2023 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

Let \(u_1 = a\). The terms of the arithmetic sequence are \(u_1 = a\), \(u_4 = a + 3d\), and \(u_{13} = a + 12d\). Since these three terms are in geometric progression, we have: \(\frac{u_4}{u_1} = \frac{u_{13}}{u_4} \implies (a+3d)^2 = a(a+12d)\). Expanding both sides gives \(a^2 + 6ad + 9d^2 = a^2 + 12ad\). Simplifying by subtracting \(a^2\) from both sides gives \(9d^2 = 6ad\). Since \(d \neq 0\), we can divide both sides by \(3d\), which yields \(3d = 2a \implies a = \frac{3}{2}d\). The common ratio \(r\) of the geometric sequence is given by: \(r = \frac{u_4}{u_1} = \frac{a+3d}{a} = 1 + \frac{3d}{a}\). Substituting \(a = \frac{3}{2}d\), we get \(r = 1 + \frac{3d}{\frac{3}{2}d} = 1 + 2 = 3\).

M1 for setting up the geometric sequence condition: \((a+3d)^2 = a(a+12d)\). A1 for expanding and simplifying to obtain \(9d^2 = 6ad\). M1 for solving for \(a\) in terms of \(d\) (or vice versa): \(a = 1.5d\). M1 for setting up the expression for the common ratio \(r = \frac{a+3d}{a}\). A1 for substituting and obtaining the final answer \(r = 3\).

Let \(y = (f \circ g)^{-1}(\ln 5)\). By the definition of the inverse function, this is equivalent to finding the value of \(y\) such that \((f \circ g)(y) = \ln 5\). First, we find the composite function \((f \circ g)(y) = f(g(y)) = f(e^{3y}) = \ln(2e^{3y} - 3)\). Setting this equal to \(\ln 5\) gives \(\ln(2e^{3y} - 3) = \ln 5\). Taking the exponential of both sides gives \(2e^{3y} - 3 = 5\). Solving for \(e^{3y}\) yields \(2e^{3y} = 8 \implies e^{3y} = 4\). Taking the natural logarithm of both sides gives \(3y = \ln 4\). Therefore, \(y = \frac{1}{3}\ln 4 = \frac{1}{3}\ln(2^2) = \frac{2}{3}\ln 2\).

M1 for setting up the equation \(f(g(y)) = \ln 5\). M1 for substituting \(g(y)\) into \(f\) to get \(\ln(2e^{3y} - 3) = \ln 5\). A1 for removing logarithms to get \(2e^{3y} - 3 = 5\). M1 for solving for \(e^{3y} = 4\). A1 for using log laws to find \(y = \frac{2}{3}\ln 2\) (or equivalent).

Using the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\), we substitute this into the equation: \(2(1 - \cos^2 x) - 3 \cos x = 0\). Expanding and rearranging terms gives a quadratic equation in terms of \(\cos x\): \(2\cos^2 x + 3\cos x - 2 = 0\). Let \(u = \cos x\). The equation becomes \(2u^2 + 3u - 2 = 0\). Factoring the quadratic expression: \((2u - 1)(u + 2) = 0\). This gives two possible solutions: \(u = \frac{1}{2}\) or \(u = -2\). Since \(-1 \le \cos x \le 1\), there are no real solutions for \(\cos x = -2\). For \(\cos x = \frac{1}{2}\) in the interval \(0 \le x \le 2\pi\), the solutions are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).

M1 for using the identity \(\sin^2 x = 1 - \cos^2 x\). A1 for obtaining the quadratic equation \(2\cos^2 x + 3\cos x - 2 = 0\). M1 for factoring or solving the quadratic to get \(\cos x = \frac{1}{2}\) or \(\cos x = -2\). R1 for rejecting \(\cos x = -2\). A1 for \(x = \frac{\pi}{3}\), A1 for \(x = \frac{5\pi}{3}\).

First, find the \(y\)-coordinate at \(x = 1\): \(y = (1)^2 \ln(2(1) - 1) = 1 \cdot \ln(1) = 0\). The point of tangency is \((1, 0)\). Next, we find the derivative \(\frac{dy}{dx}\) using the product rule: \(\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln(2x - 1) + x^2 \cdot \frac{d}{dx}(\ln(2x - 1)) = 2x \ln(2x - 1) + x^2 \cdot \frac{2}{2x - 1}\). Evaluating the derivative at \(x = 1\): \(\frac{dy}{dx}\Big|_{x=1} = 2(1)\ln(1) + 1^2 \cdot \frac{2}{2(1) - 1} = 0 + 2 = 2\). The gradient of the tangent at \(x = 1\) is \(2\). Therefore, the gradient of the normal is \(m = -\frac{1}{2}\). The equation of the normal is: \(y - 0 = -\frac{1}{2}(x - 1)\), which simplifies to \(2y = -x + 1\), or \(x + 2y - 1 = 0\).

A1 for finding the y-coordinate is 0. M1 for applying the product rule to find the derivative. A1 for the correct derivative expression. A1 for finding the gradient of the tangent is 2. M1 for finding the normal gradient is -1/2. A1 for the correct equation of the normal in the required form.

