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

(a) Use the cosine rule on triangle ABC:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 120^2 + 150^2 - 2(120)(150)\cos(64°)\)
\(AC^2 = 14400 + 22500 - 36000(0.43837)\)
\(AC^2 = 36900 - 15781.3 = 21118.7\)
\(AC = \sqrt{21118.7} \approx 145.32\text{ m}\).
To 3 significant figures, AC = 145 m.

(b) Area of triangle ABC = \(\frac{1}{2} a b \sin(C)\)
Area = \(\frac{1}{2} (120)(150)\sin(64°)\)
Area = \(9000 \times 0.89879 \approx 8089.15\text{ m}^2\).
To 3 significant figures, Area = 8090 m².

(c)(i) The bearing of B from A is 045°. The back-bearing from B to A is \(045° + 180° = 225°\).
Since angle \(ABC = 64°\) and the bearing of C from B is between 90° and 180° (towards the south-southeast), we subtract 64° from the back-bearing:
Bearing of C from B = \(225° - 64° = 161°\).

(ii) Let \(h\) be the height of the vertical tower at A. The triangle formed by the base of the tower A, the top of the tower, and C is a right-angled triangle at A.
\(\tan(8°) = \frac{h}{AC}\)
\(h = 145.32 \times \tan(8°) = 145.32 \times 0.14054 = 20.42\text{ m}\).
To 3 significant figures, the height of the tower is 20.4 m.

(a)
M1: for substitution into cosine rule: \(120^2 + 150^2 - 2(120)(150)\cos(64°)\)
A1: for \(AC^2 = 21100\) to \(21120\)
A1: for 145 or 145.3... (accept 145 to 145.3)
(b)
M1: for area formula substitution: \(0.5 \times 120 \times 150 \times \sin(64°)\)
A1: for 8090 or 8089... (accept 8090 to 8089)
(c)(i)
M1: for back-bearing \(45 + 180 = 225\)
M1: for \(225 - 64\)
A1: for 161°
(c)(ii)
M1: for writing \(\tan(8°) = \frac{h}{\text{their } AC}\)
M1: for \(h = \text{their } AC \times \tan(8°)\)
A1: for 20.4 or 20.41 to 20.43

(a) Let us analyze the patterns:
Dots (D): 5, 12, 23, 38. The differences are 7, 11, 15... The next difference is 19. So for Diagram 5, dots = \(38 + 19 = 57\).
Lines (L): 8, 13, 18, 23. This is an arithmetic sequence increasing by 5. For Diagram 5, lines = \(23 + 5 = 28\).
Regions (R): 1, 3, 7, 15. The pattern is \(2^n - 1\). For Diagram 5, regions = \(2^5 - 1 = 31\).

(b)(i) Lines: \(L_n\) is arithmetic with first term 8 and common difference 5. So, \(L_n = 8 + 5(n-1) = 5n + 3\).

(b)(ii) Dots: \(D_n\) is quadratic because second differences are constant (4). Let \(D_n = a n^2 + b n + c\).
\(2a = 4 \implies a = 2\).
Using \(n=1\): \(2(1)^2 + b(1) + c = 5 \implies b + c = 3\).
Using \(n=2\): \(2(2)^2 + b(2) + c = 12 \implies 8 + 2b + c = 12 \implies 2b + c = 4\).
Subtracting the equations: \(b = 1\), which means \(c = 2\).
So, \(D_n = 2n^2 + n + 2\).

(b)(iii) Regions: The terms are \(2^1-1, 2^2-1, 2^3-1, 2^4-1\). So, \(R_n = 2^n - 1\).

(c) Compute \(D_n - L_n + R_n\):
\(D_n - L_n + R_n = (2n^2 + n + 2) - (5n + 3) + (2^n - 1)\)
\(= 2n^2 + n + 2 - 5n - 3 + 2^n - 1\)
\(= 2n^2 - 4n - 2 + 2^n\). This completes the proof.

