2025 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

Use the sine trigonometric ratio: \(\sin(68^\circ) = \frac{\text{height}}{6.5}\). Rearranging this gives: \(\text{height} = 6.5 \times \sin(68^\circ) \approx 6.0267\text{ m}\). Correct to 3 significant figures, the height is \(6.03\text{ m}\).

M1 for setting up the equation \(\sin(68^\circ) = \frac{h}{6.5}\) or \(6.5 \times \sin(68^\circ)\). A1 for \(6.03\) (accept answers in the range \(6.02\) to \(6.03\)).

We factorise the quadratic equation by finding two numbers that multiply to \(3 \times (-5) = -15\) and add to \(-14\). These numbers are \(-15\) and \(1\). This allows us to write: \(3x^2 - 15x + x - 5 = 0 \Rightarrow 3x(x - 5) + 1(x - 5) = 0 \Rightarrow (3x + 1)(x - 5) = 0\). Solving each linear factor gives \(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\) or \(x - 5 = 0 \Rightarrow x = 5\).

M1 for a correct factorisation of the quadratic expression into two linear brackets, e.g., \((3x + 1)(x - 5) = 0\) (or equivalent use of the quadratic formula with at most one sign error). A1 for both correct values: \(x = -1/3\) (or recurring decimal \(-0.333\dots\)) and \(x = 5\).

A decrease of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price. Therefore, \(0.85 \times \text{Original Price} = 357\). We find the original price by calculating \(\text{Original Price} = \frac{357}{0.85} = 420\).

M1 for \(357 \div 0.85\) or an equivalent equation such as \(0.85x = 357\). A1 for \(420\) (accept \(\text{£}420\)).

The gradient \(m\) is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the given coordinates: \(m = \frac{9 - (-1)}{-2 - 3} = \frac{10}{-5} = -2\).

M1 for a correct method to find the gradient, e.g., \(\frac{9 - (-1)}{-2 - 3}\) or \(\frac{-1 - 9}{3 - (-2)}\). A1 for \(-2\).

First, address the negative index by taking the reciprocal of the fraction: \(\left(\frac{125}{8}\right)^{\frac{2}{3}}\). Next, apply the fractional index of \(\frac{1}{3}\) by taking the cube root of both numerator and denominator: \(\left(\sqrt[3]{\frac{125}{8}}\right)^2 = \left(\frac{5}{2}\right)^2\). Finally, square the fraction: \(\frac{25}{4}\).

M1 for applying either the negative reciprocal power or the cube root correctly, e.g., showing \(\left(\frac{125}{8}\right)^{\frac{2}{3}}\) or \(\left(\frac{2}{5}\right)^{-2}\). A1 for \(\frac{25}{4}\) (accept \(6\frac{1}{4}\) or \(6.25\)).

By Pythagoras' Theorem, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse. Let the height be \(h\): \(h^2 + 7^2 = 15^2 \Rightarrow h^2 + 49 = 225\). Subtracting \(49\) from both sides gives \(h^2 = 176\). Taking the square root gives \(h = \sqrt{176} \approx 13.266\text{ cm}\). To 3 significant figures, this is \(13.3\text{ cm}\).

M1 for applying Pythagoras' Theorem to find the shorter side: \(15^2 - 7^2\) or \(\sqrt{15^2 - 7^2}\). A1 for \(13.3\) (accept answers in the range \(13.2\) to \(13.3\)).

The value of the investment after 3 years is given by \(8000 \times (1 + 0.025)^3 = 8000 \times 1.025^3 = 8615.125\). The total interest earned is the final amount minus the original principal: \(8615.125 - 8000 = 615.125\). Rounded to the nearest penny, this is \(\text{£}615.13\).

M1 for a correct method to find the total value or total interest after 3 years, e.g., \(8000 \times 1.025^3\) (giving \(8615.125\)) or a complete year-by-year calculation. A1 for \(615.13\).

First, find the length \(AM\). Since \(ABC\) is isosceles with \(AB = AC\) and \(M\) is the midpoint of \(BC\), \(AM\) is perpendicular to \(BC\). Using Pythagoras' theorem in triangle \(ABM\): \(AM^2 = AB^2 - BM^2 = 10^2 - 6^2 = 64\), so \(AM = 8\text{ cm}\). Since \(VM\) is perpendicular to the base plane \(ABC\), the angle between \(VA\) and the base is \(\angle VAM\). In the right-angled triangle \(VMA\), \(\tan(\angle VAM) = \frac{VM}{AM} = \frac{15}{8}\). Thus, \(\angle VAM = \arctan(1.875) \approx 61.927^\circ\). To 1 decimal place, the angle is \(61.9^\circ\).

M1 for using Pythagoras to find \(AM = 8\). M1 for identifying the correct angle \(\angle VAM\). M1 for \(\tan(\theta) = 15/8\) or equivalent trig ratio. A1 for 61.9 (accept 61.9 degrees).

Using Pythagoras' theorem, \((2x - 17)^2 + (x + 12)^2 = (2x + 1)^2\). Expanding the terms gives \((4x^2 - 68x + 289) + (x^2 + 24x + 144) = 4x^2 + 4x + 1\). Simplifying both sides: \(5x^2 - 44x + 433 = 4x^2 + 4x + 1\). Rearranging into standard quadratic form: \(x^2 - 48x + 432 = 0\). Factoring the quadratic: \((x - 12)(x - 36) = 0\). This gives two possible values: \(x = 12\) and \(x = 36\). Both values produce positive physical lengths for all sides.

M1 for setting up Pythagoras' equation: \((2x - 17)^2 + (x + 12)^2 = (2x + 1)^2\). M1 for expanding correctly to get \(5x^2 - 44x + 433 = 4x^2 + 4x + 1\) or equivalent. M1 for reducing to \(x^2 - 48x + 432 = 0\) and attempting to solve. A1 for 12 and 36.

Let \(P\) be the original price of the laptop. The price after the first reduction is \(0.85P\). The price after the second reduction is \(0.85P \times 0.90 = 0.765P\). We are given \(0.765P = 535.50\). Solving for \(P\): \(P = 535.50 / 0.765 = 700\). After the sale, the original price is increased by \(5\%\), so the new price is \(700 \times 1.05 = 735\).

M1 for setting up the relation \(P \times 0.85 \times 0.9 = 535.50\) or equivalent. M1 for finding the combined multiplier \(0.765\). A1 for finding the original price of 700. A1 for the final answer 735.

The gradient of \(\mathbf{L}_1\) is \(m_1 = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\). Since \(\mathbf{L}_2\) is perpendicular to \(\mathbf{L}_1\), its gradient is \(m_2 = -\frac{1}{2}\). The midpoint of \(AB\) is \(M = \left(\frac{2+6}{2}, \frac{5+13}{2}\right) = (4, 9)\). The equation of \(\mathbf{L}_2\) is \(y - 9 = -\frac{1}{2}(x - 4)\). Multiplying by 2: \(2y - 18 = -x + 4\). Rearranging gives \(2y + x = 22\).

M1 for finding the gradient of \(\mathbf{L}_1\) as 2 or the perpendicular gradient as -0.5. M1 for finding the midpoint of \(AB\) as (4, 9). M1 for substituting their midpoint and perpendicular gradient into a line equation. A1 for 2y + x = 22 (or x + 2y = 22).

