2026 Cambridge IAL Mathematics (9709) Practice Paper with Answers

(a) For a geometric progression, the common ratio \(r\) is constant:
\[ \frac{k}{k + 12} = \frac{k - 8}{k} \]
Cross-multiplying gives:
\[ k^2 = (k + 12)(k - 8) \]
\[ k^2 = k^2 + 4k - 96 \]
\[ 4k - 96 = 0 \implies 4k = 96 \implies k = 24 \]

(b) Using \(k = 24\):
First term, \(a = 24 + 12 = 36\)
Common ratio, \(r = \frac{24}{36} = \frac{2}{3}\)

Since \(|r| < 1\), the sum to infinity exists:
\[ S_\infty = \frac{a}{1 - r} = \frac{36}{1 - \frac{2}{3}} = \frac{36}{\frac{1}{3}} = 108 \]

(a)
M1: Set up a correct equation using the property of common ratio, e.g. \(\frac{k}{k + 12} = \frac{k - 8}{k}\) or \(k^2 = (k + 12)(k - 8)\)
A1: Obtain \(k = 24\)

(b)
M1: Apply \(S_\infty = \frac{a}{1 - r}\) with their \(a = k + 12\) and \(r = \frac{k}{k + 12}\) provided \(|r| < 1\)
A1: Obtain \(108\)

First, expand \((1 - 3x)^5\) up to the term in \(x^2\) using the binomial theorem:
\[ (1 - 3x)^5 = 1 + \binom{5}{1}(-3x) + \binom{5}{2}(-3x)^2 + \dots \]
\[ = 1 + 5(-3x) + 10(9x^2) + \dots \]
\[ = 1 - 15x + 90x^2 + \dots \]

Now multiply by \((a + 2x)\):
\[ (a + 2x)(1 - 15x + 90x^2 + \dots) \]
The terms contributing to \(x^2\) are:
\[ a(90x^2) + 2x(-15x) = (90a - 30)x^2 \]

Given that the coefficient of \(x^2\) is \(240\):
\[ 90a - 30 = 240 \]
\[ 90a = 270 \]
\[ a = 3 \]

B1: Correct expansion of \((1 - 3x)^5\) up to the \(x^2\) term, giving \(-15x\) and \(90x^2\)
M1: Correctly identifying and combining the two products that give the coefficient of \(x^2\), i.e. \(a(90) + 2(-15)\)
M1: Equating their expression for the coefficient of \(x^2\) to \(240\) and attempting to solve for \(a\)
A1: Obtain \(a = 3\)

(a) Completing the square:
\[ \mathrm{f}(x) = 2(x^2 - 6x) + 13 \]
\[ = 2\left[(x - 3)^2 - 9\right] + 13 \]
\[ = 2(x - 3)^2 - 18 + 13 \]
\[ = 2(x - 3)^2 - 5 \]
Hence \(a = 2\), \(h = 3\), \(k = -5\).

(b) Starting from \(y = x^2\):
1. Stretch vertically with scale factor 2 (parallel to the \(y\)-axis), giving \(y = 2x^2\).
2. Translation by the vector \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\), giving \(y = 2(x - 3)^2 - 5\).
(Alternatively, translation by \(\begin{pmatrix} 3 \\ -2.5 \end{pmatrix}\) followed by a stretch with scale factor 2 parallel to the \(y\)-axis).

(a)
B1: State \(2(x - 3)^2\) (i.e. \(a = 2\) and \(h = 3\))
B1: State \(-5\) (i.e. \(k = -5\))

(b)
B1: Stretch with scale factor 2 parallel to the \(y\)-axis (or vertical stretch by scale factor 2)
B1: Translation by vector \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\) (or translation of 3 units in positive \(x\)-direction and 5 units in negative \(y\)-direction) in a correct sequence

(a) Starting with the left-hand side:
\[ \frac{2\sin \theta - \tan \theta}{2\sin \theta + \tan \theta} = \frac{2\sin \theta - \frac{\sin \theta}{\cos \theta}}{2\sin \theta + \frac{\sin \theta}{\cos \theta}} \]
Factorise \(\sin \theta\) from the numerator and denominator:
\[ = \frac{\sin \theta \left(2 - \frac{1}{\cos \theta}\right)}{\sin \theta \left(2 + \frac{1}{\cos \theta}\right)} = \frac{2 - \frac{1}{\cos \theta}}{2 + \frac{1}{\cos \theta}} \]
Multiply the numerator and denominator by \(\cos \theta\):
\[ = \frac{2\cos \theta - 1}{2\cos \theta + 1} \]
This completes the proof.

(b) Using the identity with \(\theta = 2x\):
\[ \frac{2\cos 2x - 1}{2\cos 2x + 1} = 1 - 2\cos 2x \]
Let \(c = \cos 2x\):
\[ \frac{2c - 1}{2c + 1} = 1 - 2c \]
\[ 2c - 1 = (1 - 2c)(1 + 2c) = 1 - 4c^2 \]
\[ 4c^2 + 2c - 2 = 0 \implies 2c^2 + c - 1 = 0 \]
Factorising gives:
\[ (2c - 1)(c + 1) = 0 \]
So \(\cos 2x = \frac{1}{2}\) or \(\cos 2x = -1\).

For \(0^\circ \le x \le 180^\circ\), the interval for \(2x\) is \(0^\circ \le 2x \le 360^\circ\).
- If \(\cos 2x = \frac{1}{2}\):
\(2x = 60^\circ\) or \(2x = 300^\circ\) \(\implies x = 30^\circ\) or \(x = 150^\circ\).
- If \(\cos 2x = -1\):
\(2x = 180^\circ \implies x = 90^\circ\).
However, when \(2x = 180^\circ\), \(\tan 2x = 0\) and \(\sin 2x = 0\), which makes the denominator \(2\sin 2x + \tan 2x = 0\) in the original equation. Therefore, \(x = 90^\circ\) is undefined and must be rejected.

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

(a)
M1: Express \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\) and attempt to combine or factorise \(\sin \theta\).
M1: Put numerator and denominator over a common denominator of \(\cos \theta\) (or multiply top and bottom by \(\cos \theta\)).
A1: Obtain the given expression correctly with full, clear steps shown (WWW).

(b)
M1: Apply the result from (a) with \(2x\) and form a quadratic equation in \(\cos 2x\).
A1: Obtain \(2\cos^2 2x + \cos 2x - 1 = 0\) (or equivalent 3-term quadratic).
M1: Solve the quadratic to find \(\cos 2x = \frac{1}{2}, -1\) and find values for \(x\).
A1: Obtain \(x = 30^\circ, 150^\circ\) only (rejecting \(x = 90^\circ\) or noting denominator is 0).

(a) The length of the arc \(AB\) is:
\[ s = r\theta = 10 \times 1.2 = 12\text{ cm} \]
The length of \(AP\) is:
\[ AP = OA - OP = 10 - 6 = 4\text{ cm} \]
In triangle \(OPB\), \(OP = 6\text{ cm}\), \(OB = 10\text{ cm}\), and angle \(POB = 1.2\text{ radians}\).
Using the cosine rule in triangle \(OPB\):
\[ BP^2 = OP^2 + OB^2 - 2(OP)(OB)\cos(1.2) \]
\[ BP^2 = 6^2 + 10^2 - 2(6)(10)\cos(1.2) = 36 + 100 - 120(0.362358) = 136 - 43.4829 = 92.5171 \]
\[ BP = \sqrt{92.5171} \approx 9.6186\text{ cm} \]
Perimeter of the region:
\[ \text{Perimeter} = AP + BP + \text{arc } AB = 4 + 9.6186 + 12 = 25.6186 \approx 25.6\text{ cm} \]

(b) The area of sector \(OAB\) is:
\[ \text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 10^2 \times 1.2 = 60\text{ cm}^2 \]
The area of triangle \(OPB\) is:
\[ \text{Area}_{\triangle OPB} = \frac{1}{2}(OP)(OB)\sin(1.2) = \frac{1}{2} \times 6 \times 10 \times \sin(1.2) = 30 \times 0.932039 = 27.9612\text{ cm}^2 \]
Area of the shaded region:
\[ \text{Area} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle OPB} = 60 - 27.9612 = 32.0388 \approx 32.0\text{ cm}^2 \]

(a)
B1: State or use \(\text{arc } AB = 10 \times 1.2 = 12\text{ cm}\).
M1: Apply cosine rule correctly to find \(BP^2\) in triangle \(OPB\).
A1: Obtain \(BP \approx 9.619\text{ cm}\) (or \(\sqrt{92.5}\)).
A1: Add \(AP (4) + BP + \text{arc } AB\) to obtain \(25.6\text{ cm}\) (allow 25.62).

(b)
M1: Use \(\frac{1}{2}r^2\theta\) to find the sector area \((60\text{ cm}^2)\).
M1: Use \(\frac{1}{2}ab\sin\theta\) to find the area of \(\triangle OPB\) \((27.96\text{ cm}^2)\).
A1: Subtract triangle area from sector area to obtain \(32.0\text{ cm}^2\) (allow 32.04).

(a) Differentiating \(y = (2x - 1)^{\frac{1}{2}}\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(2x - 1)^{-\frac{1}{2}} \times 2 = (2x - 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x - 1}} \]
At \(x = 5\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{2(5) - 1}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \]
The equation of the tangent at \(A(5, 3)\) is:
\[ y - 3 = \frac{1}{3}(x - 5) \implies y = \frac{1}{3}x + \frac{4}{3} \]
At the point \(B\), \(y = 0\):
\[ 0 = \frac{1}{3}x + \frac{4}{3} \implies \frac{1}{3}x = -\frac{4}{3} \implies x = -4 \]
Thus, the coordinates of \(B\) are \((-4, 0)\).

(b) The region is bounded by the line from \(x = -4\) to \(x = 5\), and the curve from its \(x\)-intercept \(x = \frac{1}{2}\) to \(x = 5\).

The area of the right-angled triangle formed by the tangent, the \(x\)-axis from \(x = -4\) to \(x = 5\), and the vertical line \(x = 5\) is:
\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (5 - (-4)) \times 3 = \frac{1}{2} \times 9 \times 3 = \frac{27}{2} = 13.5 \]
The area under the curve from \(x = \frac{1}{2}\) to \(x = 5\) is:
\[ \int_{\frac{1}{2}}^{5} (2x - 1)^{\frac{1}{2}} \mathrm{d}x = \left[ \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2} \times 2} \right]_{\frac{1}{2}}^{5} = \left[ \frac{1}{3}(2x - 1)^{\frac{3}{2}} \right]_{\frac{1}{2}}^{5} \]
\[ = \frac{1}{3}(9)^{\frac{3}{2}} - \frac{1}{3}(0)^{\frac{3}{2}} = \frac{1}{3}(27) - 0 = 9 \]
The shaded area is the difference between these two areas:
\[ \text{Area} = \text{Area}_{\text{triangle}} - \text{Area}_{\text{curve}} = \frac{27}{2} - 9 = \frac{9}{2} = 4.5 \]

(a)
M1: Differentiate to obtain \(k(2x - 1)^{-\frac{1}{2}}\).
A1: Obtain gradient \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3}\) at \(x = 5\).
M1: Form equation of tangent through \((5, 3)\) and set \(y = 0\).
A1: Correctly deduce \(x = -4\) and state \(B(-4, 0)\).

(b)
M1: Calculate area of triangle from \(x = -4\) to \(x = 5\) (or integrate line equation) to get \(13.5\).
M1: Integrate \((2x - 1)^{\frac{1}{2}}\) to obtain \(\frac{1}{3}(2x - 1)^{\frac{3}{2}}\) and evaluate between \(\frac{1}{2}\) and \(5\).
A1: Subtract to obtain exact area \(\frac{9}{2}\) (or \(4.5\)).

(a) Integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(x\):
\[ y = \int \left(2x - 16x^{-2}\right) \mathrm{d}x = x^2 - \frac{16x^{-1}}{-1} + c = x^2 + \frac{16}{x} + c \]
Substitute the coordinates of the given point \((1, 20)\):
\[ 20 = (1)^2 + \frac{16}{1} + c \implies 20 = 17 + c \implies c = 3 \]
Thus, the equation of the curve is:
\[ y = x^2 + \frac{16}{x} + 3 \]

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 2x - \frac{16}{x^2} = 0 \implies 2x = \frac{16}{x^2} \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2 \]
Substitute \(x = 2\) into the equation of the curve to find the \(y\)-coordinate:
\[ y = (2)^2 + \frac{16}{2} + 3 = 4 + 8 + 3 = 15 \]
So the coordinates of the stationary point are \((2, 15)\).

To determine the nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(2x - 16x^{-2}\right) = 2 + 32x^{-3} = 2 + \frac{32}{x^3} \]
At \(x = 2\):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + \frac{32}{2^3} = 2 + \frac{32}{8} = 2 + 4 = 6 \]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point is a local minimum.

(a)
M1: Integrate at least one term correctly (increasing power of \(x\) to 2 or \(-1\)).
A1: Obtain \(y = x^2 + \frac{16}{x} + c\).
A1: Substitute \((1, 20)\) and find \(c = 3\) to state \(y = x^2 + \frac{16}{x} + 3\).

(b)
M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve for \(x\).
A1: Obtain \(x = 2\) and \(y = 15\).
M1: Find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + \frac{32}{x^3}\) (or test sign of first derivative either side of \(x = 2\)).
A1: Correctly evaluate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6 > 0\) and conclude that \((2, 15)\) is a minimum.

(a) In the right-angled triangle \(OCA\):
\(AC = r\sin\theta\)
\(OC = r\cos\theta\)
Since \(OB = r\), the length \(BC = OB - OC = r - r\cos\theta = r(1 - \cos\theta)\).
The length of the arc \(AB = r\theta\).
Therefore, the perimeter \(P\) of the shaded region is:
\(P = \text{arc } AB + AC + BC = r\theta + r\sin\theta + r(1 - \cos\theta) = r(\theta + \sin\theta - \cos\theta + 1)\).

(b) Area of sector \(OAB = \frac{1}{2}r^2\theta = \frac{1}{2}(6^2)\left(\frac{\pi}{3}\right) = \frac{1}{2}(36)\left(\frac{\pi}{3}\right) = 6\pi\).
Area of triangle \(OCA = \frac{1}{2} \times OC \times AC = \frac{1}{2}(r\cos\theta)(r\sin\theta) = \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{4}r^2\sin(2\theta)\).
For \(r = 6\) and \(\theta = \frac{\pi}{3}\):
\(OC = 6\cos\left(\frac{\pi}{3}\right) = 6 \times \frac{1}{2} = 3\)
\(AC = 6\sin\left(\frac{\pi}{3}\right) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\)
Area of triangle \(OCA = \frac{1}{2} \times 3 \times 3\sqrt{3} = \frac{9\sqrt{3}}{2}\).
Area of the shaded region \(= \text{Area of sector } OAB - \text{Area of } \Delta OCA = 6\pi - \frac{9\sqrt{3}}{2}\).

(a)
B1: State \(AC = r\sin\theta\) and \(OC = r\cos\theta\) or \(BC = r - r\cos\theta\).
B1: State \(\text{arc } AB = r\theta\).
M1: Sum the three components \(r\theta + r\sin\theta + r - r\cos\theta\) to complete the proof convincingly.

(b)
M1: Use sector area formula \(\frac{1}{2}r^2\theta\) with \(r = 6\) and \(\theta = \frac{\pi}{3}\) to obtain \(6\pi\).
M1: Use valid method for the area of triangle \(OCA\), e.g. \(\frac{1}{2} \times OC \times AC\) or \(\frac{1}{2}r^2\sin\theta\cos\theta\).
A1: Correct triangle area \(\frac{9\sqrt{3}}{2}\).
A1: Correct exact area \(6\pi - \frac{9\sqrt{3}}{2}\) or \(\frac{12\pi - 9\sqrt{3}}{2}\).

