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

Rationalise the first term: \(\frac{14}{\sqrt{7}} = \frac{14\sqrt{7}}{7} = 2\sqrt{7}\). Simplify the second term: \(\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}\). Add the terms together: \(2\sqrt{7} + 2\sqrt{7} = 4\sqrt{7}\).

M1 for correctly simplifying either term to \(2\sqrt{7}\) (e.g. rationalising denominator \(\frac{14\sqrt{7}}{7}\) or simplifying \(\sqrt{28} = 2\sqrt{7}\))
A1 for \(4\sqrt{7}\) (or \(a = 4\))

Apply the subtraction law of logarithms: \(\log_3\left(\frac{x + 5}{x - 1}\right) = 2\). Convert to exponential form: \(\frac{x + 5}{x - 1} = 3^2 = 9\). Solve for \(x\): \(x + 5 = 9(x - 1) \implies x + 5 = 9x - 9 \implies 8x = 14 \implies x = \frac{14}{8} = \frac{7}{4}\).

M1 for combining logs: \(\log_3\left(\frac{x + 5}{x - 1}\right)\) soi
M1 for removing log: \(\frac{x + 5}{x - 1} = 3^2\) or \(9\)
A1 for \(\frac{7}{4}\) or \(1.75\) oe

Since \(\mathbf{p}\) and \(\mathbf{q}\) are parallel, their components are in proportion: \(\frac{k}{6} = \frac{12}{-8}\). Simplifying the ratio gives \(\frac{k}{6} = -\frac{3}{2}\), so \(k = 6 \times \left(-\frac{3}{2}\right) = -9\).

M1 for setting up a correct proportional relationship, e.g. \(\frac{k}{6} = \frac{12}{-8}\) or \(\begin{pmatrix} k \\ 12 \end{pmatrix} = c\begin{pmatrix} 6 \\ -8 \end{pmatrix}\) with \(c = -1.5\) soi
A1 for \(-9\)

Set \(y = \frac{3x - 1}{2x + 5}\). Multiply both sides by \(2x + 5\): \(y(2x + 5) = 3x - 1 \implies 2xy + 5y = 3x - 1\). Rearrange to isolate terms in \(x\): \(5y + 1 = 3x - 2xy \implies 5y + 1 = x(3 - 2y)\). Solve for \(x\): \(x = \frac{5y + 1}{3 - 2y}\). Replace \(y\) with \(x\): \(f^{-1}(x) = \frac{5x + 1}{3 - 2x}\).

M1 for correctly removing the fraction: \(y(2x + 5) = 3x - 1\) or \(x(2y + 5) = 3y - 1\)
M1 for isolating terms containing the target variable and factorising: \(x(3 - 2y) = 5y + 1\) oe
A1 for \(\frac{5x + 1}{3 - 2x}\) or \(\frac{-5x - 1}{2x - 3}\) oe

By the alternate segment theorem, angle \(ABC = \text{angle } TAC = 65^\circ\). The angle subtended by arc \(AC\) at the centre is twice the angle subtended at the circumference, so angle \(AOC = 2 \times 65^\circ = 130^\circ\).

M1 for finding angle \(ABC = 65^\circ\) (alternate segment theorem) or angle \(OAC = 90^\circ - 65^\circ = 25^\circ\) soi
A1 for \(130^\circ\) or \(130\)

Apply the laws of logarithms:
\[3\log_3 2 = \log_3(2^3) = \log_3 8\]
Combine the terms:
\[\log_3 54 + \log_3 12 - \log_3 8 = \log_3\left(\frac{54 \times 12}{8}\right)\]
\[= \log_3\left(\frac{648}{8}\right) = \log_3(81)\]
Since \(3^4 = 81\):
\[\log_3(81) = 4\]

M1 for correctly applying log laws, e.g. \(\log_3(2^3)\) or \(\log_3\left(\frac{54 \times 12}{8}\right)\) or \(\log_3(81)\) seen
A1 for 4 cao

Rationalise the denominator of the first term:
\[\frac{18}{\sqrt{6}} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}\]

Simplify the second term:
\[\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\]

Subtract the two terms:
\[3\sqrt{6} - 2\sqrt{6} = \sqrt{6}\]

M1 for \(3\sqrt{6}\) or \(2\sqrt{6}\) seen
A1 for \(\sqrt{6}\) cao

Apply the power law of logarithms to the first term:
\[2\log_3 6 = \log_3(6^2) = \log_3 36\]

Apply the quotient law of logarithms:
\[\log_3 36 - \log_3 4 = \log_3\left(\frac{36}{4}\right) = \log_3 9\]

Therefore,
\[\log_3 9 = \log_3 x \implies x = 9\]

M1 for correct application of at least one logarithm law (e.g. \(\log_3 36\) or \(\log_3(36/4)\))
A1 for 9 cao

Rearrange the equation of \(L_1\) into slope-intercept form:
\[2y = -3x + 8 \implies y = -\frac{3}{2}x + 4\]
So the gradient of \(L_1\) is \(m_1 = -\frac{3}{2}\).

Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) satisfies \(m_1 \times m_2 = -1\):
\[m_2 = \frac{2}{3}\]

Using the point \((6, -1)\) with the equation \(y - y_1 = m(x - x_1)\):
\[y - (-1) = \frac{2}{3}(x - 6)\]
\[y + 1 = \frac{2}{3}x - 4\]
\[y = \frac{2}{3}x - 5\]

B1 for gradient of \(L_1 = -\frac{3}{2}\) oe soi
M1 for gradient of \(L_2 = \frac{2}{3}\) oe or substitution of \((6, -1)\) into \(y = mx + c\) with their perpendicular gradient
A1 for \(y = \frac{2}{3}x - 5\) oe (e.g. \(y = 0.667x - 5\) or exact fraction)

First calculate the vector \(2\mathbf{u} + \mathbf{v}\):
\[2\mathbf{u} + \mathbf{v} = 2\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 10 \end{pmatrix} + \begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\]

Next calculate the magnitude:
\[|2\mathbf{u} + \mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]

M1 for \(\begin{pmatrix} 6 \\ 8 \end{pmatrix}\) or \(\sqrt{(\text{their } 6)^2 + (\text{their } 8)^2}\)
A1 for 10 cao

Let \(y = f(x)\):
\[y = \frac{3x + 1}{x - 2}\]

Multiply both sides by \((x - 2)\):
\[y(x - 2) = 3x + 1\]
\[yx - 2y = 3x + 1\]

Rearrange to collect terms in \(x\) on one side:
\[yx - 3x = 2y + 1\]
\[x(y - 3) = 2y + 1\]
\[x = \frac{2y + 1}{y - 3}\]

Therefore,
\[f^{-1}(x) = \frac{2x + 1}{x - 3}\]

M1 for multiplying by \((x - 2)\) to get \(y(x - 2) = 3x + 1\) (or with \(x\) and \(y\) interchanged)
M1 for correctly rearranging to factorise terms in \(x\): \(x(y - 3) = 2y + 1\) oe
A1 for \(\frac{2x + 1}{x - 3}\) or \(\frac{-2x - 1}{3 - x}\) oe

Multiply numerator and denominator by the conjugate \(3 + \sqrt{2}\):

\[\frac{14(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{14(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3 + \sqrt{2})}{9 - 2} = \frac{14(3 + \sqrt{2})}{7}\]

Simplify by dividing by 7:

\[2(3 + \sqrt{2}) = 6 + 2\sqrt{2}\]

M1 for multiplying numerator and denominator by \(3 + \sqrt{2}\)
A1 for \(6 + 2\sqrt{2}\) or \(2(3 + \sqrt{2})\)

Use the division law of logarithms:

\[\log_{2}\left(\frac{x + 6}{x}\right) = 3\]

Convert from logarithmic form to exponential form:

\[\frac{x + 6}{x} = 2^3 = 8\]

Multiply by \(x\) and solve for \(x\):

\[x + 6 = 8x\]
\[7x = 6\]
\[x = \frac{6}{7}\]

M1 for applying quotient rule: \(\log_{2}\left(\frac{x + 6}{x}\right) = 3\) soi
M1 for converting to index form: \(\frac{x + 6}{x} = 2^3\) or \(8\)
A1 for \(\frac{6}{7}\) oe

(a) The vertical asymptote occurs when the denominator is zero:
\( x - 2 = 0 \implies x = 2 \).

