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The general term in the binomial expansion is given by:
\[ T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{k}{x}\right)^r \]
\[ = \binom{6}{r} 2^{6-r} (-k)^r x^{12-2r-r} \]
\[ = \binom{6}{r} 2^{6-r} (-k)^r x^{12-3r} \]
For the constant term, the power of \(x\) must be zero:
\[ 12-3r = 0 \Rightarrow r = 4 \]
Substituting \(r=4\) into the term:
\[ T_5 = \binom{6}{4} 2^{6-4} (-k)^4 x^0 = 15 \times 4 \times k^4 = 60k^4 \]
We are given that the constant term is 240:
\[ 60k^4 = 240 \]
\[ k^4 = 4 \]
Since \(k \in \mathbb{R}^+\):
\[ k = \sqrt{2} \]

M1: For identifying the general term (or binomial coefficients).
A1: For the correct expression for the power of \(x\) (e.g., \(12-3r\)).
M1: For setting \(12-3r = 0\) to find \(r=4\).
A1: For the correct equation \(60k^4 = 240\) (accept \(15 \times 4 \times k^4 = 240\)).
M1: For solving \(k^4 = 4\).
A1: For \(k = \sqrt{2}\) (must reject negative root or complex roots).

(a) Let \(y = f(x)\):
\[ y = \frac{ax+3}{2x-5} \]
\[ y(2x-5) = ax+3 \]
\[ 2xy - 5y = ax+3 \]
\[ 2xy - ax = 5y+3 \]
\[ x(2y - a) = 5y+3 \]
\[ x = \frac{5y+3}{2y-a} \]
Therefore, \(f^{-1}(x) = \frac{5x+3}{2x-a}\).

(b) If \(f(x) = f^{-1}(x)\) for all \(x\) in the domain, then:
\[ \frac{ax+3}{2x-5} = \frac{5x+3}{2x-a} \]
Comparing the coefficients or denominators directly:
The denominator of \(f(x)\) has the term \(-5\), while the denominator of \(f^{-1}(x)\) has the term \(-a\).
Similarly, the numerator of \(f(x)\) has the \(x\) coefficient as \(a\), while the numerator of \(f^{-1}(x)\) has the \(x\) coefficient as \(5\).
Thus, \(a = 5\).

Part (a):
M1: Attempt to interchange \(x\) and \(y\) (or to make \(x\) the subject of \(y = f(x)\)).
M1: Correctly expanding to get \(2xy - 5y = ax + 3\).
M1: Rearranging to group terms in \(x\): \(x(2y - a) = 5y + 3\).
A1: Correct final expression for \(f^{-1}(x)\): \(f^{-1}(x) = \frac{5x+3}{2x-a}\).

Part (b):
M1: Equating \(f(x)\) to \(f^{-1}(x)\) or comparing coefficients.
A1: Correct value \(a = 5\).

Using the Pythagorean identity \(\cos^2 x = 1 - \sin^2 x\):
\[ 2(1 - \sin^2 x) + \sin x - 1 = 0 \]
\[ 2 - 2\sin^2 x + \sin x - 1 = 0 \]
\[ 2\sin^2 x - \sin x - 1 = 0 \]
Let \(u = \sin x\):
\[ 2u^2 - u - 1 = 0 \]
\[ (2u + 1)(u - 1) = 0 \]
This gives:
\[ \sin x = -\frac{1}{2} \quad \text{or} \quad \sin x = 1 \]
For \(0 \le x \le 2\pi\):
If \(\sin x = 1\), then \(x = \frac{\pi}{2}\).
If \(\sin x = -\frac{1}{2}\), then \(x = \frac{7\pi}{6}\) or \(x = \frac{11\pi}{6}\).
Thus, the solutions are \(x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

M1: Substituting \(\cos^2 x = 1 - \sin^2 x\).
A1: For obtaining the correct quadratic equation in terms of \(\sin x\): \(2\sin^2 x - \sin x - 1 = 0\).
M1: For attempting to factorize or solve the quadratic equation.
A1: For obtaining the correct roots \(\sin x = 1\) and \(\sin x = -\frac{1}{2}\).
A1: For \(x = \frac{\pi}{2}\).
A1: For \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\) (award 1 mark for both correct angles, deduct 1 mark for any extra incorrect angles within the range).

Let \(u = 3x^2 + 1\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 6x \Rightarrow x \, \mathrm{d}x = \frac{1}{6} \, \mathrm{d}u\).
Find the new integration limits:
When \(x = 0\), \(u = 3(0)^2 + 1 = 1\).
When \(x = 1\), \(u = 3(1)^2 + 1 = 4\).
Substitute these into the integral:
\[ \int_{0}^{1} x \sqrt{3x^2 + 1} \, \mathrm{d}x = \int_{1}^{4} \sqrt{u} \cdot \frac{1}{6} \, \mathrm{d}u \]
\[ = \frac{1}{6} \int_{1}^{4} u^{1/2} \, \mathrm{d}u \]
\[ = \frac{1}{6} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{4} \]
\[ = \frac{1}{9} \left( 4^{3/2} - 1^{3/2} \right) \]
Since \(4^{3/2} = 8\) and \(1^{3/2} = 1\):
\[ = \frac{1}{9} (8 - 1) = \frac{7}{9} \]

M1: Attempting to use integration by substitution with \(u = 3x^2 + 1\).
A1: Correctly finding \(x \, \mathrm{d}x = \frac{1}{6} \, \mathrm{d}u\).
A1: Correctly changing the limits of integration to \(1\) and \(4\) (or returning to \(x\) later with correct limits).
M1: Correctly integrating \(u^{1/2}\) to obtain \(\frac{2}{3}u^{3/2}\).
M1: Substituting limits into their integrated expression.
A1: Correct final answer: \(\frac{7}{9}\).

(a) Since the sum of all probabilities is 1:
\[ \sum P(X=x) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 \]
\[ k(1) + k(2) + k(3) + k(4-2) = 1 \]
\[ k + 2k + 3k + 2k = 1 \]
\[ 8k = 1 \Rightarrow k = \frac{1}{8} \]

(b) The expected value is given by \(\mathrm{E}(X) = \sum x \cdot P(X=x)\):
\[ \mathrm{E}(X) = 1(k) + 2(2k) + 3(3k) + 4(2k) \]
\[ = k + 4k + 9k + 8k \]
\[ = 22k \]
Substitute \(k = \frac{1}{8}\):
\[ \mathrm{E}(X) = 22 \times \frac{1}{8} = \frac{11}{4} = 2.75 \]

Part (a):
M1: Stating \(\sum P(X=x) = 1\).
A1: Correct equation in terms of \(k\): \(k + 2k + 3k + 2k = 1\).
A1: Correct value \(k = \frac{1}{8}\) (or 0.125).

Part (b):
M1: Correct formula for expected value \(\sum x \cdot P(X=x)\).
A1: Correct algebraic expression for expected value: \(22k\).
A1: Correct final value: \(\frac{11}{4}\) (or 2.75).

Using the logarithmic property \(\log_a(b) + \log_a(c) = \log_a(bc)\):
\[ \log_2(x(x-3)) = 2 \]
Convert the logarithmic equation to exponential form:
\[ x(x-3) = 2^2 \]
\[ x^2 - 3x = 4 \]
Rearrange to set the quadratic equation to 0:
\[ x^2 - 3x - 4 = 0 \]
Factorize the quadratic equation:
\[ (x - 4)(x + 1) = 0 \]
This gives two possible solutions:
\[ x = 4 \quad \text{or} \quad x = -1 \]
Since the equation is only defined for \(x > 3\), we reject \(x = -1\).
Therefore, the solution is \(x = 4\).

M1: For applying the log rule to write the left-hand side as \(\log_2(x(x-3))\).
M1: For converting to exponential form: \(x(x-3) = 2^2\).
A1: Correct quadratic equation \(x^2 - 3x - 4 = 0\).
M1: Attempting to solve their quadratic equation by factoring or formula.
A1: Finding both roots \(x = 4\) and \(x = -1\).
R1: Correctly rejecting \(x = -1\) with a valid reason (e.g. \(x > 3\)) to give the final answer \(x = 4\).