(a) Since the sum of all probabilities in a probability distribution must equal 1, we have: \(6k^2 + 6k^2 + k = 1 \implies 12k^2 + k - 1 = 0\). Factoring this quadratic equation: \((4k - 1)(3k + 1) = 0\). This gives \(k = \frac{1}{4}\) or \(k = -\frac{1}{3}\). Since \(k\) must be a positive constant, we reject the negative solution, so \(k = \frac{1}{4}\). (b) Substituting \(k = \frac{1}{4}\) back into the probability distribution: \(P(X=1) = 6(\frac{1}{16}) = \frac{3}{8}\), \(P(X=2) = 6(\frac{1}{16}) = \frac{3}{8}\), and \(P(X=3) = \frac{1}{4} = \frac{2}{8}\). Now, we calculate the expected value: \(E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) = 1(\frac{3}{8}) + 2(\frac{3}{8}) + 3(\frac{2}{8}) = \frac{3 + 6 + 6}{8} = \frac{15}{8}\).

M1 for setting up the equation \(12k^2 + k - 1 = 0\). A1 for solving the quadratic to find \(k = 1/4\) and rejecting \(k = -1/3\). M1 for finding the individual probabilities: \(3/8, 3/8, 2/8\). M1 for the formula of expectation \(E(X) = \sum x P(X=x)\). A1 for obtaining the final answer \(E(X) = 15/8\).

The general term in the expansion is given by: \(T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(\frac{a}{x}\right)^r = \binom{6}{r} 2^{6-r} a^r x^{12-2r-r} = \binom{6}{r} 2^{6-r} a^r x^{12-3r}\). For this to be the constant term, the exponent of \(x\) must be 0: \(12 - 3r = 0 \implies r = 4\). Now, we set the coefficient of this term equal to \(960\): \(\binom{6}{4} 2^{6-4} a^4 = 960\). Calculating the binomial coefficient and power of 2: \(15 \cdot 2^2 \cdot a^4 = 960 \implies 60 a^4 = 960\). Dividing by 60: \(a^4 = 16\). Since \(a > 0\), we take the positive fourth root: \(a = 2\).

M1 for expressing the general term with binomial coefficients and powers of x. M1 for setting the exponent of x to zero: \(12 - 3r = 0\) and solving to get \(r = 4\). A1 for identifying the coefficient expression \(\binom{6}{4} 2^2 a^4\). A1 for simplifying this to \(60 a^4\). M1 for setting \(60 a^4 = 960\) and solving for \(a^4 = 16\). A1 for the final answer \(a = 2\).

To evaluate this integral, we can use the substitution method. Let \(u = x^2 + 1\). Then the derivative is \(du = 2x \, dx \implies x \, dx = \frac{1}{2} \, du\). Next, we find the new limits of integration: when \(x = 0\), \(u = 0^2 + 1 = 1\); when \(x = \sqrt{3}\), \(u = (\sqrt{3})^2 + 1 = 4\). Substituting these into the integral: \(\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{x^2 + 1}} \, dx = \int_{1}^{4} \frac{1}{2\sqrt{u}} \, du = \left[ \sqrt{u} \right]_{1}^{4}\). Evaluating at the limits: \(\sqrt{4} - \sqrt{1} = 2 - 1 = 1\).

M1 for choosing a suitable substitution, e.g., \(u = x^2 + 1\). A1 for finding \(du = 2x \, dx\). M1 for converting the limits of integration to 1 and 4. A1 for the integrated expression \(\sqrt{u}\) or \(\sqrt{x^2+1}\). M1 for substituting the limits. A1 for obtaining the final value of 1.

We can express the quadratic function in vertex form: \(f(x) = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Since the vertex is \((3, -4)\), the function is \(f(x) = a(x - 3)^2 - 4\). Since the graph passes through \((1, 8)\), we substitute these coordinates into the equation to solve for \(a\): \(8 = a(1 - 3)^2 - 4 \implies 12 = a(-2)^2 \implies 12 = 4a \implies a = 3\). Thus, the function is \(f(x) = 3(x - 3)^2 - 4\). To find \(p\), \(q\), and \(r\), we expand the vertex form: \(f(x) = 3(x^2 - 6x + 9) - 4 = 3x^2 - 18x + 27 - 4 = 3x^2 - 18x + 23\). Comparing this with \(f(x) = px^2 + qx + r\), we get \(p = 3\), \(q = -18\), and \(r = 23\).

M1 for writing the equation in vertex form: \(f(x) = a(x-3)^2 - 4\). M1 for substituting the point \((1, 8)\) to find \(a\). A1 for finding \(a = 3\). M1 for expanding \(3(x-3)^2 - 4\). A1 for correct expansion \(3x^2 - 18x + 23\). A1 for identifying the individual coefficients: \(p = 3\), \(q = -18\), and \(r = 23\).

(a) To find \(f'(x)\), we use the product rule where \(u = x^2\) and \(v = \ln x\).

\(u' = 2x\) and \(v' = \frac{1}{x}\)

\(f'(x) = u'v + uv' = 2x \ln x + x^2 \left(\frac{1}{x}\right) = 2x \ln x + x\).

(b) To find the stationary points, set \(f'(x) = 0\):

\(x(2\ln x + 1) = 0\)

Since \(x > 0\), we have:

\(2\ln x + 1 = 0 \implies \ln x = -\frac{1}{2} \implies x = e^{-1/2}\).