(a)
B1: 57 dots
B1: 28 lines
B1: 31 regions
(b)(i)
M1: for identifying common difference of 5 (e.g., \(5n + k\))
A1: for \(5n + 3\)
(b)(ii)
M1: for recognizing quadratic and finding 2nd difference = 4 (or \(a = 2\))
M1: for attempting to solve simultaneous equations for \(b\) and \(c\)
A1: for \(2n^2 + n + 2\)
(b)(iii)
B1: for \(2^n - 1\)
(c)
M1: for substituting their expressions into the formula: \((2n^2 + n + 2) - (5n + 3) + (2^n - 1)\)
A1: for fully correct algebraic simplification leading to \(2n^2 - 4n - 2 + 2^n\)

(a) Under a reflection in \(y = -x\), the transformation rule is \((x, y) \to (-y, -x)\).
Applying this to the vertices:
\(A(1, 1) \to A'(-1, -1)\)
\(B(3, 1) \to B'(-1, -3)\)
\(C(1, 4) \to C'(-4, -1)\).

(b) For a shear with the x-axis invariant and shear factor \(k = 2\), the rule is \((x, y) \to (x + ky, y) = (x + 2y, y)\).
Applying this to the vertices:
\(A(1, 1) \to A''(1 + 2(1), 1) = A''(3, 1)\)
\(B(3, 1) \to B''(3 + 2(1), 1) = B''(5, 1)\)
\(C(1, 4) \to C''(1 + 2(4), 4) = C''(9, 4)\).

(c)(i) The matrix \(M = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) transforms the basis vectors: \(\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} -1 \\ 0 \end{pmatrix}\). This is a rotation of 90° anticlockwise about the origin \((0,0)\).

(ii) Using matrix multiplication:
\(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y \\ x \end{pmatrix}\).
Applying this to each point:
\(A(1, 1) \to A'''(-1, 1)\)
\(B(3, 1) \to B'''(-1, 3)\)
\(C(1, 4) \to C'''(-4, 1)\).

(a)
B1: for \(A'(-1, -1)\)
B1: for \(B'(-1, -3)\)
B1: for \(C'(-4, -1)\)
(b)
M1: for applying correct shear rule \((x+2y, y)\) to at least one point
A1: for two correct vertices of \(T_2\)
A1: for all three correct vertices of \(T_2\): \((3, 1), (5, 1), (9, 4)\)
(c)(i)
B1: for 'rotation' and '90° anticlockwise' (or '270° clockwise')
B1: for 'about the origin' (or 'about (0,0)')
(c)(ii)
B1: for \(A'''(-1, 1)\)
B1: for \(B'''(-1, 3)\)
B1: for \(C'''(-4, 1)\)

(a) Total volume of the toy = Volume of hemisphere + Volume of cone
\(V_{\text{total}} = \frac{2}{3}\pi R^3 + \frac{1}{3}\pi R^2 h = 252\pi\)
Substitute \(R = 6\):
\(\frac{2}{3}\pi(6)^3 + \frac{1}{3}\pi(6)^2 h = 252\pi\)
\(\frac{2}{3}\pi(216) + \frac{1}{3}\pi(36)h = 252\pi\)
\(144\pi + 12\pi h = 252\pi\)
Divide through by \(\pi\):
\(144 + 12h = 252\)
\(12h = 108 \implies h = 9\text{ cm}\).

(b) Total surface area of the toy = Curved surface area of hemisphere + Curved surface area of cone
Hemisphere curved surface area = \(2\pi R^2 = 2\pi(6)^2 = 72\pi \approx 226.19\text{ cm}^2\).
To find curved surface of cone, we need slant height \(l\):
\(l = \sqrt{R^2 + h^2} = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117} \approx 10.8166\text{ cm}\).
Cone curved surface area = \(\pi R l = \pi(6)(\sqrt{117}) \approx 64.8997\pi \approx 203.89\text{ cm}^2\).
Total surface area = \(72\pi + 6\sqrt{117}\pi = \pi(72 + 6\sqrt{117}) \approx 430.08\text{ cm}^2\).
To 3 significant figures, total surface area is 430 cm².

