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To find where the line and curve intersect, we equate their equations: \(kx - 5 = x^2 - 3x + 4\). Rearranging into a standard quadratic equation form gives \(x^2 - (3+k)x + 9 = 0\). For the line and curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be negative: \(b^2 - 4ac < 0\), which gives \((-(3+k))^2 - 4(1)(9) < 0\). Expanding and simplifying, we get \((3+k)^2 - 36 < 0\), which means \((3+k)^2 < 36\). This yields \(-6 < 3+k < 6\). Subtracting 3 from all parts of the inequality gives \(-9 < k < 3\).

**M1**: Equate the line and the curve and rearrange into a three-term quadratic equation in \(x\) (e.g. \(x^2 - (3+k)x + 9 = 0\)). **M1**: Use the discriminant \(b^2 - 4ac < 0\) on their quadratic equation to obtain a quadratic inequality in terms of \(k\). **A1**: Obtain the correct range \(-9 < k < 3\) (or equivalent).

Since the terms are in a geometric progression, the ratio between consecutive terms is constant: \(r = \frac{a-4}{a} = \frac{a-6}{a-4}\). Multiplying both sides by \(a(a-4)\) gives \((a-4)^2 = a(a-6)\), which expands to \(a^2 - 8a + 16 = a^2 - 6a\). Subtracting \(a^2\) from both sides gives \(-8a + 16 = -6a\), which simplifies to \(2a = 16\), so \(a = 8\). The first term is \(a = 8\) and the common ratio is \(r = \frac{8-4}{8} = 0.5\). Since \(|r| < 1\), we use the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) to get \(S_{\infty} = \frac{8}{1 - 0.5} = 16\).

**M1**: Set up a correct equation for the common ratio (e.g., \((a-4)^2 = a(a-6)\)) and solve to find \(a = 8\). **M1**: Find the common ratio \(r = 0.5\) and use the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\) (provided \(|r| < 1\)). **A1**: Obtain the correct sum to infinity of 16.

First, we find the coordinates of the midpoint, \(M\), of \(AB\): \(M = (\frac{2+6}{2}, \frac{-1+7}{2}) = (4, 3)\). Next, we find the gradient, \(m\), of the line segment \(AB\): \(m = \frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2\). The gradient of the perpendicular bisector is the negative reciprocal of this gradient, which is \(m_{\perp} = -\frac{1}{2}\). Finally, we find the equation of the line with gradient \(-\frac{1}{2}\) passing through the midpoint \((4, 3)\): \(y - 3 = -\frac{1}{2}(x - 4)\), which simplifies to \(y - 3 = -\frac{1}{2}x + 2\), and thus \(y = -\frac{1}{2}x + 5\).

**M1**: Find the midpoint of \(AB\) as \((4, 3)\) and the gradient of \(AB\) as \(2\). **M1**: Use the perpendicular gradient relationship \(m_1 m_2 = -1\) to get \(m_{\perp} = -0.5\) and set up the equation of the line passing through their midpoint. **A1**: Obtain the correct equation \(y = -\frac{1}{2}x + 5\) (or equivalent).

To find the stationary points, we first find the derivative: \(\frac{dy}{dx} = 6x^2 - 18x + 12\). We set this derivative equal to 0 for stationary points: \(6x^2 - 18x + 12 = 0\). Dividing by 6 gives \(x^2 - 3x + 2 = 0\), which factorises to \((x - 1)(x - 2) = 0\), giving \(x = 1\) or \(x = 2\). Substituting these \(x\)-values back into the original curve equation: for \(x = 1\), \(y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 0\), giving the point \((1, 0)\); for \(x = 2\), \(y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = -1\), giving the point \((2, -1)\). Thus, the stationary points are \((1, 0)\) and \((2, -1)\).

**M1**: Differentiate to obtain \(\frac{dy}{dx} = 6x^2 - 18x + 12\) and set \(\frac{dy}{dx} = 0\). **M1**: Solve the quadratic equation to find \(x = 1\) and \(x = 2\) and substitute at least one of these into the original equation to find a \(y\)-coordinate. **A1**: Obtain the correct coordinates \((1, 0)\) and \((2, -1)\).

To find the points of intersection, we equate the equations of the line and the curve:
\(x^2 + 4x + 2 = kx + 1\)

Rearranging into a standard quadratic equation:
\(x^2 + (4 - k)x + 1 = 0\)

For the line and the curve to not intersect, the discriminant of this quadratic equation must be less than zero:
\(\Delta = b^2 - 4ac < 0\)

Substituting \(a = 1\), \(b = 4 - k\), and \(c = 1\):
\((4 - k)^2 - 4(1)(1) < 0\)
\(16 - 8k + k^2 - 4 < 0\)
\(k^2 - 8k + 12 < 0\)

Factorising the quadratic inequality:
\((k - 2)(k - 6) < 0\)

Therefore, the set of values of \(k\) is:
\(2 < k < 6\)

M1: For equating the line and the curve and forming a 3-term quadratic equation in \(x\).
M1: For using the discriminant \(\Delta < 0\).
A1: For obtaining the correct quadratic inequality in terms of \(k\), \(k^2 - 8k + 12 < 0\).
M1: For solving the quadratic inequality to find critical values (2 and 6) and choosing the inside region.
A1: For the correct final range \(2 < k < 6\).

(i) Let the first term of the arithmetic progression (AP) be \(a\) and the common difference be \(d\).
The terms of the AP are:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_{11} = a + 10d\)

The terms of the geometric progression (GP) with first term \(a\) and common ratio \(r\) are:
\(G_1 = a\)
\(G_2 = ar\)
\(G_3 = ar^2\)

Since \(T_1 = G_1 = a\), we have:
\(a + 2d = ar \implies 2d = a(r - 1) \implies d = \frac{a(r - 1)}{2}\) (Equation 1)

\(a + 10d = ar^2 \implies 10d = a(r^2 - 1)\) (Equation 2)

Substituting Equation 1 into Equation 2:
\(10\left(\frac{a(r - 1)}{2}\right) = a(r^2 - 1)\)
\(5a(r - 1) = a(r - 1)(r + 1)\)

Since \(d \neq 0\), we have \(a \neq 0\) and \(r \neq 1\). Thus, we can divide both sides by \(a(r - 1)\):
\(5 = r + 1 \implies r = 4\)

(ii) From Equation 1, since \(r = 4\):
\(d = \frac{a(4 - 1)}{2} = 1.5a\)

The sum of the first 10 terms of the AP is given by:
\(S_{10} = \frac{10}{2}[2a + 9d] = 310\)
\(5[2a + 9(1.5a)] = 310\)
\(2a + 13.5a = 62\)
\(15.5a = 62 \implies a = 4\)

For part (i):
M1: For expressing the terms of AP and GP in terms of \(a\), \(d\) and \(r\), and setting up two equations.
M1: For eliminating \(d\) to obtain an equation in \(a\) and \(r\).
A1: For obtaining \(r = 4\).

For part (ii):
M1: For relating \(d\) to \(a\) using their value of \(r\), and substituting into the AP sum formula.
A1: For obtaining \(a = 4\).

We begin by using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) to rewrite the numerator:
\(\sin \theta \tan \theta = \frac{\sin^2 \theta}{\cos \theta}\)

Using the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(\frac{\sin^2 \theta}{\cos \theta} = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{\cos \theta}\)

Substitute this back into the original equation:
\(\frac{\frac{(1 - \cos \theta)(1 + \cos \theta)}{\cos \theta}}{1 - \cos \theta} = 3 + \cos \theta\)

Since \(\cos \theta \neq 1\) for the expression to be defined, we can cancel the factor \((1 - \cos \theta)\):
\(\frac{1 + \cos \theta}{\cos \theta} = 3 + \cos \theta\)

Multiplying both sides by \(\cos \theta\):
\(1 + \cos \theta = 3\cos \theta + \cos^2 \theta\)

Rearranging into a quadratic equation in \(\cos \theta\):
\(\cos^2 \theta + 2\cos \theta - 1 = 0\)

Letting \(u = \cos \theta\), we solve the quadratic equation \(u^2 + 2u - 1 = 0\):
\(u = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}\)

Since \(-1 \le \cos \theta \le 1\), we reject \(-1 - \sqrt{2} \approx -2.414\).
Thus, \(\cos \theta = \sqrt{2} - 1 \approx 0.4142\).

Solving for \(\theta\) in the range \(0^\circ \le \theta \le 360^\circ\):
\(\theta = \cos^{-1}(0.4142) \approx 65.5^\circ\)
\(\theta = 360^\circ - 65.53^\circ \approx 294.5^\circ\)

M1: For substituting \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and using \(\sin^2 \theta = 1 - \cos^2 \theta\).
M1: For simplifying the fraction to obtain \(\frac{1+\cos\theta}{\cos\theta} = 3 + \cos\theta\) (or equivalent).
A1: For obtaining the correct quadratic equation in terms of \(\cos \theta\): \(\cos^2 \theta + 2\cos \theta - 1 = 0\).
M1: For solving the quadratic equation to find a valid value of \(\cos \theta = \sqrt{2}-1\) (approx 0.414) and attempting to find at least one angle.
A1: For both correct angles \(\theta = 65.5^\circ\) and \(\theta = 294.5^\circ\) (accept answers rounding to 1 decimal place; penalise extra solutions in the range).

(i) The formula for the total surface area \(A\) of a solid cylinder is:
\(A = 2\pi r^2 + 2\pi r h\)

Given that \(A = 150\pi\):
\(2\pi r^2 + 2\pi r h = 150\pi\)

Dividing by \(2\pi\):
\(r^2 + rh = 75 \implies h = \frac{75 - r^2}{r}\)

The volume \(V\) of the cylinder is:
\(V = \pi r^2 h\)

Substituting the expression for \(h\):
\(V = \pi r^2 \left(\frac{75 - r^2}{r}\right) = \pi r(75 - r^2) = 75\pi r - \pi r^3\) (as required).

