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

First, multiply the coefficients: \(3 \times 8 = 24\). Next, multiply the powers of 10: \(10^5 \times 10^{-9} = 10^{5-9} = 10^{-4}\). Combine these to get \(24 \times 10^{-4}\). To write this in standard form, adjust it to have a single non-zero digit before the decimal point: \(2.4 \times 10^1 \times 10^{-4} = 2.4 \times 10^{-3}\).

M1 for writing the answer as \(24 \times 10^{-4}\) or \(0.0024\). A1 for \(2.4 \times 10^{-3}\).

Convert the logarithmic equation to its exponential form: \(2x - 1 = 3^2\). This simplifies to \(2x - 1 = 9\). Add 1 to both sides: \(2x = 10\). Divide by 2: \(x = 5\).

M1 for converting to exponential form: \(2x - 1 = 3^2\) or \(2x - 1 = 9\). A1 for \(5\).

The gradient of the given line is \(m_1 = 2\). The gradient of a line perpendicular to it is \(m_2 = -\frac{1}{2}\). Using the point-slope form with the point \((4, 3)\): \(y - 3 = -\frac{1}{2}(x - 4)\). Expanding and simplifying: \(y - 3 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 5\).

M1 for identifying the perpendicular gradient as \(-\frac{1}{2}\). A1 for the correct equation \(y = -\frac{1}{2}x + 5\) (or equivalent).

Find the highest common factor of \(12x^2 y\) and \(18xy^2\), which is \(6xy\). Dividing both terms by \(6xy\) gives the remaining terms inside the brackets: \(6xy(2x - 3y)\).

M1 for finding a partial common factor such as \(3xy(4x - 6y)\) or \(6x(2xy - 3y^2)\). A1 for \(6xy(2x - 3y)\).

The relation is \(y = \frac{k}{\sqrt{x}}\) for a constant \(k\). Substitute \(x = 16\) and \(y = 3\): \(3 = \frac{k}{\sqrt{16}} = \frac{k}{4} \implies k = 12\). Now, substitute \(x = 36\) into the formula: \(y = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\).

M1 for setting up the equation \(y = \frac{k}{\sqrt{x}}\) and finding \(k = 12\). A1 for \(2\).

First, find the size of the union: \(\text{n}(A \cup B) = \text{n}(\mathcal{U}) - \text{n}(A \cup B)' = 30 - 4 = 26\). Use the inclusion-exclusion principle: \(\text{n}(A \cup B) = \text{n}(A) + \text{n}(B) - \text{n}(A \cap B)\). Substitute the known values: \(26 = 18 + 15 - \text{n}(A \cap B) \implies 26 = 33 - \text{n}(A \cap B) \implies \text{n}(A \cap B) = 7\).

M1 for finding \(\text{n}(A \cup B) = 26\). A1 for \(7\).

First, use the negative exponent rule to write the reciprocal of the fraction: \(\left( \frac{27}{8} \right)^{\frac{2}{3}}\). Next, evaluate the fractional power by finding the cube root of the terms and then squaring: \(\left(\sqrt[3]{\frac{27}{8}}\right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}\).

M1 for correctly evaluating either the reciprocal or the cube root, e.g., \(\left(\frac{27}{8}\right)^{\frac{2}{3}}\) or \(\left(\frac{2}{3}\right)^{-2}\). A1 for \(\frac{9}{4}\) (or \(2\frac{1}{4}\) or \(2.25\)).

From the first equation, we get \(y = 13 - 3x\). Substitute this into the second equation: \(2x - 3(13 - 3x) = 16 \implies 2x - 39 + 9x = 16 \implies 11x - 39 = 16 \implies 11x = 55 \implies x = 5\). Substitute \(x = 5\) back to find \(y\): \(y = 13 - 3(5) = -2\).

M1 for a correct method to eliminate one variable (e.g. multiplying the first equation by 3 and adding it to the second, or correct substitution). A1 for both \(x = 5\) and \(y = -2\).

Let us analyze each shape in the list: 1. Kite has 1 line of symmetry and rotational symmetry of order 1. 2. Isosceles trapezium has 1 line of symmetry and rotational symmetry of order 1. 3. Parallelogram has 0 lines of symmetry and rotational symmetry of order 2. 4. Rhombus has 2 lines of symmetry (its diagonals) and rotational symmetry of order 2. Therefore, the quadrilateral is a Rhombus.

B1 for Rhombus (accept rhombus)

An isosceles triangle has two equal angles. Since one angle is \(110^\circ\) (which is obtuse), the other two angles must be the equal angles because a triangle cannot have more than one obtuse angle. The sum of the angles in a triangle is \(180^\circ\). Thus, the sum of the other two angles is \(180^\circ - 110^\circ = 70^\circ\). Since they are equal, each of the other angles is \(70^\circ \div 2 = 35^\circ\).

B1 for 35 (accept 35\(^\circ\))

Using the laws of logarithms, we can combine the terms: \(\log_2((x + 5)(x - 1)) = 4\). Converting this logarithmic equation into its exponential form gives: \((x + 5)(x - 1) = 2^4\), which simplifies to \(x^2 + 4x - 5 = 16\). Rearranging this quadratic equation into standard form, we get \(x^2 + 4x - 21 = 0\). Factoring the quadratic expression, we have \((x + 7)(x - 3) = 0\), which yields the possible solutions \(x = -7\) or \(x = 3\). Since the argument of a logarithm must be strictly positive, we require both \(x + 5 > 0\) and \(x - 1 > 0\), which means \(x > 1\). Therefore, we reject \(x = -7\). The only valid solution is \(x = 3\).

