2026 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

Apply the negative index rule to invert the fraction: \(\left(\dfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\dfrac{125}{64}\right)^{\frac{2}{3}}\). Take the cube root of the numerator and denominator: \(\sqrt[3]{\dfrac{125}{64}} = \dfrac{5}{4}\). Square the result: \(\left(\dfrac{5}{4}\right)^2 = \dfrac{25}{16}\).

M1 for \(\left(\dfrac{125}{64}\right)^{\frac{2}{3}}\) or \(\left(\dfrac{4}{5}\right)^{-2}\) or \(\dfrac{5}{4}\) seen. A0.5 for \(\dfrac{25}{16}\) or \(1\dfrac{9}{16}\) or \(1.5625\).

First, evaluate the addition inside the brackets using a common denominator of 20: \(\dfrac{7}{10} + \dfrac{3}{4} = \dfrac{14}{20} + \dfrac{15}{20} = \dfrac{29}{20}\). Next, convert the mixed number to an improper fraction: \(3\dfrac{1}{5} = \dfrac{16}{5}\). Finally, divide the fractions: \(\dfrac{16}{5} \div \dfrac{29}{20} = \dfrac{16}{5} \times \dfrac{20}{29} = \dfrac{16 \times 4}{29} = \dfrac{64}{29}\).

M1 for \(\dfrac{29}{20}\) seen or \(\dfrac{16}{5} \times \dfrac{20}{29}\). A0.5 for \(\dfrac{64}{29}\) or \(2\dfrac{6}{29}\).

Calculate the numerical part: \(\dfrac{3.6 \times 5}{1.5} = \dfrac{18}{1.5} = 12\). Calculate the power of 10: \(\dfrac{10^5 \times 10^{-2}}{10^{-4}} = 10^{5 + (-2) - (-4)} = 10^{3 + 4} = 10^7\). Combine the parts: \(12 \times 10^7 = 1.2 \times 10^8\).

M1 for \(12 \times 10^7\) or \(\dfrac{1.8 \times 10^4}{1.5 \times 10^{-4}}\) or correct intermediate working. A0.5 for \(1.2 \times 10^8\).

Apply the power to each factor: \(81^{-\frac{3}{4}} = (3^4)^{-\frac{3}{4}} = 3^{-3} = \dfrac{1}{27}\); \((x^{12})^{-\frac{3}{4}} = x^{12 \times (-\frac{3}{4})} = x^{-9}\); \((y^{-8})^{-\frac{3}{4}} = y^{(-8) \times (-\frac{3}{4})} = y^6\). Combine using positive indices: \(\dfrac{y^6}{27x^9}\).

M1 for correctly finding at least two of \(\dfrac{1}{27}\), \(x^{-9}\), \(y^6\) in an unsimplified product. A0.5 for \(\dfrac{y^6}{27x^9}\).

Evaluate each term: \(32^{\frac{3}{5}} = (\sqrt[5]{32})^3 = 2^3 = 8\); \(9^{-\frac{1}{2}} = \dfrac{1}{\sqrt{9}} = \dfrac{1}{3}\); \(27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9\); \(6^0 = 1\). Substitute back: \(8 - \left(\dfrac{1}{3} \times 9\right) + 1 = 8 - 3 + 1 = 6\).

M1 for correctly evaluating at least two of the components (8, \(\dfrac{1}{3}\), 9, or 1). A0.5 for 6.

Write all terms with a common base of 2: \(4^{n+1} = (2^2)^{n+1} = 2^{2n+2}\); \(8^{2n-1} = (2^3)^{2n-1} = 2^{6n-3}\); \(32^3 = (2^5)^3 = 2^{15}\). Equate powers of 2: \(2^{(2n+2) + (6n-3)} = 2^{15} \implies (2n+2) + (6n-3) = 15\). Simplify and solve for \(n\): \(8n - 1 = 15 \implies 8n = 16 \implies n = 2\).

M1 for expressing all terms with base 2: \(2^{2n+2} \times 2^{6n-3} = 2^{15}\) or setting up \(2(n+1) + 3(2n-1) = 15\). A0.5 for \(n = 2\).

Group the terms to factorise: \(6ax - 9ay + 4bx - 6by = 3a(2x - 3y) + 2b(2x - 3y)\). Factor out the common binomial factor: \(= (2x - 3y)(3a + 2b)\).

M1 for grouping terms correctly, e.g. \(3a(2x - 3y) + 2b(2x - 3y)\) or \(2x(3a + 2b) - 3y(3a + 2b)\). A1.5 for fully correct factorisation \((2x - 3y)(3a + 2b)\).

Find a common denominator: \(\dfrac{5(x + 4) - 2(2x - 3)}{(2x - 3)(x + 4)}\). Expand the numerator: \(5x + 20 - 4x + 6 = x + 26\). Thus, the fraction is \(\dfrac{x + 26}{(2x - 3)(x + 4)}\).

M1 for writing with a common denominator \(5(x + 4) - 2(2x - 3)\). A1.5 for \(\dfrac{x + 26}{(2x - 3)(x + 4)}\) or \(\dfrac{x + 26}{2x^2 + 5x - 12}\).

Multiply both sides by \((2 - p)\): \(w(2 - p) = 3p + 4\). Expand the brackets: \(2w - pw = 3p + 4\). Collect terms in \(p\) on one side: \(2w - 4 = 3p + pw\). Factor out \(p\): \(2w - 4 = p(3 + w)\). Divide by \((3 + w)\): \(p = \dfrac{2w - 4}{w + 3}\) (or \(p = \dfrac{4 - 2w}{-w - 3}\)).

M1 for multiplying by \((2 - p)\) and expanding correctly: \(2w - pw = 3p + 4\). M0.5 for correctly grouping \(p\) terms: \(p(w + 3) = 2w - 4\). A1 for \(p = \dfrac{2w - 4}{w + 3}\) or equivalent.

Factorise the numerator: \(2x^2 - x - 15 = (2x + 5)(x - 3)\). Factorise the denominator as a difference of two squares: \(4x^2 - 25 = (2x - 5)(2x + 5)\). Cancel the common factor \((2x + 5)\): \(\dfrac{(2x + 5)(x - 3)}{(2x - 5)(2x + 5)} = \dfrac{x - 3}{2x - 5}\).

M1 for factorising numerator \((2x + 5)(x - 3)\). M0.5 for factorising denominator \((2x - 5)(2x + 5)\). A1 for final answer \(\dfrac{x - 3}{2x - 5}\).

Multiply the entire equation by \(x(x - 2)\): \(6(x - 2) - x = x(x - 2)\). Expand both sides: \(6x - 12 - x = x^2 - 2x \implies 5x - 12 = x^2 - 2x\). Rearrange into standard quadratic form: \(x^2 - 7x + 12 = 0\). Factorise: \((x - 3)(x - 4) = 0\). Therefore, \(x = 3\) or \(x = 4\).

M1 for clearing fractions correctly to get \(6(x - 2) - x = x(x - 2)\). M0.5 for reducing to \(x^2 - 7x + 12 = 0\). A1 for both solutions \(x = 3\) and \(x = 4\).

Apply the negative index by taking the reciprocal: \(\left(\dfrac{y^9}{27x^6}\right)^{\frac{2}{3}}\). Apply the power \(\frac{2}{3}\) to each component: \(\dfrac{(y^9)^{\frac{2}{3}}}{(27)^{\frac{2}{3}} (x^6)^{\frac{2}{3}}} = \dfrac{y^{9 \times \frac{2}{3}}}{(\sqrt[3]{27})^2 x^{6 \times \frac{2}{3}}} = \dfrac{y^6}{3^2 x^4} = \dfrac{y^6}{9x^4}\).

