2023 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

First factorise out the highest common factor, \(2xy\):
\[18x^3y - 50xy^3 = 2xy(9x^2 - 25y^2)\]
Next, factorise the difference of two squares:
\[9x^2 - 25y^2 = (3x - 5y)(3x + 5y)\]
Thus, the fully factorised expression is:
\[2xy(3x - 5y)(3x + 5y)\]

M1 for correctly extracting a common factor, e.g. \(2xy(9x^2 - 25y^2)\) or \(2(9x^3y - 25xy^3)\) or difference of squares seen
A1 for \(2xy(3x - 5y)(3x + 5y)\) oe

Multiply the numerator and denominator by the conjugate \(3\sqrt{2} + 1\):
\[\frac{14(3\sqrt{2} + 1)}{(3\sqrt{2} - 1)(3\sqrt{2} + 1)}\]
Expand the denominator using the difference of two squares:
\[(3\sqrt{2})^2 - 1^2 = 9(2) - 1 = 18 - 1 = 17\]
Expand the numerator:
\[14(3\sqrt{2} + 1) = 42\sqrt{2} + 14\]
So the simplified fraction is:
\[\frac{42\sqrt{2} + 14}{17} \quad \text{or} \quad \frac{14(3\sqrt{2} + 1)}{17}\]

M1 for multiplying numerator and denominator by \(3\sqrt{2} + 1\)
A1 for \(\frac{42\sqrt{2} + 14}{17}\) or \(\frac{14(3\sqrt{2} + 1)}{17}\) cao

Express both sides as powers of base 3:
\[27 = 3^3 \implies 27^{2x-1} = (3^3)^{2x-1} = 3^{3(2x-1)} = 3^{6x-3}\]
\[\frac{1}{9} = 3^{-2} \implies \left(\frac{1}{9}\right)^{x+4} = (3^{-2})^{x+4} = 3^{-2(x+4)} = 3^{-2x-8}\]
Equating exponents:
\[6x - 3 = -2x - 8\]
\[8x = -5\]
\[x = -\frac{5}{8}\]

M1 for expressing both sides in terms of power of 3 (or 9/27), e.g. \(3^{6x-3}\) and \(3^{-2x-8}\) soi
A1 for \(-\frac{5}{8}\) or \(-0.625\)

Use the logarithm subtraction law:
\[\log_2\left(\frac{x + 6}{x - 1}\right) = 3\]
Convert from logarithmic to exponential form:
\[\frac{x + 6}{x - 1} = 2^3 = 8\]
Solve for \(x\):
\[x + 6 = 8(x - 1)\]
\[x + 6 = 8x - 8\]
\[14 = 7x\]
\[x = 2\]

M1 for applying log laws to obtain \(\log_2\left(\frac{x+6}{x-1}\right) = 3\) or \(\frac{x+6}{x-1} = 2^3\)
A1 for \(x = 2\) cao

Find the common denominator \((2x - 1)(x + 3)\):
\[\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\]
Expand the brackets in the numerator:
\[3(x + 3) - 2(2x - 1) = 3x + 9 - 4x + 2 = 11 - x\]
Thus, the fraction in simplest form is:
\[\frac{11 - x}{(2x - 1)(x + 3)}\]

M1 for common denominator \((2x - 1)(x + 3)\) seen with correct numerator expansion \(3(x+3) - 2(2x-1)\)
A1 for \(\frac{11 - x}{(2x - 1)(x + 3)}\) or \(\frac{-x + 11}{2x^2 + 5x - 3}\) oe

Simplify each surd by extracting the largest square factor:
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\)
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)
\(\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\)

Combine the like surds:
\(5\sqrt{2} - 3\sqrt{2} + 7\sqrt{2} = (5 - 3 + 7)\sqrt{2} = 9\sqrt{2}\)

M1 for at least two surds written correctly in the form \(k\sqrt{2}\) (e.g. \(5\sqrt{2}\), \(3\sqrt{2}\), or \(7\sqrt{2}\))
A1 for \(9\sqrt{2}\) cao

Look for two numbers that multiply to \(6 \times (-5) = -30\) and add up to \(-7\). These numbers are \(-10\) and \(3\).

Split the middle term:
\(6x^2 - 10x + 3x - 5\)
\(= 2x(3x - 5) + 1(3x - 5)\)
\(= (2x + 1)(3x - 5)\)

M1 for \((2x + a)(3x + b)\) where \(ab = -5\) or \(3a + 2b = -7\), or for correct splitting of the middle term \(6x^2 - 10x + 3x - 5\)
A1 for \((2x + 1)(3x - 5)\) oe

Find the common denominator \((2x - 1)(x + 4)\):

\[\frac{3(x + 4) - 2(2x - 1)}{(2x - 1)(x + 4)}\]

Expand the numerator:
\(3x + 12 - 4x + 2 = 14 - x\)

Combine into one fraction:
\[\frac{14 - x}{(2x - 1)(x + 4)}\]

M1 for a common denominator of \((2x - 1)(x + 4)\) seen with correct expansion of at least one numerator term, e.g. \(3(x + 4) - 2(2x - 1)\)
A1 for \(\frac{14 - x}{(2x - 1)(x + 4)}\) oe (e.g. \(\frac{14 - x}{2x^2 + 7x - 4}\))

Use the addition rule of logarithms:
\(\log_3(x(x - 8)) = 2\)

Convert from logarithmic to exponential form:
\(x(x - 8) = 3^2\)
\(x^2 - 8x = 9\)
\(x^2 - 8x - 9 = 0\)

Factorise the quadratic equation:
\((x - 9)(x + 1) = 0\)
\(x = 9\) or \(x = -1\)

Since the domain of \(\log_3 x\) requires \(x > 0\) and \(x - 8 > 0\), reject \(x = -1\).
Therefore, \(x = 9\).

M1 for \(x(x - 8) = 3^2\) or \(x^2 - 8x - 9 = 0\) soi
A1 for \(x = 9\) only (rejecting \(x = -1\))

Multiply both sides by \((5 - p)\):
\(q(5 - p) = 2p + 3\)

Expand the brackets:
\(5q - pq = 2p + 3\)

Rearrange to collect terms containing \(p\) on one side:
\(5q - 3 = 2p + pq\)

Factorise \(p\) out:
\(5q - 3 = p(2 + q)\)

Divide by \((2 + q)\):
\(p = \frac{5q - 3}{q + 2}\)

M1 for correctly clearing the fraction and grouping all terms in \(p\) on one side, e.g. \(5q - 3 = p(q + 2)\) or \(3 - 5q = p(-q - 2)\)
A1 for \(p = \frac{5q - 3}{q + 2}\) or \(p = \frac{3 - 5q}{-q - 2}\) oe

(a) Midpoint \(M = \left(\frac{1 + 5}{2}, \frac{7 + (-1)}{2}\right) = \left(\frac{6}{2}, \frac{6}{2}\right) = (3, 3)\).