(ii) To find the stationary value, we differentiate \(V\) with respect to \(r\):
\(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\)

At a stationary point, \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\):
\(75\pi - 3\pi r^2 = 0 \implies 3\pi r^2 = 75\pi \implies r^2 = 25\)

Since \(r > 0\), we have \(r = 5\) cm.

The stationary value of the volume is:
\(V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi\) \(\text{cm}^3\).

To determine the nature, we find the second derivative:
\(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\)

At \(r = 5\):
\(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi(5) = -30\pi\)

Since \(-30\pi < 0\), the stationary value is a maximum.

For part (i):
M1: For setting up the surface area equation \(2\pi r^2 + 2\pi rh = 150\pi\) and expressing \(h\) in terms of \(r\).
A1: For substituting into \(V = \pi r^2 h\) and correctly simplifying to show \(V = 75\pi r - \pi r^3\).

For part (ii):
M1: For differentiating \(V\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\), setting to 0, and finding \(r = 5\).
M1: For finding the second derivative and substituting \(r = 5\) (or using alternative method) to determine the nature.
A1: For finding the stationary value \(V = 250\pi\) (or approx 785) and concluding it is a maximum.

The volume \(V\) of the solid generated by rotating the region through \(360^\circ\) about the \(x\)-axis is given by:
\(V = \pi \int_{a}^{b} y^2 \mathrm{d}x\)

Here, \(y = 4(2x + 1)^{-3/4}\), so:
\(y^2 = \left[4(2x + 1)^{-3/4}\right]^2 = 16(2x + 1)^{-3/2}\)

Therefore, the volume is:
\(V = \pi \int_{0}^{4} 16(2x + 1)^{-3/2} \mathrm{d}x\)

Integrating with respect to \(x\):
\(\int 16(2x + 1)^{-3/2} \mathrm{d}x = 16 \left[ \frac{(2x + 1)^{-1/2}}{-\frac{1}{2} \times 2} \right] = -16(2x + 1)^{-1/2}\)

Now we evaluate the definite integral from 0 to 4:
\(V = \pi \left[ -16(2x + 1)^{-1/2} \right]_{0}^{4}\)

Substituting the limits:
At \(x = 4\): \(-16(9)^{-1/2} = -\frac{16}{3}\)
At \(x = 0\): \(-16(1)^{-1/2} = -16\)

Subtracting the lower limit from the upper limit:
\(V = \pi \left( -\frac{16}{3} - (-16) \right) = \pi \left( 16 - -\frac{16}{3} \right) = \frac{32}{3}\pi\)

M1: For squaring \(y\) correctly to obtain \(y^2 = 16(2x + 1)^{-3/2}\).
M1: For attempting to integrate \(y^2\) using the reverse chain rule (raising power by 1 and dividing by the new power and the coefficient of \(x\)).
A1: For obtaining the correct integrated expression \(-16(2x+1)^{-1/2}\) (ignoring \(\pi\)).
M1: For substituting the limits 0 and 4 correctly into their integrated expression and subtracting.
A1: For the correct exact answer \(\frac{32}{3}\pi\) (or equivalent exact fraction times \(\pi\)).

### Part (a) First, we find the center and radius of the circle \(C\) by completing the square: \(x^2 - 4x + y^2 - 2y - 20 = 0\) which simplifies to \((x - 2)^2 - 4 + (y - 1)^2 - 1 - 20 = 0\), giving \((x - 2)^2 + (y - 1)^2 = 25\). Thus, the center of the circle is \(P(2, 1)\) and the radius is \(R = 5\). For the line \(L\): \(3x + 4y - k = 0\) to be a tangent to the circle, the perpendicular distance from the center \(P(2, 1)\) to the line must equal the radius \(R = 5\). Using the perpendicular distance formula: \(\frac{|3(2) + 4(1) - k|}{\sqrt{3^2 + 4^2}} = 5\) which simplifies to \(\frac{|10 - k|}{5} = 5\), leading to \(|10 - k| = 25\). This gives two possible cases: 1) \(10 - k = 25 \implies k = -15\), 2) \(10 - k = -25 \implies k = 35\). Therefore, the two possible values of \(k\) are \(-15\) and \(35\). ### Part (b) For the larger value of \(k\), we have \(k = 35\), so the tangent line equation is \(3x + 4y = 35\). The line perpendicular to this tangent (the normal) passes through the center of the circle \((2, 1)\). The gradient of the tangent line is \(-\frac{3}{4}\), so the gradient of the normal line is \(\frac{4}{3}\). The equation of the normal line is: \(y - 1 = \frac{4}{3}(x - 2)\) which simplifies to \(3(y - 1) = 4(x - 2)\), giving \(3y - 3 = 4x - 8\) or \(4x - 3y = 5\). To find the point of contact, we solve the simultaneous equations of the tangent and the normal: 1) \(3x + 4y = 35\), 2) \(4x - 3y = 5\). Multiplying equation (1) by 3 and equation (2) by 4 gives: \(9x + 12y = 105\) and \(16x - 12y = 20\). Adding these two equations yields \(25x = 125 \implies x = 5\). Substituting \(x = 5\) into equation (1) gives \(3(5) + 4y = 35 \implies 15 + 4y = 35 \implies 4y = 20 \implies y = 5\). So, the coordinates of the point of contact are \((5, 5)\).

### Part (a) * **M1**: For attempting to find the center and radius of the circle by completing the square. * **A1**: For obtaining the correct center \((2, 1)\) and radius \(5\). * **M1**: For applying the perpendicular distance formula from the center to the line and setting it equal to the radius (or substituting the line equation into the circle equation and setting the discriminant \(\Delta = 0\)). * **A1**: For both correct values: \(k = -15\) and \(k = 35\). ### Part (b) * **M1**: For determining the equation of the normal line passing through the center \((2, 1)\) with gradient \(\frac{4}{3}\) (or attempting to substitute \(k = 35\) into the circle equation and solving the resulting quadratic). * **A1**: For obtaining the correct normal line equation \(4x - 3y = 5\). * **M1**: For attempting to solve the simultaneous equations representing the tangent and normal lines. * **A1**: For finding the correct point of contact \((5, 5)\).

### Part (a) The 1st, 2nd, and 5th terms of the arithmetic progression (AP) are: \(u_1 = a\), \(u_2 = a + d\), and \(u_5 = a + 4d\). Since these three terms form a geometric progression, the common ratio is constant: \(\frac{a+d}{a} = \frac{a+4d}{a+d}\). Cross-multiplying gives \((a + d)^2 = a(a + 4d)\), which expands to \(a^2 + 2ad + d^2 = a^2 + 4ad\). Subtracting \(a^2 + 2ad\) from both sides gives \(d^2 - 2ad = 0\) or \(d(d - 2a) = 0\). Since \(r \neq 1\), the common difference \(d\) cannot be 0 (as \(d=0\) would mean all terms are equal, leading to \(r=1\)). Therefore, we must have \(d = 2a\). ### Part (b) The common ratio \(r\) is given by \(r = \frac{u_2}{u_1} = \frac{a + d}{a}\). Substituting \(d = 2a\) into this expression yields \(r = \frac{a + 2a}{a} = \frac{3a}{a} = 3\). Thus, \(r = 3\). ### Part (c) The formula for the sum of the first \(n\) terms of an arithmetic progression is \(S_n = \frac{n}{2} [2a + (n - 1)d]\). For \(n = 15\), we have \(S_{15} = \frac{15}{2} [2a + 14d]\). Substituting \(d = 2a\) into this equation yields \(S_{15} = \frac{15}{2} [2a + 14(2a)] = \frac{15}{2} [30a] = 225a\). We are given that the sum is 675, so \(225a = 675\), which gives \(a = 3\).

### Part (a) * **M1**: For writing down the algebraic expressions for the three terms: \(a\), \(a+d\), and \(a+4d\). * **M1**: For setting up the geometric progression equation \((a+d)^2 = a(a+4d)\) and expanding it correctly. * **A1**: For solving to obtain \(d^2 - 2ad = 0\) and explaining why \(d = 2a\) (rejecting \(d=0\) since \(r \neq 1\)). ### Part (b) * **M1**: For expressing the common ratio \(r\) as \(\frac{a+d}{a}\) or \(\frac{a+4d}{a+d}\) and substituting \(d = 2a\). * **A1**: For obtaining \(r = 3\). ### Part (c) * **M1**: For using the AP sum formula with \(n = 15\). * **M1**: For substituting \(d = 2a\) and setting the sum equal to 675. * **A1**: For obtaining \(a = 3\).

Let \(u = e^x\). The equation becomes \(u^2 - 5u + 2 = 0\). Using the quadratic formula, we find the roots for \(u\): \(u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{17}}{2}\). Since both values are positive, we solve for \(x\) by taking the natural logarithm of both sides: \(e^x = \frac{5 \pm \sqrt{17}}{2} \implies x = \ln\left(\frac{5 \pm \sqrt{17}}{2}\right)\).

M1: Substitute \(u = e^x\) to form a quadratic equation and attempt to solve it. A1: Obtain the correct roots for \(e^x\), which are \(\frac{5 \pm \sqrt{17}}{2}\). A1: State both correct exact solutions in the required logarithmic form: \(x = \ln\left(\frac{5 \pm \sqrt{17}}{2}\right)\).

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we rewrite the equation as \((1 + \tan^2 \theta) + 2\tan\theta = 4\). This simplifies to the quadratic equation \(\tan^2 \theta + 2\tan\theta - 3 = 0\). Factorizing this quadratic gives \(\left(\tan\theta + 3\right)\left(\tan\theta - 1\right) = 0\), which has solutions \(\tan\theta = 1\) or \(\tan\theta = -3\). Within the interval \(0^\circ \le \theta \le 180^\circ\): For \(\tan\theta = 1\), we have \(\theta = 45^\circ\). For \(\tan\theta = -3\), we have \(\theta = 180^\circ - 71.57^\circ = 108.4^\circ\) (to 1 decimal place).

M1: Use the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to obtain a quadratic equation in \(\tan\theta\) and attempt to solve it. A1: Obtain the correct values \(\tan\theta = 1\) and \(\tan\theta = -3\). A1: Obtain both correct angles: \(\theta = 45^\circ\) and \(\theta = 108.4^\circ\), with no extra values in the given range.