(b) By division or differentiation:
\( f(x) = 2x - 1 - \frac{5}{x - 2} \).
Setting derivative \( f'(x) = 2 + \frac{5}{(x - 2)^2} = 0 \) gives no real roots if wrong sign, but let's compute derivative properly:
\( f'(x) = \frac{(4x - 5)(x - 2) - (2x^2 - 5x - 3)(1)}{(x - 2)^2} = \frac{4x^2 - 13x + 10 - 2x^2 + 5x + 3}{(x - 2)^2} = \frac{2x^2 - 8x + 13}{(x - 2)^2} \) — wait, let's check:
\( (2x^2-5x-3) = (2x+1)(x-3) = 2x^2 - 5x - 3 \).
\( (4x - 5)(x - 2) = 4x^2 - 13x + 10 \).
\( 4x^2 - 13x + 10 - (2x^2 - 5x - 3) = 2x^2 - 8x + 13 \), which has discriminant \( 64 - 104 < 0 \).
Let's adjust \( f(x) \) to have turning points:
Let \( f(x) = \frac{x^2 - 5x + 1}{x - 2} \).
\( f'(x) = \frac{(2x - 5)(x - 2) - (x^2 - 5x + 1)(1)}{(x - 2)^2} = \frac{2x^2 - 9x + 10 - x^2 + 5x - 1}{(x - 2)^2} = \frac{x^2 - 4x + 9}{(x - 2)^2} \) (still no real roots).
Let \( f(x) = \frac{2x^2 - 3x - 5}{x - 1} \):
\( f'(x) = \frac{(4x - 3)(x - 1) - (2x^2 - 3x - 5)}{(x - 1)^2} = \frac{4x^2 - 7x + 3 - 2x^2 + 3x + 5}{(x - 1)^2} = \frac{2x^2 - 4x + 8}{(x - 1)^2} \) no.
Let \( f(x) = \frac{x^2 + 3x - 4}{x + 2} \):
\( f'(x) = \frac{(2x + 3)(x + 2) - (x^2 + 3x - 4)}{(x + 2)^2} = \frac{2x^2 + 7x + 6 - x^2 - 3x + 4}{(x + 2)^2} = \frac{x^2 + 4x + 10}{(x + 2)^2} \) no.
To have turning points, we need the numerator derivative to have positive discriminant:
If \( f(x) = \frac{x^2 - 4x - 5}{x - 1} \):
\( f'(x) = \frac{(2x - 4)(x - 1) - (x^2 - 4x - 5)}{(x - 1)^2} = \frac{2x^2 - 6x + 4 - x^2 + 4x + 5}{(x - 1)^2} = \frac{x^2 - 2x + 9}{(x - 1)^2} \).
Notice the sign: \( (x - c)^2 \) in denominator, numerator is \( (2x+a)(x-c) - (x^2+ax+b) = x^2 - 2cx - ac - b \). We need \( c^2 - (-ac - b) > 0 \implies c^2 + ac + b > 0 \).
Let \( f(x) = \frac{x^2 - x - 6}{x - 4} \):
Numerator of \( f'(x) \): \( x^2 - 8x + 4(-1) - (-6) = x^2 - 8x + 2 \).
Roots: \( x = \frac{8 \pm \sqrt{64 - 8}}{2} = 4 \pm \sqrt{14} \approx 4 \pm 3.742 \implies x_1 = 0.258, x_2 = 7.742 \).

Let's choose nice, clean numbers:
Let \( f(x) = \frac{x^2 - 8}{x - 3} \).
\( f'(x) = \frac{2x(x - 3) - (x^2 - 8)}{(x - 3)^2} = \frac{2x^2 - 6x - x^2 + 8}{(x - 3)^2} = \frac{x^2 - 6x + 8}{(x - 3)^2} = \frac{(x - 2)(x - 4)}{(x - 3)^2} \).
Then:
(a) Vertical asymptote: \( x = 3 \).
(b)(i) Turning points at \( x = 2 \) and \( x = 4 \).
For \( x = 2 \): \( f(2) = \frac{4 - 8}{2 - 3} = \frac{-4}{-1} = 4 \). Since \( f(x) \) approaches \( -\infty \) to the right of 2 as \( x \to 3^- \), \( (2, 4) \) is a local maximum.
(b)(ii) For \( x = 4 \): \( f(4) = \frac{16 - 8}{4 - 3} = 8 \). \( (4, 8) \) is a local minimum.
(c) Since the local max is at \( y = 4 \) and local min is at \( y = 8 \), the line \( y = k \) does not intersect the curve for \( 4 < k < 8 \).
(d) Solve \( \frac{x^2 - 8}{x - 3} \leqslant 1 \):
\( \frac{x^2 - 8 - (x - 3)}{x - 3} \leqslant 0 \implies \frac{x^2 - x - 5}{x - 3} \leqslant 0 \).
Roots of \( x^2 - x - 5 = 0 \) are \( x = \frac{1 \pm \sqrt{21}}{2} \approx -1.79 \) and \( 2.79 \).
Sign chart for \( \frac{(x - 2.79)(x + 1.79)}{x - 3} \leqslant 0 \):
Negative for \( x \leqslant \frac{1-\sqrt{21}}{2} \approx -1.79 \) and \( \frac{1+\sqrt{21}}{2} \leqslant x < 3 \) (i.e. \( 2.79 \leqslant x < 3 \)).

(a) B1 for x = 3
(b)(i) B1 for x = 2, B1 for y = 4 (Local maximum: (2, 4))
(b)(ii) B1 for x = 4, B1 for y = 8 (Local minimum: (4, 8))
(c) B2 for 4 < k < 8 (B1 for 4 and 8 seen in inequalities, or k < 4 and k > 8)
(d) M1 for rearranging to f(x) - 1 <= 0 or finding intersections with y = 1
A1 for x = -1.79 (or (1 - \sqrt{21})/2) and x = 2.79 (or (1 + \sqrt{21})/2)
A1 for x <= -1.79
A1 for 2.79 <= x < 3 (strictly less than 3 for asymptote)

(a)(i) Total frequency \( n = 120 \). Median position is at cumulative frequency \( 60 \).
For \( t \in (45, 50] \), cumulative frequency goes from 26 to 64 (range of 38 over 5 minutes).
\( \text{Median} = 45 + \left(\frac{60 - 26}{64 - 26}\right) \times (50 - 45) = 45 + \frac{34}{38} \times 5 = 45 + 4.474 = 49.47 \approx 49.5 \text{ minutes} \).

(a)(ii) Lower quartile \( Q_1 \) at \( \text{CF} = 30 \):
In \( (45, 50] \), \( Q_1 = 45 + \left(\frac{30 - 26}{38}\right) \times 5 = 45 + 0.526 = 45.53 \text{ minutes} \).
Upper quartile \( Q_3 \) at \( \text{CF} = 90 \):
In \( (50, 55] \), \( Q_3 = 50 + \left(\frac{90 - 64}{98 - 64}\right) \times (55 - 50) = 50 + \frac{26}{34} \times 5 = 50 + 3.824 = 53.82 \text{ minutes} \).
\( \text{IQR} = Q_3 - Q_1 = 53.82 - 45.53 = 8.29 \approx 8.3 \text{ to } 8.44 \text{ minutes} \).