Expand the terms in the numerator: \((3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}\). \((3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}\). Subtracting these gives: \((14 + 6\sqrt{5}) - (14 - 6\sqrt{5}) = 12\sqrt{5}\). Simplify the denominator: \(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\). Dividing the simplified numerator by the simplified denominator: \(\frac{12\sqrt{5}}{4\sqrt{5}} = 3\).

M1 for expanding at least one of the squared brackets correctly. M1 for simplifying the numerator to \(12\sqrt{5}\). M1 for simplifying \(\sqrt{80}\) to \(4\sqrt{5}\). A1 for 3.

First, calculate the length of \(AC\) using the Cosine Rule: \(AC^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(60^\circ) = 49 + 100 - 70 = 79\), so \(AC = \sqrt{79} \approx 8.888\text{ cm}\). Second, find the area of triangle \(ABC\): \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(60^\circ) = \frac{1}{2} \times 7 \times 10 \times \frac{\sqrt{3}}{2} = 17.5\sqrt{3} \approx 30.31\text{ cm}^2\). The area is also given by \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BD\). Thus, \(30.31 = \frac{1}{2} \times 8.888 \times BD \Rightarrow BD = \frac{60.62}{8.888} \approx 6.82\text{ cm}\).

M1 for using the Cosine Rule to find \(AC = \sqrt{79}\) or 8.89. M1 for calculating the area of the triangle as \(17.5\sqrt{3}\) or 30.3. M1 for equating the area formulas: \(\frac{1}{2} \times \sqrt{79} \times BD = 17.5\sqrt{3}\). A1 for 6.82.

Originally, the number of female employees is \(0.60 \times 240 = 144\). The number of male employees is \(240 - 144 = 96\). After the recruitment drive, the number of male employees increases by \(25\%\), so the new number of male employees is \(96 \times 1.25 = 120\). Since the new total number of employees is 282, the new number of female employees is \(282 - 120 = 162\). The increase in female employees is \(162 - 144 = 18\). The percentage increase is \(\frac{18}{144} \times 100 = 12.5\%\).

M1 for finding original females (144) and original males (96). M1 for finding new males as \(96 \times 1.25 = 120\). M1 for finding new females as \(282 - 120 = 162\) and the increase of 18. A1 for 12.5 or 12.5%.

Multiply the entire equation by \(x(x-3)\) to clear the fractions: \(x + 4(x-3) = x(x-3)\). Expand both sides: \(x + 4x - 12 = x^2 - 3x\). Simplify the left side: \(5x - 12 = x^2 - 3x\). Rearrange into standard quadratic form: \(x^2 - 8x + 12 = 0\). Factorize the quadratic: \((x - 2)(x - 6) = 0\). This gives the solutions \(x = 2\) and \(x = 6\).

M1 for multiplying through by \(x(x-3)\) to obtain \(x + 4(x-3) = x(x-3)\). M1 for expanding and simplifying to a 3-term quadratic, e.g., \(x^2 - 8x + 12 = 0\). M1 for factorizing as \((x-2)(x-6) = 0\) or using the quadratic formula correctly. A1 for 2 and 6.

First, use the Cosine Rule to find the length of the side \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\). This gives \(AC^2 = 8.5^2 + 12.4^2 - 2 \times 8.5 \times 12.4 \times \cos(64^\circ) = 72.25 + 153.76 - 210.8 \times 0.438371... = 133.6012...\), so \(AC = \sqrt{133.6012...} \approx 11.5586\text{ cm}\). Next, calculate the area of triangle \(ABC\) using the sine area formula: \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC) = \frac{1}{2} \times 8.5 \times 12.4 \times \sin(64^\circ) = 52.7 \times 0.898794... \approx 47.3664\text{ cm}^2\). The area of a triangle can also be written in terms of its base \(AC\) and perpendicular height \(BD\): \(\text{Area} = \frac{1}{2} \times AC \times BD\). Thus, \(47.3664 = \frac{1}{2} \times 11.5586 \times BD\), which gives \(BD = \frac{2 \times 47.3664}{11.5586} \approx 8.1958\text{ cm}\). Rounding to 3 significant figures gives \(8.20\text{ cm}\).

M1: For a correct substitution into the Cosine Rule to find \(AC^2\) or \(AC\), e.g., \(8.5^2 + 12.4^2 - 2 \times 8.5 \times 12.4 \times \cos(64^\circ)\). M1: For a correct method to find the area of the triangle, e.g., \(\frac{1}{2} \times 8.5 \times 12.4 \times \sin(64^\circ)\) (yielding \(\approx 47.4\)). M1: For setting up an equation linking the area of the triangle and the perpendicular height \(BD\), e.g., \(\frac{1}{2} \times \text{their } AC \times BD = \text{their Area}\). A1: For a correct answer in the range \(8.19\) to \(8.21\) (or \(8.20\)).

Multiply all terms by the common denominator \((2x+1)(x-3)\) to clear the fractions: \(3(x-3) + 5(2x+1) = 2(2x+1)(x-3)\). Expand both sides: LHS is \(3x - 9 + 10x + 5 = 13x - 4\) and RHS is \(2(2x^2 - 5x - 3) = 4x^2 - 10x - 6\). Equate the sides and rearrange into a standard quadratic form \(ax^2 + bx + c = 0\): \(4x^2 - 23x - 2 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve: \(x = \frac{23 \pm \sqrt{(-23)^2 - 4(4)(-2)}}{8} = \frac{23 \pm \sqrt{561}}{8}\). Calculating the two values gives \(x \approx 5.84\) and \(x \approx -0.0857\).

M1: Correct method to eliminate fractions by multiplying by \((2x+1)(x-3)\) to obtain \(3(x-3) + 5(2x+1) = 2(2x+1)(x-3)\) or equivalent. M1: Correct expansion of both sides to get \(13x - 4 = 4x^2 - 10x - 6\) or equivalent. M1: Rearranging to get a correct 3-term quadratic equation, e.g., \(4x^2 - 23x - 2 = 0\), and attempting to solve using the quadratic formula (with at most one sign error). A1: Both solutions correct to 3 significant figures: \(5.84\) and \(-0.0857\).

First, write down an expression for the value of Company A's investment after 6 years: \(V_A = 6000 \times (1.035)^6 \approx \$7375.53\). Next, write down an expression for the value of Company B's investment after 6 years: \(V_B = 5500 \times (1 + \frac{r}{100})^6\). Since the values are equal after 6 years, set \(V_B = V_A\): \(5500 \times (1 + \frac{r}{100})^6 = 7375.5323...\). Divide by 5500: \((1 + \frac{r}{100})^6 \approx 1.34100588...\). Take the 6th root of both sides: \(1 + \frac{r}{100} \approx 1.0501145...\). Subtract 1 and multiply by 100 to find \(r\): \(r \approx 5.01145...\). To 2 decimal places, \(r = 5.01\).