(c) Volume of the recast sphere = Total volume of the toy = \(252\pi\)
\(\frac{4}{3}\pi r^3 = 252\pi\)
Divide both sides by \(\pi\):
\(\frac{4}{3} r^3 = 252\)
\(r^3 = 252 \times \frac{3}{4} = 189\)
\(r = \sqrt[3]{189} \approx 5.7388\text{ cm}\).
To 3 significant figures, \(r = 5.74\text{ cm}\).

(a)
M1: for hemisphere volume formula: \(\frac{2}{3}\pi (6)^3\) or cone volume formula: \(\frac{1}{3}\pi (6)^2 h\)
M1: for setting up equation: \(144\pi + 12\pi h = 252\pi\) (or equivalent with \(\pi\) omitted)
A1: for leading to \(12h = 108\) and showing \(h=9\) with no errors
(b)
M1: for curved surface area of hemisphere: \(2\pi (6)^2\) (or \(72\pi\) or 226.2)
M1: for finding slant height of cone: \(l = \sqrt{6^2 + 9^2}\) (or \(\sqrt{117}\) or 10.8...)
M1: for curved surface area of cone: \(\pi(6)(\sqrt{117})\) (or 203.9)
M1: for adding the two surface areas together
A1: for 430 or 430.0 to 430.1 (or exact equivalent \(72\pi + 6\pi\sqrt{117}\))
(c)
M1: for setting \(\frac{4}{3}\pi r^3 = 252\pi\)
M1: for rearranging to find \(r^3 = 189\)
A1: for 5.74 or 5.738 to 5.739

(a) To find the stationary points, find the derivative and set it to 0:
\(f'(x) = 3x^2 - 6x - 9\)
Set \(3x^2 - 6x - 9 = 0\)
\(3(x^2 - 2x - 3) = 0 \implies 3(x - 3)(x + 1) = 0\)
So stationary points occur at \(x = 3\) and \(x = -1\).

Find the y-coordinates:
For \(x = -1\): \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\).
For \(x = 3\): \(f(3) = 3^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\).

Determine their nature using the second derivative:
\(f''(x) = 6x - 6\)
At \(x = -1\): \(f''(-1) = 6(-1) - 6 = -12 < 0\) (Local Maximum).
At \(x = 3\): \(f''(3) = 6(3) - 6 = 12 > 0\) (Local Minimum).
So, the local maximum is \((-1, 10)\) and the local minimum is \((3, -22)\).

(b) At \(x = 1\), find the y-coordinate:
\(y = f(1) = 1^3 - 3(1)^2 - 9(1) + 5 = 1 - 3 - 9 + 5 = -6\). Point is \((1, -6)\).
Find the gradient \(m\) at \(x = 1\):
\(m = f'(1) = 3(1)^2 - 6(1) - 9 = 3 - 6 - 9 = -12\).
Using the point-slope formula for the tangent line:
\(y - y_1 = m(x - x_1) \implies y - (-6) = -12(x - 1)\)
\(y + 6 = -12x + 12 \implies y = -12x + 6\).

(c) For \(f(x) = k\) to have exactly three real roots, the horizontal line \(y = k\) must intersect the curve in three places. This occurs between the local minimum y-value and the local maximum y-value.
Therefore, \(-22 < k < 10\).