Using the factor theorem, since \(2x - 1\) is a factor, \(p\left(\frac{1}{2}\right) = 0\).
Substituting \(x = \frac{1}{2}\):
\(2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) - 6 = 0\)
\(\frac{1}{4} + \frac{a}{4} + \frac{b}{2} - 6 = 0\)
Multiplying by 4:
\(1 + a + 2b - 24 = 0 \implies a + 2b = 23\) [Equation 1]

Using the remainder theorem, since the remainder when divided by \(x + 2\) is \(-30\), we have \(p(-2) = -30\).
Substituting \(x = -2\):
\(2(-2)^3 + a(-2)^2 + b(-2) - 6 = -30\)
\(-16 + 4a - 2b - 6 = -30\)
\(4a - 2b = -8 \implies 2a - b = -4\) [Equation 2]

Solving the simultaneous equations:
From [Equation 2], \(b = 2a + 4\).
Substitute into [Equation 1]:
\(a + 2(2a + 4) = 23 \implies 5a + 8 = 23 \implies 5a = 15 \implies a = 3\).
Substitute \(a = 3\) back into the expression for \(b\):
\(b = 2(3) + 4 = 10\).
Thus, the values are \(a = 3\) and \(b = 10\).

* M1: Attempt to apply the factor theorem by setting \(p\left(\frac{1}{2}\right) = 0\)
* A1: Obtain a correct equation in \(a\) and \(b\), e.g., \(a + 2b = 23\)
* M1: Attempt to apply the remainder theorem by setting \(p(-2) = -30\)
* A1: Obtain a second correct equation, e.g., \(2a - b = -4\)
* A1: Solve simultaneously to obtain \(a = 3\) and \(b = 10\) (both required)

Let \(y = 5^x\). The equation \(5^{2x} - 5^{x+1} + 4 = 0\) can be written as:
\((5^x)^2 - 5(5^x) + 4 = 0 \implies y^2 - 5y + 4 = 0\)

Factorising the quadratic equation:
\((y - 1)(y - 4) = 0\)
So \(y = 1\) or \(y = 4\).

Case 1: \(5^x = 1\)
\(x = 0\)

Case 2: \(5^x = 4\)
Taking natural logarithms on both sides:
\(\ln(5^x) = \ln(4) \implies x \ln(5) = \ln(4)\)
\(x = \frac{\ln(4)}{\ln(5)} \approx 0.86135\)

Correct to 3 significant figures, the solutions are \(x = 0\) and \(x = 0.861\).

* M1: Substitute \(y = 5^x\) to obtain a quadratic equation of the form \(y^2 - 5y + 4 = 0\) or equivalent
* A1: Solve the quadratic equation to find \(5^x = 1\) or \(5^x = 4\)
* M1: Apply logarithms correctly to solve \(5^x = k\) for a positive constant \(k\)
* A1: Obtain \(x = 0\)
* A1: Obtain \(x = 0.861\) (accept 0.86)

We find the indefinite integral first:
\(\int (2\sec^2(x) - \sin(2x)) \, dx = 2\tan(x) - \left(-\frac{1}{2}\cos(2x)\right) + C = 2\tan(x) + \frac{1}{2}\cos(2x) + C\)

Now we apply the limits \(0\) and \(\frac{\pi}{3}\):
\(\left[ 2\tan(x) + \frac{1}{2}\cos(2x) \right]_{0}^{\frac{\pi}{3}}\)

At the upper limit \(x = \frac{\pi}{3}\):
\(2\tan\left(\frac{\pi}{3}\right) + \frac{1}{2}\cos\left(\frac{2\pi}{3}\right) = 2(\sqrt{3}) + \frac{1}{2}\left(-\frac{1}{2}\right) = 2\sqrt{3} - \frac{1}{4}\)

At the lower limit \(x = 0\):
\(2\tan(0) + \frac{1}{2}\cos(0) = 2(0) + \frac{1}{2}(1) = \frac{1}{2}\)

Subtracting the lower limit value from the upper limit value:
\(\left(2\sqrt{3} - \frac{1}{4}\right) - \frac{1}{2} = 2\sqrt{3} - \frac{3}{4}\)

Thus, the exact value of the integral is \(2\sqrt{3} - \frac{3}{4}\).

* B1: State \(2\tan(x)\) as the integral of \(2\sec^2(x)\)
* M1: Integrate \(\sin(2x)\) to obtain \(\pm k \cos(2x)\) where \(k = \frac{1}{2}\)
* A1: Obtain the correct integrated expression \(2\tan(x) + \frac{1}{2}\cos(2x)\)
* M1: Correctly substitute limits \(\frac{\pi}{3}\) and \(0\) into their integrated expression
* A1: Obtain the exact value \(2\sqrt{3} - \frac{3}{4}\) (or equivalent single fraction like \(\frac{8\sqrt{3} - 3}{4}\))

We first use the double-angle identity for \(\sin^2(3x)\):
\[\sin^2(3x) = \frac{1 - \cos(6x)}{2}\]

Substituting this into the integrand:
\[8\sin^2(3x) = 4(1 - \cos(6x)) = 4 - 4\cos(6x)\]

Thus, the integral is:
\[\int_0^{\frac{\pi}{6}} \left( 4 - 4\cos(6x) + \sec^2(2x) \right) \text{d}x\]

Now we integrate each term:
\[\int 4 \, \text{d}x = 4x\]
\[\int -4\cos(6x) \, \text{d}x = -\frac{4}{6}\sin(6x) = -\frac{2}{3}\sin(6x)\]
\[\int \sec^2(2x) \, \text{d}x = \frac{1}{2}\tan(2x)\]

So the antiderivative is:
\[F(x) = 4x - \frac{2}{3}\sin(6x) + \frac{1}{2}\tan(2x)\]

Now we evaluate from \(0\) to \(\frac{\pi}{6}\):
At the upper limit \(x = \frac{\pi}{6}\):
\[F\left(\frac{\pi}{6}\right) = 4\left(\frac{\pi}{6}\right) - \frac{2}{3}\sin\left(6 \cdot \frac{\pi}{6}\right) + \frac{1}{2}\tan\left(2 \cdot \frac{\pi}{6}\right)\]
\[= \frac{2\pi}{3} - \frac{2}{3}\sin(\pi) + \frac{1}{2}\tan\left(\frac{\pi}{3}\right)\]
Since \(\sin(\pi) = 0\) and \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\):
\[F\left(\frac{\pi}{6}\right) = \frac{2\pi}{3} + \frac{1}{2}\sqrt{3}\]

At the lower limit \(x = 0\):
\[F(0) = 4(0) - \frac{2}{3}\sin(0) + \frac{1}{2}\tan(0) = 0\]

Subtracting the lower limit value from the upper limit value gives the exact value:
\[\frac{2\pi}{3} + \frac{\sqrt{3}}{2}\]

M1: Attempt to use double angle formula to write \(\sin^2(3x)\) in the form \(a + b\cos(6x)\)
A1: Obtain correct integrand \(4 - 4\cos(6x) + \sec^2(2x)\)
M1: Integrate to obtain an expression of the form \(px + q\sin(6x) + r\tan(2x)\) where \(p, q, r \neq 0\)
A1: Obtain correct antiderivative \(4x - \frac{2}{3}\sin(6x) + \frac{1}{2}\tan(2x)\) (allow unsimplified)
M1: Correctly substitute limits \(0\) and \(\frac{\pi}{6}\) into their integrated expression
A1: Obtain exact answer \(\frac{2}{3}\pi + \frac{1}{2}\sqrt{3}\) or equivalent single fraction

To solve the equation \(|3e^y - 5| = e^y + 3\), we can set up two equations representing the two cases of the modulus:

**Case 1:**
\[3e^y - 5 = e^y + 3\]
\[2e^y = 8\]
\[e^y = 4\]
Taking the natural logarithm of both sides:
\[y = \ln 4\]
Which can also be written as \(y = 2\ln 2\).

**Case 2:**
\[3e^y - 5 = -(e^y + 3)\]
\[3e^y - 5 = -e^y - 3\]
\[4e^y = 2\]
\[e^y = \frac{1}{2}\]
Taking the natural logarithm of both sides:
\[y = \ln\left(\frac{1}{2}\right)\]
Which can also be written as \(y = -\ln 2\).

**Alternative method (squaring both sides):**
Let \(u = e^y\). The equation is \(|3u - 5| = u + 3\).
Squaring both sides:
\[(3u - 5)^2 = (u + 3)^2\]
\[9u^2 - 30u + 25 = u^2 + 6u + 9\]
\[8u^2 - 36u + 16 = 0\]
Dividing by 4:
\[2u^2 - 9u + 4 = 0\]
Factoring the quadratic:
\[(2u - 1)(u - 4) = 0\]
So \(u = \frac{1}{2}\) or \(u = 4\).
Since \(u = e^y\), we have:
\(e^y = 4 \implies y = \ln 4 = 2\ln 2\)
\(e^y = \frac{1}{2} \implies y = \ln\left(\frac{1}{2}\right) = -\ln 2\).

Both values satisfy the original equation as both lead to positive values on the right-hand side (since \(e^y > 0\)).

M1: State a correct method for removing the modulus signs (either by setting up two linear equations or squaring both sides)
A1: Obtain the simplified linear equations \(2e^y = 8\) and \(4e^y = 2\) (or the quadratic equation \(2u^2 - 9u + 4 = 0\) where \(u = e^y\))
A1: Solve to find \(e^y = 4\) and \(e^y = \frac{1}{2}\)
M1: Use natural logarithms correctly to solve \(e^y = k\) for \(y\) (where \(k > 0\))
A1: Obtain \(y = \ln 4\) (or \(2\ln 2\))
A1: Obtain \(y = -\ln 2\) (or \(\ln\frac{1}{2}\)) and no other solutions

First, express the left-hand side as a single logarithm using the division law: \(\ln\left(\frac{2x+1}{x}\right) = \ln(3) + 1\). Since \(1 = \ln(\mathrm{e})\), we can write the equation as \(\ln\left(\frac{2x+1}{x}\right) = \ln(3) + \ln(\mathrm{e}) = \ln(3\mathrm{e})\). Taking exponentials of both sides gives \(\frac{2x+1}{x} = 3\mathrm{e}\). Multiply by \(x\) to obtain \(2x + 1 = 3\mathrm{e}x\). Rearranging the terms to solve for \(x\) yields \(x(3\mathrm{e} - 2) = 1\), which gives \(x = \frac{1}{3\mathrm{e} - 2}\).