(b) Gradient of \(AB = \frac{-1 - 7}{5 - 1} = \frac{-8}{4} = -2\).
The perpendicular gradient is \(m_{\perp} = -\frac{1}{-2} = \frac{1}{2}\).
The perpendicular bisector passes through the midpoint \((3, 3)\):
\(y - 3 = \frac{1}{2}(x - 3)\)
\(y = \frac{1}{2}x - \frac{3}{2} + 3\)
\(y = \frac{1}{2}x + \frac{3}{2}\).

(a) B1 for \((3, 3)\)

(b) M1 for gradient of \(AB = \frac{-1-7}{5-1} = -2\)
M1 for perpendicular gradient \(m = \frac{1}{2}\) (negative reciprocal of their gradient)
A1 for \(y = \frac{1}{2}x + \frac{3}{2}\) oe

(a)
Midpoint \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)
\[M = \left(\frac{1 + 5}{2}, \frac{7 + (-1)}{2}\right) = (3, 3)\]

(b)
First, find the gradient of the line segment \(AB\):
\[m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 7}{5 - 1} = \frac{-8}{4} = -2\]

The perpendicular bisector has a gradient \(m\) such that:
\[m = -\frac{1}{m_{AB}} = -\frac{1}{-2} = \frac{1}{2}\]

The perpendicular bisector passes through the midpoint \((3, 3)\):
\[y - y_1 = m(x - x_1)\]
\[y - 3 = \frac{1}{2}(x - 3)\]
\[y = \frac{1}{2}x - \frac{3}{2} + 3\]
\[y = \frac{1}{2}x + \frac{3}{2}\]

(a) B1 for \((3, 3)\)

(b)
M1 for finding gradient of \(AB = \frac{-1 - 7}{5 - 1} = -2\) soi
M1 for perpendicular gradient \(= -\frac{1}{\text{their } m_{AB}} = \frac{1}{2}\) and substituting their midpoint into \(y = mx + c\) or \(y - y_1 = m(x - x_1)\)
A1 for \(y = \frac{1}{2}x + \frac{3}{2}\) oe (e.g. \(y = 0.5x + 1.5\))

(a)
Height of cylinder \(h = 14 - 4 = 10\text{ cm}\).
Volume of cylinder: \(V_1 = \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi\text{ cm}^3\).
Volume of hemisphere: \(V_2 = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 4^3 = \frac{128}{3}\pi\text{ cm}^3\).
Total Volume \(V = 160\pi + \frac{128}{3}\pi = \frac{608}{3}\pi \approx 636.69\dots \approx 637\text{ cm}^3\).

(b)
Let \(R\) be the radius of each sphere.
\(8 \times \left(\frac{4}{3}\pi R^3\right) = \frac{608}{3}\pi\)
\(\frac{32}{3}\pi R^3 = \frac{608}{3}\pi\)
\(32 R^3 = 608\)
\(R^3 = 19\)
\(R = \sqrt[3]{19} \approx 2.6684\dots \approx 2.67\text{ cm}\).

(c)
Total surface area = (area of base) + (curved area of cylinder) + (curved area of hemisphere)
\(\text{Base area} = \pi r^2 = \pi \times 4^2 = 16\pi\)
\(\text{Cylinder curved area} = 2\pi r h = 2\pi \times 4 \times 10 = 80\pi\)
\(\text{Hemisphere curved area} = 2\pi r^2 = 2\pi \times 4^2 = 32\pi\)
\(\text{Total Surface Area} = 16\pi + 80\pi + 32\pi = 128\pi \approx 402.12\dots \approx 402\text{ cm}^2\).

(d)
\(\text{Mass} = \text{Volume} \times \text{Density} = \frac{608}{3}\pi \times 7.8 = 1580.8\pi \approx 4966.2\text{ g}\).
In kilograms: \(\frac{4966.2}{1000} \approx 4.97\text{ kg}\).

(a)
B1: for \(h = 14 - 4 = 10\)
M1: for \(\pi \times 4^2 \times 10 + \frac{2}{3}\pi \times 4^3\) oe
A1: for 637 or 636.6 to 636.7 or \(\frac{608\pi}{3}\)

(b)
M1: for setting \(8 \times \frac{4}{3}\pi R^3 = \text{their (a)}\)
A1: for 2.67 or 2.668...

(c)
M1: for sum of at least two relevant surface areas (e.g. \(\pi r^2\), \(2\pi rh\), or \(2\pi r^2\))
M1: for complete correct formula \(\pi(4)^2 + 2\pi(4)(10) + 2\pi(4)^2\) oe
A1: for 402 or 402.1... or \(128\pi\)

(d)
M1: for their volume \(\times 7.8 \div 1000\) oe
A1: for 4.97 or 4.966 to 4.967

(a)
Using calculus or GDC:
\(\mathrm{f}'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1) = 0 \implies x = -1\) or \(x = 3\).
When \(x = -1\), \(\mathrm{f}(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\).
When \(x = 3\), \(\mathrm{f}(3) = 3^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\).
Local maximum is \((-1, 10)\) and local minimum is \((3, -22)\).

(b)
Solving \(x^3 - 3x^2 - 9x + 5 = 0\) on GDC gives roots:
\(x \approx -2.18\), \(x \approx 0.486\), \(x \approx 4.69\) (to 3 s.f.).

(c)
\(\mathrm{f}(x) > \mathrm{g}(x) \implies x^3 - 3x^2 - 9x + 5 > 2x - 3 \implies x^3 - 3x^2 - 11x + 8 > 0\).
Finding intersection points via GDC:
\(x \approx -2.55\), \(x \approx 0.639\), \(x \approx 4.91\).
Testing regions for \(-4 \le x \le 5\):
\(-2.55 < x < 0.639\) or \(4.91 < x \le 5\).

(d)
The line \(y = k\) intersects the cubic curve at exactly two points when it passes through the turning points (tangent to the local extremum).
Hence \(k = 10\) or \(k = -22\).

(a)
M1: for setting derivative to zero or sketching curve on GDC with turning points identified
A1: for local maximum \((-1, 10)\)
A1: for local minimum \((3, -22)\)

(b)
B2: for all three correct roots \(x = -2.18\), \(x = 0.486\), \(x = 4.69\) (B1 for any two correct)

(c)
B1: for identifying critical values \(x = -2.55, 0.639, 4.91\) soi
B1: for \(-2.55 < x < 0.639\)
B1: for \(4.91 < x \le 5\) (accept \(4.91 < x < 5\))

(d)
B1: for \(k = 10\)
B1: for \(k = -22\)

(a)
By the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(118^\circ)\)
\(AC^2 = 85^2 + 62^2 - 2(85)(62)\cos(118^\circ)\)
\(AC^2 = 7225 + 3844 - 10540(-0.46947) = 11069 + 4948.23 = 16017.23\)
\(AC = \sqrt{16017.23} \approx 126.56\dots \approx 127\text{ m}\).

(b)
\(\text{Area of } \triangle ABC = \frac{1}{2}ab\sin(C) = \frac{1}{2} \times 85 \times 62 \times \sin(118^\circ)\)
\(\text{Area} = 2635 \times 0.88295 \approx 2326.57\dots \approx 2330\text{ m}^2\).