M1 for dealing with negative index or fractional power, e.g. \(27^{-\frac{2}{3}} = \frac{1}{9}\) or \(x^{6 \times -\frac{2}{3}} = x^{-4}\) or \(y^{-9 \times -\frac{2}{3}} = y^6\). A1.5 for \(\dfrac{y^6}{9x^4}\) or \(\frac{1}{9}x^{-4}y^6\).

Multiply the first equation by 3 and the second equation by 2: \(9x + 6y = 39\) and \(10x - 6y = 18\). Add the two equations together: \(19x = 57 \implies x = 3\). Substitute \(x = 3\) into \(3x + 2y = 13\): \(3(3) + 2y = 13 \implies 9 + 2y = 13 \implies 2y = 4 \implies y = 2\).

M1 for a valid method to eliminate one variable (multiplying equations and adding/subtracting). A1 for \(x = 3\). A0.5 for \(y = 2\).

Find the midpoint of \(AB\):
\(\text{Midpoint} = \left(\frac{-3 + 5}{2}, \frac{5 + (-1)}{2}\right) = (1, 2)\).

Find the gradient of line \(AB\):
\(m_{AB} = \frac{-1 - 5}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4}\).

The gradient of the perpendicular bisector is the negative reciprocal:
\(m_{\perp} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}\).

Use the point-slope form with \((1, 2)\):
\(y - 2 = \frac{4}{3}(x - 1)\)
\(y - 2 = \frac{4}{3}x - \frac{4}{3}\)
\(y = \frac{4}{3}x + \frac{2}{3}\).

M1 for finding midpoint \((1, 2)\)
M1 for gradient of perpendicular \(m = \frac{4}{3}\) from gradient of \(AB = -\frac{3}{4}\)
A1 for \(y = \frac{4}{3}x + \frac{2}{3}\) or exact equivalent in form \(y = mx + c\)

In triangle \(OAD\), \(OA = OD\) because both are radii of the circle, so triangle \(OAD\) is isosceles.
Therefore, \(\angle ODA = \angle OAD = 28^\circ\).
Angle at centre \(\angle COD = 2 \times \angle CAD = 2 \times 28^\circ = 56^\circ\) (or \(\angle AOD = 180^\circ - 2(28^\circ) = 124^\circ\), giving \(\angle DOT = 180^\circ - 124^\circ = 56^\circ\)).

Since \(DT\) is a tangent at \(D\), \(\angle ODT = 90^\circ\).

In right-angled triangle \(ODT\):
\(\angle CTD = 180^\circ - 90^\circ - 56^\circ = 34^\circ\).

M1 for identifying \(\angle ODT = 90^\circ\) (tangent perpendicular to radius)
M1 for calculating \(\angle DOT = 56^\circ\) or \(\angle AOD = 124^\circ\)
A1 for \(34^\circ\)

Angles on the straight line \(EAD\) sum to \(180^\circ\):
\(\angle BAD = 180^\circ - 112^\circ = 68^\circ\).

Opposite angles in cyclic quadrilateral \(ABCD\) sum to \(180^\circ\):
\(\angle BCD = 180^\circ - \angle BAD = 180^\circ - 68^\circ = 112^\circ\)
(alternatively, the exterior angle of a cyclic quadrilateral equals the interior opposite angle, so \(\angle BCD = \angle BAE = 112^\circ\)).

In triangle \(BCD\), the sum of angles is \(180^\circ\):
\(\angle BDC = 180^\circ - \angle BCD - \angle CBD = 180^\circ - 112^\circ - 38^\circ = 30^\circ\).

M1 for finding \(\angle BAD = 68^\circ\) or stating exterior angle rule \(\angle BCD = 112^\circ\)
M1 for \(\angle BCD = 112^\circ\)
A1 for \(30^\circ\)

Gradient of \(L_1\):
\(m_1 = \frac{-1 - 7}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\).

Since \(L_2\) is perpendicular to \(L_1\), its gradient is:
\(m_2 = -\frac{1}{m_1} = \frac{3}{4}\).

Equation of line \(L_2\) passing through \(C(2, 5)\):
\(y - 5 = \frac{3}{4}(x - 2)\)
\(y = \frac{3}{4}x - \frac{3}{2} + 5 = \frac{3}{4}x + \frac{7}{2}\).

At the \(x\)-axis, \(y = 0\):
\(0 = \frac{3}{4}x + \frac{7}{2}\)
\(\frac{3}{4}x = -\frac{7}{2}\)
\(x = -\frac{7}{2} \times \frac{4}{3} = -\frac{14}{3}\).

Thus, the coordinates of \(P\) are \(\left(-\frac{14}{3}, 0\right)\).

M1 for gradient of perpendicular \(m_2 = \frac{3}{4}\) derived from gradient of \(L_1 = -\frac{4}{3}\)
M1 for substituting \(y = 0\) into their equation of \(L_2\)
A1 for \(\left(-\frac{14}{3}, 0\right)\) or \(\left(-4\frac{2}{3}, 0\right)\)

The tangents \(TP\) and \(TQ\) meet radii \(OP\) and \(OQ\) at \(90^\circ\):
\(\angle OPT = 90^\circ\) and \(\angle OQT = 90^\circ\).

In quadrilateral \(OPTQ\), the sum of interior angles is \(360^\circ\):
\(\angle POQ = 360^\circ - 90^\circ - 90^\circ - 52^\circ = 180^\circ - 52^\circ = 128^\circ\).

The angle subtended by the arc \(PQ\) at the circumference is half the angle subtended at the centre:
\(\angle PRQ = \frac{1}{2} \angle POQ = \frac{1}{2} \times 128^\circ = 64^\circ\).

M1 for finding obtuse angle \(\angle POQ = 128^\circ\) using tangent properties in quadrilateral \(OPTQ\)
M1 for using \(\angle PRQ = \frac{1}{2} \angle POQ\)
A1 for \(64^\circ\)

(a) By the cosine rule:
\[XZ^2 = XY^2 + YZ^2 - 2(XY)(YZ)\cos(120^\circ)\]
Since \(\cos(120^\circ) = -\cos(60^\circ) = -\frac{1}{2}\):
\[XZ^2 = 5^2 + 8^2 - 2(5)(8)\left(-\frac{1}{2}\right)\]
\[XZ^2 = 25 + 64 + 40 = 129\]
\[XZ = \sqrt{129}\text{ cm}\]

(b) Using the area formula:
\[\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(5)(8)\sin(120^\circ)\]
Since \(\sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[\text{Area} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\]

(a)
M1 for \(5^2 + 8^2 - 2(5)(8)\left(-\frac{1}{2}\right)\) or stating \(\cos(120^\circ) = -\frac{1}{2}\)
A1 for fully correct working leading to \(XZ = \sqrt{129}\)

(b)
M1 for \(\frac{1}{2} \times 5 \times 8 \times \sin(120^\circ)\) with \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\)
A1 for \(10\sqrt{3}\)

(a) Using Pythagoras' theorem on right-angled triangle \(ABC\):
\[AC^2 = AB^2 + BC^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2 = 32 + 32 = 64\]
\[AC = \sqrt{64} = 8\text{ cm}\]

(b) The centre \(O\) is the midpoint of \(AC\), so \(OA = \frac{1}{2}AC = 4\text{ cm}\).
In right-angled triangle \(VOA\):
\[VO = \sqrt{VA^2 - OA^2} = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}\text{ cm}\]

(c) Let \(\theta\) be the angle between \(VA\) and the base \(ABCD\), which is \(\angle VAO\):
\[\cos\theta = \frac{OA}{VA} = \frac{4}{8} = \frac{1}{2}\]
\[\theta = 60^\circ\]

(a)
B1 for \((4\sqrt{2})^2 + (4\sqrt{2})^2 = 64\) leading to \(AC = 8\)

(b)
B1 for \(\sqrt{8^2 - 4^2} = 4\sqrt{3}\) or \(\sqrt{48}\)

(c)
M1 for \(\cos\theta = \frac{4}{8}\) or \(\sin\theta = \frac{4\sqrt{3}}{8}\) or \(\tan\theta = \frac{4\sqrt{3}}{4}\)
A1 for \(60^\circ\)

Let \(u = \sin x\). The equation becomes:
\[2u^2 + 3u - 2 = 0\]
Factorising:
\[(2u - 1)(u + 2) = 0\]
So \(u = \frac{1}{2}\) or \(u = -2\).