(a) Complete the square for the equation of \(C\):
\((x - 2)^2 - 4 + (y - 1)^2 - 1 - 4 = 0\)
\((x - 2)^2 + (y - 1)^2 = 9\)
So the centre is \((2, 1)\) and the radius is \(r = \sqrt{9} = 3\).

(b) Distance from the centre \((2, 1)\) to \(P(6, 4)\):
\(CP^2 = (6 - 2)^2 + (4 - 1)^2 = 4^2 + 3^2 = 25\)
Since \(CP^2 = 25 > 9 = r^2\) (or \(CP = 5 > 3\)), \(P\) lies outside the circle \(C\).
Length of each tangent \(= \sqrt{CP^2 - r^2} = \sqrt{25 - 9} = \sqrt{16} = 4\).

(c) A line through \(P(6, 4)\) with gradient \(m\) has equation:
\(y - 4 = m(x - 6) \implies mx - y + (4 - 6m) = 0\)
The perpendicular distance from the centre \((2, 1)\) to this line equals the radius \(3\):
\(\frac{|m(2) - 1(1) + 4 - 6m|}{\sqrt{m^2 + (-1)^2}} = 3\)
\(\frac{|3 - 4m|}{\sqrt{m^2 + 1}} = 3\)
Squaring both sides:
\((3 - 4m)^2 = 9(m^2 + 1)\)
\(9 - 24m + 16m^2 = 9m^2 + 9\)
\(7m^2 - 24m = 0 \implies m(7m - 24) = 0\)
So \(m = 0\) or \(m = \frac{24}{7}\).
For \(m = 0\): \(y - 4 = 0(x - 6) \implies y = 4\).
For \(m = \frac{24}{7}\): \(y - 4 = \frac{24}{7}(x - 6) \implies 7y - 28 = 24x - 144 \implies 24x - 7y - 116 = 0\).

(d) The quadrilateral \(CAPB\) is formed by two congruent right-angled triangles, \(\triangle CAP\) and \(\triangle CBP\), right-angled at \(A\) and \(B\) respectively.
Area \(= 2 \times \left(\frac{1}{2} \times r \times \text{tangent length}\right) = 2 \times \left(\frac{1}{2} \times 3 \times 4\right) = 12\).

(a)
B1: State correct centre \((2, 1)\)
B1: State correct radius \(r = 3\)

(b)
B1: Show \(CP = 5 > 3\) or \(CP^2 = 25 > 9\) to confirm \(P\) is outside \(C\)
B1: Correct tangent length \(= 4\)

(c)
M1: Form equation of line through \((6, 4)\) in terms of gradient \(m\) and apply perpendicular distance to \((2, 1)\) equals 3, or substitute into circle equation and set discriminant \(\Delta = 0\)
M1: Simplify to a quadratic equation in \(m\), e.g. \(7m^2 - 24m = 0\)
A1: Find \(m = 0\) and \(m = \frac{24}{7}\)
A1: State both equations correctly: \(y = 4\) and \(24x - 7y - 116 = 0\) (or \(y = \frac{24}{7}x - \frac{116}{7}\))

(d)
M1: Use formula for area of two right-angled triangles \(2 \times \left(\frac{1}{2} \times 3 \times 4\right)\) or equivalent valid method
A1: Correct area \(12\)

(a) When \(x = 1\), \(y = \frac{6}{(2(1) + 1)^2} = \frac{6}{9} = \frac{2}{3}\).
\(y = 6(2x + 1)^{-2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(-2)(2x + 1)^{-3}(2) = -24(2x + 1)^{-3} = -\frac{24}{(2x + 1)^3}\)
At \(x = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{24}{3^3} = -\frac{24}{27} = -\frac{8}{9}\).
Equation of the tangent:
\(y - \frac{2}{3} = -\frac{8}{9}(x - 1) \implies y = -\frac{8}{9}x + \frac{14}{9}\) (or \(8x + 9y - 14 = 0\)).

(b) Area under the curve from \(x = 0\) to \(x = 1\):
\(\int_0^1 6(2x + 1)^{-2}\,\mathrm{d}x = \left[ \frac{6(2x + 1)^{-1}}{-1 \times 2} \right]_0^1 = \left[ -\frac{3}{2x + 1} \right]_0^1\)
\(= \left(-\frac{3}{3}\right) - \left(-\frac{3}{1}\right) = -1 - (-3) = 2\).
Area under the tangent from \(x = 0\) to \(x = 1\):
\(\int_0^1 \left(-\frac{8}{9}x + \frac{14}{9}\right)\,\mathrm{d}x = \left[ -\frac{4}{9}x^2 + \frac{14}{9}x \right]_0^1 = -\frac{4}{9} + \frac{14}{9} = \frac{10}{9}\)
(Alternatively, using the area of a trapezium: \(\frac{1}{2} \left(\frac{14}{9} + \frac{2}{3}\right) \times 1 = \frac{10}{9}\)).
Required area \(= 2 - \frac{10}{9} = \frac{8}{9}\).

(c) Volume of revolution about the \(x\)-axis:
\(V = \pi \int_0^1 y^2\,\mathrm{d}x = \pi \int_0^1 \left(\frac{6}{(2x + 1)^2}\right)^2\,\mathrm{d}x = \pi \int_0^1 36(2x + 1)^{-4}\,\mathrm{d}x\)
\(\int 36(2x + 1)^{-4}\,\mathrm{d}x = \frac{36(2x + 1)^{-3}}{-3 \times 2} = -\frac{6}{(2x + 1)^3}\)
\(V = \pi \left[ -\frac{6}{(2x + 1)^3} \right]_0^1 = \pi \left( -\frac{6}{3^3} - \left(-\frac{6}{1^3}\right) \right) = \pi \left( -\frac{6}{27} + 6 \right) = \pi \left( -\frac{2}{9} + 6 \right) = \frac{52}{9}\pi\).

(a)
M1: Differentiate to obtain \(k(2x+1)^{-3}\) and evaluate at \(x = 1\)
A1: Obtain gradient \(-\frac{8}{9}\) and \(y\)-coordinate \(\frac{2}{3}\)
A1: Correct equation of tangent: \(y = -\frac{8}{9}x + \frac{14}{9}\) or \(8x + 9y - 14 = 0\)

(b)
M1: Integrate \(6(2x + 1)^{-2}\) correctly to obtain \(-\frac{3}{2x + 1}\) and substitute limits 0 and 1 to get 2
M1: Find area under tangent line from 0 to 1 (by integration or area of trapezium) to obtain \(\frac{10}{9}\)
A1: Subtract to obtain \(\frac{8}{9}\)

(c)
M1: Use \(V = \pi \int y^2\,\mathrm{d}x\) and square \(y\) to get \(36(2x+1)^{-4}\)
M1: Integrate to obtain \(a(2x+1)^{-3}\) where \(a = -6\)
DM1: Correctly apply limits \(x = 0\) and \(x = 1\)
A1: Obtain \(\frac{52}{9}\pi\) (or exact equivalent in form \(k\pi\))

(a) The centre of \(C_1\) is the midpoint of \(AB\):
\(\text{Centre} = \left(\frac{-2 + 4}{2}, \frac{3 + (-5)}{2}\right) = (1, -1)\).
The diameter is the distance \(AB\):
\(AB = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).
So the radius \(r_1 = \frac{10}{2} = 5\).
The equation of \(C_1\) is:
\((x - 1)^2 + (y + 1)^2 = 25\) (or \(x^2 + y^2 - 2x + 2y - 23 = 0\)).

(b) For \(L\) to be a tangent to \(C_1\), the perpendicular distance from the centre \((1, -1)\) to the line \(3x - 4y - c = 0\) must equal the radius \(5\):
\(\frac{|3(1) - 4(-1) - c|}{\sqrt{3^2 + (-4)^2}} = 5\)
\(\frac{|7 - c|}{5} = 5\)
\(|7 - c| = 25\)
This gives two linear equations:
\(7 - c = 25 \implies c = -18\)
\(7 - c = -25 \implies c = 32\).

(c)(i) Let the centre of \(C_1\) be \(O_1(1, -1)\) and the centre of \(C_2\) be \(O_2(7, 7)\).
The distance between the centres is:
\(O_1 O_2 = \sqrt{(7 - 1)^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\).
Since the two circles touch externally, \(r_1 + r_2 = O_1 O_2\):
\(5 + r_2 = 10 \implies r_2 = 5\).
The equation of \(C_2\) is:
\((x - 7)^2 + (y - 7)^2 = 25\) (or \(x^2 + y^2 - 14x - 14y + 73 = 0\)).

(c)(ii) Since \(r_1 = r_2 = 5\), the point of contact is the midpoint of the line segment joining the centres \(O_1(1, -1)\) and \(O_2(7, 7)\):
\(\text{Point of contact} = \left(\frac{1 + 7}{2}, \frac{-1 + 7}{2}\right) = (4, 3)\).

(a)
M1: Find the midpoint of \(AB\) to determine the centre \((1, -1)\)
M1: Calculate the radius \(r_1 = 5\) or \(r_1^2 = 25\) using distance formula
A1: Correct equation of circle: \((x - 1)^2 + (y + 1)^2 = 25\) or expanded equivalent

(b)
M1: Use perpendicular distance formula from \((1, -1)\) to \(3x - 4y - c = 0\), or substitute \(y = \frac{3x - c}{4}\) into the circle equation to form a quadratic in \(x\)
M1: Set distance equal to 5 (or set discriminant \(\Delta = 0\))
A1: Obtain \(|7 - c| = 25\) or quadratic \(c^2 - 14c - 576 = 0\)
A1: Correct values \(c = -18\) and \(c = 32\)

(c)(i)
M1: Calculate distance between centres \(\sqrt{(7-1)^2 + (7+1)^2} = 10\) and deduce \(r_2 = 10 - 5 = 5\)
A1: Correct equation of \(C_2\): \((x - 7)^2 + (y - 7)^2 = 25\)

(c)(ii)
B1: State correct coordinates of point of contact \((4, 3)\)

To solve \(|3x - 5| < 2|x + 1|\), square both sides since both sides are non-negative:
\[(3x - 5)^2 < 4(x + 1)^2\]

Expand both expressions:
\[9x^2 - 30x + 25 < 4(x^2 + 2x + 1)\]
\[9x^2 - 30x + 25 < 4x^2 + 8x + 4\]

Rearrange into a single quadratic inequality:
\[5x^2 - 38x + 21 < 0\]

Find the critical values by solving \(5x^2 - 38x + 21 = 0\):
\[(5x - 3)(x - 7) = 0\]
\[x = \frac{3}{5} \quad \text{or} \quad x = 7\]

Since the inequality is \(< 0\), the solution corresponds to the region between the roots:
\[\frac{3}{5} < x < 7\]

M1: Valid method to solve the modulus inequality, such as squaring both sides or solving linear equations for critical values.
A1: Correct 3-term quadratic expression \(5x^2 - 38x + 21 < 0\) (or correct critical values \(x = \frac{3}{5}\) and \(x = 7\)).
M1: Identifying the correct inner interval for the strict inequality.
A1: Final correct inequality \(\frac{3}{5} < x < 7\) (or \(0.6 < x < 7\)).

Let \(u = \mathrm{e}^x\). The equation becomes a quadratic in \(u\):
\[3u^2 - 11u + 6 = 0\]

Factorise the quadratic:
\[(3u - 2)(u - 3) = 0\]

Thus:
\[u = \frac{2}{3} \quad \text{or} \quad u = 3\]

Substitute back \(u = \mathrm{e}^x\):
1. \(\mathrm{e}^x = \frac{2}{3} \implies x = \ln\left(\frac{2}{3}\right)\) (or \(\ln 2 - \ln 3\))
2. \(\mathrm{e}^x = 3 \implies x = \ln 3\)

Both values of \(u\) are positive, so both give valid real solutions.

M1: Recognise the equation as quadratic in \(\mathrm{e}^x\) and attempt to solve/factorise.
A1: Obtain correct roots \(\mathrm{e}^x = \frac{2}{3}\) and \(\mathrm{e}^x = 3\).
M1: Apply natural logarithms correctly to solve \(\mathrm{e}^x = k\) for \(k > 0\).
A1: State both exact solutions \(x = \ln\left(\frac{2}{3}\right)\) (or \(\ln 2 - \ln 3\)) and \(x = \ln 3\) correctly.

(a) Interval width \(h = \frac{2 - 0}{4} = 0.5\).

Calculate the function values \(y = \frac{6}{\sqrt{2x + 1}}\) at each ordinate:
- \(x_0 = 0.0 \implies y_0 = \frac{6}{\sqrt{1}} = 6\)
- \(x_1 = 0.5 \implies y_1 = \frac{6}{\sqrt{2}} \approx 4.24264\)
- \(x_2 = 1.0 \implies y_2 = \frac{6}{\sqrt{3}} \approx 3.46410\)
- \(x_3 = 1.5 \implies y_3 = \frac{6}{\sqrt{4}} = 3\)
- \(x_4 = 2.0 \implies y_4 = \frac{6}{\sqrt{5}} \approx 2.68328\)

Applying the trapezium rule formula:
\[\text{Area} \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right]\]
\[= \frac{0.5}{2} \left[ 6 + 2.68328 + 2(4.24264 + 3.46410 + 3) \right]\]
\[= 0.25 \left[ 8.68328 + 2(10.70674) \right]\]
\[= 0.25 \left[ 8.68328 + 21.41348 \right] = 0.25 \times 30.09676 = 7.52419 \approx 7.52\]

(b) Overestimate, because the curve is convex (concave upwards) on the interval \([0, 2]\), so the straight line chords (trapezia tops) lie entirely above the curve.

(a) B1: Correct value of \(h = 0.5\) and correct ordinates calculated to at least 3 significant figures.
M1: Correct application of the trapezium rule formula with 4 intervals.
A1: Obtain \(7.52\) (accept 7.524... rounded to 2 d.p.).
(b) B1: State 'overestimate' and give a valid justification referring to the chords lying above the curve or the curve being concave upwards / convex.

(a)
Expanding \( R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha \).
Equating coefficients:
\( R\cos\alpha = 8 \)
\( R\sin\alpha = 15 \)

\( R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \)
\( \tan\alpha = \frac{15}{8} = 1.875 \implies \alpha = \tan^{-1}(1.875) \approx 61.93^\circ \)

Thus, \( 8\cos\theta - 15\sin\theta = 17\cos(\theta + 61.93^\circ) \).

(b)
Using the result from (a):
\( 17\cos(\theta + 61.93^\circ) = 8.5 \)
\( \cos(\theta + 61.93^\circ) = \frac{8.5}{17} = 0.5 \)

Since \( 0^\circ \le \theta \le 360^\circ \), the interval for \( \theta + 61.93^\circ \) is \( 61.93^\circ \le \theta + 61.93^\circ \le 421.93^\circ \).