M1: Apply the subtraction law of logarithms to write the left-hand side as a single logarithm, or write the right-hand side as \(\ln(3\mathrm{e})\). M1: Remove logarithms correctly to obtain a linear equation in \(x\). A1: Obtain the correct exact answer \(x = \frac{1}{3\mathrm{e} - 2}\).

Rewrite the expression as \((4 - x)(1 + 2x)^{-2}\). The binomial expansion of \((1 + 2x)^{-2}\) is given by \(1 + (-2)(2x) + \frac{(-2)(-3)}{2!}(2x)^2 + \dots = 1 - 4x + 12x^2 - \dots\). Multiplying this expansion by \(4 - x\) gives \((4 - x)(1 - 4x + 12x^2) = 4(1 - 4x + 12x^2) - x(1 - 4x) + \dots\). The term in \(x^2\) is \(4(12x^2) - x(-4x) = 48x^2 + 4x^2 = 52x^2\). Thus, the coefficient of \(x^2\) is 52.

M1: Attempt to expand \((1+2x)^{-2}\) up to the term in \(x^2\) with at least one correct binomial coefficient. A1: Obtain the correct three-term expansion \(1 - 4x + 12x^2\). A1: Obtain the correct final coefficient of 52.

Multiply the numerator and denominator of \(u\) by the conjugate of the denominator, \(1 - 2\mathrm{i}\): \(u = \frac{(a + 3\mathrm{i})(1 - 2\mathrm{i})}{(1 + 2\mathrm{i})(1 - 2\mathrm{i})} = \frac{a - 2a\mathrm{i} + 3\mathrm{i} + 6}{1^2 + 2^2} = \frac{(a + 6) + (3 - 2a)\mathrm{i}}{5}\). The real part of \(u\) is \(\frac{a+6}{5}\). Since the real part is 2, we have \(\frac{a+6}{5} = 2\), which gives \(a + 6 = 10\), so \(a = 4\). Substituting \(a = 4\) back into the expression for \(u\) gives \(u = \frac{10 - 5\mathrm{i}}{5} = 2 - \mathrm{i}\).

M1: Multiply numerator and denominator by the conjugate \(1-2\mathrm{i}\) to separate real and imaginary parts. A1: Equate the real part to 2 and solve to find \(a = 4\). A1: Obtain the final form \(u = 2 - \mathrm{i}\).

Using the product rule to differentiate \(y = \mathrm{e}^{2x} \tan(x)\), we get \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}\tan(x) + \mathrm{e}^{2x}\sec^2(x)\). Substituting \(x = \frac{\pi}{4}\): since \(2x = \frac{\pi}{2}\), \(\tan\left(\frac{\pi}{4}\right) = 1\), and \(\sec^2\left(\frac{\pi}{4}\right) = 2\), we find the gradient is \(2\mathrm{e}^{\pi/2}(1) + \mathrm{e}^{\pi/2}(2) = 4\mathrm{e}^{\pi/2}\).

M1: Apply the product rule to obtain a derivative of the form \(a\mathrm{e}^{2x}\tan(x) + b\mathrm{e}^{2x}\sec^2(x)\). A1: Obtain the correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}\tan(x) + \mathrm{e}^{2x}\sec^2(x)\). A1: Obtain the correct exact gradient \(4\mathrm{e}^{\pi/2}\).

Let \( u = 2^x \).

Then \( 2^{2x+1} = 2 \cdot (2^x)^2 = 2u^2 \).

The equation becomes:
\( 2u^2 - 7u + 3 = 0 \)

Factoring the quadratic equation:
\( (2u - 1)(u - 3) = 0 \)

This gives:
\( u = 0.5 \) or \( u = 3 \)

Case 1: \( 2^x = 0.5 \)
\( x = -1 \)

Case 2: \( 2^x = 3 \)
\( x = \frac{\ln 3}{\ln 2} \approx 1.58 \) (to 3 s.f.)

Thus, the solutions are \( x = -1 \) and \( x \approx 1.58 \).

M1: Substitute \( u = 2^x \) or equivalent to obtain a quadratic equation in terms of \( 2^x \).
A1: Obtain the correct quadratic equation \( 2u^2 - 7u + 3 = 0 \).
M1: Solve the quadratic equation to find \( u = 0.5 \) and \( u = 3 \).
A1: Obtain \( x = -1 \).
A1: Obtain \( x = 1.58 \) (accept \( \log_2 3 \) or exact equivalents).

First, find the coordinates of the point where \( t = 1 \):
\( x = 1^2 + 3(1) = 4 \)
\( y = \ln(2(1) + 1) = \ln 3 \)

Next, differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = 2t + 3 \)
At \( t = 1 \), \( \frac{dx}{dt} = 2(1) + 3 = 5 \).

\( \frac{dy}{dt} = \frac{2}{2t+1} \)
At \( t = 1 \), \( \frac{dy}{dt} = \frac{2}{2(1)+1} = \frac{2}{3} \).

Use the chain rule to find the gradient of the tangent, \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2/3}{5} = \frac{2}{15} \).

Now, find the equation of the tangent line using \( y - y_1 = m(x - x_1) \):
\( y - \ln 3 = \frac{2}{15}(x - 4) \)

Multiply through by 15:
\( 15y - 15\ln 3 = 2x - 8 \)
\( 15y = 2x - 8 + 15\ln 3 \)

Here, \( a = 15 \), \( b = 2 \), \( c = -8 \), \( p = 15 \), and \( d = 3 \).

B1: Identify coordinates \( (4, \ln 3) \) when \( t = 1 \).
M1: Attempt differentiation of parametric equations with respect to \( t \).
A1: Obtain correct gradient \( \frac{dy}{dx} = \frac{2}{15} \) at \( t = 1 \).
M1: Correctly apply tangent equation formula using their point and gradient.
A1: Obtain final equation in the required form: \( 15y = 2x - 8 + 15\ln 3 \) (or any integer multiple).

Let \( u = \sqrt{2x+5} \). Then \( u^2 = 2x+5 \), which gives \( x = \frac{u^2-5}{2} \).
Differentiating with respect to \( u \) gives \( dx = u \, du \).

Change the limits of integration:
When \( x = 0 \), \( u = \sqrt{5} \).
When \( x = 2 \), \( u = 3 \).

Substitute these into the integral:
\( \int_{0}^{2} \frac{x}{\sqrt{2x+5}} \, dx = \int_{\sqrt{5}}^{3} \frac{\frac{u^2-5}{2}}{u} (u \, du) \)
\( = \int_{\sqrt{5}}^{3} \frac{u^2-5}{2} \, du \)
\( = \left[ \frac{u^3}{6} - \frac{5u}{2} \right]_{\sqrt{5}}^{3} \)

Evaluate at the upper limit \( u = 3 \):
\( \left(\frac{27}{6} - \frac{15}{2}\right) = \frac{9}{2} - \frac{15}{2} = -3 \).

Evaluate at the lower limit \( u = \sqrt{5} \):
\( \left(\frac{5\sqrt{5}}{6} - \frac{5\sqrt{5}}{2}\right) = \frac{5\sqrt{5}}{6} - \frac{15\sqrt{5}}{6} = -\frac{10\sqrt{5}}{6} = -\frac{5\sqrt{5}}{3} \).

Subtracting limits:
\( -3 - \left(-\frac{5\sqrt{5}}{3}\right) = \frac{5\sqrt{5}}{3} - 3 \).

Thus, \( a = \frac{5}{3} \) and \( b = -3 \).

M1: For using substitution \( u = \sqrt{2x+5} \) to get \( dx = u \, du \) (or equivalent integration by parts method).
A1: Transform integral correctly to \( \int \frac{u^2-5}{2} \, du \).
M1: Integrate polynomial correctly to obtain \( \frac{u^3}{6} - \frac{5u}{2} \) (or equivalent expression in \( x \)).
M1: Substitute limits (either in terms of \( u \) or \( x \)) into their integrated expression.
A1: Obtain the correct final exact value \( \frac{5}{3}\sqrt{5} - 3 \).

Let \( z = x + \mathrm{i}y \) where \( x \) and \( y \) are real. Then \( z^* = x - \mathrm{i}y \).

Substitute into the equation:
\( 3(x + \mathrm{i}y) + \mathrm{i}(x - \mathrm{i}y) = 10 + 14\mathrm{i} \)
\( 3x + 3\mathrm{i}y + \mathrm{i}x + y = 10 + 14\mathrm{i} \)

Equating real and imaginary parts:
Real part: \( 3x + y = 10 \)
Imaginary part: \( x + 3y = 14 \)

From \( 3x + y = 10 \), we have \( y = 10 - 3x \). Substitute this into the second equation:
\( x + 3(10 - 3x) = 14 \)
\( x + 30 - 9x = 14 \)
\( -8x = -16 \implies x = 2 \)

Then \( y = 10 - 3(2) = 4 \).
So, \( z = 2 + 4\mathrm{i} \).

Modulus \( |z| \):
\( |z| = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \)

Argument \( \arg(z) \):
Since \( z \) lies in the first quadrant:
\( \arg(z) = \tan^{-1}\left(\frac{4}{2}\right) = \tan^{-1}(2) \approx 1.11 \) radians.

M1: Substitute \( z = x + \mathrm{i}y \) and \( z^* = x - \mathrm{i}y \) and equate real and imaginary parts.
A1: Obtain correct simultaneous equations \( 3x + y = 10 \) and \( x + 3y = 14 \).
A1: Solve equations to find \( z = 2 + 4\mathrm{i} \).
A1: Find modulus \( 2\sqrt{5} \) (or equivalent exact form, accept decimal \( 4.47 \)).
A1: Find argument \( 1.11 \) radians.

