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(a) Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\). The 2nd, 5th and 11th terms are given by \(T_2 = a + d\), \(T_5 = a + 4d\) and \(T_{11} = a + 10d\). Since these form a geometric progression, we have \((a + 4d)^2 = (a + d)(a + 10d)\). Expanding both sides gives \(a^2 + 8ad + 16d^2 = a^2 + 11ad + 10d^2\). Simplifying this equation leads to \(6d^2 = 3ad\). Since \(d \neq 0\), we can divide by \(3d\) to obtain \(a = 2d\). The common ratio \(r\) of the geometric progression is \(r = \frac{T_5}{T_2} = \frac{a + 4d}{a + d} = \frac{2d + 4d}{2d + d} = \frac{6d}{3d} = 2\). (b) The sum of the first 12 terms of the arithmetic progression is \(S_{12} = \frac{12}{2}[2a + 11d] = 180\). This simplifies to \(6(2a + 11d) = 180\), which gives \(2a + 11d = 30\). Substituting \(a = 2d\) into this equation yields \(2(2d) + 11d = 30 \Rightarrow 15d = 30 \Rightarrow d = 2\). Since \(a = 2d\), we find \(a = 4\).

M1: Set up the geometric progression equation using terms of the arithmetic progression. A1: Correctly simplify the quadratic equation to obtain \(a = 2d\). M1: Use the expression to find the common ratio and show \(r = 2\). M1: Use the sum formula for an arithmetic progression to set up \(6(2a+11d)=180\). A1: Find the correct value \(d = 2\). A1: Find the correct value \(a = 4\).

(a) To find when \(f(x)\) has an inverse, we express it in completed square form: \(f(x) = 2(x^2 - 6x) + 13 = 2((x-3)^2 - 9) + 13 = 2(x-3)^2 - 5\). The vertex of the parabola is at \((3, -5)\). For \(f\) to be one-to-one, the domain must be restricted to one side of the line of symmetry \(x = 3\). Since the domain is defined for \(x \le k\), the greatest possible value of \(k\) is 3. (b) To find the inverse, let \(y = 2(x-3)^2 - 5\). Rearranging for \(x\): \(y + 5 = 2(x-3)^2 \Rightarrow \frac{y+5}{2} = (x-3)^2\). Taking the square root, and noting that \(x \le 3\) requires the negative root: \(x - 3 = -\sqrt{\frac{y+5}{2}} \Rightarrow x = 3 - \sqrt{\frac{y+5}{2}}\). Thus, \(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\). The domain of \(f^{-1}\) is the range of \(f\). For \(x \le 3\), the range of \(f\) is \(f(x) \ge -5\), so the domain of \(f^{-1}\) is \(x \ge -5\).

M1: Attempt to complete the square for \(2x^2 - 12x + 13\). A1: Obtain \(2(x-3)^2 - 5\) and deduce \(k = 3\). M1: Set \(y = 2(x-3)^2 - 5\) and attempt to make \(x\) the subject. A1: Correctly choose the negative square root due to the domain restriction. A1: State the correct inverse function \(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\). A1: Correctly state the domain of \(f^{-1}\) as \(x \ge -5\).

(a) Rewrite the circle equation by completing the square for both \(x\) and \(y\): \((x-2)^2 - 4 + (y+3)^2 - 9 - 12 = 0 \Rightarrow (x-2)^2 + (y+3)^2 = 25\). Therefore, the coordinates of the centre \(C\) are \((2, -3)\) and the radius is \(\sqrt{25} = 5\). (b) The gradient of the radius from the centre \(C(2, -3)\) to the point \(P(5, 1)\) is \(m_{CP} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\). Since the tangent is perpendicular to the radius at the point of contact, the gradient \(m\) of the tangent is \(-\frac{1}{m_{CP}} = -\frac{3}{4}\). Using the point-slope form with \(P(5, 1)\): \(y - 1 = -\frac{3}{4}(x - 5) \Rightarrow y = -\frac{3}{4}x + \frac{15}{4} + 1 \Rightarrow y = -\frac{3}{4}x + \frac{19}{4}\). Comparing with \(y = mx + c\), we find \(m = -\frac{3}{4}\) and \(c = \frac{19}{4}\).

M1: Attempt to complete the square for \(x\) and \(y\). A1: Obtain centre \((2, -3)\) and radius 5. M1: Calculate the gradient of the radius \(CP\). A1: Deduce that the gradient of the tangent is \(m = -\frac{3}{4}\). M1: Substitute \(P(5, 1)\) and gradient \(m\) to find the equation of the line. A1: Obtain \(c = \frac{19}{4}\).

(a) Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to substitute into the given equation: \(3(1 - \sin^2 \theta) + 2\sin \theta = 2\). This expands to \(3 - 3\sin^2 \theta + 2\sin \theta = 2\). Rearranging all terms to one side gives \(3\sin^2 \theta - 2\sin \theta - 1 = 0\). (b) Solve the quadratic equation by factoring: \((3\sin \theta + 1)(\sin \theta - 1) = 0\). This gives two cases: \(\sin \theta = 1\) or \(\sin \theta = -\frac{1}{3}\). Case 1: \(\sin \theta = 1 \Rightarrow \theta = 90^\circ\). Case 2: \(\sin \theta = -\frac{1}{3}\). The basic angle is \(\alpha = \sin^{-1}(\frac{1}{3}) \approx 19.47^\circ\). Since sine is negative, \(\theta\) lies in the third and fourth quadrants: \(\theta = 180^\circ + 19.47^\circ = 199.5^\circ\) (to 1 d.p.) and \(\theta = 360^\circ - 19.47^\circ = 340.5^\circ\) (to 1 d.p.).

M1: Substitute \(1 - \sin^2 \theta\) for \(\cos^2 \theta\). A1: Correctly show the quadratic form. M1: Attempt to solve the quadratic equation to find values for \(\sin \theta\). A1: Identify \(\sin \theta = 1\) and \(\sin \theta = -1/3\). A1: Obtain \(\theta = 90^\circ\). A1: Obtain \(\theta = 199.5^\circ\) and \(\theta = 340.5^\circ\) (allow rounding to 1 d.p.).

(a) The perimeter of the sector is given by \(P = 2r + r\theta = 24\). Rearranging this formula for \(\theta\): \(r\theta = 24 - 2r \Rightarrow \theta = \frac{24 - 2r}{r}\). (b) The area of the sector is \(A = \frac{1}{2}r^2\theta\). Substituting the expression for \(\theta\) from part (a): \(A = \frac{1}{2}r^2\left(\frac{24 - 2r}{r}\right) = \frac{1}{2}r(24 - 2r) = 12r - r^2\). (c) Setting \(A = 32\) gives the quadratic equation \(12r - r^2 = 32 \Rightarrow r^2 - 12r + 32 = 0\). Factoring gives \((r - 4)(r - 8) = 0\), which yields the solutions \(r = 4\) and \(r = 8\). For \(r = 4\), \(\theta = \frac{24 - 2(4)}{4} = 4\) radians. For \(r = 8\), \(\theta = \frac{24 - 2(8)}{8} = 1\) radian.

M1: Set up the perimeter equation \(2r + r\theta = 24\). A1: Express \(\theta\) correctly in terms of \(r\). M1: Substitute \(\theta\) into the area formula \(A = \frac{1}{2}r^2\theta\) and simplify. A1: Correctly show \(A = 12r - r^2\). M1: Solve \(12r - r^2 = 32\) to find values of \(r\). A1: Obtain \(r = 4, \theta = 4\) and \(r = 8, \theta = 1\) (both pairs must be correct).

(a) Rewrite the equation as \(y = 4x + 9(x-1)^{-1}\). Differentiating with respect to \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4 - 9(x-1)^{-2} = 4 - \frac{9}{(x-1)^2}\). Differentiating again: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 18(x-1)^{-3} = \frac{18}{(x-1)^3}\). (b) To find the stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\): \(4 - \frac{9}{(x-1)^2} = 0 \Rightarrow (x-1)^2 = \frac{9}{4}\). Since \(x > 1\), we take the positive square root: \(x-1 = 1.5 \Rightarrow x = 2.5\). Substituting \(x = 2.5\) into the original equation: \(y = 4(2.5) + \frac{9}{2.5-1} = 10 + \frac{9}{1.5} = 16\). The coordinates of the stationary point are \((2.5, 16)\). (c) To find the nature of the stationary point, substitute \(x = 2.5\) into the second derivative: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{18}{(2.5-1)^3} = \frac{18}{3.375} = \frac{16}{3} > 0\). Since the second derivative is positive, the stationary point is a local minimum.

M1: Correctly differentiate \(y\) to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\). A1: Correctly differentiate to find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\). M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve for \(x\). A1: Find the correct value of \(x = 2.5\) (rejecting the negative root). A1: Find the correct \(y\)-coordinate, \(y = 16\). A1: Evaluate second derivative and state that the stationary point is a minimum.

(a) Differentiating \(y = (3x-2)^{1/2}\) with respect to \(x\) using the chain rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(3x-2)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x-2}}\). At \(P(2, 2)\), \(x = 2\), so the gradient of the curve is \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2\sqrt{4}} = \frac{3}{4}\). The gradient of the normal is \(-\frac{1}{3/4} = -\frac{4}{3}\). The equation of the normal at \(P(2, 2)\) is \(y - 2 = -\frac{4}{3}(x - 2)\), which simplifies to \(y = -\frac{4}{3}x + \frac{14}{3}\). (b) The curve meets the \(x\)-axis when \(y = 0 \Rightarrow 3x - 2 = 0 \Rightarrow x = \frac{2}{3}\). The area is given by the integral of \(y\) from \(x = \frac{2}{3}\) to \(x = 6\): \(A = \int_{2/3}^6 (3x-2)^{1/2} \mathrm{d}x\). Integrating gives \(A = \left[ \frac{2}{3} \cdot \frac{(3x-2)^{3/2}}{3} \right]_{2/3}^6 = \left[ \frac{2}{9}(3x-2)^{3/2} \right]_{2/3}^6\). Substituting the limits: \(A = \frac{2}{9}(3(6)-2)^{3/2} - \frac{2}{9}(3(2/3)-2)^{3/2} = \frac{2}{9}(16)^{3/2} - 0 = \frac{2}{9}(64) = \frac{128}{9}\).

M1: Differentiate the curve equation to find the gradient. A1: Find the correct tangent gradient and then normal gradient \(-4/3\). A1: Correctly write the normal line equation. M1: Find the lower limit of integration by setting \(y = 0\). M1: Integrate the function using reverse chain rule. A1: Correctly evaluate the definite integral to get \(128/9\).

To find where the line and curve intersect, we equate their expressions: \(x^2 + kx + 5 = 2kx - 4\). Rearranging this equation into the standard quadratic form: \(x^2 - kx + 9 = 0\). Since the line and curve do not intersect, this quadratic equation has no real solutions. This means the discriminant \(\Delta\) must be strictly less than 0. The discriminant is \(\Delta = b^2 - 4ac = (-k)^2 - 4(1)(9) = k^2 - 36\). Setting \(\Delta < 0\) gives \(k^2 - 36 < 0 \Rightarrow (k - 6)(k + 6) < 0\). Therefore, the set of values of \(k\) is \(-6 < k < 6\).

M1: Equate the line and curve equations and rearrange into standard quadratic form. A1: Obtain the correct quadratic equation \(x^2 - kx + 9 = 0\). M1: Set up the discriminant expression \((-k)^2 - 4(1)(9)\). M1: Set the discriminant to be less than zero. A1: Correctly solve the quadratic inequality to find \(-6 < k < 6\) (allow interval notation).

(i) Let the terms of the arithmetic progression be \( T_n = a + (n-1)d \).
The 2nd, 5th, and 14th terms are:
\( T_2 = a + d \)
\( T_5 = a + 4d \)
\( T_{14} = a + 13d \)

Since these three terms are the first three terms of a geometric progression, we have:
\( \frac{a + 4d}{a + d} = \frac{a + 13d}{a + 4d} \)

Cross-multiplying:
\( (a + 4d)^2 = (a + d)(a + 13d) \)
\( a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \)

Subtracting \( a^2 \) from both sides and rearranging:
\( 3d^2 - 6ad = 0 \)
\( 3d(d - 2a) = 0 \)

Since \( d \neq 0 \), we can divide by \( 3d \) to obtain:
\( d = 2a \).