Substitute \(x = e^{-1/2}\) back into \(f(x)\) to find the \(y\)-coordinate:

\(f(e^{-1/2}) = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1} \left(-\frac{1}{2}\right) = -\frac{1}{2e}\).

To confirm this is a local minimum, we find the second derivative:

\(f''(x) = 2\ln x + 2 + 1 = 2\ln x + 3\).

At \(x = e^{-1/2}\):

\(f''(e^{-1/2}) = 2\left(-\frac{1}{2}\right) + 3 = 2 > 0\).

Since the second derivative is positive, \(\left(e^{-1/2}, -\frac{1}{2e}\right)\) is indeed a local minimum.

(a)

M1 for attempting product rule.

A1 for correct derivative: \(f'(x) = 2x \ln x + x\).

(b)

M1 for setting \(f'(x) = 0\).

A1 for finding \(x = e^{-1/2}\) (or equivalent).

A1 for finding \(y = -\frac{1}{2e}\) (or equivalent).

R1 for valid justification that it is a local minimum (e.g., second derivative test showing \(f''(e^{-1/2}) = 2 > 0\) or sign diagram of the first derivative).

(a) (i) Using the quotient rule with u = \ln(x) and v = x :
f'(x) = \frac{(\frac{1}{x})(x) - (\ln(x))(1)}{x^2} = \frac{1 - \ln(x)}{x^2}

(ii) For a local maximum, set f'(x) = 0 :
\frac{1 - \ln(x)}{x^2} = 0 \implies 1 - \ln(x) = 0 \implies x = e
Substituting x = e into f(x) :
f(e) = \frac{\ln(e)}{e} = \frac{1}{e}
So, the coordinates of the local maximum are (e, \frac{1}{e}) .

(b) At x = 1 , the y -coordinate is f(1) = \frac{\ln(1)}{1} = 0 .
The gradient of the tangent is f'(1) = \frac{1 - \ln(1)}{1^2} = 1 .
The gradient of the normal, m , is the negative reciprocal: m = -1 .
The equation of the normal is:
y - 0 = -1(x - 1) \implies y = -x + 1 .

(c) Using the quotient rule on f'(x) = \frac{1 - \ln(x)}{x^2} :
f''(x) = \frac{(-\frac{1}{x})(x^2) - (1 - \ln(x))(2x)}{(x^2)^2}
f''(x) = \frac{-x - 2x + 2x\ln(x)}{x^4}
f''(x) = \frac{2x\ln(x) - 3x}{x^4} = \frac{2\ln(x) - 3}{x^3} .

(d) The area A is given by the integral:
A = \int_{1}^{e^2} \frac{\ln(x)}{x} \, dx
Using substitution, let u = \ln(x) , then du = \frac{1}{x} \, dx .
When x = 1 , u = 0 .
When x = e^2 , u = 2 .
A = \int_{0}^{2} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{2} = \frac{4}{2} - 0 = 2 .

(a) (i)
M1 for attempting quotient rule.
A1 for \frac{1 - \ln(x)}{x^2} .
(ii)
M1 for setting f'(x) = 0 \implies x = e .
A1 for y = \frac{1}{e} .
[5 marks]

(b)
A1 for y = 0 .
M1 for evaluating f'(1) = 1 .
M1 for finding normal gradient m = -1 .
A1 for y = -x + 1 .
[4 marks]

(c)
M1 for attempting quotient rule on f' .
A1 for \frac{-x - 2x + 2x\ln(x)}{x^4} or equivalent.
A1 for factorizing out x and simplifying to show the given expression.
[3 marks]

(d)
M1 for setting up the integral \int_{1}^{e^2} \frac{\ln(x)}{x} \, dx .
M1 for attempting integration by substitution (identifying u = \ln(x) ).
A1 for correct limits u = 0 and u = 2 (or equivalent back-substitution).
M1 for integrating to \frac{u^2}{2} (or \frac{(\ln(x))^2}{2} ).
A1 for final answer of 2 .
[5 marks]

(a) Since f(x) is a probability density function, the total area under the curve must be 1 :
\int_{0}^{3} a x^2 (3 - x) \, dx = 1
ax \int_{0}^{3} (3x^2 - x^3) \, dx = 1
a \left[ x^3 - \frac{x^4}{4} \right]_{0}^{3} = 1
a \left( 27 - \frac{81}{4} - 0 \right) = 1
a \left( \frac{108 - 81}{4} \right) = 1 \implies a \left( \frac{27}{4} \right) = 1 \implies a = \frac{4}{27} .

(b) The expected value E(X) is defined as:
E(X) = \int_{0}^{3} x f(x) \, dx = \frac{4}{27} \int_{0}^{3} x^3 (3 - x) \, dx
E(X) = \frac{4}{27} \int_{0}^{3} (3x^3 - x^4) \, dx
E(X) = \frac{4}{27} \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right]_{0}^{3}
E(X) = \frac{4}{27} \left( \frac{3(81)}{4} - \frac{243}{5} - 0 \right)
Factoring out 243 :
E(X) = \frac{4 \times 243}{27} \left( \frac{1}{4} - \frac{1}{5} ight) = 4 \times 9 \times \frac{1}{20} = \frac{36}{20} = \frac{9}{5} (or 1.8 ).