1. \(\sin x = \frac{1}{2}\):
In the first quadrant, \(x = 30^\circ\).
In the second quadrant, \(x = 180^\circ - 30^\circ = 150^\circ\).

2. \(\sin x = -2\):
Since \(-1 \le \sin x \le 1\), there are no solutions.

Thus, the solutions are \(x = 30^\circ\) and \(x = 150^\circ\).

M1 for factorising into \((2\sin x - 1)(\sin x + 2) = 0\) or using the quadratic formula
B1 for indicating that \(\sin x = -2\) has no solutions
M1 for \(\sin x = \frac{1}{2} \implies x = 30^\circ\) or \(x = 150^\circ\)
A1 for both \(x = 30^\circ\) and \(x = 150^\circ\) with no additional solutions in the range

(a) In right-angled triangle \(PSR\):
\[\tan(60^\circ) = \frac{PR}{SR} \implies \sqrt{3} = \frac{\sqrt{3}}{SR} \implies SR = 1\text{ cm}\]

(b) In right-angled triangle \(PQR\):
\[\tan(30^\circ) = \frac{PR}{QR} \implies \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{QR} \implies QR = 3\text{ cm}\]

(c) The length \(QS = QR - SR = 3 - 1 = 2\text{ cm}\).
Triangle \(PQS\) has base \(QS = 2\text{ cm}\) and perpendicular height \(PR = \sqrt{3}\text{ cm}\).
\[\text{Area of } \triangle PQS = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}\text{ cm}^2\]

(a)
B1 for \(SR = 1\)

(b)
B1 for \(QR = 3\)

(c)
M1 for finding \(QS = 3 - 1 = 2\) and using \(\frac{1}{2} \times QS \times PR\) or \(\text{Area}(\triangle PQR) - \text{Area}(\triangle PSR) = \frac{1}{2}(3)(\sqrt{3}) - \frac{1}{2}(1)(\sqrt{3})\)
A1 for fully correct working leading to \(\sqrt{3}\)

(a) Applying the cosine rule at angle \(KLM\):
\[KM^2 = KL^2 + LM^2 - 2(KL)(LM)\cos(\angle KLM)\]
\[(\sqrt{39})^2 = x^2 + (x + 3)^2 - 2(x)(x + 3)\cos(120^\circ)\]
Since \(\cos(120^\circ) = -\frac{1}{2}\):
\[39 = x^2 + (x^2 + 6x + 9) - 2x(x + 3)\left(-\frac{1}{2}\right)\]
\[39 = 2x^2 + 6x + 9 + x(x + 3)\]
\[39 = 2x^2 + 6x + 9 + x^2 + 3x\]
\[39 = 3x^2 + 9x + 9\]
\[3x^2 + 9x - 30 = 0\]
Dividing the entire equation by 3:
\[x^2 + 3x - 10 = 0\]

(b) Factorising the quadratic equation:
\[(x + 5)(x - 2) = 0\]
So \(x = -5\) or \(x = 2\).
Since a side length must be positive, \(x = 2\).

(a)
B1 for \(\cos(120^\circ) = -\frac{1}{2}\)
M1 for substituting correctly into the cosine rule: \((\sqrt{39})^2 = x^2 + (x + 3)^2 - 2(x)(x + 3)\cos(120^\circ)\)
A1 for correct expansion and simplifying convincingly to \(x^2 + 3x - 10 = 0\)

(b)
B1 for \(x = 2\) (must reject \(x = -5\) if mentioned)

(a)(i) Total frequency \(N = 200\).
Median corresponds to cumulative frequency \(0.5 \times 200 = 100\).
This lies in the interval \(100 < d \leqslant 150\), where cumulative frequency rises from 58 to 130.
$$\text{Median} = 100 + \frac{100 - 58}{130 - 58} \times (150 - 100) = 100 + \frac{42}{72} \times 50 = 100 + 29.167 = 129.167\text{ km} \approx 129\text{ km}$$

(a)(ii) The 80th percentile corresponds to cumulative frequency \(0.80 \times 200 = 160\).
This lies in the interval \(150 < d \leqslant 200\), where cumulative frequency rises from 130 to 182.
$$\text{80th percentile} = 150 + \frac{160 - 130}{182 - 130} \times (200 - 150) = 150 + \frac{30}{52} \times 50 = 150 + 28.846 = 178.846\text{ km} \approx 179\text{ km}$$

(b) For \(d = 175\text{ km}\), the value lies halfway through the interval \(150 < d \leqslant 200\).
$$\text{Cumulative frequency at } 175 = 130 + \frac{175 - 150}{200 - 150} \times (182 - 130) = 130 + \frac{25}{50} \times 52 = 130 + 26 = 156$$
Number of vans travelling more than 175 km is \(200 - 156 = 44\).
$$\text{Percentage} = \frac{44}{200} \times 100\% = 22\%$$

(a)(i) B1 for 129 or 129.2
(a)(ii) B1 for 179 or 178.8
(b) M1 for finding cumulative frequency of 156 or \(200 - 156 = 44\) vans
A1 for 22%

(a) Use midpoints \(x\) for each class: 10, 30, 50, 70, 90.
$$\sum fx = (10 \times 15) + (30 \times 28) + (50 \times 42) + (70 \times 25) + (90 \times 10)$$
$$\sum fx = 150 + 840 + 2100 + 1750 + 900 = 5740$$
$$\text{Mean} = \frac{5740}{120} = 47.833... \approx 47.8$$

(b) Cumulative frequencies are:
\(s \leqslant 20: 15\)
\(s \leqslant 40: 43\)
\(s \leqslant 60: 85\)
\(s \leqslant 80: 110\)
\(s \leqslant 100: 120\)

Lower quartile \(Q_1\) at position \(0.25 \times 120 = 30\) (lies in \(20 < s \leqslant 40\)):
$$Q_1 = 20 + \frac{30 - 15}{43 - 15} \times 20 = 20 + \frac{15}{28} \times 20 = 20 + 10.714 = 30.714$$

Upper quartile \(Q_3\) at position \(0.75 \times 120 = 90\) (lies in \(60 < s \leqslant 80\)):
$$Q_3 = 60 + \frac{90 - 85}{110 - 85} \times 20 = 60 + \frac{5}{25} \times 20 = 60 + 4 = 64$$

$$\text{Interquartile range} = Q_3 - Q_1 = 64 - 30.714 = 33.286 \approx 33.3$$

(a) M1 for correct use of class midpoints with frequencies: \(\frac{\sum fx}{120}\)
A1 for 47.8 or 47.83...
(b) B1 for \(Q_1 \approx 30.7\) and \(Q_3 = 64\) leading to \(\text{IQR} = 33.3\) (or 33.29)

(a)
Find the first differences:
\(11 - 3 = 8\)
\(25 - 11 = 14\)
\(45 - 25 = 20\)

Find the second differences:
\(14 - 8 = 6\)
\(20 - 14 = 6\)

The second difference is constant and equal to 6, so the sequence is quadratic of the form \(an^2 + bn + c\).
\(2a = 6 \implies a = 3\)

Subtract \(3n^2\) from each term:
For \(n = 1\): \(3 - 3(1)^2 = 0\)
For \(n = 2\): \(11 - 3(2)^2 = 11 - 12 = -1\)
For \(n = 3\): \(25 - 3(3)^2 = 25 - 27 = -2\)
For \(n = 4\): \(45 - 3(4)^2 = 45 - 48 = -3\)

The linear part has first term 0 and common difference \(-1\):
\(bn + c = -n + 1\)

Therefore, the \(n\)th term is \(3n^2 - n + 1\).