(ii) The sum of the first 10 terms of the arithmetic progression is given by:
\( S_{10} = \frac{10}{2} [2a + 9d] = 300 \)
\( 5[2a + 9d] = 300 \implies 2a + 9d = 60 \)

Substitute \( d = 2a \) into the equation:
\( 2a + 9(2a) = 60 \)
\( 20a = 60 \implies a = 3 \)

Hence, \( d = 2(3) = 6 \).

The first three terms of the geometric progression are:
\( g_1 = T_2 = a + d = 3 + 6 = 9 \)
\( g_2 = T_5 = a + 4d = 3 + 24 = 27 \)
\( g_3 = T_{14} = a + 13d = 3 + 78 = 81 \)

The common ratio is \( r = \frac{27}{9} = 3 \).

The 6th term of the geometric progression is:
\( g_6 = g_1 r^5 = 9 \times 3^5 = 9 \times 243 = 2187 \).

(i)
M1: For writing down the expressions for \( T_2, T_5, T_{14} \) in terms of \( a \) and \( d \) and setting up the geometric progression relationship (e.g., \( (a+4d)^2 = (a+d)(a+13d) \)).
A1: For expanding correctly to obtain \( 3d^2 = 6ad \) or equivalent.
A1: For concluding \( d = 2a \) with clear justification that \( d \neq 0 \).

(ii)
M1: For using the sum formula of AP to set up \( 5(2a + 9d) = 300 \) and substituting \( d = 2a \) to find \( a \) and/or \( d \).
A1: For obtaining \( a = 3 \) and \( d = 6 \).
M1: For finding the first term \( g_1 = 9 \) and common ratio \( r = 3 \) of the GP, and using the formula \( g_6 = g_1 r^5 \).
A1: For obtaining the correct final answer 2187.

(i) Expressing \( \mathrm{f}(x) \) in completed square form:
\( \mathrm{f}(x) = 2(x^2 - 6x) + 13 \)
\( = 2[(x - 3)^2 - 9] + 13 \)
\( = 2(x - 3)^2 - 18 + 13 \)
\( = 2(x - 3)^2 - 5 \)

For a quadratic function to have an inverse, its domain must be restricted such that it is a one-to-one function.
The line of symmetry of the curve is at \( x = 3 \).
Since the domain is \( x \le k \), the function is one-to-one if \( k \le 3 \).
Thus, the greatest value of \( k \) is 3.

(ii) For \( k = 3 \), the domain of \( \mathrm{f} \) is \( x \le 3 \).
Let \( y = 2(x - 3)^2 - 5 \).
Rearranging to make \( x \) the subject:
\( y + 5 = 2(x - 3)^2 \)
\( \frac{y + 5}{2} = (x - 3)^2 \)
Taking the square root:
\( \pm\sqrt{\frac{y + 5}{2}} = x - 3 \)

Since the domain is \( x \le 3 \), we have \( x - 3 \le 0 \). Hence, we must choose the negative square root:
\( x - 3 = -\sqrt{\frac{y + 5}{2}} \)
\( x = 3 - \sqrt{\frac{y + 5}{2}} \)

Thus, the inverse function is:
\( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}} \)

The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \).
Since the vertex is at \( (3, -5) \) and the coefficient of \( x^2 \) is positive, the minimum value of \( \mathrm{f}(x) \) is \( -5 \).
For the domain \( x \le 3 \), the range of \( \mathrm{f} \) is \( \mathrm{f}(x) \ge -5 \).
Therefore, the domain of \( \mathrm{f}^{-1} \) is \( x \ge -5 \).

(i)
B1: For completed square form \( 2(x - 3)^2 - 5 \).
M1: For identifying the line of symmetry is at \( x = 3 \).
A1: For identifying the greatest value of \( k \) is 3.

(ii)
M1: For attempting to make \( x \) the subject of \( y = 2(x - 3)^2 - 5 \) or equivalent.
A1: For obtaining \( x = 3 \pm \sqrt{\frac{y+5}{2}} \) or equivalent.
A1: For selecting the negative root and writing the correct inverse function \( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}} \).
B1: For stating the domain of \( \mathrm{f}^{-1} \) is \( x \ge -5 \).

(i) Given:
\( 3 \tan \theta \sin \theta = 8 \)

Substitute \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\( 3 \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta = 8 \)
\( 3 \frac{\sin^2 \theta}{\cos \theta} = 8 \)

Multiply both sides by \( \cos \theta \):
\( 3 \sin^2 \theta = 8 \cos \theta \)

Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( 3(1 - \cos^2 \theta) = 8 \cos \theta \)
\( 3 - 3 \cos^2 \theta = 8 \cos \theta \)

Rearranging all terms to one side:
\( 3\cos^2\theta + 8\cos\theta - 3 = 0 \) (Shown)

(ii) Let \( y = \cos \theta \). The equation becomes:
\( 3y^2 + 8y - 3 = 0 \)

Factorizing the quadratic:
\( (3y - 1)(y + 3) = 0 \)

This gives:
\( \cos \theta = \frac{1}{3} \) or \( \cos \theta = -3 \)

Since \( -1 \le \cos \theta \le 1 \), the equation \( \cos \theta = -3 \) has no real solutions.

For \( \cos \theta = \frac{1}{3} \):
In the interval \( 0 \le \theta \le 2\pi \):
The principal value is \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 1.2309 \dots \approx 1.23 \) radians.
The second solution is \( \theta = 2\pi - 1.2309 \approx 5.052 \dots \approx 5.05 \) radians.

Thus, the solutions are \( \theta = 1.23 \) and \( \theta = 5.05 \).

(i)
M1: For substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
M1: For using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to get a quadratic in \( \cos \theta \).
A1: For obtaining the correct quadratic equation with no errors shown.

(ii)
M1: For attempting to solve the quadratic equation \( 3\cos^2\theta + 8\cos\theta - 3 = 0 \) for \( \cos \theta \).
A1: For obtaining \( \cos \theta = \frac{1}{3} \) and identifying \( \cos \theta = -3 \) has no solutions.
M1: For finding the first solution \( \theta \approx 1.23 \).
A1: For both solutions \( \theta = 1.23 \) and \( \theta = 5.05 \) correct to 3 significant figures.

Part (a): To solve \(|3x - 5| < |2x + 1|\), we square both sides to get \((3x-5)^2 < (2x+1)^2\). Expanding both sides gives \(9x^2 - 30x + 25 < 4x^2 + 4x + 1\). Rearranging yields \(5x^2 - 34x + 24 < 0\). Factoring the quadratic expression gives \((5x - 4)(x - 6) < 0\). Thus, the critical values are \(x = 0.8\) and \(x = 6\). Since we want the expression to be less than zero, the solution is \(0.8 < x < 6\). Part (b): Let \(x = e^{-y}\). Substituting this into the inequality from part (a) gives \(0.8 < e^{-y} < 6\). Taking the natural logarithm of all parts yields \(\ln(0.8) < -y < \ln 6\). Multiplying by \(-1\) reverses the inequality signs, giving \(-\ln 6 < y < -\ln(0.8)\). Since \(-\ln(0.8) = -\ln(4/5) = \ln(5/4) = \ln(1.25)\), the exact range is \(-\ln 6 < y < \ln 1.25\).

M1: Squaring both sides and expanding correctly to obtain a 3-term quadratic inequality. A1: Obtaining \(5x^2 - 34x + 24 < 0\) (or equivalent). A1: Finding the correct interval \(0.8 < x < 6\) (or equivalent). M1: Substituting \(e^{-y}\) for \(x\) and applying logarithms correctly. A1: Performing inequality sign reversal correctly. A1: Obtaining the final exact range \(-\ln 6 < y < \ln 1.25\) (or equivalent exact values).

Part (a): Using exponential laws, we can rewrite the first term: \(3(2^{2x+1}) = 3(2^1 \cdot 2^{2x}) = 6(2^x)^2\). Substituting \(u = 2^x\) into the equation gives \(6u^2 - 17u + 10 = 0\), which is a quadratic in \(u\). Part (b): To solve \(6u^2 - 17u + 10 = 0\), we can factorize the quadratic: \((6u - 5)(u - 2) = 0\). This yields the solutions \(u = 2\) or \(u = \frac{5}{6}\). Re-substituting \(u = 2^x\): For \(2^x = 2\), we find \(x = 1\). For \(2^x = \frac{5}{6}\), taking the natural logarithm of both sides gives \(x \ln 2 = \ln(\frac{5}{6})\), which simplifies to \(x = \frac{\ln(5/6)}{\ln 2} \approx -0.26303\). To 3 significant figures, this root is \(-0.263\).

M1: Applying exponential laws to rewrite \(2^{2x+1}\) as \(2 \cdot (2^x)^2\). A1: Showing clearly the substitution leads to \(6u^2 - 17u + 10 = 0\). M1: Factorizing or solving the quadratic equation in \(u\). A1: Finding both critical values \(u = 2\) and \(u = 5/6\). B1: Obtaining the exact root \(x = 1\). M1: Using logarithms correctly to solve \(2^x = 5/6\). A1: Obtaining \(x = -0.263\) correct to 3 significant figures.

Using the double angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\), we substitute this into the equation: \(3(1 - 2\sin^2\theta) + 7\sin\theta = 5\). Expanding gives \(3 - 6\sin^2\theta + 7\sin\theta = 5\). Rearranging all terms to one side yields the quadratic equation: \(6\sin^2\theta - 7\sin\theta + 2 = 0\). Factoring this quadratic gives \((2\sin\theta - 1)(3\sin\theta - 2) = 0\). This gives two possible cases: Case 1: \(2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2}\). Within the range \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = 30^\circ\) and \(\theta = 180^\circ - 30^\circ = 150^\circ\). Case 2: \(3\sin\theta - 2 = 0 \implies \sin\theta = \frac{2}{3}\). The principal value is \(\theta = \sin^{-1}(2/3) \approx 41.8^\circ\). The second solution in the range is \(\theta = 180^\circ - 41.8^\circ = 138.2^\circ\). Hence, the complete set of solutions is \(\theta = 30^\circ, 41.8^\circ, 138.2^\circ, 150^\circ\).

M1: Using the identity \(\cos(2\theta) = 1 - 2\sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\) only. A1: Correctly obtaining the quadratic equation \(6\sin^2\theta - 7\sin\theta + 2 = 0\). M1: Factorizing or solving the quadratic to find values for \(\sin\theta\). A1: Identifying the values \(\sin\theta = 1/2\) and \(\sin\theta = 2/3\). B1: Finding solutions \(\theta = 30^\circ, 150^\circ\). A1: Finding solution \(\theta = 41.8^\circ\) (accept 41.8). A1: Finding solution \(\theta = 138.2^\circ\) (accept 138.2).

Part (a): Using the product rule \(\frac{d}{dx}(uv) = u'v + uv'\) with \(u = e^{-2x}\) and \(v = 2x^2 - x\): We have \(u' = -2e^{-2x}\) and \(v' = 4x - 1\). Therefore, \(\frac{dy}{dx} = -2e^{-2x}(2x^2 - x) + e^{-2x}(4x - 1)\). Factoring out \(e^{-2x}\) gives \(\frac{dy}{dx} = e^{-2x}[-2(2x^2 - x) + (4x - 1)] = e^{-2x}(-4x^2 + 6x - 1)\). Part (b): For stationary points, set \(\frac{dy}{dx} = 0\). Since \(e^{-2x} \neq 0\) for all real \(x\), we solve \(-4x^2 + 6x - 1 = 0\), which is equivalent to \(4x^2 - 6x + 1 = 0\). Using the quadratic formula: \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(4)(1)}}{2(4)} = \frac{6 \pm \sqrt{20}}{8} = \frac{3 \pm \sqrt{5}}{4}\). To determine the nature of the stationary points, consider the sign of the derivative on either side of each root. Since \(e^{-2x} > 0\), the sign of \(\frac{dy}{dx}\) is determined by the quadratic expression \(Q(x) = -4x^2 + 6x - 1\). The parabola opens downwards, meaning \(Q(x) < 0\) for \(x < \frac{3-\sqrt{5}}{4}\), \(Q(x) > 0\) for \(\frac{3-\sqrt{5}}{4} < x < \frac{3+\sqrt{5}}{4}\), and \(Q(x) < 0\) for \(x > \frac{3+\sqrt{5}}{4}\). Thus, the derivative changes from negative to positive at \(x = \frac{3 - \sqrt{5}}{4}\), making it a local minimum. The derivative changes from positive to negative at \(x = \frac{3 + \sqrt{5}}{4}\), making it a local maximum.