(a) To find \(f'(x)\), use the product rule with \(u = x\) and \(v = e^{-2x}\):
\[ u' = 1, \quad v' = -2e^{-2x} \]
\[ f'(x) = 1 \cdot e^{-2x} + x \cdot (-2e^{-2x}) \]
\[ = e^{-2x}(1 - 2x) \]

(b) For a local maximum, we set \(f'(x) = 0\):
\[ e^{-2x}(1 - 2x) = 0 \]
Since \(e^{-2x} > 0\) for all \(x \in \mathbb{R}\):
\[ 1 - 2x = 0 \Rightarrow x = \frac{1}{2} \]
To find the corresponding \(y\)-coordinate, substitute \(x = \frac{1}{2}\) back into the original function \(f(x)\):
\[ y = f\left(\frac{1}{2}\right) = \frac{1}{2} e^{-2(1/2)} = \frac{1}{2} e^{-1} = \frac{1}{2e} \]
Therefore, the coordinates of the local maximum point are \(\left(\frac{1}{2}, \frac{1}{2e}\right)\).

Part (a):
M1: Attempting to apply the product rule.
A1: Correct derivative of \(e^{-2x}\) (which is \(-2e^{-2x}\)).
A1: Correct simplified derivative: \(f'(x) = e^{-2x}(1-2x)\).

Part (b):
M1: Setting \(f'(x) = 0\) and solving for \(x\).
A1: Correct \(x\)-coordinate: \(x = \frac{1}{2}\).
A1: Correct \(y\)-coordinate: \(\frac{1}{2e}\), and coordinates expressed as \(\left(\frac{1}{2}, \frac{1}{2e}\right)\).

(a) Using the area formula of a triangle:
\[ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(B\hat{A}C) \]
\[ 12\sqrt{3} = \frac{1}{2} \times 6 \times 8 \times \sin(B\hat{A}C) \]
\[ 12\sqrt{3} = 24 \sin(B\hat{A}C) \]
\[ \sin(B\hat{A}C) = \frac{12\sqrt{3}}{24} = \frac{\sqrt{3}}{2} \]

(b) We use the identity \(\sin^2(B\hat{A}C) + \cos^2(B\hat{A}C) = 1\):
\[ \cos^2(B\hat{A}C) = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4} \]
Since the angle \(B\hat{A}C\) is obtuse, the cosine must be negative:
\[ \cos(B\hat{A}C) = -\sqrt{\frac{1}{4}} = -\frac{1}{2} \]

(c) Using the cosine rule to find the length of \(BC\):
\[ BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(B\hat{A}C) \]
\[ BC^2 = 6^2 + 8^2 - 2(6)(8)\left(-\frac{1}{2}\right) \]
\[ BC^2 = 36 + 64 + 48 \]
\[ BC^2 = 148 \]
\[ BC = \sqrt{148} = 2\sqrt{37}\text{ cm} \]

Part (a):
M1: Substituting correct values into the triangle area formula.
A1: Correct value: \(\sin(B\hat{A}C) = \frac{\sqrt{3}}{2}\).

Part (b):
M1: Using the Pythagorean identity to find \(\cos^2 A = \frac{1}{4}\).
A1: Correct value: \(\cos(B\hat{A}C) = -\frac{1}{2}\) (must justify or select the negative sign due to 'obtuse').

Part (c):
M1: Correct substitution of values into the cosine rule.
A1: Correct simplified answer: \(BC = \sqrt{148}\) or \(2\sqrt{37}\) (units not required).

(a) To find the indefinite integral, we use integration by substitution.

Let \( u = x^2 + 9 \).
Then \( \frac{du}{dx} = 2x \), which means \( x \, dx = \frac{1}{2} \, du \).

Substituting these into the integral gives:
\( \int x \sqrt{x^2 + 9} \, dx = \int \sqrt{u} \cdot \left(\frac{1}{2} \, du\right) \)
\( = \frac{1}{2} \int u^{1/2} \, du \)

Integrate with respect to \( u \):
\( = \frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C \)
\( = \frac{1}{3} u^{3/2} + C \)

Substitute back \( u = x^2 + 9 \):
\( = \frac{1}{3} (x^2 + 9)^{3/2} + C \)

(b) The area of the region is given by the definite integral:
\( \text{Area} = \int_{0}^{4} x \sqrt{x^2 + 9} \, dx \)

Using the antiderivative from part (a):
\( \text{Area} = \left[ \frac{1}{3} (x^2 + 9)^{3/2} \right]_{0}^{4} \)
\( = \frac{1}{3} (4^2 + 9)^{3/2} - \frac{1}{3} (0^2 + 9)^{3/2} \)
\( = \frac{1}{3} (25)^{3/2} - \frac{1}{3} (9)^{3/2} \)
\( = \frac{1}{3} (125) - \frac{1}{3} (27) \)
\( = \frac{125 - 27}{3} \)
\( = \frac{98}{3} \)

**(a)**
* **M1**: For attempting integration by substitution with \( u = x^2 + 9 \) (or equivalent, e.g., integration by inspection).
* **A1**: For obtaining \( \frac{1}{2} \int u^{1/2} \, du \) (or equivalent expression in terms of \( u \)).
* **A1**: For integrating to obtain \( \frac{1}{3} u^{3/2} \) (condone omission of constant of integration at this stage).
* **A1**: For the correct final expression: \( \frac{1}{3}(x^2 + 9)^{3/2} + C \) (must include \( +C \)).

**(b)**
* **M1**: For attempting to evaluate their antiderivative from part (a) at limits \( 4 \) and \( 0 \).
* **A1**: For the correct exact area of \( \frac{98}{3} \) (or equivalent, such as \( 32\frac{2}{3} \)).

(a) We apply the quotient rule to \(f(x) = \frac{\ln(x)}{x^2}\):
\(f'(x) = \frac{\frac{d}{dx}(\ln(x)) \cdot x^2 - \ln(x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2}\)
\(f'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln(x) \cdot 2x}{x^4} = \frac{x - 2x\ln(x)}{x^4} = \frac{1 - 2\ln(x)}{x^3}\). [3 marks]

(b) To find the local maximum, set \(f'(x) = 0\):
\(1 - 2\ln(x) = 0 \implies \ln(x) = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}\).
The \(y\)-coordinate is \(f(\sqrt{e}) = \frac{\ln(\sqrt{e})}{(\sqrt{e})^2} = \frac{1/2}{e} = \frac{1}{2e}\).
To justify that this is a local maximum, we check the sign of \(f'(x)\) around \(x = \sqrt{e}\):
- For \(0 < x < \sqrt{e}\), \(\ln(x) < 1/2 \implies 1 - 2\ln(x) > 0 \implies f'(x) > 0\).
- For \(x > \sqrt{e}\), \(\ln(x) > 1/2 \implies 1 - 2\ln(x) < 0 \implies f'(x) < 0\).
Since \(f'(x)\) changes from positive to negative, the point \((\sqrt{e}, \frac{1}{2e})\) is a local maximum. [4 marks]

(c) We find the second derivative by differentiating \(f'(x) = \frac{1 - 2\ln(x)}{x^3}\) using the quotient rule:
\(f''(x) = \frac{\left(-\frac{2}{x}\right) \cdot x^3 - (1 - 2\ln(x)) \cdot 3x^2}{(x^3)^2}\)
\(f''(x) = \frac{-2x^2 - 3x^2 + 6x^2\ln(x)}{x^6} = \frac{6\ln(x) - 5}{x^4}\).
To find the point of inflection, set \(f''(x) = 0 \implies 6\ln(x) - 5 = 0 \implies \ln(x) = \frac{5}{6} \implies x = e^{5/6}\).
The \(y\)-coordinate is \(f(e^{5/6}) = \frac{\ln(e^{5/6})}{(e^{5/6})^2} = \frac{5/6}{e^{5/3}} = \frac{5}{6e^{5/3}}\).
Since \(6\ln(x) - 5\) changes sign at \(x = e^{5/6}\), the point of inflection is \((e^{5/6}, \frac{5}{6e^{5/3}})\). [4 marks]

(d)(i) Since \(f(x) > 0\) for \(x \ge \sqrt{e} > 1\), the area of \(R\) is given by:
\(\text{Area} = \int_{\sqrt{e}}^{a} \frac{\ln(x)}{x^2} \, dx\).
Using integration by parts, let \(u = \ln(x) \implies du = \frac{1}{x} \, dx\) and \(dv = x^{-2} \, dx \implies v = -\frac{1}{x}\).
\(\int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} - \int -\frac{1}{x^2} \, dx = -\frac{\ln(x)}{x} - \frac{1}{x} = -\frac{\ln(x) + 1}{x}\).
Evaluating from \(\sqrt{e}\) to \(a\):
\(\left[ -\frac{\ln(x) + 1}{x} \right]_{\sqrt{e}}^{a} = -\frac{\ln(a) + 1}{a} - \left( -\frac{\ln(\sqrt{e}) + 1}{\sqrt{e}} \right) = \frac{1/2 + 1}{\sqrt{e}} - \frac{\ln(a) + 1}{a} = \frac{3}{2\sqrt{e}} - \frac{\ln(a) + 1}{a}\).
(ii) Taking the limit as \(a \to \infty\):
\(\lim_{a \to \infty} \left( \frac{3}{2\sqrt{e}} - \frac{\ln(a) + 1}{a} \right) = \frac{3}{2\sqrt{e}} - 0 = \frac{3}{2\sqrt{e}}\). [7 marks]