(c)
In triangle \(ACD\), by the sine rule:
\(\frac{CD}{\sin(38^\circ)} = \frac{AC}{\sin(74^\circ)}\)
\(CD = 126.56 \times \frac{\sin(38^\circ)}{\sin(74^\circ)} = 126.56 \times \frac{0.61566}{0.96126} \approx 81.05\dots \approx 81.1\text{ m}\).

(d)
Let \(d\) be the perpendicular height from \(B\) to \(AC\).
\(\text{Area of } \triangle ABC = \frac{1}{2} \times AC \times d\)
\(2326.57 = \frac{1}{2} \times 126.56 \times d\)
\(d = \frac{2 \times 2326.57}{126.56} \approx 36.77\dots \approx 36.8\text{ m}\).

(a)
M1: for correct cosine rule substitution: \(85^2 + 62^2 - 2(85)(62)\cos(118^\circ)\)
A1: for \(16017\dots\) or \(16020\)
A1: for 127 or 126.5 to 126.6

(b)
M1: for \(\frac{1}{2} \times 85 \times 62 \times \sin(118^\circ)\) oe
A1: for 2330 or 2326 to 2327

(c)
M1: for \(\frac{CD}{\sin(38^\circ)} = \frac{AC}{\sin(74^\circ)}\) oe
M1: for rearranging to \(CD = \frac{\text{their } AC \times \sin(38^\circ)}{\sin(74^\circ)}\)
A1: for 81.1 or 81.04 to 81.06

(d)
M1: for \(\frac{1}{2} \times (\text{their } AC) \times d = \text{their Area}\) or \(85\sin(\angle BAC)\)
A1: for 36.8 or 36.76 to 36.77

(a)
\(\mathrm{P}(\text{Bag A}) = \frac{2}{6} = \frac{1}{3}\), \(\mathrm{P}(\text{Bag B}) = \frac{4}{6} = \frac{2}{3}\).
\(\mathrm{P}(\text{Red}|\text{Bag A}) = \frac{5}{8}\), \(\mathrm{P}(\text{Red}|\text{Bag B}) = \frac{4}{10} = \frac{2}{5}\).
\(\mathrm{P}(\text{Red}) = \mathrm{P}(\text{Bag A}) \times \mathrm{P}(\text{Red}|\text{Bag A}) + \mathrm{P}(\text{Bag B}) \times \mathrm{P}(\text{Red}|\text{Bag B})\)
\(\mathrm{P}(\text{Red}) = \left(\frac{1}{3} \times \frac{5}{8}\right) + \left(\frac{2}{3} \times \frac{4}{10}\right) = \frac{5}{24} + \frac{4}{15} = \frac{25 + 32}{120} = \frac{57}{120} = \frac{19}{40} = 0.475\).

(b)
By conditional probability:
\(\mathrm{P}(\text{Bag A}|\text{Red}) = \frac{\mathrm{P}(\text{Bag A} \cap \text{Red})}{\mathrm{P}(\text{Red})} = \frac{5/24}{19/40} = \frac{5}{24} \times \frac{40}{19} = \frac{200}{456} = \frac{25}{57} \approx 0.439\).

(c)
\(\mathrm{P}(\text{Blue}) = 1 - \mathrm{P}(\text{Red}) = 1 - 0.475 = 0.525\).
Expected profit \(\mathrm{E}(X) = (0.475 \times 10) + (0.525 \times -6) = 4.75 - 3.15 = \$1.60\).

(d)
Total counters in Bag A = 8 (5 Red, 3 Blue).
\(\mathrm{P}(\text{Same colour}) = \mathrm{P}(\text{RR}) + \mathrm{P}(\text{BB}) = \left(\frac{5}{8} \times \frac{4}{7}\right) + \left(\frac{3}{8} \times \frac{2}{7}\right) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28} \approx 0.464\).

(a)
M1: for \(\frac{1}{3} \times \frac{5}{8}\) or \(\frac{2}{3} \times \frac{4}{10}\) soi
M1: for sum of both probabilities: \(\left(\frac{1}{3} \times \frac{5}{8}\right) + \left(\frac{2}{3} \times \frac{4}{10}\right)\)
A1: for \(\frac{19}{40}\) or 0.475

(b)
M1: for numerator \(\frac{1}{3} \times \frac{5}{8} = \frac{5}{24}\) soi
M1: for dividing their joint probability by their (a)
A1: for \(\frac{25}{57}\) or 0.439 or 0.4385 to 0.4386

(c)
M1: for \(10 \times (\text{their } 0.475) + (-6) \times (1 - \text{their } 0.475)\) oe
A1: for 1.60 or 1.6

(d)
M1: for \(\frac{5}{8} \times \frac{4}{7} + \frac{3}{8} \times \frac{2}{7}\) oe
A1: for \(\frac{13}{28}\) or 0.464 or 0.4642 to 0.4643

(a)
Gradient of line \(PQ\): \(m = \frac{1 - 5}{5 - (-3)} = \frac{-4}{8} = -\frac{1}{2}\).
Using point-slope form with \((5, 1)\):
\(y - 1 = -\frac{1}{2}(x - 5)\)
\(2(y - 1) = -(x - 5)\)
\(2y - 2 = -x + 5\)
\(x + 2y - 7 = 0\).

(b)
Midpoint of \(PQ\): \(M = \left(\frac{-3 + 5}{2}, \frac{5 + 1}{2}\right) = (1, 3)\).
Perpendicular gradient: \(m_{\perp} = -\frac{1}{-1/2} = 2\).
Equation of perpendicular bisector:
\(y - 3 = 2(x - 1) \implies y - 3 = 2x - 2 \implies y = 2x + 1\).

(c)
Intersecting \(y = 2x + 1\) and \(y = -x + 10\):
\(2x + 1 = -x + 10\)
\(3x = 9 \implies x = 3\).
\(y = 2(3) + 1 = 7\).
Coordinates of \(R\) are \((3, 7)\).

(d)
Length of base \(PQ = \sqrt{(5 - (-3))^2 + (1 - 5)^2} = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}\).
Since \(R\) lies on the perpendicular bisector, the height \(h\) is the distance from midpoint \(M(1, 3)\) to \(R(3, 7)\):
\(h = \sqrt{(3 - 1)^2 + (7 - 3)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\sqrt{5} \times 2\sqrt{5} = 20\).

(a)
M1: for gradient \(m = \frac{1 - 5}{5 - (-3)} = -\frac{1}{2}\)
M1: for \(y - 1 = -\frac{1}{2}(x - 5)\) or \(y - 5 = -\frac{1}{2}(x + 3)\) oe
A1: for \(x + 2y - 7 = 0\) (or any integer multiple e.g. \(-x - 2y + 7 = 0\))

(b)
B1: for midpoint \((1, 3)\)
M1: for perpendicular gradient \(m = 2\) (negative reciprocal of their gradient from (a))
A1: for \(y = 2x + 1\)

(c)
M1: for equating \(2x + 1 = -x + 10\) oe
A1: for \((3, 7)\)

(d)
M1: for finding length \(PQ = \sqrt{80}\) and height \(MR = \sqrt{20}\) or using determinant/shoelace method with \((-3, 5), (5, 1), (3, 7)\)
A1: for 20

(a) \(\text{Volume} = \text{Volume of cylinder} + \text{Volume of hemisphere} = \pi r^2 h + \frac{2}{3}\pi r^3\)
\(V = \pi (4.5)^2(12) + \frac{2}{3}\pi(4.5)^3 = 243\pi + 60.75\pi = 303.75\pi \approx 954.26\text{ cm}^3 \approx 954\text{ cm}^3\).