M1 for applying the product rule: \(\log_2((x + 5)(x - 1))\) M1 for converting to exponential form: \((x + 5)(x - 1) = 16\) M1 for solving the quadratic equation to get factors: \((x + 7)(x - 3) = 0\) A1 for \(x = 3\) only (must reject \(x = -7\))

Let \(\theta = \angle ACB\). The area of a triangle is given by \(\text{Area} = \frac{1}{2} a b \sin(\theta)\). Substituting the given values: \(3\sqrt{3} = \frac{1}{2} \cdot 4 \cdot 3 \cdot \sin(\theta)\), which simplifies to \(3\sqrt{3} = 6 \sin(\theta)\). This gives \(\sin(\theta) = \frac{\sqrt{3}}{2}\). Since \(\theta\) is an obtuse angle (between \(90^\circ\) and \(180^\circ\)), we must have \(\theta = 120^\circ\). Thus, \(\cos(\theta) = \cos(120^\circ) = -\frac{1}{2}\). Using the cosine rule to find the length of \(AB\): \(AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\theta)\). Substituting our known values: \(AB^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \left(-\frac{1}{2}\right)\), which simplifies to \(AB^2 = 9 + 16 + 12 = 37\). Taking the square root, we find the exact length of \(AB\) is \(\sqrt{37}\text{ cm}\).

M1 for substituting into the area formula: \(3\sqrt{3} = \frac{1}{2} \cdot 4 \cdot 3 \cdot \sin(\theta)\) A1 for obtaining \(\sin(\theta) = \frac{\sqrt{3}}{2}\) and identifying \(\cos(\theta) = -\frac{1}{2}\) (or \(\theta = 120^\circ\)) M1 for substituting into the cosine rule: \(AB^2 = 3^2 + 4^2 - 2(3)(4)\cos(\theta)\) A1 for \(AB = \sqrt{37}\)

The total number of marbles in the bag is \(n + 6\). The probability of drawing a blue marble on the first draw is \(\frac{6}{n+6}\). Since the marble is not replaced, there are now \(n + 5\) marbles remaining, of which \(5\) are blue. The probability of drawing a second blue marble is \(\frac{5}{n+5}\). The probability of drawing two blue marbles is the product of these probabilities: \(\frac{6}{n+6} \times \frac{5}{n+5} = \frac{1}{3}\). This simplifies to \(\frac{30}{(n+6)(n+5)} = \frac{1}{3}\). Multiplying both sides by \(3(n+6)(n+5)\) yields \(90 = (n+6)(n+5)\). Expanding the right-hand side, we get \(90 = n^2 + 11n + 30\). Rearranging into standard quadratic form gives \(n^2 + 11n - 60 = 0\). Factoring the quadratic, we obtain \((n + 15)(n - 4) = 0\), which gives solutions \(n = -15\) or \(n = 4\). Since the number of red marbles \(n\) must be a positive integer, we reject \(n = -15\). Therefore, \(n = 4\).

M1 for setting up the probability product: \(\frac{6}{n+6} \times \frac{5}{n+5}\) M1 for setting the product equal to \(\frac{1}{3}\) and deriving the quadratic equation: \(n^2 + 11n - 60 = 0\) (or equivalent) M1 for factorizing the quadratic equation: \((n+15)(n-4) = 0\) A1 for \(n = 4\) (must reject the negative root)

The area of a sector is given by the formula \(A = \frac{\theta}{360} \pi R^2\). Given \(A = 24\pi\), we have \(\frac{\theta}{360} \pi R^2 = 24\pi\), which simplifies to \(\frac{\theta}{360} R^2 = 24\) (Equation 1). The arc length of a sector is given by the formula \(L = \frac{\theta}{360} 2\pi R\). Given \(L = 4\pi\), we have \(\frac{\theta}{360} 2\pi R = 4\pi\), which simplifies to \(\frac{\theta}{360} R = 2\) (Equation 2). Dividing Equation 1 by Equation 2 yields \(R = \frac{24}{2} = 12\). Substituting \(R = 12\) back into Equation 2 gives \(\frac{\theta}{360} \cdot 12 = 2\), which simplifies to \(\frac{\theta}{30} = 2\), so \(\theta = 60\). Finally, calculating \(R + \theta\) gives \(12 + 60 = 72\).

M1 for setting up equations for area and arc length: \(\frac{\theta}{360} \pi R^2 = 24\pi\) and \(\frac{\theta}{360} 2\pi R = 4\pi\) M1 for a valid method to solve for \(R\) (e.g. dividing the two equations or substitution) A1 for \(R = 12\) and \(\theta = 60\) A1 for \(R + \theta = 72\)

Using the laws of logarithms: \(\log_3((x-2)(x+4)) = 3\). Converting this to exponential form gives: \((x-2)(x+4) = 3^3\), which simplifies to \(x^2 + 2x - 8 = 27\). Rearranging the terms, we get the quadratic equation: \(x^2 + 2x - 35 = 0\). Factoring the quadratic gives: \((x+7)(x-5) = 0\), which yields solutions \(x = -7\) or \(x = 5\). Since the domain of \(\log_3(x-2)\) requires \(x - 2 > 0\) (so \(x > 2\)), we reject \(x = -7\). Thus, the only valid solution is \(x = 5\).

M1 for applying logarithm laws to get \(\log_3((x-2)(x+4)) = 3\) or \((x-2)(x+4) = 27\). M1 for solving the quadratic equation to find \(x = 5\) and \(x = -7\). A1 for identifying \(x = 5\) as the only valid solution and rejecting \(x = -7\).

We use the exact values of the trigonometric ratios: \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\tan 45^\circ = 1\), and \(\sin 90^\circ = 1\). Substituting these into the expression, we get: \(4 \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - (1)(1) = 4 \left(\frac{3}{4}\right) - 1 = 3 - 1 = 2\).

M1 for correctly substituting at least two exact trigonometric values. M1 for simplifying the terms to obtain \(3 - 1\). A1 for the correct final answer of 2.