The principal value is \( \cos^{-1}(0.5) = 60^\circ \).
In the given interval:
\( \theta + 61.928^\circ = 360^\circ - 60^\circ = 300^\circ \implies \theta = 300^\circ - 61.928^\circ = 238.07^\circ \approx 238.1^\circ \)
\( \theta + 61.928^\circ = 360^\circ + 60^\circ = 420^\circ \implies \theta = 420^\circ - 61.928^\circ = 358.07^\circ \approx 358.1^\circ \)

(c)
The expression is \( \frac{68}{17\cos(\theta + 61.93^\circ) + 21} \).
Since \( -1 \le \cos(\theta + 61.93^\circ) \le 1 \):
- The maximum value occurs when the denominator is at its minimum, which is when \( \cos(\theta + 61.93^\circ) = -1 \):
\( \text{Maximum} = \frac{68}{17(-1) + 21} = \frac{68}{4} = 17 \)
- The minimum value occurs when the denominator is at its maximum, which is when \( \cos(\theta + 61.93^\circ) = 1 \):
\( \text{Minimum} = \frac{68}{17(1) + 21} = \frac{68}{38} = \frac{34}{19} \)

(a)
B1: State \( R = 17 \)
M1: Use \( \tan\alpha = \frac{15}{8} \) or equivalent to find \( \alpha \)
A1: Obtain \( \alpha = 61.93^\circ \) (accept 61.9° or 1.08 rad)

(b)
M1: Equate their \( R\cos(\theta + \alpha) \) to 8.5 and evaluate \( \cos^{-1}\left(\frac{8.5}{R}\right) \)
M1: Add/subtract appropriately within the range \( [0^\circ, 360^\circ] \) to find values of \( \theta \)
A1: Obtain \( \theta = 238.1^\circ \) and \( 358.1^\circ \) (allow 238° and 358°)

(c)
B1: State maximum value is 17
B1: State minimum value is \( \frac{34}{19} \) (or exact equivalent fraction)

(a)
By the Remainder Theorem, \( P\left(\frac{1}{2}\right) = -5 \).
\( 4\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) - 3 = -5 \)
\( 4\left(\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) + \frac{1}{2}k - 3 = -5 \)
\( \frac{1}{2} - 2 + \frac{1}{2}k - 3 = -5 \)
\( \frac{1}{2}k - 4.5 = -5 \implies \frac{1}{2}k = -0.5 \implies k = -1 \).

(b)
Dividing \( 4x^3 - 8x^2 - x - 3 \) by \( (x - 2) \):
\( 4x^3 - 8x^2 - x - 3 = 4x^2(x - 2) + 0x(x - 2) - 1(x - 2) - 5 \)
\( = (x - 2)(4x^2 - 1) - 5 \)
Thus, the quotient is \( 4x^2 - 1 \) and the remainder is \( -5 \).

(c)
Let \( f(x) = 4x^3 - 8x^2 - x - 3 \).
\( f(2.2) = 4(2.2)^3 - 8(2.2)^2 - (2.2) - 3 = 4(10.648) - 8(4.84) - 5.2 = 42.592 - 38.72 - 5.2 = -1.328 < 0 \)
\( f(2.4) = 4(2.4)^3 - 8(2.4)^2 - (2.4) - 3 = 4(13.824) - 8(5.76) - 5.4 = 55.296 - 46.08 - 5.4 = 3.816 > 0 \)
Since there is a change of sign and \( f(x) \) is continuous, there is a root in the interval \( (2.2, 2.4) \).

(d)
Using \( x_{n+1} = \sqrt[3]{\frac{8x_n^2 + x_n + 3}{4}} \) with \( x_1 = 2.3 \):
\( x_2 = \sqrt[3]{\frac{8(2.3)^2 + 2.3 + 3}{4}} = \sqrt[3]{11.905} \approx 2.2834 \)
\( x_3 = \sqrt[3]{\frac{8(2.2834)^2 + 2.2834 + 3}{4}} \approx 2.2734 \)
\( x_4 = \sqrt[3]{\frac{8(2.2734)^2 + 2.2734 + 3}{4}} \approx 2.2672 \)
\( x_5 = \sqrt[3]{\frac{8(2.2672)^2 + 2.2672 + 3}{4}} \approx 2.2635 \)
\( x_6 = \sqrt[3]{\frac{8(2.2635)^2 + 2.2635 + 3}{4}} \approx 2.2612 \)
\( x_7 \approx 2.2598 \)
Rounding to 2 decimal places gives \( x = 2.26 \).

(a)
M1: Substitute \( x = \frac{1}{2} \) into \( P(x) \) and set equal to \( -5 \)
A1: Complete fully justified working to arrive at \( k = -1 \)

(b)
M1: Attempt polynomial division or algebraic inspection to divide by \( (x - 2) \)
A1: Obtain quotient \( 4x^2 - 1 \) and remainder \( -5 \)

(c)
M1: Evaluate \( f(2.2) \) and \( f(2.4) \)
A1: Obtain correct numerical values with differing signs and make a concluding statement referring to sign change

(d)
M1: Correctly apply the iterative formula at least once
A1: Show iterations converging, with values given to at least 4 decimal places
A1: State final root as 2.26

(a)
Using the double-angle identity for \(\sin^2(2x)\):
\[ \sin^2(2x) = \frac{1 - \cos(4x)}{2} = \frac{1}{2} - \frac{1}{2}\cos(4x) \]

Using the double-angle identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\):
\[ \cos^4(x) = \left( \frac{1 + \cos(2x)}{2} \right)^2 = \frac{1 + 2\cos(2x) + \cos^2(2x)}{4} \]

Using \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\):
\[ \cos^4(x) = \frac{1}{4} + \frac{1}{2}\cos(2x) + \frac{1}{4}\left(\frac{1 + \cos(4x)}{2}\right) = \frac{1}{4} + \frac{1}{2}\cos(2x) + \frac{1}{8} + \frac{1}{8}\cos(4x) \]
\[ \cos^4(x) = \frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \]

Adding the two expressions together:
\[ \sin^2(2x) + \cos^4(x) = \left(\frac{1}{2} - \frac{1}{2}\cos(4x)\right) + \left(\frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) \]
\[ = \left(\frac{1}{2} + \frac{3}{8}\right) + \frac{1}{2}\cos(2x) + \left(-\frac{1}{2} + \frac{1}{8}\right)\cos(4x) \]
\[ = \frac{7}{8} + \frac{1}{2}\cos(2x) - \frac{3}{8}\cos(4x) \]

(b)
Using the identity established in part (a):
\[ \int_0^{\frac{\pi}{4}} \left( \sin^2(2x) + \cos^4(x) \right) \mathrm{d}x = \int_0^{\frac{\pi}{4}} \left( \frac{7}{8} + \frac{1}{2}\cos(2x) - \frac{3}{8}\cos(4x) \right) \mathrm{d}x \]

Integrating term by term:
\[ = \left[ \frac{7}{8}x + \frac{1}{4}\sin(2x) - \frac{3}{32}\sin(4x) \right]_0^{\frac{\pi}{4}} \]

Evaluating at the upper limit \(x = \frac{\pi}{4}\):
\[ \frac{7}{8}\left(\frac{\pi}{4}\right) + \frac{1}{4}\sin\left(\frac{\pi}{2}\right) - \frac{3}{32}\sin(\pi) = \frac{7\pi}{32} + \frac{1}{4}(1) - \frac{3}{32}(0) = \frac{7\pi}{32} + \frac{1}{4} \]

Evaluating at the lower limit \(x = 0\):
\[ \frac{7}{8}(0) + \frac{1}{4}\sin(0) - \frac{3}{32}\sin(0) = 0 \]

Subtracting the lower limit from the upper limit:
\[ \frac{7\pi}{32} + \frac{8}{32} = \frac{7\pi + 8}{32} \]

(a)
* B1: Expressing \(\sin^2(2x)\) as \(\frac{1 - \cos(4x)}{2}\)
* M1: Applying \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) to write \(\cos^4(x) = \frac{(1 + \cos(2x))^2}{4}\)
* M1: Expanding and using \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\)
* A1: Obtaining \(\cos^4(x) = \frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\)
* A1: Combining both expressions to reach \(\frac{7}{8} + \frac{1}{2}\cos(2x) - \frac{3}{8}\cos(4x)\) with complete and clear working shown

(b)
* M1: Attempting integration of each trigonometric term in the form \(p x + q\sin(2x) + r\sin(4x)\)
* A1: Correct integral \(\left[ \frac{7}{8}x + \frac{1}{4}\sin(2x) - \frac{3}{32}\sin(4x) \right]\)
* M1: Correct substitution of limits \(0\) and \(\frac{\pi}{4}\) with correct evaluation of \(\sin(\frac{\pi}{2}) = 1\) and \(\sin(\pi) = 0\)
* A1: Correct exact answer in the required form: \(\frac{7\pi + 8}{32}\)

(a)
Differentiating \( x \) with respect to \( \theta \):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta - 2\sin 2\theta \]
Using the double-angle identity \( \sin 2\theta = 2\sin\theta\cos\theta \):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta - 4\sin\theta\cos\theta = -2\sin\theta(1 + 2\cos\theta) \]

Differentiating \( y \) with respect to \( \theta \):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta - 2\cos 2\theta \]
Using the double-angle identity \( \cos 2\theta = 2\cos^2\theta - 1 \):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta - 2(2\cos^2\theta - 1) = 2 + 2\cos\theta - 4\cos^2\theta = 2(1 - \cos\theta)(1 + 2\cos\theta) \]

Using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{2(1 - \cos\theta)(1 + 2\cos\theta)}{-2\sin\theta(1 + 2\cos\theta)} = -\frac{1 - \cos\theta}{\sin\theta} \]

Applying half-angle identities \( 1 - \cos\theta = 2\sin^2\left(\frac{1}{2}\theta\right) \) and \( \sin\theta = 2\sin\left(\frac{1}{2}\theta\right)\cos\left(\frac{1}{2}\theta\right) \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2\sin^2\left(\frac{1}{2}\theta\right)}{2\sin\left(\frac{1}{2}\theta\right)\cos\left(\frac{1}{2}\theta\right)} = -\frac{\sin\left(\frac{1}{2}\theta\right)}{\cos\left(\frac{1}{2}\theta\right)} = -\tan\left(\frac{1}{2}\theta\right) \]

(b)
At \( \theta = \frac{1}{3}\pi \):
\[ x = 2\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) = 2\left(\frac{1}{2}\right) + \left(-\frac{1}{2}\right) = \frac{1}{2} \]
\[ y = 2\sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

The gradient of the tangent is:
\[ m = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \]

The equation of the tangent is:
\[ y - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{3}\left(x - \frac{1}{2}\right) \]
\[ y = -\frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{6} + \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{3}x + \frac{2\sqrt{3}}{3} \]

(a)
- B1: Correct derivative \( \frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta - 2\sin 2\theta \) and \( \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta - 2\cos 2\theta \).
- M1: Uses double-angle formula for \( \sin 2\theta \) and \( \cos 2\theta \) to factorise numerator and denominator.
- A1: Obtains \( \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1 - \cos\theta}{\sin\theta} \) or equivalent unsimplified single-fraction form.
- M1: Applies half-angle identities \( 1 - \cos\theta = 2\sin^2(\theta/2) \) and \( \sin\theta = 2\sin(\theta/2)\cos(\theta/2) \).
- A1 (CWO): Fully justifies simplification to reach \( -\tan\left(\frac{1}{2}\theta\right) \) with all steps clearly shown.

(b)
- B1: Finds correct coordinates \( x = \frac{1}{2} \) and \( y = \frac{\sqrt{3}}{2} \).
- B1: Obtains gradient \( m = -\frac{\sqrt{3}}{3} \) (or \( -\frac{1}{\sqrt{3}} \)).
- M1: Forms the linear equation of the tangent line using their point and gradient.
- A1: Correct final equation \( y = -\frac{\sqrt{3}}{3}x + \frac{2\sqrt{3}}{3} \) (or \( y = -\frac{1}{\sqrt{3}}x + \frac{2}{\sqrt{3}} \)).

Square both sides to eliminate the modulus signs:
\[ (3x - 5)^2 < 4(x + 1)^2 \]

Expand both expressions:
\[ 9x^2 - 30x + 25 < 4(x^2 + 2x + 1) \]
\[ 9x^2 - 30x + 25 < 4x^2 + 8x + 4 \]

Rearrange into a single quadratic inequality:
\[ 5x^2 - 38x + 21 < 0 \]

Factorise the quadratic:
\[ (5x - 3)(x - 7) < 0 \]

The critical values are \(x = \frac{3}{5}\) and \(x = 7\).

Since the quadratic expression is less than zero, the solution is:
\[ \frac{3}{5} < x < 7 \]

M1: State or imply squaring both sides \((3x - 5)^2 < 4(x + 1)^2\) or set up correct linear equations \(3x - 5 = \pm 2(x + 1)\)
A1: Obtain correct expanded quadratic \(5x^2 - 38x + 21 < 0\) (or correct linear equations)
M1: Attempt to solve the 3-term quadratic or linear equations to find critical values
A1: Obtain critical values \(x = \frac{3}{5}\) (or \(0.6\)) and \(x = 7\)
A1: State the correct inequality \(\frac{3}{5} < x < 7\) (or \(0.6 < x < 7\))

Take the natural logarithm of both sides of \(y = A \mathrm{e}^{k x^2}\):
\[ \ln y = \ln A + \ln\left(\mathrm{e}^{k x^2}\right) \]
\[ \ln y = \ln A + kx^2 \]

Let \(Y = \ln y\) and \(X = x^2\). The relationship is linear with gradient \(m = k\) and vertical intercept \(c = \ln A\).

Calculate the gradient \(k\) using the given coordinates \((2, 1.34)\) and \((6, 2.58)\):
\[ k = \frac{2.58 - 1.34}{6 - 2} = \frac{1.24}{4} = 0.31 \]

Substitute \(k = 0.31\) and the point \((2, 1.34)\) to find \(\ln A\):
\[ 1.34 = \ln A + 0.31(2) \]
\[ \ln A = 1.34 - 0.62 = 0.72 \]

Find \(A\):
\[ A = \mathrm{e}^{0.72} \approx 2.0544... \approx 2.05 \]

B1: State or imply \(\ln y = \ln A + kx^2\)
M1: Use given coordinates to find gradient \(k = \frac{2.58 - 1.34}{6 - 2}\)
A1: Obtain \(k = 0.31\)
M1: Substitute \(k\) and a point to find \(\ln A\) and calculate \(A = \mathrm{e}^{\ln A}\)
A1: Obtain \(A = 2.05\) (accept 2.054)

(a) Since all coefficients of the polynomial are real, complex roots occur in conjugate pairs. Therefore, the other complex root is \(2 + \mathrm{i}\sqrt{3}\).

(b) The quadratic factor corresponding to the two complex conjugate roots is:
\[ (z - (2 - \mathrm{i}\sqrt{3}))(z - (2 + \mathrm{i}\sqrt{3})) = (z - 2)^2 - (\mathrm{i}\sqrt{3})^2 = z^2 - 4z + 4 + 3 = z^2 - 4z + 7 \]

Let the cubic polynomial be factorised as \((2z - c)(z^2 - 4z + 7)\).
Expanding this gives:
\[ (2z - c)(z^2 - 4z + 7) = 2z^3 - (8 + c)z^2 + (14 + 4c)z - 7c \]

Equating coefficients with \(2z^3 - 7z^2 + pz + q\):
For \(z^2\):
\[ -(8 + c) = -7 \implies 8 + c = 7 \implies c = -1 \]

So the linear factor is \((2z + 1)\).

Now find \(p\) and \(q\):
\[ p = 14 + 4(-1) = 10 \]
\[ q = -7(-1) = 7 \]

(a) B1: State \(2 + \mathrm{i}\sqrt{3}\)
(b) M1: Form the quadratic factor \((z - (2 - \mathrm{i}\sqrt{3}))(z - (2 + \mathrm{i}\sqrt{3})) = z^2 - 4z + 7\) or find third root \(z = -\frac{1}{2}\) using sum of roots
M1: Multiply quadratic factor by \((2z + 1)\) or compare coefficients using root relationships
A1: Obtain \(p = 10\)
A1: Obtain \(q = 7\)

Let the square root be \(x + \mathrm{i}y\), where \(x, y \in \mathbb{R}\).