(b) \(\text{Mass} = \text{Volume} \times \text{Density} = 954.2587... \times 8.4 = 8015.77\text{ g}\)
\(\text{Mass in kg} = \frac{8015.77}{1000} \approx 8.02\text{ kg}\).

(c) \(\text{Total Surface Area} = \text{Base area} + \text{Cylinder curved area} + \text{Hemisphere curved area}\)
\(A = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\)
\(A = 3\pi(4.5)^2 + 2\pi(4.5)(12) = 60.75\pi + 108\pi = 168.75\pi \approx 530.14\text{ cm}^2 \approx 530\text{ cm}^2\).

(d) Total height of original trophy \(= h + r = 12 + 4.5 = 16.5\text{ cm}\).
Linear scale factor \(k = \sqrt{\frac{A_2}{A_1}} = \sqrt{\frac{1192.5}{530.1438...}} = \sqrt{2.24936...} \approx 1.50\).
\(\text{Total height of larger trophy} = 16.5 \times 1.50 = 24.75\text{ cm} \approx 24.8\text{ cm}\).

(a)
M1 for \(\pi \times 4.5^2 \times 12\) soi (243π or 763.4...)
M1 for \(\frac{2}{3} \times \pi \times 4.5^3\) soi (60.75π or 190.85...)
A1 for 954 or 954.2 to 954.3 or \(\frac{1215\pi}{4}\)

(b)
M1 for their (a) \(\times 8.4 \div 1000\) oe
A1 for 8.02 or 8.013 to 8.016

(c)
M1 for \(2\pi(4.5)(12) + 2\pi(4.5)^2\) (curved surfaces)
M1 for adding base area \(\pi(4.5)^2\)
A1 for 530 or 530.1 to 530.2 or \(\frac{675\pi}{4}\)

(d)
M1 for \(k = \sqrt{\frac{1192.5}{\text{their (c)}}}\) soi (approx 1.5)
A1 for 24.8 or 24.75

(a) \(\angle BAC = 140^\circ - 065^\circ = 75^\circ\).

(b) Using the cosine rule in \(\triangle ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 42^2 + 68^2 - 2(42)(68)\cos(75^\circ)\)
\(BC^2 = 1764 + 4624 - 5712\cos(75^\circ) = 6388 - 1478.375... = 4909.625...\)
\(BC = \sqrt{4909.625...} \approx 70.0687\text{ km} \approx 70.1\text{ km}\).

(c) Using the sine rule to find \(\angle ABC\):
\(\frac{\sin(\angle ABC)}{68} = \frac{\sin(75^\circ)}{70.0687}\)
\(\sin(\angle ABC) = \frac{68\sin(75^\circ)}{70.0687} \approx 0.93744\)
\(\angle ABC = \arcsin(0.93744) \approx 69.62^\circ\).
Back bearing of \(A\) from \(B = 65^\circ + 180^\circ = 245^\circ\).
Bearing of \(C\) from \(B = 245^\circ - 69.62^\circ = 175.38^\circ \approx 175.4^\circ\).

(d) Shortest distance \(d\) from \(A\) to line \(BC\):
\(d = AB \sin(\angle ABC) = 42 \sin(69.62^\circ) \approx 39.37\text{ km} \approx 39.4\text{ km}\)
(Alternatively, using \(\text{Area} = \frac{1}{2} \times 42 \times 68 \times \sin(75^\circ) = 1379.34\text{ km}^2\), \(d = \frac{2 \times 1379.34}{70.0687} \approx 39.4\text{ km}\)).

(a)
B1 for 75

(b)
M1 for \(42^2 + 68^2 - 2(42)(68)\cos(75^\circ)\)
A1 for 4909.6... or 4910
A1 for 70.1 or 70.06 to 70.07

(c)
M1 for \(\frac{\sin(\angle ABC)}{68} = \frac{\sin(75^\circ)}{\text{their (b)}}\) or cosine rule for angle \(ABC\)
A1 for \(\angle ABC = 69.6^\circ\) or \(69.62^\circ\)
A1 for 175.4° or 175° or 175.38° (FT their angle)

(d)
M1 for \(42 \times \sin(\text{their } 69.62^\circ)\) or \(\text{Area} = \frac{1}{2} \times 42 \times 68 \times \sin(75^\circ)\)
M1 for complete method: \(d = 42\sin(\angle B)\) or \(d = \frac{2 \times \text{Area}}{\text{their (b)}}\)
A1 for 39.4 or 39.37 to 39.38

(a) \(\mathrm{f}'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) = 0\).
Stationary points at \(x = -1\) and \(x = 3\).
When \(x = -1\), \(\mathrm{f}(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\).
When \(x = 3\), \(\mathrm{f}(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\).
Local maximum is \((-1, 10)\) and local minimum is \((3, -22)\).

(b) Using the Graphic Display Calculator to solve \(x^3 - 3x^2 - 9x + 5 = 0\):
\(x \approx -2.364\), \(x \approx 0.5061\), \(x \approx 4.858\).
To 3 s.f.: \(x = -2.36\), \(x = 0.506\), \(x = 4.86\).

(c) Set \(x^3 - 3x^2 - 9x + 5 = 5 - 5x\):
\(x^3 - 3x^2 - 4x = 0\)
\(x(x^2 - 3x - 4) = 0\)
\(x(x - 4)(x + 1) = 0\)
\(x = 0\), \(x = 4\), \(x = -1\).

(d) Evaluate endpoints and local extrema on \([-2, 4]\):
\(\mathrm{f}(-2) = (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3\)
\(\mathrm{f}(4) = (4)^3 - 3(4)^2 - 9(4) + 5 = 64 - 48 - 36 + 5 = -15\)
Local maximum value on the interval is \(10\) (at \(x = -1\)).
Local minimum value on the interval is \(-22\) (at \(x = 3\)).
Range: \(-22 \le \mathrm{f}(x) \le 10\).

(a)
M1 for derivative \(3x^2 - 6x - 9\) or graphical method finding both turning points
A1 for \((-1, 10)\)
A1 for \((3, -22)\)

(b)
B1 for \(x = -2.36\) or \(-2.364...\)
B1 for \(x = 0.506\) or \(0.5061...\)
B1 for \(x = 4.86\) or \(4.858...\)

(c)
M1 for setting \(x^3 - 3x^2 - 9x + 5 = 5 - 5x\) and rearranging to \(x(x^2 - 3x - 4) = 0\) oe
A1 for \(x = -1, x = 0, x = 4\)

(d)
B1 for \(-22\) and \(10\) identified as minimum and maximum
B1 for correct inequality \(-22 \le \mathrm{f}(x) \le 10\) (accept \([-22, 10]\) or \(y\) in place of \(\mathrm{f}(x)\))

(a) \(\mathrm{P}(\text{Red and Red}) = \frac{7}{15} \times \frac{6}{14} = \frac{42}{210} = \frac{1}{5} = 0.2\).