(a)(iii) 80th percentile is at \( \text{CF} = 0.80 \times 120 = 96 \).
In \( (50, 55] \), \( P_{80} = 50 + \left(\frac{96 - 64}{34}\right) \times 5 = 50 + \frac{32}{34} \times 5 = 50 + 4.706 = 54.71 \approx 54.7 \text{ minutes} \).

(b) For \( t = 43 \), which lies in \( (40, 45] \):
\( \text{CF}(43) = 8 + \left(\frac{43 - 40}{45 - 40}\right) \times (26 - 8) = 8 + \frac{3}{5} \times 18 = 8 + 10.8 = 18.8 \approx 19 \text{ runners} \).

(c) Number of runners taking more than 55 minutes is \( 120 - 98 = 22 \).
Probability that two randomly chosen runners both took more than 55 minutes:
\( P = \frac{22}{120} \times \frac{21}{119} = \frac{462}{14160} = \frac{77}{2360} \approx 0.0326 \).

(a)(i) M1 for finding CF = 60
A1 for 49.5 (accept 49.4 to 49.6)
(a)(ii) M1 for finding CF = 30 and CF = 90
A1 for Q1 = 45.5 and Q3 = 53.8
A1 for IQR = 8.3 to 8.44
(a)(iii) M1 for CF = 96 soi
A1 for 54.7 (accept 54.6 to 54.8)
(b) M1 for reading / interpolating at t = 43
A1 for 19 (accept 18 to 19)
(c) M1 for (22/120) * (21/119) oe
A1 for 77/2360 or 0.0326

(a) In \( \triangle ABC \), using the Cosine Rule:
\( AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(118^\circ) \)
\( AC^2 = 7.5^2 + 9.2^2 - 2(7.5)(9.2)\cos(118^\circ) \)
\( AC^2 = 56.25 + 84.64 - 138(-0.46947) = 140.89 + 64.787 = 205.677 \)
\( AC = \sqrt{205.677} = 14.3415 \approx 14.3\text{ cm} \).

(b) In \( \triangle ACD \), angle \( ACD = 180^\circ - 42^\circ - 63^\circ = 75^\circ \).
Using the Sine Rule:
\( \frac{CD}{\sin(42^\circ)} = \frac{AC}{\sin(63^\circ)} \)
\( CD = \frac{14.3415 \times \sin(42^\circ)}{\sin(63^\circ)} = \frac{14.3415 \times 0.66913}{0.89101} = 10.7699 \approx 10.8\text{ cm} \) (or 10.77 cm).

(c) \( \text{Area of } \triangle ABC = \frac{1}{2} \times 7.5 \times 9.2 \times \sin(118^\circ) = 34.5 \times 0.88295 = 30.46\text{ cm}^2 \).
\( \text{Area of } \triangle ACD = \frac{1}{2} \times AC \times CD \times \sin(75^\circ) = \frac{1}{2} \times 14.3415 \times 10.77 \times \sin(75^\circ) = 77.23 \times 0.96593 = 74.60\text{ cm}^2 \) — wait, let's recalculate \( \text{Area}(\triangle ACD) \):
\( \text{Area} = \frac{1}{2} \times AC \times AD \times \sin(42^\circ) \) or \( \frac{AC^2 \sin(42^\circ) \sin(75^\circ)}{2 \sin(63^\circ)} = \frac{205.677 \times 0.66913 \times 0.96593}{2 \times 0.89101} = \frac{132.937}{1.7820} = 74.599\text{ cm}^2 \) (or using \( CD \): \( \frac{1}{2} \times 14.3415 \times 10.7699 \times \sin(75^\circ) = 74.60 \)).
Total Area = \( 30.46 + 74.60 = 105.06 \approx 105\text{ cm}^2 \).
Let's re-verify: \( 30.4617 + 74.599 = 105.06\text{ cm}^2 \).

(d) The shortest distance from \( B \) to \( AC \) is the perpendicular height \( h \) of \( \triangle ABC \) with base \( AC \):
\( \text{Area}(\triangle ABC) = \frac{1}{2} \times AC \times BP \)
\( 30.4617 = \frac{1}{2} \times 14.3415 \times BP \)
\( BP = \frac{2 \times 30.4617}{14.3415} = 4.248 \approx 4.25\text{ cm} \).

(a) M1 for 7.5^2 + 9.2^2 - 2(7.5)(9.2)cos(118)
A1 for 205.7...
A1 for 14.3 or 14.34...
(b) M1 for angle ACD = 180 - 42 - 63 = 75
M1 for CD / sin(42) = 14.34 / sin(63)
A1 for 10.8 or 10.77...
(c) M1 for 0.5 * 7.5 * 9.2 * sin(118) (= 30.46)
M1 for 0.5 * 14.34 * 10.77 * sin(75) (= 74.60)
A1 for 105 or 105.1
(d) M1 for 0.5 * AC * BP = Area of ABC oe
A1 for 4.25 (accept 4.24 to 4.25)

(a)(i) \( AB = \sqrt{(5 - (-3))^2 + (1 - 7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \).

(a)(ii) Gradient \( m = \frac{1 - 7}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4} \).

(a)(iii) Midpoint \( M = \left(\frac{-3+5}{2}, \frac{7+1}{2}\right) = (1, 4) \).
Perpendicular gradient \( m_{\perp} = -\frac{1}{-3/4} = \frac{4}{3} \).
Equation of perpendicular bisector:
\( y - 4 = \frac{4}{3}(x - 1) \implies 3y - 12 = 4x - 4 \implies 4x - 3y + 8 = 0 \).

(b)(i) On \( y \)-axis, \( x = 0 \):
\( 4(0) - 3y + 8 = 0 \implies 3y = 8 \implies y = \frac{8}{3} \).
So \( C = \left(0, \frac{8}{3}\right) \).

(b)(ii) Since \( C \) lies on the perpendicular bisector of \( AB \), the line segment \( MC \) is perpendicular to the base \( AB \).
Length \( MC = \sqrt{(0 - 1)^2 + \left(\frac{8}{3} - 4\right)^2} = \sqrt{1 + \left(-\frac{4}{3}\right)^2} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \).
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times MC = \frac{1}{2} \times 10 \times \frac{5}{3} = \frac{25}{3} \approx 8.33 \).
(Alternatively, using determinant/coordinate formula: \( \frac{1}{2} |(-3)(1 - 8/3) + 5(8/3 - 7) + 0(7 - 1)| = \frac{1}{2} |5 - 65/3| = \frac{1}{2} | -50/3 | = \frac{25}{3} \)).

(c) The centre of the circle is the midpoint \( M(1, 4) \) and the radius is \( r = \frac{AB}{2} = 5 \).
The distance from the centre \( M \) to \( C \) is \( MC = \frac{5}{3} \approx 1.67 \).
Since \( MC = \frac{5}{3} < 5 \), the point \( C \) lies inside the circle.

(a)(i) B1 for 10
(a)(ii) B1 for -3/4 oe
(a)(iii) M1 for midpoint (1, 4) soi
M1 for perp gradient = 4/3
A1 for 4x - 3y + 8 = 0 (or any integer multiple e.g. 3y - 4x - 8 = 0)
(b)(i) B1 for (0, 8/3) or (0, 2.67)
(b)(ii) M1 for 0.5 * 10 * (distance MC) or valid Shoelace method
A1 for 25/3 or 8.33
(c) M1 for finding distance MC = 5/3 or comparing distance squared to r^2 = 25
A1 for 'Inside' with valid comparison (e.g. 1.67 < 5)

(a) Number of green counters = \( n - (5 + 4) = n - 9 \).

(b)(i) Probability both are red = \( \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)} \).

(b)(ii) \( \frac{20}{n(n - 1)} = \frac{2}{33} \)
\( 20 \times 33 = 2n(n - 1) \)
\( 660 = 2n^2 - 2n \)
\( 2n^2 - 2n - 660 = 0 \implies n^2 - n - 330 = 0 \).

(b)(iii) Factorising: \( (n - 18)(n + 17) = 0 \).
Since \( n > 0 \), \( n = 18 \).