M1: For a correct expression or calculation for Company A's investment after 6 years, e.g., \(6000 \times (1.035)^6\) or \(7375.53\). M1: For setting up a correct equation equating both investments, e.g., \(5500 \times (1 + \frac{r}{100})^6 = 6000 \times (1.035)^6\). M1: For a correct algebraic process to find \(1 + \frac{r}{100}\), e.g., \(\left(\frac{6000 \times (1.035)^6}{5500}\right)^{1/6}\). A1: For \(5.01\) (accept 5 or 5.0 or 5.01).

(a) Factor out 3 from the first two terms: \(3x^2 - 12x + 7 = 3(x^2 - 4x) + 7\). Complete the square inside the bracket: \(x^2 - 4x = (x - 2)^2 - 4\). Substitute this back: \(3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 12 + 7 = 3(x - 2)^2 - 5\). So, \(a = 3\), \(b = 2\), and \(c = -5\). (b) For a curve in the form \(y = a(x - b)^2 + c\), the turning point is at \((b, c)\). Therefore, the turning point of the curve is \((2, -5)\).

M1: For factoring out 3 or starting the completing square process, e.g., \(3(x^2 - 4x) ...\). M1: For completing the square correctly inside the bracket, e.g., \((x-2)^2 - 4\). A1: For the correct completed square form \(3(x - 2)^2 - 5\). B1: For the correct coordinates of the turning point \((2, -5)\) (follow through from their \(b\) and \(c\)).

Express all terms as powers of base 3: \(27 = 3^3\) and \(9 = 3^2\). Substituting these gives \((3^3)^{2x-1} = (3^2)^{x+4} \times 3^{x-2}\). Applying laws of indices: LHS becomes \(3^{3(2x-1)} = 3^{6x-3}\). RHS becomes \(3^{2(x+4)} \times 3^{x-2} = 3^{2x+8} \times 3^{x-2} = 3^{(2x+8)+(x-2)} = 3^{3x+6}\). Equating the indices gives \(6x - 3 = 3x + 6\). Solving for \(x\) yields \(3x = 9\), so \(x = 3\).

M1: For expressing 27 as \(3^3\) or 9 as \(3^2\) (or both) in the equation. M1: For correctly expanding the exponents on at least one side, e.g., \(3^{6x-3}\) or \(3^{2x+8}\). M1: For writing the RHS as a single power of 3, e.g., \(3^{3x+6}\). A1: For \(x = 3\) with correct working shown.

To prove that triangle \(ABE\) is similar to triangle \(DCE\), we find three equal corresponding angles: 1. \(\angle ABE = \angle DCE\) because they are angles in the same segment, subtended by the same arc \(AD\). 2. \(\angle BAE = \angle CDE\) because they are angles in the same segment, subtended by the same arc \(BC\). 3. \(\angle AEB = \angle DEC\) because they are vertically opposite angles. Since all three pairs of corresponding angles are equal, triangle \(ABE\) is similar to triangle \(DCE\) (by AAA similarity). Because the triangles are similar, the ratios of their corresponding sides must be equal: \(\frac{AE}{ED} = \frac{BE}{EC}\). Multiplying both sides of this equation by \(ED \times EC\), we obtain the required proof: \(AE \times EC = BE \times ED\).

M1: Identifies one pair of equal angles with a correct geometric reason (e.g., \(\angle ABE = \angle DCE\) because they are angles subtended by the same arc). M1: Identifies a second pair of equal angles with a correct geometric reason (e.g., \(\angle AEB = \angle DEC\) because they are vertically opposite). A1: Correctly concludes that triangle \(ABE\) is similar to triangle \(DCE\), citing AAA. M1: Sets up the correct ratio of corresponding sides from similar triangles: \(\frac{AE}{ED} = \frac{BE}{EC}\). A1: Rearranges the ratio to correctly prove \(AE \times EC = BE \times ED\).

First, find the vector \(\overrightarrow{OP}\). Since \(OABC\) is a parallelogram, \(\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}\). Since \(P\) lies on \(AB\) such that \(AP : PB = 2 : 1\), we have \(\overrightarrow{AP} = \frac{2}{3}\mathbf{c}\). Therefore, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{2}{3}\mathbf{c}\). Since \(Q\) lies on the line \(OP\), \(\overrightarrow{OQ} = k\overrightarrow{OP} = k\left(\mathbf{a} + \frac{2}{3}\mathbf{c}\right) = k\mathbf{a} + \frac{2}{3}k\mathbf{c}\) for some scalar \(k\). Since \(Q\) also lies on the diagonal \(AC\), we can write another pathway: \(\overrightarrow{OQ} = \overrightarrow{OA} + m\overrightarrow{AC}\) for some scalar \(m\). Since \(\overrightarrow{AC} = \mathbf{c} - \mathbf{a}\), we have: \(\overrightarrow{OQ} = \mathbf{a} + m(\mathbf{c} - \mathbf{a}) = (1 - m)\mathbf{a} + m\mathbf{c}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{c}\) from both expressions for \(\overrightarrow{OQ}\): 1) \(k = 1 - m\) and 2) \(\frac{2}{3}k = m\). Substituting (2) into (1) gives \(k = 1 - \frac{2}{3}k \implies \frac{5}{3}k = 1 \implies k = \frac{3}{5}\). Substituting \(k = \frac{3}{5}\) back into \(\overrightarrow{OQ} = k\mathbf{a} + \frac{2}{3}k\mathbf{c}\) yields: \(\overrightarrow{OQ} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{c}\). This completes the proof.

M1: Expresses \(\overrightarrow{OP}\) correctly in terms of \(\mathbf{a}\) and \(\mathbf{c}\) as \(\mathbf{a} + \frac{2}{3}\mathbf{c}\). M1: Writes two different vector expressions for \(\overrightarrow{OQ}\) involving unknown scalars, e.g., \(\overrightarrow{OQ} = k\left(\mathbf{a} + \frac{2}{3}\mathbf{c}\right)\) and \(\overrightarrow{OQ} = (1-m)\mathbf{a} + m\mathbf{c}\). M1: Sets up a system of simultaneous equations by equating the coefficients of \(\mathbf{a}\) and \(\mathbf{c}\). M1: Solves the simultaneous equations to find the value of either scalar (e.g., \(k = \frac{3}{5}\) or \(m = \frac{2}{5}\)). A1: Successfully obtains the exact target expression \(\overrightarrow{OQ} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{c}\) with full supporting working.

We are given that the area of triangle \(ABC\) is \(3\sqrt{3}\text{ cm}^2\). Using the area formula \(\text{Area} = \frac{1}{2}bc\sin A\): \(\frac{1}{2} \times x \times (x+2) \times \sin(60^\circ) = 3\sqrt{3}\). Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have: \(\frac{1}{2}x(x+2)\frac{\sqrt{3}}{2} = 3\sqrt{3}\) which simplifies to \(\frac{\sqrt{3}}{4}(x^2 + 2x) = 3\sqrt{3}\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives: \(x^2 + 2x = 12\). Next, apply the Cosine Rule to find the length of \(BC\): \(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(BAC)\). \(BC^2 = x^2 + (x+2)^2 - 2x(x+2)\cos(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), this simplifies to: \(BC^2 = x^2 + (x^2 + 4x + 4) - x(x+2)\) \(BC^2 = 2x^2 + 4x + 4 - x^2 - 2x\) \(BC^2 = x^2 + 2x + 4\). We already proved that \(x^2 + 2x = 12\). Substituting this into the simplified Cosine Rule expression: \(BC^2 = 12 + 4 = 16\). Therefore, \(BC = \sqrt{16} = 4\text{ cm}\) (since length must be positive). This completes the proof.