Then:
\[ (x + \mathrm{i}y)^2 = -15 + 8\mathrm{i} \]
\[ x^2 - y^2 + 2xy\mathrm{i} = -15 + 8\mathrm{i} \]

Equating real and imaginary parts:
\[ x^2 - y^2 = -15 \quad \text{--- (1)} \]
\[ 2xy = 8 \implies xy = 4 \implies y = \frac{4}{x} \quad \text{--- (2)} \]

Also, equating moduli:
\[ x^2 + y^2 = \sqrt{(-15)^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \quad \text{--- (3)} \]

Adding equation (1) and equation (3):
\[ 2x^2 = -15 + 17 = 2 \implies x^2 = 1 \implies x = \pm 1 \]

Using equation (2) to find corresponding values of \(y\):
When \(x = 1\), \(y = \frac{4}{1} = 4\).
When \(x = -1\), \(y = \frac{4}{-1} = -4\).

Therefore, the two square roots are \(1 + 4\mathrm{i}\) and \(-1 - 4\mathrm{i}\).

M1: Set \((x + \mathrm{i}y)^2 = -15 + 8\mathrm{i}\) and expand
A1: Equate real and imaginary parts to obtain \(x^2 - y^2 = -15\) and \(2xy = 8\)
M1: Eliminate one variable to obtain a quadratic equation in \(x^2\) or \(y^2\), or use modulus \(x^2 + y^2 = 17\)
A1: Find correct values \(x = \pm 1\) and \(y = \pm 4\)
A1: State both roots correctly as \(1 + 4\mathrm{i}\) and \(-1 - 4\mathrm{i}\) (or \(\pm(1 + 4\mathrm{i})\))

(a) Using the identity \(\cot 2\theta = \frac{1}{\tan 2\theta} = \frac{1 - \tan^2\theta}{2\tan\theta}\):

\(\frac{1 - \tan^2\theta}{2\tan\theta} + 2\tan\theta = 3\)

Multiply through by \(2\tan\theta\):

\(1 - \tan^2\theta + 4\tan^2\theta = 6\tan\theta\)

\(3\tan^2\theta - 6\tan\theta + 1 = 0\)

(b) Solving the quadratic equation in \(\tan\theta\):

\(\tan\theta = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(1)}}{2(3)} = \frac{6 \pm \sqrt{24}}{6} = 1 \pm \frac{\sqrt{6}}{3}\)

This gives two values for \(\tan\theta\):
1. \(\tan\theta = 1 + \frac{\sqrt{6}}{3} \approx 1.8165 \implies \theta = \tan^{-1}(1.8165) \approx 61.17^\circ \approx 61.2^\circ\)
2. \(\tan\theta = 1 - \frac{\sqrt{6}}{3} \approx 0.1835 \implies \theta = \tan^{-1}(0.1835) \approx 10.40^\circ \approx 10.4^\circ\)

In the interval \(0^\circ < \theta < 180^\circ\), there are no other solutions since \(\tan\theta > 0\) occurs only in the first quadrant for this range.

Therefore, \(\theta = 10.4^\circ\) or \(\theta = 61.2^\circ\).

(a)
- M1: Uses \(\cot 2\theta = \frac{1}{\tan 2\theta}\) and applies the double angle formula \(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}\)
- M1: Clears fractions correctly to form a single quadratic equation in \(\tan\theta\)
- A1: Obtains \(3\tan^2\theta - 6\tan\theta + 1 = 0\) with full and convincing working shown (CWO)

(b)
- M1: Solves the quadratic equation to obtain at least one value for \(\tan\theta\)
- A1: Obtains \(\tan\theta = 1 \pm \frac{\sqrt{6}}{3}\) or numerical equivalents \(\approx 1.8165, 0.1835\)
- A1: Obtains one correct angle (e.g. \(10.4^\circ\) or \(61.2^\circ\))
- A1: Obtains second correct angle and no extra solutions in the range (must be to 1 d.p.)

(a) The inequality \(|z - (3 + 2\mathrm{i})| \le 2\) represents the circular disc of radius 2 centered at \((3, 2)\).

The inequality \(0 \le \arg(z - 1) \le \frac{\pi}{4}\) represents the region bounded by two half-lines starting from \((1, 0)\):
1. A horizontal ray extending to the right along the line \(y = 0\) (for \(x \ge 1\)).
2. A half-line making an angle of \(\frac{\pi}{4}\) with the positive real axis, which has Cartesian equation \(y = x - 1\) (for \(x \ge 1\)).

Notice that the center of the circle \((3, 2)\) lies on the line \(y = x - 1\) since \(2 = 3 - 1\).

The shaded region is the intersection of the circular disc and the sector formed between the two rays.

(b) The expression \(|z - 1|\) represents the distance from the point \(A(1, 0)\) to the point \(z\).

The maximum distance from \(A(1, 0)\) to any point on the circular disc occurs at the point furthest along the line passing through \(A(1, 0)\) and the center of the circle \(C(3, 2)\).

The line passing through \(A\) and \(C\) is precisely the line with \(\arg(z - 1) = \frac{\pi}{4}\), which is within the allowed region.

The distance from \(A(1, 0)\) to the center \(C(3, 2)\) is:
\[ AC = \sqrt{(3 - 1)^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

The maximum distance to the boundary of the circle is:
\[ AC + R = 2\sqrt{2} + 2 \]

Therefore, the exact maximum value of \(|z - 1|\) is \(2\sqrt{2} + 2\).

(a)
- B1: Draws a circle with center \((3, 2)\) and radius 2
- B1: Draws a half-line starting from \((1, 0)\) in the positive real direction
- B1: Draws a half-line starting from \((1, 0)\) at angle \(\frac{\pi}{4}\) passing through the center of the circle \((3, 2)\)
- B1: Correctly identifies and shades the intersection region satisfying both inequalities

(b)
- M1: Identifies that the maximum occurs along the line connecting \((1, 0)\) and the center \((3, 2)\)
- M1: Calculates the distance from \((1, 0)\) to \((3, 2)\) as \(\sqrt{(3-1)^2 + 2^2} = 2\sqrt{2}\)
- A1: Obtains the exact maximum value \(2\sqrt{2} + 2\) (or \(\sqrt{8} + 2\))

(a) Equating the components of \(l_1\) and \(l_2\):

\(1 + 2s = 4 + t \implies 2s - t = 3\) --- (1)

\(4 - s = 5 + 2t \implies s + 2t = -1\) --- (2)

\(-2 + 3s = -1 - 2t \implies 3s + 2t = 1\) --- (3)

From (1), \(t = 2s - 3\). Substituting into (2):

\(s + 2(2s - 3) = -1 \implies 5s - 6 = -1 \implies 5s = 5 \implies s = 1\)

Then \(t = 2(1) - 3 = -1\).

Check consistency in equation (3):

\(3(1) + 2(-1) = 3 - 2 = 1\), which matches the right-hand side.

Since all three equations are satisfied, the lines intersect.

Substituting \(s = 1\) into \(l_1\):

\(\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} + 1\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix}\).

(b) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\).

The scalar product is:

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(2) + (3)(-2) = 2 - 2 - 6 = -6\).

The magnitudes are:

\(|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}\)

\(|\mathbf{d}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\)

Let \(\theta\) be the acute angle between the two lines:

\(\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-6|}{3\sqrt{14}} = \frac{6}{3\sqrt{14}} = \frac{2}{\sqrt{14}}\)

\(\theta = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \approx 57.688^\circ \approx 57.7^\circ\).

(a)
- M1: Equates components of \(l_1\) and \(l_2\) to form a system of simultaneous equations in \(s\) and \(t\)
- A1: Solves a pair of equations to obtain \(s = 1\) and \(t = -1\)
- A1: Demonstrates consistency by substituting into the third equation (CWO)
- A1: States the correct position vector \(\begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix}\) or \(3\mathbf{i} + 3\mathbf{j} + \mathbf{k}\)

(b)
- M1: Uses the scalar product formula \(\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}\) with direction vectors
- M1: Evaluates scalar product and moduli correctly: \(\mathbf{d}_1 \cdot \mathbf{d}_2 = -6\), \(|\mathbf{d}_1| = \sqrt{14}\), \(|\mathbf{d}_2| = 3\)
- A1: Obtains \(57.7^\circ\) (must be acute and rounded to 1 d.p.)

(a) Expanding the trigonometric compound angle terms:
\(\cos(x + 60^\circ) = \cos x \cos 60^\circ - \sin x \sin 60^\circ = \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x\)
\(\sin(x - 30^\circ) = \sin x \cos 30^\circ - \cos x \sin 30^\circ = \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x\)

Substitute these expansions into the given equation:
\(\left(\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x\right) + 2\left(\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x\right) = 3\cos x\)
\(\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x + \sqrt{3}\sin x - \cos x = 3\cos x\)
\(-\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = 3\cos x\)
\(\frac{\sqrt{3}}{2}\sin x = \frac{7}{2}\cos x\)

Dividing both sides by \(\cos x\) (where \(\cos x \neq 0\)) and by \(\frac{\sqrt{3}}{2}\):
\(\tan x = \frac{7/2}{\sqrt{3}/2} = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}\).

(b) Using the result from part (a) with \(x = 2\theta\):
\(\tan 2\theta = \frac{7\sqrt{3}}{3}\)
For \(0^\circ \le \theta \le 180^\circ\), the interval for \(2\theta\) is \(0^\circ \le 2\theta \le 360^\circ\).
\(2\theta = \tan^{-1}\left(\frac{7}{\sqrt{3}}\right) \approx 76.102^\circ\)
or \(2\theta = 76.102^\circ + 180^\circ = 256.102^\circ\).

Dividing by 2 gives:
\(\theta = 38.051^\circ \approx 38.1^\circ\)
\(\theta = 128.051^\circ \approx 128.1^\circ\).

(a)
M1: Apply correct compound angle expansions to both \(\cos(x + 60^\circ)\) and \(\sin(x - 30^\circ)\).
B1: State correct exact trigonometric values \(\cos 60^\circ = \sin 30^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\).
M1: Collect like terms in \(\sin x\) and \(\cos x\) and divide to obtain an equation in \(\tan x\).
A1: Obtain \(\tan x = \frac{7\sqrt{3}}{3}\) or \(\frac{7}{\sqrt{3}}\) (exact form required).

(b)
M1: Set \(2\theta = \tan^{-1}(k)\) and find at least one value for \(2\theta\) in the correct range.
A1: Obtain \(\theta = 38.1^\circ\).
A1: Obtain \(\theta = 128.1^\circ\) and no extra solutions within the range.

(a) Substituting \(t = 0\), \(m = 0\) and \(\frac{\mathrm{d}m}{\mathrm{d}t} = 4\) into the differential equation:
\[ 4 = k(100 - 0)(20 + 0) = 2000k \implies k = \frac{4}{2000} = \frac{1}{500} = 0.002. \]

(b) Separating the variables:
\[ \int \frac{1}{(100 - m)(20 + m)} \, \mathrm{d}m = \int k \, \mathrm{d}t. \]
Expressing in partial fractions:
\[ \frac{1}{(100 - m)(20 + m)} = \frac{A}{100 - m} + \frac{B}{20 + m} \implies 1 = A(20 + m) + B(100 - m). \]
Setting \(m = 100 \implies 1 = 120A \implies A = \frac{1}{120}\).
Setting \(m = -20 \implies 1 = 120B \implies B = \frac{1}{120}\).
So:
\[ \frac{1}{120} \int \left( \frac{1}{100 - m} + \frac{1}{20 + m} \right) \mathrm{d}m = \int \frac{1}{500} \, \mathrm{d}t \]
\[ \frac{1}{120} \left[ -\ln(100 - m) + \ln(20 + m) \right] = \frac{1}{500}t + C \]
\[ \ln\left( \frac{20 + m}{100 - m} \right) = \frac{120}{500}t + 120C = 0.24t + c_1. \]
Using the initial condition \(t = 0\), \(m = 0\):
\[ \ln\left( \frac{20}{100} \right) = c_1 \implies c_1 = \ln\left(\frac{1}{5}\right) = -\ln 5. \]
Thus:
\[ \ln\left( \frac{20 + m}{100 - m} \right) = 0.24t - \ln 5 \implies \frac{20 + m}{100 - m} = \frac{1}{5}\mathrm{e}^{0.24t}. \]
Rearranging for \(m\):
\[ 5(20 + m) = (100 - m)\mathrm{e}^{0.24t} \implies 100 + 5m = 100\mathrm{e}^{0.24t} - m\mathrm{e}^{0.24t} \]
\[ m\left(5 + \mathrm{e}^{0.24t}\right) = 100\left(\mathrm{e}^{0.24t} - 1\right) \implies m = \frac{100\left(\mathrm{e}^{0.24t} - 1\right)}{\mathrm{e}^{0.24t} + 5} = \frac{100\left(1 - \mathrm{e}^{-0.24t}\right)}{1 + 5\mathrm{e}^{-0.24t}}. \]

(c) As \(t \to \infty\), \(\mathrm{e}^{-0.24t} \to 0\), so \(m \to \frac{100(1 - 0)}{1 + 0} = 100\) grams.

(a)
M1: Substitute \(m=0\) and \(\frac{\mathrm{d}m}{\mathrm{d}t}=4\) into the given equation.
A1: Obtain \(k = 0.002\) or \(k = \frac{1}{500}\).

(b)
B1: State correct partial fractions \(\frac{1/120}{100-m} + \frac{1/120}{20+m}\).
M1: Separate variables and attempt integration of both sides, producing \(-\ln(100-m)\) and \(\ln(20+m)\).
A1: Correct integrated form with constant of integration \(\frac{1}{120}\ln\left(\frac{20+m}{100-m}\right) = \frac{1}{500}t + c\).
M1: Substitute \(t = 0, m = 0\) to evaluate the constant of integration.
M1: Remove logarithms correctly to obtain an equation in \(m\) and \(\mathrm{e}^{0.24t}\).
A1: Obtain \(m = \frac{100(\mathrm{e}^{0.24t}-1)}{\mathrm{e}^{0.24t}+5}\) or \(m = \frac{100(1-\mathrm{e}^{-0.24t})}{1+5\mathrm{e}^{-0.24t}}\) or equivalent.

(c)
B1: State that \(m\) approaches 100 (or mass tends to 100 g).

(a) Using integration by parts:
Let \(u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1\),
and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \sin\left(\frac{1}{2}x\right) \implies v = -2\cos\left(\frac{1}{2}x\right)\).

Then:
\[ \int x \sin\left(\frac{1}{2}x\right) \mathrm{d}x = -2x \cos\left(\frac{1}{2}x\right) - \int \left( -2\cos\left(\frac{1}{2}x\right) \right) \mathrm{d}x \]
\[ = -2x \cos\left(\frac{1}{2}x\right) + 2 \int \cos\left(\frac{1}{2}x\right) \mathrm{d}x \]
\[ = -2x \cos\left(\frac{1}{2}x\right) + 4 \sin\left(\frac{1}{2}x\right) + c. \]

(b) To evaluate \(\int_0^{\pi} x^2 \cos\left(\frac{1}{2}x\right) \mathrm{d}x\), apply integration by parts:
Let \(u = x^2 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2x\),
and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \cos\left(\frac{1}{2}x\right) \implies v = 2\sin\left(\frac{1}{2}x\right)\).