(a) First, find the midpoints of each of the intervals: 250, 750, 1250, 1750, 2250, 2750. Multiply each midpoint by its corresponding frequency and sum them up: \(250 \times 20 + 750 \times 30 + 1250 \times 100 + 1750 \times 30 + 2250 \times 15 + 2750 \times 5 = 5000 + 22500 + 125000 + 52500 + 33750 + 13750 = 252500\). Divide this sum of products by the total frequency: \(\frac{252500}{200} = 1262.5\) hours (or 1260 hours to 3 significant figures). (b) Add the frequencies sequentially to find cumulative frequencies: \(h \le 500\) is 20; \(h \le 1000\) is \(20 + 30 = 50\); \(h \le 1500\) is \(50 + 100 = 150\); \(h \le 2000\) is \(150 + 30 = 180\); \(h \le 2500\) is \(180 + 15 = 195\); \(h \le 3000\) is \(195 + 5 = 200\). (c)(i) The median value corresponds to a cumulative frequency of 100. This falls within the interval \(1000 < h \le 1500\) where cumulative frequency rises from 50 to 150. By linear interpolation: \(Median = 1000 + \frac{100 - 50}{150 - 50} \times (1500 - 1000) = 1000 + 0.5 \times 500 = 1250\) hours. (c)(ii) For \(h = 1800\), the interval is \(1500 < h \le 2000\) where cumulative frequency rises from 150 to 180. The estimated cumulative frequency is: \(CF = 150 + \frac{1800 - 1500}{2000 - 1500} \times (180 - 150) = 150 + 0.6 \times 30 = 150 + 18 = 168\). This means 168 bulbs lasted 1800 hours or less. The number of bulbs that lasted more than 1800 hours is \(200 - 168 = 32\).

(a) [3 marks] M1 for finding at least 4 correct midpoints. M1 for calculating the sum of products (allow one arithmetic slip). A1 for 1262.5 or 1260. (b) [1 mark] B1 for all cumulative frequencies correct: 20, 50, 150, 180, 195, 200. (c)(i) [2 marks] M1 for a valid interpolation calculation, e.g. \(1000 + \frac{50}{100} \times 500\). A1 for 1250. (c)(ii) [2 marks] M1 for setting up the interpolation for cumulative frequency at 1800 to get 168, or for writing \(200 - \text{their CF}\). A1 for 32.

(a) Enter the scores in List 1 and frequencies in List 2 on your graphic display calculator, then execute 1-Variable Statistics: (i) The mean score \(\bar{x} = 13.24\) (or 13.2 to 3 significant figures). (ii) The population standard deviation \(\sigma_x \approx 1.64389...\) which is 1.64 to 3 significant figures. (If sample standard deviation is computed, it is 1.66). (iii) The quartiles from the calculator are: Lower quartile \(Q_1 = 12\), and Upper quartile \(Q_3 = 14\). The Interquartile Range \(IQR = Q_3 - Q_1 = 14 - 12 = 2\). (b) When adding a constant value of 2 to all scores in a distribution: (i) The new mean is shifted up by 2: \(13.24 + 2 = 15.24\) (or 15.2 to 3 significant figures). (ii) The measures of spread remain unchanged, so the population standard deviation is still 1.64 (or 1.66). (c) The sum of scores for the first class is \(50 \times 13.24 = 662\). The sum of scores for the second class is \(30 \times 14.5 = 435\). The combined sum of scores for all 80 students is \(662 + 435 = 1097\). The combined mean is \(\frac{1097}{80} = 13.7125\) which is 13.7 to 3 significant figures.

(a)(i) [1 mark] B1 for 13.24 or 13.2. (a)(ii) [1 mark] B1 for 1.64 (accept 1.66). (a)(iii) [2 marks] M1 for identifying both quartiles \(Q_1 = 12\) and \(Q_3 = 14\). A1 for 2 (accept 2.25 if using linear percentile interpolation). (b) [2 marks] B1 for 15.24 or 15.2. B1 for 1.64 (or 1.66, or must match their answer to a(ii)). (c) [2 marks] M1 for calculating total scores: \(50 \times 13.24 + 30 \times 14.5\) (or \(662 + 435\)) divided by 80. A1 for 13.7 or 13.71.

Let \(t\) be the number of years.
The value of the savings account after \(t\) years is given by:
\(120\,000 \times (1.054)^t\)

The value of the delivery vehicle after \(t\) years is given by:
\(180\,000 \times (1 - 0.082)^t = 180\,000 \times (0.918)^t\)

We want to find the smallest integer \(t\) such that the savings exceed the vehicle value:
\(120\,000 \times (1.054)^t > 180\,000 \times (0.918)^t\)

Divide both sides by \(120\,000\) and by \((0.918)^t\):
\(\left(\frac{1.054}{0.918}\right)^t > \frac{180\,000}{120\,000}\)

\((1.148148\dots)^t > 1.5\)

Taking the logarithm of both sides:
\(t \log(1.148148\dots) > \log(1.5)\)

\(t > \frac{\log(1.5)}{\log(1.148148\dots)}\)

\(t > 2.935\)

Since \(t\) must be a number of complete years, we round up to the next integer:
\(t = 3\) years.

We can verify this with the values:
At \(t = 2\):
Savings = \(120\,000 \times 1.054^2 = \$133\,310.40\)
Vehicle = \(180\,000 \times 0.918^2 = \$151\,689.12\) (Savings < Vehicle)

At \(t = 3\):
Savings = \(120\,000 \times 1.054^3 = \$140\,509.16\)
Vehicle = \(180\,000 \times 0.918^3 = \$139\,250.61\) (Savings > Vehicle)

M1 for setting up the equation or inequality: \(120\,000(1.054)^t = 180\,000(0.918)^t\)
M1 for simplifying to \(\left(\frac{1.054}{0.918}\right)^t = 1.5\) or equivalent
M1 for using logarithms correctly to solve for \(t\), or for a systematic trial and error method showing evaluation for at least two values of \(t\)
A1 for obtaining \(t \approx 2.93\dots\) (or equivalent)
M1 for testing/interpreting the integer values around the boundary (e.g., evaluating \(t=2\) and \(t=3\) to show the crossover)
A1 for the final answer of 3

First, calculate Liam's total amount after 5 years.
Interest is compounded monthly, so there are \(5 \times 12 = 60\) periods, with a monthly interest rate of \(\frac{4.2\%}{12} = 0.35\%\).