M1: Applying product rule to differentiate the function. A1: Correctly obtaining derivative parts \(-2e^{-2x}\) and \(4x-1\). A1: Correctly simplifying to get \(\frac{dy}{dx} = e^{-2x}(-4x^2 + 6x - 1)\). M1: Setting their derivative equal to zero and attempting to solve the quadratic equation. A1: Finding the exact roots \(x = \frac{3 \pm \sqrt{5}}{4}\). M1: Attempting a valid method (sign of first derivative or second derivative test) to find nature. A1: Correctly identifying \(x = \frac{3 - \sqrt{5}}{4}\) as a minimum and \(x = \frac{3 + \sqrt{5}}{4}\) as a maximum.

Part (a): Integrating the cosine term: \(\int \cos(3x + \frac{\pi}{6}) \, dx = \frac{1}{3} \sin(3x + \frac{\pi}{6})\). Evaluating between the limits \(0\) and \(\frac{\pi}{6}\): Upper limit: \(\frac{1}{3} \sin(3(\frac{\pi}{6}) + \frac{\pi}{6}) = \frac{1}{3} \sin(\frac{\pi}{2} + \frac{\pi}{6}) = \frac{1}{3} \sin(\frac{2\pi}{3}) = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}\). Lower limit: \(\frac{1}{3} \sin(3(0) + \frac{\pi}{6}) = \frac{1}{3} \sin(\frac{\pi}{6}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}\). Subtracting the lower limit from the upper limit: \(\frac{\sqrt{3}}{6} - \frac{1}{6} = \frac{\sqrt{3} - 1}{6}\). Part (b): The width of each interval is \(h = \frac{3 - 0}{3} = 1\). The ordinates are evaluated at \(x = 0, 1, 2, 3\). Let \(y = \frac{4}{2^x + 1}\): For \(x = 0\), \(y_0 = \frac{4}{2^0 + 1} = \frac{4}{2} = 2\). For \(x = 1\), \(y_1 = \frac{4}{2^1 + 1} = \frac{4}{3}\). For \(x = 2\), \(y_2 = \frac{4}{2^2 + 1} = \frac{4}{5} = 0.8\). For \(x = 3\), \(y_3 = \frac{4}{2^3 + 1} = \frac{4}{9}\). Using the trapezium rule: \(\text{Area} \approx \frac{1}{2} h [y_0 + y_3 + 2(y_1 + y_2)] = \frac{1}{2} (1) [2 + \frac{4}{9} + 2(\frac{4}{3} + 0.8)]\). Simplifying inside the brackets: \(\text{Area} \approx 0.5 [2 + 0.44444 + 2(1.33333 + 0.8)] = 0.5 [2.44444 + 4.26667] = 0.5 [6.71111] \approx 3.3556\). To 3 significant figures, this is \(3.36\).

M1: Integrating to get \(k \sin(3x + \pi/6)\) with some constant \(k\). A1: Identifying the correct antiderivative \(\frac{1}{3} \sin(3x + \frac{\pi}{6})\). A1: Correctly evaluating limits to show the exact result \(\frac{\sqrt{3} - 1}{6}\). B1: Finding the interval width \(h = 1\) and listing all four correct \(y\)-values. M1: Applying the trapezium rule formula with their values. A1: Correctly computing the final value as \(3.36\) (to 3 s.f.).

Part (a): Let \(f(x) = e^x - 5x + 1\). Evaluating \(f(x)\) at the boundaries: \(f(2.3) = e^{2.3} - 5(2.3) + 1 \approx 9.97418 - 11.5 = -0.52582 < 0\). \(f(2.4) = e^{2.4} - 5(2.4) + 1 \approx 11.02318 - 12 = 0.02318 > 0\). Since there is a change of sign and \(f(x)\) is a continuous function, a root \(\alpha\) must lie in the interval \((2.3, 2.4)\). Part (b): Applying the iterative formula \(x_{n+1} = \ln(5x_n - 1)\): \(x_1 = 2.35\). \(x_2 = \ln(5(2.35) - 1) = \ln(10.75) \approx 2.37491\). \(x_3 = \ln(5(2.37491) - 1) = \ln(10.87453) \approx 2.38639\). \(x_4 = \ln(5(2.38639) - 1) = \ln(10.93195) \approx 2.39165\). \(x_5 = \ln(5(2.39165) - 1) = \ln(10.95825) \approx 2.39405\). \(x_6 = \ln(5(2.39405) - 1) = \ln(10.97025) \approx 2.39515\). \(x_7 = \ln(5(2.39515) - 1) = \ln(10.97575) \approx 2.39565\). \(x_8 = \ln(5(2.39565) - 1) = \ln(10.97825) \approx 2.39588\). The values are converging to \(2.396\) correct to 3 decimal places. Part (c): To verify that \(\alpha = 2.396\) correct to 3 decimal places, we evaluate \(f(x)\) at the upper and lower bounds of this rounding interval, which are \(2.3955\) and \(2.3965\): \(f(2.3955) = e^{2.3955} - 5(2.3955) + 1 \approx 10.97375 - 11.9775 = -0.00375 < 0\). \(f(2.3965) = e^{2.3965} - 5(2.3965) + 1 \approx 10.98473 - 11.9825 = 0.00223 > 0\). Because there is a sign change between \(2.3955\) and \(2.3965\), the root \(\alpha\) must lie in \((2.3955, 2.3965)\), which means \(\alpha\) rounds to \(2.396\) correct to 3 decimal places.

B1: Showing sign change with calculated values \(f(2.3)\) and \(f(2.4)\). M1: Commencing iteration process correctly. A1: Calculating at least two iterations correctly to 5 decimal places. A1: Finding the root converges to \(2.396\). M1: Evaluating \(f(x)\) at bounds \(2.3955\) and \(2.3965\). A1: Concluding root is \(2.396\) based on the sign change.

Part (a): Since \(x+2\) is a factor of \(P(x)\), by the Factor Theorem we have \(P(-2) = 0\). Substitute \(x = -2\): \(a(-2)^3 + 7(-2)^2 + b(-2) - 6 = 0 \implies -8a + 28 - 2b - 6 = 0 \implies -8a - 2b + 22 = 0 \implies 4a + b = 11\) (Equation 1). By the Remainder Theorem, since dividing by \(x-1\) gives remainder 9, we have \(P(1) = 9\). Substitute \(x = 1\): \(a(1)^3 + 7(1)^2 + b(1) - 6 = 9 \implies a + 7 + b - 6 = 9 \implies a + b = 8\) (Equation 2). Subtract Equation 2 from Equation 1: \((4a + b) - (a + b) = 11 - 8 \implies 3a = 3 \implies a = 1\). Substitute \(a = 1\) into Equation 2: \(1 + b = 8 \implies b = 7\). Thus, \(a=1, b=7\). Part (b): With \(a=1\) and \(b=7\), \(P(x) = x^3 + 7x^2 + 7x - 6\). Since \(x+2\) is a factor, we can write \(P(x) = (x+2)(x^2 + kx - 3)\) for some constant \(k\). Expanding this product: \((x+2)(x^2 + kx - 3) = x^3 + (k+2)x^2 + (2k-3)x - 6\). Comparing the coefficient of \(x^2\): \(k+2 = 7 \implies k = 5\). So \(P(x) = (x+2)(x^2 + 5x - 3) = 0\). This gives \(x+2 = 0 \implies x = -2\) or \(x^2 + 5x - 3 = 0\). Solving the quadratic part using the quadratic formula: \(x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-3)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 12}}{2} = \frac{-5 \pm \sqrt{37}}{2}\). Hence, the exact roots are \(x = -2\) and \(x = \frac{-5 \pm \sqrt{37}}{2}\).

M1: Using \(P(-2) = 0\) to obtain a linear equation in \(a\) and \(b\). A1: Obtaining \(4a + b = 11\) (or equivalent). M1: Using \(P(1) = 9\) to obtain a second linear equation in \(a\) and \(b\). A1: Solving the simultaneous equations to find \(a=1, b=7\). M1: Performing algebraic division or equating coefficients to express \(P(x)\) as \((x+2)(x^2+5x-3)\). A1: Solving the quadratic equation to obtain the remaining exact roots \(x = \frac{-5 \pm \sqrt{37}}{2}\).

(i) The locus of \( z \) satisfying \( |z - (3 + \mathrm{i})| = 2 \) is a circle on the Argand diagram. The center of the circle is at the point \( (3, 1) \) representing the complex number \( 3 + \mathrm{i} \), and the radius of the circle is \( 2 \).

(ii) Let \( C \) be the center of the circle \( (3, 1) \) and \( O \) be the origin. The distance from the origin to the center is:
\[ OC = \sqrt{3^2 + 1^2} = \sqrt{10} \]

Since the radius \( R = 2 \) is less than \( \sqrt{10} \), the origin lies outside the circle. The maximum argument of \( z \) on the circle occurs at the point of tangency where a line from the origin to the circle is tangent to the upper side of the circle.

Let \( \alpha \) be the angle that the line \( OC \) makes with the positive real axis:
\[ \tan \alpha = \frac{1}{3} \implies \alpha \approx 0.32175 \text{ radians} \]

Let \( \beta \) be the angle between the tangent line and \( OC \). In the right-angled triangle formed by the origin, the tangent point, and the center \( C \):
\[ \sin \beta = \frac{R}{OC} = \frac{2}{\sqrt{10}} \implies \beta \approx 0.68472 \text{ radians} \]

The maximum value of \( \arg z \) is:
\[ \theta_{\max} = \alpha + \beta = 0.32175 + 0.68472 = 1.00647 \approx 1.01 \text{ radians} \]

- M1: Sketching a circle centered in the first quadrant.
- A1: Circle correctly centered at \( (3, 1) \) with radius \( 2 \) (must clearly show position relative to axes).
- M1: Recognizing that the maximum argument is achieved at a tangent from the origin, and using trigonometry with the center \( (3,1) \) and radius \( 2 \).
- A1: Finding \( \alpha = \arctan(1/3) \) or \( \beta = \arcsin(2/\sqrt{10}) \) (or equivalent angles).
- A1.8: Correctly calculating the sum of the angles to get \( 1.01 \) radians (allow equivalent answer in degrees, \( 57.7^{\circ} \), only if degrees are clearly specified, but radians is the standard default).

Separate the variables of the differential equation:
\[ \int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x \]

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

Integrating the right-hand side using substitution (let \( u = \ln x \), so \( \mathrm{d}u = \frac{1}{x} \mathrm{d}x \)):
\[ \int \frac{\ln x}{x} \mathrm{d}x = \int u \, \mathrm{d}u = \frac{1}{2} u^2 + c = \frac{1}{2} (\ln x)^2 + c \]

Equating both integrated sides:
\[ -\frac{1}{y} = \frac{1}{2} (\ln x)^2 + c \]

Substitute the initial conditions \( y = -1 \) when \( x = 1 \) to find \( c \):
\[ -\frac{1}{-1} = \frac{1}{2} (\ln 1)^2 + c \implies 1 = 0 + c \implies c = 1 \]

Substitute \( c = 1 \) back into the equation:
\[ -\frac{1}{y} = \frac{1}{2} (\ln x)^2 + 1 = \frac{(\ln x)^2 + 2}{2} \]

Solve for \( y \):
\[ \frac{1}{y} = -\frac{2 + (\ln x)^2}{2} \implies y = -\frac{2}{2 + (\ln x)^2} \]

- M1: Attempting to separate variables and integrating both sides.
- A1: Correct integration of LHS to obtain \( -1/y \).
- A1: Correct integration of RHS to obtain \( \frac{1}{2}(\ln x)^2 \).
- M1: Substituting the given initial conditions \( y = -1, x = 1 \) to find the constant of integration.
- A1: Finding \( c = 1 \) (or equivalent constant depending on the arrangement).
- A1.8: Rearranging the equation correctly to express \( y \) as a function of \( x \): \( y = -\frac{2}{2 + (\ln x)^2} \).