(a)
M1: for derivative \(f'(x) = 3x^2 - 6x - 9\)
M1: for setting \(f'(x) = 0\) and solving quadratic to find \(x = 3\) and \(x = -1\)
A1: for y-coordinates \(10\) and \(-22\)
M1: for showing nature of points (either using 2nd derivative or checking values)
A1: for correct matching: Max at \((-1, 10)\) and Min at \((3, -22)\)
(b)
M1: for finding the point \((1, -6)\) by substituting \(x = 1\) into \(f(x)\)
M1: for finding the gradient \(m = -12\) by substituting \(x = 1\) into \(f'(x)\)
M1: for using tangent line equation formula with their point and gradient
A1: for \(y = -12x + 6\) (or equivalent integer equation)
(c)
M1: for recognizing that \(k\) lies between the y-values of the stationary points
A1: for \(-22 < k < 10\) (accept inequality written in words or interval notation)

(a) \(\bar{x} = \frac{2+4+5+7+9+10+12+15}{8} = \frac{64}{8} = 8\)
\(\bar{y} = \frac{45+52+58+63+75+76+82+93}{8} = \frac{544}{8} = 68\).

(b) Using a calculator to find the least-squares regression line:
\(y = mx + c\)
\(m = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}\)
Let's check the sums:
\(\sum(x-\bar{x})^2 = (-6)^2 + (-4)^2 + (-3)^2 + (-1)^2 + 1^2 + 2^2 + 4^2 + 7^2 = 36+16+9+1+1+4+16+49 = 132\)
\(\sum(x-\bar{x})(y-\bar{y}) = (-6)(-23) + (-4)(-16) + (-3)(-10) + (-1)(-5) + (1)(7) + (2)(8) + (4)(14) + (7)(25) = 138+64+30+5+7+16+56+175 = 491\)
\(m = \frac{491}{132} \approx 3.71969\)
\(c = \bar{y} - m\bar{x} = 68 - 3.71969(8) = 68 - 29.7575 = 38.2424\)
Rounding to 3 significant figures:
\(y = 3.72x + 38.2\).

(c)(i) Substituting \(x = 11\) into the regression line:
\(y = 3.71969(11) + 38.2424 = 40.9166 + 38.2424 = 79.159\)
Using the 3 s.f. equation: \(y = 3.72(11) + 38.2 = 79.1\). (Accept either 79.1 or 79.2).

(ii) The estimate is reliable because 11 hours is within the range of the given data (2 to 15 hours), which is interpolation, and the correlation is very strong.

(d) The product-moment correlation coefficient \(r\) can be calculated or retrieved from the calculator:
\(r = \frac{491}{\sqrt{132 \times 1844}} \approx 0.995\).
This indicates a strong positive linear correlation.

(a)
B1: for \(\bar{x} = 8\)
B1: for \(\bar{y} = 68\)
(b)
M1: for attempting to calculate gradient \(m\) (using formula or GDC)
A1: for \(m = 3.72\) (accept 3.720)
M1: for attempting to calculate intercept \(c\) (using formula or GDC)
A1: for \(c = 38.2\) (accept 38.24)
(c)(i)
B1: for 79.1 or 79.2 (or value from unrounded calculation \(79.159\))
(c)(ii)
B1: for mentioning that 11 is inside the data range (interpolation) and there is a very strong correlation
(d)
M1: for attempting to calculate \(r\) (or using GDC)
A1: for \(r = 0.995\) (accept 0.9952)
B1: for 'strong positive' correlation

Total number of balls = \(5 + 4 + 3 = 12\).

(a) The three balls can be all Red, all Blue, or all Green.
\(P(\text{All Red}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\)
\(P(\text{All Blue}) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\)
\(P(\text{All Green}) = \frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)
\(P(\text{Same Color}) = \frac{60 + 24 + 6}{1320} = \frac{90}{1320} = \frac{3}{44}\).
To 3 s.f., this is 0.0682.

(b) \(P(\text{at least one Red}) = 1 - P(\text{No Red balls})\).
There are \(4 + 3 = 7\) non-red balls.
\(P(\text{No Red}) = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{210}{1320} = \frac{7}{44}\).
\(P(\text{at least one Red}) = 1 - \frac{7}{44} = \frac{37}{44}\).
To 3 s.f., this is 0.841.