\(A_{\text{Liam}} = 8000 \times \left(1 + \frac{0.042}{12}\right)^{60}\)
\(A_{\text{Liam}} = 8000 \times (1.0035)^{60} \approx 9865.72\)

Sophia has \(\$150\) more than Liam:
\(A_{\text{Sophia}} = 9865.72 + 150 = 10015.72\)

Sophia's interest is compounded quarterly, so there are \(5 \times 4 = 20\) periods over 5 years.
\(8000 \times \left(1 + \frac{r}{400}\right)^{20} = 10015.72\)

Divide by \(8000\):
\(\left(1 + \frac{r}{400}\right)^{20} = \frac{10015.72}{8000}\)
\(\left(1 + \frac{r}{400}\right)^{20} \approx 1.251965\)

Take the 20th root of both sides:
\(1 + \frac{r}{400} = (1.251965)^{1/20}\)
\(1 + \frac{r}{400} \approx 1.011281\)

Subtract 1:
\[\frac{r}{400} \approx 0.011281\\]
\(r \approx 0.011281 \times 400 \approx 4.5126\)

Correct to 3 significant figures, \(r = 4.51\).

M1 for Liam's formula: \(8000 \times \left(1 + \frac{0.042}{12}\right)^{60}\)
A1 for Liam's amount \(9865.72\) (or \(9866\))
M1 for adding 150 to get Sophia's amount: \(10015.72\) (or \(10016\))
M1 for setting up Sophia's formula: \(8000 \times \left(1 + \frac{r}{400}\right)^{20} = \text{Sophia's amount}\)
M1 for rearranging the equation to solve for \(r\) (e.g. taking the 20th root)
A1 for \(4.51\) (accept answers in the range \(4.51\) to \(4.52\))

(a) The total number of marbles is \( 6 + 4 = 10 \). The probability of drawing different colours is the probability of drawing Red then Blue, plus Blue then Red. \( P(\text{Different}) = P(R, B) + P(B, R) = \left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15} \approx 0.533 \). (b) The probability of getting two heads on the coin is \( P(H, H) = 0.7 \times 0.7 = 0.49 \). The probability that the marbles are of different colours and two heads are obtained is: \( P(\text{Different and HH}) = P(\text{Different}) \times P(H, H) = \frac{8}{15} \times 0.49 = \frac{8}{15} \times \frac{49}{100} = \frac{392}{1500} = \frac{98}{375} \approx 0.261 \). (c) The probability that the marbles are of the same colour is \( P(\text{Same}) = 1 - P(\text{Different}) = 1 - \frac{8}{15} = \frac{7}{15} \). If they are of the same colour, a fair six-sided die is rolled. The prime numbers on a six-sided die are 2, 3, and 5. So, \( P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \). The probability of getting same colour and a prime number is: \( P(\text{Same and Prime}) = \frac{7}{15} \times \frac{1}{2} = \frac{7}{30} \). The probability of getting different colours and two heads was calculated in part (b) as \( \frac{98}{375} \). These two events are mutually exclusive, so the total probability is: \( P(\text{Prime or 2 Heads}) = P(\text{Same and Prime}) + P(\text{Different and HH}) = \frac{7}{30} + \frac{98}{375} = \frac{175}{750} + \frac{196}{750} = \frac{371}{750} \approx 0.495 \).

(a) M1 for \( \frac{6}{10} \times \frac{4}{9} \) or \( \frac{4}{10} \times \frac{6}{9} \) or equivalent tree diagram branches. M1 for adding the two product terms. A1 for \( \frac{8}{15} \) or exact equivalent decimal/percentage (e.g. 0.533). [3 marks] (b) M1 for \( 0.7 \times 0.7 \) or \( 0.49 \) seen. A1 for \( \frac{98}{375} \) or exact equivalent decimal (e.g. 0.261). [2 marks] (c) B1 for \( P(\text{Same}) = \frac{7}{15} \) or equivalent. B1 for \( P(\text{Prime}) = \frac{1}{2} \) or finding \( P(\text{Same and Prime}) = \frac{7}{30} \). A1 for adding \( \frac{7}{30} \) and the answer from part (b) to get \( \frac{371}{750} \) or 0.495. [3 marks]

(a)
- The bearing of \(B\) from \(A\) is \(065^\circ\). Therefore, the back-bearing (bearing of \(A\) from \(B\)) is:
\(065^\circ + 180^\circ = 245^\circ\).
- The bearing of \(C\) from \(B\) is \(140^\circ\).
- The angle \(\angle ABC\) is the difference between these two bearings:
\(\angle ABC = 245^\circ - 140^\circ = 105^\circ\).

*Alternative Method using parallel lines:*
- The north-south line through \(A\) is parallel to the north-south line through \(B\).
- The interior angle between the north line at \(B\) and the line \(BA\) is \(180^\circ - 065^\circ = 115^\circ\).
- The bearing of \(C\) from \(B\) is \(140^\circ\).
- Sum of angles around a point is \(360^\circ\), so:
\(\angle ABC = 360^\circ - (115^\circ + 140^\circ) = 105^\circ\).

(b)
Let \(BC = x\).

*Method 1: Using the Sine Rule to find other angles first*
- By the sine rule in triangle \(ABC\):
\(\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle ABC)}{AC}\)
\(\sin(\angle ACB) = \frac{15.5 \cdot \sin(105^\circ)}{22.0}\)
\(\sin(\angle ACB) \approx 0.68053\)
\(\angle ACB \approx 42.886^\circ\)

- Now find \(\angle BAC\):
\(\angle BAC = 180^\circ - 105^\circ - 42.886^\circ = 32.114^\circ\)

- Find \(BC\) using the sine rule again:
\(\frac{BC}{\sin(32.114^\circ)} = \frac{22.0}{\sin(105^\circ)}\)
\(BC = \frac{22.0 \cdot \sin(32.114^\circ)}{\sin(105^\circ)} \approx 12.108\) km
\(BC \approx 12.1\) km (to 3 s.f.)

*Method 2: Using the Cosine Rule to form a quadratic equation*
- By the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(22.0^2 = 15.5^2 + x^2 - 2 \cdot 15.5 \cdot x \cdot \cos(105^\circ)\)
\(484 = 240.25 + x^2 - 31x \cos(105^\circ)\)
Since \(\cos(105^\circ) \approx -0.258819\):
\(x^2 + 8.023x - 243.75 = 0\)

- Solving the quadratic equation:
\(x = \frac{-8.023 + \sqrt{8.023^2 - 4(1)(-243.75)}}{2}\)
\(x \approx 12.108\) km
\(BC \approx 12.1\) km (to 3 s.f.)