M1: Sets up the triangle area equation correctly: \(\frac{1}{2} x(x+2) \sin(60^\circ) = 3\sqrt{3}\). A1: Simplifies the area equation to show that \(x^2 + 2x = 12\) (or \(x^2 + 2x - 12 = 0\)). M1: Applies the Cosine Rule correctly for \(BC^2\): \(BC^2 = x^2 + (x+2)^2 - 2x(x+2)\cos(60^\circ)\). M1: Simplifies the Cosine Rule algebraic expression down to \(BC^2 = x^2 + 2x + 4\). A1: Substitutes \(12\) for \(x^2 + 2x\) to obtain \(BC^2 = 16\) and concludes with \(BC = 4\).

To find the stationary points, differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - 6x - 9\). Set \(\frac{dy}{dx} = 0\) to find the \(x\)-coordinates: \(3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x-3)(x+1) = 0\). Thus, the stationary points occur at \(x = 3\) and \(x = -1\). Now substitute these values back into the original curve equation to find the corresponding \(y\)-coordinates: For \(x = 3\): \(y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\). So \(P\) is \((3, -22)\). For \(x = -1\): \(y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\). So \(Q\) is \((-1, 10)\). Now determine the equation of the line passing through \(P(3, -22)\) and \(Q(-1, 10)\). Find the gradient \(m\): \(m = \frac{10 - (-22)}{-1 - 3} = \frac{32}{-4} = -8\). Using the point-gradient equation of a line with point \(Q(-1, 10)\): \(y - 10 = -8(x - (-1)) \implies y - 10 = -8x - 8 \implies 8x + y - 2 = 0\). This completes the proof.

M1: Differentiates the curve equation correctly to find \(\frac{dy}{dx} = 3x^2 - 6x - 9\). M1: Sets the derivative to 0 and solves the quadratic equation to get \(x = 3\) and \(x = -1\). M1: Correctly evaluates the \(y\)-coordinates to find the points \(P(3, -22)\) and \(Q(-1, 10)\). M1: Finds the gradient of the line passing through \(P\) and \(Q\) as \(-8\). A1: Formulates the equation of the line and simplifies it to the target form \(8x + y - 2 = 0\) with no errors.

The volume of a sphere of radius \(r\) is \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). The volume of a cone of base radius \(r\) and height \(h\) is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\). Since the metal sphere is recast into the cone, their volumes are equal: \(\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h\). Dividing both sides by \(\frac{1}{3}\pi r^2\) yields: \(h = 4r\). Next, find the slant height \(l\) of the cone using Pythagoras' theorem: \(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (4r)^2} = \sqrt{r^2 + 16r^2} = \sqrt{17r^2} = r\sqrt{17}\). The total surface area (\(A\)) of a cone is the sum of its base area and curved surface area: \(A = \pi r^2 + \pi r l\). Substitute \(l = r\sqrt{17}\) into this formula: \(A = \pi r^2 + \pi r (r\sqrt{17}) = \pi r^2 + \pi r^2 \sqrt{17}\). Factoring out \(\pi r^2\) gives: \(A = \pi r^2(1 + \sqrt{17})\). This completes the proof.

M1: Writes correct volume formulas for both the sphere and the cone in terms of \(r\) and \(h\). M1: Equates the two volumes and solves to express the height \(h\) in terms of \(r\) (obtaining \(h = 4r\)). M1: Uses Pythagoras' theorem to set up an expression for the slant height: \(l = \sqrt{r^2 + h^2}\). M1: Simplifies the slant height to \(l = r\sqrt{17}\) and substitutes it into the total surface area formula \(\text{TSA} = \pi r^2 + \pi r l\). A1: Correctly factors out \(\pi r^2\) to arrive at the exact expression \(\pi r^2(1 + \sqrt{17})\).

Let the length of the third side be \(x\text{ cm}\).
Using Pythagoras' theorem:
\(x^2 + 10^2 = 26^2\)
\(x^2 + 100 = 676\)
\(x^2 = 576\)
\(x = \sqrt{576} = 24\)

M1 for \(26^2 - 10^2\) or \(x^2 + 10^2 = 26^2\)
A1 for \(24\)

Factorising the quadratic expression:
\((x - 7)(x + 4) = 0\)
This gives \(x = 7\) or \(x = -4\).
The positive solution is \(x = 7\).

M1 for factorising as \((x - 7)(x + 4)\) or correct substitution into the quadratic formula
A1 for \(7\)

Let the original price be \(P\).
A loss of \(15\%\) means the sale price is \(85\%\) of the original price.
\(0.85 \times P = 340\)
\(P = \frac{340}{0.85} = 400\)
The original price was \(\$400\).

M1 for \(340 \div 0.85\) or an equivalent equation
A1 for \(400\)

The gradient \(m\) is given by:
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{23 - 5}{4 - (-2)}\)
\(m = \frac{18}{6} = 3\)

M1 for a correct fraction representing change in \(y\) over change in \(x\), e.g., \(\frac{23 - 5}{4 - (-2)}\)
A1 for \(3\)

Apply the fractional power to both parts of the term inside the bracket:
\(\left(64x^{12}\right)^{\frac{2}{3}} = 64^{\frac{2}{3}} \times \left(x^{12}\right)^{\frac{2}{3}}\)
\(64^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\)
\(\left(x^{12}\right)^{\frac{2}{3}} = x^{12 \times \frac{2}{3}} = x^8\)
This simplifies to \(16x^8\).

M1 for writing \(16\) as part of the term, or \(x^8\) as part of the term, or showing the correct root and power of both parts
A1 for \(16x^8\)

First, find the length of the hypotenuse \(AC\):
\(AC = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\).
Using the definition of cosine:
\(\cos(A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5}{13}\).

M1 for finding the hypotenuse is \(13\)
A1 for \(5/13\) or equivalent exact fraction

The depreciation factor is \(1 - 0.08 = 0.92\).
After 2 years (from the start of 2021 to the start of 2023), the value is:
\(15000 \times 0.92^2\)
\(= 15000 \times 0.8464 = 12696\)
The value of the car is \(\$12,696\).

M1 for \(15000 \times 0.92^2\) or \(15000 \times 0.92 = 13800\) followed by \(13800 \times 0.92\)
A1 for \(12696\)

Find the highest common factor of the coefficients and algebraic terms:
\(\text{HCF of } 12 \text{ and } 18 = 6\)
\(\text{HCF of } x^2y \text{ and } xy^2 = xy\)
Factorising out \(6xy\):
\(12x^2y - 18xy^2 = 6xy(2x - 3y)\)

M1 for any partial factorisation with at least a common variable or integer factorised out (e.g. \(6(2x^2y - 3xy^2)\) or \(xy(12x - 18y)\))
A1 for \(6xy(2x-3y)\)

1. Find the interior angle \(\angle ABC\):
The bearing from \(A\) to \(B\) is \(070^\circ\).
The back bearing from \(B\) to \(A\) is \(70^\circ + 180^\circ = 250^\circ\).
The bearing from \(B\) to \(C\) is \(150^\circ\).
Therefore, the interior angle \(\angle ABC = 250^\circ - 150^\circ = 100^\circ\).