(b)
Set the \(n\)th term equal to 2671:
\(3k^2 - k + 1 = 2671\)
\(3k^2 - k - 2670 = 0\)

Using the quadratic formula or graphic display calculator:
\(k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-2670)}}{2(3)}\)
\(k = \frac{1 \pm \sqrt{1 + 32040}}{6} = \frac{1 \pm \sqrt{32041}}{6} = \frac{1 \pm 179}{6}\)

Since \(k\) must be a positive integer:
\(k = \frac{1 + 179}{6} = \frac{180}{6} = 30\)

(a)
M1 for finding constant second difference = 6 or setting up a system of equations for \(an^2 + bn + c\)
M1 for identifying \(a = 3\) and attempting to find linear component \(bn + c\)
A1 for \(3n^2 - n + 1\) oe

(b)
M1 for setting their \(n\)th term \(= 2671\) and rearranging to a 3-term quadratic \(3k^2 - k - 2670 = 0\) oe
A1 for \(30\)

(a)
Using the compound interest formula \(A = P\left(1 + \frac{r}{100}\right)^t\):
\(A = 6500 \times (1 + 0.042)^5 = 6500 \times (1.042)^5\)
\(A = 6500 \times 1.228396... = 7984.57...\)
To the nearest dollar, the amount is $7985.

(b)
Let \(n\) be the number of years.
Amount in Maya's account: \(6500 \times (1.042)^n\)
Amount in Lucas's account: \(7200 \times (1.028)^n\)

We need:
\(6500 \times (1.042)^n > 7200 \times (1.028)^n\)
\(\left(\frac{1.042}{1.028}\right)^n > \frac{7200}{6500}\)
\((1.0136186...)^n > 1.107692...\)

Taking logarithms:
\(n \log\left(\frac{1.042}{1.028}\right) > \log\left(\frac{72}{65}\right)\)
\(n > \frac{\log\left(\frac{72}{65}\right)}{\log\left(\frac{1.042}{1.028}\right)}\)
\(n > \frac{0.04442...}{0.005869...} \approx 7.57\)

Checking integer values:
At \(n = 7\):
Maya: \(6500 \times (1.042)^7 \approx 8670.36\)
Lucas: \(7200 \times (1.028)^7 \approx 8731.83\) (Lucas is greater)

At \(n = 8\):
Maya: \(6500 \times (1.042)^8 \approx 9034.52\)
Lucas: \(7200 \times (1.028)^8 \approx 8976.32\) (Maya is greater)

Therefore, the number of complete years is 8.

(a)
M1 for \(6500 \times (1.042)^5\) oe
A1 for 7985 (must be rounded to the nearest dollar)

(b)
M1 for setting up inequality or equation \(6500(1.042)^n = 7200(1.028)^n\) or \(6500(1.042)^n > 7200(1.028)^n\) oe
M1 for correct use of logarithms, graph intersection, or trial showing evaluation at \(n = 7\) and \(n = 8\)
A1 for 8

(a)
\(\frac{u_2}{u_1} = \frac{460}{500} = 0.92\)
\(\frac{u_3}{u_2} = \frac{423.2}{460} = 0.92\)
Since the ratio of consecutive terms is constant, the common ratio is 0.92.

(b)
The \(n\)th term of a geometric sequence is given by \(u_n = a r^{n-1}\).
With \(a = 500\) and \(r = 0.92\):
\(u_n = 500 \times (0.92)^{n-1}\)

(c)
We solve the inequality:
\(500 \times (0.92)^{n-1} < 50\)
\((0.92)^{n-1} < \frac{50}{500}\)
\((0.92)^{n-1} < 0.1\)

Taking logarithms of both sides:
\(\log(0.92^{n-1}) < \log(0.1)\)
\((n - 1) \log(0.92) < \log(0.1)\)

Since \(\log(0.92) < 0\), dividing reverses the inequality sign:
\(n - 1 > \frac{\log(0.1)}{\log(0.92)}\)
\(n - 1 > \frac{-1}{-0.036212...} \approx 27.615\)
\(n > 28.615\)

Since \(n\) must be an integer, the smallest value is \(n = 29\).

(a)
B1 for showing \(\frac{460}{500} = 0.92\) and \(\frac{423.2}{460} = 0.92\) clearly

(b)
B1 for \(500 \times (0.92)^{n-1}\) oe

(c)
M1 for \(500 \times (0.92)^{n-1} < 50\) or \(= 50\) oe
M1 for correct method of solving using logs or GDC table/sketch leading to \(n - 1 \approx 27.6\) or \(n \approx 28.6\)
A1 for 29

(a) \(\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} -2 \\ 8 \end{pmatrix} - \begin{pmatrix} 7 \\ -4 \end{pmatrix} = \begin{pmatrix} -2 - 7 \\ 8 - (-4) \end{pmatrix} = \begin{pmatrix} -9 \\ 12 \end{pmatrix}\)

(b) \(|\vec{PQ}| = \sqrt{(-9)^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\)

(c) First, determine \(\vec{PR}\):
\(\vec{PR} = 3\begin{pmatrix} -9 \\ 12 \end{pmatrix} + 2\begin{pmatrix} 4 \\ -1 \end{pmatrix} = \begin{pmatrix} -27 \\ 36 \end{pmatrix} + \begin{pmatrix} 8 \\ -2 \end{pmatrix} = \begin{pmatrix} -19 \\ 34 \end{pmatrix}\)

Then find \(\vec{OR}\):
\(\vec{OR} = \vec{OP} + \vec{PR} = \begin{pmatrix} 7 \\ -4 \end{pmatrix} + \begin{pmatrix} -19 \\ 34 \end{pmatrix} = \begin{pmatrix} -12 \\ 30 \end{pmatrix}\)

(a) M1 for \(\vec{OQ} - \vec{OP}\) or for one correct component in \(\begin{pmatrix} -9 \\ 12 \end{pmatrix}\)
A1 for correct column vector \(\begin{pmatrix} -9 \\ 12 \end{pmatrix}\)

(b) B1 for 15 (or FT \(\sqrt{x^2 + y^2}\) from their (a))

(c) M1 for correct method to find \(\vec{PR}\) or for \(\vec{OR} = \vec{OP} + \vec{PR}\)
A1 for \(\begin{pmatrix} -12 \\ 30 \end{pmatrix}\)

(a) Reflection in the line \(y = -x\) maps each point \((x, y)\) to \((-y, -x)\).
- \((1, 2) \to (-2, -1)\)
- \((4, 2) \to (-2, -4)\)
- \((1, 5) \to (-5, -1)\)

(b) Comparing the vertices of \(T\) to the image:
- \((1, 2) \to (-2, -4) = -2(1, 2)\)
- \((4, 2) \to (-8, -4) = -2(4, 2)\)
- \((1, 5) \to (-2, -10) = -2(1, 5)\)
Each coordinate is multiplied by \(-2\). The transformation is an enlargement with scale factor \(-2\) and centre of enlargement at the origin \((0, 0)\).