(c)
- The bearing of \(C\) from \(A\) is calculated by adding the angle \(\angle BAC\) to the bearing of \(B\) from \(A\) because \(C\) lies to the clockwise side of the line \(AB\).
- From part (b), \(\angle BAC \approx 32.114^\circ\).
- Therefore, the bearing of \(C\) from \(A\) is:
\(065^\circ + 32.114^\circ = 97.114^\circ\)
Which rounds to \(097.1^\circ\) (or \(97.1^\circ\)) to 1 decimal place.

(a) [2 marks]
- M1: For finding the back-bearing of \(A\) from \(B\) as \(245^\circ\) OR for finding the interior angle between the north line at \(B\) and \(AB\) as \(115^\circ\).
- A1: For full correct working leading to \(105^\circ\) (e.g., \(245^\circ - 140^\circ = 105^\circ\) or \(360^\circ - 115^\circ - 140^\circ = 105^\circ\)).

(b) [4 marks]
- M1: For substituting correctly into the sine rule to find \(\angle ACB\): \(\frac{\sin(\angle ACB)}{15.5} = \frac{\sin(105^\circ)}{22.0}\), OR for substituting correctly into the cosine rule: \(22.0^2 = 15.5^2 + BC^2 - 2(15.5)(BC)\cos(105^\circ)\).
- A1: For finding \(\angle ACB \approx 42.9^\circ\) (or \(42.886^\circ\)) OR for simplifying to a correct quadratic equation: \(BC^2 + 8.02BC - 243.75 = 0\).
- M1: For calculating \(\angle BAC \approx 32.1^\circ\) (or \(32.114^\circ\)) and setting up the sine rule to find \(BC\), OR for substituting correctly into the quadratic formula to solve for \(BC\).
- A1: For \(BC \approx 12.1\) (accept answers in the range \(12.0\) to \(12.2\) resulting from early rounding).

(c) [5 marks]
- M1: For identifying that Bearing of \(C\) from \(A\) is \(065^\circ + \angle BAC\).
- M1: For setting up an equation to find \(\angle BAC\) using the sine or cosine rule (if not already done in part b), e.g., \(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(105^\circ)}{22}\).
- A1: For finding \(\angle BAC \approx 32.1^\circ\) (or \(32.114^\circ\)).
- M1: For adding \(065^\circ\) to their angle \(\angle BAC\).
- A1: For final bearing of \(097.1^\circ\) (or \(97.1^\circ\)), accept answers in the range \(97.0^\circ\) to \(97.2^\circ\).

\(f(x) = \frac{x^2 - 2x + 5}{x - 1} = \frac{(x-1)^2 + 4}{x - 1} = x - 1 + \frac{4}{x - 1}\).

(a) To find the stationary points, we differentiate \(f(x)\):
\(f'(x) = 1 - \frac{4}{(x-1)^2}\).

Setting \(f'(x) = 0\) gives:
\((x-1)^2 = 4 \implies x - 1 = \pm 2\).

For the local minimum (where \(x > 1\)):
\(x - 1 = 2 \implies x = 3\).
\(f(3) = \frac{3^2 - 2(3) + 5}{3 - 1} = \frac{8}{2} = 4\).

So the local minimum is at \((3, 4)\).

(b) For the local maximum (where \(x < 1\)):
\(x - 1 = -2 \implies x = -1\).
\(f(-1) = \frac{(-1)^2 - 2(-1) + 5}{-1 - 1} = \frac{8}{-2} = -4\).

So the local maximum is at \((-1, -4)\).

(c) The function is undefined at \(x = 1\), and as \(x \to 1\), \(f(x) \to \pm\infty\). Thus, the vertical asymptote is \(x = 1\).

(d) Expressing \(f(x)\) as \(x - 1 + \frac{4}{x - 1}\), as \(x \to \pm\infty\), the term \(\frac{4}{x - 1} \to 0\).

Therefore, the oblique asymptote is \(y = x - 1\).

(e) The graph of \(f(x)\) consists of two branches: one for \(x < 1\) where the maximum value is \(-4\), and one for \(x > 1\) where the minimum value is \(4\).

Thus, the range of \(f(x)\) is \(y \le -4\) or \(y \ge 4\).

The equation \(f(x) = k\) has no real solutions when \(k\) lies strictly between these two branches, which is \(-4 < k < 4\).

(a)
M1 for differentiating \(f(x)\) or using GDC.
A1 for coordinates \((3, 4)\).

(b)
M1 for finding the other stationary point.
A1 for coordinates \((-1, -4)\).

(c)
B1 for \(x = 1\).

(d)
M1 for algebraic division or splitting the fraction.
A1 for \(y = x - 1\).

(e)
M1 for recognizing that the equation has no solutions in the interval between the local maximum and local minimum.
A1 for identifying the boundaries \(-4\) and \(4\).
A1 for the correct inequality: \(-4 < k < 4\) (or equivalent).

Let \(f(x) = e^{-x} \sin(2x)\).

(a) To find the local maximum, we find \(f'(x)\):

Using the product rule:
\(f'(x) = -e^{-x} \sin(2x) + 2e^{-x} \cos(2x) = e^{-x}(2\cos(2x) - \sin(2x))\).

Setting \(f'(x) = 0 \implies 2\cos(2x) - \sin(2x) = 0 \implies \tan(2x) = 2\).

For \(0 \le x \le \pi \implies 0 \le 2x \le 2\pi\).

\(2x = \arctan(2) \approx 1.1071 \implies x \approx 0.5536\).

(The other solution \(2x = 1.1071 + \pi \approx 4.2487 \implies x \approx 2.124\) gives a local minimum).

Substitute \(x = 0.5536\) into \(f(x)\):
\(f(0.5536) = e^{-0.5536} \sin(1.1071) \approx 0.5749 \times 0.8944 \approx 0.5144\).