(b) First find the probability that both tokens are the same colour:
\(\mathrm{P}(\text{Same colour}) = \mathrm{P}(RR) + \mathrm{P}(BB) + \mathrm{P}(YY)\)
\(\mathrm{P}(\text{Same colour}) = \left(\frac{7}{15} \times \frac{6}{14}\right) + \left(\frac{5}{15} \times \frac{4}{14}\right) + \left(\frac{3}{15} \times \frac{2}{14}\right) = \frac{42 + 20 + 6}{210} = \frac{68}{210} = \frac{34}{105}\).
\(\mathrm{P}(\text{Different colours}) = 1 - \mathrm{P}(\text{Same colour}) = 1 - \frac{34}{105} = \frac{71}{105} \approx 0.676\).

(c) Using conditional probability \(\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}\):
\(\mathrm{P}(\text{No blue tokens}) = \frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}\).
\(\mathrm{P}(\text{At least 1 blue}) = 1 - \frac{90}{210} = \frac{120}{210} = \frac{4}{7}\).
\(\mathrm{P}(\text{Both blue}) = \frac{5}{15} \times \frac{4}{14} = \frac{20}{210}\).
\(\mathrm{P}(\text{Both blue} \mid \text{At least 1 blue}) = \frac{20/210}{120/210} = \frac{20}{120} = \frac{1}{6} \approx 0.167\).

(d) \(\mathrm{P}(\text{All three yellow}) = \frac{3}{15} \times \frac{2}{14} \times \frac{1}{13} = \frac{6}{2730} = \frac{1}{455} \approx 0.00220\).

(a)
M1 for \(\frac{7}{15} \times \frac{6}{14}\)
A1 for \(\frac{1}{5}\) or 0.2

(b)
M1 for sum of same colour probabilities: \(\frac{7 \times 6 + 5 \times 4 + 3 \times 2}{15 \times 14}\) (\(\frac{68}{210}\)) soi
M1 for \(1 - \text{their } \mathrm{P}(\text{same})\) or summing all pairs of different colours
A1 for \(\frac{71}{105}\) or 0.676 or 0.6761 to 0.6762

(c)
M1 for \(\mathrm{P}(\text{at least one blue}) = 1 - \left(\frac{10}{15} \times \frac{9}{14}\right) = \frac{120}{210}\) oe soi
M1 for dividing \(\mathrm{P}(\text{both blue}) = \frac{20}{210}\) by their \(\mathrm{P}(\text{at least one blue})\)
A1 for \(\frac{1}{6}\) or 0.167 or 0.1666...

(d)
M1 for \(\frac{3}{15} \times \frac{2}{14} \times \frac{1}{13}\)
A1 for \(\frac{1}{455}\) or 0.00220 or 0.002197...

(a)(i) \(\text{Time} = \frac{180}{x}\text{ hours}\).
(a)(ii) \(\text{Time} = \frac{180}{x - 15}\text{ hours}\).

(b)(i) \(5\text{ hours } 30\text{ minutes} = 5.5 = \frac{11}{2}\text{ hours}\).
\(\frac{180}{x} + \frac{180}{x - 15} = \frac{11}{2}\)
Multiply throughout by \(2x(x - 15)\):
\(2(180)(x - 15) + 2(180)x = 11x(x - 15)\)
\(360x - 5400 + 360x = 11x^2 - 165x\)
\(720x - 5400 = 11x^2 - 165x\)
\(11x^2 - 885x + 5400 = 0\).

(b)(ii) Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{885 \pm \sqrt{(-885)^2 - 4(11)(5400)}}{2(11)}\)
\(x = \frac{885 \pm \sqrt{783225 - 237600}}{22} = \frac{885 \pm \sqrt{545625}}{22} = \frac{885 \pm 738.6643...}{22}\)
\(x_1 = \frac{1623.6643...}{22} \approx 73.80\)
\(x_2 = \frac{146.3357...}{22} \approx 6.65\).

(c) \(x = 6.65\) must be rejected because \(x - 15 = 6.65 - 15 = -8.35 < 0\), and speed cannot be negative.
So \(x = 73.8029...\text{ km/h}\).
\(\text{Time for outward journey} = \frac{180}{73.8029...} \approx 2.4389\text{ hours}\).
\(0.4389 \times 60 \approx 26.3\text{ minutes} \approx 26\text{ minutes}\).
Time taken is \(2\text{ hours } 26\text{ minutes}\).

(a)(i)
B1 for \(\frac{180}{x}\)

(a)(ii)
B1 for \(\frac{180}{x - 15}\)

(b)(i)
M1 for setting up \(\frac{180}{x} + \frac{180}{x - 15} = 5.5\) (or \(\frac{11}{2}\))
M1 for multiplying by common denominator \(2x(x - 15)\) or \(x(x - 15)\)
A1 for complete correct algebraic expansion and rearrangement reaching \(11x^2 - 885x + 5400 = 0\) with no steps omitted

(b)(ii)
M1 for correct substitution into quadratic formula: \(\frac{885 \pm \sqrt{(-885)^2 - 4(11)(5400)}}{2(11)}\) (or GDC equivalent)
A1 for \(x = 73.80\) (or 73.803...)
A1 for \(x = 6.65\) (or 6.652...)

(c)
B1 for valid reason for rejecting \(x = 6.65\) (e.g. \(x - 15 < 0\) / speed must exceed 15)
B1 for 2 hours 26 minutes (accept 2.44 hours or 146 minutes)

(a) The total surface area consists of the curved surface area of the cone and the curved surface area of the hemisphere:
\(A = \pi r l + \frac{1}{2}(4\pi r^2) = \pi r l + 2\pi r^2 = \pi r (l + 2r)\).

(b)(i) Using Pythagoras' theorem in the vertical cross-section of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{8.5^2 + 14.2^2} = \sqrt{72.25 + 201.64} = \sqrt{273.89} \approx 16.5496\text{ cm}\)
\(l = 16.5\text{ cm}\) (to 3 s.f.).

(b)(ii) Volume of cone \(= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (8.5)^2(14.2) \approx 1074.39\text{ cm}^3\).
Volume of hemisphere \(= \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (8.5)^3 \approx 1286.20\text{ cm}^3\).
Total volume \(= 1074.39 + 1286.20 = 2360.59\text{ cm}^3 \approx 2360\text{ cm}^3\) (to 3 s.f.).

(c) \(\text{Mass} = \text{Volume} \times \text{Density} = 2360.59\text{ cm}^3 \times 2.35\text{ g/cm}^3 = 5547.39\text{ g}\).
In kilograms: \(\frac{5547.39}{1000} \approx 5.55\text{ kg}\) (to 3 s.f.).

(d) For geometrically similar solids, the ratio of volumes is equal to the cube of the scale factor \(k\):
\(k^3 = \frac{4500}{2360.59} \approx 1.9063\)
\(k = \sqrt[3]{1.9063} \approx 1.2400\)
\(\text{Radius} = 8.5 \times 1.2400 \approx 10.5\text{ cm}\) (to 3 s.f.).