(a)
- M1 for attempting quotient rule (or product rule)
- A1 for correct numerator components: \(x\) (or \(1/x \cdot x^2\)) and \(2x\ln(x)\)
- A1 for algebraic simplification to obtain the given expression

(b)
- M1 for setting \(f'(x) = 0\)
- A1 for finding \(x = \sqrt{e}\) and \(y = \frac{1}{2e}\)
- R1 for clear justification using first derivative test (or second derivative test showing \(f''(\sqrt{e}) < 0\))

(c)
- M1 for attempting second derivative
- A1 for obtaining \(f''(x) = \frac{6\ln(x) - 5}{x^4}\)
- A1 for solving \(f''(x) = 0\) to get \(x = e^{5/6}\)
- A1 for \(y = \frac{5}{6e^{5/3}}\)

(d)(i)
- M1 for setting up the integral \(\int_{\sqrt{e}}^{a} \frac{\ln(x)}{x^2} \, dx\)
- M1 for integration by parts setup (defined \(u\) and \(dv\))
- A1 for the antiderivative \(-\frac{\ln(x) + 1}{x}\)
- A1 for substituting limits to obtain \(\frac{3}{2\sqrt{e}} - \frac{\ln(a) + 1}{a}\)

(d)(ii)
- M1 for attempting the limit as \(a \to \infty\)
- A1 for stating \(\lim_{a \to \infty} \frac{\ln(a)+1}{a} = 0\)
- A1 for the final exact limit \(\frac{3}{2\sqrt{e}}\)

(a) We equate \(\sqrt{3} \sin(2x) - \cos(2x)\) to \(R \sin(2x - \alpha) = R \sin(2x) \cos(\alpha) - R \cos(2x) \sin(\alpha)\).
This gives \(R \cos(\alpha) = \sqrt{3}\) and \(R \sin(\alpha) = 1\).
Squaring and adding both equations:
\(R^2 = (\sqrt{3})^2 + 1^2 = 4 \implies R = 2\).
Dividing the equations:
\(\tan(\alpha) = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6}\).
Thus, \(f(x) = 2 \sin(2x - \frac{\pi}{6})\). [4 marks]

(b) On the domain \(0 \le x \le \pi\), the argument \(2x - \frac{\pi}{6}\) ranges from \(-\frac{\pi}{6}\) to \(\frac{11\pi}{6}\), which spans a full period of \(2\pi\).
Since the sine function completes a full period, its range is \([-1, 1]\).
Therefore, the range of \(f(x) = 2 \sin(2x - \frac{\pi}{6})\) is \([-2, 2]\). [2 marks]

(c) Set \(f(x) = 0 \implies 2 \sin(2x - \frac{\pi}{6}) = 0\).
Within the interval \(-\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6}\), the solutions are:
\(2x - \frac{\pi}{6} = 0 \implies 2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}\).
\(2x - \frac{\pi}{6} = \pi \implies 2x = \frac{7\pi}{6} \implies x = \frac{7\pi}{12}\).
The \(x\)-intercepts are at \(x = \frac{\pi}{12}\) and \(x = \frac{7\pi}{12}\). [4 marks]

(d)(i) Note that \(g(x) = \frac{1}{f(x) + 3}\).
Since the range of \(f(x)\) is \([-2, 2]\), the range of \(f(x) + 3\) is \([1, 5]\).
- The maximum of \(g(x)\) occurs when \(f(x) + 3\) is at its minimum:
\(g_{\text{max}} = \frac{1}{-2 + 3} = 1\).
- The minimum of \(g(x)\) occurs when \(f(x) + 3\) is at its maximum:
\(g_{\text{min}} = \frac{1}{2 + 3} = \frac{1}{5}\). [3 marks]

(ii) Solve \(g(x) \ge \frac{1}{2} \implies \frac{1}{f(x) + 3} \ge \frac{1}{2}\).
Since \(f(x) + 3 > 0\), we can take the reciprocal (reversing the inequality):
\(f(x) + 3 \le 2 \implies f(x) \le -1\).
This means \(2 \sin(2x - \frac{\pi}{6}) \le -1 \implies \sin(2x - \frac{\pi}{6}) \le -\frac{1}{2}\).
Let \(\theta = 2x - \frac{\pi}{6}\), where \(-\frac{\pi}{6} \le \theta \le \frac{11\pi}{6}\).
Using the unit circle, \(\sin(\theta) \le -\frac{1}{2}\) in this interval for:
1. The single boundary point: \(\theta = -\frac{\pi}{6}\)
2. The continuous interval: \(\frac{7\pi}{6} \le \theta \le \frac{11\pi}{6}\).
Converting back to \(x\):
- For \(2x - \frac{\pi}{6} = -\frac{\pi}{6} \implies 2x = 0 \implies x = 0\).
- For \(\frac{7\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6} \implies \frac{8\pi}{6} \le 2x \le \frac{12\pi}{6} \implies \frac{2\pi}{3} \le x \le \pi\).
Combining these, the solution is \(x = 0\) or \(\frac{2\pi}{3} \le x \le \pi\). [5 marks]

(a)
- M1 for attempting to expand \(R\sin(2x-\alpha)\) and equating coefficients
- A1 for \(R\cos\alpha = \sqrt{3}\) and \(R\sin\alpha = 1\)
- A1 for \(R = 2\)
- A1 for \(\alpha = \frac{\pi}{6}\)

(b)
- M1 for recognizing that the domain covers a full period
- A1 for range \([-2, 2]\) (accept \(-2 \le f(x) \le 2\))

(c)
- M1 for setting \(f(x) = 0\)
- A1 for identifying argument values \(0\) and \(\pi\)
- A1 for \(x = \frac{\pi}{12}\)
- A1 for \(x = \frac{7\pi}{12}\)

(d)(i)
- M1 for relating the min/max of \(g(x)\) to those of \(f(x)\)
- A1 for Maximum \(= 1\)
- A1 for Minimum \(= \frac{1}{5}\)

(d)(ii)
- M1 for converting inequality to \(f(x) \le -1\) or \(\sin(2x - \frac{\pi}{6}) \le -\frac{1}{2}\)
- M1 for identifying the core interval \(\frac{7\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6}\)
- A1 for solving this interval to get \(\frac{2\pi}{3} \le x \le \pi\)
- M1 for identifying the boundary solution at \(x = 0\)
- A1 for the correct complete solution: \(x = 0\) or \(\frac{2\pi}{3} \le x \le \pi\)

(a) Since \(u_n\) is arithmetic, \(u_2 = a + d\) and \(u_4 = a + 3d\).
Since \(v_n\) is geometric, \(v_2 = ar\) and \(v_3 = ar^2\).
From \(u_2 = v_2\):
\(a + d = ar \implies d = ar - a = a(r - 1)\).
From \(u_4 = v_3\):
\(a + 3d = ar^2 \implies 3d = ar^2 - a = a(r^2 - 1)\). [3 marks]

(b)(i) We substitute the expression for \(d\) into the second equation:
\(3a(r - 1) = a(r^2 - 1)\).
Since \(a > 0\) and \(r \neq 1\), we can divide both sides by \(a(r - 1)\):
\(3 = r + 1 \implies r = 2\). [2 marks]

(ii) Substituting \(r = 2\) into \(d = a(r - 1)\):
\(d = a(2 - 1) = a\). [2 marks]