So the local maximum is at \((0.554, 0.514)\).

(b) The curve meets the \(x\)-axis when \(f(x) = 0 \implies e^{-x} \sin(2x) = 0\).

Since \(e^{-x} > 0\) for all \(x\), we have \(\sin(2x) = 0\).

For \(0 \le x \le \pi \implies 0 \le 2x \le 2\pi\), so:
\(2x = 0, \pi, 2\pi \implies x = 0, \frac{\pi}{2}, \pi\).

In decimal form to 3 s.f.: \(x = 0\), \(x = 1.57\), \(x = 3.14\).

(c) The gradient of the tangent at \(x = 0\) is \(f'(0)\):
\(f'(0) = e^{0}(2\cos(0) - \sin(0)) = 1(2 - 0) = 2\).

At \(x = 0\), \(y = f(0) = e^{0}\sin(0) = 0\).

Using \(y - y_1 = m(x - x_1)\):
\(y - 0 = 2(x - 0) \implies y = 2x\).

(d) We need to solve \(e^{-x} \sin(2x) > 0.2\) for \(0 \le x \le \pi\).

Using a GDC to find the intersections of \(y = e^{-x}\sin(2x)\) and \(y = 0.2\):

The intersection points on the interval are:
\(x \approx 0.113\) and \(x \approx 1.21\).

Since the local maximum of \(0.514\) occurs at \(x \approx 0.554\), the curve is above \(y = 0.2\) between these two intersection points.

Thus, the solution is \(0.113 < x < 1.21\).

(a)
M1 for differentiating \(f(x)\) or setting up GDC.
A1 for \(x \approx 0.554\).
A1 for \(y \approx 0.514\).

(b)
M1 for setting \(\sin(2x) = 0\).
A1 for \(x = 0, 1.57, 3.14\) (accept exact values \(0, \frac{\pi}{2}, \pi\)).

(c)
M1 for substituting \(x = 0\) into derivative to find \(m = 2\).
B1 for finding \(y = 0\) at \(x = 0\).
A1 for \(y = 2x\).

(d)
M1 for finding the intersection points of the curve with \(y = 0.2\) using GDC (0.113 and 1.21).
A1 for the correct interval \(0.113 < x < 1.21\).

(a) \(P\) has coordinates \((x, x^2 - 3x + 4)\). The distance \(d\) from \(O(0,0)\) to \(P\) is given by \(d^2 = x^2 + y^2\). Substituting \(y\), we get \(d^2 = x^2 + (x^2 - 3x + 4)^2\). Since \((x^2 - 3x + 4)^2 = x^4 - 6x^3 + 17x^2 - 24x + 16\), adding \(x^2\) gives \(d^2 = x^4 - 6x^3 + 18x^2 - 24x + 16\). (b) Graphing \(d(x) = \sqrt{x^4 - 6x^3 + 18x^2 - 24x + 16}\) on a GDC, we find the minimum occurs at \(x \approx 1.18\). (c) Substituting \(x = 1.1804\) into \(y = x^2 - 3x + 4\) gives \(y \approx 1.85\). Thus, the coordinates of \(P\) are \((1.18, 1.85)\). (d) The minimum distance is \(d = \sqrt{1.1804^2 + 1.8521^2} \approx 2.20\).

M1 for \(d^2 = x^2 + (x^2 - 3x + 4)^2\). A1 for expanding and simplifying to show the required expression. M1 for using GDC to find the minimum of the function. A1 for \(x \approx 1.18\). M1 for substituting their \(x\) into the curve equation. A1 for \(y \approx 1.85\). A1 for coordinates \((1.18, 1.85)\). M1 for calculating \(d = \sqrt{1.18^2 + 1.85^2}\). A1 for \(d \approx 2.20\).

(a) At time \(t\), particle \(A\) has moved \(3t\) units vertically upwards, so \(A(0, 3t)\). Particle \(B\) has moved \(2t\) units in the negative x-direction, so \(B(12 - 2t, 5)\). (b) Using the distance formula: \(d^2 = (12 - 2t - 0)^2 + (5 - 3t)^2 = (144 - 48t + 4t^2) + (25 - 30t + 9t^2) = 13t^2 - 78t + 169\). Taking the square root gives \(d = \sqrt{13t^2 - 78t + 169}\). (c) The minimum of the quadratic \(13t^2 - 78t + 169\) occurs at \(t = -\frac{-78}{2(13)} = 3\). (d) At \(t = 3\), \(d = \sqrt{13(3)^2 - 78(3) + 169} = \sqrt{52} = 2\sqrt{13} \approx 7.21\).

B1 for \(A(0, 3t)\). B1 for \(B(12 - 2t, 5)\). M1 for substituting into the distance formula. A1 for expanding correctly. A1 for showing \(d = \sqrt{13t^2 - 78t + 169}\). M1 for attempting to find the minimum of the quadratic. A1 for \(t = 3\). B1 for \(\sqrt{52}\) or \(2\sqrt{13}\). B1 for \(7.21\).

(a) The width of the rectangle is \(2x\) and the height is \(12 - x^2\). The area \(A\) is \(2x(12 - x^2) = 24x - 2x^3\). (b) Differentiating \(A\) with respect to \(x\) gives \(\frac{dA}{dx} = 24 - 6x^2\). Setting this to 0: \(24 - 6x^2 = 0 \Rightarrow x^2 = 4\). Since \(x > 0\), we get \(x = 2\). (c) Substituting \(x = 2\) into the area formula gives \(A = 24(2) - 2(2)^3 = 32\). (d) When \(x = 2\), the width is \(4\) and the height is \(8\). The perimeter is \(2(4 + 8) = 24\).

M1 for identifying width as \(2x\) and height as \(12 - x^2\). A1 for \(A = 24x - 2x^3\). M1 for differentiating \(A\) to get \(24 - 6x^2\). M1 for setting their derivative to 0 and solving for \(x\). A1 for \(x = 2\) (rejecting \(x = -2\)). M1 for substituting \(x = 2\) into the area expression. A1 for Area = \(32\). M1 for substituting \(x = 2\) into the perimeter expression. A1 for Perimeter = \(24\).