(a) B2 for all three correct coordinate pairs: \((-2, -1)\), \((-2, -4)\), \((-5, -1)\)
(B1 for two correct coordinate pairs)

(b) B1 for Enlargement (award 0 marks if more than one transformation is stated)
B1 for Scale factor \(-2\)
B1 for Centre \((0, 0)\) or origin

(a) Let \( y = \frac{4x + 1}{x - 3} \).
\[ y(x - 3) = 4x + 1 \]
\[ xy - 3y = 4x + 1 \]
\[ xy - 4x = 3y + 1 \]
\[ x(y - 4) = 3y + 1 \]
\[ x = \frac{3y + 1}{y - 4} \]
Therefore, \( \mathrm{f}^{-1}(x) = \frac{3x + 1}{x - 4} \).

(b) \( \mathrm{g}(\mathrm{f}(x)) = 7 \implies 2\mathrm{f}(x) - 5 = 7 \)
\[ 2\mathrm{f}(x) = 12 \implies \mathrm{f}(x) = 6 \]
\[ \frac{4x + 1}{x - 3} = 6 \]
\[ 4x + 1 = 6(x - 3) \]
\[ 4x + 1 = 6x - 18 \]
\[ 19 = 2x \implies x = 9.5 \]

(a) [2 marks]
M1 for correct algebraic rearrangement steps to make \( x \) the subject (e.g. \( y(x - 3) = 4x + 1 \) leading to collecting terms in \( x \)).
A1 for \( \frac{3x + 1}{x - 4} \) (or equivalent form in terms of \( x \)).

(b) [1.5 marks]
M0.5 for setting \( 2\mathrm{f}(x) - 5 = 7 \) leading to \( \mathrm{f}(x) = 6 \) (or for expanding \( 2\left(\frac{4x+1}{x-3}\right) - 5 = 7 \)).
A1 for \( x = 9.5 \) or \( \frac{19}{2} \).

(a) \( y \propto \frac{1}{\sqrt{x}} \implies y = \frac{k}{\sqrt{x}} \)
Substitute \( x = 36 \) and \( y = 2.5 \):
\[ 2.5 = \frac{k}{\sqrt{36}} = \frac{k}{6} \]
\[ k = 2.5 \times 6 = 15 \]
So, \( y = \frac{15}{\sqrt{x}} \).

(b) When \( y = 0.75 \):
\[ 0.75 = \frac{15}{\sqrt{x}} \]
\[ \sqrt{x} = \frac{15}{0.75} = 20 \]
\[ x = 20^2 = 400 \]

(a) [2 marks]
M1 for \( y = \frac{k}{\sqrt{x}} \) and substituting \( (36, 2.5) \).
A1 for \( y = \frac{15}{\sqrt{x}} \).

(b) [1.5 marks]
M0.5 for substituting \( y = 0.75 \) into their formula and rearranging to find \( \sqrt{x} \).
A1 for \( 400 \).

(a) The vertical asymptote occurs where the denominator is zero, so \( x - b = 0 \implies x = b \). Hence, \( b = 4 \).
The horizontal asymptote as \( x \to \pm\infty \) is \( y = c \). Hence, \( c = -2 \).

(b) The equation is \( y = \frac{a}{x - 4} - 2 \).
Substitute the coordinates \( (7, 3) \):
\[ 3 = \frac{a}{7 - 4} - 2 \]
\[ 3 = \frac{a}{3} - 2 \]
\[ 5 = \frac{a}{3} \]
\[ a = 15 \]

(a) [2 marks]
B1 for \( b = 4 \).
B1 for \( c = -2 \).

(b) [1.5 marks]
M0.5 for substituting \( x = 7 \), \( y = 3 \) and their \( b, c \) into the equation.
A1 for \( a = 15 \).

(a) The region lies below or on \( y = 2x + 1 \) (or to the right of it), below or on \( x + y = 7 \), and above or on \( y = 1 \).
Hence, the inequalities are:
\[ y \leqslant 2x + 1 \quad (\text{or } 2x - y + 1 \geqslant 0) \]
\[ x + y \leqslant 7 \]
\[ y \geqslant 1 \]

(b) Find the vertices of the triangular region \( R \):
1. Intersection of \( L_1 \) and \( L_3 \):
\( y = 1 \implies 1 = 2x + 1 \implies x = 0 \). Vertex is \( (0, 1) \).

2. Intersection of \( L_2 \) and \( L_3 \):
\( y = 1 \implies x + 1 = 7 \implies x = 6 \). Vertex is \( (6, 1) \).

3. Intersection of \( L_1 \) and \( L_2 \):
\( y = 2x + 1 \) and \( x + (2x + 1) = 7 \implies 3x = 6 \implies x = 2, y = 5 \). Vertex is \( (2, 5) \).

Evaluate the objective function \( P = 3x - 2y \) at each vertex:
- At \( (0, 1) \): \( P = 3(0) - 2(1) = -2 \)
- At \( (6, 1) \): \( P = 3(6) - 2(1) = 18 - 2 = 16 \)
- At \( (2, 5) \): \( P = 3(2) - 2(5) = 6 - 10 = -4 \)

Thus, the maximum value is 16.

(a) [2 marks]
B1 for two correct inequalities from \( y \leqslant 2x + 1 \), \( x + y \leqslant 7 \), \( y \geqslant 1 \).
B1 for all three correct inequalities.

(b) [1.5 marks]
M0.5 for finding the coordinates of the vertices \( (0, 1), (6, 1), (2, 5) \) and evaluating \( 3x - 2y \).
A1 for 16.

(a) Set \( 2^x - 4 = 6 - x^2 \), which is equivalent to \( 2^x + x^2 - 10 = 0 \).
Using a graphic display calculator to find the intersection points of \( y = 2^x - 4 \) and \( y = 6 - x^2 \):
For the negative root: \( x \approx -3.1508... \to -3.15 \) (3 s.f.)
For the positive root: \( x \approx 2.4496... \to 2.45 \) (3 s.f.)

(b) The curve \( y = \mathrm{f}(x) \) is strictly below \( y = \mathrm{g}(x) \) between the two intersection points.
Therefore, the inequality \( 2^x - 4 < 6 - x^2 \) is satisfied for:
\[ -3.15 < x < 2.45 \]

(a) [2.5 marks]
B1 for \( x = -3.15 \) (or value in range \( [-3.152, -3.150] \)).
B1.5 for \( x = 2.45 \) (or value in range \( [2.449, 2.450] \)).

(b) [1 mark]
B1 FT for \( \text{their lower root} < x < \text{their upper root} \).

(a) In the horizontal rectangular base \(ABCD\):
\(AC = \sqrt{14^2 + 10^2} = \sqrt{196 + 100} = \sqrt{296}\text{ cm}\)
Since \(M\) is the midpoint of diagonal \(AC\):
\(AM = \frac{1}{2}\sqrt{296} = \sqrt{74} \approx 8.6023\text{ cm}\)
In the right-angled triangle \(VMA\):
\(VM = \sqrt{VA^2 - AM^2} = \sqrt{15^2 - (\sqrt{74})^2} = \sqrt{225 - 74} = \sqrt{151} \approx 12.288\text{ cm} \approx 12.3\text{ cm}\)

(b) The angle between \(VA\) and the base \(ABCD\) is \(\angle VAM\).
\(\cos(\angle VAM) = \frac{AM}{VA} = \frac{\sqrt{74}}{15} \approx 0.57349\)
\(\angle VAM = \arccos(0.57349) \approx 55.003^\circ \approx 55.0^\circ\)

(c) Total surface area = area of base + area of 4 triangular faces.
Area of base \(ABCD = 14 \times 10 = 140\text{ cm}^2\).