Let \( \frac{5x-1}{(1-2x)(1+x)} = \frac{A}{1-2x} + \frac{B}{1+x} \).

Multiplying through by the denominator:
\( 5x - 1 = A(1+x) + B(1-2x) \)

Let \( x = -1 \):
\( -6 = B(3) \implies B = -2 \).

Let \( x = 0.5 \):
\( 1.5 = A(1.5) \implies A = 1 \).

So, \( \frac{5x-1}{(1-2x)(1+x)} = (1-2x)^{-1} - 2(1+x)^{-1} \).

Now, expand each part using the binomial theorem:
\( (1-2x)^{-1} = 1 + (-1)(-2x) + \frac{(-1)(-2)}{2!}(-2x)^2 + \dots = 1 + 2x + 4x^2 + \dots \)

\( -2(1+x)^{-1} = -2(1 - x + x^2 - \dots) = -2 + 2x - 2x^2 + \dots \)

Adding these expansions together:
\( (1 + 2x + 4x^2) + (-2 + 2x - 2x^2) = -1 + 4x + 2x^2 \).

M1: Attempt to express the fraction in partial fraction form and solve for constants \( A \) and \( B \).
A1: Obtain correct partial fractions \( \frac{1}{1-2x} - \frac{2}{1+x} \).
M1: Differentiate or expand either term up to \( x^2 \) using binomial expansion.
A1: Find correct expansions for both terms (e.g. \( 1+2x+4x^2 \) and \( -2+2x-2x^2 \)).
A1: Combine terms correctly to find the final expansion \( -1 + 4x + 2x^2 \).

First, separate the variables:
\[ \int \frac{1}{y^2 + 1} \mathrm{d}y = \int \frac{1}{x \ln x} \mathrm{d}x \]

Integrate the left-hand side:
\[ \int \frac{1}{y^2 + 1} \mathrm{d}y = \arctan y \]

Integrate the right-hand side using the substitution \(u = \ln x\), which gives \(\mathrm{d}u = \frac{1}{x} \mathrm{d}x\):
\[ \int \frac{1}{x \ln x} \mathrm{d}x = \int \frac{1}{u} \mathrm{d}u = \ln|u| = \ln(\ln x) \] (since \(x > 1\), we have \
\ln x > 0\).

Combine the results and include a constant of integration \(C\):
\[ \arctan y = \ln(\ln x) + C \]

Using the boundary condition \(y = 0\) when \(x = \mathrm{e}\):
\[ \arctan 0 = \ln(\ln \mathrm{e}) + C \]
\[ 0 = \ln 1 + C \implies C = 0 \]

Thus, the solution is:
\[ \arctan y = \ln(\ln x) \]

Expressing \(y\) explicitly in terms of \(x\):
\[ y = \tan(\ln(\ln x)) \]

M1: For separating variables and attempting to integrate both sides
A1: For obtaining \(\arctan y\) (or equivalent)
M1: For attempting to integrate \(\frac{1}{x \ln x}\) (e.g., using substitution \(u = \ln x\))
A1: For obtaining \(\ln(\ln x)\)
B1: For including a constant of integration at some stage
M1: For substituting \(y = 0\) and \(x = \mathrm{e}\) into an equation containing a constant of integration
A1: For finding \(C = 0\)
A1: For obtaining the final correct explicit expression \(y = \tan(\ln(\ln x))\)

(a) To express \(w\) in the form \(x + \mathrm{i}y\), multiply the numerator and denominator by the conjugate of the denominator, which is \(2 + \mathrm{i}\):
\[ w = \frac{(6 + 7\mathrm{i})(2 + \mathrm{i})}{(2 - \mathrm{i})(2 + \mathrm{i})} \]
\[ w = \frac{12 + 6\mathrm{i} + 14\mathrm{i} - 7}{4 + 1} \]
\[ w = \frac{5 + 20\mathrm{i}}{5} = 1 + 4\mathrm{i} \]

(b) The quadratic equation is:
\[ z^2 + 4z + (1 + 4\mathrm{i}) = 0 \]

We complete the square:
\[ (z + 2)^2 - 4 + 1 + 4\mathrm{i} = 0 \]
\[ (z + 2)^2 = 3 - 4\mathrm{i} \]

Let \(p + \mathrm{i}q\) be a square root of \(3 - 4\mathrm{i}\), where \(p\) and \(q\) are real:
\[ (p + \mathrm{i}q)^2 = 3 - 4\mathrm{i} \implies p^2 - q^2 + 2pq\mathrm{i} = 3 - 4\mathrm{i} \]

Equating real and imaginary parts:
\[ p^2 - q^2 = 3 \]
\[ 2pq = -4 \implies q = -\frac{2}{p} \]

Substitute \(q = -\frac{2}{p}\) into \(p^2 - q^2 = 3\):
\[ p^2 - \frac{4}{p^2} = 3 \implies p^4 - 3p^2 - 4 = 0 \]
\[ (p^2 - 4)(p^2 + 1) = 0 \]

Since \(p\) is real, \(p^2 = 4\), which gives \(p = \pm 2\).
- If \(p = 2\), then \(q = -1\).
- If \(p = -2\), then \(q = 1\).

So the square roots of \(3 - 4\mathrm{i}\) are \(\pm(2 - \mathrm{i})\).

Thus:
\[ z + 2 = \pm(2 - \mathrm{i}) \]

Using the positive sign:
\[ z = -2 + 2 - \mathrm{i} = -\mathrm{i} \]

Using the negative sign:
\[ z = -2 - 2 + \mathrm{i} = -4 + \mathrm{i} \]

So the solutions are \(z = -\mathrm{i}\) and \(z = -4 + \mathrm{i}\).

Part (a):
M1: For multiplying the numerator and denominator by \(2 + \mathrm{i}\)
A1: For obtaining the correct form \(1 + 4\mathrm{i}\)

Part (b):
M1: For attempting to complete the square or use the quadratic formula on \(z^2 + 4z + w = 0\)
A1: For obtaining \((z + 2)^2 = 3 - 4\mathrm{i}\) or the equivalent discriminant \(\Delta = 12 - 16\mathrm{i}\)
M1: For setting up equations for the real and imaginary parts of the square root of \(3 - 4\mathrm{i}\) (or \(12 - 16\mathrm{i}\)) and solving for \(p\) or \(q\)
A1: For finding the correct values \(p = \pm 2, q = \mp 1\) (or \(p = \pm 4, q = \mp 2\))
A1: For obtaining the square roots \(\pm (2 - \mathrm{i})\) (or \(\pm (4 - 2\mathrm{i})\))
A1: For obtaining both correct final solutions: \(z = -\mathrm{i}\) and \(z = -4 + \mathrm{i}\)

Let the direction of the initial velocity of \(A\) be the positive direction.

Before the collision:
- Velocity of \(A\), \(u_A = 6\text{ m s}^{-1}\)
- Velocity of \(B\), \(u_B = -2\text{ m s}^{-1}\)

After the collision:
- Velocity of \(A\), \(v_A = -1.5\text{ m s}^{-1}\) (since its direction is reversed)
- Let the velocity of \(B\) be \(v_B\).

Using the principle of conservation of momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)

Substitute the known values:
\(0.3(6) + 0.5(-2) = 0.3(-1.5) + 0.5 v_B\)
\(1.8 - 1.0 = -0.45 + 0.5 v_B\)
\(0.8 = -0.45 + 0.5 v_B\)

Solve for \(v_B\):
\(0.5 v_B = 1.25\)
\(v_B = 2.5\text{ m s}^{-1}\)

Therefore, the speed of \(B\) after the collision is \(2.5\text{ m s}^{-1}\).

M1: For attempting to use the principle of conservation of momentum with correct signs for opposing directions.
A1: For a correct equation, e.g., \(0.3(6) - 0.5(2) = 0.3(-1.5) + 0.5 v_B\) (or equivalent).
A1: For obtaining speed \(2.5\text{ m s}^{-1}\) (or \(2.5\)).

Let \(R\) be the normal reaction force and \(F\) be the frictional force.

Since the particle is on the point of sliding up the slope, the frictional force acts down the slope.

Resolving forces perpendicular to the inclined plane:
\(R = mg \cos 30^\circ\)
\(R = 4 \times 10 \cos 30^\circ = 40 \cos 30^\circ = 20\sqrt{3}\text{ N} \approx 34.641\text{ N}\)

The friction is limiting, so:
\(F = \mu R = 0.3 \times 20\sqrt{3} = 6\sqrt{3}\text{ N} \approx 10.392\text{ N}\)

Resolving forces parallel to the inclined plane:
\(P - mg \sin 30^\circ - F = 0\)
\(P = 40 \sin 30^\circ + 6\sqrt{3}\)
\(P = 20 + 6\sqrt{3} \approx 30.4\text{ N}\) (to 3 significant figures).

M1: For resolving forces perpendicular to the plane to find \(R\) and using \(F = \mu R\).
M1: For resolving forces parallel to the plane to obtain an equation containing \(P\), \(mg \sin 30^\circ\), and \(F\).
A1: For obtaining \(P = 30.4\) (accept \(20 + 6\sqrt{3}\)).

The power of the engine is \(P = 36\text{ kW} = 36\text{ }000\text{ W}\).
At speed \(v = 15\text{ m s}^{-1}\), the driving force \(F\) of the engine is:
\(F = \frac{P}{v} = \frac{36\text{ }000}{15} = 2400\text{ N}\).

Applying Newton's second law for the motion of the car up the incline:
\(F - R - mg\sin\theta = ma\)

Substitute the known values into the equation, with \(m = 1200\text{ kg}\), \(g = 10\text{ m s}^{-2}\), \(\sin\theta = 0.05\), and \(a = 0.2\text{ m s}^{-2}\):
\(2400 - R - 1200 \times 10 \times 0.05 = 1200 \times 0.2\)
\(2400 - R - 600 = 240\)
\(1800 - R = 240\)
\(R = 1560\).