Let \( I = \int e^{2x} \sin x \, \mathrm{d}x \).

Use integration by parts:
Let \( u = \sin x \implies \mathrm{d}u = \cos x \, \mathrm{d}x \)
and \( \mathrm{d}v = e^{2x} \, \mathrm{d}x \implies v = \frac{1}{2} e^{2x} \).

Then:
\[ I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \int e^{2x} \cos x \, \mathrm{d}x \]

Apply integration by parts a second time on the remaining integral:
Let \( u = \cos x \implies \mathrm{d}u = -\sin x \, \mathrm{d}x \)
and \( \mathrm{d}v = e^{2x} \, \mathrm{d}x \implies v = \frac{1}{2} e^{2x} \).

Then:
\[ \int e^{2x} \cos x \, \mathrm{d}x = \frac{1}{2} e^{2x} \cos x - \int \left( \frac{1}{2} e^{2x} \right) (-\sin x) \, \mathrm{d}x = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I \]

Substitute this back into the expression for \( I \):
\[ I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I \right) \]
\[ I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I \]

Add \( \frac{1}{4} I \) to both sides:
\[ \frac{5}{4} I = \frac{1}{4} e^{2x} (2 \sin x - \cos x) \implies I = \frac{1}{5} e^{2x} (2 \sin x - \cos x) \]

Now, evaluate this from limit \( 0 \) to \( \pi/2 \):
\[ \int_0^{\pi/2} e^{2x} \sin x \, \mathrm{d}x = \left[ \frac{1}{5} e^{2x} (2 \sin x - \cos x) \right]_0^{\pi/2} \]

At \( x = \pi/2 \):
\[ \frac{1}{5} e^{\pi} (2 \sin(\pi/2) - \cos(\pi/2)) = \frac{1}{5} e^{\pi} (2(1) - 0) = \frac{2}{5} e^{\pi} \]

At \( x = 0 \):
\[ \frac{1}{5} e^{0} (2 \sin(0) - \cos(0)) = \frac{1}{5} (1) (0 - 1) = -\frac{1}{5} \]

Subtracting the two values:
\[ \frac{2}{5} e^{\pi} - \left( -\frac{1}{5} \right) = \frac{2e^{\pi} + 1}{5} \]

- M1: Attempting integration by parts once, choosing appropriate terms.
- A1: Correctly completing first stage of integration by parts.
- M1: Attempting integration by parts a second time on the resulting integral.
- A1: Correctly obtaining the cyclic expression for \( I \) and solving for it to get \( I = \frac{1}{5} e^{2x} (2 \sin x - \cos x) \).
- M1: Applying the limits \( 0 \) and \( \pi/2 \) correctly to their integrated expression.
- A1.8: Obtaining the exact final simplified answer \( \frac{2e^{\pi} + 1}{5} \) (or equivalent form).

(i) Write the equations of the lines in vector parametric form:
\[ l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]
\[ l_2: \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \]

If the lines intersect, there must exist parameter values \( \lambda \) and \( \mu \) such that:
1) \( 1 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 1 \)
2) \( 2 - \lambda = -1 + \mu \implies \lambda + \mu = 3 \)
3) \( -1 + 2\lambda = 3 - \mu \implies 2\lambda + \mu = 4 \)

From equations 1 and 2, subtract equation 1 from 2:
\[ 3\mu = 2 \implies \mu = \frac{2}{3} \]
Then substitute into equation 2:
\[ \lambda = 3 - \frac{2}{3} = \frac{7}{3} \]

Check if these values satisfy equation 3:
\[ \text{LHS of (3)} = 2\left(\frac{7}{3}\right) + \frac{2}{3} = \frac{16}{3} \]
\[ \text{RHS of (3)} = 4 \]
Since \( \frac{16}{3} \neq 4 \), the system of equations has no consistent solution, so the lines do not intersect.

(ii) The shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\mathbf{a} - \mathbf{b}) \cdot \mathbf{n}|}{|\mathbf{n}|} \]
where \( \mathbf{a} - \mathbf{b} \) is the vector between points \( A \) and \( B \), and \( \mathbf{n} \) is the common perpendicular vector to both lines.

Calculate \( \mathbf{a} - \mathbf{b} \):
\[ \mathbf{a} - \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} \]

Find \( \mathbf{n} \) by calculating the vector product of the direction vectors \( \mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \):
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{pmatrix} = \mathbf{i}(1 - 2) - \mathbf{j}(-1 - 4) + \mathbf{k}(1 - (-2)) = -\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \]

Calculate the scalar product \( (\mathbf{a} - \mathbf{b}) \cdot \mathbf{n} \):
\[ (-\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}) \cdot (-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = 1 + 15 - 12 = 4 \]

Calculate the magnitude \( |\mathbf{n}| \):
\[ |\mathbf{n}| = \sqrt{(-1)^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35} \]

Calculate the shortest distance:
\[ d = \frac{4}{\sqrt{35}} \approx 0.676 \]

- M1: Setting up parametric equations for coordinates on both lines.
- A1: Solving a pair of equations to obtain \( \lambda = 7/3 \) and \( \mu = 2/3 \).
- A1.8: Showing inconsistency in the third equation and concluding that the lines do not intersect.
- M1: Finding the vector product of the two direction vectors.
- A1: Obtaining correct perpendicular vector \( -\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \) (or equivalent scalar multiple).
- A1: Correctly applying the shortest distance formula to get \( \frac{4}{\sqrt{35}} \) or \( 0.676 \).

(i) Let \( f(x) = x^3 - 5x + 1 \). Since \( f(x) \) is a continuous polynomial function, we evaluate it at the endpoints of the interval:
\[ f(0) = (0)^3 - 5(0) + 1 = 1 > 0 \]
\[ f(1) = (1)^3 - 5(1) + 1 = 1 - 5 + 1 = -3 < 0 \]

There is a change of sign between \( f(0) \) and \( f(1) \). Therefore, by the Intermediate Value Theorem, there exists at least one root \( \alpha \) in the interval \( (0, 1) \).

(ii) Using the iterative formula \( x_{n+1} = \frac{x_n^3 + 1}{5} \) with \( x_1 = 0.2 \):
\[ x_2 = \frac{0.2^3 + 1}{5} = \frac{0.008 + 1}{5} = 0.20160 \]
\[ x_3 = \frac{0.20160^3 + 1}{5} \approx \frac{0.008194 + 1}{5} = 0.20164 \]
\[ x_4 = \frac{0.20164^3 + 1}{5} \approx \frac{0.008199 + 1}{5} = 0.20164 \]

Since the successive values of the iterations have converged to \( 0.2016 \) up to 4 decimal places, the root correct to 3 decimal places is \( 0.202 \).

- M1: Calculating values of the function at \( x=0 \) and \( x=1 \).
- A1: Correct evaluation of \( f(0)=1 \) and \( f(1)=-3 \) with a conclusion mentioning the sign change.
- M1: Attempting to use the given iteration formula to calculate at least \( x_2 \).
- A1: Obtaining \( x_2 = 0.20160 \) correctly.
- A1: Obtaining subsequent iterations \( x_3 = 0.20164 \) and \( x_4 = 0.20164 \) correct to 5 decimal places.
- A1.8: Giving final answer as \( 0.202 \) and showing sufficient iterations to justify this precision.

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

Now differentiating \( y \) with respect to \( \theta \):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\sin(2\theta) \]
Using the double-angle identity \( \sin(2\theta) = 2\sin\theta\cos\theta \):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 4\sin\theta\cos\theta \]

Now, use the chain rule to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{4\sin\theta\cos\theta}{4\sin^2\theta} = \frac{\cos\theta}{\sin\theta} = \cot\theta \]

(ii) At the point where \( \theta = \frac{\pi}{4} \):
\[ x = 2\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1 \]
\[ y = 1 - \cos\left(\frac{\pi}{2}\right) = 1 \]

The gradient of the tangent at \( \theta = \frac{\pi}{4} \) is:
\[ m = \cot\left(\frac{\pi}{4}\right) = 1 \]

Using the point-slope form, the equation of the tangent line is:
\[ y - 1 = 1 \left( x - \left(\frac{\pi}{2} - 1\right) \right) \]
\[ y - 1 = x - \frac{\pi}{2} + 1 \]
\[ y = x - \frac{\pi}{2} + 2 \]

- M1: Differentiating both parametric equations to find \( \frac{\mathrm{d}x}{\mathrm{d}\theta} \) and \( \frac{\mathrm{d}y}{\mathrm{d}\theta} \).
- A1: Finding correct derivatives: \( 2 - 2\cos(2\theta) \) and \( 2\sin(2\theta) \).
- M1: Applying double-angle trigonometric identities to simplify the derivatives.
- A0.8: Successfully completing the proof to show \( \frac{\mathrm{d}y}{\mathrm{d}x} = \cot\theta \).
- M1: Finding the coordinates \( (x, y) \) and the gradient at \( \theta = \frac{\pi}{4} \).
- A1: Finding coordinates \( (\pi/2 - 1, 1) \) and gradient \( m = 1 \).
- A1: Finding the correct tangent line equation \( y = x - \frac{\pi}{2} + 2 \) (or equivalent form).

(i) We want to express \( 3\sin x - 2\cos x \) in the form \( R\sin(x - \alpha) \):
\[ R\sin(x - \alpha) = R\sin x\cos\alpha - R\cos x\sin\alpha \]

Equating the coefficients:
\[ R\cos\alpha = 3 \]
\[ R\sin\alpha = 2 \]

To find \( R \):
\[ R^2 = 3^2 + 2^2 = 13 \implies R = \sqrt{13} \approx 3.61 \]

To find \( \alpha \):
\[ \tan\alpha = \frac{2}{3} \implies \alpha = \arctan\left(\frac{2}{3}\right) \approx 0.588 \text{ radians} \]

So:
\[ 3\sin x - 2\cos x = \sqrt{13}\sin(x - 0.588) \]

(ii) Substitute this form into the equation:
\[ \sqrt{13}\sin(x - 0.588) = 1 \]
\[ \sin(x - 0.588) = \frac{1}{\sqrt{13}} \]

Find the principal value of the inverse sine:
\[ x - 0.588 = \arcsin\left(\frac{1}{\sqrt{13}}\right) \approx 0.2810 \text{ radians} \]

We need solutions in the range \( 0 \le x \le 2\pi \), which means \( -0.588 \le x - 0.588 \le 5.695 \).

The possible values for \( x - 0.588 \) are:
1) \( x - 0.588 = 0.2810 \implies x = 0.2810 + 0.588 = 0.869 \text{ radians} \)
2) \( x - 0.588 = \pi - 0.2810 \approx 2.8606 \implies x = 2.8606 + 0.588 = 3.449 \approx 3.45 \text{ radians} \)

Thus, the solutions in the interval are \( x = 0.869 \) and \( x = 3.45 \).

- M1: Setting up the simultaneous equations for \( R \) and \( \alpha \) by expanding \( R\sin(x-\alpha) \).
- A1: Finding \( R = \sqrt{13} \) (or equivalent exact form).
- A1: Finding \( \alpha = 0.588 \) (or \( 0.5880 \) to 4 d.p.).
- M1: Solving \( \sin(x - 0.588) = \frac{1}{\sqrt{13}} \) to find a principal value.
- A1: Finding one correct root \( x = 0.869 \).
- A1.8: Finding the second correct root \( x = 3.45 \) and ensuring no extra values are in the interval.