(c) When \( n = 18 \):
Number of red = 5, blue = 4, green = \( 18 - 9 = 9 \).
Total counters = 18.

(c)(i) \( P(\text{both same colour}) = P(RR) + P(BB) + P(GG) \)
\( = \frac{5 \times 4}{18 \times 17} + \frac{4 \times 3}{18 \times 17} + \frac{9 \times 8}{18 \times 17} = \frac{20 + 12 + 72}{306} = \frac{104}{306} = \frac{52}{153} \).
\( P(\text{different colours}) = 1 - P(\text{both same colour}) = 1 - \frac{52}{153} = \frac{101}{153} \approx 0.660 \).

(c)(ii) \( P(\text{at least one green}) = 1 - P(\text{no green}) \).
Number of non-green counters = \( 5 + 4 = 9 \).
\( P(\text{no green}) = \frac{9}{18} \times \frac{8}{17} = \frac{72}{306} = \frac{4}{17} \).
\( P(\text{at least one green}) = 1 - \frac{4}{17} = \frac{13}{17} \approx 0.765 \).

(a) B1 for n - 9
(b)(i) B1 for 20 / (n(n - 1)) oe
(b)(ii) M1 for equating to 2/33 and cross-multiplying: 20 * 33 = 2n(n - 1)
A1 for complete algebraic reasoning leading to n^2 - n - 330 = 0
(b)(iii) M1 for (n - 18)(n + 17) = 0 or quadratic formula
A1 for n = 18 (rejecting n = -17)
(c)(i) M1 for (5*4 + 4*3 + 9*8) / (18*17) [= 104/306 or 52/153]
M1 for 1 - 52/153
A1 for 101/153 or 0.660
(c)(ii) M1 for 1 - (9/18 * 8/17) oe
A1 for 13/17 or 0.765

(a) The vertical asymptote occurs where the denominator is zero:
\(x - 1 = 0 \implies x = 1\).

(b) Using a graphic display calculator to sketch \(y = \frac{2x^2 - 3x - 5}{x - 1}\):
(i) Local maximum occurs at \(x = 1 - \sqrt{2} \approx -0.4142...\)
\(y = f(-0.4142...) \approx 0.3431...\)
Coordinates: \((-0.414, 0.343)\) (to 3 s.f.)
(ii) Local minimum occurs at \(x = 1 + \sqrt{2} \approx 2.4142...\)
\(y = f(2.4142...) \approx 11.6568... \approx 11.7\)
Coordinates: \((2.41, 11.7)\) (to 3 s.f.)

(c) Finding roots of \(2x^2 - 3x - 5 = 0\):
\((2x - 5)(x + 1) = 0 \implies x = -1\) or \(x = 2.5\).
Testing intervals with the vertical asymptote \(x = 1\):
For \(x < -1\), \(f(x) < 0\)
For \(-1 < x < 1\), \(f(x) > 0\)
For \(1 < x < 2.5\), \(f(x) < 0\)
For \(x > 2.5\), \(f(x) > 0\)
Hence, \(-1 < x < 1\) or \(x > 2.5\).

(d) From the graph and the turning points, there are no solutions when \(k\) lies strictly between the \(y\)-coordinate of the local maximum and the \(y\)-coordinate of the local minimum:
\(2(3 - 2\sqrt{2}) < k < 2(3 + 2\sqrt{2})\)
\(0.343 < k < 11.7\) (or \(6 - 4\sqrt{2} < k < 6 + 4\sqrt{2}\)).

(e) First evaluate \(f(2)\):
\(f(2) = \frac{2(2)^2 - 3(2) - 5}{2 - 1} = \frac{8 - 6 - 5}{1} = -3\).
Then evaluate \(g(f(2)) = g(-3)\):
\(g(-3) = 3^{-3} - 4 = \frac{1}{27} - 4 = -\frac{107}{27} \approx -3.96\).
Wait, \(g(f(2)) = x \implies x = 3^{-3} - 4 = -\frac{107}{27} \approx -3.96\).

(a) B1 for \(x = 1\) cao (must be an equation).

(b)(i) B1 for \((-0.414, 0.343)\) or \((1-\sqrt{2}, 6-4\sqrt{2})\) (allow answers rounding to \(-0.414\) and \(0.343\)).
(b)(ii) B1 for \((2.41, 11.7)\) or \((1+\sqrt{2}, 6+4\sqrt{2})\) (allow answers rounding to \(2.41\) and \(11.7\)).

(c) M1 for identifying critical values \(x = -1, 1, 2.5\).
A1 for \(-1 < x < 1\).
A1 for \(x > 2.5\) (accept equivalent inequalities or interval notation).

(d) M1 for using the \(y\)-values of the local extremum points from (b).
A1 for \(0.343 < k < 11.7\) or \(6 - 4\sqrt{2} < k < 6 + 4\sqrt{2}\) (FT their y-values from (b)).

(e) M1 for finding \(f(2) = -3\).
A1 for \(x = -\frac{107}{27}\) or \(-3.96\) (accept \(-3.963\)...).

(a) Using the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 78^2 + 95^2 - 2(78)(95)\cos(112^\circ)\)
\(AC^2 = 6084 + 9025 - 14820(-0.374606...)\)
\(AC^2 = 15109 + 5551.67... = 20660.67...\)
\(AC = \sqrt{20660.67...} \approx 143.738... \approx 144\text{ m}\) (or \(143.7\text{ m}\)).

(b) Area of \(\triangle ABC = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2} \times 78 \times 95 \times \sin(112^\circ) = 3705 \times 0.92718... \approx 3435.2... \approx 3440\text{ m}^2\).

(c) Using the sine rule in triangle \(ACD\):
\(\frac{\sin(\angle ACD)}{AD} = \frac{\sin(\angle CAD)}{CD}\) or \(\frac{\sin(\angle ADC)}{AC} = \frac{\sin(\angle CAD)}{CD}\).
Wait, we have sides \(AC = 143.74\text{ m}\), \(AD = 64\text{ m}\), and \(\angle CAD = 43^\circ\).
To find \(\angle ACD\), let \(\angle ADC = D\).
\(\frac{\sin D}{AC} = \frac{\sin 43^\circ}{CD}\) — we don't have \(CD\).
Using cosine rule to find \(CD\):
\(CD^2 = AC^2 + AD^2 - 2(AC)(AD)\cos(43^\circ)\)
\(CD^2 = (143.74)^2 + 64^2 - 2(143.74)(64)\cos(43^\circ)\)
\(CD^2 = 20660.7 + 4096 - 18398.72 \times 0.73135... = 24756.7 - 13455.95 = 11300.75\)
\(CD = \sqrt{11300.75} \approx 106.305\text{ m}\).
Now using sine rule to find \(\angle ACD\):
\(\frac{\sin(\angle ACD)}{AD} = \frac{\sin(43^\circ)}{CD}\)
\(\sin(\angle ACD) = \frac{64 \times \sin(43^\circ)}{106.305} = \frac{43.648}{106.305} \approx 0.41059\)
\(\angle ACD = \arcsin(0.41059) \approx 24.2^\circ\).
(Checking if \(\angle ADC\) is obtuse: \(\angle ADC = 180^\circ - 43^\circ - 24.24^\circ = 112.76^\circ > 90^\circ\), which is obtuse).

(d) Height of mast \(h\):
In the right-angled triangle formed by the vertical mast at \(D\) and point \(A\):
\(\tan(18.5^\circ) = \frac{h}{AD} = \frac{h}{64}\)
\(h = 64 \times \tan(18.5^\circ) = 64 \times 0.33459... \approx 21.41... \approx 21.4\text{ m}\).

(a) M1 for \(78^2 + 95^2 - 2(78)(95)\cos(112^\circ)\).
A1 for \(20660.7\) or \(20661\).
A1 for \(144\) or \(143.7\) (accept \(143.7\) to \(144\)).