For triangles \(VAB\) and \(VCD\) (isosceles with base 14 cm and legs 15 cm):
Slant height \(h_1 = \sqrt{15^2 - 7^2} = \sqrt{225 - 49} = \sqrt{176} = 4\sqrt{11} \approx 13.2665\text{ cm}\)
Area of 2 faces \(= 2 \times \left(\frac{1}{2} \times 14 \times \sqrt{176}\right) = 14\sqrt{176} \approx 185.731\text{ cm}^2\)

For triangles \(VBC\) and \(VAD\) (isosceles with base 10 cm and legs 15 cm):
Slant height \(h_2 = \sqrt{15^2 - 5^2} = \sqrt{225 - 25} = \sqrt{200} = 10\sqrt{2} \approx 14.1421\text{ cm}\)
Area of 2 faces \(= 2 \times \left(\frac{1}{2} \times 10 \times \sqrt{200}\right) = 10\sqrt{200} \approx 141.421\text{ cm}^2\)

Total surface area \(= 140 + 185.731 + 141.421 = 467.152\text{ cm}^2 \approx 467\text{ cm}^2\).

(a) M1 for \(AM = \frac{1}{2}\sqrt{14^2 + 10^2} = \sqrt{74}\) or equivalent
A1 for 12.3 or 12.28... or \(\sqrt{151}\)

(b) B1 for 55.0° or 55.00...

(c) M1 for a method to find slant height of at least one triangular face (e.g. \(\sqrt{15^2 - 7^2}\) or \(\sqrt{15^2 - 5^2}\))
M1 for complete method for total surface area: \(140 + 2\left(\frac{1}{2} \times 14 \times \sqrt{176}\right) + 2\left(\frac{1}{2} \times 10 \times \sqrt{200}\right)\)
A1 for 467 or 467.1 to 467.2

(a) Using the cosine rule in triangle \(PQR\):
\(QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\angle QPR)\)
\(QR^2 = 85^2 + 110^2 - 2(85)(110)\cos(64^\circ)\)
\(QR^2 = 7225 + 12100 - 18700\cos(64^\circ)\)
\(QR^2 = 19325 - 18700(0.438371...)\)
\(QR^2 = 19325 - 8197.54... = 11127.46...\)
\(QR = \sqrt{11127.46...} \approx 105.487\text{ m} \approx 105\text{ m}\) (or \(105.5\text{ m}\)).

(b) The line \(PS\) divides triangle \(PQR\) into two triangles, \(PQS\) and \(PRS\), each with an angle of \(32^\circ\) at \(P\).
Area of \(\triangle PQR = \text{Area of } \triangle PQS + \text{Area of } \triangle PRS\)
\(\frac{1}{2} \times 85 \times 110 \times \sin(64^\circ) = \frac{1}{2} \times 85 \times PS \times \sin(32^\circ) + \frac{1}{2} \times 110 \times PS \times \sin(32^\circ)\)
\(4675 \sin(64^\circ) = \frac{1}{2} PS \sin(32^\circ) (85 + 110)\)
\(4675 \sin(64^\circ) = 97.5 PS \sin(32^\circ)\)
\(PS = \frac{4675 \sin(64^\circ)}{97.5 \sin(32^\circ)}\)
Using \(\sin(64^\circ) = 2\sin(32^\circ)\cos(32^\circ)\) or direct evaluation:
\(PS = \frac{4675 \times 0.898794}{97.5 \times 0.529919} = \frac{4201.86}{51.667} \approx 81.325\text{ m} \approx 81.3\text{ m}\).

(a) M1 for correct substitution into cosine rule: \(85^2 + 110^2 - 2(85)(110)\cos(64^\circ)\)
A1 for 11127... or 11130
A1 for 105 or 105.5

(b) M1 for angle \(QPS = \text{angle } RPS = 32^\circ\) seen or implied
M1 for setting up the area relation: \(\frac{1}{2}(85)(110)\sin(64^\circ) = \frac{1}{2}(85)(PS)\sin(32^\circ) + \frac{1}{2}(110)(PS)\sin(32^\circ)\) or using sine rule on triangles
A1 for 81.3 or 81.32 to 81.33

(a) Arc length of the sector \(= \frac{140}{360} \times 2 \times \pi \times 18 = \frac{7}{18} \times 36\pi = 14\pi\text{ cm}\).
The arc length equals the circumference of the circular base of the cone:
\(2\pi r = 14\pi \implies r = 7\text{ cm}\).

(b) The slant height of the cone is \(l = 18\text{ cm}\) and base radius is \(r = 7\text{ cm}\).
Height of the cone, \(h = \sqrt{l^2 - r^2} = \sqrt{18^2 - 7^2} = \sqrt{324 - 49} = \sqrt{275} \approx 16.5831\text{ cm}\).
Volume of the cone \(= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7^2) \sqrt{275} = \frac{49\sqrt{275}\pi}{3} \approx 850.93\text{ cm}^3 \approx 851\text{ cm}^3\).

(c) Let the radius of the hemisphere be \(R\).
Volume of hemisphere \(= \frac{2}{3}\pi R^3 = 850.93\text{ cm}^3\)
\(\frac{2}{3}\pi R^3 = \frac{49\sqrt{275}\pi}{3} \implies 2R^3 = 49\sqrt{275} \implies R^3 = \frac{49\sqrt{275}}{2} \approx 406.2866\)
\(R = \sqrt[3]{406.2866} \approx 7.4067\text{ cm}\).
Total surface area of solid hemisphere \(= 2\pi R^2 + \pi R^2 = 3\pi R^2\)
\(\text{Total surface area} = 3\pi (7.4067)^2 \approx 3\pi (54.859) \approx 516.99\text{ cm}^2 \approx 517\text{ cm}^2\).

(a) M1 for \(\frac{140}{360} \times 2\pi(18)\) or \(14\pi\) or \(43.98...\)
A1 for completing the argument to show \(r = 7\) convincingly

(b) M1 for finding vertical height \(h = \sqrt{18^2 - 7^2} = \sqrt{275}\) (or \(16.58...\)) and substituting into \(\frac{1}{3}\pi r^2 h\)
A1 for 851 or 850.9 to 851.0

(c) M1 for \(\frac{2}{3}\pi R^3 = \text{their (b)}\) leading to \(R \approx 7.41\) or using \(3\pi R^2\)
A1 for 517 or 516.9 to 517.1

For (a), the last number in Row \(n\) is \(n^2\). Since there are \(2n - 1\) numbers in Row \(n\), the middle number is \(n - 1\) positions before the last number:
$$\text{Middle number} = n^2 - (n - 1) = n^2 - n + 1$$

For (b), set the middle number equal to 421:
$$k^2 - k + 1 = 421$$
$$k^2 - k - 420 = 0$$
$$(k - 21)(k + 20) = 0$$
Since \(k > 0\), \(k = 21\).

The first number in Row \(k\) is one more than the last number of Row \(k-1\):
$$\text{First number} = (k-1)^2 + 1 = 20^2 + 1 = 401$$

B1: Correct expression \(n^2 - n + 1\) (or equivalent simplified expression) for (a)
M1: Setting \(k^2 - k + 1 = 421\) and solving quadratic equation for \(k\)
A1: Both \(k = 21\) and first number \(= 401\) correct

For (a), each tile has 4 edges, giving a total of \(4 T_n = 4 \times \frac{n(n+1)}{2} = 2n(n+1) = 2n^2 + 2n\) edges.
Each interior grid line is shared by 2 tiles, and each perimeter edge belongs to 1 tile:
$$2 I_n + P_n = 4 T_n$$
$$2 I_n + 4n = 2n^2 + 2n$$
$$2 I_n = 2n^2 - 2n \implies I_n = n^2 - n = n(n-1)$$

For (b), set \(I_n = 156\):
$$n^2 - n = 156$$
$$n^2 - n - 156 = 0$$
$$(n - 13)(n + 12) = 0$$
Since \(n > 0\), \(n = 13\).