(c) The arithmetic sequence \(u_n\) has first term \(a\) and common difference \(d = a\).
Thus, its sum \(S_n\) is:
\(S_n = \frac{n}{2}(2a + (n - 1)a) = \frac{n(n + 1)}{2}a\).
The geometric sequence \(v_n\) has first term \(a\) and common ratio \(r = 2\).
Thus, its sum \(G_n\) is:
\(G_n = \frac{a(2^n - 1)}{2 - 1} = a(2^n - 1)\).
We want \(G_n - S_n > 100a\):
\(a(2^n - 1) - \frac{n(n + 1)}{2}a > 100a\).
Since \(a > 0\), dividing by \(a\) yields:
\(2^n - 1 - \frac{n(n + 1)}{2} > 100 \implies 2^n - \frac{n(n + 1)}{2} > 101\).
We test integer values of \(n\):
- For \(n = 6\): \(2^6 - \frac{6 \times 7}{2} = 64 - 21 = 43 \le 101\).
- For \(n = 7\): \(2^7 - \frac{7 \times 8}{2} = 128 - 28 = 100 \le 101\).
- For \(n = 8\): \(2^8 - \frac{8 \times 9}{2} = 256 - 36 = 220 > 101\).
Thus, the smallest value of \(n\) is \(n = 8\). [6 marks]

(d)(i) We write the general term \(v_n\):
\(v_n = a \cdot 2^{n-1}\).
Taking the natural logarithm:
\(w_n = \ln(v_n) = \ln(a \cdot 2^{n-1}) = \ln(a) + \ln(2^{n-1}) = \ln(a) + (n-1)\ln(2)\).
This is of the form of an arithmetic sequence with first term \(\ln(a)\) and common difference \(\ln(2)\).
Alternatively, the difference between consecutive terms is constant:
\(w_{n+1} - w_n = \ln(v_{n+1}) - \ln(v_n) = \ln\left(\frac{v_{n+1}}{v_n}\right) = \ln(2)\).
So \(w_n\) is arithmetic with common difference \(\ln(2)\). [3 marks]

(ii) The sum of the first 10 terms of this arithmetic sequence is:
\(S_{10} = \frac{10}{2}(2w_1 + 9d') = 5(2\ln(a) + 9\ln(2)) = 10\ln(a) + 45\ln(2)\).
Thus, \(p = 10\) and \(q = 45\). [2 marks]

(a)
- M1 for writing general terms: \(u_2 = a+d\), \(v_2=ar\), \(u_4=a+3d\), \(v_3=ar^2\)
- A1 for showing \(d = a(r-1)\)
- A1 for showing \(3d = a(r^2-1)\)

(b)(i)
- M1 for substituting \(d\) into the second equation and expanding
- A1 for correct division by \(a(r-1)\) (justifying \(r \neq 1\)) to get \(r = 2\)

(b)(ii)
- M1 for substituting \(r=2\) back
- A1 for \(d = a\)

(c)
- M1 for finding \(S_n = \frac{n(n+1)}{2}a\)
- M1 for finding \(G_n = a(2^n-1)\)
- M1 for setting up inequality and dividing by \(a\)
- A1 for obtaining \(2^n - \frac{n(n+1)}{2} > 101\)
- M1 for showing testing of \(n = 7\) and \(n = 8\)
- A1 for concluding \(n = 8\)

(d)(i)
- M1 for writing \(w_n = \ln(a) + (n-1)\ln(2)\) or testing \(w_{n+1} - w_n\)
- A1 for demonstrating a constant difference
- A1 for stating the common difference is \(\ln(2)\)

(d)(ii)
- M1 for applying the arithmetic sum formula for 10 terms
- A1 for the correct simplified expression \(10\ln(a) + 45\ln(2)\)

To find the maximum volume, we find when the rate of change \(\frac{\text{d}V}{\text{d}t} = 0\) and changes sign from positive to negative.

Set \(12 \cos(0.4t) - 5 = 0 \implies \cos(0.4t) = \frac{5}{12}\).

For \(0 \le t \le 12\), the first solution is \(0.4t = \arccos(5/12) \approx 1.1410\), which gives \(t \approx 2.8525\).
Since \(\frac{\text{d}V}{\text{d}t} > 0\) for \(t < 2.8525\) and \(\frac{\text{d}V}{\text{d}t} < 0\) for \(t > 2.8525\), this represents a local maximum.

The volume at any time \(t\) is given by:
\[V(t) = 150 + \int_{0}^{t} (12 \cos(0.4u) - 5) \text{d}u\]

Using the GDC to evaluate this at \(t \approx 2.8525\):
\[V(2.8525) = 150 + \left[ \frac{12}{0.4} \sin(0.4u) - 5u \right]_0^{2.8525}\]
\[V(2.8525) = 150 + 30 \sin(1.1410) - 5(2.8525) \approx 150 + 27.272 - 14.263 = 163.009\text{ m}^3\]

Comparing with the boundary values:
- At \(t = 0\), \(V(0) = 150\text{ m}^3\).
- At \(t = 12\), \(V(12) \approx 60.1\text{ m}^3\).

Thus, the maximum volume is approximately \(163.0\text{ m}^3\).

M1: Setting \(\frac{\text{d}V}{\text{d}t} = 0\) and attempting to solve for \(t\).
A1: Finding the correct critical value \(t \approx 2.85\) (or exact equivalent).
M1: Recognizing that the volume is the integral of the rate of change added to the initial volume.
A1: Writing the correct integral expression: \(V(t) = 150 + \int_{0}^{2.8525} (12 \cos(0.4t) - 5) \text{d}t\).
A2: Evaluating the volume to obtain \(163.0\) (award 1 mark for \(163\) or \(163.01\)).

The total distance traveled by the particle is given by the integral of the speed (absolute value of velocity) over the given time interval:
\[\text{Distance} = \int_{0}^{6} |v(t)| \text{d}t = \int_{0}^{6} |\text{e}^{0.5t} \sin(2t) - 1.5| \text{d}t\]

Using a graphic display calculator (GDC) to evaluate this definite integral directly:
\[\int_{0}^{6} |\text{e}^{0.5t} \sin(2t) - 1.5| \text{d}t \approx 27.863\text{ m}\]

To three significant figures, the total distance traveled is \(27.9\text{ m}\).

M1: Recognizing that the total distance is the integral of the absolute value of the velocity.
A2: Correct integral expression \(\int_{0}^{6} |\text{e}^{0.5t} \sin(2t) - 1.5| \text{d}t\) (award M1 A1 if limits are correct, A1 for the absolute value structure).
A3: Correct GDC evaluation to get \(27.9\) (award 2 marks for \(27.86\), full marks for correct 3 s.f. rounding).

(a) As \(t \to \infty\), \(T(t) \to a\). Since the limiting temperature is \(20\text{ }{^\circ}\text{C}\), we have \(a = 20\).
At \(t = 0\), \(T(0) = a + b = 85\text{ }{^\circ}\text{C}\). Substituting \(a = 20\) gives \(b = 65\).

(b) At \(t = 5\), \(T(5) = 20 + 65 \text{e}^{-5k} = 60\).
\[65 \text{e}^{-5k} = 40 \implies \text{e}^{-5k} = \frac{40}{65} = \frac{8}{13}\]
\[-5k = \ln\left(\frac{8}{13}\right) \implies k = -\frac{1}{5}\ln\left(\frac{8}{13}\right) \approx 0.0971\]

(c) Set \(T(t) = 40\):
\[20 + 65 \text{e}^{-kt} = 40 \implies 65 \text{e}^{-kt} = 20 \implies \text{e}^{-kt} = \frac{40}{130} = \frac{4}{13}\]
Using \(k \approx 0.09712\):
\[-kt = \ln\left(\frac{4}{13}\right) \implies t = \frac{\ln(13/4)}{0.09712} \approx 12.1\text{ minutes}\]

(a) A1: Finding \(a = 20\). A1: Finding \(b = 65\).
(b) M1: Substituting \(t = 5, T = 60\) into their equation. A1: Correctly obtaining \(k \approx 0.0971\) (or exact \(-\frac{1}{5}\ln(8/13)\)).
(c) M1: Setting \(T(t) = 40\) and attempting to solve for \(t\). A1: Correctly finding \(t \approx 12.1\) (accept 12).

(a) To find where \(f(x) \ge g(x)\), we use a GDC to find the intersection of the curves \(y = \ln(x^2 - 3x + 4)\) and \(y = 0.5x + 1\) in the interval \(0 \le x \le 6\).

Setting \(\ln(x^2 - 3x + 4) = 0.5x + 1\) yields one intersection point in the domain at \(x \approx 0.307\).

Comparing the graphs, \(f(x) \ge g(x)\) for \(0 \le x \le 0.307\).

(b) Let \(h(x) = f(x) - g(x) = \ln(x^2 - 3x + 4) - 0.5x - 1\).
We find the minimum value of \(h(x)\) on \([0, 6]\) using GDC.