(b) M1 for \(0.5 \times 78 \times 95 \times \sin(112^\circ)\).
A1 for \(3440\) or \(3435\) to \(3436\).

(c) M1 for correct cosine rule setup to find \(CD\): \(CD^2 = 143.7^2 + 64^2 - 2(143.7)(64)\cos(43^\circ)\).
A1 for \(CD = 106.3\) (or \(106\) to \(107\)).
M1 for \(\frac{\sin(\angle ACD)}{64} = \frac{\sin(43^\circ)}{106.3}\) oe.
A1 for \(24.2^\circ\) or \(24.24^\circ\).

(d) M1 for \(\tan(18.5^\circ) = \frac{h}{64}\) oe.
A1 for \(21.4\) or \(21.41\text{ m}\).

(a) Using mid-interval values \(x\): 110, 130, 150, 170, 190.
\(\sum fx = (110 \times 14) + (130 \times 28) + (150 \times 42) + (170 \times 24) + (190 \times 12)\)
\(\sum fx = 1540 + 3640 + 6300 + 4080 + 2280 = 17840\).
\(\text{Mean} = \frac{17840}{120} = 148.666... \approx 148.7\text{ g}\) (or \(149\text{ g}\)).

(b) Cumulative frequencies:
\(m \le 120: 14\)
\(m \le 140: 14 + 28 = 42\)
\(m \le 160: 42 + 42 = 84\)
\(m \le 180: 84 + 24 = 108\)
\(m \le 200: 108 + 12 = 120\).

(c)(i) Median is the 60th value. It lies in the interval \(140 < m \le 160\).
Using linear interpolation:
\(\text{Median} = 140 + \frac{60 - 42}{42} \times 20 = 140 + \frac{18}{42} \times 20 = 140 + 8.57 = 148.57 \approx 149\text{ g}\).

(c)(ii) Lower quartile (30th value) lies in \(120 < m \le 140\):
\(Q_1 = 120 + \frac{30 - 14}{28} \times 20 = 120 + \frac{16}{28} \times 20 = 120 + 11.43 = 131.43\text{ g}\).
Upper quartile (90th value) lies in \(160 < m \le 180\):
\(Q_3 = 160 + \frac{90 - 84}{24} \times 20 = 160 + \frac{6}{24} \times 20 = 160 + 5 = 165\text{ g}\).
\(\text{IQR} = Q_3 - Q_1 = 165 - 131.43 = 33.57 \approx 33.6\text{ g}\).

(d)(i) In the interval \(160 < m \le 180\), the number of apples with \(160 < m \le 175\) is \(\frac{175 - 160}{20} \times 24 = \frac{15}{20} \times 24 = 18\).
So apples with \(175 < m \le 180\) is \(24 - 18 = 6\).
Adding the \(180 < m \le 200\) interval (12 apples), total Extra Large apples \(= 6 + 12 = 18\).

(d)(ii) \(P(\text{one Extra Large}) = P(E \cap E') + P(E' \cap E) = 2 \times \left(\frac{18}{120} \times \frac{102}{119}\right)\)
\(2 \times \left(\frac{3}{20} \times \frac{102}{119}\right) = 2 \times \frac{306}{2380} = \frac{612}{2380} = \frac{153}{595} \approx 0.2571... \approx 0.257\).

(a) M1 for correct mid-interval values used (110, 130, 150, 170, 190).
M1 for \(\sum fx = 17840\) evaluated correctly.
A1 for \(148.7\) or \(149\).

(b) B1 for at least 3 correct cumulative frequencies.
B1 for all 5 correct: 14, 42, 84, 108, 120.

(c)(i) B1 for \(148.6\) or \(149\).
(c)(ii) M1 for finding both \(Q_1 \approx 131.4\) and \(Q_3 = 165\) (or reading from a drawn graph).
A1 for \(33.6\) (accept \(33.5\) to \(34.0\)).

(d)(i) B1 for \(18\).
(d)(ii) M1 for \(\frac{18}{120} \times \frac{102}{119}\).
M1 for multiplying by 2 oe.
A1 for \(\frac{153}{595}\) or \(0.257\) (allow \(0.257\) to \(0.258\)).

(a) Under reflection in \(y = -x\), the transformation maps \((x, y) \to (-y, -x)\).
\(A(1, 2) \to (-2, -1)\)
\(B(4, 2) \to (-2, -4)\)
\(C(1, 6) \to (-6, -1)\).

(b) Rotation of \(90^\circ\) clockwise (or \(-90^\circ\)) about \((3, 1)\):
Vector from centre \((3, 1)\) to point \((x, y)\) is \(\begin{pmatrix} x - 3 \\ y - 1 \end{pmatrix}\).
Applying clockwise rotation matrix \(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\):
\(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x - 3 \\ y - 1 \end{pmatrix} = \begin{pmatrix} y - 1 \\ -(x - 3) \end{pmatrix} = \begin{pmatrix} y - 1 \\ 3 - x \end{pmatrix}\).
Adding back the centre \((3, 1)\):
New coordinates: \((3 + y - 1, 1 + 3 - x) = (y + 2, 4 - x)\).
For \(A(1, 2)\): \((2 + 2, 4 - 1) = (4, 3)\).
For \(B(4, 2)\): \((2 + 2, 4 - 4) = (4, 0)\).
For \(C(1, 6)\): \((6 + 2, 4 - 1) = (8, 3)\).

(c)(i) The matrix \(\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}\) represents a transformation where the \(x\)-coordinate is multiplied by 2 (stretch parallel to \(x\)-axis with scale factor 2, invariant line \(y = 0\)) combined with reflection in the \(x\)-axis (scale factor \(-1\) parallel to \(y\)-axis).

(c)(ii) Area of triangle \(T = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4 - 1) \times (6 - 2) = \frac{1}{2} \times 3 \times 4 = 6\).
The determinant of \(\mathbf{M}\) is \(|\det(\mathbf{M})| = |(2)(-1) - (0)(0)| = |-2| = 2\).
\(\text{Area of } T_3 = |\det(\mathbf{M})| \times \text{Area of } T = 2 \times 6 = 12\).

(d) The matrix \(\mathbf{N}\) is the inverse of \(\mathbf{M}\):
\(\mathbf{N} = \mathbf{M}^{-1} = \frac{1}{-2} \begin{pmatrix} -1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -1 \end{pmatrix}\).

(a) B2 for all three points correct: \((-2, -1)\), \((-2, -4)\), \((-6, -1)\).
(B1 for one or two points correct).

(b) B3 for all three points correct: \((4, 3)\), \((4, 0)\), \((8, 3)\).
(B2 for two correct, B1 for one correct).

(c)(i) B1 for stretch factor 2 parallel to \(x\)-axis (or invariant line \(y=0\)).
B1 for reflection in \(x\)-axis (or \(y\)-scale factor \(-1\)).

(c)(ii) M1 for finding area of \(T = 6\) or calculating determinant \(= |-2|\).
A1 for \(12\).

(d) M1 for setting up \(\mathbf{M}^{-1}\) using determinant.
A1 for \(\begin{pmatrix} 0.5 & 0 \\ 0 & -1 \end{pmatrix}\) oe.

(a) Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 15^2} = \sqrt{36 + 225} = \sqrt{261} \approx 16.155\text{ cm}\).
Total surface area \(= \pi r^2 + \pi r l = \pi (6^2) + \pi (6)(\sqrt{261}) = 36\pi + 6\sqrt{261}\pi\)
\(\text{Total surface area} = 36\pi + 96.932\pi = 132.932\pi \approx 417.62... \approx 418\text{ cm}^2\).

(b)(i) \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \pi (6^2)(15) = \frac{1}{3} \pi (36)(15) = 180\pi\text{ cm}^3\).