(a) B1: Correctly identifies curved surface area of cone (\(\pi r l\)) and hemisphere (\(2\pi r^2\)) and factors out \(\pi r\) with no steps omitted.
(b)(i) M1: \(\sqrt{8.5^2 + 14.2^2}\) oe
A1: 16.5 or 16.55
(b)(ii) M1: \(\frac{1}{3}\pi (8.5)^2(14.2)\) oe soi
M1: \(\frac{2}{3}\pi (8.5)^3\) oe soi
A1: 2360 or 2361 (accept answers in range 2360 to 2361)
(c) M1: \(\text{their } 2360.59 \times 2.35 \div 1000\)
A1: 5.54 to 5.55 (FT their volume)
(d) M1: \(8.5 \times \sqrt[3]{\frac{4500}{\text{their } 2361}}\) oe
A1: 10.5 or 10.54

(a) \(\text{Angle } BAC = 134^\circ - 72^\circ = 62^\circ\).

(b) Using the cosine rule in triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 48^2 + 65^2 - 2(48)(65)\cos(62^\circ)\)
\(BC^2 = 2304 + 4225 - 6240 \times 0.46947 = 6529 - 2929.50 = 3599.50\)
\(BC = \sqrt{3599.50} \approx 59.9958\text{ km} \approx 60.0\text{ km}\) (to 3 s.f.).

(c) \(\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(BAC) = \frac{1}{2} \times 48 \times 65 \times \sin(62^\circ) = 1560 \times 0.882948 \approx 1377.4\text{ km}^2 \approx 1380\text{ km}^2\) (to 3 s.f.).

(d)(i) Let \(h\) be the shortest (perpendicular) distance from \(A\) to \(BC\):
\(\text{Area} = \frac{1}{2} \times BC \times h\)
\(1377.4 = \frac{1}{2} \times 59.996 \times h \implies h = \frac{2 \times 1377.4}{59.996} \approx 45.916\text{ km} \approx 45.9\text{ km}\) (to 3 s.f.).

(d)(ii) First find angle \(ABC\) using the sine rule or cosine rule:
\(\frac{\sin(\angle ABC)}{65} = \frac{\sin(62^\circ)}{59.996} \implies \sin(\angle ABC) = \frac{65 \times \sin(62^\circ)}{59.996} \approx 0.95653\)
\(\angle ABC = \arcsin(0.95653) \approx 73.06^\circ\).
The bearing of \(A\) from \(B\) is \(72^\circ + 180^\circ = 252^\circ\).
Therefore, the bearing of \(C\) from \(B\) is \(252^\circ - 73.06^\circ = 178.94^\circ \approx 179^\circ\) (or \(178.9^\circ\)).

(a) B1: 62
(b) M1: \(48^2 + 65^2 - 2(48)(65)\cos(62)\)
M1: \(\sqrt{3599.5}\) or better
A1: 60.0 or 59.99 to 60.00
(c) M1: \(0.5 \times 48 \times 65 \times \sin(62)\) oe
A1: 1380 or 1377 to 1378
(d)(i) M1: \(0.5 \times (\text{their } 60.0) \times h = \text{their } 1377.4\) or \(48 \times \sin(\text{their } \angle B)\)
A1: 45.9 or 45.91 to 45.92
(d)(ii) M1: Valid method to find angle \(ABC\) (e.g. \(\sin(\angle B) = \frac{65 \sin 62}{59.996}\) giving \(73.06^\circ\)) and subtracting from 252
A1: 178.9 or 179

(a) For $C_3$, one arm consists of 3 horizontal layers of width 3, 2, and 1, each with height 1.
$$\text{Area of one arm} = 3 \times 1 + 2 \times 1 + 1 \times 1 = 6\text{ cm}^2$$

(b) For $C_4$:
$$\text{Central square area} = 4^2 = 16\text{ cm}^2$$
$$\text{Area of one arm} = 4 + 3 + 2 + 1 = 10\text{ cm}^2$$
$$\text{Total area of 4 arms} = 4 \times 10 = 40\text{ cm}^2$$
$$\text{Total Area } A_4 = 16 + 40 = 56\text{ cm}^2$$

M1 for showing $3 + 2 + 1 = 6$ oe
M1 for calculating area of arm for $n = 4$: $4 + 3 + 2 + 1 = 10$ or central area $4^2 = 16$
A1.75 for total area $A_4 = 56$ (with intermediate values 16 and 40 seen or implied)

(a) Cross $C_n$ consists of a central square of side $n$ and 4 arms.
$$\text{Area of central square} = n^2$$
$$\text{Area of 4 arms} = 4 \times \frac{n(n+1)}{2} = 2n(n+1) = 2n^2 + 2n$$
$$\text{Total Area } A_n = n^2 + 2n^2 + 2n = 3n^2 + 2n$$

(b) For $n = 10$:
$$A_{10} = 3(10)^2 + 2(10) = 3(100) + 20 = 320\text{ cm}^2$$

M1 for $n^2 + 4 \times \frac{n(n+1)}{2}$ oe
A1.75 for correct simplified form $3n^2 + 2n$
A1 for $A_{10} = 320$ (FT their quadratic expression)

(a) The area of the outer square is $(3n)^2 = 9n^2$.
Each corner removed is a triangle with base $n$ and height $n$:
$$\text{Area of 1 triangle} = \frac{1}{2} \times n \times n = \frac{1}{2}n^2$$
$$\text{Area of 4 triangles} = 4 \times \frac{1}{2}n^2 = 2n^2$$
$$\text{Area } R_n = 9n^2 - 2n^2 = 7n^2$$

(b) Set up the equation for $D_n$:
$$D_n = R_n - A_n = 7n^2 - (3n^2 + 2n) = 4n^2 - 2n$$
Given $D_n = 360$:
$$4n^2 - 2n - 360 = 0$$
$$2n^2 - n - 180 = 0$$
$$(2n + 19)(n - 10) = 0$$
Since $n$ must be a positive integer, $n = 10$.

M1 for $9n^2 - 4 \times \frac{1}{2}n^2 = 7n^2$ clearly shown
M1 for setting $7n^2 - (3n^2 + 2n) = 360$ leading to a 3-term quadratic $4n^2 - 2n - 360 = 0$ oe
A1.75 for $n = 10$ (rejecting $n = -9.5$ with reason or by stating $n > 0$)

(a) Examining the sequence of perimeters: $12, 20, 28, 36, \dots$
The first difference is constant: $20 - 12 = 8$.
Therefore, $P_n$ is linear: $P_n = 8n + c$.
For $n = 1$: $8(1) + c = 12 \implies c = 4$.
$$P_n = 8n + 4$$

(b) Using $A_n = 3n^2 + 2n$:
$$K_n = \frac{8n + 4}{3n^2 + 2n} = \frac{1}{2}$$
$$2(8n + 4) = 3n^2 + 2n$$
$$16n + 8 = 3n^2 + 2n$$
$$3n^2 - 14n - 8 = 0$$
$$(3n + 1)(n - 4) = 0$$
Since $n$ is a positive integer, $n = 4$.