M1: Use of \(P = Fv\) to find the driving force.
A1: Obtain \(F = 2400\text{ N}\).
M1: Attempt Newton's second law along the slope, including driving force, resistance, and component of weight.
A1: Correct substitution of values into the equation of motion: \(2400 - R - 600 = 240\).
A1: Correct final answer \(R = 1560\).

First, resolve forces perpendicular to the plane to find the normal reaction force \(R_N\):
\(R_N = mg\cos 30^\circ = 8 \times 10 \times \cos 30^\circ = 40\sqrt{3}\text{ N} \approx 69.28\text{ N}\).

The maximum available frictional force is:
\(F_{\text{max}} = \mu R_N = 0.4 \times 40\sqrt{3} = 16\sqrt{3}\text{ N} \approx 27.71\text{ N}\).

The component of the weight acting down the incline is:
\(W_{\parallel} = mg\sin 30^\circ = 8 \times 10 \times \sin 30^\circ = 40\text{ N}\).

Case 1: The block is on the point of slipping down the plane. The frictional force acts up the plane:
\(P_{\text{min}} + F_{\text{max}} = mg\sin 30^\circ\)
\(P_{\text{min}} = 40 - 16\sqrt{3} \approx 12.3\text{ N}\).

Case 2: The block is on the point of slipping up the plane. The frictional force acts down the plane:
\(P_{\text{max}} = mg\sin 30^\circ + F_{\text{max}}\)
\(P_{\text{max}} = 40 + 16\sqrt{3} \approx 67.7\text{ N}\).

Therefore, the range of possible values for equilibrium is \(12.3 \le P \le 67.7\).

M1: Resolve forces perpendicular to the slope to find the normal reaction \(R_N = 80\cos 30^\circ\).
A1: Correct maximum friction force \(F_{\text{max}} = 0.4 \times 40\sqrt{3} = 27.71\text{ N}\).
M1: Set up equilibrium equation for minimum force (friction acting up-slope): \(P_{\text{min}} = mg\sin 30^\circ - F_{\text{max}}\).
A1: Obtain \(P_{\text{min}} = 12.3\) (3 s.f.).
A1: Set up and solve equilibrium equation for maximum force (friction acting down-slope), obtaining \(P_{\text{max}} = 67.7\) (3 s.f.) and write the final range.

First, find the times when the particle is instantaneously at rest by setting \(v(t) = 0\):
\(3t^2 - 12t + 9 = 0\)
\(3(t^2 - 4t + 3) = 0\)
\(3(t - 1)(t - 3) = 0\)
So the particle is at rest at \(t = 1\) and \(t = 3\).

Since we need the total distance travelled in the interval \([0, 4]\), we must integrate the absolute value of velocity, splitting the integration at the turning points:
\(d = \int_{0}^{1} v(t)\,dt + \left|\int_{1}^{3} v(t)\,dt\right| + \int_{3}^{4} v(t)\,dt\).

Find the general antiderivative for displacement \(s(t)\):
\(s(t) = \int (3t^2 - 12t + 9)\,dt = t^3 - 6t^2 + 9t + C\).
Taking \(C = 0\), we evaluate \(s(t)\) at the boundaries:
\(s(0) = 0\)
\(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\)
\(s(3) = 3^3 - 6(3)^2 + 9(3) = 0\)
\(s(4) = 4^3 - 6(4)^2 + 9(4) = 4\).

Calculate the distances in each sub-interval:
- From \(t = 0\) to \(t = 1\): distance is \(|s(1) - s(0)| = |4 - 0| = 4\text{ m}\).
- From \(t = 1\) to \(t = 3\): distance is \(|s(3) - s(1)| = |0 - 4| = 4\text{ m}\).
- From \(t = 3\) to \(t = 4\): distance is \(|s(4) - s(3)| = |4 - 0| = 4\text{ m}\).

The total distance travelled is \(4 + 4 + 4 = 12\text{ m}\).

M1: Set \(v(t) = 0\) and solve to find \(t = 1\) and \(t = 3\).
M1: Integrate \(v(t)\) to find displacement function \(s(t) = t^3 - 6t^2 + 9t\) (ignore constant).
A1: Evaluate the displacement at key times correctly: \(s(0)=0, s(1)=4, s(3)=0, s(4)=4\).
M1: Show a clear method of summing the absolute changes in position to find total distance.
A1: Obtain the correct total distance of \(12\text{ m}\).

**(i)**
At a constant speed, the acceleration of the car is zero.
Let the driving force be \(F\text{ N}\).
Resolving parallel to the slope:
\(F - mg \sin \alpha - R = 0\)
\(F = (1200 \times 10 \times 0.05) + 400\)
\(F = 600 + 400 = 1000\text{ N}\)

Using \(P = Fv\), where \(P = 20\text{ kW} = 20000\text{ W}\):
\(20000 = 1000 \times v \implies v = 20\text{ m s}^{-1}\).

**(ii)**
When the engine's power is increased to \(24\text{ kW} = 24000\text{ W}\) and the speed is \(16\text{ m s}^{-1}\):
The new driving force \(F'\) is:
\(F' = \frac{P}{v} = \frac{24000}{16} = 1500\text{ N}\).

Applying Newton's second law up the slope:
\(F' - mg \sin \alpha - R = ma\)
\(1500 - 600 - 400 = 1200 \times a\)
\(500 = 1200 a\)
\(a = \frac{500}{1200} = \frac{5}{12} \approx 0.417\text{ m s}^{-2}\) (to 3 s.f.).

**Part (i)**
* **M1**: For resolving forces parallel to the slope to find the driving force \(F = mg \sin \alpha + R\).
* **A1**: For obtaining \(F = 1000\text{ N}\).
* **A1**: For finding \(v = 20\text{ m s}^{-1}\) using \(P = Fv\).

**Part (ii)**
* **M1**: For finding the new driving force \(F'\) using \(F' = \frac{P}{v}\).
* **M1**: For setting up Newton's second law along the slope: \(F' - mg \sin \alpha - R = ma\).
* **A1**: For obtaining \(a = 0.417\text{ m s}^{-2}\) (or \(\frac{5}{12}\text{ m s}^{-2}\)).

**(i)**
Let \(T\) be the tension in the string and \(a\) be the acceleration of the system.
For particle \(A\):
Normal reaction \(R_A = m_A g = 0.6 \times 10 = 6\text{ N}\).
Friction force \(F = \mu R_A = 0.5 \times 6 = 3\text{ N}\).
Applying Newton's second law horizontally for \(A\):
\(T - F = m_A a \implies T - 3 = 0.6 a\) --- (Equation 1)

For particle \(B\):
Applying Newton's second law vertically for \(B\):
\(m_B g - T = m_B a \implies (0.4 \times 10) - T = 0.4 a \implies 4 - T = 0.4 a\) --- (Equation 2)

Adding Equation 1 and Equation 2:
\((T - 3) + (4 - T) = 0.6 a + 0.4 a\)
\(1 = 1.0 a \implies a = 1\text{ m s}^{-2}\) (as required).

Substituting \(a = 1\) into Equation 1:
\(T - 3 = 0.6(1) \implies T = 3.6\text{ N}\).

**(ii)**
First phase (before string breaks, for \(t = 1.2\text{ s}\)):
Distance traveled by \(A\):
\(s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1)(1.2)^2 = 0.72\text{ m}\).
Speed attained by \(A\):
\(v = u + at = 0 + (1)(1.2) = 1.2\text{ m s}^{-1}\).

Second phase (after string breaks):
The tension in the string becomes zero. The only horizontal force acting on \(A\) is the friction force \(F = 3\text{ N}\) opposing the motion.
Applying Newton's second law for \(A\):
\(-F = m_A a' \implies -3 = 0.6 a' \implies a' = -5\text{ m s}^{-2}\).

Let \(s_2\) be the distance traveled by \(A\) before coming to rest:
\(v_{\text{final}}^2 = v^2 + 2a's_2\)
\(0 = (1.2)^2 + 2(-5)s_2\)
\(0 = 1.44 - 10s_2\)
\(s_2 = 0.144\text{ m}\).

Total distance traveled by \(A\):
\(s = s_1 + s_2 = 0.72 + 0.144 = 0.864\text{ m}\).

**Part (i)**
* **B1**: For calculating the friction force acting on \(A\), \(F = 3\text{ N}\).
* **M1**: For writing the equations of motion for both particles (at least one correct).
* **A1**: For adding the equations to correctly show that \(a = 1\text{ m s}^{-2}\).
* **A1**: For finding \(T = 3.6\text{ N}\).

**Part (ii)**
* **M1**: For calculating the distance \(s_1 = 0.72\text{ m}\) and velocity \(v = 1.2\text{ m s}^{-1}\) before the string breaks, and using Newton's second law to find the new deceleration \(a' = -5\text{ m s}^{-2}\).
* **A1**: For obtaining the correct total distance of \(0.864\text{ m}\) (or \(\frac{108}{125}\text{ m}\)).

Method 1: The total number of arrangements of the 7 letters of OCTOBER (which contains 2 'O's and 5 other distinct letters) is \(\frac{7!}{2!} = 2520\).

If the two 'O's are together, we treat them as a single block 'OO'. The number of arrangements of the 6 entities (OO, C, T, B, E, R) is \(6! = 720\).

Therefore, the number of arrangements where the two 'O's are not together is \(2520 - 720 = 1800\).

Method 2: First, arrange the 5 distinct letters (C, T, B, E, R) in \(5! = 120\) ways.

There are 6 possible spaces (including the ends) to insert the two 'O's: _ C _ T _ B _ E _ R _.

Choosing 2 of these spaces for the 'O's can be done in \(\binom{6}{2} = 15\) ways.

Total number of arrangements is \(120 \times 15 = 1800\).

M1: For finding the total number of arrangements of OCTOBER (2520) OR for finding the number of arrangements of the 5 distinct letters (120).
M1: For finding the number of arrangements with 'O's together (720) and subtracting from their total, OR for multiplying their 5-letter arrangement by \(\binom{6}{2}\).
A1: For the correct final answer of 1800.