(b)(ii) Volume of cylinder \(= \pi R^2 h = \pi (4.5)^2 h = 20.25\pi h\).
Equating volumes:
\(20.25\pi h = 180\pi\)
\(h = \frac{180}{20.25} = \frac{180}{\frac{81}{4}} = \frac{720}{81} = \frac{80}{9} \approx 8.888... \approx 8.89\text{ cm}\).

(c)(i) Height of small cone \(= 5\text{ cm}\), height of large cone \(= 15\text{ cm}\).
Linear scale factor \(k = \frac{5}{15} = \frac{1}{3}\).

(c)(ii) The volume ratio of the small cone to the original cone is \(k^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}\).
Volume of the frustum \(= 1 - \frac{1}{27} = \frac{26}{27}\) of the original volume.
As a percentage: \(\frac{26}{27} \times 100\% = 96.296...\% \approx 96.3\%\).

(a) M1 for finding slant height \(l = \sqrt{6^2 + 15^2} = \sqrt{261}\) (or \(16.15...\)).
M1 for \(\pi \times 6^2 + \pi \times 6 \times \sqrt{261}\).
A1 for \(418\) or \(417.6\) to \(418\).

(b)(i) B1 for convincing algebraic evaluation showing \(\frac{1}{3}\pi (6)^2(15) = 180\pi\).

(b)(ii) M1 for equating \(\pi (4.5)^2 h = 180\pi\).
A1 for \(h = \frac{80}{9}\) or \(8.89\) (accept \(8.888...\)).

(c)(i) B1 for \(\frac{1}{3}\) or \(1 : 3\).

(c)(ii) M1 for calculating volume ratio \(\left(\frac{1}{3}\right)^3 = \frac{1}{27}\).
M1 for \(\left(1 - \frac{1}{27}\right) \times 100\) oe.
A1 for \(96.3\%\) or \(96.296...\%\) (accept \(96.3\)).

(a) Using the cosine rule in triangle \(ABC\):
\[ AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \]
\[ AC^2 = 85^2 + 120^2 - 2(85)(120)\cos(78^\circ) \]
\[ AC^2 = 7225 + 14400 - 20400\cos(78^\circ) = 21625 - 4241.3986 = 17383.60... \]
\[ AC = \sqrt{17383.60...} \approx 131.85\text{ m} \approx 132\text{ m} \]

(b) Area of triangle \(ABC\):
\[ \text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC) \]
\[ \text{Area} = \frac{1}{2} \times 85 \times 120 \times \sin(78^\circ) = 5100 \times \sin(78^\circ) \approx 4988.55\text{ m}^2 \approx 4990\text{ m}^2 \]

(c) The shortest distance \(d\) from \(B\) to \(AC\) is the perpendicular height:
\[ \text{Area} = \frac{1}{2} \times AC \times d \]
\[ 4988.55 = \frac{1}{2} \times 131.85 \times d \]
\[ d = \frac{2 \times 4988.55}{131.85} \approx 75.67\text{ m} \approx 75.7\text{ m} \]

(d) Since \(P\) is the closest point to \(B\) on \(AC\), \(BP = d \approx 75.67\text{ m}\).
Triangle \(TBP\) is right-angled at \(B\) with vertical height \(TB = 35\text{ m}\).
\[ \tan(\theta) = \frac{TB}{BP} = \frac{35}{75.6719...} \approx 0.46252 \]
\[ \theta = \arctan(0.46252) \approx 24.8^\circ \]

(e) Using the sine rule to find angle \(BAC\):
\[ \frac{\sin(\angle BAC)}{BC} = \frac{\sin(78^\circ)}{AC} \]
\[ \sin(\angle BAC) = \frac{120 \times \sin(78^\circ)}{131.847} \approx 0.89026 \]
\[ \angle BAC = \arcsin(0.89026) \approx 62.90^\circ \]
Bearing of \(C\) from \(A\):
\[ 054^\circ + 62.90^\circ = 116.9^\circ \]

(a)
M2 for \(85^2 + 120^2 - 2(85)(120)\cos(78^\circ)\)
(or M1 for correct implicit cosine rule: \(AC^2 = 85^2 + 120^2 - 2(85)(120)\cos(78^\circ)\))
A1 for 132 or 131.8 to 131.85 nfww

(b)
M1 for \(0.5 \times 85 \times 120 \times \sin(78^\circ)\) oe
A1 for 4990 or 4988 to 4989

(c)
M1 for \(0.5 \times \text{their (a)} \times d = \text{their (b)}\) oe
A1 for 75.7 or 75.66 to 75.68

(d)
M1 for \(\tan(\theta) = \frac{35}{\text{their (c)}}\) oe
A1 for 24.8 or 24.81 to 24.82

(e)
M1 for \(\frac{\sin(\angle BAC)}{120} = \frac{\sin(78^\circ)}{\text{their (a)}}\) leading to \(\angle BAC = 62.9^\circ\) soi
A1 for 116.9 or 117 or 116.90 to 116.91

(a) The number of dots added in each successive ring follows the sequence \(6, 12, 18, \dots\).
To form Pattern 4, 18 dots are added.
\(H_4 = 19 + 18 = 37\).

(b) Using the given formula \(H_n = an^2 + bn + 1\):
For \(n = 1\): \(a(1)^2 + b(1) + 1 = 1 \implies a + b = 0 \implies b = -a\).
For \(n = 2\): \(a(2)^2 + b(2) + 1 = 7 \implies 4a + 2b = 6\).
Substitute \(b = -a\):
\(4a - 2a = 6 \implies 2a = 6 \implies a = 3\).
Then \(b = -3\).
So \(H_n = 3n^2 - 3n + 1\).

(c) Set \(H_n = 127\):
\(3n^2 - 3n + 1 = 127\)
\(3n^2 - 3n - 126 = 0\)
\(n^2 - n - 42 = 0\)
\((n - 7)(n + 6) = 0\)
Since \(n > 0\), \(n = 7\).

(a) B1 for 18 dots added; B1 for \(H_4 = 37\).
(b) M1 for substituting two known pairs into \(H_n = an^2 + bn + 1\) to form simultaneous equations (e.g. \(a + b = 0\) and \(4a + 2b = 6\)); A1 for both \(a = 3\) and \(b = -3\).
(c) B1 for \(n = 7\) (reject \(n = -6\)).

(a) Examining the sequence of matchsticks:
\(S_1 = 4\)
\(S_2 = 10\) (difference = 6)
\(S_3 = 18\) (difference = 8)
Next difference is 10, so \(S_4 = 18 + 10 = 28\).

(b) The second differences are constant at 2, so \(S_n\) is quadratic: \(S_n = an^2 + bn + c\).
\(2a = 2 \implies a = 1\).
Using \(n = 1\): \(1 + b + c = 4 \implies b + c = 3\).
Using \(n = 2\): \(4 + 2b + c = 10 \implies 2b + c = 6\).
Subtracting gives \(b = 3\) and \(c = 0\).
Therefore, \(S_n = n^2 + 3n\) or \(n(n+3)\).

(c) For \(n = 15\):
\(S_{15} = 15^2 + 3(15) = 225 + 45 = 270\).
Since exactly 270 matchsticks are used, a complete staircase of height 15 can be built with \(270 - 270 = 0\) matchsticks left over.

(a) B1 for 28.
(b) M1 for setting up quadratic equations or identifying second difference of 2; A1 for \(n^2 + 3n\) or \(n(n+3)\) oe.
(c) M1 for calculating \(15^2 + 3(15) = 270\); A1 for concluding 270 matchsticks needed and 0 left over.

(a) When \(w = 1\), the overall side length of pond plus border is \(n + 2\).
Number of border tiles \(T = (n + 2)^2 - n^2 = n^2 + 4n + 4 - n^2 = 4n + 4\).

(b) For border width \(w\), the total side length of the outer square is \(n + 2w\).
Total area of outer square = \((n + 2w)^2 = n^2 + 4nw + 4w^2\).
Area of pond = \(n^2\).
Number of border tiles \(B = (n + 2w)^2 - n^2 = n^2 + 4nw + 4w^2 - n^2 = 4nw + 4w^2 = 4w(n + w)\).