2. Use the Cosine Rule to find the length \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 12^2 + 18^2 - 2(12)(18)\cos(100^\circ)\)
\(AC^2 = 144 + 324 - 432\cos(100^\circ)\)
Since \(\cos(100^\circ) \approx -0.173648\):
\(AC^2 = 468 - 432(-0.173648) \approx 468 + 75.016 = 543.016\)
\(AC = \sqrt{543.016} \approx 23.3027\text{ km}\).

Rounding to 3 significant figures gives \(23.3\text{ km}\).

M1 for finding the angle \(\angle ABC = 100^\circ\)
M1 for substituting correctly into the Cosine Rule: \(AC^2 = 12^2 + 18^2 - 2 \times 12 \times 18 \times \cos(100^\circ)\)
A1 for evaluating \(AC^2 \approx 543\) or \(AC \approx 23.3...\)
A1 for the final answer of \(23.3\) (accept answers in the range \(23.3\) to \(23.4\))

1. Apply Pythagoras' theorem:
\(x^2 + (2x + 2)^2 = (2x + 3)^2\)

2. Expand the brackets:
\(x^2 + (4x^2 + 8x + 4) = 4x^2 + 12x + 9\)

3. Simplify and rearrange into standard quadratic form:
\(5x^2 + 8x + 4 = 4x^2 + 12x + 9\)
\(x^2 - 4x - 5 = 0\)

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

5. This gives \(x = 5\) or \(x = -1\).
Since a length must be positive, we reject \(x = -1\). Therefore, \(x = 5\).

M1 for setting up Pythagoras' theorem: \(x^2 + (2x + 2)^2 = (2x + 3)^2\)
M1 for expanding correctly: \(x^2 + 4x^2 + 8x + 4 = 4x^2 + 12x + 9\)
M1 for rearranging to \(x^2 - 4x - 5 = 0\) and attempting to solve (by factoring or formula)
A1 for \(x = 5\) (rejecting \(x = -1\) since length must be positive)

1. Form an equation for the profit in 2023:
\(40000 \left(1 + \frac{x}{100}\right)\left(1 + \frac{2x}{100}\right) = 52800\)

2. Divide both sides by 40,000:
\(\left(1 + \frac{x}{100}\right)\left(1 + \frac{2x}{100}\right) = 1.32\)

3. Let \(k = \frac{x}{100}\). Then:
\((1 + k)(1 + 2k) = 1.32\)
\(1 + 3k + 2k^2 = 1.32\)
\(2k^2 + 3k - 0.32 = 0\)

4. Multiply by 100 to clear decimals:
\(200k^2 + 300k - 32 = 0\)
Divide by 4:
\(50k^2 + 75k - 8 = 0\)

5. Factorise the quadratic:
\((10k - 1)(5k + 8) = 0\)
This gives \(k = 0.1\) or \(k = -1.6\).

6. Since the profit increased, \(x\) (and thus \(k\)) must be positive.
Therefore, \(k = 0.1\), which means:
\(\frac{x}{100} = 0.1 \implies x = 10\).

M1 for writing down a correct initial equation, e.g. \(40000(1 + \frac{x}{100})(1 + \frac{2x}{100}) = 52800\)
M1 for expanding and simplifying to a 3-term quadratic equation, e.g. \(2k^2 + 3k - 0.32 = 0\) or \(2x^2 + 300x - 3200 = 0\)
M1 for a valid method to solve their quadratic equation
A1 for \(x = 10\) (with clear algebraic working shown, rejecting the negative root)

1. Express \(27\) and \(9\) as powers of \(3\):
\(27 = 3^3\) and \(9 = 3^2\).

2. Substitute these into the equation:
\((3^3)^{2x - 1} = \frac{(3^2)^{x + 4}}{3^{x - 1}}\)

3. Apply the laws of indices:
\(3^{3(2x - 1)} = \frac{3^{2(x + 4)}}{3^{x - 1}}\)
\(3^{6x - 3} = \frac{3^{2x + 8}}{3^{x - 1}}\)

4. Simplify the division on the right-hand side by subtracting exponents:
\(3^{6x - 3} = 3^{(2x + 8) - (x - 1)}
\)3^{6x - 3} = 3^{x + 9}\)

5. Equate the exponents:
\(6x - 3 = x + 9\)
\(5x = 12\)
\(x = \frac{12}{5} = 2.4\)

M1 for writing \(27^{2x - 1}\) as \(3^{3(2x-1)}\) or \(3^{6x-3}\), or \(9^{x+4}\) as \(3^{2(x+4)}\) or \(3^{2x+8}\)
M1 for the correct division rule on the RHS: \(3^{(2x + 8) - (x - 1)}\) leading to \(3^{x+9}\)
M1 for equating the indices: \(6x - 3 = x + 9\) (or equivalent equation from their indices)
A1 for \(2.4\) (or \(\frac{12}{5}\) or \(2\frac{2}{5}\))

1. Find the gradient of the radius from the origin \((0,0)\) to the point \(P(3, -5)\):
\(m_{\text{radius}} = \frac{-5 - 0}{3 - 0} = -\frac{5}{3}\)

2. The tangent is perpendicular to the radius, so its gradient \(m_{\text{tangent}}\) is the negative reciprocal:
\(m_{\text{tangent}} = -\frac{1}{-5/3} = \frac{3}{5}\)

3. Write down the equation of the tangent line using point-slope form with \(P(3, -5)\):
\(y - (-5) = \frac{3}{5}(x - 3)\)
\(y + 5 = \frac{3}{5}(x - 3)\)

4. Rearrange this into the form \(ax + by + c = 0\):
Multiply both sides by 5:
\(5(y + 5) = 3(x - 3)\)
\(5y + 25 = 3x - 9\)
Subtract \(5y\) and \(25\) from both sides:
\(3x - 5y - 34 = 0\)

M1 for finding the gradient of the radius: \(-\frac{5}{3}\)
M1 for finding the gradient of the tangent using the perpendicular gradient rule: \(m = \frac{3}{5}\)
M1 for substituting their gradient and the point \((3, -5)\) into a linear equation formula, e.g., \(y - (-5) = \frac{3}{5}(x - 3)\)
A1 for \(3x - 5y - 34 = 0\) (or any integer multiple, e.g., \(-3x + 5y + 34 = 0\))

1. Substitute \(y = 2x - 3\) into the second equation:
\(x^2 + (2x - 3)^2 = 41\)

2. Expand the bracketed term:
\(x^2 + (4x^2 - 12x + 9) = 41\)

3. Simplify and rearrange into standard quadratic form:
\(5x^2 - 12x - 32 = 0\)

4. Solve the quadratic equation using the quadratic formula:
\(x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-32)}}{2(5)}\)
\(x = \frac{12 \pm \sqrt{144 + 640}}{10}\)
\(x = \frac{12 \pm \sqrt{784}}{10}\)
Since \(\sqrt{784} = 28\):
\(x = \frac{12 \pm 28}{10}\)
So, \(x = 4\) or \(x = -1.6\).