Then:
\[ \int_0^{\pi} x^2 \cos\left(\frac{1}{2}x\right) \mathrm{d}x = \left[ 2x^2 \sin\left(\frac{1}{2}x\right) \right]_0^{\pi} - \int_0^{\pi} 4x \sin\left(\frac{1}{2}x\right) \mathrm{d}x. \]

Evaluating the bracketed term:
\[ \left[ 2x^2 \sin\left(\frac{1}{2}x\right) \right]_0^{\pi} = \left( 2\pi^2 \sin\left(\frac{\pi}{2}\right) \right) - (0) = 2\pi^2(1) = 2\pi^2. \]

Using the result from part (a) for the integral term:
\[ \int_0^{\pi} 4x \sin\left(\frac{1}{2}x\right) \mathrm{d}x = 4 \left[ -2x \cos\left(\frac{1}{2}x\right) + 4 \sin\left(\frac{1}{2}x\right) \right]_0^{\pi} \]
\[ = 4 \left[ \left( -2\pi \cos\left(\frac{\pi}{2}\right) + 4 \sin\left(\frac{\pi}{2}\right) \right) - \left( -2(0)\cos(0) + 4\sin(0) \right) \right] \]
\[ = 4 \left[ \left( -2\pi(0) + 4(1) \right) - (0) \right] = 4(4) = 16. \]

Combining the results:
\[ \int_0^{\pi} x^2 \cos\left(\frac{1}{2}x\right) \mathrm{d}x = 2\pi^2 - 16. \]

(a)
M1: Apply integration by parts to \(\int x \sin\left(\frac{1}{2}x\right) \mathrm{d}x\), choosing \(u=x\) and \(v'=\sin\left(\frac{1}{2}x\right)\) to reach the form \(-2x\cos\left(\frac{1}{2}x\right) - \int -2\cos\left(\frac{1}{2}x\right) \mathrm{d}x\).
A1: Obtain \(-2x\cos\left(\frac{1}{2}x\right) + 2\int\cos\left(\frac{1}{2}x\right)\mathrm{d}x\).
A1: Obtain final expression \(-2x\cos\left(\frac{1}{2}x\right) + 4\sin\left(\frac{1}{2}x\right) + c\).

(b)
M1: Apply integration by parts to \(\int x^2 \cos\left(\frac{1}{2}x\right) \mathrm{d}x\), obtaining \(2x^2\sin\left(\frac{1}{2}x\right) - \int 4x\sin\left(\frac{1}{2}x\right)\mathrm{d}x\).
A1: Correct first term \(2x^2\sin\left(\frac{1}{2}x\right)\) and correct integral term.
M1: Correctly substitute limits \(0\) and \(\pi\) into \(\left[2x^2\sin\left(\frac{1}{2}x\right)\right]_0^\pi\) to obtain \(2\pi^2\).
M1: Substitute limits \(0\) and \(\pi\) into the result from part (a) multiplied by 4.
A1: Correct evaluation of the integral term as 16.
A1: Obtain exact final answer \(2\pi^2 - 16\).

(a) Separating the variables:
\[ \int \frac{1}{4 - m}\,\mathrm{d}m = \int -k\mathrm{e}^{-0.1t}\,\mathrm{d}t \]
\[ -\ln(4 - m) = 10k\mathrm{e}^{-0.1t} + c \]

Using the initial condition \(t = 0, m = 1\):
\[ -\ln(3) = 10k + c \implies c = -10k - \ln 3 \]

Substituting \(c\) back into the equation:
\[ -\ln(4 - m) = 10k\mathrm{e}^{-0.1t} - 10k - \ln 3 \]
\[ \ln\left(\frac{4 - m}{3}\right) = -10k(1 - \mathrm{e}^{-0.1t}) \]
\[ 4 - m = 3\mathrm{e}^{-10k(1 - \mathrm{e}^{-0.1t})} \]
\[ m = 4 - 3\mathrm{e}^{-10k(1 - \mathrm{e}^{-0.1t})} \]

Using the condition \(t = 10, m = 2\):
\[ 2 = 4 - 3\mathrm{e}^{-10k(1 - \mathrm{e}^{-1})} \]
\[ 3\mathrm{e}^{-10k(1 - \mathrm{e}^{-1})} = 2 \implies \mathrm{e}^{-10k(1 - \mathrm{e}^{-1})} = \frac{2}{3} \]
\[ -10k(1 - \mathrm{e}^{-1}) = \ln\left(\frac{2}{3}\right) = -\ln(1.5) \]
\[ 10k = \frac{\ln(1.5)}{1 - \mathrm{e}^{-1}} \approx 0.6414 \]

Hence,
\[ m = 4 - 3\exp\left(-\frac{\ln(1.5)(1 - \mathrm{e}^{-0.1t})}{1 - \mathrm{e}^{-1}}\right) \]

(b) As \(t \to \infty\), \(\mathrm{e}^{-0.1t} \to 0\):
\[ m \to 4 - 3\exp(-10k) = 4 - 3\exp\left(-\frac{\ln 1.5}{1 - \mathrm{e}^{-1}}\right) \]
\[ m \to 4 - 3\left(\frac{2}{3}\right)^{\frac{1}{1 - \mathrm{e}^{-1}}} \approx 4 - 3(0.5265) = 2.4205 \approx 2.42 \]

(a)
- M1: Separate variables and attempt integration of at least one side.
- A1: Obtain \(-\ln(4 - m)\) and \(10k\mathrm{e}^{-0.1t}\) (or correct signs).
- M1: Evaluate constant of integration using \(t = 0, m = 1\).
- M1: Use \(t = 10, m = 2\) to form an equation in \(k\).
- A1: Obtain correct exact or 3 s.f. value for \(k\) or \(10k\) (e.g. \(10k = \frac{\ln 1.5}{1 - \mathrm{e}^{-1}} \approx 0.641\) or \(k \approx 0.0641\)).
- M1: Remove logarithms to make \(m\) the subject.
- A1: Obtain a correct expression for \(m\) in terms of \(t\).

(b)
- M1: Let \(t \to \infty\) (or \(\mathrm{e}^{-0.1t} \to 0\)) in their expression for \(m\).
- A1: Obtain \(2.42\) (accept \(2.420\) to \(2.421\)).

(a)
Resolving forces parallel to the \(x\)-axis in the positive direction:
\[ 15 - X\cos 30^\circ - Y\cos 60^\circ = 0 \]
\[ 15 - X\left(\frac{\sqrt{3}}{2}\right) - Y\left(\frac{1}{2}\right) = 0 \implies \frac{\sqrt{3}}{2}X + \frac{1}{2}Y = 15 \quad \text{--- (1)} \]

Resolving forces parallel to the \(y\)-axis in the positive direction:
\[ 20 + X\sin 30^\circ - Y\sin 60^\circ = 0 \]
\[ 20 + X\left(\frac{1}{2}\right) - Y\left(\frac{\sqrt{3}}{2}\right) = 0 \implies \frac{1}{2}X - \frac{\sqrt{3}}{2}Y = -20 \quad \text{--- (2)} \]

(b)
Multiplying equation (1) by 2 gives:
\[ \sqrt{3}X + Y = 30 \implies Y = 30 - \sqrt{3}X \]
Multiplying equation (2) by 2 gives:
\[ X - \sqrt{3}Y = -40 \]
Substitute \(Y = 30 - \sqrt{3}X\) into this equation:
\[ X - \sqrt{3}(30 - \sqrt{3}X) = -40 \]
\[ X - 30\sqrt{3} + 3X = -40 \]
\[ 4X = 30\sqrt{3} - 40 \implies X = \frac{15\sqrt{3}}{2} - 10 \quad (\text{or } 7.5\sqrt{3} - 10) \]
Now substitute \(X\) back to find \(Y\):
\[ Y = 30 - \sqrt{3}\left(\frac{15\sqrt{3}}{2} - 10\right) = 30 - \frac{45}{2} + 10\sqrt{3} = 10\sqrt{3} + \frac{15}{2} \quad (\text{or } 10\sqrt{3} + 7.5) \]

(a)
* M1: Resolving forces in the \(x\)-direction with correct number of terms and trig functions.
* A1: Correct equation in \(x\)-direction, e.g. \(15 - X\cos 30^\circ - Y\cos 60^\circ = 0\) or \(\sqrt{3}X + Y = 30\).
* A1: Resolving in \(y\)-direction to obtain a correct equation, e.g. \(20 + X\sin 30^\circ - Y\sin 60^\circ = 0\) or \(X - \sqrt{3}Y = -40\).

(b)
* M1: Valid method to eliminate one variable from the simultaneous equations.
* A1: Correct exact value for \(X = \frac{15}{2}\sqrt{3} - 10\) (or \(7.5\sqrt{3} - 10\)).
* A1: Correct exact value for \(Y = 10\sqrt{3} + \frac{15}{2}\) (or \(10\sqrt{3} + 7.5\)).

(a)
Using \(P = F_D v\), the driving force \(F_D\) exerted by the engine is:
\[ F_D = \frac{P}{v} = \frac{36000}{20} = 1800\text{ N} \]
The component of the weight acting down the plane is:
\[ mg\sin\alpha = 1200 \times 10 \times \frac{1}{15} = 800\text{ N} \]
Since the car is moving up the hill at constant speed, the acceleration is \(0\). Resolving forces parallel to the inclined plane:
\[ F_D - mg\sin\alpha - R = 0 \]
\[ 1800 - 800 - R = 0 \implies R = 1000\text{ N} \]

(b)
When travelling down the hill with power \(P = 24000\text{ W}\) at speed \(v = 25\text{ ms}^{-1}\), the driving force is:
\[ F_D = \frac{P}{v} = \frac{24000}{25} = 960\text{ N} \]
The driving force and the component of weight act down the slope, while resistance \(R\) acts up the slope. Applying Newton's second law down the slope:
\[ F_D + mg\sin\alpha - R = ma \]
\[ 960 + 800 - 1000 = 1200a \]
\[ 760 = 1200a \]
\[ a = \frac{760}{1200} = \frac{19}{30} \approx 0.633\text{ ms}^{-2} \]

(a)
* B1: Calculates driving force \(F_D = \frac{36000}{20} = 1800\text{ N}\).
* M1: Sets up equilibrium equation along the slope: \(F_D - mg\sin\alpha - R = 0\).
* A1: Obtains \(R = 1000\text{ N}\).

(b)
* B1: Calculates new driving force \(F_D = \frac{24000}{25} = 960\text{ N}\).
* M1: Applies Newton's second law down the slope: \(F_D + mg\sin\alpha - R = ma\) with their \(R\).
* A1: Obtains \(a = \frac{19}{30}\text{ ms}^{-2}\) or \(0.633\text{ ms}^{-2}\) (to 3 s.f.).

(a) To find the maximum velocity, differentiate \(v\) with respect to \(t\):
\[ a = \frac{\mathrm{d}v}{\mathrm{d}t} = 3.6t - 0.6t^2 \]
Set \(\frac{\mathrm{d}v}{\mathrm{d}t} = 0\):
\[ 0.6t(6 - t) = 0 \]
For \(t > 0\), this gives \(t = 6\).
Substituting \(t = 6\) into \(v\):
\[ v_{\max} = 1.8(6)^2 - 0.2(6)^3 = 1.8(36) - 0.2(216) = 64.8 - 43.2 = 21.6\text{ m s}^{-1} \]

(b) The displacement \(s\) at \(t = 7\) is found by integrating \(v\) with respect to \(t\):
\[ s(7) = \int_0^7 (1.8t^2 - 0.2t^3)\,\mathrm{d}t = \left[ 0.6t^3 - 0.05t^4 \right]_0^7 \]
\[ s(7) = 0.6(7)^3 - 0.05(7)^4 = 0.6(343) - 0.05(2401) = 205.8 - 120.05 = 85.75\text{ m} \]

(c) The velocity at \(t = 7\) is:
\[ v(7) = 1.8(7)^2 - 0.2(7)^3 = 88.2 - 68.6 = 19.6\text{ m s}^{-1} \]
For \(t > 7\), using \(v = u + at\) with \(u = 19.6\), \(a = -2.8\), and final velocity \(v = 0\):
\[ 0 = 19.6 - 2.8(T - 7) \]
\[ 2.8(T - 7) = 19.6 \implies T - 7 = 7 \implies T = 14 \]

(d) During the second phase of motion (from \(t = 7\) to \(t = 14\)), the distance travelled \(s_2\) under constant deceleration is:
\[ s_2 = \frac{u + v}{2} \times (T - 7) = \frac{19.6 + 0}{2} \times 7 = 9.8 \times 7 = 68.6\text{ m} \]
Total distance travelled from \(t = 0\) to \(t = 14\):
\[ \text{Total distance} = 85.75 + 68.6 = 154.35\text{ m} \quad (\text{or } 154\text{ m to 3 s.f.}) \]

(a)
- M1: Differentiates \(v\) with respect to \(t\) and sets equal to 0.
- A1: Finds \(t = 6\).
- A1: Obtains \(v_{\max} = 21.6\text{ m s}^{-1}\).

(b)
- M1: Integrates \(v(t)\) with respect to \(t\) with limits \(0\) to \(7\).
- A1: Obtains \(85.75\text{ m}\) (or \(\frac{343}{4}\)).

(c)
- M1: Evaluates \(v(7) = 19.6\) and applies constant acceleration formula \(0 = 19.6 - 2.8(T - 7)\) (or equivalent).
- A1: Obtains \(T = 14\).

(d)
- M1: Uses appropriate method (e.g. \(v^2 = u^2 + 2as\) or \(s = \frac{u+v}{2}t\)) to find distance in the second phase.
- A1: Obtains total distance \(154.35\text{ m}\) (or \(154\text{ m}\)).

(a) For the inclined plane with \(\tan \alpha = \frac{3}{4}\), we have \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\).
For particle \(A\) (mass \(0.5\text{ kg}\)):
Resolving perpendicular to the plane:
\[ R = 0.5g \cos \alpha = 0.5(10)(0.8) = 4\text{ N} \]
Frictional force on \(A\):
\[ F = \mu R = 0.25 \times 4 = 1.0\text{ N} \]
Component of weight down the plane:
\[ 0.5g \sin \alpha = 0.5(10)(0.6) = 3.0\text{ N} \]
Equation of motion for \(A\) moving up the plane:
\[ T - 3.0 - 1.0 = 0.5a \implies T - 4.0 = 0.5a \quad \text{--- (1)} \]
For particle \(B\) (mass \(0.7\text{ kg}\)) moving downwards:
\[ 0.7g - T = 0.7a \implies 7.0 - T = 0.7a \quad \text{--- (2)} \]
Adding equations (1) and (2):
\[ 7.0 - 4.0 = (0.5 + 0.7)a \implies 3.0 = 1.2a \implies a = 2.5\text{ m s}^{-2} \]
Substituting \(a = 2.5\) into equation (1):
\[ T = 4.0 + 0.5(2.5) = 4.0 + 1.25 = 5.25\text{ N} \]

(b) Particle \(B\) accelerates from rest over a distance \(s = 0.8\text{ m}\) with \(a = 2.5\text{ m s}^{-2}\):
\[ v^2 = u^2 + 2as = 0 + 2(2.5)(0.8) = 4 \]
\[ v = \sqrt{4} = 2.0\text{ m s}^{-1} \]

(c) When \(B\) hits the floor, \(A\) is moving up the plane with speed \(u = 2.0\text{ m s}^{-1}\).
The string is slack, so \(T = 0\). The forces acting on \(A\) parallel to the plane (both opposing motion) are the weight component and friction:
\[ - (0.5g \sin \alpha + F) = 0.5 a_2 \implies -(3.0 + 1.0) = 0.5 a_2 \implies a_2 = -8\text{ m s}^{-2} \]
Let \(s_2\) be the additional distance travelled by \(A\) up the plane until \(v = 0\):
\[ 0^2 = (2.0)^2 + 2(-8)s_2 \implies 16 s_2 = 4 \implies s_2 = 0.25\text{ m} \]
Total distance travelled by \(A\) from release:
\[ \text{Total distance} = 0.8 + 0.25 = 1.05\text{ m} \]

(a)
- B1: Correctly calculates normal contact force \(R = 4\text{ N}\) and friction \(F = 1\text{ N}\).
- M1: Sets up two equations of motion using Newton's second law for \(A\) and \(B\).
- A1: Adds/solves equations correctly to show \(a = 2.5\text{ m s}^{-2}\).
- A1: Obtains \(T = 5.25\text{ N}\).