(c) For three different colors, we must select one of each color (Red, Blue, and Green) in any order.
The number of possible orders is \(3! = 6\).
The probability of one specific order (e.g., Red then Blue then Green) is:
\(P(R \cap B \cap G) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\).
Since there are 6 distinct orders, the total probability is:
\(P(\text{three different colors}) = 6 \times \frac{1}{22} = \frac{3}{11}\).
To 3 s.f., this is 0.273.

(a)
M1: for correct product for any one color (e.g., \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\))
M1: for finding the three products: \(\frac{60}{1320}\), \(\frac{24}{1320}\), and \(\frac{6}{1320}\)
M1: for adding their three probabilities together
A1: for \(\frac{3}{44}\) (or equivalent fraction, or 0.0682)
(b)
M1: for identifying \(P(\text{at least one Red}) = 1 - P(\text{no Reds})\)
M1: for finding \(P(\text{no Reds}) = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{7}{44}\) (or \(\frac{210}{1320}\))
A1: for \(\frac{37}{44}\) (or equivalent fraction, or 0.841)
(c)
M1: for product of one order of three different colors, e.g. \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)
M1: for multiplying by 6 (for the 6 permutations)
A1: for \(\frac{3}{11}\) (or equivalent fraction, or 0.273)

(a) The perimeter of the garden is \(2x + 2y = 54\).
Dividing by 2 gives \(x + y = 27\). This completes the proof.

(b) Since there is a path of width 1 m all around the outside, the new width is \(x + 2\) and the new length is \(y + 2\).
The combined area is \((x + 2)(y + 2) = 240\).
Expanding the brackets gives:
\(xy + 2x + 2y + 4 = 240\).
We can rewrite this as \(xy + 2(x + y) + 4 = 240\).
Substitute \(x + y = 27\) into the equation:
\(xy + 2(27) + 4 = 240\)
\(xy + 54 + 4 = 240\)
\(xy + 58 = 240 \implies xy = 182\). This completes the proof.

(c) From (a), \(y = 27 - x\). Substitute this into the equation \(xy = 182\):
\(x(27 - x) = 182\)
\(27x - x^2 = 182\)
\(x^2 - 27x + 182 = 0\).
Factorizing the quadratic equation:
We need two numbers that multiply to 182 and add to -27. These are -13 and -14.
\((x - 13)(x - 14) = 0\).
So \(x = 13\) or \(x = 14\).
If \(x = 13\), then \(y = 14\). If \(x = 14\), then \(y = 13\).
Therefore, the dimensions of the garden are 13 metres by 14 metres.

(d) Let \(d\) be the length of the diagonal of the garden. By Pythagoras' theorem:
\(d = \sqrt{x^2 + y^2} = \sqrt{13^2 + 14^2} = \sqrt{169 + 196} = \sqrt{365} \approx 19.1049\text{ m}\).
To 3 significant figures, the length of the diagonal is 19.1 m.

(a)
M1: for writing \(2x + 2y = 54\)
A1: for leading to \(x + y = 27\) with no errors
(b)
M1: for expressing the new dimensions as \(x + 2\) and \(y + 2\) and setting \((x+2)(y+2) = 240\)
A1: for expanding to get \(xy + 2x + 2y + 4 = 240\)
A1: for substituting \(x + y = 27\) and leading to \(xy = 182\) with no errors
(c)
M1: for substituting \(y = 27 - x\) (or vice versa) into \(xy = 182\)
M1: for creating a standard quadratic equation: \(x^2 - 27x + 182 = 0\) (or in \(y\))
M1: for attempting to factorize or use quadratic formula on their 3-term quadratic
A1: for dimensions 13 m and 14 m
(d)
M1: for \(\sqrt{13^2 + 14^2}\)
A1: for 19.1 or 19.10 to 19.11