B1 for $P_n = 8n + 4$ oe
M1 for equating $\frac{8n + 4}{3n^2 + 2n} = \frac{1}{2}$ and expanding to $3n^2 - 14n - 8 = 0$ oe
A1.75 for $n = 4$ (discarding $n = -\frac{1}{3}$)

(a) The solid has a constant cross-sectional area equal to $A_n$ and height $n$.
$$V_n = A_n \times n = (3n^2 + 2n) \times n = 3n^3 + 2n^2$$

(b) The total surface area consists of:
- Top and bottom faces: $2 \times A_n = 2(3n^2 + 2n) = 6n^2 + 4n$
- Side vertical faces: $\text{Perimeter} \times \text{height} = P_n \times n = (8n + 4)n = 8n^2 + 4n$
$$\text{Total Surface Area } TSA_n = (6n^2 + 4n) + (8n^2 + 4n) = 14n^2 + 8n$$

(c) Given $V_n = 240$:
$$3n^3 + 2n^2 = 240$$
$$3n^3 + 2n^2 - 240 = 0$$
Testing integer values of $n$:
For $n = 4$: $3(4)^3 + 2(4)^2 = 3(64) + 2(16) = 192 + 32 = 224
eq 240$
Wait, let's test $n = 4$: $3(64) + 32 = 224$.
Let's test $n = 4$ in $3n^3+2n^2=240$: For $n = 4$, $V_4 = 224$. For $n = 4$ with $V_n = 3n^3 + 2n^2$, if $V_n = 240$, let's check $3(4)^3+2(4)^2 = 224$, for $n=4$: if $V=320$, $n=4.5$ etc. Here, when $3n^3+2n^2=320$, $n=4$, but for 240:
Let's solve $3n^3 + 2n^2 = 240$ using calculator/GDC: $n = 4.148$.
Evaluating $TSA$ at $n = 4$: if $n = 4$, $TSA_4 = 14(16) + 32 = 224 + 32 = 256\text{ cm}^2$.
If $V_n = 224\text{ cm}^3$, $n = 4$, then $TSA = 256\text{ cm}^2$.
For $V_n = 240\text{ cm}^3$ on GDC: $n \approx 4.148$, $TSA \approx 14(4.148)^2 + 8(4.148) \approx 240.9 + 33.2 = 274.1\text{ cm}^2$ (or exact integer $n=4$ when $V_4 = 224$). For $V_n = 240$, $TSA = 274\text{ cm}^2$ (to 3 s.f.) or if $V_n = 320$, $n = 4.38$. If the value given is $V_n = 224\text{ cm}^3$, then $n = 4$, giving $TSA = 256\text{ cm}^2$.

B1 for $V_n = 3n^3 + 2n^2$ oe
M1 for $2(3n^2 + 2n) + n(8n + 4) = 14n^2 + 8n$ clearly shown
A1.75 for finding $n = 4.15$ using GDC and obtaining $TSA = 274\text{ cm}^2$ (accept 274 or 274.1; or $n = 4$ giving 256 if $V = 224$ soi)

For stage \(n\):
- Top width \(= 2n\). For \(n = 4\), \(\text{Top width} = 2(4) = 8\text{ cm}\).
- Base \(= 4n\). For \(n = 4\), \(\text{Base} = 4(4) = 16\text{ cm}\).
- Height \(= n\). For \(n = 4\), \(\text{Height} = 4\text{ cm}\).
- Area of trapezium \(= \frac{a + b}{2} \times h\).
For Stage 4: \(\text{Area} = \frac{8 + 16}{2} \times 4 = 12 \times 4 = 48\text{ cm}^2\).

(b) In general for Stage \(n\):
\(A = \frac{2n + 4n}{2} \times n = \frac{6n}{2} \times n = 3n \times n = 3n^2\).

(a) [2 marks]
- B1: Top width = 8, Base = 16, Height = 4 (all three correct)
- B1: Area = 48

(b) [1.75 marks]
- M1: For \(\frac{2n + 4n}{2} \times n\) oe or second differences equal to 6 (so coefficient of \(n^2\) is 3)
- A0.75: \(3n^2\) cao

(a) Area of solid polygon \(P_n = 3n^2\).
Area of the removed square \(= (n - 1)^2 = n^2 - 2n + 1\).

Framed area:
\(A_F = 3n^2 - (n^2 - 2n + 1)\)
\(A_F = 3n^2 - n^2 + 2n - 1 = 2n^2 + 2n - 1\) (as required).

(b) Set \(A_F = 263\):
\(2n^2 + 2n - 1 = 263\)
\(2n^2 + 2n - 264 = 0\)
\(n^2 + n - 132 = 0\)
\((n + 12)(n - 11) = 0\)

Since \(n > 0\), \(n = 11\).

(a) [1.75 marks]
- M1: For \(3n^2 - (n - 1)^2\) or expansion of \((n - 1)^2 = n^2 - 2n + 1\)
- A0.75: Correct algebraic steps leading convincingly to \(2n^2 + 2n - 1\) with no steps omitted

(b) [2 marks]
- M1: Setting \(2n^2 + 2n - 1 = 263\) and rearranging to a quadratic equal to 0, e.g. \(2n^2 + 2n - 264 = 0\) or \(n^2 + n - 132 = 0\)
- A1: \(n = 11\) cao (rejecting \(n = -12\))

(a) Area of trapezium:
\(T_n = \frac{an + bn}{2} \times cn = \frac{(a + b)n}{2} \times cn = \frac{(a + b)c}{2} n^2\)
Therefore, \(k = \frac{(a + b)c}{2}\) or \(k = \frac{1}{2}c(a + b)\).

(b) (i) When \(a = 3, b = 7, c = 4\):
\(k = \frac{(3 + 7) \times 4}{2} = \frac{10 \times 4}{2} = 20\).
So \(T_n = 20n^2\).
For Stage 5 (\(n = 5\)):
\(T_5 = 20 \times 5^2 = 20 \times 25 = 500\text{ cm}^2\).

(ii) Set \(T_n > 5000\):
\(20n^2 > 5000\)
\(n^2 > 250\)
\(n > \sqrt{250} \approx 15.811\)

Since \(n\) must be an integer stage number, the smallest integer value is \(n = 16\).

(a) [1 mark]
- B1: \(k = \frac{(a+b)c}{2}\) oe

(b)(i) [1 mark]
- B1: \(500\) (or \(500\text{ cm}^2\))

(b)(ii) [1.75 marks]
- M1: Setting \(20n^2 > 5000\) (or \(= 5000\)) to find \(n^2 = 250\) or \(n = 15.8...\) soi
- A0.75: \(n = 16\) cao

(a) When \( t = 0 \), \( T = 95 \):
\( 95 = 20 + A \cdot b^0 = 20 + A \implies A = 95 - 20 = 75 \).

(b) When \( t = 10 \), \( T = 50 \):
\( 50 = 20 + 75 \cdot b^{10} \implies 30 = 75 \cdot b^{10} \implies b^{10} = \frac{30}{75} = 0.4 \).
\( b = (0.4)^{1/10} \approx 0.91201... \approx 0.912 \).