The total number of tiles is:
$$T_{13} = \frac{13(14)}{2} = 91$$

B1: Correct expression \(n^2 - n\) or \(n(n-1)\) for \(I_n\)
M1: Equating their expression for \(I_n\) to 156 and solving for positive integer \(n\) (giving \(n = 13\))
A1: 91 tiles

For (a), expand each product:
$$c \times g = (x + 2)(x + 2w) = x^2 + 2wx + 2x + 4w$$
$$a \times i = x(x + 2w + 2) = x^2 + 2wx + 2x$$
Subtracting the two expressions:
$$(c \times g) - (a \times i) = (x^2 + 2wx + 2x + 4w) - (x^2 + 2wx + 2x) = 4w$$

For (b), given \(4w = 68 \implies w = 17\).
The sum of the four corners is:
$$a + c + g + i = x + (x + 2) + (x + 2w) + (x + 2w + 2) = 4x + 4w + 4$$
Substitute \(w = 17\) and sum \(= 260\):
$$4x + 4(17) + 4 = 260$$
$$4x + 72 = 260$$
$$4x = 188 \implies x = 47$$

M1: Correct expansion of \((x+2)(x+2w)\) and \(x(x+2w+2)\) leading convincingly to \(4w\)
B1: \(w = 17\)
A1: \(x = 47\)

(a) \(D_n = T_n - B_n = (2n^2 + 3n) - (n^2 - n + 4) = 2n^2 + 3n - n^2 + n - 4 = n^2 + 4n - 4\).

(b) Substitute \(n+1\) into \(D_n\):
\(D_{n+1} = (n+1)^2 + 4(n+1) - 4 = (n^2 + 2n + 1) + (4n + 4) - 4 = n^2 + 6n + 1\).
Now subtract \(D_n\):
\(D_{n+1} - D_n = (n^2 + 6n + 1) - (n^2 + 4n - 4) = n^2 + 6n + 1 - n^2 - 4n + 4 = 2n + 5\).

(c) For \(n = 3\):
\(D_3 = 3^2 + 4(3) - 4 = 9 + 12 - 4 = 17\).
For \(n = 4\):
\(D_4 = 4^2 + 4(4) - 4 = 16 + 16 - 4 = 28\).
\(D_4 - D_3 = 28 - 17 = 11\).
Using the formula with \(n = 3\): \(2(3) + 5 = 6 + 5 = 11\), which verifies the result.

(a) B1: for \(n^2 + 4n - 4\) oe
(b) M1: for expanding \(D_{n+1} = (n+1)^2 + 4(n+1) - 4\) to at least \(n^2 + 6n + 1\)
A1: for fully correct algebraic subtraction leading convincingly to \(2n + 5\)
(c) C1: for showing both \(D_3 = 17\) and \(D_4 = 28\) leading to \(28 - 17 = 11\) (or equating to \(2(3)+5=11\))

(a)(i) The number directly above is \(n - w\).
(a)(ii) The number directly below is \(n + w\).

(b) \(H = (n - 1)(n + 1) = n^2 - 1\).
\(V = (n - w)(n + w) = n^2 - w^2\).
\(H - V = (n^2 - 1) - (n^2 - w^2) = n^2 - 1 - n^2 + w^2 = w^2 - 1\).

(c) \(w^2 - 1 = 224 \implies w^2 = 225 \implies w = 15\) (since width \(w > 0\)).

(a) B1: for both \(n - w\) and \(n + w\) correct
(b) M1: for correct expansion of both \(H = n^2 - 1\) and \(V = n^2 - w^2\)
A1: for correct subtraction \((n^2 - 1) - (n^2 - w^2)\) leading convincingly to \(w^2 - 1\)
(c) B1: for \(15\) (ignore \(-15\) if seen but positive selected)

(a) \(P_1 = n(n+3) = n^2 + 3n\).
\(P_2 = (n+1)(n+2) = n^2 + 3n + 2\).
\(P_2 - P_1 = (n^2 + 3n + 2) - (n^2 + 3n) = 2\).

(b) The product of all four integers plus 1 is:
\(P_1 P_2 + 1 = P_1(P_1 + 2) + 1 = P_1^2 + 2P_1 + 1 = (P_1 + 1)^2\).
Since \(P_1 = n^2 + 3n\), we have \((P_1 + 1)^2 = (n^2 + 3n + 1)^2\).

(c) Here \(n = 20\).
\(\sqrt{20 \times 21 \times 22 \times 23 + 1} = \sqrt{(20^2 + 3(20) + 1)^2} = 20^2 + 3(20) + 1 = 400 + 60 + 1 = 461\).

(a) B1: for expanding both products \(n^2 + 3n\) and \(n^2 + 3n + 2\) and showing the difference is 2
(b) M1: for writing \(P_1(P_1 + 2) + 1 = (P_1 + 1)^2\) oe
A1: for substituting \(P_1 = n^2 + 3n\) to reach \((n^2 + 3n + 1)^2\) convincingly
(c) B1: for \(461\)

(a) Replacing \(n\) with \(n-1\): \(T_{n-1} = \dfrac{(n-1)((n-1)+1)}{2} = \dfrac{n(n-1)}{2}\).

(b) \((T_n)^2 - (T_{n-1})^2 = \left(\dfrac{n(n+1)}{2}\right)^2 - \left(\dfrac{n(n-1)}{2}\right)^2\)
\(= \dfrac{n^2(n+1)^2}{4} - \dfrac{n^2(n-1)^2}{4}\)
\(= \dfrac{n^2}{4}\left[(n^2 + 2n + 1) - (n^2 - 2n + 1)\right]\)
\(= \dfrac{n^2}{4}[4n] = n^3\).

(c) \(1^3 + 2^3 + \dots + 10^3 = (T_{10})^2\).
\(T_{10} = \dfrac{10(11)}{2} = 55\).
\((T_{10})^2 = 55^2 = 3025\).

(a) B1: for \(\dfrac{n(n-1)}{2}\) or \(\dfrac{n^2 - n}{2}\)
(b) M1: for correct factorisation or expansion \(\dfrac{n^2}{4}[(n+1)^2 - (n-1)^2]\) oe
A1: for fully correct algebraic simplification leading convincingly to \(n^3\)
(c) B1: for \(3025\)

(a) Using the similarity ratio of corresponding linear dimensions from the vertical cross-section through the apex:
\[ \frac{r}{R} = \frac{h}{H} \implies r = \frac{R h}{H} \]

(b) The volume of the frustum is the difference between the volume of the large cone and the volume of the small cone:
\[ V_{\text{frustum}} = \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi r^2 h \]
Substitute \( r = \frac{R h}{H} \):
\[ V_{\text{frustum}} = \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi \left(\frac{Rh}{H}\right)^2 h \]
\[ V_{\text{frustum}} = \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi \frac{R^2 h^3}{H^2} \]
Factor out \( \frac{1}{3}\pi R^2 H \):
\[ V_{\text{frustum}} = \frac{1}{3}\pi R^2 H \left( 1 - \frac{h^3}{H^3} \right) \]

(a)
B1: for \( r = \frac{Rh}{H} \) oe.