We are given that \(W \sim \text{N}(\mu, 45^2)\) and \(P(W < 250) = 0.123\).

Standardizing the variable:
\(P\left(Z < \frac{250 - \mu}{45}\right) = 0.123\)

Since the probability is less than 0.5, the z-score must be negative.

Let \(z = \frac{250 - \mu}{45}\).
Using the normal distribution table:
\(\Phi(-z) = 0.123 \implies \Phi(z) = 1 - 0.123 = 0.877\)

From the tables, \(z \approx 1.160\).

Thus, we set:
\(\frac{250 - \mu}{45} = -1.160\)
\(250 - \mu = -1.160 \times 45\)
\(250 - \mu = -52.2\)
\(\mu = 302.2 \approx 302\) (to 3 significant figures).

M1: For standardizing and setting up an equation of the form \(\frac{250 - \mu}{45} = z\), where \(z\) is a non-zero z-value.
M1: For finding the correct z-value of \(-1.160\) (or \(-1.16\)) and equating it to \(\frac{250 - \mu}{45}\).
A1: For the correct value of \(\mu = 302\) (or 302.2; accept 302).

To find the number of playlists, we first select the tracks and then arrange them. There are two possible cases for the selection of 5 tracks containing at least 2 classical and at least 2 jazz tracks:

Case 1: 2 classical tracks and 3 jazz tracks
Number of ways to choose: \(\binom{6}{2} \times \binom{5}{3} = 15 \times 10 = 150\)

Case 2: 3 classical tracks and 2 jazz tracks
Number of ways to choose: \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\)

Total number of selections = \(150 + 200 = 350\)

Since the songs in the playlist are ordered, we must multiply the number of selections by \(5!\) (the number of ways to arrange 5 tracks):

Total number of playlists = \(350 \times 5! = 350 \times 120 = 42\,000\).

M1 for identifying the two valid combinations: (2C, 3J) and (3C, 2J)
M1 for calculating the number of selections for both cases: \(\binom{6}{2}\times\binom{5}{3}\) and \(\binom{6}{3}\times\binom{5}{2}\)
A1 for getting 350 total selections
M1 for multiplying the total selections by \(5!\)
A1 for the correct final answer of 42000

(i) The probability that at least one of the events occurs is given by \(P(A \cup B)\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.15 + 0.20 - 0.05 = 0.3\)

(ii) The conditional probability is \(P(A | A \cup B)\):
\(P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)} = \frac{0.15}{0.3} = 0.5\)

(iii) For \(A\) and \(B\) to be independent, we must have \(P(A \cap B) = P(A) \times P(B)\).
\(P(A) \times P(B) = 0.15 \times 0.20 = 0.03\)
Since \(P(A \cap B) = 0.05 \neq 0.03\), the events \(A\) and \(B\) are not independent.

(i) B1 for \(P(A \cup B) = 0.3\)
(ii) M1 for using the conditional probability formula \(\frac{P(A)}{P(A \cup B)}\), A1 for 0.5
(iii) M1 for calculating \(P(A) \times P(B)\) and comparing with \(P(A \cap B)\), A1 for a correct conclusion with comparison shown (\(0.03 \neq 0.05\))

Let \(X\) be the mass of a packet of tea. We are given:
\(P(X < 242) = 0.10\) and \(P(X > 255) = 0.05\)

Standardising these values:
\(P\left(Z < \frac{242 - \mu}{\sigma}\right) = 0.10 \implies \frac{242 - \mu}{\sigma} = -1.282\)
\(P\left(Z > \frac{255 - \mu}{\sigma}\right) = 0.05 \implies \frac{255 - \mu}{\sigma} = 1.645\)

This gives us two simultaneous equations:
1) \(\mu - 1.282\sigma = 242\)
2) \(\mu + 1.645\sigma = 255\)

Subtracting equation (1) from equation (2):
\(2.927\sigma = 13 \implies \sigma \approx 4.44\text{ (to 3 sf)}\)

Substituting \(\sigma = 4.4414\) back into equation (1):
\(\mu = 242 + 1.282(4.4414) \approx 248\text{ (to 3 sf)}\)

M1 for writing standardisation equations with correct z-values (\(-1.282\) and \(1.645\))
A1 for obtaining two correct simultaneous equations (condone slight differences in z-value decimal places, e.g., 1.28 or 1.64)
M1 for solving the simultaneous equations to find one variable
A1 for \(\sigma = 4.44\) (3 sf)
A1 for \(\mu = 248\) (3 sf)

(a) The committee must contain at least 4 women. Since the total committee size is 6, we consider the mutually exclusive cases:

- Case 1: 4 women and 2 men
Number of ways = \(\binom{7}{4} \times \binom{8}{2} = 35 \times 28 = 980\)

- Case 2: 5 women and 1 man
Number of ways = \(\binom{7}{5} \times \binom{8}{1} = 21 \times 8 = 168\)

- Case 3: 6 women and 0 men
Number of ways = \(\binom{7}{6} \times \binom{8}{0} = 7 \times 1 = 7\)

Total number of different committees = \(980 + 168 + 7 = 1155\).

(b) Method 1 (Subtractive):
Total ways of choosing any 6 people from the 15 without restriction:
\(\binom{15}{6} = 5005\)

Number of committees that include both Alan and Bob (which means choosing the 4 remaining members from the other 13 people):
\(\binom{13}{4} = 715\)

Number of acceptable committees where they are not both included:
\(5005 - 715 = 4290\).

Method 2 (Additive):
- Case 1: Neither Alan nor Bob is on the committee. Choose 6 from 13:
\(\binom{13}{6} = 1716\)
- Case 2: Exactly one of Alan or Bob is on the committee. Choose 1 of them, then 5 from 13:
\(\binom{2}{1} \times \binom{13}{5} = 2 \times 1287 = 2574\)

Total = \(1716 + 2574 = 4290\).

(a)
M1: For attempting to find the sum of three scenarios with 4, 5, or 6 women (at least two scenarios correct in form)
A1: For correct terms: 980, 168, and 7 (or unsimplified forms \(\binom{7}{4} \times \binom{8}{2}\), \(\binom{7}{5} \times \binom{8}{1}\), \(\binom{7}{6}\))
A1: For correct final answer of 1155

(b)
M1: For a correct strategy, e.g., finding total combinations \(\binom{15}{6}\) and subtracting combinations with both Alan and Bob \(\binom{13}{4}\), or summing the cases where neither or only one is chosen
A1: For correct calculation of components, e.g., 5005 and 715, or 1716 and 2574
A1: For correct final answer of 4290

Let \(X\) represent the mass of an apple, where \(X \sim N(\mu, \sigma^2)\).

From the first condition, we are given:
\(P(X > 185) = 0.15 \implies P(X < 185) = 0.85\)

Standardizing this value:
\(\frac{185 - \mu}{\sigma} = z_1\), where \(\Phi(z_1) = 0.85\).
Using the normal table, we find \(z_1 = 1.036\).
So, \(\frac{185 - \mu}{\sigma} = 1.036 \implies \mu + 1.036\sigma = 185\) (Equation 1)

From the second condition, we are given:
\(P(X < 140) = 0.30\)

Since \(0.30 < 0.5\), the corresponding z-value is negative:
\(\frac{140 - \mu}{\sigma} = -z_2\), where \(\Phi(z_2) = 1 - 0.30 = 0.70\).
Using the normal table, we find \(z_2 = 0.524\).
So, \(\frac{140 - \mu}{\sigma} = -0.524 \implies \mu - 0.524\sigma = 140\) (Equation 2)

Subtract Equation 2 from Equation 1 to eliminate \(\mu\):
\((\mu + 1.036\sigma) - (\mu - 0.524\sigma) = 185 - 140\)
\(1.560\sigma = 45\)
\(\sigma = \frac{45}{1.560} \approx 28.846\text{ g}\)

Substitute \(\sigma = 28.846\) back into Equation 2 to find \(\mu\):
\(\mu = 140 + 0.524(28.846) = 140 + 15.115 = 155.115\text{ g}\)

Thus, to 3 significant figures, we have:
\(\sigma = 28.8\) and \(\mu = 155\).

M1: For standardizing 185 and equating to a positive z-value (derived from area 0.85)
B1: For identifying the correct z-value 1.036 (or 1.037)
M1: For standardizing 140 and equating to a negative z-value (derived from area 0.30)
B1: For identifying the correct z-value -0.524 (or -0.525)
M1: For setting up and solving two linear simultaneous equations in \(\mu\) and \(\sigma\)
A1: For finding both \(\sigma = 28.8\) and \(\mu = 155\) (accept answers rounding to 28.8 and 155, or exact 3 s.f. equivalent)

Using the formula for the Poisson distribution probability:

\[P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}\]

Given \(P(X = 3) = 1.5 \times P(X = 2)\):

\[\frac{e^{-\lambda} \lambda^3}{3!} = 1.5 \times \frac{e^{-\lambda} \lambda^2}{2!}\]

\[\frac{e^{-\lambda} \lambda^3}{6} = 1.5 \times \frac{e^{-\lambda} \lambda^2}{2}\]

Since \(e^{-\lambda} \neq 0\) and \(\lambda > 0\), we can divide both sides by \(e^{-\lambda} \lambda^2\):

\[\frac{\lambda}{6} = \frac{1.5}{2}\]

\[\frac{\lambda}{6} = 0.75 \implies \lambda = 4.5\]

Now, calculate \(P(X = 4)\):

\[P(X = 4) = \frac{e^{-4.5} \times 4.5^4}{4!} = \frac{e^{-4.5} \times 410.0625}{24} \approx 0.1898\]

To 3 significant figures, the probability is \(0.190\).

- **M1**: Sets up a correct algebraic equation for \(\lambda\) using the Poisson probability formula and attempts to solve.
- **A1**: Correctly finds \(\lambda = 4.5\).
- **A1**: Obtains \(P(X = 4) \approx 0.190\) (accept \(0.19\)).