To find \(S(4)\), observe the differences between consecutive terms of \(S(k)\):
\(S(2) - S(1) = 5 - 1 = 4\)
\(S(3) - S(2) = 13 - 5 = 8\)
The differences are \(4, 8, \dots\) which increase by 4 at each stage. Thus, \(S(4) - S(3) = 12\), which means \(S(4) = 13 + 12 = 25\).

To find \(P(4)\), observe the differences of \(P(k)\):
\(P(2) - P(1) = 12 - 4 = 8\)
\(P(3) - P(2) = 20 - 12 = 8\)
This is an arithmetic sequence with a common difference of 8.
Thus, \(P(4) = 20 + 8 = 28\).

- Method to find the next term of \(S(k)\), e.g., identifying the difference increases by 4 [1 mark]
- \(S(4) = 25\) [1 mark]
- Method to find the next term of \(P(k)\), e.g., identifying a common difference of 8 [2 marks]
- \(P(4) = 28\) [1 mark]

Let \(S(k) = ak^2 + bk + c\) since the second differences are constant.
First differences: 4, 8, 12.
Second difference: 4.
Thus, \(2a = 4 \implies a = 2\).

Using \(S(1) = 1\):
\(2(1)^2 + b(1) + c = 1 \implies b + c = -1\).

Using \(S(2) = 5\):
\(2(2)^2 + b(2) + c = 5 \implies 8 + 2b + c = 5 \implies 2b + c = -3\).

Subtracting the first equation from the second:
\((2b + c) - (b + c) = -3 - (-1) \implies b = -2\).

Then \(c = -1 - b = -1 - (-2) = 1\).

So, the formula is \(S(k) = 2k^2 - 2k + 1\).

- Identifying that the second difference is constant (4) [1 mark]
- Deducing that the coefficient of \(k^2\) is \(a = 2\) [1 mark]
- Setting up a system of linear equations for \(b\) and \(c\) [1 mark]
- Finding correct values of \(b = -2\) and \(c = 1\) [1 mark]
- Correct final expression: \(2k^2 - 2k + 1\) [1 mark]

(a) The perimeter sequence is \(4, 12, 20, 28, \dots\) which is an arithmetic sequence with first term \(a = 4\) and common difference \(d = 8\).
Thus, \(P(k) = 4 + (k-1) \times 8 = 8k - 4\).

(b) Let's compare \(P(2n)\) and \(2P(n)\):
\(P(2n) = 8(2n) - 4 = 16n - 4\).
\(2P(n) = 2(8n - 4) = 16n - 8\).

Since \(16n - 4 \neq 16n - 8\), the perimeter of \(D_{2n}\) is not twice the perimeter of \(D_n\).

- Correctly finding \(P(k) = 8k - 4\) [2 marks]
- Finding the expression for \(P(2n) = 16n - 4\) [1 mark]
- Finding the expression for \(2P(n) = 16n - 8\) [1 mark]
- Concluding that the expressions are unequal with a clear explanation [1 mark]

We can count the internal edges using the relation that each unit square has 4 edges.
Sum of all edges of the \(S(k)\) individual squares is \(4 S(k)\).
These edges are either boundary edges (which belong to 1 square) or internal edges (which are shared by 2 squares).
Thus, \(4 S(k) = 2 E(k) + P(k)\), which means \(E(k) = 2 S(k) - \frac{P(k)}{2}\).

For \(k=3\):
\(E(3) = 2 S(3) - \frac{P(3)}{2} = 2(13) - \frac{20}{2} = 26 - 10 = 16\).

For \(k=4\):
\(E(4) = 2 S(4) - \frac{P(4)}{2} = 2(25) - \frac{28}{2} = 50 - 14 = 36\).

Both 16 and 36 are perfect squares:
\(E(3) = 16 = 4 \times 2^2 = 4(3-1)^2\)
\(E(4) = 36 = 4 \times 3^2 = 4(4-1)^2\)

Thus, \(E(k) = 4(k-1)^2\) with \(a = 4\).

- Correctly finding \(E(3) = 16\) [1 mark]
- Correctly finding \(E(4) = 36\) [1 mark]
- Showing that \(16 = 4(2)^2\) and \(36 = 4(3)^2\) [1 mark]
- Generalizing to the correct formula \(E(k) = 4(k-1)^2\) [1 mark]
- Identifying that \(a = 4\) [1 mark]

Let's calculate the Left-Hand Side (LHS):
\(LHS = 4S(k) = 4(2k^2 - 2k + 1) = 8k^2 - 8k + 4\).

Now let's calculate the Right-Hand Side (RHS):
\(RHS = 2E(k) + P(k) = 2[4(k-1)^2] + (8k - 4)\)
\(RHS = 8(k^2 - 2k + 1) + 8k - 4\)
\(RHS = 8k^2 - 16k + 8 + 8k - 4\)
\(RHS = 8k^2 - 8k + 4\).

Since LHS = RHS, the relationship is proven for all \(k \ge 1\).

- Correctly expanding \(4S(k)\) to \(8k^2 - 8k + 4\) [1 mark]
- Correctly expanding \(E(k) = 4(k^2 - 2k + 1)\) [1 mark]
- Multiplying \(E(k)\) by 2 to get \(8k^2 - 16k + 8\) [1 mark]
- Correctly adding \(P(k)\) to get \(8k^2 - 8k + 4\) [1 mark]
- Establishing that LHS = RHS with a final concluding statement [1 mark]

(a) We are given \(E(k) = 576\).
Using the formula:
\(4(k-1)^2 = 576\)
\((k-1)^2 = 144\)
Taking the positive square root (since \(k \ge 1\)):
\(k-1 = 12 \implies k = 13\).

(b) Using the value \(k = 13\):
\(P(13) = 8(13) - 4 = 104 - 4 = 100\).
\(S(13) = 2(13)^2 - 2(13) + 1 = 2(169) - 26 + 1 = 338 - 26 + 1 = 313\).