(b)
M1: for setting up the volume of the frustum as \( \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi r^2 h \).
M1: for substituting their expression for \( r \) into the volume formula.
A1: for fully correct algebraic manipulation leading convincingly to \( \frac{1}{3}\pi R^2 H \left( 1 - \frac{h^3}{H^3} \right) \) with no steps omitted.

(a) The triangular cross-section of the water is similar to the triangular cross-section of the channel because corresponding angles are equal.
Therefore, the ratio of widths equals the ratio of depths:
\[ \frac{w}{W} = \frac{h}{D} \]
Multiplying both sides by \( W \):
\[ w = \frac{W h}{D} \]

(b) The cross-sectional area of the water is the area of a triangle with base \( w \) and height \( h \):
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} w h \]
The volume \( V \) of water in the channel is the cross-sectional area multiplied by the length \( L \):
\[ V = \frac{1}{2} w h L \]
Substituting \( w = \frac{W h}{D} \):
\[ V = \frac{1}{2} \left( \frac{W h}{D} \right) h L = \frac{L W h^2}{2D} \]

(a)
M1: for stating ratio of similar triangles, e.g. \( \frac{w}{W} = \frac{h}{D} \) or \( \frac{w/2}{W/2} = \frac{h}{D} \).
A1: for concluding \( w = \frac{Wh}{D} \) with clear justification.

(b)
M1: for writing \( V = \frac{1}{2}wh \times L \) and substituting \( w = \frac{Wh}{D} \).
A1: for completing the proof to reach \( V = \frac{L W h^2}{2D} \).

(a) In the vertical cross-section, the cone has height \( H \) and half-width \( R \). The upper unshaded cone above the top of the cylinder has height \( H - h \) and radius \( r \).
By similar triangles:
\[ \frac{H - h}{H} = \frac{r}{R} \]
\[ 1 - \frac{h}{H} = \frac{r}{R} \]
\[ \frac{h}{H} = 1 - \frac{r}{R} \]
Multiplying both sides by \( H \):
\[ h = H \left( 1 - \frac{r}{R} \right) \]

(b) The curved surface area of the cylinder is:
\[ S = 2\pi r h \]
Substitute \( h = H \left( 1 - \frac{r}{R} \right) \) into the formula:
\[ S = 2\pi r \left[ H \left( 1 - \frac{r}{R} \right) \right] \]
\[ S = 2\pi H r \left( 1 - \frac{r}{R} \right) \]
Expand the bracket:
\[ S = 2\pi H \left( r - \frac{r^2}{R} \right) \]

(a)
M1: for setting up the similarity ratio \( \frac{H-h}{H} = \frac{r}{R} \) or \( \frac{H-h}{r} = \frac{H}{R} \).
A1: for clear algebraic steps leading to \( h = H\left(1 - \frac{r}{R}\right) \).

(b)
M1: for substituting the expression for \( h \) into \( S = 2\pi r h \).
A1: for expanding to obtain \( S = 2\pi H \left( r - \frac{r^2}{R} \right) \) with all working shown.

(a) Using a graphic display calculator (GDC), plot \(y = 2x^2 + \frac{2000}{x}\) on the window \(1 \le x \le 20\), \(0 \le y \le 1000\).
The sketch must show a smooth, continuous U-shaped curve in the first quadrant, decreasing steeply from large values near \(x = 1\) to a single local minimum, and then rising quadratically as \(x\) increases towards \(20\).

(b) Using the minimum feature on the GDC for \(S(x) = 2x^2 + \frac{2000}{x}\):
\(x = 7.9370... \approx 7.94\text{ cm}\)
\(S(7.9370) = 377.976... \approx 378\text{ cm}^2\).

(c) To find when \(S(x) \le 420\), solve \(2x^2 + \frac{2000}{x} = 420\) using the intersection feature on the GDC with \(y = 420\):
First intersection: \(x = 5.5966... \approx 5.60\)
Second intersection: \(x = 10.858... \approx 10.9\)

Therefore, the surface area does not exceed \(420\text{ cm}^2\) for \(5.60 \le x \le 10.9\).

(a) B1: Correct shape of graph (smooth curve with a single minimum, decreasing then increasing, no sharp corners).

(b) B1: \(x = 7.94\) or \(7.937...\)
B1: Minimum surface area \(= 378\) or \(377.9...\) to \(378.0\)

(c) B1: \(5.60 \le x \le 10.9\) (accept \(5.59\) to \(5.60\) and \(10.8\) to \(10.9\), or \([5.60, 10.9]\)).

(a) Graph \(y = 1.2v + \frac{480}{v}\) on a GDC over the domain \(5 \le v \le 60\).
The curve begins at \(v = 5, F(5) = 102\), curves downwards to a local minimum, and rises smoothly to \(v = 60, F(60) = 80\).

(b) Using the GDC minimum finder:
Minimum occurs at \(v = 20\text{ km/h}\).
Minimum fuel consumed is \(F(20) = 1.2(20) + \frac{480}{20} = 24 + 24 = 48\text{ litres}\).

(c) Solve \(1.2v + \frac{480}{v} \le 55\):
Plot \(y = 55\) and find points of intersection on the GDC:
\(1.2v^2 - 55v + 480 = 0\)
\(v_1 = \frac{55 - \sqrt{55^2 - 4(1.2)(480)}}{2.4} = \frac{55 - \sqrt{721}}{2.4} \approx 11.728... \approx 11.7\text{ km/h}\)
\(v_2 = \frac{55 + \sqrt{721}}{2.4} \approx 34.104... \approx 34.1\text{ km/h}\)

Thus, the speed range is \(11.7 \le v \le 34.1\text{ km/h}\).

(a) B1: Smooth sketched curve with correct minimum point and end-point trends on \(5 \le v \le 60\).

(b) B1: Speed \(= 20\text{ km/h}\)
B1: Minimum fuel \(= 48\text{ litres}\)

(c) B1: \(11.7 \le v \le 34.1\) (accept \([11.7, 34.1]\) or answers rounding to \(11.7\) and \(34.1\)).

(a) For a physical box, the side lengths must be positive:
\(x > 0\), \(30 - 2x > 0 \implies x < 15\), and \(20 - 2x > 0 \implies x < 10\).
Therefore, the domain is \(0 < x < 10\).

(b) Using a GDC, plot \(y = x(30 - 2x)(20 - 2x)\) for \(0 \le x \le 10\).
The sketch starts at \((0,0)\), increases to a single local maximum, and descends back to the \(x\)-axis at \((10,0)\).

(c) Using the maximum function on the GDC:
\(x = 3.9237... \approx 3.92\text{ cm}\)
\(V(3.9237) = 1056.3... \approx 1060\text{ cm}^3\) (or \(1056\text{ cm}^3\) to 4 s.f.).

(d) Using the GDC, find the points of intersection between \(y = V(x)\) and \(y = 800\) within the domain \(0 < x < 10\):
\(4x^3 - 100x^2 + 600x = 800\)
First value: \(x = 1.879... \approx 1.88\text{ cm}\)
Second value: \(x = 6.361... \approx 6.36\text{ cm}\).

(a) B1: \(0 < x < 10\) (accept \(0 \le x \le 10\) or \(]0, 10[\)).

(b) B1: Correct sketch showing a single peak starting at the origin and crossing the \(x\)-axis at \(x = 10\).

(c) B1: \(x = 3.92\) or \(3.923...\)
B1: Maximum volume \(= 1060\) or \(1056\) (accept \(1056\) to \(1060\)).

(d) B1: \(x = 1.88\) and \(x = 6.36\) (allow \(1.879...\) and \(6.361...\); both values required for 1 mark).