Alternatively, solving \(h'(x) = 0\):
\[\frac{2x - 3}{x^2 - 3x + 4} - 0.5 = 0 \implies 2x - 3 = 0.5x^2 - 1.5x + 2 \implies 0.5x^2 - 3.5x + 5 = 0\]
This factorizes to \(0.5(x-2)(x-5) = 0\).

The stationary points are at \(x = 2\) (local minimum) and \(x = 5\) (local maximum).
Comparing values:
- At \(x = 0\), \(h(0) = \ln 4 - 1 \approx 0.386\)
- At \(x = 2\), \(h(2) = \ln 2 - 2 \approx -1.31\)
- At \(x = 6\), \(h(6) = \ln 22 - 4 \approx -0.909\)

Thus, the global minimum value is approximately \(-1.31\) (at \(x = 2\)).

(a) M1: Writing down or showing the inequality/equation \(\ln(x^2-3x+4) = 0.5x+1\).
A1: Finding the critical boundary value \(x \approx 0.307\).
A1: Expressing the correct interval \(0 \le x \le 0.307\) (accept \([0, 0.307]\)).

(b) M1: Defining the function \(h(x) = f(x) - g(x)\) and attempting to find its minimum using GDC or calculus.
A1: Finding the x-coordinate of the minimum at \(x = 2\).
A1: Finding the correct minimum value of \(-1.31\) (accept \(\ln 2 - 2\)).

First, write down the expressions for \(S_n\) and \(G_n\):
\[S_n = \frac{n}{2} [2(5) + (n-1)(4.5)] = \frac{n}{2} (4.5n + 5.5) = 2.25n^2 + 2.75n\]
\[G_n = \frac{3(1.15^n - 1)}{1.15 - 1} = 20(1.15^n - 1)\]

We require \(G_n > S_n \implies 20(1.15^n - 1) > 2.25n^2 + 2.75n\).

Using a GDC to create a table of values or graph both functions:
- For \(n = 35\):
\(S_{35} = 2.25(35)^2 + 2.75(35) = 2852.5\)
\(G_{35} = 20(1.15^{35} - 1) \approx 2643.5\)
Here \(G_{35} < S_{35}\).

- For \(n = 36\):
\(S_{36} = 2.25(36)^2 + 2.75(36) = 3015\)
\(G_{36} = 20(1.15^{36} - 1) \approx 3043.0\)
Here \(G_{36} > S_{36}\).

Thus, the smallest value of \(n\) is \(36\).

A1: Correct formula for \(S_n\) in terms of \(n\).
A1: Correct formula for \(G_n\) in terms of \(n\).
M1: Setting up the inequality \(20(1.15^n - 1) > 2.25n^2 + 2.75n\).
M1: Utilizing GDC table or intersection finding methods.
A1: Stating the values at \(n = 35\) or \(n = 36\) to justify selection.
A1: Final correct integer answer \(n = 36\).

(a) Using the area formula for a triangle:
\[\text{Area} = \frac{1}{2} c b \sin A \implies 28 = \frac{1}{2} (7)(9) \sin(B\widehat{A}C)\]
\[28 = 31.5 \sin(B\widehat{A}C) \implies \sin(B\widehat{A}C) = \frac{28}{31.5} = \frac{8}{9} \approx 0.8889\]

Thus, the two possible angles are:
\[B\widehat{A}C = \arcsin\left(\frac{8}{9}\right) \approx 62.729^\circ \approx 62.7^\circ\]
\[B\widehat{A}C = 180^\circ - 62.729^\circ \approx 117.27^\circ \approx 117.3^\circ\]

(b) Since \(B\widehat{A}C\) is obtuse, \(B\widehat{A}C \approx 117.27^\circ\).
Using the Cosine Rule:
\[BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(B\widehat{A}C)\]
\[BC^2 = 7^2 + 9^2 - 2(7)(9) \cos(117.27^\circ)\]
\[BC^2 = 49 + 81 - 126 (-0.4581) \approx 130 + 57.72 = 187.72\]
\[BC \approx \sqrt{187.72} \approx 13.7\text{ cm}\]

(a) M1: Substituting given values into the triangle area formula.
A1: Correct value for \\sin(A) = 8/9\\.
A1: First angle \(62.7^\circ\).
A1: Second angle \(117.3^\circ\).

(b) M1: Choosing the obtuse angle and substituting into the Cosine Rule.
A1: Correct final length \(13.7\text{ cm}\) (accept 13.7 or exact equivalent).

(a) Let us model this with a triangle \(PAB\).
The bearing of \(A\) from \(P\) is \(075^\circ\).
The bearing of \(B\) from \(A\) is \(160^\circ\).

The interior angle \(P\widehat{A}B\) can be found using the parallel lines representing the North direction:
- The angle from North clockwise to \(AP\) (the back-bearing from \(A\) to \(P\)) is \(75^\circ + 180^\circ = 255^\circ\).
- The bearing to \(B\) is \(160^\circ\).
- Thus, the interior angle \(P\widehat{A}B = 255^\circ - 160^\circ = 95^\circ\).

Using the Cosine Rule:
\[PB^2 = PA^2 + AB^2 - 2(PA)(AB) \cos(P\widehat{A}B)\]
\[PB^2 = 15^2 + 22^2 - 2(15)(22) \cos(95^\circ)\]
\[PB^2 = 225 + 484 - 660(-0.08715) \approx 709 + 57.52 = 766.52\]
\[PB \approx \sqrt{766.52} \approx 27.686 \approx 27.7\text{ km}\]

(b) To find the bearing, we first find the angle \(A\widehat{P}B\) using the Sine Rule:
\[\frac{\sin(A\widehat{P}B)}{22} = \frac{\sin(95^\circ)}{27.686}\]
\[\sin(A\widehat{P}B) = \frac{22 \sin(95^\circ)}{27.686} \approx 0.7916 \implies A\widehat{P}B \approx 52.34^\circ\]

Since the bearing of \(A\) from \(P\) was \(075^\circ\), the bearing of \(B\) from \(P\) is:
\[\text{Bearing} = 75^\circ + 52.34^\circ = 127.34^\circ \approx 127^\circ\]

(a) M1: Calculating the interior angle \(P\widehat{A}B = 95^\circ\).
M1: Substituting values into the Cosine Rule.
A1: Distance \(PB \approx 27.7\text{ km}\).

(b) M1: Utilizing the Sine Rule to calculate the angle \(A\widehat{P}B\).
A1: Finding \(A\widehat{P}B \approx 52.3^\circ\).
A1: Calculating the final bearing as \(127^\circ\) (accept \(127.3^\circ\)).

(a) Let \(X\) be the mass of an apple. \(X \sim N(150, 18^2)\).
Using GDC: \(P(130 < X < 170) = \text{normalCDF}(130, 170, 150, 18) \approx 0.733\).

(b) We are given \(P(X < w) = 0.12\).
Using the inverse normal function on the GDC:
\(w = \text{invNorm}(0.12, 150, 18) \approx 128.85\text{ g}\).
Rounding to 3 significant figures gives \(w = 129\text{ g}\).

(c) Let \(Y\) be the number of apples in the box with a mass between \(130\text{ g}\) and \(170\text{ g}\).
\(Y\) follows a binomial distribution: \(Y \sim B(15, p)\), where \(p = 0.73349\).

We want to find \(P(Y \ge 10)\):
\[P(Y \ge 10) = 1 - P(Y \le 9)\]
Using the binomial CDF on GDC:
\[P(Y \le 9) \approx 0.21371\]
\[P(Y \ge 10) = 1 - 0.21371 = 0.78629 \approx 0.786\]

(a) M1: Setting up correct normal probability expression.
A1: Correct probability \(0.733\).

(b) M1: Translating the problem into \(P(X < w) = 0.12\).
A1: Correct value \(w = 129\).

(c) M1: Recognizing the binomial setup \(Y \sim B(15, 0.733)\).
M1: Expressing the required probability as \(1 - P(Y \le 9)\).
A1: Correct binomial probability \(0.786\).

(a) The particle is momentarily at rest when its velocity is zero:
\(5 e^{-0.1t} \sin(t) = 0\)

Since \(5 e^{-0.1t} \ne 0\) for all real \(t\), we solve:
\(\sin(t) = 0\)

In the interval \(0 < t < 6\), the only solution is:
\(t = \pi \approx 3.14\) seconds.