- Setting up the equation \(4(k-1)^2 = 576\) [1 mark]
- Solving to find \(k = 13\) [1 mark]
- Correct method/substitution to find \(P(13)\) or \(S(13)\) [1 mark]
- Correct value of \(P(13) = 100\) [1 mark]
- Correct value of \(S(13) = 313\) [1 mark]

1. Find the value of \(a\): As \(t \to \infty\), the temperature of the coffee approaches room temperature, so \(a = 20\).
2. Find the value of \(b\): At \(t = 0\), the temperature is \(85^\circ\text{C}\). \(T(0) = 20 + b e^0 = 85 \implies 20 + b = 85 \implies b = 65\).
3. Find the value of \(k\): At \(t = 5\), the temperature is \(55^\circ\text{C}\). \(20 + 65 e^{-5k} = 55 \implies 65 e^{-5k} = 35 \implies e^{-5k} = \frac{35}{65} = \frac{7}{13}\). Taking the natural logarithm on both sides: \(-5k = \ln\left(\frac{7}{13}\right) \approx -0.619039 \implies k \approx 0.123808\). To 4 decimal places, \(k = 0.1238\).
4. Calculate \(T(12)\): Using \(k = 0.1238\), \(T(12) = 20 + 65 e^{-12 \times 0.1238} = 20 + 65 e^{-1.4856} \approx 20 + 14.714 = 34.714^\circ\text{C}\). Rounded to 1 decimal place, the temperature is \(34.7^\circ\text{C}\).

Award [1.5 marks] for correctly identifying room temperature \(a = 20\). Award [1.5 marks] for finding \(b = 65\) using the initial temperature. Award [2 marks] for calculating \(k \approx 0.1238\) (showing the logarithmic step). Award [2.5 marks] for substituting \(t = 12\) and correctly calculating \(34.7\) (accept answers in the range \(34.6\) to \(34.8\) depending on early rounding of \(k\)).

1. Determine the practical domain: All dimensions of the box must be strictly positive. Hence, \(x > 0\), \(30 - 2x > 0 \implies x < 15\), and \(20 - 2x > 0 \implies x < 10\). Combining these inequalities, the practical domain is \(0 < x < 10\).
2. Expand the volume function: \(V(x) = x(600 - 100x + 4x^2) = 4x^3 - 100x^2 + 600x\).
3. Find the derivative to locate stationary points: \(V'(x) = 12x^2 - 200x + 600 = 0\). Dividing by 4 gives \(3x^2 - 50x + 150 = 0\).
4. Solve using the quadratic formula: \(x = \frac{50 \pm \sqrt{(-50)^2 - 4(3)(150)}}{6} = \frac{50 \pm \sqrt{2500 - 1800}}{6} = \frac{50 \pm \sqrt{700}}{6}\). This gives two values: \(x \approx 12.7\text{ cm}\) or \(x \approx 3.92\text{ cm}\). Since \(12.7\) is outside our practical domain of \(0 < x < 10\), the value of \(x\) that maximises the volume is \(3.92\text{ cm}\).

Award [2 marks] for identifying the correct practical domain \(0 < x < 10\). Award [3 marks] for differentiating \(V(x)\) or setting up the GDC graph window with appropriate domain/range. Award [2.5 marks] for identifying \(x \approx 3.92\) as the valid maximum location (correctly rejecting \(12.7\)).

1. Establish the quadratic model using the vertex form: \(y = -k(x - p)^2 + q\), where \((p, q) = (10, 14.5)\) is the vertex. Thus, \(y = -k(x - 10)^2 + 14.5\).
2. Substitute the starting point \((0, 2)\) to solve for the constant \(k\): \(2 = -k(0 - 10)^2 + 14.5 \implies 2 = -100k + 14.5 \implies 100k = 12.5 \implies k = 0.125\). Thus, the model is \(y = -0.125(x - 10)^2 + 14.5\).
3. Find when the ball hits the ground (set \(y = 0\)): \(0 = -0.125(x - 10)^2 + 14.5 \implies 0.125(x - 10)^2 = 14.5 \implies (x - 10)^2 = \frac{14.5}{0.125} = 116\).
4. Solve for \(x\): \(x - 10 = \pm\sqrt{116} \implies x = 10 + \sqrt{116}\text{ or } x = 10 - \sqrt{116}\). Since \(x > 0\), we take \(x = 10 + \sqrt{116} \approx 10 + 10.77 = 20.77\text{ m}\). To 3 significant figures, the distance is \(20.8\text{ m}\).

Award [1.5 marks] for correctly formulating the vertex equation form. Award [1.5 marks] for finding \(k = 0.125\). Award [2 marks] for setting up the equation \(y = 0\) and showing the step to solve for \(x\). Award [2.5 marks] for obtaining \(20.8\) (accept \(20.8\) or \(20.7\) if correct working is shown).

1. At high tide (\(t = 0\)), depth \(d = 12\): \(d(0) = a \cos(0) + c = a + c = 12\).
2. Low tide is the minimum, which occurs at \(t = 6\), so \(d(6) = 4\). For a cosine function of the form \(\cos(b t^\circ)\) with \(b > 0\), the first minimum occurs when the angle is \(180^\circ\). Thus, \(6b = 180 \implies b = 30\).
3. At \(t = 6\), \(d(6) = a \cos(180^\circ) + c = -a + c = 4\).
4. Solve the simultaneous equations:
\(a + c = 12\)
\(-a + c = 4\)
Adding the equations gives \(2c = 16 \implies c = 8\). Substituting \(c = 8\) back gives \(a = 4\).
5. Substitute \(t = 4\) into the model \(d(t) = 4 \cos(30 t^\circ) + 8\): \(d(4) = 4 \cos(120^\circ) + 8 = 4(-0.5) + 8 = -2 + 8 = 6\text{ m}\).

Award [1.5 marks] for writing the system equation \(a + c = 12\). Award [1.5 marks] for identifying \(b = 30\) from the period of the wave. Award [2 marks] for finding \(a = 4\) and \(c = 8\). Award [2.5 marks] for evaluating the depth at \(t = 4\) to obtain exactly \(6\).