(c) For 0 \le x \le 3 , the cumulative distribution function F(x) is:
F(x) = \int_{0}^{x} f(t) \, dt = \frac{4}{27} \int_{0}^{x} (3t^2 - t^3) \, dt
F(x) = \frac{4}{27} \left[ t^3 - \frac{t^4}{4} \right]_{0}^{x} = \frac{4}{27} \left( x^3 - \frac{x^4}{4} ight) = \frac{4x^3 - x^4}{27} .

(d) Using the definition of conditional probability:
P(X > 2 \mid X > 1) = \frac{P(X > 2 \cap X > 1)}{P(X > 1)} = \frac{P(X > 2)}{P(X > 1)}
First, calculate the required probabilities using F(x) :
P(X > 2) = 1 - F(2) = 1 - \frac{4(2)^3 - 2^4}{27} = 1 - \frac{32 - 16}{27} = 1 - \frac{16}{27} = \frac{11}{27}
P(X > 1) = 1 - F(1) = 1 - \frac{4(1)^3 - 1^4}{27} = 1 - \frac{3}{27} = 1 - \frac{1}{9} = \frac{8}{9} = \frac{24}{27}
Now, substitute these into the conditional probability formula:
P(X > 2 \mid X > 1) = \frac{11/27}{24/27} = \frac{11}{24} .

(a)
M1 for setting up the integral equation \int_{0}^{3} f(x) \, dx = 1 .
M1 for correct integration of 3x^2 - x^3 .
A1 for correct substitution of limits to get \frac{27}{4} a = 1 .
A1 for completing the proof to show a = \frac{4}{27} .
[4 marks]

(b)
M1 for setting up the expected value integral \int_{0}^{3} x f(x) \, dx .
A1 for correct integrated expression \frac{4}{27} \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right] .
M1 for substitute limits 3 and 0 .
A1 for correct final answer \frac{9}{5} (or 1.8 ).
[4 marks]

(c)
M1 for setting up F(x) = \int_{0}^{x} f(t) \, dt .
A1 for correct integration of 3t^2 - t^3 \implies t^3 - \frac{t^4}{4} .
A1 for applying limits 0 and x .
A1 for final simplified expression \frac{4x^3 - x^4}{27} .
[4 marks]

(d)
M1 for recognizing conditional probability formula \frac{P(X > 2)}{P(X > 1)} .
A1 for finding P(X > 2) = \frac{11}{27} .
A1 for finding P(X > 1) = \frac{24}{27} (or \frac{8}{9} ).
M1 for dividing the two probabilities.
A1 for correct final answer \frac{11}{24} .
[5 marks]

(a) The area of triangle ABC is given by:
Area = \frac{1}{2} \times AB \times AC \times \sin(\widehat{BAC})
Area = \frac{1}{2} \times x \times 2x \times \sin \theta = x^2 \sin \theta .

(b) Setting the area equal to \frac{\sqrt{3}}{2} x^2 :
f(x) = x^2 \sin \theta = \frac{\sqrt{3}}{2} x^2 \implies \sin \theta = \frac{\sqrt{3}}{2}
Since 0 < \theta < \pi , we have:
\theta = \frac{\pi}{3} or \theta = \frac{2\pi}{3} .

(c) (i) The larger value of \theta is \frac{2\pi}{3} .
Using the cosine rule on triangle ABC :
BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos \theta
BC^2 = x^2 + (2x)^2 - 2(x)(2x)\cos\left(\frac{2\pi}{3}\right)
BC^2 = x^2 + 4x^2 - 4x^2\left(-\frac{1}{2}\right)
BC^2 = 5x^2 + 2x^2 = 7x^2
Taking the positive square root (as length is positive):
BC = \sqrt{7} x .

(ii) Let \beta = \widehat{ABC} . Using the cosine rule:
AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos \beta
(2x)^2 = x^2 + (\sqrt{7}x)^2 - 2(x)(\sqrt{7}x)\cos \beta
4x^2 = x^2 + 7x^2 - 2\sqrt{7}x^2\cos \beta
4x^2 = 8x^2 - 2\sqrt{7}x^2\cos \beta \implies 2\sqrt{7}x^2\cos \beta = 4x^2
Dividing both sides by 2x^2 :
\sqrt{7}\cos \beta = 2 \implies \cos \beta = \frac{2}{\sqrt{7}} .
Thus, a = 2 .

(d) By the sine rule for the circumcircle of triangle ABC :
\frac{BC}{\sin(\widehat{BAC})} = 2R
Substituting BC = \sqrt{7}x and \widehat{BAC} = \frac{2\pi}{3} :
\frac{\sqrt{7}x}{\sin(2\pi/3)} = 2R
Since \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} :
2R = \frac{\sqrt{7}x}{\sqrt{3}/2} = \frac{2\sqrt{7}x}{\sqrt{3}}
Dividing both sides by 2 :
R = \frac{\sqrt{7}}{\sqrt{3}}x = \sqrt{\frac{7}{3}} x .