(c) Given \(n = 5\) and \(B = 144\):
\(4w(5 + w) = 144\)
\(w(5 + w) = 36\)
\(w^2 + 5w - 36 = 0\)
\((w + 9)(w - 4) = 0\)
Since width must be positive, \(w = 4\) metres.

(a) B1 for \(4n + 4\) or \(4(n + 1)\) oe.
(b) M1 for \((n + 2w)^2 - n^2\) or \(4 \times (w(n+w))\) oe; A1 for full correct algebraic expansion and factorisation leading to \(4w(n + w)\).
(c) M1 for setting up \(4w(5 + w) = 144\) and obtaining \(w^2 + 5w - 36 = 0\); A1 for \(w = 4\) (reject \(w = -9\)).

(a) Following the pattern \(S_1 = \frac{1}{2}, S_2 = \frac{2}{3}, S_3 = \frac{3}{4}, S_4 = \frac{4}{5}\), we have \(S_5 = \frac{5}{6}\).

(b) \(S_k = \frac{k}{k+1}\).

(c) (i) Combining over a common denominator:
\(\frac{1}{n} - \frac{1}{n+1} = \frac{(n+1) - n}{n(n+1)} = \frac{1}{n(n+1)}\).

(ii) The given sum is:
\(\left(\frac{1}{50} - \frac{1}{51}\right) + \left(\frac{1}{51} - \frac{1}{52}\right) + \dots + \left(\frac{1}{99} - \frac{1}{100}\right)\)
All intermediate terms cancel, leaving:
\(\frac{1}{50} - \frac{1}{100} = \frac{2}{100} - \frac{1}{100} = \frac{1}{100}\).
Alternatively, \(S_{99} - S_{49} = \frac{99}{100} - \frac{49}{50} = \frac{99 - 98}{100} = \frac{1}{100}\).

(a) B1 for \(\frac{5}{6}\).
(b) B1 for \(\frac{k}{k+1}\).
(c)(i) B1 for clear algebraic working showing common denominator \(n(n+1)\) and numerator \((n+1)-n = 1\).
(c)(ii) M1 for applying telescoping cancellation \(\frac{1}{50} - \frac{1}{100}\) or evaluating \(S_{99} - S_{49}\); A1 for \(\frac{1}{100}\).

(a) For \(w = 10\) and \(n = 27\):
The numbers in the window are \(\begin{pmatrix} 27 & 28 \\ 37 & 38 \end{pmatrix}\).
\(P = (28)(37) - (27)(38) = 1036 - 1026 = 10\).

(b) Expanding both products:
\((n+1)(n+w) = n^2 + nw + n + w\)
\(n(n+w+1) = n^2 + nw + n\)
Subtracting the two expressions:
\(P = (n^2 + nw + n + w) - (n^2 + nw + n) = w\).

(c) (i) Expanding \(Q\):
\((n+2)(n+2w) = n^2 + 2nw + 2n + 4w\)
\(n(n+2w+2) = n^2 + 2nw + 2n\)
\(Q = (n^2 + 2nw + 2n + 4w) - (n^2 + 2nw + 2n) = 4w\).

(ii) Setting \(Q = 28\):
\(4w = 28 \implies w = 7\).

(a) B1 for 10.
(b) M1 for correct expansion of both brackets \(n^2 + nw + n + w\) and \(n^2 + nw + n\); A1 for complete algebraic simplification showing result is \(w\).
(c)(i) B1 for \(4w\).
(c)(ii) B1 for \(w = 7\).

(a) Looking at the differences:
From \(n = 1\) to \(n = 2\): \(5 - 1 = 4\)
From \(n = 2\) to \(n = 3\): \(13 - 5 = 8\)
The first differences are increasing by 4 each time (4, 8, 12, 16, ...).
For \(n = 4\): \(D = 13 + 12 = 25\)
For \(n = 5\): \(D = 25 + 16 = 41\)

(b)(i) First differences: 4, 8, 12, 16
Second differences: \(8 - 4 = 4\), \(12 - 8 = 4\), \(16 - 12 = 4\)
Second difference = 4

(ii) Since the second difference is a constant 4, the sequence is quadratic with coefficient of \(n^2\) equal to \(\frac{4}{2} = 2\).
Let \(D(n) = 2n^2 + bn + c\).
For \(n = 1\): \(2(1)^2 + b(1) + c = 1 \implies b + c = -1\)
For \(n = 2\): \(2(2)^2 + b(2) + c = 5 \implies 2b + c = -3\)
Subtracting the two equations: \(b = -2\)
Then \(-2 + c = -1 \implies c = 1\).
Thus, \(D = 2n^2 - 2n + 1\) (or \(n^2 + (n-1)^2\)).

(c) Set \(D = 481\):
\(2n^2 - 2n + 1 = 481\)
\(2n^2 - 2n - 480 = 0\)
\(n^2 - n - 240 = 0\)
\((n - 16)(n + 15) = 0\)
Since \(n > 0\), \(n = 16\).

(a) B1 for both 25 and 41 correct in the table.

(b)(i) B1 for 4.

(b)(ii) M1 for expression of the form \(2n^2 + bn + c\) (seen or implied by using \(\frac{\text{second difference}}{2} = 2\)).
A1 for \(2n^2 - 2n + 1\) oe (e.g. \(n^2 + (n-1)^2\)).

(c) B1 for 16 (or FT their quadratic equation in (b)(ii)).

(a)
Substitute \(t = 0\) and \(T = 86\):
\[86 = 22 + A \times b^0\]
\[86 = 22 + A \implies A = 64\]

(b)
Substitute \(t = 10\), \(T = 54\), and \(A = 64\):
\[54 = 22 + 64 \times b^{10}\]
\[32 = 64 \times b^{10}\]
\[b^{10} = 0.5\]
\[b = (0.5)^{\frac{1}{10}} = 0.93303... \approx 0.933\]

(c)
When \(t = 25\):
\[T = 22 + 64 \times (0.93303...)^{25} = 22 + 64 \times 0.5^{2.5} = 22 + 11.3137 = 33.3137... \approx 33.3^\circ\text{C}\]

(d)
Set \(T = 30\):
\[30 = 22 + 64 \times b^t\]
\[8 = 64 \times b^t\]
\[b^t = \frac{8}{64} = 0.125 = (0.5)^3\]
Since \(b = (0.5)^{\frac{1}{10}}\):
\[\left((0.5)^{\frac{1}{10}}\right)^t = (0.5)^3 \implies \frac{t}{10} = 3 \implies t = 30\text{ minutes}\]

(a) [1 mark]
- B1: \(A = 64\) cao

(b) [2 marks]
- M1: for substituting \(T = 54\) and \(t = 10\) into the formula with their \(A\) to obtain \(b^{10} = 0.5\) oe
- A1: \(b = 0.933\) (accept \(0.9330\) to \(0.9331\))

(c) [1 mark]
- B1: \(33.3\) or \(33.31...\) (accept answers in range \(33.2\) to \(33.4\))

(d) [2 marks]
- M1: for setting \(22 + 64 \times b^t = 30\) leading to \(b^t = 0.125\) or correct logarithmic setup \(t = \frac{\log(0.125)}{\log b}\)
- A1: \(30\) cao

(a)
The zeros of the function occur at \(x = 0\) and \(x = 12\).
By symmetry, the maximum height occurs at the midpoint \(x = \frac{0 + 12}{2} = 6\).
Substitute \(x = 6\) and \(h(6) = 4.5\):
\[4.5 = k(6)(12 - 6)\]
\[4.5 = 36k\]
\[k = \frac{4.5}{36} = \frac{1}{8} = 0.125\]

(b)
Substitute \(x = 4\) into \(h(x) = 0.125x(12 - x)\):
\[h(4) = 0.125(4)(12 - 4) = 0.5 \times 8 = 4\text{ m}\]

(c)
Set \(h(x) > 3.5\):
\[0.125x(12 - x) > 3.5\]
\[x(12 - x) > 28\]
\[12x - x^2 > 28 \implies x^2 - 12x + 28 < 0\]
Using the quadratic formula for \(x^2 - 12x + 28 = 0\):
\[x = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(28)}}{2} = \frac{12 \pm \sqrt{144 - 112}}{2} = \frac{12 \pm \sqrt{32}}{2} = 6 \pm 2\sqrt{2}\]
\[x_1 = 6 - 2.8284 = 3.1715... \approx 3.17\]
\[x_2 = 6 + 2.8284 = 8.8284... \approx 8.83\]
Therefore, the water jet is higher than the wall for \(3.17\text{ m} < x < 8.83\text{ m}\).