5. Find the corresponding values of \(y\) using \(y = 2x - 3\):
If \(x = 4\):
\(y = 2(4) - 3 = 5\)
If \(x = -1.6\):
\(y = 2(-1.6) - 3 = -6.2\)

Thus, the solutions are \(x = 4, y = 5\) and \(x = -1.6, y = -6.2\).

M1 for substituting \(2x - 3\) into \(x^2 + y^2 = 41\)
M1 for expanding and simplifying to a 3-term quadratic equation, e.g., \(5x^2 - 12x - 32 = 0\)
M1 for a valid method to solve their quadratic equation, yielding \(x = 4\) and \(x = -1.6\) (or \(y = 5\) and \(y = -6.2\))
A1 for both pairs of correct solutions: \(x = 4, y = 5\) and \(x = -1.6, y = -6.2\) (or equivalent fractions)

1. Let the original price of the laptop be \(P\).

2. Apply the 15% discount:
The price after discount is \(P \times (1 - 0.15) = 0.85P\).

3. Apply the 20% sales tax to the discounted price:
The final price is \(0.85P \times (1 + 0.20) = 0.85P \times 1.20\).

4. Set up the equation with the final price:
\(0.85 \times 1.20 \times P = 561\)

5. Simplify the multipliers:
\(1.02P = 561\)

6. Solve for \(P\):
\(P = \frac{561}{1.02} = 550\)

Therefore, the original price of the laptop was £550.

M1 for representing the 15% discount as a multiplier of \(0.85\) or setting up an expression like \(0.85P\)
M1 for representing the 20% sales tax as a multiplier of \(1.2\) or multiplying their expression by \(1.2\)
M1 for setting up the equation \(0.85 \times 1.2 \times P = 561\) (or equivalent, e.g., finding the price before tax first: \(561 \div 1.2 = 467.50\))
A1 for \(550\) (with clear working)

1. Simplify \(\sqrt{12}\) as a surd:
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
So the expression becomes:
\(\frac{4 + 2\sqrt{3}}{\sqrt{3} - 1}\)

2. Rationalise the denominator by multiplying the numerator and denominator by \(\sqrt{3} + 1\):
\(\frac{(4 + 2\sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)

3. Expand the numerator:
\((4 + 2\sqrt{3})(\sqrt{3} + 1) = 4\sqrt{3} + 4 + 2\sqrt{3}\sqrt{3} + 2\sqrt{3}\)
\(= 4\sqrt{3} + 4 + 2(3) + 2\sqrt{3}\)
\(= 4\sqrt{3} + 4 + 6 + 2\sqrt{3}\)
\(= 10 + 6\sqrt{3}\)

4. Expand the denominator:
\((\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2\)

5. Simplify the fraction:
\(\frac{10 + 6\sqrt{3}}{2} = 5 + 3\sqrt{3}\)

Thus, \(a = 5\) and \(b = 3\).

M1 for simplifying \(\sqrt{12}\) to \(2\sqrt{3}\)
M1 for multiplying numerator and denominator by \((\sqrt{3} + 1)\)
M1 for correctly expanding the numerator to \(10 + 6\sqrt{3}\) or the denominator to \(2\)
A1 for both \(a = 5\) and \(b = 3\) (or writing the form \(5 + 3\sqrt{3}\) with clear values of \(a\) and \(b\))

The formula for the area of a triangle is:
\(\text{Area} = \frac{1}{2} a b \sin C\)

Substitute the given values into the formula:
\(10\sqrt{3} = \frac{1}{2} \times x \times (x + 3) \times \sin(60^\circ)\)

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\(10\sqrt{3} = \frac{1}{2} \times x(x+3) \times \frac{\sqrt{3}}{2}\)
\(10\sqrt{3} = \frac{\sqrt{3}}{4} (x^2 + 3x)\)

Divide both sides by \(\sqrt{3}\) and multiply by 4:
\(40 = x^2 + 3x\)
\(x^2 + 3x - 40 = 0\)

Factorise the quadratic equation:
\((x - 5)(x + 8) = 0\)

Since the side length \(x\) must be positive, we have:
\(x = 5\)

This means \(AB = 5\text{ cm}\) and \(BC = 5 + 3 = 8\text{ cm}\).

Now, use the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times \cos(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\):
\(AC^2 = 25 + 64 - 80 \times 0.5\)
\(AC^2 = 89 - 40 = 49\)
\(AC = \sqrt{49} = 7\text{ cm}\).

M1: Set up an algebraic equation for the area of the triangle, e.g., \(\frac{1}{2} x(x+3)\sin(60^\circ) = 10\sqrt{3}\).
M1: Simplify to a quadratic equation of the form \(x^2 + 3x - 40 = 0\) and solve to find the positive root \(x = 5\).
M1: Correctly substitute \(AB = 5\) and \(BC = 8\) into the Cosine Rule.
A1: Correct length of \(AC = 7\).

Using Pythagoras' theorem, we can set up the equation:
\((x - 1)^2 + (3x)^2 = (3x + 1)^2\)

Expand the brackets:
\(x^2 - 2x + 1 + 9x^2 = 9x^2 + 6x + 1\)

Subtract \(9x^2 + 1\) from both sides to simplify:
\(x^2 - 2x = 6x\)

Rearrange to form a quadratic equation:
\(x^2 - 8x = 0\)

Factorise the equation:
\(x(x - 8) = 0\)

Since side lengths must be positive, \(x\) cannot be 0 (as \(x - 1 > 0 \implies x > 1\)). Therefore:
\(x = 8\)

Now we calculate each side length:
- First side: \(x - 1 = 8 - 1 = 7\text{ cm}\)
- Second side: \(3x = 3 \times 8 = 24\text{ cm}\)
- Hypotenuse: \(3x + 1 = 3 \times 8 + 1 = 25\text{ cm}\)

The perimeter is the sum of these three sides:
\(\text{Perimeter} = 7 + 24 + 25 = 56\text{ cm}\).

M1: Establish the Pythagoras' equation \((x - 1)^2 + (3x)^2 = (3x + 1)^2\).
M1: Expand and simplify to a quadratic equation, e.g., \(x^2 - 8x = 0\).
M1: Find the positive solution \(x = 8\) and use it to determine the side lengths: 7, 24, and 25.
A1: Correctly state the perimeter as 56.

Let \(y\) represent the multiplier for the compound interest, where \(y = 1 + \frac{r}{100}\).

At the end of 2 years:
\(P \times y^2 = 8820\) (Equation 1)

At the end of 4 years:
\(P \times y^4 = 9724.05\) (Equation 2)

Divide Equation 2 by Equation 1:
\(\frac{P \times y^4}{P \times y^2} = \frac{9724.05}{8820}\)
\(y^2 = 1.1025\)

Take the square root of both sides to find \(y\) (since \(y > 0\)):
\(y = \sqrt{1.1025} = 1.05\)

Now, substitute \(y^2 = 1.1025\) back into Equation 1 to find \(P\):
\(P \times 1.1025 = 8820\)
\(P = \frac{8820}{1.1025} = 8000\)

Thus, the initial investment was £8000.