(b)
- M1: Uses \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 2.5\), \(s = 0.8\).
- A1: Obtains \(v = 2\text{ m s}^{-1}\).

(c)
- M1: Applies Newton's second law to \(A\) with string slack to find new deceleration \(a = -8\text{ m s}^{-2}\).
- M1: Uses constant acceleration formula to find additional distance \(s_2 = 0.25\text{ m}\).
- A1: Obtains total distance \(1.05\text{ m}\).

Given \(\tan \alpha = \frac{5}{12}\), we have \(\sin \alpha = \frac{5}{13}\) and \(\cos \alpha = \frac{12}{13}\).

The mass is \(m = 2.5\text{ kg}\), so the weight is \(W = mg = 2.5 \times 10 = 25\text{ N}\).

Resolving perpendicular to the plane:
\[R = mg \cos \alpha = 25 \times \frac{12}{13} = \frac{300}{13}\text{ N}\]

The maximum frictional force is:
\[F_{\max} = \mu R = 0.2 \times \frac{300}{13} = \frac{60}{13}\text{ N}\]

The component of weight acting down the plane is:
\[mg \sin \alpha = 25 \times \frac{5}{13} = \frac{125}{13}\text{ N}\]

(a)(i)
When the particle is on the point of slipping down the plane, friction \(F\) acts up the plane with limiting value \(F = \frac{60}{13}\text{ N}\).

Resolving parallel to the plane in equilibrium:
\[T + F = mg \sin \alpha\]
\[T + \frac{60}{13} = \frac{125}{13}\]
\[T = \frac{125 - 60}{13} = \frac{65}{13} = 5\text{ N}\]

(a)(ii)
When the particle accelerates up the plane, friction \(F\) acts down the plane with \(F = \frac{60}{13}\text{ N}\).

Applying Newton's second law up the plane:
\[T - mg \sin \alpha - F = ma\]
\[T - \frac{125}{13} - \frac{60}{13} = 2.5 \times 1.5\]
\[T - \frac{185}{13} = 3.75\]
\[T = \frac{185}{13} + \frac{15}{4} = \frac{740 + 195}{52} = \frac{935}{52} \approx 18.0\text{ N}\]

(b)
When the pulling force \(T\) is removed, the particle continues moving up the plane until it comes to rest.
During this motion, both the component of weight and the frictional force act down the plane.

Total retarding force down the plane:
\[F_{\text{net}} = mg \sin \alpha + F = \frac{125}{13} + \frac{60}{13} = \frac{185}{13}\text{ N}\]

Using \(F_{\text{net}} = ma_{\text{retard}}\):
\[a_{\text{retard}} = \frac{\frac{185}{13}}{2.5} = \frac{74}{13}\text{ m s}^{-2} \approx 5.692\text{ m s}^{-2}\]

Using the constant acceleration formula \(v^2 = u^2 - 2as\) with \(v = 0\), \(u = 6\text{ m s}^{-1}\), and \(a = \frac{74}{13}\text{ m s}^{-2}\):
\[0 = 6^2 - 2 \left(\frac{74}{13}\right) s\]
\[36 = \frac{148}{13} s\]
\[s = \frac{36 \times 13}{148} = \frac{9 \times 13}{37} = \frac{117}{37} \approx 3.16\text{ m}\]

(a)(i)
- B1: For correct normal reaction \(R = \frac{300}{13}\text{ N}\) (or \(23.1\)) and limiting friction \(F = \frac{60}{13}\text{ N}\) (or \(4.62\)).
- M1: For resolving parallel to the plane with friction directed up the plane: \(T + F - mg \sin \alpha = 0\).
- A1: For \(T = 5\text{ N}\).

(a)(ii)
- M1: For applying Newton's second law parallel to the plane with 4 terms and friction acting down the plane: \(T - mg \sin \alpha - F = ma\).
- A1: For a correct unsimplified equation: \(T - \frac{185}{13} = 2.5 \times 1.5\).
- A1: For \(T = \frac{935}{52}\text{ N}\) or \(18.0\text{ N}\) (accept \(17.98\)).

(b)
- M1: For resolving down the plane to find the new deceleration: \(a = g \sin \alpha + \frac{F}{m}\) or \(F_{\text{net}} = \frac{185}{13}\).
- A1: For correct deceleration \(a = \frac{74}{13}\text{ m s}^{-2}\) (or \(5.69\text{ m s}^{-2}\)).
- M1: For using a suitable kinematic equation (e.g. \(v^2 = u^2 - 2as\)) with \(v = 0\) and \(u = 6\).
- A1: For \(s = \frac{117}{37}\text{ m}\) or \(3.16\text{ m}\) (awarded for 3 s.f.).

(a)(i)
Applying the principle of conservation of linear momentum to the collision between \(A\) and \(B\):
\[m_A u_A + m_B u_B = m_A v_A + m_B v_B\]
\[0.4(6) + 0.6(0) = 0.4(1.5) + 0.6 v_B\]
\[2.4 = 0.6 + 0.6 v_B\]
\[0.6 v_B = 1.8 \implies v_B = 3\text{ m s}^{-1}\]

(a)(ii)
Initial kinetic energy of the system:
\[E_k(\text{initial}) = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(0.4)(6^2) = 0.2 \times 36 = 7.2\text{ J}\]

Final kinetic energy of the system:
\[E_k(\text{final}) = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2}(0.4)(1.5^2) + \frac{1}{2}(0.6)(3^2)\]
\[E_k(\text{final}) = 0.2(2.25) + 0.3(9) = 0.45 + 2.7 = 3.15\text{ J}\]

Loss of kinetic energy:
\[\Delta E_k = 7.2 - 3.15 = 4.05\text{ J}\]

(b)
Applying the principle of conservation of linear momentum to the coalescence of \(B\) and \(C\):
\[m_B v_B + m_C u_C = (m_B + m_C) v_D\]
\[0.6(3) + 0 = (0.6 + 0.8) v_D\]
\[1.8 = 1.4 v_D\]
\[v_D = \frac{1.8}{1.4} = \frac{9}{7}\text{ m s}^{-1} \approx 1.29\text{ m s}^{-1}\]

(c)(i)
For particle \(D\) on the rough surface:
Mass of \(D = 1.4\text{ kg}\).
Normal reaction \(R = m_D g = 1.4 \times 10 = 14\text{ N}\).
Frictional force \(F = \mu R = 0.2 \times 14 = 2.8\text{ N}\).

Deceleration \(a\) is given by:
\[a = \frac{F}{m_D} = \frac{2.8}{1.4} = 2\text{ m s}^{-2}\]

Using \(v^2 = u^2 - 2as\) with \(v = 0\), \(u = \frac{9}{7}\text{ m s}^{-1}\), and \(a = 2\text{ m s}^{-2}\):
\[0 = \left(\frac{9}{7}\right)^2 - 2(2)s\]
\[4s = \frac{81}{49}\]
\[s = \frac{81}{196}\text{ m} \approx 0.413\text{ m}\]

(c)(ii)
Using \(v = u - at\):
\[0 = \frac{9}{7} - 2t\]
\[t = \frac{9}{14}\text{ s} \approx 0.643\text{ s}\]

(a)(i)
- M1: For applying conservation of momentum to the direct collision of \(A\) and \(B\): \(0.4(6) = 0.4(1.5) + 0.6 v_B\).
- A1: For \(v_B = 3\text{ m s}^{-1}\).

(a)(ii)
- M1: For finding the total initial and final KE and subtracting: \(\frac{1}{2}(0.4)(6^2) - [\frac{1}{2}(0.4)(1.5^2) + \frac{1}{2}(0.6)(3^2)]\).
- A1: For loss in kinetic energy \(= 4.05\text{ J}\).

(b)
- M1: For applying conservation of momentum for coalescence of \(B\) and \(C\): \(0.6(3) = (0.6 + 0.8) v_D\).
- A1: For \(v_D = \frac{9}{7}\text{ m s}^{-1}\) (or \(1.29\text{ m s}^{-1}\)).

(c)(i)
- M1: For finding the friction force \(F = 0.2(14) = 2.8\text{ N}\) and deceleration \(a = 2\text{ m s}^{-2}\).
- M1: For applying \(v^2 = u^2 - 2as\) with \(v = 0\) and their \(u = v_D\).
- A1: For \(s = \frac{81}{196}\text{ m}\) or \(0.413\text{ m}\).

(c)(ii)
- B1: For \(t = \frac{9}{14}\text{ s}\) or \(0.643\text{ s}\).

(a)
To obtain a sum of 3, the two cards chosen must be a 1 and a 2 (in either order):
\[\text{P}(X = 3) = \text{P}(1, 2) + \text{P}(2, 1) = \left(\frac{3}{6} \times \frac{2}{5}\right) + \left(\frac{2}{6} \times \frac{3}{5}\right) = \frac{6}{30} + \frac{6}{30} = \frac{12}{30} = \frac{2}{5}\]

(b)
The possible sums are:
- \(1 + 1 = 2\): \(\text{P}(X = 2) = \frac{3}{6} \times \frac{2}{5} = \frac{6}{30} = \frac{1}{5}\)
- \(1 + 2 = 3\): \(\text{P}(X = 3) = \frac{2}{5}\)
- \(2 + 2 = 4\): \(\text{P}(X = 4) = \frac{2}{6} \times \frac{1}{5} = \frac{2}{30} = \frac{1}{15}\)
- \(1 + 5 = 6\): \(\text{P}(X = 6) = 2 \times \left(\frac{3}{6} \times \frac{1}{5}\right) = \frac{6}{30} = \frac{1}{5}\)
- \(2 + 5 = 7\): \(\text{P}(X = 7) = 2 \times \left(\frac{2}{6} \times \frac{1}{5}\right) = \frac{4}{30} = \frac{2}{15}\)

The probability distribution table is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 6 & 7 \\ \hline \text{P}(X = x) & \frac{1}{5} & \frac{2}{5} & \frac{1}{15} & \frac{1}{5} & \frac{2}{15} \\ \hline \end{array} \]

(c)
First, find \(\text{E}(X)\):
\[\text{E}(X) = 2\left(\frac{6}{30}\right) + 3\left(\frac{12}{30}\right) + 4\left(\frac{2}{30}\right) + 6\left(\frac{6}{30}\right) + 7\left(\frac{4}{30}\right) = \frac{12 + 36 + 8 + 36 + 28}{30} = \frac{120}{30} = 4\]

Next, find \(\text{E}(X^2)\):
\[\text{E}(X^2) = 2^2\left(\frac{6}{30}\right) + 3^2\left(\frac{12}{30}\right) + 4^2\left(\frac{2}{30}\right) + 6^2\left(\frac{6}{30}\right) + 7^2\left(\frac{4}{30}\right)\]
\[\text{E}(X^2) = \frac{24 + 108 + 32 + 216 + 196}{30} = \frac{576}{30} = 19.2\]

Then calculate the variance:
\[\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 19.2 - 4^2 = 19.2 - 16 = 3.2 = \frac{16}{5}\]

(a)
M1: For selecting cards 1 and 2 in either order without replacement, e.g. \(2 \times \left(\frac{3}{6} \times \frac{2}{5}\right)\) or \(\frac{\binom{3}{1} \times \binom{2}{1}}{\binom{6}{2}}\).
A1: For obtaining \(\frac{2}{5}\) (or \(\frac{12}{30}\)) correctly with sufficient working shown.

(b)
B1: For all correct values of \(X\) (2, 3, 4, 6, 7) in the table.
B1: For all probabilities correct (accept fractions over 30 or simplified fractions).

(c)
M1: For correct method to evaluate \(\text{Var}(X) = \sum x^2 p - (\sum x p)^2\) using their distribution table.
A1: For \(3.2\) or \(\frac{16}{5}\) (CWO).

(a)(i)
\[\text{P}(L \cup E) = \text{P}(L) + \text{P}(E) - \text{P}(L \cap E) = \frac{40}{120} + \frac{60}{120} - \frac{16}{120} = \frac{84}{120} = \frac{7}{10} = 0.7\]
Alternatively, sum the mutually exclusive frequencies: \(\frac{24 + 16 + 20 + 24}{120} = \frac{84}{120} = \frac{7}{10}\).

(a)(ii)
\[\text{P}(L \mid E) = \frac{\text{P}(L \cap E)}{\text{P}(E)} = \frac{16}{60} = \frac{4}{15} \approx 0.267\]

(b)
Evaluate \(\text{P}(L \cap M)\):
\[\text{P}(L \cap M) = \frac{24}{120} = \frac{1}{5} = 0.2\]
Evaluate \(\text{P}(L) \times \text{P}(M)\):
\[\text{P}(L) = \frac{40}{120} = \frac{1}{3}\]
\[\text{P}(M) = \frac{60}{120} = \frac{1}{2}\]
\[\text{P}(L) \times \text{P}(M) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \approx 0.167\]
Since \(\text{P}(L \cap M) \neq \text{P}(L) \times \text{P}(M)\) (as \(0.2 \neq 0.167\)), the events \(L\) and \(M\) are not independent.
(Alternatively, compare \(\text{P}(M \mid L) = \frac{24}{40} = 0.6\) with \(\text{P}(M) = 0.5\); since \(\text{P}(M \mid L) \neq \text{P}(M)\), they are not independent).

(c)
There are 60 students who prefer studying in the Evening: 16 in Library, 20 at Home, and 24 in Café.
Method 1 (Complement):
\[\text{P}(\text{same environment}) = \frac{\binom{16}{2} + \binom{20}{2} + \binom{24}{2}}{\binom{60}{2}} = \frac{120 + 190 + 276}{1770} = \frac{586}{1770} = \frac{293}{885}\]
\[\text{P}(\text{different environments}) = 1 - \frac{293}{885} = \frac{592}{885} \approx 0.669\]

Method 2 (Direct pairs):
\[\text{P}(\text{different}) = \frac{(16 \times 20) + (16 \times 24) + (20 \times 24)}{\binom{60}{2}} = \frac{320 + 384 + 480}{1770} = \frac{1184}{1770} = \frac{592}{885} \approx 0.669\]

(a)(i)
M1: For a valid method to find \(\text{P}(L \cup E)\), e.g. \(\frac{40 + 60 - 16}{120}\) or adding four disjoint cell counts.
A1: Correct answer \(\frac{7}{10}\) or \(0.7\) oe.

(a)(ii)
B1: Correct conditional probability \(\frac{16}{60}\) or \(\frac{4}{15}\) or \(0.267\) (3 s.f.).

(b)
M1: For finding and comparing \(\text{P}(L \cap M)\) and \(\text{P}(L) \times \text{P}(M)\), or comparing \(\text{P}(M \mid L)\) with \(\text{P}(M)\).
A1: For correct values stated and correct conclusion that the events are not independent.

(c)
M1: For using combinations or probability products without replacement from the 60 evening students, either via complement \(1 - \frac{\binom{16}{2} + \binom{20}{2} + \binom{24}{2}}{\binom{60}{2}}\) or sum of pairs \(\frac{16 \times 20 + 16 \times 24 + 20 \times 24}{\binom{60}{2}}\).
A1: Correct probability \(\frac{592}{885}\) or \(0.669\) (3 s.f.).

(a)
For Station A, \(n = 15\).
Ordered data: 12, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 28, 29, 31, 34.
- Position of median: \(\frac{15 + 1}{2} = 8\text{th}\) value \(\Rightarrow \text{Median} = 22^\circ\text{C}\).
- Position of lower quartile \(Q_1\): \(\frac{15 + 1}{4} = 4\text{th}\) value \(\Rightarrow Q_1 = 17^\circ\text{C}\).
- Position of upper quartile \(Q_3\): \(\frac{3(15 + 1)}{4} = 12\text{th}\) value \(\Rightarrow Q_3 = 28^\circ\text{C}\).
- Interquartile range: \(\text{IQR} = Q_3 - Q_1 = 28 - 17 = 11^\circ\text{C}\).