(a)
M1 for applying the area formula \frac{1}{2}ab\sin C .
A1 for simplifying to show x^2 \sin \theta .
[2 marks]

(b)
M1 for setting up \sin \theta = \frac{\sqrt{3}}{2} .
A1 for \theta = \frac{\pi}{3} .
A1 for \theta = \frac{2\pi}{3} .
[3 marks]

(c) (i)
M1 for applying the cosine rule to find BC^2 .
A1 for substituting \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} .
A1 for BC^2 = 7x^2 .
A1 for taking the square root to show BC = \sqrt{7} x .
(ii)
M1 for rearranging the cosine rule to find \cos(\widehat{ABC}) .
A1 for substituting correct values into the rearranged cosine rule.
A1 for \cos(\widehat{ABC}) = \frac{2}{\sqrt{7}} .
[7 marks]

(d)
M1 for recognizing the relationship \frac{BC}{\sin A} = 2R .
M1 for substituting \sqrt{7}x and \sin\left(\frac{2\pi}{3}\right) .
A1 for \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} .
M1 for simplifying the fraction to find 2R = \frac{2\sqrt{7}x}{\sqrt{3}} .
A1 for obtaining the correct expression for R .
[5 marks]

(a) Let \(X\) be the mass of an apple, where \(X \sim N(150, \sigma^2)\).
We are given that \(P(X > 185) = 0.15\).
Standardizing the variable:
\(P\left(Z > \frac{185 - 150}{\sigma}\right) = 0.15 \implies P\left(Z > \frac{35}{\sigma}\right) = 0.15\)
Using the inverse normal cumulative distribution function on a GDC:
\(\frac{35}{\sigma} \approx 1.03643...\)
\(\sigma \approx \frac{35}{1.03643...} \approx 33.7697...\)
So, \(\sigma \approx 33.8\).

(b) We want to find \(P(X < 140)\) where \(X \sim N(150, 33.7697^2)\).
Using the normal cumulative distribution function on a GDC:
\(P(X < 140) \approx 0.38359...\)
So, \(P(X < 140) \approx 0.384\).

(a)
- \(P\left(Z > \frac{35}{\sigma}\right) = 0.15\) (or equivalent statement) (M1)
- \(\frac{35}{\sigma} \approx 1.036\) (A1)
- \(\sigma \approx 33.8\) (or \(33.77\)) (A1)

(b)
- Attempt to set up the probability using their standard deviation (M1)
- Correct calculation or GDC input: \(P(X < 140)\) with \(\sigma \approx 33.77\) (M1)
- \(0.384\) (or \(0.383\) from premature rounding) (A1)

The volume \(V\) of a solid of revolution about the \(x\)-axis is given by:
\(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)

Substituting the function and the boundaries:
\(V = \pi \int_{0}^{3} \left(\sqrt{x} e^{-0.1x^2}\right)^2 \, dx = \pi \int_{0}^{3} x e^{-0.2x^2} \, dx\)

To integrate analytically:
Let \(u = -0.2x^2 \implies du = -0.4x \, dx \implies x \, dx = -2.5 \, du\)
At \(x = 0\), \(u = 0\).
At \(x = 3\), \(u = -1.8\).

\(V = \pi \int_{0}^{-1.8} e^u (-2.5) \, du = 2.5\pi \int_{-1.8}^{0} e^u \, du\)
\(V = 2.5\pi [e^u]_{-1.8}^{0} = 2.5\pi (1 - e^{-1.8})\)

Using GDC to calculate the value:
\(V \approx 2.5\pi (1 - 0.165299...) \approx 2.5\pi (0.834701...) \approx 6.5557...\)
So, \(V \approx 6.56\).

- Correct formulation of the volume integral (M1)
- Correct integrand: \(x e^{-0.2x^2}\) (A1)
- Correct integration method (analytical with substitution or GDC setup) (M1)
- Obtaining the antiderivative \(-2.5 e^{-0.2x^2}\) or equivalent analytical step (A1)
- Correct substitute of limits or analytical evaluation \(2.5\pi (1 - e^{-1.8})\) (A1)
- Final value \(6.56\) (A1)

(a) The starting salary is \(u_1 = 45000\) with a common ratio \(r = 1.035\).
The salary in Year 10 is the 10th term of this geometric sequence:
\(u_{10} = 45000 \times (1.035)^9\)
\(u_{10} \approx 61330.39\)
So, the salary is \(\$61,330\) (to the nearest dollar).

(b) Let \(S_{1, n}\) be the total earnings from the first company after \(n\) years:
\(S_{1, n} = \frac{45000(1.035^n - 1)}{1.035 - 1} = \frac{45000(1.035^n - 1)}{0.035}\)

Let \(S_{2, n}\) be the total earnings from the second company after \(n\) years (which forms an arithmetic series with \(v_1 = 48000\) and \(d = 1200\)):
\(S_{2, n} = \frac{n}{2} [ 2(48000) + (n-1)1200 ] = \frac{n}{2} [ 94800 + 1200n ] = 47400n + 600n^2\)

We require \(S_{1, n} > S_{2, n}\). Using a GDC table or solver:
At \(n = 11\):
\(S_{1, 11} \approx 591,385\)
\(S_{2, 11} = 594,000\)
At \(n = 12\):
\(S_{1, 12} \approx 657,083\)
\(S_{2, 12} = 655,200\)

Since \(S_{1, 12} > S_{2, 12}\), the minimum number of years required is 12.