(b) The total distance traveled by the particle is given by the integral of the speed (absolute value of velocity) over the given time interval:
\(\text{Distance} = \int_{0}^{6} |v(t)| \, dt = \int_{0}^{6} |5 e^{-0.1t} \sin(t)| \, dt\)

Using a Graphic Display Calculator (GDC) to evaluate this integral:
\(\int_{0}^{6} |5 e^{-0.1t} \sin(t)| \, dt \approx 14.7149\dots\text{ m}\)

To three significant figures, the total distance traveled is \(14.7\text{ m}\).

*Alternatively, splitting the integral at the boundary point \(t = \pi\):*
\(\text{Distance} = \int_{0}^{\pi} v(t) \, dt - \int_{\pi}^{6} v(t) \, dt \approx 8.5663 - (-6.1486) = 14.7149\dots\text{ m}\)

(a)
**M1** for attempting to solve \(v(t) = 0\) (or \(\sin(t) = 0\)).
**A1** for \(t = \pi\) (or \(3.14\)).

(b)
**M2** for writing a correct integral expression for total distance (either using absolute value or splitting the intervals correctly).
*Note: Award **M1** for an incorrect split or missing absolute value, e.g., \(\int_{0}^{6} v(t) \, dt\).*
**A1** for at least one correct intermediate calculated area, e.g., \(\int_{0}^{\pi} v(t) \, dt \approx 8.57\) or \(\int_{\pi}^{6} v(t) \, dt \approx -6.15\).
**A1** for the final answer \(14.7\text{ m}\) (accept \(14.7149\dots\)).

(a) The particle is at rest when \( v(t) = 0 \).
Using GDC to solve \( 5 \ln(t+1) \cos(t) - 1 = 0 \) in the interval \( 0 \le t \le 5 \):
\( t \approx 0.246 \text{ s} \), \( t \approx 1.34 \text{ s} \), \( t \approx 4.82 \text{ s} \).

(b) The acceleration is the derivative of velocity, \( a(t) = v'(t) \).
Using the GDC numerical derivative function at \( t = 3 \):
\( a(3) = v'(3) \approx -2.22 \text{ m s}^{-2} \).

(c) The total distance traveled is the integral of the absolute value of the velocity:
\( \text{Distance} = \int_{0}^{5} |v(t)| \, dt \).
Evaluating this on a GDC:
\( \text{Distance} \approx 18.3 \text{ m} \).

(d) (i) Since \( s(t) = s(0) + \int_{0}^{t} v(x) \, dx \) and \( s(0) = 2 \):
\( s(t) = 2 + \int_{0}^{t} (5 \ln(x+1) \cos(x) - 1) \, dx \).

(ii) To find the displacement at \( t = 5 \):
\( s(5) = 2 + \int_{0}^{5} (5 \ln(t+1) \cos(t) - 1) \, dt \).
Using the GDC to compute the definite integral:
\( s(5) \approx 2 - 16.5 = -14.5 \text{ m} \).

(iii) Extrema of displacement occur where \( s'(t) = v(t) = 0 \) or at the boundaries.
Let's evaluate \( s(t) \) at the critical points and endpoints:
At \( t = 0 \): \( s(0) = 2 \)
At \( t \approx 0.246 \): \( s(0.246) = 2 + \int_{0}^{0.246} v(t) \, dt \approx 1.88 \)
At \( t \approx 1.34 \): \( s(1.34) = 2 + \int_{0}^{1.34} v(t) \, dt \approx 2.65 \)
At \( t \approx 4.82 \): \( s(4.82) = 2 + \int_{0}^{4.82} v(t) \, dt \approx -14.8 \)
At \( t = 5 \): \( s(5) \approx -14.5 \)
Comparing the absolute values (distances from the origin):
\( |s(0)| = 2 \), \( |s(0.246)| \approx 1.88 \), \( |s(1.34)| \approx 2.65 \), \( |s(4.82)| \approx 14.8 \), \( |s(5)| \approx 14.5 \).
Thus, the maximum distance from the origin is \( 14.8 \text{ m} \).

(a) M1 for setting \( v(t) = 0 \).
A2 for correct values of \( t \) (award A1 for any two correct, A2 for all three). (Accept 3 s.f. values: 0.246, 1.34, 4.82).

(b) M1 for recognized relation \( a(3) = v'(3) \) (can be shown via derivative expression or GDC setup).
A2 for correct value \( -2.22 \text{ m s}^{-2} \) (accept \( -2.22 \)).

(c) M2 for the integral expression \( \int_{0}^{5} |v(t)| \, dt \).
A2 for correct value \( 18.3 \text{ m} \).

(d) (i) M1 for using the initial condition, A1 for correct integral expression.
(ii) M1 for setting up the integral for \( s(5) \), A1 for correct answer \( -14.5 \text{ m} \).
(iii) M1 for identifying that maximum distance occurs at critical points or endpoints.
M1 for calculating position at critical points (especially \( t \approx 4.82 \)).
A1.7 for correct final answer of \( 14.8 \text{ m} \) (accept \( 14.8 \)).

(a) Let \( X \) be the weight of a randomly selected apple, where \( X \sim N(145, 12^2) \).
(i) Using GDC Normal CDF function:
\( P(X > 160) \approx 0.106 \).
(ii) Using GDC Normal CDF function:
\( P(130 < X < 150) \approx 0.556 \).

(b) We are given \( P(X < w) = 0.15 \).
Using GDC inverse Normal function with area = 0.15, \( \mu = 145 \), \( \sigma = 12 \):
\( w \approx 133 \text{ g} \) (to 3 s.f.).

(c) Let \( Y \) be the number of small apples in the box. Then \( Y \sim B(20, 0.15) \).
(i) Using GDC Binomial PDF with \( n = 20, p = 0.15, x = 4 \):
\( P(Y = 4) \approx 0.182 \).
(ii) We need \( P(Y \ge 2) = 1 - P(Y \le 1) \).
Using GDC Binomial CDF with \( n = 20, p = 0.15, x = 1 \):
\( P(Y \le 1) \approx 0.176 \).
\( P(Y \ge 2) = 1 - 0.1756... \approx 0.824 \).

(d) Let \( W \) be the number of small apples in a sample of 5. Then \( W \sim B(5, 0.15) \).
We are looking for the conditional probability \( P(W = 2 \mid W \ge 1) \):
\( P(W = 2 \mid W \ge 1) = \frac{P(W = 2 \cap W \ge 1)}{P(W \ge 1)} = \frac{P(W = 2)}{P(W \ge 1)} \).
Using GDC Binomial PDF:
\( P(W = 2) = \binom{5}{2}(0.15)^2(0.85)^3 \approx 0.13818 \).
Using GDC Binomial CDF (or direct calculation):
\( P(W \ge 1) = 1 - P(W = 0) = 1 - (0.85)^5 \approx 1 - 0.44371 = 0.55629 \).
Thus:
\( P(W = 2 \mid W \ge 1) = \frac{0.13818}{0.55629} \approx 0.248 \).

(a) (i) M1 for writing/visualizing standard normal setup, A1 for \( 0.106 \).
(ii) M1 for setting interval \( P(130 < X < 150) \), A1 for \( 0.556 \).

(b) M1 for equation \( P(X < w) = 0.15 \).
M1 for standardizing or setting up invNormal on GDC.
A1 for \( w = 133 \).

(c) (i) M1 for identifying binomial model \( B(20, 0.15) \).
A2 for \( 0.182 \) (M1 A1 if intermediate formula shown).
(ii) M1 for expressing the probability as \( 1 - P(Y \le 1) \) or similar.
A2 for \( 0.824 \).

(d) M1 for setting up the conditional probability formula \( \frac{P(W=2)}{P(W \ge 1)} \).
M1 for finding \( P(W = 2) \approx 0.138 \).
M1 for finding \( P(W \ge 1) \approx 0.556 \).
A1.7 for correct final answer \( 0.248 \).

(a) (i) The amplitude \( a \) is half the difference between maximum and minimum depths:
\( a = \frac{14.6 - 8.2}{2} = 3.2 \).

(ii) The vertical shift \( d \) is the average of the maximum and minimum depths:
\( d = \frac{14.6 + 8.2}{2} = 11.4 \).

(iii) High tide to low tide takes half of a complete cycle (period \( T \)):
\( \frac{T}{2} = 6.2 \implies T = 12.4 \text{ hours} \).
\( b = \frac{2\pi}{T} = \frac{2\pi}{12.4} \approx 0.5067 \approx 0.507 \text{ (to 3 s.f.)} \).

