
- The flat circular base of the cylinder: \(\pi r^2 = \pi (5)^2 = 25\pi\)
- The curved surface area of the cylinder: \(2\pi r h = 2\pi (5)(12) = 120\pi\)
- The curved surface area of the cone: \(\pi r l = \pi (5)(13) = 65\pi\)

\(\text{Total Surface Area} = 25\pi + 120\pi + 65\pi = 210\pi \approx 659.73\text{ cm}^2\).
To 3 significant figures, the total surface area is \(660\text{ cm}^2\).

(d) The angle \(\theta\) that the slant edge makes with the base is:
\(\tan\theta = \frac{H}{r} = \frac{12}{5} = 2.4\)
\(\theta = \arctan(2.4) \approx 67.380^\circ\).
To 3 significant figures, the angle is \(67.4^\circ\).

(a) [3 Marks]
- M1 for finding the volume of the cylinder: \(300\pi\).
- M1 for setting up the equation \(\frac{1}{3}\pi (5)^2 H = 100\pi\).
- A1 for fully showing \(H = 12\) with clear algebraic steps.

(b) [2 Marks]
- M1 for Pythagoras' theorem: \(\sqrt{5^2 + 12^2}\).
- A1 for \(13\).

(c) [3 Marks]
- M1 for sum of any two correct components of the surface area (e.g., \(25\pi\), \(120\pi\), or \(65\pi\)).
- A1 for the exact total surface area \(210\pi\).
- A1 for \(660\) (accept \(659.7\) to \(660\)).

(d) [2 Marks]
- M1 for \(\tan\theta = \frac{12}{5}\) or equivalent trigonometric expression.
- A1 for \(67.4^\circ\) (accept \(67.38^\circ\)).

(a) In the right-angled triangle \(TPA\), the angle of elevation is \(\angle TAP = 18^\circ\) and the adjacent side is \(PA = 80\text{ m}\).
\(\tan(18^\circ) = \frac{h}{80}\)
\(h = 80 \tan(18^\circ) \approx 25.9935\text{ m}\).
To 3 significant figures, \(h = 26.0\text{ m}\).

(b) The bearings of \(A\) and \(C\) from \(P\) are \(200^\circ\) and \(120^\circ\) respectively.
The angle between them at \(P\) is:
\(\angle APC = 200^\circ - 120^\circ = 80^\circ\).

Using the Cosine Rule in triangle \(APC\):
\(AC^2 = PA^2 + PC^2 - 2(PA)(PC)\cos(\angle APC)\)
\(AC^2 = 80^2 + 110^2 - 2(80)(110)\cos(80^\circ)\)
\(AC^2 = 6400 + 12100 - 17600 \cos(80^\circ)\)
\(AC^2 = 18500 - 17600(0.173648)\)
\(AC^2 = 18500 - 3056.21 = 15443.79\)
\(AC = \sqrt{15443.79} \approx 124.273\text{ m}\).
To 3 significant figures, \(AC = 124\text{ m}\).

(c) Using the Sine Rule in triangle \(APC\) to find the interior angle \(\angle PAC\):
\(\frac{\sin(\angle PAC)}{PC} = \frac{\sin(\angle APC)}{AC}\)
\(\frac{\sin(\angle PAC)}{110} = \frac{\sin(80^\circ)}{124.273}\)
\(\sin(\angle PAC) = \frac{110 \sin(80^\circ)}{124.273} \approx 0.87169\)
\(\angle PAC = \arcsin(0.87169) \approx 60.652^\circ\).

The bearing of \(A\) from \(P\) is \(200^\circ\), which means the bearing of \(P\) from \(A\) is:
\(200^\circ - 180^\circ = 20^\circ\).

Since \(C\) lies to the east of the line \(AP\), the bearing of \(C\) from \(A\) is:
\(\text{Bearing} = 20^\circ + \angle PAC = 20^\circ + 60.652^\circ = 80.652^\circ\).
To 3 significant figures (bearing notation), this is \(080.7^\circ\).

(a) [2 Marks]
- M1 for \(80 \tan(18^\circ)\).
- A1 for \(26.0\) (accept \(26\)).

(b) [4 Marks]
- B1 for stating or using \(\angle APC = 80^\circ\).
- M1 for substituting correct values into the Cosine Rule: \(80^2 + 110^2 - 2(80)(110)\cos(80^\circ)\).
- A1 for \(15444\) or equivalent.
- A1 for \(124\) (accept \(124.2\) to \(124.3\)).

(c) [4 Marks]
- M1 for using the Sine Rule to find \(\angle PAC\): \(\frac{\sin(\angle PAC)}{110} = \frac{\sin(80^\circ)}{AC}\).
- A1 for \(\angle PAC \approx 60.7^\circ\) (or \(60.65^\circ\)).
- M1 for calculating the back-bearing of \(P\) from \(A\) as \(20^\circ\).
- A1 for \(080.7^\circ\) or \(80.7^\circ\) (accept \(80.6^\circ\) to \(80.7^\circ\)).