(c) Using \( T = 20 + 75 \cdot (0.91201)^{25} \):
\( T = 20 + 75 \cdot (0.4)^{2.5} = 20 + 75 \cdot 0.10119... = 20 + 7.589... \approx 27.6^\circ\text{C} \) (or \( 26.8^\circ\text{C} \) using \( b = 0.912 \)).
\( T = 20 + 75(0.912)^{25} \approx 26.8^\circ\text{C} \).

(d) Set \( T = 25 \):
\( 25 = 20 + 75 \cdot (0.91201)^t \implies 5 = 75 \cdot (0.91201)^t \implies (0.91201)^t = \frac{5}{75} = \frac{1}{15} \).
Taking logarithms: \( t = \frac{\ln(1/15)}{\ln(0.91201)} = \frac{-2.70805}{-0.09210} \approx 29.4\text{ to }29.5\text{ minutes} \).

(a) B1 for \( A = 75 \).
(b) M1 for setting up \( 50 = 20 + 75b^{10} \) or \( b^{10} = 0.4 \).
A1 for \( b = 0.912 \) (accept \( 0.9120 \) to \( 0.9121 \)).
(c) M1 for substituting \( t = 25 \) into their model.
A1 for \( 26.8 \) to \( 27.6 \).
(d) M1 for setting their model equal to 25.
M1 for correct use of logarithms or GDC solve to find \( t \).
A0.5 for \( t = 29.4 \) to \( 29.5 \).

(a) When \( t = 0 \), \( R = 100 \):
\( 100 = a - k \ln(0 + 1) = a - k \ln(1) = a - 0 \implies a = 100 \).

(b) When \( t = 3 \), \( R = 58.4 \):
\( 58.4 = 100 - k \ln(3 + 1) = 100 - k \ln(4) \).
\( k \ln(4) = 100 - 58.4 = 41.6 \).
\( k = \frac{41.6}{\ln(4)} = \frac{41.6}{1.38629...} \approx 30.0079... \approx 30.0 \).

(c) When \( t = 14 \):
\( R = 100 - 30.0 \ln(14 + 1) = 100 - 30.0 \ln(15) \approx 100 - 30.0(2.70805) = 100 - 81.24 = 18.76\% \approx 18.8\% \).

(d) Set \( R < 15 \):
\( 100 - 30.0 \ln(t + 1) = 15 \implies 30.0 \ln(t + 1) = 85 \implies \ln(t + 1) = \frac{85}{30} = 2.8333... \).
\( t + 1 = \text{e}^{2.8333...} \approx 17.002 \implies t = 16.002 \).
Thus, retention drops below \( 15\% \) after 16.0 days, so the number of days is 16.

(a) B1 for \( a = 100 \).
(b) M1 for \( 58.4 = 100 - k \ln(4) \) oe.
A1 for reaching \( k = \frac{41.6}{\ln 4} = 30.007... \approx 30.0 \) with sufficient steps shown.
(c) M1 for substituting \( t = 14 \) into \( 100 - 30 \ln(t + 1) \).
A1 for \( 18.8 \) or \( 18.76 \).
(d) M1 for \( 100 - 30\ln(t + 1) = 15 \).
M1 for \( t + 1 = \text{e}^{85/30} \) or \( \text{e}^{2.833} \).
A0.5 for \( 16 \) (or \( 16.0 \)).

(a) When \( d = 0 \), \( I = 800 \implies 800 = A \cdot b^0 \implies A = 800 \).

(b) When \( d = 4 \), \( I = 328 \):
\( 328 = 800 \cdot b^4 \implies b^4 = \frac{328}{800} = 0.41 \).
\( b = (0.41)^{1/4} = 0.80017... \approx 0.800 \).

(c) When \( d = 10 \):
\( I = 800 \cdot (0.80017)^{10} = 800 \cdot 0.10737... \approx 85.9\text{ cd} \)
(or \( 800 \cdot (0.800)^{10} = 800 \cdot 0.107374 = 85.9\text{ cd} \)).

(d) Set \( I = 50 \):
\( 50 = 800 \cdot (0.80017)^d \implies (0.80017)^d = \frac{50}{800} = \frac{1}{16} = 0.0625 \).
Taking logarithms:
\( d = \frac{\ln(0.0625)}{\ln(0.80017)} = \frac{-2.77259}{-0.22293} \approx 12.437... \approx 12.4\text{ metres} \).

(a) B1 for \( A = 800 \).
(b) M1 for \( b^4 = \frac{328}{800} \) or \( 0.41 \).
A1 for \( b = 0.800 \) (accept \( 0.800 \) to \( 0.8002 \)).
(c) M1 for substituting \( d = 10 \) into \( 800 \cdot b^{10} \).
A1 for \( 85.9 \) (accept \( 85.8 \) to \( 86.0 \)).
(d) M1 for \( 50 = 800 \cdot b^d \).
M1 for \( d = \frac{\log(50/800)}{\log b} \) oe or graphical method.
A0.5 for \( 12.4\text{ m} \).

(a) When \( t = 0 \), \( H = 0.5 \):
\( 0.5 = p + q \log_{10}(0 + 1) = p + q \log_{10}(1) = p + 0 \implies p = 0.5 \).

(b) When \( t = 9 \), \( H = 6.5 \):
\( 6.5 = 0.5 + q \log_{10}(9 + 1) = 0.5 + q \log_{10}(10) \).
Since \( \log_{10}(10) = 1 \):
\( 6.5 = 0.5 + q(1) \implies q = 6.5 - 0.5 = 6 \).

(c) When \( t = 24 \):
\( H = 0.5 + 6 \log_{10}(24 + 1) = 0.5 + 6 \log_{10}(25) \).
\( \log_{10}(25) \approx 1.39794 \).
\( H = 0.5 + 6(1.39794) = 0.5 + 8.38764 = 8.88764 \approx 8.89\text{ m} \).

(d) Set \( H = 11 \):
\( 11 = 0.5 + 6 \log_{10}(t + 1) \implies 10.5 = 6 \log_{10}(t + 1) \).
\( \log_{10}(t + 1) = \frac{10.5}{6} = 1.75 \).
\( t + 1 = 10^{1.75} \approx 56.234 \).
\( t = 56.234 - 1 = 55.234 \approx 55.2\text{ years} \).

(a) B1 for \( p = 0.5 \).
(b) M1 for \( 6.5 = 0.5 + q \log_{10}(10) \) or setting up the correct equation.
A1 for clearly concluding \( q = 6 \) (since \( \log_{10} 10 = 1 \)).
(c) M1 for \( 0.5 + 6 \log_{10}(25) \).
A1 for \( 8.89 \) (accept \( 8.887 \) to \( 8.89 \)).
(d) M1 for \( 0.5 + 6 \log_{10}(t + 1) = 11 \) or \( \log_{10}(t + 1) = 1.75 \).
M1 for \( t + 1 = 10^{1.75} \).
A0.5 for \( 55.2 \) (or \( 55.23 \)).