(a)
- Correct formula for Year 10 salary: \(45000 \times 1.035^9\) (M1)
- \(\$61,330\) (A1)

(b)
- Correct expression for \(S_{1, n}\) (geometric sum) (A1)
- Correct expression for \(S_{2, n}\) (arithmetic sum) (A1)
- Attempt to solve the inequality \(S_{1, n} > S_{2, n}\) (e.g. comparing table of values or graphing) (M1)
- \(12\) years (A1)

(a) Amplitude \(A = \frac{\text{Max} - \text{Min}}{2} = \frac{32 - 16}{2} = 8\).

Vertical shift \(D = \frac{\text{Max} + \text{Min}}{2} = \frac{32 + 16}{2} = 24\).

The period is 24 hours. Therefore, \(B = \frac{2\pi}{24} = \frac{\pi}{12}\).

The maximum temperature of \(32^\circ\text{C}\) occurs at \(t = 14\):
\(8 \sin\left(\frac{\pi}{12}(14 - C)\right) + 24 = 32\)
\(\sin\left(\frac{\pi}{12}(14 - C)\right) = 1\)
\(\frac{\pi}{12}(14 - C) = \frac{\pi}{2}\)
\(14 - C = 6 \implies C = 8\).

So, the model is \(T(t) = 8 \sin\left(\frac{\pi}{12}(t - 8)\right) + 24\).

(b) We want to solve \(T(t) > 28\):
\(8 \sin\left(\frac{\pi}{12}(t - 8)\right) + 24 > 28 \implies \sin\left(\frac{\pi}{12}(t - 8)\right) > 0.5\)

Since \(\sin(\theta) > 0.5\) for \(\frac{\pi}{6} < \theta < \frac{5\pi}{6}\):
\(\frac{\pi}{6} < \frac{\pi}{12}(t - 8) < \frac{5\pi}{6}\)
\(2 < t - 8 < 10\)
\(10 < t < 18\)

This occurs between \(t = 10\) and \(t = 18\). The total number of hours is \(18 - 10 = 8\) hours.

(a)
- \(A = 8\) (A1)
- \(D = 24\) (A1)
- \(B = \frac{\pi}{12}\) (A1)
- \(C = 8\) (A1)

(b)
- Setting up the inequality or equation: \(T(t) > 28\) (M1)
- Resolving to the interval \(10 < t < 18\) or finding critical values \(10\) and \(18\) (A1)
- Awarding full marks for the correct answer \(8\) hours (A1)

Let \(F\) be the base of the tower and \(h\) be the height of the tower.

In triangle \(ABT\):
- \(\angle TAB = 28^\circ\)
- \(\angle ABT = 180^\circ - 43^\circ = 137^\circ\)
- \(\angle ATB = 180^\circ - (28^\circ + 137^\circ) = 15^\circ\)

Using the sine rule in \(\triangle ABT\):
\(\frac{BT}{\sin(28^\circ)} = \frac{AB}{\sin(15^\circ)}\)
\(BT = \frac{50 \sin(28^\circ)}{\sin(15^\circ)}\)
\(BT \approx \frac{50 \times 0.46947}{0.25882} \approx 90.716\text{ m}\)

In right-angled triangle \(\triangle BFT\):
\(\sin(43^\circ) = \frac{h}{BT}\)
\(h = BT \sin(43^\circ)\)
\(h \approx 90.716 \times \sin(43^\circ) \approx 61.868\text{ m}\)

Giving the answer to three significant figures, the height of the tower is \(61.9\text{ m}\).

- Correct identification of \(\angle ABT = 137^\circ\) or \(\angle ATB = 15^\circ\) (A1)
- Appropriate application of the sine rule in \(\triangle ABT\) (M1)
- Correct expression for \(BT\) (or \(AT\)) (A1)
- Correct calculation of \(BT \approx 90.7\text{ m}\) (or \(AT \approx 131.7\text{ m}\)) (A1)
- Correct trigonometric equation linking \(h\) and their found side (M1)
- \(61.9\text{ m}\) (A1)

(a) Entering the data into the statistics lists on a GDC and performing a linear regression (ax+b) yields:
Slope \(a \approx 2.78885...\)
\(y\)-intercept \(b \approx 41.7505...\)

Rounding to 3 significant figures, the equation of the regression line of \(y\) on \(x\) is:
\(y = 2.79x + 41.8\)

(b) To estimate the score for \(x = 14\), we substitute \(x = 14\) into the regression equation:
Using the unrounded values:
\(y = 2.78885... \times 14 + 41.7505... \approx 80.794\)

Rounding to 3 significant figures gives \(80.8\) (or \(81\) to the nearest integer).
Note: If using the rounded equation: \(y = 2.79 \times 14 + 41.8 = 80.86 \approx 80.9\).

(a)
- Correct slope \(2.79\) (or \(2.8\)) (A1)
- Correct intercept \(41.8\) (A1)
- Correct equation of the regression line: \(y = 2.79x + 41.8\) (A2)

(b)
- Substitution of \(x = 14\) into the regression equation (M1)
- Final estimate \(80.8\) or \(80.9\) (accept \(81\)) (A1)

The general term in the expansion of \((a + b)^n\) is given by:
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)

Substituting \(a = 2x^2\), \(b = \frac{k}{x}\), and \(n = 8\):
\(T_{r+1} = \binom{8}{r} (2x^2)^{8-r} \left(\frac{k}{x}\right)^r\)
\(T_{r+1} = \binom{8}{r} 2^{8-r} (x^2)^{8-r} k^r x^{-r}\)
\(T_{r+1} = \binom{8}{r} 2^{8-r} k^r x^{16-2r-r} = \binom{8}{r} 2^{8-r} k^r x^{16-3r}\)