First, we find the constant \(k\) using the property that the total area under the probability density function is 1:

\[\int_{0}^{2} k(4 - x^2) \, dx = 1\]

\[k \left[ 4x - \frac{x^3}{3} \right]_0^2 = 1\]

\[k \left( 8 - \frac{8}{3} - 0 \right) = 1\]

\[k \left( \frac{16}{3} \right) = 1 \implies k = \frac{3}{16}\]

Next, we calculate the expectation \(E(X)\):

\[E(X) = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} \frac{3}{16} x(4 - x^2) \, dx\]

\[E(X) = \frac{3}{16} \int_{0}^{2} (4x - x^3) \, dx\]

\[E(X) = \frac{3}{16} \left[ 2x^2 - \frac{x^4}{4} \right]_0^2\]

\[E(X) = \frac{3}{16} \left( 2(4) - \frac{16}{4} - 0 \right)\]

\[E(X) = \frac{3}{16} (8 - 4) = \frac{3}{16} \times 4 = \frac{3}{4} \text{ (or } 0.75\)\]

- **M1**: For integrating the pdf from 0 to 2, setting it equal to 1, and solving for \(k = \frac{3}{16}\).
- **M1**: For a correct attempt to integrate \(x f(x)\) from 0 to 2 to find \(E(X)\).
- **A1**: For obtaining the correct final answer \(\frac{3}{4}\) (or 0.75).

Let \(\lambda\) be the mean number of errors per page. We state the hypotheses: \(H_0: \lambda = 1.2\) and \(H_1: \lambda < 1.2\). Under \(H_0\), the total number of errors in 40 pages, \(X\), follows a Poisson distribution: \(X \sim \text{Po}(48)\). Since the mean \(\mu = 48 > 15\), we can approximate \(X\) using a normal distribution: \(X \approx \text{N}(48, 48)\). We test the observed value of 38 using a one-tailed test with a continuity correction: \(P(X \le 38) \approx P(Y \le 38.5)\) where \(Y \sim \text{N}(48, 48)\). Standardizing this value: \(z = \frac{38.5 - 48}{\sqrt{48}} = \frac{-9.5}{6.9282} = -1.371\). The critical value at the 5% significance level for a one-tailed test is \(-1.645\). Since \(-1.371 > -1.645\) (or the p-value is \(P(Z < -1.371) = 1 - 0.9148 = 0.0852 > 0.05\)), we fail to reject \(H_0\). There is insufficient evidence to suggest that the mean number of errors has decreased.

B1: State both hypotheses correctly in terms of \(\lambda\) or the sample total. M1: For identifying and using the normal approximation with mean and variance both equal to 48. M1: For applying the correct continuity correction (using 38.5). A1: For obtaining the correct z-value of -1.37 (or -1.371) or the correct probability of 0.0852. A1: For a correct comparison with the critical value (-1.645 or 0.05) leading to a consistent non-assertive conclusion in context.

(i) Since \(f(x)\) is a probability density function, the total area under the curve is 1: \(\int_{1}^{3} k(x-1)(3-x) \, dx = 1\). Expanding the integrand gives \(k \int_{1}^{3} (4x - x^2 - 3) \, dx = 1\). Integrating gives \(k \left[ 2x^2 - \frac{x^3}{3} - 3x \right]_{1}^{3} = 1\). Evaluating at the limits: \(k \left( (18 - 9 - 9) - (2 - \frac{1}{3} - 3) \right) = 1 \implies k \left( 0 - (-\frac{4}{3}) \right) = 1 \implies \frac{4}{3}k = 1 \implies k = \frac{3}{4}\). (ii) To find \(P(X > 2.5)\): \(P(X > 2.5) = \int_{2.5}^{3} \frac{3}{4}(4x - x^2 - 3) \, dx = \frac{3}{4} \left[ 2x^2 - \frac{x^3}{3} - 3x \right]_{2.5}^{3}\). Evaluating at the limits: \(\frac{3}{4} \left( 0 - (2(2.5)^2 - \frac{2.5^3}{3} - 3(2.5)) \right) = \frac{3}{4} \left( 0 - (12.5 - 5.20833 - 7.5) \right) = \frac{3}{4} \left( 0.20833 \right) = \frac{5}{32} = 0.15625\), which is \(0.156\) to 3 significant figures.

(i) M1: Integrate \(k(x-1)(3-x)\) over the limits 1 to 3 and equate to 1. A1: Correctly evaluate the definite integral and show \(k = \frac{3}{4}\). (ii) M1: Set up the correct integral for \(P(X > 2.5)\) with correct limits. A1: Correctly integrate and substitute limits. A1: Obtain \(\frac{5}{32}\) or \(0.156\) (3 s.f.).

Let \(T = G_1 + G_2 + G_3 - 3B\). We want to find \(P(T > 0)\). Since \(G_1, G_2, G_3\) and \(B\) are independent normal variables, \(T\) is also normally distributed. First, find the mean of \(T\): \(E(T) = E(G_1) + E(G_2) + E(G_3) - 3E(B) = 3(250) - 3(240) = 750 - 720 = 30\). Next, find the variance of \(T\): \(Var(T) = Var(G_1) + Var(G_2) + Var(G_3) + 3^2 Var(B) = 3(15^2) + 9(12^2) = 3(225) + 9(144) = 675 + 1296 = 1971\). Thus, \(T \sim \text{N}(30, 1971)\). Now, standardise to find \(P(T > 0)\): \(P(T > 0) = P\left(Z > \frac{0 - 30}{\sqrt{1971}}\right) = P(Z > -0.6757) = P(Z < 0.676) = \Phi(0.676)\). From the normal tables, \(\Phi(0.676) \approx 0.7505\), which rounds to \(0.751\) (3 s.f.).

M1: Find the mean of the linear combination, \(E(T) = 30\). M1: Attempt to find the variance of \(T\) using \(3Var(G) + 9Var(B)\). A1: Obtain the correct variance of 1971. M1: Standardise with their mean and variance to calculate \(P(T > 0)\). A1: Correct final probability of 0.751 (accept 0.750 from non-interpolated tables).

Let \(p\) be the probability of a treated seed germinating.

We set up the hypotheses:
\(H_0: p = 0.6\)
\(H_1: p > 0.6\)

Under the null hypothesis, the number of germinated seeds, \(X\), follows a binomial distribution:
\(X \sim \text{B}(18, 0.6)\)

To perform the test at the 5% level of significance, we calculate the probability of obtaining a result at least as extreme as the observed value of 14:
\(P(X \ge 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)\)

Using the binomial probability formula \(P(X = k) = \binom{18}{k} p^k (1-p)^{18-k}\):
- \(P(X = 14) = \binom{18}{14} (0.6)^{14} (0.4)^4 = 3060 \times 0.00078364 \times 0.0256 \approx 0.06139\)
- \(P(X = 15) = \binom{18}{15} (0.6)^{15} (0.4)^3 = 816 \times 0.00047018 \times 0.064 \approx 0.02455\)
- \(P(X = 16) = \binom{18}{16} (0.6)^{16} (0.4)^2 = 153 \times 0.00028211 \times 0.16 \approx 0.00691\)
- \(P(X = 17) = \binom{18}{17} (0.6)^{17} (0.4)^1 = 18 \times 0.00016927 \times 0.4 \approx 0.00122\)
- \(P(X = 18) = (0.6)^{18} \approx 0.00010\)

Summing these probabilities:
\(P(X \ge 14) = 0.06139 + 0.02455 + 0.00691 + 0.00122 + 0.00010 = 0.0942\) (to 3 s.f.)

Since \(0.0942 > 0.05\), the result is not significant at the 5% level.

Therefore, we do not reject \(H_0\). There is insufficient evidence to suggest that the treatment increases the germination rate.

**B1**: For both hypotheses stated correctly (using \(p\) or 'probability/proportion of germinating').
**M1**: For attempting to calculate \(P(X \ge 14)\) (at least 3 correct binomial terms summed, or 1 - \(P(X \le 13)\)).
**A1**: For correct calculation of \(P(X \ge 14) \approx 0.0942\) (accept 0.0941 to 0.0943).
**M1**: For comparing their calculated probability with 0.05 (or comparing 14 with the critical value of 15).
**A1**: For correctly stating that \(H_0\) is not rejected.
**A1**: For concluding in context, stating there is insufficient evidence that the treatment increases the germination rate.

To find the mean of \(X\):
\[ E(X) = \int_{0}^{1} x f(x) \, dx = \int_{0}^{1} 12x(x^2 - x^3) \, dx \]
\[ E(X) = 12 \int_{0}^{1} (x^3 - x^4) \, dx = 12 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_{0}^{1} \]
\[ E(X) = 12 \left( \frac{1}{4} - \frac{1}{5} \right) = 12 \left( \frac{1}{20} \right) = 0.6 \]

To find the variance of \(X\), we first find \(E(X^2)\):
\[ E(X^2) = \int_{0}^{1} x^2 f(x) \, dx = \int_{0}^{1} 12x^2(x^2 - x^3) \, dx \]
\[ E(X^2) = 12 \int_{0}^{1} (x^4 - x^5) \, dx = 12 \left[ \frac{x^5}{5} - \frac{x^6}{6} \right]_{0}^{1} \]
\[ E(X^2) = 12 \left( \frac{1}{5} - \frac{1}{6} \right) = 12 \left( \frac{1}{30} \right) = 0.4 \]

Now, we calculate the variance using the formula:
\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]
\[ \text{Var}(X) = 0.4 - (0.6)^2 = 0.4 - 0.36 = 0.04 \]

**M1**: For setting up the integral for \(E(X)\) with limits \(0\) to \(1\).
**A1**: For correct integration and obtaining \(E(X) = 0.6\) (or \(\frac{3}{5}\)).
**M1**: For setting up the integral for \(E(X^2)\) with limits \(0\) to \(1\).
**A1**: For correct integration and obtaining \(E(X^2) = 0.4\) (or \(\frac{2}{5}\)).
**M1**: For using the formula \(\text{Var}(X) = E(X^2) - [E(X)]^2\).
**A1**: For obtaining \(\text{Var}(X) = 0.04\) (or \(\frac{1}{25}\)).