(b)
1. Comparison of average/centre: On average, the daily maximum temperature was higher at Station B than at Station A (since the median of Station B is \(24^\circ\text{C}\), which is greater than the median of Station A, \(22^\circ\text{C}\)).
2. Comparison of spread/dispersion: The daily maximum temperatures were more spread out / variable at Station A than at Station B (since the IQR of Station A is \(11^\circ\text{C}\), which is greater than the IQR of Station B, \(9^\circ\text{C}\)).

(c)
Calculate the outlier boundaries for Station A:
- Lower boundary: \(Q_1 - 1.5 \times \text{IQR} = 17 - 1.5(11) = 17 - 16.5 = 0.5^\circ\text{C}\).
- Upper boundary: \(Q_3 + 1.5 \times \text{IQR} = 28 + 1.5(11) = 28 + 16.5 = 44.5^\circ\text{C}\).

The minimum recorded value at Station A is \(12^\circ\text{C}\) (which is \(> 0.5^\circ\text{C}\)) and the maximum recorded value is \(34^\circ\text{C}\) (which is \(< 44.5^\circ\text{C}\)).
Since all values lie within the range \([0.5, 44.5]\), there are no outliers.

(a)
B1: For correct median = \(22\).
M1: For identifying lower quartile \(Q_1 = 17\) and upper quartile \(Q_3 = 28\).
A1: For correct \(\text{IQR} = 11\).

(b)
B1: For comparing centres in context (e.g. Station B is warmer on average than Station A as median \(24 > 22\)).
B1: For comparing spread in context (e.g. Station A has greater variability in temperature than Station B as IQR \(11 > 9\)).

(c)
M1: For calculating both boundary thresholds: \(17 - 1.5 \times 11 = 0.5\) and \(28 + 1.5 \times 11 = 44.5\) (FT their \(Q_1, Q_3, \text{IQR}\)).
A1: For concluding that there are no outliers with reference to minimum value \(12 > 0.5\) and maximum value \(34 < 44.5\).

(a) (i) The word \(\text{RESTORATION}\) has 11 letters with repeated letters: \(2\) R's, \(2\) T's, \(2\) O's, \(1\) E, \(1\) S, \(1\) A, \(1\) I, \(1\) N.
\[\text{Total arrangements} = \frac{11!}{2! \times 2! \times 2!} = \frac{39\,916\,800}{8} = 4\,989\,600\]

(a) (ii) There are 6 consonants (\(\text{R, R, T, T, S, N}\)) and 5 vowels (\(\text{E, O, O, A, I}\)).
For consonants and vowels to alternate in an 11-letter word, the pattern must be:
\[\text{C V C V C V C V C V C}\]
Number of ways to arrange the 6 consonants in the 6 consonant positions:
\[\frac{6!}{2! \times 2!} = \frac{720}{4} = 180\]
Number of ways to arrange the 5 vowels in the 5 vowel positions:
\[\frac{5!}{2!} = \frac{120}{2} = 60\]
Total alternating arrangements \(= 180 \times 60 = 10\,800\).

(b) (i) A committee of 5 with at least 2 men and at least 2 women can be formed in two mutually exclusive ways:
- Case 1: 2 men and 3 women
\[\binom{8}{2} \times \binom{6}{3} = 28 \times 20 = 560\]
- Case 2: 3 men and 2 women
\[\binom{8}{3} \times \binom{6}{2} = 56 \times 15 = 840\]
Total number of committees \(= 560 + 840 = 1400\).

(b) (ii) Find the number of valid committees containing both David and Edward:
If both David and Edward are chosen, 2 men are already selected.
- Subcase 1: Exactly 2 men (David and Edward) and 3 women:
\[\binom{6}{3} = 20\]
- Subcase 2: Exactly 3 men (David, Edward, and 1 other from the remaining 6 men) and 2 women:
\[\binom{6}{1} \times \binom{6}{2} = 6 \times 15 = 90\]
Total committees containing both \(= 20 + 90 = 110\).

Number of committees where David and Edward are not together \(= 1400 - 110 = 1290\).

(a)(i)
B1: Correct answer \(4\,989\,600\) [1]

(a)(ii)
M1: For \(\frac{6!}{2!2!}\) or \(\frac{5!}{2!}\) seen
M1: For multiplying their consonant arrangements by their vowel arrangements
A1: Correct answer \(10\,800\) [3]

(b)(i)
M1: For identifying both cases \(\binom{8}{2}\binom{6}{3}\) and \(\binom{8}{3}\binom{6}{2}\)
M1: For calculating and adding the two products
A1: Correct answer \(1400\) [3]

(b)(ii)
M1: For finding the number of committees containing both David and Edward using appropriate combinations
M1: For subtracting their forbidden count from their (b)(i) answer
A1: Correct answer \(1290\) [3]

(a) Let \(X\) be the number of customers choosing express delivery. \(X \sim \text{B}(12, 0.18)\).
\[\text{P}(X = 3) = \binom{12}{3} (0.18)^3 (0.82)^9 = 220 \times 0.005832 \times 0.167618 = 0.215\text{ (3 s.f.)}\]

(b) (i) For \(n = 250\) and \(p = 0.18\):
\[np = 250 \times 0.18 = 45 > 5\]
\[n(1 - p) = 250 \times 0.82 = 205 > 5\]
Since both \(np > 5\) and \(n(1-p) > 5\), the binomial distribution \(\text{B}(250, 0.18)\) can be approximated by a normal distribution \(Y \sim \text{N}(\mu, \sigma^2)\), where:
\[\mu = np = 45\]
\[\sigma^2 = np(1-p) = 250 \times 0.18 \times 0.82 = 36.9\]

(b) (ii) We require \(\text{P}(40 \leqslant X \leqslant 55)\).
Applying the continuity correction:
\[\text{P}(39.5 \leqslant Y \leqslant 55.5)\]
Standardising:
\[z_1 = \frac{39.5 - 45}{\sqrt{36.9}} = \frac{-5.5}{6.0745} \approx -0.905\]
\[z_2 = \frac{55.5 - 45}{\sqrt{36.9}} = \frac{10.5}{6.0745} \approx 1.729\]
Using standard normal tables:
\[\Phi(1.729) = 0.9581\]
\[\Phi(-0.905) = 1 - \Phi(0.905) = 1 - 0.8172 = 0.1828\]
\[\text{P}(39.5 \leqslant Y \leqslant 55.5) = 0.9581 - 0.1828 = 0.7753 \approx 0.775\text{ (3 s.f.)}\]

(b) (iii) We require \(\text{P}(X \leqslant 52 \mid X > 42)\).
With continuity correction, this corresponds to:
\[\frac{\text{P}(42.5 < Y \leqslant 52.5)}{\text{P}(Y > 42.5)}\]
Standardising the boundaries:
\[z_{42.5} = \frac{42.5 - 45}{6.0745} = \frac{-2.5}{6.0745} \approx -0.412\]
\[z_{52.5} = \frac{52.5 - 45}{6.0745} = \frac{7.5}{6.0745} \approx 1.235\]
Values from normal tables:
\[\Phi(1.235) = 0.8916\]
\[\Phi(-0.412) = 1 - \Phi(0.412) = 1 - 0.6598 = 0.3402\]
Numerator: \(0.8916 - 0.3402 = 0.5514\)
Denominator: \(\text{P}(Y > 42.5) = 1 - 0.3402 = 0.6598\)
\[\text{Probability} = \frac{0.5514}{0.6598} = 0.8357 \approx 0.836\text{ (3 s.f.)}\]

(a)
M1: For \(\binom{12}{3} (0.18)^3 (0.82)^9\) seen
A1: Correct answer \(0.215\) [2]

(b)(i)
B1: State \(\text{N}(45, 36.9)\) with correct mean and variance
B1: Justify with \(np = 45 > 5\) and \(n(1-p) = 205 > 5\) [2]

(b)(ii)
M1: For using continuity correction \(39.5\) and/or \(55.5\)
M1: For standardising with their mean and standard deviation: \(\frac{39.5 - 45}{\sqrt{36.9}}\) and \(\frac{55.5 - 45}{\sqrt{36.9}}\)
M1: For calculating \(\Phi(z_2) - \Phi(z_1)\) correctly using tables
A1: Correct answer \(0.775\) [4]

(b)(iii)
M1: For conditional probability fraction with correct continuity corrections \(\frac{\text{P}(42.5 < Y \leqslant 52.5)}{\text{P}(Y > 42.5)}\)
A1: Correct answer \(0.836\) [2]

(a) (i) Total number of ways to choose 4 tiles from 14 is:
\[\binom{14}{4} = 1001\]
We need exactly 2 red tiles (from 5) and 2 non-red tiles (from 4 blue and 5 green) with at least 1 blue tile.
The choices for the 2 non-red tiles are:
- 1 blue and 1 green: \(\binom{4}{1} \times \binom{5}{1} = 4 \times 5 = 20\)
- 2 blue and 0 green: \(\binom{4}{2} \times \binom{5}{0} = 6 \times 1 = 6\)
Total favourable selections \(= \binom{5}{2} \times (20 + 6) = 10 \times 26 = 260\).
\[\text{Probability} = \frac{260}{1001} = \frac{20}{77} \approx 0.260\text{ (3 s.f.)}\]

(a) (ii) There are 5 non-red tiles (3 blue, 2 green) and 4 red tiles.
Number of ways to arrange the 5 non-red tiles:
\[\frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10\]
This arrangement creates 6 available spaces (including the ends) for the red tiles:
\[\_ \text{ N } \_ \text{ N } \_ \text{ N } \_ \text{ N } \_ \text{ N } \_\]
Number of ways to choose 4 distinct spaces out of 6 for the identical red tiles:
\[\binom{6}{4} = 15\]
Total arrangements \(= 10 \times 15 = 150\).

(b) (i) Let \(L\) be the lifespan of a lamp. \(L \sim \text{N}(1200, 85^2)\).
\[\text{P}(L > 1320) = \text{P}\left(Z > \frac{1320 - 1200}{85}\right) = \text{P}\left(Z > \frac{120}{85}\right) = \text{P}(Z > 1.412)\]
\[\text{P}(Z > 1.412) = 1 - \Phi(1.412) = 1 - 0.9210 = 0.0790\text{ (3 s.f.)}\]

(b) (ii) Let \(K\) be the number of lamps lasting more than 1320 hours in the batch of 200. \(K \sim \text{B}(200, 0.0790)\).
\[\mu = np = 200 \times 0.0790 = 15.8\]
\[\sigma^2 = np(1 - p) = 200 \times 0.0790 \times 0.9210 = 14.5518\]
Since \(np > 5\) and \(n(1-p) > 5\), approximate by \(W \sim \text{N}(15.8, 14.5518)\).
We require \(\text{P}(K \geqslant 22)\).
Applying continuity correction:
\[\text{P}(W \geqslant 21.5) = \text{P}\left(Z \geqslant \frac{21.5 - 15.8}{\sqrt{14.5518}}\right) = \text{P}\left(Z \geqslant \frac{5.7}{3.8147}\right) = \text{P}(Z \geqslant 1.494)\]
\[\text{P}(Z \geqslant 1.494) = 1 - \Phi(1.494) = 1 - 0.9324 = 0.0676\text{ (3 s.f.)}\]

(a)(i)
M1: For \(\binom{14}{4} = 1001\) seen as denominator
M1: For calculating favourable cases: \(\binom{5}{2} \times [\binom{4}{1}\binom{5}{1} + \binom{4}{2}]\)
A1: Correct answer \(\frac{260}{1001}\) or \(\frac{20}{77}\) or \(0.260\) [3]

(a)(ii)
M1: For \(\frac{5!}{3!2!} = 10\)
M1: For \(\binom{6}{4} = 15\) and multiplying by their arrangement of non-red tiles
A1: Correct answer \(150\) [3]

(b)(i)
M1: For standardising \(\frac{1320 - 1200}{85}\) and finding \(1 - \Phi(z)\)
A1: Correct answer \(0.0790\) [2]

(b)(ii)
M1: For using normal approximation with \(\mu = 15.8\), \(\sigma = \sqrt{14.5518}\) and continuity correction \(21.5\)
A1: Correct answer \(0.0676\) [2]

(a) Let \(T = B_1 + B_2 + B_3\).
\(\text{E}(T) = 3 \times 28.6 = 85.8\)
\(\text{Var}(T) = 3 \times 0.81 = 2.43\)
So \(T \sim \text{N}(85.8, 2.43)\).
\(\text{P}(T < 85) = \text{P}\left(Z < \frac{85 - 85.8}{\sqrt{2.43}}\right) = \text{P}(Z < -0.5132) = 1 - \Phi(0.513) = 1 - 0.6960 = 0.304\).

(b) We require \(\text{P}(A > 1.5B) = \text{P}(A - 1.5B > 0)\).
Let \(W = A - 1.5B\).
\(\text{E}(W) = \text{E}(A) - 1.5\text{E}(B) = 45.2 - 1.5(28.6) = 45.2 - 42.9 = 2.3\)
\(\text{Var}(W) = \text{Var}(A) + 1.5^2 \text{Var}(B) = 1.44 + 2.25(0.81) = 1.44 + 1.8225 = 3.2625\)
So \(W \sim \text{N}(2.3, 3.2625)\).
\(\text{P}(W > 0) = \text{P}\left(Z > \frac{0 - 2.3}{\sqrt{3.2625}}\right) = \text{P}\left(Z > \frac{-2.3}{1.8062}\right) = \text{P}(Z > -1.273) = \Phi(1.273) = 0.8985 \approx 0.899\).

(a)
M1: For stating or using \(\text{E}(T) = 3 \times 28.6 = 85.8\) and \(\text{Var}(T) = 3 \times 0.81 = 2.43\) with correct standardisation method.
A1: For correct answer \(0.304\) (accept \(0.304\) to \(0.305\)).

(b)
M1: For considering \(A - 1.5B\) and finding \(\text{E}(A - 1.5B) = 2.3\) and \(\text{Var}(A - 1.5B) = 1.44 + 1.5^2(0.81) = 3.2625\).
M1: For standardising \(0\) with their mean and standard deviation: \(\frac{0 - 2.3}{\sqrt{3.2625}}\) (or \(\frac{2.3 - 0}{\sqrt{3.2625}}\)).
A1: For correct answer \(0.899\) (accept \(0.898\) to \(0.899\)).

(a) Null hypothesis \(H_0: \lambda = 3.2\)
Alternative hypothesis \(H_1: \lambda < 3.2\)

Under \(H_0\), \(X \sim \text{Po}(3.2)\).
\(\text{P}(X = 0) = \text{e}^{-3.2} \approx 0.0408\)
\(\text{P}(X \le 1) = \text{e}^{-3.2}(1 + 3.2) = 4.2\text{e}^{-3.2} \approx 0.1712\)
Since \(\text{P}(X = 0) = 0.0408 < 0.05\) and \(\text{P}(X \le 1) = 0.1712 > 0.05\), the rejection region is \(X = 0\).

(b) A Type II error occurs when \(H_0\) is not rejected given that \(H_1\) is true (i.e. \(\lambda = 1.4\)).
Acceptance region is \(X \ge 1\).
When \(\lambda = 1.4\), \(X \sim \text{Po}(1.4)\).
\(\text{P}(\text{Type II error}) = \text{P}(X \ge 1 \mid \lambda = 1.4) = 1 - \text{P}(X = 0 \mid \lambda = 1.4) = 1 - \text{e}^{-1.4} = 1 - 0.24660 = 0.7534 \approx 0.753\).