We require the term in \(x^7\):
\(16 - 3r = 7 \implies 3r = 9 \implies r = 3\)

Now, substitute \(r = 3\) to find the coefficient:
\(\text{Coefficient} = \binom{8}{3} 2^{8-3} k^3 = 56 \times 32 \times k^3 = 1792 k^3\)

We are given that the coefficient is \(3584\):
\(1792 k^3 = 3584\)
\(k^3 = 2\)
\(k = \sqrt[3]{2}\)

- General term written with correct powers (M1)
- Simplifying the power of \(x\) to \(16-3r\) (A1)
- Finding \(r = 3\) (A1)
- Correct term coefficient expression \(\binom{8}{3} 2^5 k^3\) or \(1792 k^3\) (M1)
- Setting their coefficient equal to \(3584\) (M1)
- \(k = \sqrt[3]{2}\) (A1)

(a) The particle is at rest when \(v(t) = 0\).
\[3t^2 e^{-0.5t} - 2t = 0\]
\[t(3t e^{-0.5t} - 2) = 0\]

Since \(t > 0\), we solve \(3t e^{-0.5t} - 2 = 0\) using a GDC.
Using numerical solver or graphing function:
\(t \approx 1.244...\) and \(t \approx 3.027...\)

So the values are \(t \approx 1.24\) and \(t \approx 3.03\) seconds.

(b) The total distance travelled \(D\) is the integral of the speed:
\[D = \int_{0}^{4} |v(t)| \, dt = \int_{0}^{4} |3t^2 e^{-0.5t} - 2t| \, dt\]

Using GDC numerical integration directly:
\[D \approx 1.4955...\text{ m}\]
Rounding to 3 significant figures, \(D \approx 1.50\text{ m}\).

(a)
- Setting \(v(t) = 0\) (M1)
- \(t \approx 1.24\) (A1)
- \(t \approx 3.03\) (A1)

(b)
- Correct formulation of the distance integral: \(\int_{0}^{4} |v(t)| \, dt\) (M1)
- Attempt to evaluate the integral using GDC (M1)
- \(1.50\text{ m}\) (or \(1.5\)) (A1)

(a) Let \(L\) be the length of a metal rod, where \(L \sim N(25.0, \sigma^2)\).

We are given \(P(L > 25.8) = 0.15\), which means \(P(L \le 25.8) = 0.85\).

Standardizing the variable:
\(P\left(Z \le \frac{25.8 - 25.0}{\sigma}\right) = 0.85\)

Using the inverse normal function on a GDC:
\(\frac{0.8}{\sigma} \approx 1.03643...\)

Solving for \(\sigma\):
\(\sigma = \frac{0.8}{1.03643...} \approx 0.771878...\)

\(\sigma \approx 0.772\text{ cm}\) (to 3 significant figures).

(b) Let \(X\) be the number of rods with a length greater than \(25.8\text{ cm}\) out of the 3 selected.

Since each rod is selected independently, \(X\) follows a binomial distribution:
\(X \sim B(3, 0.15)\)

We want to find \(P(X = 2)\):
\(P(X = 2) = \binom{3}{2} (0.15)^2 (0.85)^1\)

\(P(X = 2) = 3 \times 0.0225 \times 0.85 = 0.057375\)

\(P(X = 2) \approx 0.0574\) (to 3 significant figures).

(a)
- **M1** for attempting to set up a normal probability equation or standardizing, e.g., \(P(L < 25.8) = 0.85\) or \(\frac{25.8 - 25.0}{\sigma} = z\).
- **A1** for finding the correct z-value: \(\frac{0.8}{\sigma} = 1.03643...\)
- **A1** for \(\sigma \approx 0.772\) (accept \(0.771878...\)).

(b)
- **M1** for recognizing the binomial distribution, e.g., writing \(X \sim B(3, 0.15)\) or using a binomial probability formula with \(n = 3\) and \(p = 0.15\).
- **A1** for a correct expression or setup, e.g., \(\binom{3}{2} (0.15)^2 (0.85)\).
- **A1** for \(0.0574\) (accept \(0.057375\)).

(a) Let \(L\) be the length of a metal rod, where \(L \sim N(25.0, \sigma^2)\).

We are given \(P(L > 25.8) = 0.15\), which means \(P(L \le 25.8) = 0.85\).

Standardizing the variable:
\(P\left(Z \le \frac{25.8 - 25.0}{\sigma}\right) = 0.85\)

Using the inverse normal function on a GDC:
\(\frac{0.8}{\sigma} \approx 1.03643...\)

Solving for \(\sigma\):
\(\sigma = \frac{0.8}{1.03643...} \approx 0.771878...\)

\(\sigma \approx 0.772\text{ cm}\) (to 3 significant figures).

(b) Let \(X\) be the number of rods with a length greater than \(25.8\text{ cm}\) out of the 3 selected.

Since each rod is selected independently, \(X\) follows a binomial distribution:
\(X \sim B(3, 0.15)\)

We want to find \(P(X = 2)\):
\(P(X = 2) = \binom{3}{2} (0.15)^2 (0.85)^1\)

\(P(X = 2) = 3 \times 0.0225 \times 0.85 = 0.057375\)

\(P(X = 2) \approx 0.0574\) (to 3 significant figures).