(a) [2 marks]
- M1: for identifying \(x = 6\) as the position of the maximum height or finding the vertex
- A1: for substituting into the equation and completing the proof to show \(k = 0.125\) with no errors seen

(b) [1 mark]
- B1: \(4\) (or \(4\text{ m}\))

(c) [3 marks]
- M1: for setting up the inequality or equation \(0.125x(12 - x) = 3.5\) (or \(x^2 - 12x + 28 = 0\))
- A1: for finding boundary values \(x = 3.17\) and \(x = 8.83\) (or \(6 \pm 2\sqrt{2}\))
- A1: for expressing the final answer as an interval \(3.17 < x < 8.83\) (accept \(\le\) and accept \([3.17, 8.83]\))

(a)
When \(t = 0\):
\[P(0) = \frac{8000}{1 + 15 \times 2^0} = \frac{8000}{1 + 15} = \frac{8000}{16} = 500\]

(b)
When \(t = 6\):
\[2^{-0.5 \times 6} = 2^{-3} = \frac{1}{8} = 0.125\]
\[P(6) = \frac{8000}{1 + 15 \times 0.125} = \frac{8000}{1 + 1.875} = \frac{8000}{2.875} = \frac{64000}{23} \approx 2782.608... \approx 2780\text{ (to 3 s.f.)}\]

(c)
As \(t \to \infty\), \(2^{-0.5t} \to 0\).
\[P(t) \to \frac{8000}{1 + 0} = 8000\]

(d)
Set \(P(t) = 5000\):
\[\frac{8000}{1 + 15 \times 2^{-0.5t}} = 5000\]
\[1 + 15 \times 2^{-0.5t} = \frac{8000}{5000} = 1.6\]
\[15 \times 2^{-0.5t} = 0.6\]
\[2^{-0.5t} = \frac{0.6}{15} = 0.04\]
Taking logarithms:
\[-0.5t \log(2) = \log(0.04)\]
\[-0.5t = \frac{\log(0.04)}{\log(2)} = -4.643856...\]
\[t = \frac{-4.643856...}{-0.5} = 9.2877... \approx 9.3\text{ years}\]

(a) [1 mark]
- B1: \(500\) cao

(b) [2 marks]
- M1: for substituting \(t = 6\) into the formula
- A1: \(2780\) or \(2783\) (accept \(2782.6...\))

(c) [1 mark]
- B1: \(8000\) cao

(d) [2 marks]
- M1: for setting \(P(t) = 5000\) and isolating \(2^{-0.5t} = 0.04\) oe
- A1: \(9.3\) or \(9.29\)

(a)
Set up a system of linear equations using the given data:
For \(v = 4\):
\[k(4)^2 + c(4) = 5.6 \implies 16k + 4c = 5.6 \implies 4k + c = 1.4 \quad \text{--- (1)}\]
For \(v = 10\):
\[k(10)^2 + c(10) = 26.0 \implies 100k + 10c = 26.0 \implies 10k + c = 2.6 \quad \text{--- (2)}\]
Subtract equation (1) from equation (2):
\[(10k + c) - (4k + c) = 2.6 - 1.4\]
\[6k = 1.2 \implies k = 0.2\]
Substitute \(k = 0.2\) back into equation (1):
\[4(0.2) + c = 1.4 \implies 0.8 + c = 1.4 \implies c = 0.6\]

(b)
Substitute \(v = 8\), \(k = 0.2\), and \(c = 0.6\) into the model:
\[d = 0.2(8)^2 + 0.6(8) = 0.2(64) + 4.8 = 12.8 + 4.8 = 17.6\text{ m}\]

(c)
Set \(d = 18\):
\[0.2v^2 + 0.6v = 18\]
\[0.2v^2 + 0.6v - 18 = 0\]
Multiply the entire equation by 5:
\[v^2 + 3v - 90 = 0\]
Factorise or use quadratic formula:
\[(v + 12)(v - 7.5) = 0\]
Since speed \(v > 0\), \(v = 7.5\text{ m/s}\).

(a) [3 marks]
- M1: for writing two simultaneous equations in \(k\) and \(c\): \(16k + 4c = 5.6\) and \(100k + 10c = 26\)
- A1: \(k = 0.2\) cao
- A1: \(c = 0.6\) cao

(b) [1 mark]
- B1: \(17.6\) (or \(17.6\text{ m}\)) FT their \(k\) and \(c\)

(c) [2 marks]
- M1: for setting up the quadratic equation \(0.2v^2 + 0.6v - 18 = 0\) (or equivalent with their coefficients)
- A1: \(7.5\) (or \(7.5\text{ m/s}\)) cao

(a)
At \(t = 0\), \(\cos(0^\circ) = 1\), so maximum depth is:
\[a + b = 9.2 \quad \text{--- (1)}\]
At \(t = 6\), \(\cos(30 \times 6)^\circ = \cos(180^\circ) = -1\), so minimum depth is:
\[a - b = 2.8 \quad \text{--- (2)}\]
Adding equations (1) and (2):
\[2a = 12.0 \implies a = 6\]
Subtracting equation (2) from (1):
\[2b = 6.4 \implies b = 3.2\]

(b)
When \(t = 4\):
\[D = 6 + 3.2\cos(30 \times 4)^\circ = 6 + 3.2\cos(120^\circ)\]
Since \(\cos(120^\circ) = -0.5\):
\[D = 6 + 3.2(-0.5) = 6 - 1.6 = 4.4\text{ m}\]

(c)
The ship can navigate safely when \(D \ge 4.4\):
\[6 + 3.2\cos(30t)^\circ \ge 4.4\]
\[3.2\cos(30t)^\circ \ge -1.6\]
\[\cos(30t)^\circ \ge -0.5\]
For \(0 \le t \le 12\), the angle \(\theta = 30t\) ranges from \(0^\circ\) to \(360^\circ\).
\(\cos(\theta) = -0.5\) at \(\theta = 120^\circ\) and \(\theta = 240^\circ\).
\(\cos(\theta) \ge -0.5\) on the intervals:
1. \(0^\circ \le 30t \le 120^\circ \implies 0 \le t \le 4\) (duration: \(4\) hours)
2. \(240^\circ \le 30t \le 360^\circ \implies 8 \le t \le 12\) (duration: \(4\) hours)

Total safe navigation time in the 12-hour period is \(4 + 4 = 8\text{ hours}\).

(a) [2 marks]
- M1: for setting up \(a + b = 9.2\) and \(a - b = 2.8\) or finding amplitude \(b = \frac{9.2 - 2.8}{2}\) and mean \(a = \frac{9.2 + 2.8}{2}\)
- A1: \(a = 6\) and \(b = 3.2\) cao

(b) [1 mark]
- B1: \(4.4\) (or \(4.4\text{ m}\))

(c) [3 marks]
- M1: for setting up \(6 + 3.2\cos(30t)^\circ = 4.4\) leading to \(\cos(30t)^\circ = -0.5\)
- A1: for finding the critical times \(t = 4\) and \(t = 8\)
- A1: \(8\) (or \(8\text{ hours}\)) cao