(i) Express in partial fractions:
\[ \frac{x + 7}{(x + 1)(2x - 1)} = \frac{A}{x + 1} + \frac{B}{2x - 1} \]

Multiply by the denominator \( (x + 1)(2x - 1) \):
\[ x + 7 = A(2x - 1) + B(x + 1) \]

To find \( A \), let \( x = -1 \):
\[ -1 + 7 = A(2(-1) - 1) \implies 6 = -3A \implies A = -2 \]

To find \( B \), let \( x = \frac{1}{2} \):
\[ \frac{1}{2} + 7 = B\left(\frac{1}{2} + 1\right) \implies \frac{15}{2} = \frac{3}{2}B \implies B = 5 \]

Thus:
\[ f(x) = -\frac{2}{x + 1} + \frac{5}{2x - 1} \]

(ii) Rewrite \( f(x) \) using negative exponents to facilitate binomial expansion:
\[ f(x) = -2(1 + x)^{-1} + 5(2x - 1)^{-1} = -2(1 + x)^{-1} - 5(1 - 2x)^{-1} \]

Now, expand each term using the general binomial expansion \( (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots \):

For \( -2(1 + x)^{-1} \):
\[ -2(1 + x)^{-1} = -2\left(1 - x + x^2 - \dots\right) = -2 + 2x - 2x^2 \]

For \( -5(1 - 2x)^{-1} \):
\[ -5(1 - 2x)^{-1} = -5\left(1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + \dots\right) \]
\[ = -5\left(1 + 2x + 4x^2 + \dots\right) = -5 - 10x - 20x^2 \]

Sum the two expansions:
\[ f(x) = (-2 + 2x - 2x^2) + (-5 - 10x - 20x^2) \]
\[ f(x) = -7 - 8x - 22x^2 \]

- M1: Setting up correct partial fraction form and attempting to solve for coefficients.
- A1: Correctly finding either \( A = -2 \) or \( B = 5 \).
- A1: Obtaining the correct partial fractions: \( f(x) = -\frac{2}{x + 1} + \frac{5}{2x - 1} \).
- M1: Rewriting fractions with negative indices and factoring out coefficients appropriately for binomial expansion.
- A1: Correctly obtaining the expansion of \( (1+x)^{-1} \) or \( (1-2x)^{-1} \) up to the \( x^2 \) term.
- A1: Obtaining correct expansions for both components.
- A0.8: Correctly combining terms to find the final quadratic expression: \( -7 - 8x - 22x^2 \).

1. Identify the locus of \(z\):
The equation \(|z - (3 + 4\mathrm{i})| = 2\) represents a circle in the Argand diagram with center \(C(3, 4)\) and radius \(R = 2\).

2. Find the distance from the origin \(O(0,0)\) to the center \(C\):
\(OC = \sqrt{3^2 + 4^2} = 5\).

3. Calculate the angle \(\theta\) of the center \(C\) relative to the positive real axis:
\(\theta = \arg(3 + 4\mathrm{i}) = \arctan\left(\frac{4}{3}\right) \approx 0.9273\) radians.

4. Use geometry to find the maximum variation in angle, \(\alpha\), from the line \(OC\):
The tangent lines from the origin to the circle make an angle \(\alpha\) with the line \(OC\), where:
\(\sin\alpha = \frac{R}{OC} = \frac{2}{5} = 0.4\)
\(\alpha = \arcsin(0.4) \approx 0.4115\) radians.

5. Calculate the minimum and maximum values of \(\arg(z)\):
Minimum \(\arg(z) = \theta - \alpha \approx 0.9273 - 0.4115 = 0.516\) radians (to 3 d.p.).
Maximum \(\arg(z) = \theta + \alpha \approx 0.9273 + 0.4115 = 1.339\) radians (to 3 d.p.).

M1: Sketch or state that the locus of \(z\) is a circle with center \(3 + 4\mathrm{i}\) and radius 2.
M1: Find the distance from the origin to the center, \(OC = 5\).
M1: Calculate the angle of the center, \(\theta = \arctan(4/3)\) (or approx 0.927).
M1: State or use \(\sin\alpha = \frac{2}{5}\) to find the angular radius \(\alpha = \arcsin(0.4)\) (or approx 0.412).
M1: Realise that minimum arg is \(\theta - \alpha\) and maximum arg is \(\theta + \alpha\).
A1: Correct minimum value of 0.516 (allow 0.515 to 0.517).
A1: Correct maximum value of 1.339 (allow 1.338 to 1.340).

1. Separate the variables:
\[\int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x\]

2. Integrate both sides:
LHS: \[\int y^{-2} \mathrm{d}y = -\frac{1}{y}\]
RHS: Let \(u = \ln x\), then \(\mathrm{d}u = \frac{1}{x} \mathrm{d}x\).
\[\int u \mathrm{d}u = \frac{1}{2} u^2 = \frac{1}{2} (\ln x)^2\]
Combining these gives:
\[-\frac{1}{y} = \frac{1}{2} (\ln x)^2 + C\]

3. Use the boundary condition \(y = -2\) when \(x = 1\) to find \(C\):
\[-\frac{1}{-2} = \frac{1}{2} (\ln 1)^2 + C \implies \frac{1}{2} = 0 + C \implies C = \frac{1}{2}\]

4. Substitute \(C\) back into the equation:
\[-\frac{1}{y} = \frac{1}{2} (\ln x)^2 + \frac{1}{2} = \frac{(\ln x)^2 + 1}{2}\]

5. Solve for \(y\):
\[y = -\frac{2}{(\ln x)^2 + 1}\]

M1: Separate variables correctly to obtain \(\int y^{-2} \mathrm{d}y = \int x^{-1} \ln x \mathrm{d}x\).
B1: Integrate LHS to get \(-1/y\).
M1: Integrate RHS using substitution or inspection to get \(k (\ln x)^2\).
A1: Obtain correct integrated RHS: \(\frac{1}{2} (\ln x)^2 + C\).
M1: Substitute \(y = -2\) and \(x = 1\) to find the constant of integration \(C\).
A1: Obtain \(C = \frac{1}{2}\) (or equivalent depending on integration constants).
A1: Obtain the final explicit form \(y = -\frac{2}{(\ln x)^2 + 1}\) (or equivalent).

**Part (i)**
Let \[\frac{7x - 1}{(1 - x)(1 + 2x)} = \frac{A}{1 - x} + \frac{B}{1 + 2x}\]
Multiply through by \((1-x)(1+2x)\):
\[7x - 1 = A(1 + 2x) + B(1 - x)\]
- Let \(x = 1\):
\[7(1) - 1 = A(1 + 2(1)) \implies 6 = 3A \implies A = 2\]
- Let \(x = -\frac{1}{2}\):
\[7\left(-\frac{1}{2}\right) - 1 = B\left(1 - \left(-\frac{1}{2}\right)\right) \implies -\frac{9}{2} = \frac{3}{2} B \implies B = -3\]
Thus, the partial fractions are:
\[\frac{2}{1 - x} - \frac{3}{1 + 2x}\]

**Part (ii)**
We expand each term as a binomial series:
1. For \(2(1 - x)^{-1}\):
\[2(1 - x)^{-1} = 2\left(1 + x + x^2 + \dots\right) = 2 + 2x + 2x^2\]
2. For \(-3(1 + 2x)^{-1}\):
\[-3(1 + 2x)^{-1} = -3\left(1 + (-1)(2x) + \frac{(-1)(-2)}{2}(2x)^2 + \dots\right) = -3(1 - 2x + 4x^2) = -3 + 6x - 12x^2\]
Summing the two expansions:
\[(2 + 2x + 2x^2) + (-3 + 6x - 12x^2) = -1 + 8x - 10x^2\]

Part (i):
M1: State or imply the form \(\frac{A}{1-x} + \frac{B}{1+2x}\) and solve for at least one constant.
A1: Obtain \(A = 2\) and \(B = -3\).

Part (ii):
M1: Use binomial theorem to expand \((1-x)^{-1}\) or \((1+2x)^{-1}\) up to the \(x^2\) term.
A1: Correct expansion for \(2(1-x)^{-1}\) as \(2 + 2x + 2x^2\).
A1: Correct expansion for \(-3(1+2x)^{-1}\) as \(-3 + 6x - 12x^2\).
M1: Add the two expansions together.
A1: Obtain the correct final simplified expression \(-1 + 8x - 10x^2\).

Given mass \(m = 3\text{ kg}\), the weight of the particle is \(W = mg = 3 \times 10 = 30\text{ N}\). The angle of inclination \(\alpha\) satisfies \(\sin \alpha = 0.6\), which gives \(\cos \alpha = 0.8\). The normal reaction force \(R\) is perpendicular to the plane: \(R = W \cos \alpha = 30 \times 0.8 = 24\text{ N}\). The maximum frictional force is given by: \(F_{\text{max}} = \mu R = 0.4 \times 24 = 9.6\text{ N}\). The component of the weight down the plane is: \(W \sin \alpha = 30 \times 0.6 = 18\text{ N}\). We consider two limiting cases of equilibrium: 1) The particle is on the point of sliding down the slope. In this case, the frictional force acts up the slope: \(P + F_{\text{max}} = W \sin \alpha \implies P + 9.6 = 18 \implies P = 8.4\text{ N}\). 2) The particle is on the point of sliding up the slope. In this case, the frictional force acts down the slope: \(P = W \sin \alpha + F_{\text{max}} \implies P = 18 + 9.6 = 27.6\text{ N}\). Thus, the range of values for \(P\) to maintain equilibrium is \(8.4 \le P \le 27.6\).

M1: Attempt to resolve forces perpendicular to the plane to find \(R\). A1: Correct normal reaction \(R = 24\text{ N}\). M1: Use of \(F = \mu R\) to find limiting friction. A1: Correct \(F_{\text{max}} = 9.6\text{ N}\). M1: Set up equilibrium equations for both limiting cases. A1: Correct lower limit \(P = 8.4\text{ N}\). A1.1: Correct upper limit \(P = 27.6\text{ N}\).

Let \(T\) be the tension in the string and \(a\) be the acceleration of the system before \(B\) hits the floor. For particle \(A\) on the rough horizontal table, the normal reaction is \(R = m_A g = 0.8 \times 10 = 8\text{ N}\). The friction force is \(F = \mu R = 0.25 \times 8 = 2\text{ N}\). The equation of motion for \(A\) is \(T - 2 = 0.8 a\). For particle \(B\), the equation of motion is \(m_B g - T = m_B a \implies 4 - T = 0.4 a\). Adding the two equations: \(2 = 1.2 a \implies a = 5/3\text{ m s}^{-2}\). The distance travelled by \(B\) before hitting the floor is \(s_1 = 1.5\text{ m}\). The speed \(v\) of both particles at the instant \(B\) hits the floor is given by \(v^2 = u^2 + 2 a s_1 = 0 + 2 \times (5/3) \times 1.5 = 5\text{ m}^2\text{ s}^{-2}\). Once \(B\) hits the floor, the string becomes slack, so the tension \(T = 0\). The only horizontal force acting on \(A\) is friction, so its new acceleration \(a'\) is given by \(-F = m_A a' \implies -2 = 0.8 a' \implies a' = -2.5\text{ m s}^{-2}\). Let \(s_2\) be the distance \(A\) travels while decelerating to rest: \(0 = v^2 + 2 a' s_2 \implies 0 = 5 + 2(-2.5)s_2 \implies s_2 = 1\text{ m}\). The total distance travelled by \(A\) is \(s_{\text{total}} = s_1 + s_2 = 1.5 + 1 = 2.5\text{ m}\).

M1: Use of \(F = \mu R\) to find the frictional force on \(A\). A1: Correct friction force \(F = 2\text{ N}\). M1: Write equations of motion for both particles and solve for acceleration \(a\). A1: Correct acceleration \(a = 5/3\text{ m s}^{-2}\). M1: Find velocity when \(B\) hits the floor. M1: Calculate deceleration of \(A\) after \(B\) hits the floor and find the further distance \(s_2\). A1.1: Correct total distance \(2.5\text{ m}\).