(b) At 08:00, \( t = 8 \).
\( D(8) = 3.2 \cos(0.5067 \times 8) + 11.4 \approx 3.2 \cos(4.0536) + 11.4 \approx 9.44 \text{ m} \) (using the more precise value of \( b \)).

(c) (i) We solve \( D(t) = 10.5 \) for \( 0 \le t \le 12 \):
\( 3.2 \cos(0.5067 t) + 11.4 = 10.5 \).
Using GDC to find the intersections:
\( t_1 \approx 3.66 \text{ hours} \) (or 3 hours 40 minutes)
\( t_2 \approx 8.74 \text{ hours} \) (or 8 hours 44 minutes).

(ii) The ship can enter safely when \( D(t) \ge 10.5 \).
From \( t = 0 \) to \( t_1 \approx 3.66 \), depth is above 10.5.
From \( t_1 \approx 3.66 \) to \( t_2 \approx 8.74 \), depth is below 10.5.
From \( t_2 \approx 8.74 \) to \( t = 12 \), depth is above 10.5 (since the depth rises toward the next high tide at \( t = 12.4 \)).
Total safe hours: \( 3.66 + (12 - 8.74) = 3.66 + 3.26 = 6.92 \text{ hours} \).

(d) We want \( H'(t) = -1.2 \) (rate of decrease is 1.2, so the derivative is \( -1.2 \)).
\( H'(t) = -3.2 \times 0.507 \sin(0.507 t) - 0.15 = -1.6224 \sin(0.507 t) - 0.15 \).
Set \( -1.6224 \sin(0.507 t) - 0.15 = -1.2 \)
\( 1.6224 \sin(0.507 t) = 1.05 \)
\( \sin(0.507 t) \approx 0.6472 \).
For the first time \( t > 0 \):
\( 0.507 t = \arcsin(0.6472) \approx 0.7040 \)
\( t = \frac{0.7040}{0.507} \approx 1.39 \text{ hours} \).

(a) (i) M1 for substituting max and min values into \( \frac{\text{max} - \text{min}}{2} \), A1 for concluding \( a=3.2 \).
(ii) M1 for substituting max and min values into \( \frac{\text{max} + \text{min}}{2} \), A1 for concluding \( d=11.4 \).
(iii) M1 for finding period \( T = 12.4 \), A1 for showing step to get \( b = 0.507 \).

(b) M1 for substituting \( t = 8 \) into \( D(t) \), A1 for \( 9.44 \text{ m} \).

(c) (i) M1 for setting up the equation \( 3.2 \cos(0.5067 t) + 11.4 = 10.5 \).
A1 for attempting to solve via GDC, A1 for \( 3.66 \), A1 for \( 8.74 \).
(ii) M1 for identifying intervals of safety: \( [0, 3.66] \) and \( [8.74, 12] \).
M1 for setting up addition: \( 3.66 + (12 - 8.74) \).
A1 for \( 6.92 \text{ hours} \).

(d) M1 for finding derivative \( H'(t) \).
M1 for setting \( H'(t) = -1.2 \).
A0.7 for solving to get \( t = 1.39 \text{ hours} \).

Part A
(i) Using the product rule on \( f_n(x) = x^n e^{-x} \):
\( f_n'(x) = n x^{n-1} e^{-x} + x^n (-e^{-x}) = x^{n-1} e^{-x} (n - x) \).
(ii) Set \( f_n'(x) = 0 \). Since \( x > 0 \) and \( e^{-x} \neq 0 \), we have \( n - x = 0 \Rightarrow x = n \).
The corresponding y-coordinate is \( f_n(n) = n^n e^{-n} \).
Thus, \( P_n = (n, n^n e^{-n}) \).
(iii) Let \( x = n \) and \( y = n^n e^{-n} \). Substituting \( n = x \) yields \( y = x^x e^{-x} = \left(\frac{x}{e}\right)^x \).

Part B
(i) Differentiating \( f_n'(x) = (n x^{n-1} - x^n) e^{-x} \) with respect to \( x \) using the product rule:
\( f_n''(x) = (n(n-1)x^{n-2} - n x^{n-1})e^{-x} - (n x^{n-1} - x^n)e^{-x} \)
\( = x^{n-2} e^{-x} [n(n-1) - n x - n x + x^2] = x^{n-2} e^{-x} [x^2 - 2nx + n(n-1)] \).
(ii) Set \( f_n''(x) = 0 \). For \( x > 0 \), this requires:
\( x^2 - 2nx + n(n-1) = 0 \).
Using the quadratic formula:
\( x = \frac{2n \pm \sqrt{4n^2 - 4n(n-1)}}{2} = \frac{2n \pm \sqrt{4n}}{2} = n \pm \sqrt{n} \).
(iii) The horizontal distance is \( (n + \sqrt{n}) - (n - \sqrt{n}) = 2\sqrt{n} \).

Part C
(i) \( I_n = \int_{0}^{\infty} x^n e^{-x} dx \). Let \( u = x^n \) and \( dv = e^{-x} dx \). Then \( du = n x^{n-1} dx \) and \( v = -e^{-x} \).
\( I_n = [-x^n e^{-x}]_0^\infty - \int_0^\infty -e^{-x} (n x^{n-1}) dx \).
Since \( \lim_{x \to \infty} x^n e^{-x} = 0 \) and at \( x=0 \) the expression is \( 0 \) (for \( n \ge 1 \)):
\( I_n = 0 + n \int_0^\infty x^{n-1} e^{-x} dx = n I_{n-1} \).
(ii) \( I_0 = \int_0^\infty e^{-x} dx = [-e^{-x}]_0^\infty = 0 - (-1) = 1 \).
By recurrence, \( I_n = n I_{n-1} = n(n-1)I_{n-2} = \dots = n! I_0 = n! \).

Part D
(i) \( \mu_n = \int_0^\infty x h_n(x) dx = \frac{1}{n!} \int_0^\infty x^{n+1} e^{-x} dx = \frac{I_{n+1}}{n!} = \frac{(n+1)!}{n!} = n + 1 \).
(ii) \( \sigma_n^2 = \int_0^\infty (x - \mu_n)^2 h_n(x) dx = \int_0^\infty x^2 h_n(x) dx - \mu_n^2 \).
\( \int_0^\infty x^2 h_n(x) dx = \frac{1}{n!} \int_0^\infty x^{n+2} e^{-x} dx = \frac{I_{n+2}}{n!} = \frac{(n+2)!}{n!} = (n+2)(n+1) \).
Thus, \( \sigma_n^2 = (n+2)(n+1) - (n+1)^2 = (n+1)[(n+2) - (n+1)] = n+1 \).
(iii) As \( n \to \infty \), by the Central Limit Theorem, the distribution curve \( h_n(x) \) becomes increasingly symmetric and bell-shaped, approaching a normal distribution with mean \( n+1 \) and variance \( n+1 \).

Part A
(i)
M1: Attempt to apply product rule to \( f_n(x) = x^n e^{-x} \).
A1: Correct derivative showing the required form. [2 marks]
(ii)
M1: Equating derivative to 0 to find stationary point.
A1: Coordinates \( (n, n^n e^{-n}) \). [2 marks]
(iii)
M1: Substituting \( n = x \) into the y-coordinate.
A1: Concluding the locus formula \( y = (x/e)^x \). [2 marks]

Part B
(i)
M1: Attempt to differentiate \( f_n'(x) \).
A1: Correct use of product rule.
A1: Algebraic simplification to obtain the given expression. [3 marks]
(ii)
M1: Setting second derivative to 0.
A1: Finding \( x = n \pm \sqrt{n} \). [2 marks]
(iii)
A1: Subtracting roots to show \( 2\sqrt{n} \). [1 mark]

Part C
(i)
M1: Set up integration by parts with \( u = x^n \) and \( dv = e^{-x} dx \).
A1: Correct derivative \( du \) and antiderivative \( v \).
M1: Evaluating the boundary limits at \( 0 \) and \( \infty \).
A1: Showing clearly that \( I_n = n I_{n-1} \). [4 marks]
(ii)
A1: Correct evaluation of \( I_0 = 1 \).
A1: Clear inductive argument or expansion showing \( I_n = n! \). [2 marks]