(a) The total volume \(V\) of the container is the sum of the volume of the cone and the volume of the hemisphere:
\(V = V_{\text{cone}} + V_{\text{hemisphere}}\)
\(V = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3\)
Given \(h = 12\) and \(V = 288\pi\):
\(\frac{1}{3} \pi r^2 (12) + \frac{2}{3} \pi r^3 = 288\pi\)
\(4\pi r^2 + \frac{2}{3} \pi r^3 = 288\pi\)
Divide both sides by \(\pi\):
\(4r^2 + \frac{2}{3} r^3 = 288\)
Multiply by \(3\):
\(12r^2 + 2r^3 = 864\)
\(2r^3 + 12r^2 - 864 = 0\)
\(r^3 + 6r^2 - 432 = 0\)
Substitute \(r = 6\):
\((6)^3 + 6(6)^2 - 432 = 216 + 216 - 432 = 0\).
Hence, \(r = 6\).

(b) The total surface area consists of the curved surface area of the cone and the curved surface area of the hemisphere.
First, find the slant height \(l\) of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 12^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5}\) cm.
Curved surface area of the cone:
\(A_{\text{cone}} = \pi r l = \pi (6)(6\sqrt{5}) = 36\sqrt{5}\pi\) cm\(^2\).
Curved surface area of the hemisphere:
\(A_{\text{hemisphere}} = 2\pi r^2 = 2\pi (6)^2 = 72\pi\) cm\(^2\).
Total surface area:
\(A = (72 + 36\sqrt{5})\pi\) cm\(^2\).

(c) The volume of water is \(288\pi\) cm\(^3\).
When poured into a cylinder of radius \(R = 8\) cm, the volume of water is given by:
\(V = \pi R^2 H\)
\(288\pi = \pi (8)^2 H\)
\(288\pi = 64\pi H\)
\(H = \frac{288}{64} = 4.5\) cm.

(a) [4 marks]:
- M1: For correct volume formula of cone + hemisphere set equal to \(288\pi\).
- M1: For substituting \(h=12\) and simplifying to get \(4r^2 + \frac{2}{3}r^3 = 288\) (or equivalent).
- M1: For showing the cubic equation \(r^3 + 6r^2 - 432 = 0\).
- A1: For verifying \(r=6\) yields 0 or solving to show \(r=6\).

(b) [4 marks]:
- M1: For finding the slant height \(l = \sqrt{6^2 + 12^2} = \sqrt{180} = 6\sqrt{5}\).
- M1: For curved surface area of cone \(= \pi \times 6 \times 6\sqrt{5}\).
- M1: For curved surface area of hemisphere \(= 2\pi (6)^2 = 72\pi\).
- A1: For the correct exact total surface area in the required form: \((72 + 36\sqrt{5})\pi\).

(c) [3 marks]:
- M1: For setting up the cylinder volume equation: \(\pi \times 8^2 \times H = 288\pi\).
- M1: For simplifying to \(64 H = 288\).
- A1: For correct answer \(4.5\) cm.

(a) The vertical asymptote occurs where the denominator is zero:
\(x - 1 = 0 \implies x = 1\).
As \(x \to \pm\infty\), \(f(x) \to \frac{2x}{x} = 2\), so the horizontal asymptote is \(y = 2\).

(b) Y-intercept occurs when \(x = 0\):
\(f(0) = \frac{2(0) - 5}{0 - 1} = 5\), so the point is \((0, 5)\).
X-intercept occurs when \(y = 0\):
\(\frac{2x - 5}{x - 1} = 0 \implies 2x - 5 = 0 \implies x = 2.5\), so the point is \((2.5, 0)\).

(c) To solve \(f(x) = 3 - x\):
\(\frac{2x - 5}{x - 1} = 3 - x\)
\(2x - 5 = (3 - x)(x - 1)\)
\(2x - 5 = -x^2 + 4x - 3\)
\(x^2 - 2x - 2 = 0\)
Using the quadratic formula:
\(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}\)
\(x \approx 2.73\) or \(x \approx -0.732\).

(d) The equation \(f(x) = k\) has no solutions when the horizontal line \(y = k\) does not intersect the curve. This occurs at the horizontal asymptote, so \(k = 2\).