To find the maximum speed, we differentiate \(v\) with respect to \(t\) to find the acceleration \(a\): \(a = dv/dt = 0.1(18t - 3t^2) = 0.3t(6 - t)\). Setting \(a = 0\) for stationary points gives \(t = 0\) or \(t = 6\). The maximum speed occurs at \(t = 6\): \(v(6) = 0.1(9(6^2) - 6^3) = 0.1(324 - 216) = 10.8\text{ m s}^{-1}\). The particle comes to instantaneous rest when \(v = 0\): \(0.1(9t^2 - t^3) = 0 \implies t = 9\). The distance travelled \(s\) from \(t = 0\) to \(t = 9\) is given by: \(s = \int_{0}^{9} 0.1(9t^2 - t^3)\,dt = 0.1 \left[ 3t^3 - 0.25t^4 \right]_{0}^{9} = 0.1 \left( 3(729) - 1640.25 \right) = 0.1(2187 - 1640.25) = 54.675\text{ m}\).

M1: Differentiate \(v(t)\) to obtain the acceleration \(a(t)\). A1: Correct acceleration expression \(a = 0.1(18t - 3t^2)\). M1: Find the value of \(t\) where \(a = 0\) and compute the maximum speed. A1: Correct maximum speed \(10.8\text{ m s}^{-1}\). M1: Integrate \(v(t)\) with respect to \(t\). A1: Correct integration \([0.3t^3 - 0.025t^4]\). A1.1: Correct distance \(54.675\text{ m}\).

The component of the weight down the slope is: \(W_{\parallel} = mg \sin \theta = 1200 \times 10 \times 0.05 = 600\text{ N}\). (i) The power of the engine is \(P = 24000\text{ W}\). At speed \(v = 12\text{ m s}^{-1}\), the driving force \(DF\) is: \(DF = P/v = 24000/12 = 2000\text{ N}\). Using Newton's second law along the slope: \(DF - R - W_{\parallel} = ma \implies 2000 - 600 - 600 = 1200 a \implies 800 = 1200 a \implies a = 2/3 \approx 0.667\text{ m s}^{-2}\). (ii) Let \(v_{\text{steady}}\) be the steady speed under the increased power \(P' = 31200\text{ W}\). At steady speed, acceleration is 0, so the driving force \(DF'\) is: \(DF' = R + W_{\parallel} = 600 + 600 = 1200\text{ N}\). Since \(DF' = P'/v_{\text{steady}}\): \(1200 = 31200/v_{\text{steady}} \implies v_{\text{steady}} = 31200/1200 = 26\text{ m s}^{-1}\).

M1: Calculate the component of weight down the slope. A1: Correct weight component \(600\text{ N}\). M1: Use \(P = DF \times v\) to find the driving force. A1: Correct acceleration \(0.667\text{ m s}^{-2}\). M1: Apply equilibrium condition for steady speed to find the required driving force. A1: Correct steady driving force of \(1200\text{ N}\). A1.1: Correct steady speed of \(26\text{ m s}^{-1}\).

Let \(v_A\) and \(v_B\) be the velocities after collision. By conservation of momentum: \(0.3 \times 6 + 0.5 \times 2 = 0.3 v_A + 0.5 v_B \implies 3 v_A + 5 v_B = 28\) (Eq 1). The initial kinetic energy is \(KE_i = 0.5 \times 0.3 \times 6^2 + 0.5 \times 0.5 \times 2^2 = 5.4 + 1.0 = 6.4\text{ J}\). The final kinetic energy is \(KE_f = 6.4 - 0.96 = 5.44\text{ J}\). So, \(0.15 v_A^2 + 0.25 v_B^2 = 5.44 \implies 3 v_A^2 + 5 v_B^2 = 108.8\) (Eq 2). Substitute \(v_A = (28 - 5 v_B)/3\) into Eq 2: \(3((28 - 5 v_B)/3)^2 + 5 v_B^2 = 108.8 \implies 40 v_B^2 - 280 v_B + 457.6 = 0 \implies 25 v_B^2 - 175 v_B + 286 = 0\). Solving the quadratic gives \(v_B = (175 \pm 45)/50\). This yields: 1) \(v_B = 4.4\text{ m s}^{-1} \implies v_A = 2\text{ m s}^{-1}\). 2) \(v_B = 2.6\text{ m s}^{-1} \implies v_A = 5\text{ m s}^{-1}\). Since \(A\) was behind \(B\) initially, we must have \(v_A \le v_B\) after collision. The second solution is physically impossible because \(5 > 2.6\). Thus, the final speeds are \(v_A = 2\text{ m s}^{-1}\) and \(v_B = 4.4\text{ m s}^{-1}\).

M1: Use conservation of momentum to obtain an equation in \(v_A\) and \(v_B\). A1: Correct equation \(3 v_A + 5 v_B = 28\). M1: Calculate initial KE and use the lost KE to write a final KE equation. A1: Correct KE equation \(3 v_A^2 + 5 v_B^2 = 108.8\). M1: Substitute and solve the resulting quadratic equation. A1: Find the two mathematical solution pairs. A1.1: Correctly reject the physically impossible solution and state the final speeds.

(i) Using \(s = ut + 0.5at^2\) with \(u = 0\), \(s = 8\), \(t = 4\): \(8 = 0.5 a (16) \implies 8 = 8 a \implies a = 1\text{ m s}^{-2}\). (ii) Resolving perpendicular to the slope: \(R + 40 \sin 15^\circ = 5g \cos 30^\circ \implies R = 50 \cos 30^\circ - 40 \sin 15^\circ \approx 43.301 - 10.353 = 32.948\text{ N}\). Resolving parallel to the slope: \(40 \cos 15^\circ - F - 5g \sin 30^\circ = ma \implies 40 \cos 15^\circ - F - 25 = 5 \times 1 \implies F = 40 \cos 15^\circ - 30 \approx 38.637 - 30 = 8.637\text{ N}\). Using \(F = \mu R \implies \mu = F/R = 8.637/32.948 \approx 0.262\).

B1: Show \(a = 1\text{ m s}^{-2}\) using kinematics. M1: Resolve forces perpendicular to the plane, including the component of the pulling force. A1: Correct normal reaction \(R \approx 32.95\text{ N}\). M1: Resolve forces parallel to the plane. A1: Correct frictional force \(F \approx 8.64\text{ N}\). M1: Use \(\mu = F/R\). A1.1: Correct coefficient of friction \(\mu \approx 0.262\).

(i) During the acceleration phase, \(u = 0\), \(a = 0.4\text{ m s}^{-2}\) and \(t_1 = 30\text{ s}\). The constant speed \(V\) reached is: \(V = u + a t_1 = 0 + 0.4 \times 30 = 12\text{ m s}^{-1}\). (ii) The distance travelled during this phase is: \(s_1 = 0.5(u + V)t_1 = 0.5(0 + 12) \times 30 = 180\text{ m}\). During the deceleration phase, the train decelerates from \(12\text{ m s}^{-1}\) to \(0\) at \(0.5\text{ m s}^{-2}\). The time taken \(t_3\) is: \(t_3 = V / 0.5 = 12/0.5 = 24\text{ s}\). The distance travelled during this phase is: \(s_3 = 0.5(V + 0)t_3 = 0.5 \times 12 \times 24 = 144\text{ m}\). The distance travelled at constant speed is: \(s_2 = 2400 - s_1 - s_3 = 2400 - 180 - 144 = 2076\text{ m}\). The time taken for this phase is: \(t_2 = s_2 / V = 2076 / 12 = 173\text{ s}\). The total time taken is: \(T = t_1 + t_2 + t_3 = 30 + 173 + 24 = 227\text{ s}\).

B1: Find the constant speed \(V = 12\text{ m s}^{-1}\). M1: Calculate the distance during the acceleration phase. A1: Correct distance \(s_1 = 180\text{ m}\). M1: Calculate the time and distance during the deceleration phase. A1: Correct deceleration distance \(s_3 = 144\text{ m}\) and time \(t_3 = 24\text{ s}\). M1: Subtract to find the distance for the constant speed phase and calculate its duration. A1.1: Correct total time taken of \(227\text{ s}\).

(a) The committee of 5 can contain 3, 4, or 5 girls. Case 1: 3 girls and 2 boys. Number of ways = \(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\). Case 2: 4 girls and 1 boy. Number of ways = \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\). Case 3: 5 girls and 0 boys. Number of ways = \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\). Total number of ways = \(120 + 60 + 6 = 186\). (b) First, arrange the 6 girls in a line, which can be done in \(6! = 720\) ways. This creates 7 spaces (including the ends) where boys can stand: _ G _ G _ G _ G _ G _ G _ . We choose 4 of these 7 spaces for the 4 boys and arrange them, which can be done in \(\binom{7}{4} \times 4! = 35 \times 24 = 840\) ways. Total number of arrangements = \(720 \times 840 = 604,800\). (c) Treat Alice and Beatrice as a single unit (AB), which can be arranged in \(2!\) ways. To arrange the remaining students such that George and Harry are not together: First, arrange the 7 units other than George and Harry: (AB), and the other 6 individual students. The number of arrangements is \(7! \times 2! = 10,080\) ways. George and Harry can then be placed in any of the 8 available spaces between and at the ends of these 7 units. The number of ways to place them is \({}^{8}\text{P}_{2} = 56\). Total number of arrangements = \(10,080 \times 56 = 564,480\).

(a) M1 for finding and summing three appropriate cases. A1 for any two cases correct. A1 for 186. (b) M1 for arranging 6 girls in \(6!\) ways. M1 for multiplying by \(\binom{7}{4} \times 4!\) (or \({}^{7}\text{P}_{4}\)). A1 for 604,800. (c) M1 for treating Alice and Beatrice as a unit to find arrangements of \(7! \times 2!\). M1 for placing George and Harry in the 8 slots: \(10,080 \times {}^{8}\text{P}_{2}\) (or using subtraction from total together). A0.3 for 564,480.

(a) The total number of ways to choose 3 balls from 10 is \(\binom{10}{3} = 120\). The number of ways to choose 1 ball of each colour is \(\binom{5}{1} \times \binom{3}{1} \times \binom{2}{1} = 5 \times 3 \times 2 = 30\). Therefore, the probability is \(\frac{30}{120} = \frac{1}{4} = 0.25\). (b) Let \(D\) be the event that the first two balls are of different colours, and \(B_3\) be the event that the third ball is blue. The probability that the first two balls are of the same colour is: \(P(\text{same colour}) = P(RR) + P(BB) + P(GG) = \left(\frac{5}{10} \times \frac{4}{9}\right) + \left(\frac{3}{10} \times \frac{2}{9}\right) + \left(\frac{2}{10} \times \frac{1}{9}\right) = \frac{20 + 6 + 2}{90} = \frac{28}{90} = \frac{14}{45}\). Thus, \(P(D) = 1 - \frac{14}{45} = \frac{31}{45}\). The probability of drawing different colours for the first two and a blue third is: \(P(B_3 \cap D) = P(RB B) + P(BR B) + P(RG B) + P(GR B) + P(BG B) + P(GB B)\). Calculating this gives: \(P(RB B) = \frac{5}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{30}{720}\), \(P(BR B) = \frac{3}{10} \times \frac{5}{9} \times \frac{2}{8} = \frac{30}{720}\), \(P(RG B) = \frac{5}{10} \times \frac{2}{9} \times \frac{3}{8} = \frac{30}{720}\), \(P(GR B) = \frac{2}{10} \times \frac{5}{9} \times \frac{3}{8} = \frac{30}{720}\), \(P(BG B) = \frac{3}{10} \times \frac{2}{9} \times \frac{2}{8} = \frac{12}{720}\), \(P(GB B) = \frac{2}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{12}{720}\). Summing these gives: \(P(B_3 \cap D) = \frac{30 \times 4 + 12 \times 2}{720} = \frac{144}{720} = \frac{1}{5}\). Using conditional probability: \(P(B_3 \mid D) = \frac{P(B_3 \cap D)}{P(D)} = \frac{1/5}{31/45} = \frac{9}{31} \approx 0.290\). (c) The probability of drawing no red balls is the probability of drawing 3 balls from the 5 non-red balls (3 blue, 2 green): \(P(\text{no Red}) = \frac{\binom{5}{3}}{\binom{10}{3}} = \frac{10}{120} = \frac{1}{12}\). Thus, \(P(\text{at least 1 Red}) = 1 - P(\text{no Red}) = 1 - \frac{1}{12} = \frac{11}{12} \approx 0.917\).