M1: Set up two simultaneous compound interest equations, e.g., \(P y^2 = 8820\) and \(P y^4 = 9724.05\).
M1: Eliminate \(P\) by division to get \(y^2 = 1.1025\) (or equivalent).
M1: Find \(y = 1.05\) or solve for \(P\) directly by division \(P = \frac{8820}{1.1025}\).
A1: Correct initial investment of 8000 (or £8000).

There are two main methods to solve this question:

**Method 1: Completing the Square**
Since the curve has a minimum point at \((2, 4)\), we can write the equation in vertex form:
\(y = a(x - 2)^2 + 4\)

Expand the brackets:
\(y = a(x^2 - 4x + 4) + 4\)
\(y = a x^2 - 4a x + (4a + 4)\)

Comparing this with \(y = a x^2 + b x + 12\), we can equate the constant terms:
\(4a + 4 = 12\)
\(4a = 8 \implies a = 2\)

Now equate the coefficients of \(x\):
\(b = -4a = -4(2) = -8\)

**Method 2: Using Calculus**
The gradient at the minimum point is 0, so we find the derivative \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = 2ax + b\)

At \(x = 2\), \(\frac{dy}{dx} = 0\):
\(2a(2) + b = 0 \implies b = -4a\)

Since the minimum point \((2, 4)\) lies on the curve, substitute \(x = 2\) and \(y = 4\) into the original equation:
\(4 = a(2)^2 + b(2) + 12\)
\(4 = 4a + 2b + 12\)
\(4a + 2b = -8\)

Substitute \(b = -4a\) into this equation:
\(4a + 2(-4a) = -8\)
\(-4a = -8 \implies a = 2\)

Find \(b\):
\(b = -4(2) = -8\).

M1: Write the quadratic in vertex form \(a(x-2)^2 + 4\) OR use differentiation to find \(4a+b=0\).
M1: Equate the constant term \(4a+4 = 12\) OR substitute the point \((2, 4)\) into the original equation to get \(4a+2b = -8\).
M1: Solve the resulting system of equations to find either \(a\) or \(b\).
A1: Correct values: \(a = 2\) and \(b = -8\).

First, simplify \(\sqrt{18}\):
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)

So the expression becomes:
\(\frac{4 + \sqrt{2}}{3\sqrt{2} - 2}\)

To rationalise the denominator, multiply the numerator and the denominator by the conjugate of the denominator, \(3\sqrt{2} + 2\):
\(\frac{(4 + \sqrt{2})(3\sqrt{2} + 2)}{(3\sqrt{2} - 2)(3\sqrt{2} + 2)}\)

Expand the numerator:
\((4 + \sqrt{2})(3\sqrt{2} + 2) = 4(3\sqrt{2}) + 4(2) + \sqrt{2}(3\sqrt{2}) + 2\sqrt{2}\)
\(= 12\sqrt{2} + 8 + 6 + 2\sqrt{2}\)
\(= 14 + 14\sqrt{2}\)

Expand the denominator:
\((3\sqrt{2} - 2)(3\sqrt{2} + 2) = (3\sqrt{2})^2 - 2^2\)
\(= 18 - 4 = 14\)

Rewrite the fraction:
\(\frac{14 + 14\sqrt{2}}{14} = \frac{14}{14} + \frac{14\sqrt{2}}{14} = 1 + \sqrt{2}\)

Thus, \(a = 1\) and \(b = 1\), which are integers.

M1: Simplify \(\sqrt{18}\) to \(3\sqrt{2}\).
M1: Multiply the numerator and denominator by \(3\sqrt{2} + 2\) (or equivalent for their unsimplified denominator).
M1: Correct expansion of the numerator to \(14 + 14\sqrt{2}\) or denominator to \(14\).
A1: Final simplified expression \(1 + \sqrt{2}\) obtained from fully correct working.

Let's find the interior angle \(\angle PQR\) in the triangle \(PQR\).

Draw a North line at \(P\) and a North line at \(Q\).
The bearing of \(Q\) from \(P\) is \(070^\circ\).
Using co-interior angles, the angle between the line \(PQ\) and the North-to-South line at \(Q\) is \(70^\circ\).
Therefore, the bearing of \(P\) from \(Q\) is:
\(70^\circ + 180^\circ = 250^\circ\)

The bearing of \(R\) from \(Q\) is \(130^\circ\).

The interior angle \(\angle PQR\) is the difference between these two bearings:
\(\angle PQR = 250^\circ - 130^\circ = 120^\circ\)

Now, we use the Cosine Rule in triangle \(PQR\) to find the length of \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\)
\(PR^2 = 15^2 + 20^2 - 2 \times 15 \times 20 \times \cos(120^\circ)\)

Since \(\cos(120^\circ) = -0.5\):
\(PR^2 = 225 + 400 - 600 \times (-0.5)\)
\(PR^2 = 625 + 300 = 925\)

Take the square root of both sides:
\(PR = \sqrt{925} \approx 30.4138\text{ km}\)

Correct to 3 significant figures, the distance is \(30.4\text{ km}\).

M1: Correct use of bearings or co-interior angles to establish the relationship of angles at \(Q\).
M1: Find the interior angle \(\angle PQR = 120^\circ\).
M1: Substitute correctly into the Cosine Rule: \(PR^2 = 15^2 + 20^2 - 2(15)(20)\cos(120^\circ)\).
A1: Correct distance of \(30.4\) (accept answers in the range \(30.4 - 30.5\)).

The largest angle in a triangle is always opposite the longest side.
Since \(d > 0\), the longest side is \(x+d\).
By the Cosine Rule:
\[(x+d)^2 = (x-d)^2 + x^2 - 2 \cdot (x-d) \cdot x \cdot \cos\theta\]
Substitute \(\cos\theta = \frac{1}{4}\):
\[(x+d)^2 = (x-d)^2 + x^2 - 2x(x-d)\left(\frac{1}{4}\right)\]
Expand the squared terms:
\[x^2 + 2xd + d^2 = x^2 - 2xd + d^2 + x^2 - \frac{1}{2}x(x-d)\]
Simplify by subtracting \(x^2 + d^2\) from both sides:
\[2xd = -2xd + x^2 - \frac{1}{2}x^2 + \frac{1}{2}xd\]
\[2xd = -2xd + \frac{1}{2}x^2 + \frac{1}{2}xd\]
Add \(2xd\) and subtract \(\frac{1}{2}xd\) from both sides:
\[\frac{7}{2}xd = \frac{1}{2}x^2\]
Multiply both sides by 2:
\[7xd = x^2\]
Since \(x > 0\), we can divide both sides by \(x\):
\[7d = x\]
Hence, \(x = 7d\).