(a) [2 marks]:
- B1: For vertical asymptote \(x = 1\).
- B1: For horizontal asymptote \(y = 2\).

(b) [3 marks]:
- B1: For finding the y-intercept at \((0, 5)\).
- M1: For setting the numerator \(2x - 5 = 0\).
- A1: For finding the x-intercept at \((2.5, 0)\).

(c) [4 marks]:
- M1: For writing \(2x - 5 = (3 - x)(x - 1)\).
- A1: For expanding and simplifying to get \(x^2 - 2x - 2 = 0\).
- M1: For applying quadratic formula or GDC method to find roots.
- A1: For \(x \approx -0.732\) and \(x \approx 2.73\).

(d) [2 marks]:
- M1: For recognizing that no solution occurs at the horizontal asymptote.
- A1: For \(k = 2\).

(a) Using the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\)
\(AC^2 = 75^2 + 60^2 - 2 \times 75 \times 60 \times \cos(110^\circ)\)
\(AC^2 = 5625 + 3600 - 9000 \times (-0.34202)\)
\(AC^2 = 9225 + 3078.18 = 12303.18\)
\(AC = \sqrt{12303.18} \approx 110.92\) m.
To 3 significant figures, \(AC = 111\) m.

(b) The area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC)\)
\(\text{Area} = \frac{1}{2} \times 75 \times 60 \times \sin(110^\circ)\)
\(\text{Area} = 2250 \times 0.93969 = 2114.3\) m\(^2\).
To 3 significant figures, the area is \(2110\) m\(^2\).

(c) Using the cosine rule in triangle \(ACD\) with \(AC \approx 110.92\) m:
\(CD^2 = AC^2 + AD^2 - 2 \times AC \times AD \times \cos(CAD)\)
\(CD^2 = 110.92^2 + 85^2 - 2 \times 110.92 \times 85 \times \cos(45^\circ)\)
\(CD^2 = 12303.18 + 7225 - 18856.4 \times 0.70711\)
\(CD^2 = 19528.18 - 13333.58 = 6194.6\)
\(CD = \sqrt{6194.6} \approx 78.71\) m.
To 3 significant figures, \(CD = 78.7\) m.

(d) Using the sine rule in triangle \(ACD\):
\(\frac{\sin(ADC)}{AC} = \frac{\sin(CAD)}{CD}\)
\(\sin(ADC) = \frac{AC \times \sin(45^\circ)}{CD}\)
\(\sin(ADC) = \frac{110.92 \times \sin(45^\circ)}{78.71}\)
\(\sin(ADC) = \frac{110.92 \times 0.70711}{78.71} \approx 0.99646\)
\(ADC = \sin^{-1}(0.99646) \approx 85.23^\circ\).
To 3 significant figures, the acute angle \(ADC = 85.2^\circ\).

(a) [3 marks]:
- M1: For correct substitution into cosine rule: \(75^2 + 60^2 - 2 \times 75 \times 60 \times \cos(110^\circ)\).
- A1: For intermediate calculation: \(AC^2 \approx 12300\).
- A1: For final answer \(111\) m.

(b) [2 marks]:
- M1: For correct area formula \(\frac{1}{2} \times 75 \times 60 \times \sin(110^\circ)\).
- A1: For final answer \(2110\) m\(^2\).

(c) [3 marks]:
- M1: For using cosine rule on triangle \(ACD\): \(CD^2 = AC^2 + 85^2 - 2 \times AC \times 85 \times \cos(45^\circ)\).
- A1: For intermediate calculation: \(CD^2 \approx 6195\).
- A1: For final answer \(78.7\) m.

(d) [3 marks]:
- M1: For correct sine rule setup: \(\frac{\sin(ADC)}{AC} = \frac{\sin(45^\circ)}{CD}\).
- A1: For intermediate calculation: \(\text{sin}(ADC) \approx 0.996\).
- A1: For final answer \(85.2^\circ\).