(a) M1 for using combinations or multiplying probabilities with permutations. A1 for 30 ways or equivalent probability terms. A1 for 0.25. (b) M1 for finding \(P(D) = 31/45\). M1 for finding \(P(B_3 \cap D) = 1/5\). A1 for 9/31 (or 0.290). (c) M1 for finding the complement probability of no red balls. A1.3 for 11/12 (or 0.917).

(a) The sum of the probabilities must equal 1: \(\sum_{x=1}^{4} P(X=x) = k(5) + k(4) + k(3) + k(2) = 1 \implies 14k = 1 \implies k = \frac{1}{14}\). (b) The probability distribution of \(X\) is: \(P(X=1) = \frac{5}{14}\), \(P(X=2) = \frac{4}{14}\), \(P(X=3) = \frac{3}{14}\), \(P(X=4) = \frac{2}{14}\). The expectation is: \(E(X) = 1\left(\frac{5}{14}\right) + 2\left(\frac{4}{14}\right) + 3\left(\frac{3}{14}\right) + 4\left(\frac{2}{14}\right) = \frac{30}{14} = \frac{15}{7} \approx 2.14\). The expectation of \(X^2\) is: \(E(X^2) = 1^2\left(\frac{5}{14}\right) + 2^2\left(\frac{4}{14}\right) + 3^2\left(\frac{3}{14}\right) + 4^2\left(\frac{2}{14}\right) = \frac{5 + 16 + 27 + 32}{14} = \frac{80}{14} = \frac{40}{7}\). The variance is: \(\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{40}{7} - \left(\frac{15}{7}\right)^2 = \frac{280 - 225}{49} = \frac{55}{49} \approx 1.12\). (c) Let \(X_1\) and \(X_2\) be the results of the two rolls. We want to find \(P(X_1 + X_2 > 5)\). The possible outcomes where the sum is greater than 5 are: \((2,4)\) with probability \(\frac{4}{14} \times \frac{2}{14} = \frac{8}{196}\), \((3,3)\) with probability \(\frac{3}{14} \times \frac{3}{14} = \frac{9}{196}\), \((4,2)\) with probability \(\frac{2}{14} \times \frac{4}{14} = \frac{8}{196}\), \((3,4)\) with probability \(\frac{3}{14} \times \frac{2}{14} = \frac{6}{196}\), \((4,3)\) with probability \(\frac{2}{14} \times \frac{3}{14} = \frac{6}{196}\), \((4,4)\) with probability \(\frac{2}{14} \times \frac{2}{14} = \frac{4}{196}\). Summing these probabilities: \(P(X_1 + X_2 > 5) = \frac{8 + 9 + 8 + 6 + 6 + 4}{196} = \frac{41}{196} \approx 0.209\).

(a) M1 for setting sum of probabilities to 1. A1 for showing \(k = 1/14\). (b) M1 for finding \(E(X)\). M1 for finding \(E(X^2)\). A1.3 for both \(E(X) = 15/7\) (or 2.14) and \(\text{Var}(X) = 55/49\) (or 1.12). (c) M1 for listing the correct outcomes for sum > 5. M1 for calculating their probabilities and summing. A1 for 41/196 (or 0.209).

(a) Let \(X \sim N(\mu, \sigma^2)\). We are given: \(P(X > 180) = 0.15 \implies P\left(Z > \frac{180 - \mu}{\sigma}\right) = 0.15\). From normal distribution tables, the z-score corresponding to a tail probability of 0.15 is 1.036 (or 1.0364). Thus: \(\frac{180 - \mu}{\sigma} = 1.036 \implies 180 - \mu = 1.036\sigma\) [Equation 1]. We are also given: \(P(X < 130) = 0.25 \implies P\left(Z < \frac{130 - \mu}{\sigma}\right) = 0.25\). From normal distribution tables, the z-score corresponding to a left-tail probability of 0.25 is \(-0.674\) (or \(-0.6745\)). Thus: \(\frac{130 - \mu}{\sigma} = -0.674 \implies 130 - \mu = -0.674\sigma\) [Equation 2]. Subtracting Equation 2 from Equation 1 gives: \(50 = 1.710\sigma \implies \sigma \approx 29.2\). Substituting \(\sigma = 29.24\) back into Equation 1 gives: \(180 - \mu = 1.036(29.24) \implies \mu \approx 150\). (b) Let \(Y\) be the number of apples with a mass greater than 180 grams. Since 15% of the apples have a mass greater than 180 grams, we have \(Y \sim B(8, 0.15)\). We want to find: \(P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1)\). \(P(Y = 0) = (0.85)^8 \approx 0.2725\). \(P(Y = 1) = \binom{8}{1}(0.15)^1(0.85)^7 \approx 0.3847\). Thus, \(P(Y \ge 2) = 1 - 0.2725 - 0.3847 = 0.343\) (to 3 s.f.).

(a) M1 for setting up standardisation equation for either condition. A1 for correct z-values (1.036 and -0.674). M1 for solving simultaneous equations. A1 for \(\sigma = 29.2\) (accept 29.2 to 29.3). A1 for \(\mu = 150\) (accept 149.7 to 150). (b) M1 for identifying Binomial distribution \(B(8, 0.15)\). M1 for calculating \(1 - P(0) - P(1)\). A1.3 for 0.343 (accept 0.342 to 0.343).

(a) Listing the times for Club A in ascending order: 40, 43, 45, 51, 52, 55, 56, 59, 60, 61, 64, 67, 71, 72, 75. Since \(n = 15\), the median is the 8th value: 59 minutes. The lower quartile \(Q_1\) is the 4th value: 51 minutes. The upper quartile \(Q_3\) is the 12th value: 67 minutes. The interquartile range (IQR) = \(67 - 51 = 16\) minutes. (b) Listing the times for Club B in ascending order: 42, 45, 47, 50, 53, 53, 56, 58, 61, 64, 65, 68, 72, 76, 79. Since \(n = 15\), the median is the 8th value: 58 minutes. The lower quartile \(Q_1\) is the 4th value: 50 minutes. The upper quartile \(Q_3\) is the 12th value: 68 minutes. The interquartile range (IQR) = \(68 - 50 = 18\) minutes. (c) Comparison 1: The median time for Club B (58 minutes) is less than that for Club A (59 minutes), indicating that competitors from Club B were generally faster. Comparison 2: The IQR for Club A (16 minutes) is less than that for Club B (18 minutes), indicating that the times of competitors from Club A were more consistent (less varied).

(a) M1 for listing values or identifying median position for A. A1 for Median = 59. A1.3 for IQR = 16 (from \(67 - 51\)). (b) M1 for listing values or identifying median position for B. A1 for Median = 58. A1 for IQR = 18 (from \(68 - 50\)). (c) B1 for a valid comparison of medians in context. B1 for a valid comparison of IQRs in context.

(a) By the law of total probability: \(P(B) = P(B \cap A) + P(B \cap A')\). We are given: \(P(B \cap A) = P(A) \times P(B \mid A) = 0.6 \times 0.35 = 0.21\). Since \(P(A') = 1 - P(A) = 1 - 0.6 = 0.4\), we have: \(P(B \cap A') = P(A') \times P(B \mid A') = 0.4 \times 0.5 = 0.2\). Therefore: \(P(B) = 0.21 + 0.2 = 0.41\). (b) The probability of the union is: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.41 - 0.21 = 0.8\). (c) If \(A\) and \(B\) were independent, then \(P(B \mid A) = P(B)\). Here, \(P(B \mid A) = 0.35\) and \(P(B) = 0.41\). Since \(0.35 \neq 0.41\), the events \(A\) and \(B\) are not independent. (Alternatively, \(P(A \cap B) = 0.21\) while \(P(A) \times P(B) = 0.6 \times 0.41 = 0.246 \neq 0.21\).) (d) By definition: \(P(A' \mid B') = \frac{P(A' \cap B')}{P(B')}\). We know that: \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.8 = 0.2\). Also: \(P(B') = 1 - P(B) = 1 - 0.41 = 0.59\). Therefore: \(P(A' \mid B') = \frac{0.2}{0.59} = \frac{20}{59} \approx 0.339\).

(a) M1 for finding either \(P(B \cap A) = 0.21\) or \(P(B \cap A') = 0.2\). M1 for summing the two. A0.3 for 0.41. (b) M1 for applying the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). A1 for 0.8. (c) M1 for checking the condition of independence (e.g., comparing \(P(B \mid A)\) to \(P(B)\)). A1 for correct conclusion with clear explanation. (d) M1 for using the formula \(P(A' \mid B') = \frac{P(A' \cap B')}{P(B')}\). A1 for 20/59 (or 0.339).

(i) To find \(k\), we use the property that the total area under the probability density function is 1:
\[ \int_0^1 k(x^2 - x^3) \, dx = 1 \]
\[ k \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 1 \]
\[ k \left( \frac{1}{3} - \frac{1}{4} \right) = 1 \]
\[ k \left( \frac{1}{12} \right) = 1 \implies k = 12. \]

(ii) The expectation \(\text{E}(X)\) is given by:
\[ \text{E}(X) = \int_0^1 x f(x) \, dx = 12 \int_0^1 (x^3 - x^4) \, dx \]
\[ \text{E}(X) = 12 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_0^1 = 12 \left( \frac{1}{4} - \frac{1}{5} \right) = 12 \left( \frac{1}{20} \right) = \frac{3}{5} = 0.6. \]

(iii) The probability \(\text{P}(X > 0.5)\) is:
\[ \text{P}(X > 0.5) = 12 \int_{0.5}^1 (x^2 - x^3) \, dx = 12 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0.5}^1 \]
\[ = 12 \left[ \left( \frac{1}{12} \right) - \left( \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} \right) \right] \]
\[ = 12 \left[ \frac{1}{12} - \left( \frac{1}{24} - \frac{1}{64} \right) \right] = 12 \left[ \frac{1}{12} - \frac{5}{192} \right] = 12 \left[ \frac{16 - 5}{192} \right] = 12 \times \frac{11}{192} = \frac{11}{16} = 0.6875. \]

(i)
M1: Integrate \(k(x^2 - x^3)\) over the interval \([0, 1]\) and set equal to 1.
A1: Correct integration and showing \(k = 12\) convincingly.

(ii)
M1: Use \(\int x f(x) \, dx\) to find the expectation with limits 0 and 1.
A1: Correct integration and final answer of 0.6 (or \(\frac{3}{5}\)).

(iii)
M1: Integrate \(12(x^2 - x^3)\) from 0.5 to 1 (or find \(1 - \text{P}(X \le 0.5)\)).
A1: Correct substitution of limits.
A1: Correct final probability of 0.6875 (or \(\frac{11}{16}\)).

(i) Let \(X\) be the number of errors in a 250-line section of code. The mean rate is:
\[ \lambda = 1.2 \times \frac{250}{100} = 3.0 \]
Using the Poisson distribution \(X \sim \text{Po}(3)\):
\[ \text{P}(X = 4) = \frac{e^{-3} 3^4}{4!} = \frac{81 e^{-3}}{24} \approx 0.1680 \text{ (3 s.f.)} \]

(ii) Let \(Y\) be the number of errors in a 5000-line section of code. The mean rate is:
\[ \lambda = 1.2 \times \frac{5000}{100} = 60 \]
Since \(\lambda > 15\), we can approximate the Poisson distribution \(Y \sim \text{Po}(60)\) with a normal distribution:
\[ Y \approx W \sim \text{N}(60, 60) \]
We want to find \(\text{P}(Y < 52) = \text{P}(Y \le 51)\). Applying a continuity correction:
\[ \text{P}(Y \le 51) \approx \text{P}(W < 51.5) \]
Standardising:
\[ Z = \frac{51.5 - 60}{\sqrt{60}} = \frac{-8.5}{7.746} \approx -1.097 \]
\[ \text{P}(W < 51.5) = \Phi(-1.097) = 1 - \Phi(1.097) = 1 - 0.8637 = 0.1363 \approx 0.136 \text{ (3 s.f.)} \]

(i)
M1: Calculate the mean rate \(\lambda = 3.0\) for 250 lines.
M1: Use Poisson formula with \(\lambda = 3\) and \(x = 4\).
A1: Correct probability 0.168 (or 0.1680).