M1: Identifies that the largest angle \(\theta\) is opposite the largest side \(x+d\).
M1: Sets up the Cosine Rule correctly: \((x+d)^2 = (x-d)^2 + x^2 - 2x(x-d)\cos\theta\).
M1: Correctly expands both quadratic brackets: \((x+d)^2 = x^2 + 2xd + d^2\) and \((x-d)^2 = x^2 - 2xd + d^2\).
M1: Substitutes \(\cos\theta = \frac{1}{4}\) and simplifies to a single quadratic equation in terms of \(x\) and \(d\).
M1: Simplifies to \(7xd = x^2\) or equivalent.
A1: Divides by \(x\) (with justification that \(x \neq 0\)) to obtain the final proof \(x = 7d\).

Let the length of the side parallel to the wall be \(y\) metres.
There are 4 fences of length \(x\) (the 2 outer sides and the 2 internal dividing fences).
The total length of fencing is:
\[4x + y = 120\]
Rearranging for \(y\):
\[y = 120 - 4x\]
The total area, \(A\), of the rectangular enclosure is:
\[A = x \cdot y = x(120 - 4x) = 120x - 4x^2\]
To find the maximum area, we complete the square:
\[A = -4(x^2 - 30x)\]
\[A = -4((x-15)^2 - 225)\]
\[A = 900 - 4(x-15)^2\]
Since \((x-15)^2 \ge 0\) for all real \(x\), the maximum value of \(A\) is \(900\text{ m}^2\), which occurs when \(x = 15\).
When \(x = 15\), the length of the side parallel to the wall is:
\[y = 120 - 4(15) = 120 - 60 = 60\text{ metres}\]
This completes the proof.

M1: Sets up an expression for the total fencing length: \(4x + y = 120\) (or equivalent with different variable).
M1: Writes the area of the enclosure in terms of a single variable: \(A = x(120 - 4x)\) or \(A = y\left(\frac{120 - y}{4}\right)\).
M1: Expands to get a quadratic expression: \(A = 120x - 4x^2\) or \(A = 30y - \frac{1}{4}y^2\).
M1: Completes the square or uses calculus (differentiation) to find the stationary point.
M1: Identifies \(x = 15\) (or \(y = 60\)) as the value giving the maximum area.
A1: Shows clearly that the maximum area is \(900\text{ m}^2\) and that \(y = 60\text{ m}\) at this maximum.

First, express \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\):
\(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\)
\(\overrightarrow{OQ} = \frac{1}{2}\mathbf{b}\)

Since \(X\) lies on the line \(AQ\), let \(\overrightarrow{AX} = \mu \overrightarrow{AQ}\) for some scalar \(\mu\).
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\)
Therefore,
\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \mu(-\mathbf{a} + \frac{1}{2}\mathbf{b}) = (1-\mu)\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\) --- (Equation 1)

Since \(X\) lies on the line \(BP\), let \(\overrightarrow{PX} = \lambda \overrightarrow{PB}\) for some scalar \(\lambda\).
\(\overrightarrow{PB} = \overrightarrow{PO} + \overrightarrow{OB} = -\frac{2}{3}\mathbf{a} + \mathbf{b}\)
Therefore,
\(\overrightarrow{OX} = \overrightarrow{OP} + \overrightarrow{PX} = \frac{2}{3}\mathbf{a} + \lambda(-\frac{2}{3}\mathbf{a} + \mathbf{b}) = \frac{2}{3}(1-\lambda)\mathbf{a} + \lambda\mathbf{b}\) --- (Equation 2)

Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from Equation 1 and Equation 2:
From \(\mathbf{b}\): \(\frac{1}{2}\mu = \lambda \implies \mu = 2\lambda\)
From \(\mathbf{a}\): \(1-\mu = \frac{2}{3}(1-\lambda)\)

Substitute \(\mu = 2\lambda\) into the second equation:
\[1 - 2\lambda = \frac{2}{3} - \frac{2}{3}\lambda\]
Multiply by 3:
\[3 - 6\lambda = 2 - 2\lambda\]
\[1 = 4\lambda \implies \lambda = \frac{1}{4}\]
Substitute \(\lambda = \frac{1}{4}\) back to get \(\mu\):
\[\mu = 2\left(\frac{1}{4}\right) = \frac{1}{2}\]

Substitute \(\mu = \frac{1}{2}\) into Equation 1:
\[\overrightarrow{OX} = \left(1-\frac{1}{2}\right)\mathbf{a} + \frac{1}{2}\left(\frac{1}{2}\right)\mathbf{b} = \frac{1}{2}\mathbf{a} + \frac{1}{4}\mathbf{b}\]
This proves the required result.

M1: Writes down correct vector expressions for \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: Expresses \(\overrightarrow{AQ}\) as \(-\mathbf{a} + \frac{1}{2}\mathbf{b}\) and sets up a vector equation for \(\overrightarrow{OX}\) along \(AQ\): \(\overrightarrow{OX} = (1-\mu)\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\).
M1: Expresses \(\overrightarrow{PB}\) as \(-\frac{2}{3}\mathbf{a} + \mathbf{b}\) and sets up a vector equation for \(\overrightarrow{OX}\) along \(BP\): \(\overrightarrow{OX} = \frac{2}{3}(1-\lambda)\mathbf{a} + \lambda\mathbf{b}\).
M1: Equates the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations for \(\mu\) and \(\lambda\).
M1: Solves the simultaneous equations to find \(\lambda = \frac{1}{4}\) or \(\mu = \frac{1}{2}\).
A1: Substitutes the scalar value back to correctly show \(\overrightarrow{OX} = \frac{1}{2}\mathbf{a} + \frac{1}{4}\mathbf{b}\).

First, simplify the expression for the area:
\[\text{Area} = \sqrt{125} + \sqrt{45} = \sqrt{25 \times 5} + \sqrt{9 \times 5} = 5\sqrt{5} + 3\sqrt{5} = 8\sqrt{5}\]

Second, simplify the expression for the length by rationalising the denominator:
\[\text{Length} = \frac{4}{\sqrt{5}-1} = \frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\]
\[\text{Length} = \frac{4(\sqrt{5}+1)}{5 - 1} = \frac{4(\sqrt{5}+1)}{4} = \sqrt{5} + 1\]

Now, the width \(W\) is given by:
\[W = \frac{\text{Area}}{\text{Length}} = \frac{8\sqrt{5}}{\sqrt{5}+1}\]

Rationalise the denominator of \(W\):
\[W = \frac{8\sqrt{5}(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}\]
\[W = \frac{8\sqrt{5}\sqrt{5} - 8\sqrt{5}}{5 - 1}\]
\[W = \frac{8 \times 5 - 8\sqrt{5}}{4}\]
\[W = \frac{40 - 8\sqrt{5}}{4}\]
\[W = 10 - 2\sqrt{5}\]
This completes the proof.

M1: Simplifies \(\sqrt{125}\) to \(5\sqrt{5}\) or \(\sqrt{45}\) to \(3\sqrt{5}\).
M1: Correctly expresses the total Area as \(8\sqrt{5}\).
M1: Multiplies the numerator and denominator of the length by \((\sqrt{5}+1)\).
M1: Simplifies the length to \(\sqrt{5}+1\).
M1: Sets up the fraction for the width \(\frac{8\sqrt{5}}{\sqrt{5}+1}\) and multiplies numerator and denominator by \((\sqrt{5}-1)\).
A1: Shows clear algebraic steps to arrive at the final simplified width \(10 - 2\sqrt{5}\) with no errors.