(a)
B1: For stating hypotheses \(H_0: \lambda = 3.2\) and \(H_1: \lambda < 3.2\).
B1: For identifying probabilities \(\text{P}(X = 0) \approx 0.0408\) and concluding that the rejection region is \(X = 0\) (or \(\{0\}\)).

(b)
B1: For identifying the non-rejection (acceptance) region as \(X \ge 1\) (or \(X > 0\)).
M1: For calculating \(1 - \text{P}(X = 0)\) using \(\text{Po}(1.4)\) (i.e. \(1 - \text{e}^{-1.4}\)).
A1: For obtaining \(0.753\) (allow \(0.7534\)).

(a)
Since \(f(x)\) is a probability density function, \(\int_{0}^{\infty} f(x)\,\mathrm{d}x = 1\).
\[ \int_{0}^{\infty} k(x+2)^{-4}\,\mathrm{d}x = \left[ -\frac{k}{3}(x+2)^{-3} \right]_{0}^{\infty} = 0 - \left( -\frac{k}{3(2)^3} \right) = \frac{k}{24} \]
Setting \(\dfrac{k}{24} = 1\) gives \(k = 24\).

(b)
\[ \text{E}(X) = \int_{0}^{\infty} x f(x)\,\mathrm{d}x = \int_{0}^{\infty} 24x(x+2)^{-4}\,\mathrm{d}x \]
Let \(u = x + 2\), so \(x = u - 2\) and \(\mathrm{d}x = \mathrm{d}u\). When \(x = 0, u = 2\); as \(x \to \infty, u \to \infty\).
\[ \int_{2}^{\infty} 24(u-2)u^{-4}\,\mathrm{d}u = 24 \int_{2}^{\infty} (u^{-3} - 2u^{-4})\,\mathrm{d}u \]
\[ = 24 \left[ -\frac{1}{2}u^{-2} + \frac{2}{3}u^{-3} \right]_{2}^{\infty} = 24 \left[ 0 - \left( -\frac{1}{2(4)} + \frac{2}{3(8)} \right) \right] \]
\[ = 24 \left( \frac{1}{8} - \frac{1}{12} \right) = 24 \left( \frac{1}{24} \right) = 1 \]

(c)
Let \(m\) be the median of \(X\). Then \(\int_{0}^{m} 24(x+2)^{-4}\,\mathrm{d}x = 0.5\).
\[ \left[ -8(x+2)^{-3} \right]_{0}^{m} = 0.5 \implies -8(m+2)^{-3} - (-8(2)^{-3}) = 0.5 \]
\[ 1 - \frac{8}{(m+2)^3} = 0.5 \implies \frac{8}{(m+2)^3} = 0.5 \implies (m+2)^3 = 16 \]
\[ m+2 = \sqrt[3]{16} = 2\sqrt[3]{2} \implies m = 2\sqrt[3]{2} - 2 \]

(d)
\[ \text{P}(X > 1 \mid X < 3) = \frac{\text{P}(1 < X < 3)}{\text{P}(X < 3)} \]
First, calculate the probabilities:
\[ \text{P}(X < 3) = \int_{0}^{3} 24(x+2)^{-4}\,\mathrm{d}x = \left[ -8(x+2)^{-3} \right]_{0}^{3} = -\frac{8}{5^3} - \left(-\frac{8}{2^3}\right) = 1 - \frac{8}{125} = \frac{117}{125} \]
\[ \text{P}(1 < X < 3) = \left[ -8(x+2)^{-3} \right]_{1}^{3} = -\frac{8}{125} - \left(-\frac{8}{3^3}\right) = \frac{8}{27} - \frac{8}{125} = \frac{1000 - 216}{3375} = \frac{784}{3375} \]
Therefore:
\[ \text{P}(X > 1 \mid X < 3) = \frac{\frac{784}{3375}}{\frac{117}{125}} = \frac{784}{3375} \times \frac{125}{117} = \frac{784}{27 \times 117} = \frac{784}{3159} \]

(a)
- M1: Integrates \(k(x+2)^{-4}\) with respect to \(x\) to obtain \(A(x+2)^{-3}\) and sets the integral from \(0\) to \(\infty\) equal to \(1\).
- A1: Fully correct working leading without error to \(k = 24\).

(b)
- M1: Applies \(\text{E}(X) = \int_{0}^{\infty} x f(x)\,\mathrm{d}x\) using substitution or integration by parts.
- M1: Correctly evaluates the integrated expression between limits \(0\) (or \(2\)) and \(\infty\).
- A1: Obtains \(\text{E}(X) = 1\).

(c)
- M1: Sets \(\int_{0}^{m} f(x)\,\mathrm{d}x = 0.5\) and solves for \((m+2)^3\).
- A1: Obtains \(m = 2\sqrt[3]{2} - 2\) (or \(a=2, b=2\)).

(d)
- M1: Uses conditional probability formula \(\dfrac{\text{P}(1 < X < 3)}{\text{P}(X < 3)}\).
- A1: Obtains \(\dfrac{784}{3159}\) (or exact equivalent fraction).

(a)
In \(1\) hour (\(6\) periods of \(10\) minutes), the mean rate is \(\lambda = 4.2 \times 6 = 25.2\).
Let \(X\) be the number of lorries arriving in \(1\) hour. Then \(X \sim \text{Po}(25.2)\).

(b)
Since \(\lambda = 25.2 > 15\), the Poisson distribution can be approximated by a normal distribution.
The approximating distribution is \(Y \sim \text{N}(25.2, 25.2)\).

(c)
We require \(\text{P}(X < 20) = \text{P}(X \le 19)\).
Applying a continuity correction:
\[ \text{P}(X \le 19) \approx \text{P}(Y < 19.5) \]
Standardising:
\[ Z = \frac{19.5 - 25.2}{\sqrt{25.2}} = \frac{-5.7}{5.01996} \approx -1.135 \]
Using standard normal tables:
\[ \text{P}(Z < -1.135) = 1 - \Phi(1.135) = 1 - 0.8718 = 0.1282 \approx 0.128\text{ (to 3 s.f.)} \]

(d)
In a \(3\)-hour period, let \(W\) be the number of arriving lorries.
\(\lambda_{3\text{h}} = 3 \times 25.2 = 75.6\), so \(W \sim \text{Po}(75.6) \approx \text{N}(75.6, 75.6)\).
All lorries can be processed if \(W \le N\).
We require \(\text{P}(W \le N) \ge 0.975\).
Applying the continuity correction:
\[ \text{P}(W \le N) \approx \text{P}(V < N + 0.5) \ge 0.975 \]
Standardising:
\[ \frac{(N + 0.5) - 75.6}{\sqrt{75.6}} \ge z_{0.975} \]
From the critical value table, \(z_{0.975} = 1.960\).
\[ \frac{N + 0.5 - 75.6}{\sqrt{75.6}} \ge 1.960 \]
\[ N + 0.5 - 75.6 \ge 1.960 \times \sqrt{75.6} = 1.960 \times 8.6948 = 17.042 \]
\[ N \ge 75.6 + 17.042 - 0.5 = 92.142 \]
Since \(N\) must be an integer, the smallest integer value is \(N = 93\).

(a)
- B1: States \(\text{Po}(25.2)\) (accept mean \(= 25.2\)).

(b)
- B1: States \(\lambda = 25.2 > 15\) (or \(\lambda > 15\)) as justification.
- B1: States \(\text{N}(25.2, 25.2)\) with both mean and variance correct.

(c)
- M1: Uses continuity correction correctly with \(19.5\) (i.e. \(\text{P}(Y < 19.5)\)).
- M1: Standardises with their mean and standard deviation: \(\dfrac{19.5 - 25.2}{\sqrt{25.2}}\).
- A1: Obtains \(0.128\) (awarded for \(0.128\) or \(0.1282\)).

(d)
- B1: States or uses \(\text{Po}(75.6)\) or \(\text{N}(75.6, 75.6)\).
- M1: Applies continuity correction \(N + 0.5\) in standardisation equation equated to \(1.960\) (or \(1.96\)).
- M1: Solves the inequality/equation for \(N\).
- A1: Obtains \(N = 93\) as the smallest integer.

(a)
\[ \operatorname{E}(T) = \int_0^4 t \cdot \frac{3}{32}(4t - t^2) \, \mathrm{d}t = \frac{3}{32} \int_0^4 (4t^2 - t^3) \, \mathrm{d}t = \frac{3}{32} \left[ \frac{4}{3}t^3 - \frac{1}{4}t^4 \right]_0^4 \]
\[ = \frac{3}{32} \left( \frac{256}{3} - 64 \right) = \frac{3}{32} \times \frac{64}{3} = 2 \quad \text{(as required)} \]
\[ \operatorname{E}(T^2) = \int_0^4 t^2 \cdot \frac{3}{32}(4t - t^2) \, \mathrm{d}t = \frac{3}{32} \int_0^4 (4t^3 - t^4) \, \mathrm{d}t = \frac{3}{32} \left[ t^4 - \frac{1}{5}t^5 \right]_0^4 \]
\[ = \frac{3}{32} \left( 256 - \frac{1024}{5} \right) = \frac{3}{32} \times \frac{256}{5} = \frac{24}{5} = 4.8 \]
\[ \operatorname{Var}(T) = \operatorname{E}(T^2) - (\operatorname{E}(T))^2 = 4.8 - 2^2 = 0.8 \]

(b)(i)
Let \(\bar{T}\) be the mean duration of the random sample of \(n = 100\) calls.
By the Central Limit Theorem:
\[ \bar{T} \sim \mathrm{N}\left(2, \frac{0.8}{100}\right) = \mathrm{N}(2, 0.008) \]
\[ \mathrm{P}(\bar{T} > 2.15) = 1 - \Phi\left( \frac{2.15 - 2}{\sqrt{0.008}} \right) = 1 - \Phi\left( \frac{0.15}{0.089443} \right) = 1 - \Phi(1.677) \]
\[ = 1 - 0.9532 = 0.0468 \]

(b)(ii)
The Central Limit Theorem was used when assuming that \(\bar{T}\) has an approximate normal distribution.
It was necessary because the distribution of \(T\) itself is not normal (it has a polynomial density function) and the sample size is large (\(n = 100\)).

(c)
Sample mean: \(\bar{x} = \frac{208}{80} = 2.6\).
Unbiased estimate of the population variance:
\[ s^2 = \frac{1}{79} \left( 612 - \frac{208^2}{80} \right) = \frac{1}{79} (612 - 540.8) = \frac{71.2}{79} \approx 0.901266 \]
For a 95% confidence interval, \(z = 1.960\).
\[ \text{Confidence Interval} = \bar{x} \pm 1.960 \sqrt{\frac{s^2}{n}} = 2.6 \pm 1.960 \sqrt{\frac{0.901266}{80}} = 2.6 \pm 0.208 = [2.39, 2.81] \]

(a)
M1: Correct integral expression for \(\operatorname{E}(T)\) and integration to at least one correct term.
A1: Fully correct working leading to \(\operatorname{E}(T) = 2\).
M1: Correct method to find \(\operatorname{E}(T^2)\) and uses \(\operatorname{Var}(T) = \operatorname{E}(T^2) - (\operatorname{E}(T))^2\).
A1: Correct variance \(\operatorname{Var}(T) = 0.8\) (or \(\frac{4}{5}\)).

(b)(i)
M1: Standardising with mean 2 and variance \(\frac{0.8}{100}\) (or standard error \(\sqrt{0.008}\)).
M1: Correct probability structure \(1 - \Phi(z)\).
A1: \(0.0468\) (accept answers in range 0.0467 to 0.0469).

(b)(ii)
B1: States that CLT was used to justify the normal distribution of the sample mean \(\bar{T}\).
B1: States that it is necessary because the underlying distribution of \(T\) is non-normal and \(n\) is large.

(c)
M1: Correct calculation of \(\bar{x}\) and unbiased variance estimate \(s^2\) (or correct variance of sample mean), and standard form \(\bar{x} \pm 1.960 \frac{s}{\sqrt{n}}\).
A1: \([2.39, 2.81]\) (or \(2.39 \le \mu \le 2.81\)).

(a)
Any two from:
- Flaws occur independently of one another.
- Flaws occur singly (one at a time).
- Flaws occur at a constant average rate throughout the fabric.

(b)(i)
Let \(X\) be the number of flaws in a 30-metre length of fabric.
For a 30-metre length, the expected number of flaws is \(\lambda = 1.4 \times 3 = 4.2\).
\[ \mathrm{H}_0: \lambda = 4.2 \]
\[ \mathrm{H}_1: \lambda < 4.2 \]
Assuming \(\mathrm{H}_0\) is true, \(X \sim \mathrm{Po}(4.2)\).
Observed value: \(x = 1\).
\[ \mathrm{P}(X \le 1) = \mathrm{P}(X = 0) + \mathrm{P}(X = 1) = \mathrm{e}^{-4.2} + 4.2\mathrm{e}^{-4.2} = 5.2\mathrm{e}^{-4.2} \approx 0.07798 = 0.0780 \]
Since \(0.0780 > 0.05\), the result is not significant at the 5% level.
Do not reject \(\mathrm{H}_0\).
There is insufficient evidence to support the plant manager's claim that the average rate of flaws has decreased.

(b)(ii)
A Type I error occurs when \(\mathrm{H}_0\) is rejected given that \(\mathrm{H}_0\) is true.
Comparing cumulative probabilities with the significance level \(0.05\):
\[ \mathrm{P}(X = 0) = \mathrm{e}^{-4.2} \approx 0.0150 < 0.05 \]
\[ \mathrm{P}(X \le 1) \approx 0.0780 > 0.05 \]
Hence, the critical (rejection) region is \(X = 0\).
\[ \mathrm{P}(\text{Type I error}) = \mathrm{P}(X = 0 \mid \lambda = 4.2) = \mathrm{e}^{-4.2} \approx 0.0150 \]

(c)
Let \(Y\) be the number of flaws in 250 metres of fabric.
\(\lambda = 1.4 \times 25 = 35\).
Since \(\lambda = 35 > 15\), \(Y \sim \mathrm{Po}(35)\) can be approximated by \(W \sim \mathrm{N}(35, 35)\).
We require \(\mathrm{P}(Y < 30) = \mathrm{P}(Y \le 29)\).
Applying a continuity correction:
\[ \mathrm{P}(Y \le 29) \approx \mathrm{P}(W < 29.5) \]
Standardising:
\[ Z = \frac{29.5 - 35}{\sqrt{35}} = \frac{-5.5}{5.9161} \approx -0.930 \]
\[ \mathrm{P}(Z < -0.930) = 1 - \Phi(0.930) = 1 - 0.8238 = 0.1762 \approx 0.176 \]

(a)
B1: States one valid Poisson condition in context (e.g. flaws occur independently).
B1: States a second distinct valid Poisson condition (e.g. flaws occur singly / at constant rate).

(b)(i)
B1: State both hypotheses correctly in terms of \(\lambda\) or mean rate (\(\mathrm{H}_0: \lambda = 4.2\) and \(\mathrm{H}_1: \lambda < 4.2\)).
M1: Calculate \(\mathrm{P}(X \le 1)\) for \(X \sim \mathrm{Po}(4.2)\) using correct Poisson formula.
A1: Correct probability \(0.0780\) (or \(0.078\)).
A1: Correct conclusion in context comparing \(0.0780\) to \(0.05\) (no contradiction, must refer to insufficient evidence that flaw rate has decreased).

(b)(ii)
M1: Identifies critical region as \(X = 0\) by comparing tail probabilities with \(0.05\).
A1: \(0.0150\) (or \(\mathrm{e}^{-4.2}\)).

(c)
B1: States approximating normal distribution \(\mathrm{N}(35, 35)\).
M1: Standardises with continuous value \(29.5\) (continuity correction) and standard deviation \(\sqrt{35}\).
A1: \(0.176\) (accept 0.1762).