(ii)
B1: State normal approximation \(\text{N}(60, 60)\).
M1: Apply continuity correction to get \(\text{P}(Y \le 51) \to \text{P}(W < 51.5)\).
M1: Standardise with their mean and standard deviation.
A1: Correct final probability 0.136 (or 0.1363).

(i) Let \(W = Y_1 + Y_2 + Y_3 + Y_4 - X\). We want to find \(\text{P}(W > 0)\).
Since the bars are independent and normally distributed:
\[ \text{E}(W) = 4 \text{E}(Y) - \text{E}(X) = 4(51) - 202 = 204 - 202 = 2 \]
\[ \text{Var}(W) = 4 \text{Var}(Y) + \text{Var}(X) = 4(1.5^2) + 3^2 = 4(2.25) + 9 = 9 + 9 = 18 \]
Thus \(W \sim \text{N}(2, 18)\).
\[ \text{P}(W > 0) = \text{P}\left(Z > \frac{0 - 2}{\sqrt{18}}\right) = \text{P}(Z > -0.4714) \]
\[ = \Phi(0.4714) \approx 0.681 \text{ (3 s.f.)} \]

(ii) Let \(V = X - 4Y\). We want to find \(\text{P}(V > 0)\).
\[ \text{E}(V) = \text{E}(X) - 4\text{E}(Y) = 202 - 4(51) = -2 \]
\[ \text{Var}(V) = \text{Var}(X) + 16\text{Var}(Y) = 3^2 + 16(1.5^2) = 9 + 16(2.25) = 9 + 36 = 45 \]
Thus \(V \sim \text{N}(-2, 45)\).
\[ \text{P}(V > 0) = \text{P}\left(Z > \frac{0 - (-2)}{\sqrt{45}}\right) = \text{P}\left(Z > \frac{2}{6.708}\right) = \text{P}(Z > 0.2981) \]
\[ = 1 - \Phi(0.2981) = 1 - 0.6172 = 0.3828 \approx 0.383 \text{ (3 s.f.)} \]

(i)
M1: Find the mean of \(W\), i.e., \(4 \times 51 - 202 = 2\).
M1: Find the variance of \(W\), i.e., \(4 \times 1.5^2 + 3^2 = 18\).
M1: Standardise and find the probability.
A1: Correct probability 0.681.

(ii)
M1: Find the mean of \(V\), i.e., \(202 - 4 \times 51 = -2\) and the variance of \(V\), i.e., \(3^2 + 16 \times 1.5^2 = 45\).
M1: Standardise and find the correct tail probability.
A1: Correct probability 0.383.

(i) Unbiased estimate of the population mean, \(\bar{t}\):
\[ \bar{t} = \frac{\sum t}{n} = \frac{9840}{120} = 82 \text{ minutes} \]
Unbiased estimate of the population variance, \(s^2\):
\[ s^2 = \frac{1}{n-1} \left( \sum t^2 - \frac{(\sum t)^2}{n} \right) = \frac{1}{119} \left( 824600 - \frac{9840^2}{120} \right) \]
\[ s^2 = \frac{1}{119} (824600 - 806880) = \frac{17720}{119} \approx 148.908 \approx 149 \text{ (3 s.f.)} \]

(ii) For a 95% confidence interval, the critical value is \(z = 1.960\).
The confidence interval is given by:
\[ \bar{t} \pm z \times \frac{s}{\sqrt{n}} = \bar{t} \pm 1.960 \times \sqrt{\frac{148.908}{120}} \]
\[ = 82 \pm 1.960 \times \sqrt{1.2409} = 82 \pm 1.960 \times 1.114 = 82 \pm 2.183 \]
This gives the interval \((82 - 2.183, 82 + 2.183) = (79.817, 84.183)\).
To 3 s.f., the confidence interval is \((79.8, 84.2)\).

(iii) It means that if we took many samples of size 120 and constructed a 95% confidence interval from each, we would expect 95% of these intervals to contain the true population mean studying time of students at the college.

(i)
B1: Unbiased estimate of mean = 82.
M1: Use unbiased variance formula with divisor \(n-1\).
A1: Unbiased estimate of variance = 149 (or 148.9).

(ii)
B1: Use \(z = 1.96\).
M1: Correctly apply formula \(\bar{t} \pm z \sqrt{\frac{s^2}{n}}\) with their values.
A1: Correct interval \((79.8, 84.2)\).

(iii)
B1: Correct explanation mentioning replication of samples and the proportion (95%) containing the true mean.

(i) Let \(p\) be the proportion of defective microchips in the population.
\[ \text{H}_0: p = 0.08 \]
\[ \text{H}_1: p > 0.08 \]

(ii) Under the null hypothesis \(\text{H}_0\), the number of defective microchips \(X\) follows a binomial distribution:
\[ X \sim \text{B}(50, 0.08) \]
We want to find the probability of getting at least 7 defective microchips, i.e., \(\text{P}(X \ge 7)\):
\[ \text{P}(X \ge 7) = 1 - \text{P}(X \le 6) \]
\[ \text{P}(X \le 6) = \sum_{k=0}^{6} \binom{50}{k} (0.08)^k (0.92)^{50-k} \]
Calculating individual probabilities:
- \(\text{P}(X=0) = (0.92)^{50} \approx 0.0155\)
- \(\text{P}(X=1) = 50(0.08)(0.92)^{49} \approx 0.0673\)
- \(\text{P}(X=2) = \binom{50}{2}(0.08)^2(0.92)^{48} \approx 0.1434\)
- \(\text{P}(X=3) = \binom{50}{3}(0.08)^3(0.92)^{47} \approx 0.1981\)
- \(\text{P}(X=4) = \binom{50}{4}(0.08)^4(0.92)^{46} \approx 0.2024\)
- \(\text{P}(X=5) = \binom{50}{5}(0.08)^5(0.92)^{45} \approx 0.1620\)
- \(\text{P}(X=6) = \binom{50}{6}(0.08)^6(0.92)^{44} \approx 0.1013\)

\[ \text{P}(X \le 6) \approx 0.0155 + 0.0673 + 0.1434 + 0.1981 + 0.2024 + 0.1620 + 0.1013 = 0.8900 \]
\[ \text{P}(X \ge 7) = 1 - 0.8900 = 0.1100 \]
Since the \(p\)-value \(0.1100 > 0.05\) (the 5% significance level), we fail to reject \(\text{H}_0\).
There is insufficient evidence at the 5% level to support the manager's claim that the proportion of defective microchips is greater than 0.08.

(iii) Since we failed to reject \(\text{H}_0\), a Type II error (failing to reject \(\text{H}_0\) when it is false) might have been made.

(i)
B1: State both hypotheses correctly with \(p\).

(ii)
M1: State Binomial distribution with \(n=50\), \(p=0.08\).
M1: Attempt to find \(\text{P}(X \ge 7)\) or \(\text{P}(X \le 6)\).
A1: Calculate \(\text{P}(X \le 6) = 0.8900\) or \(\text{P}(X \ge 7) = 0.1100\).
M1: Compare their probability with 0.05.
A1: Correct conclusion in context (do not reject \(\text{H}_0\)).

(iii)
B1: Correctly state Type II error.

Let \(\mu\) be the population mean height of plants grown with the new fertilizer.
We set up the hypotheses:
\[ \text{H}_0: \mu = 14.2 \]
\[ \text{H}_1: \mu > 14.2 \]

Under \(\text{H}_0\), the sample mean height \(\bar{X}\) of a sample of size \(n = 16\) has distribution:
\[ \bar{X} \sim \text{N}\left(14.2, \frac{2.1^2}{16}\right) \implies \bar{X} \sim \text{N}(14.2, 0.2756) \]

The standard error of the mean is:
\[ \text{SE} = \frac{2.1}{\sqrt{16}} = 0.525 \]

Now we calculate the test statistic \(z\):
\[ z = \frac{\bar{x} - \mu}{\text{SE}} = \frac{15.3 - 14.2}{0.525} = \frac{1.1}{0.525} \approx 2.095 \]

Since this is a one-tailed test at the 5% significance level, the critical value is:
\[ z_{\text{crit}} = 1.645 \]

Comparing the test statistic to the critical value:
\[ 2.095 > 1.645 \]

Alternatively, we can find the \(p\)-value:
\[ p\text{-value} = \text{P}(Z > 2.095) = 1 - \Phi(2.095) = 1 - 0.9819 = 0.0181 < 0.05 \]

Since our test statistic is in the critical region (or \(p < 0.05\)), we reject the null hypothesis \(\text{H}_0\).
There is significant evidence at the 5% significance level to suggest that the fertilizer has increased the mean height of the plants.

B1: Correct null and alternative hypotheses with \(\mu\).
B1: State the correct standard error of the mean, 0.525.
M1: Use formula for standardising the sample mean.
A1: Correct test statistic \(z = 2.095\) (or 2.10).
B1: Correct critical value \(z = 1.645\) (or comparison of \(p\)-value with 0.05).
M1: Compare test statistic with critical value (or compare \(p\)-value with 0.05).
A1.1: Correct decision to reject \(\text{H}_0\) and state the conclusion in context.

(i) Let \(\lambda\) be the mean number of goals scored per match under the new coach.
We set up the hypotheses:
\[ \text{H}_0: \lambda = 1.8 \]
\[ \text{H}_1: \lambda > 1.8 \]

Under \(\text{H}_0\), the number of goals scored \(X \sim \text{Po}(1.8)\).
To test at the 10% significance level, we calculate the probability of getting 5 or more goals:
\[ \text{P}(X \ge 5) = 1 - \text{P}(X \le 4) \]
\[ \text{P}(X \le 4) = e^{-1.8} \left( 1 + 1.8 + \frac{1.8^2}{2} + \frac{1.8^3}{6} + \frac{1.8^4}{24} \right) \]
\[ = e^{-1.8} (1 + 1.8 + 1.62 + 0.972 + 0.4374) = e^{-1.8} (5.8294) \]
\[ \approx 0.1653 \times 5.8294 \approx 0.9636 \]

Thus:
\[ \text{P}(X \ge 5) = 1 - 0.9636 = 0.0364 \]

Since \(0.0364 < 0.10\), the result is significant at the 10% level.
We reject \(\text{H}_0\). There is sufficient evidence to suggest that the mean number of goals scored per match has increased.

(ii) The critical region consists of values of \(X\) for which the probability under \(\text{H}_0\) is less than or equal to 0.10.
Let's test other values of \(X\):
\[ \text{P}(X \ge 4) = 1 - \text{P}(X \le 3) = 1 - e^{-1.8} (1 + 1.8 + 1.62 + 0.972) = 1 - e^{-1.8} (5.392) \]
\[ = 1 - 0.8913 = 0.1087 \]

Since \(\text{P}(X \ge 4) = 0.1087 > 0.10\), the value 4 is not in the critical region.
Since \(\text{P}(X \ge 5) = 0.0364 \le 0.10\), the critical region is \(X \ge 5\).

(i)
B1: State correct hypotheses with \(\lambda\).
M1: Attempt to find \(\text{P}(X \ge 5)\) or \(\text{P}(X \le 4)\) using Poisson formula with \(\lambda = 1.8\).
A1: Correct value for \(\text{P}(X \le 4) = 0.9636\) or \(\text{P}(X \ge 5) = 0.0364\).
M1: Compare their probability with 0.10.
A1: Correctly reject \(\text{H}_0\) and state conclusion in context.

(ii)
M1: Evaluate \(\text{P}(X \ge 4)\) to determine if 4 is in the critical region.
A1.1: Conclude that the critical region is \(X \ge 5\).