Part D
(i)
M1: Set up the integral for the mean.
A1: Substitute \( I_{n+1} = (n+1)! \).
A1: Simplify factorial expression to get \( n+1 \). [3.5 marks]
(ii)
M1: Expand \( (x-\mu_n)^2 \) or use \( E[X^2] - E[X]^2 \).
M1: Evaluate the integral containing \( x^{n+2} e^{-x} \) to get \( (n+2)(n+1) \).
A1: Set up subtraction \( (n+2)(n+1) - (n+1)^2 \).
A1: Correct simplification to \( n+1 \). [4 marks]
(iii)
R1: State that the curve becomes bell-shaped and symmetric (normal-like).
R1: Connect this to the Central Limit Theorem (as the sum of independent exponential random variables). [2 marks]

Part A
(i) \( T_0(x) = \cos(0 \cdot \arccos x) = 1 \).
\( T_1(x) = \cos(1 \cdot \arccos x) = x \).
(ii) Let \( \theta = \arccos x \), which gives \( \cos \theta = x \). Thus, \( T_k(x) = \cos(k\theta) \).
Using the identity \( \cos(A+B) + \cos(A-B) = 2 \cos A \cos B \) with \( A = n\theta \) and \( B = \theta \):
\( \cos((n+1)\theta) + \cos((n-1)\theta) = 2 \cos(n\theta) \cos \theta \).
Substituting \( T_k(x) \) and \( x \):
\( T_{n+1}(x) + T_{n-1}(x) = 2x T_n(x) \).
Rearranging gives \( T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) \).
(iii) Using the recurrence:
\( T_2(x) = 2x(x) - 1 = 2x^2 - 1 \).
\( T_3(x) = 2x(2x^2 - 1) - x = 4x^3 - 3x \).

Part B
(i) Let \( P(n) \) be the statement: "The leading coefficient of \( T_n(x) \) is \( 2^{n-1} \) for \( n \ge 1 \)".
For \( n=1 \), leading coefficient of \( T_1(x)=x \) is \( 1 = 2^0 \). Thus \( P(1) \) is true.
For \( n=2 \), leading coefficient of \( T_2(x)=2x^2-1 \) is \( 2 = 2^1 \). Thus \( P(2) \) is true.
Assume \( P(k-1) \) and \( P(k) \) are true for some \( k \ge 2 \).
Then \( T_{k+1}(x) = 2x T_k(x) - T_{k-1}(x) \).
The term with the highest degree in \( 2x T_k(x) \) is \( 2x \cdot 2^{k-1}x^k = 2^k x^{k+1} \).
Since the degree of \( T_{k-1}(x) \) is \( k-1 \), it does not affect the leading coefficient of \( T_{k+1}(x) \).
Thus, the leading coefficient of \( T_{k+1}(x) \) is \( 2^k = 2^{(k+1)-1} \).
Therefore, the statement is true for all \( n \ge 1 \) by mathematical induction.
(ii) Solve \( T_n(x) = 0 \):
\( \cos(n \arccos x) = 0 \Rightarrow n \arccos x = \frac{\pi}{2} + k\pi \) for \( k \in \mathbb{Z} \).
\( \arccos x = \frac{2k+1}{2n}\pi \).
Since \( 0 \le \arccos x \le \pi \), we must have \( 0 \le k \le n-1 \).
Thus, the roots are \( x = \cos\left(\frac{2k+1}{2n}\pi\right) \) for \( k = 0, 1, \dots, n-1 \).
(iii) Since each of these cosine values is distinct, there are exactly \( n \) distinct roots.

Part C
(i) Let \( x = \cos \theta \). Then \( T_n(\cos \theta) = \cos(n\theta) \).
Differentiating both sides with respect to \( \theta \) using the chain rule:
\( T_n'(\cos\theta) \cdot (-\sin\theta) = -n \sin(n\theta) \).
Since \( x = \cos\theta \), we have \( \sin\theta = \sqrt{1-x^2} \) for \( \theta \in (0, \pi) \).
Substituting back:
\( T_n'(x) (-\sqrt{1-x^2}) = -n \sin(n \arccos x) \Rightarrow T_n'(x) = \frac{n \sin(n \arccos x)}{\sqrt{1-x^2}} \).
(ii) We can write: \( \sqrt{1-x^2} T_n'(x) = n \sin(n \arccos x) \).
Differentiating both sides with respect to \( x \):
\( T_n''(x)\sqrt{1-x^2} + T_n'(x)\frac{-x}{\sqrt{1-x^2}} = n \cos(n \arccos x) \cdot \frac{-n}{\sqrt{1-x^2}} \).
Using \( T_n(x) = \cos(n \arccos x) \):
\( T_n''(x)\sqrt{1-x^2} - \frac{x T_n'(x)}{\sqrt{1-x^2}} = -\frac{n^2 T_n(x)}{\sqrt{1-x^2}} \).
Multiplying both sides by \( \sqrt{1-x^2} \):
\( (1-x^2)T_n''(x) - x T_n'(x) = -n^2 T_n(x) \)
\( (1-x^2)T_n''(x) - x T_n'(x) + n^2 T_n(x) = 0 \).

Part D
(i) Let \( x = \cos\theta \), then \( dx = -\sin\theta d\theta \). When \( x = -1 \), \( \theta = \pi \), and when \( x = 1 \), \( \theta = 0 \).
\( I_{m,n} = \int_{\pi}^0 \frac{\cos(m\theta)\cos(n\theta)}{\sin\theta} (-\sin\theta d\theta) = \int_0^\pi \cos(m\theta)\cos(n\theta) d\theta \).
Using \( \cos(m\theta)\cos(n\theta) = \frac{1}{2}(\cos((m+n)\theta) + \cos((m-n)\theta)) \):
\( I_{m,n} = \frac{1}{2} \left[ \frac{\sin((m+n)\theta)}{m+n} + \frac{\sin((m-n)\theta)}{m-n} \right]_0^\pi = 0 \) since both sine terms evaluate to 0 at \( 0 \) and \( \pi \).
(ii) For \( m = n > 0 \):
\( I_{n,n} = \int_0^\pi \cos^2(n\theta) d\theta = \int_0^\pi \frac{1+\cos(2n\theta)}{2} d\theta = \left[ \frac{\theta}{2} + \frac{\sin(2n\theta)}{4n} \right]_0^\pi = \frac{\pi}{2} \).

Part A
(i)
A1: Correct expression for \( T_0(x) = 1 \).
A1: Correct expression for \( T_1(x) = x \). [2 marks]
(ii)
M1: Letting \( x = \cos\theta \) and defining \( T_k(x) = \cos(k\theta) \).
A1: Setting up identity with \( A = n\theta \) and \( B = \theta \).
M1: Converting trigonometric terms back to polynomial terms.
A1: Concluding the relation clearly. [4 marks]
(iii)
A1: Correctly finding \( T_2(x) = 2x^2 - 1 \).
A1: Correctly finding \( T_3(x) = 4x^3 - 3x \). [2 marks]

Part B
(i)
M1: Verifying base cases (e.g., \( n=1, 2 \)).
M1: Formulation of inductive hypothesis and considering \( T_{k+1}(x) \).
A1: Logical conclusion showing leading term is indeed \( 2^k x^{k+1} \). [3 marks]
(ii)
M1: Setting \( \cos(n \arccos x) = 0 \).
A1: Correctly writing \( n \arccos x = \frac{2k+1}{2}\pi \).
A1: Finding final root expression \( x = \cos\left(\frac{2k+1}{2n}\pi\right) \) with correct range for \( k \). [3 marks]
(iii)
A1: Identifying \( n \) distinct roots. [1 mark]

Part C
(i)
M1: Attempt to differentiate \( T_n(\cos\theta) = \cos(n\theta) \) with respect to \( \theta \).
A1: Correct chain rule application.
A1: Substituting \( \sin\theta = \sqrt{1-x^2} \) to get the required expression. [3 marks]
(ii)
M1: Rewriting derivative expression and taking derivative of both sides with respect to \( x \).
M1: Correct use of product/quotient rule and chain rule.
A0.5: Correctly substituting \( T_n(x) \) back into the equation.
M1: Eliminating the denominator and simplifying.
A1: Getting the final differential equation. [4.5 marks]

Part D
(i)
M1: Using the substitution \( x = \cos\theta \) and transforming limits.
M1: Correct trigonometric simplification to \( \int_0^\pi \cos(m\theta)\cos(n\theta) d\theta \).
A1: Concluding the integral is 0 for \( m \neq n \). [2 marks]
(ii)
M1: Setting up integral \( \int_0^\pi \cos^2(n\theta) d\theta \).
M1: Using double angle identity to integrate.
A1: Correct final value of \( \frac{\pi}{2} \). [3 marks]