(a)
- **M1** for attempting to set up a normal probability equation or standardizing, e.g., \(P(L < 25.8) = 0.85\) or \(\frac{25.8 - 25.0}{\sigma} = z\).
- **A1** for finding the correct z-value: \(\frac{0.8}{\sigma} = 1.03643...\)
- **A1** for \(\sigma \approx 0.772\) (accept \(0.771878...\)).

(b)
- **M1** for recognizing the binomial distribution, e.g., writing \(X \sim B(3, 0.15)\) or using a binomial probability formula with \(n = 3\) and \(p = 0.15\).
- **A1** for a correct expression or setup, e.g., \(\binom{3}{2} (0.15)^2 (0.85)\).
- **A1** for \(0.0574\) (accept \(0.057375\)).

(a) Let \(M \sim N(150, 12^2)\). Using GDC: \(P(140 < M < 165) \approx 0.690\).

(b) We require \(P(M < k) = 0.15\). Using GDC inverse normal function: \(k \approx 137.56 \approx 138\).

(c) Let \(Y\) be the number of small apples in a box of 20. \(Y \sim B(20, 0.15)\). We require \(P(Y \ge 4) = 1 - P(Y \le 3)\). Using GDC binomial CDF: \(P(Y \le 3) \approx 0.6477\). Thus, \(P(Y \ge 4) = 1 - 0.6477 \approx 0.352\).

(d) Let \(X \sim N(\mu, \sigma^2)\). Given \(P(X > 180) = 0.20 \implies P(X < 180) = 0.80\), so \(\frac{180 - \mu}{\sigma} = z_{0.80} \approx 0.8416\). Given \(P(X < 130) = 0.10\), so \(\frac{130 - \mu}{\sigma} = z_{0.10} \approx -1.2816\). Subtracting the two equations: \(180 - \mu = 0.8416\sigma\) and \(130 - \mu = -1.2816\sigma\) yields \(50 = 2.1232\sigma \implies \sigma \approx 23.549 \approx 23.5\). Substituting \(\sigma\) back: \(\mu \approx 180 - 0.8416(23.549) \approx 160.18 \approx 160\).

(e) We seek \(P(150 < X < 170 \mid X > 150) = \frac{P(150 < X < 170)}{P(X > 150)}\). Using \(\mu = 160.18\) and \(\sigma = 23.549\): \(P(150 < X < 170) \approx 0.32924\), and \(P(X > 150) \approx 0.66723\). The conditional probability is \(\frac{0.32924}{0.66723} \approx 0.493\).

(a) M1 for setting up the normal probability. A2 for 0.690 (or 0.6903). [3 marks]

(b) M1 for inverse normal setup \(P(M < k) = 0.15\). A2 for \(138\) (accept 137.6). [3 marks]

(c) M1 for identifying Binomial distribution \(B(20, 0.15)\). M1 for \(1 - P(Y \le 3)\). A1 for \(0.352\). [3 marks]

(d) M1 for expressing equations with z-scores: \(\frac{180 - \mu}{\sigma} = 0.8416\) and \(\frac{130 - \mu}{\sigma} = -1.2816\) (accept 3 s.f. z-scores). A1 for correct z-scores. M1 for attempting to eliminate \(\mu\) or \(\sigma\). A1 for \(\sigma \approx 23.5\). A1 for \(\mu \approx 160\). [6 marks]

(e) M1 for conditional probability formula setup. M1 for correct individual probabilities from GDC using their \(\mu\) and \(\sigma\). A1 for \(0.493\). [3 marks]

(a) At \(t = 0\), \(v(0) = 3\cos(0) + 0 - 2 = 1\text{ m s}^{-1}\).

(b) Acceleration \(a(t) = v'(t) = -3\sin(t) + 1\). At \(t = 2\), \(a(2) = -3\sin(2) + 1 \approx -1.73\text{ m s}^{-2}\).

(c) The particle is at rest when \(v(t) = 0\). Using GDC solver for \(3\cos(t) + t - 2 = 0\) in the interval \(0 \le t \le 6\) yields: \(t \approx 1.35\text{ s}\) and \(t \approx 3.99\text{ s}\).

(d) Total distance traveled is \(\int_0^6 |v(t)| \mathrm{d}t = \int_0^6 |3\cos(t) + t - 2| \mathrm{d}t\). Using GDC numeric integration: \(\text{Distance} \approx 11.98 \approx 12.0\text{ m}\).

(e) (i) \(s(t) = 3 + \int_0^t v(x) \mathrm{d}x\) (or \(3 + \int_0^t (3\cos(x) + x - 2) \mathrm{d}x\)).
(ii) The maximum displacement on a closed interval occurs at endpoints or critical points. Critical points occur when \(v(t) = 0\) (i.e., \(t \approx 1.35\) and \(t \approx 3.99\)).
Evaluate \(s(t)\) at key values using GDC:
\(s(0) = 3\)
\(s(1.35) = 3 + \int_0^{1.35} v(t)\mathrm{d}t \approx 4.14\text{ m}\)
\(s(3.99) = 3 + \int_0^{3.99} v(t)\mathrm{d}t \approx 0.73\text{ m}\)
\(s(6) = 3 + \int_0^6 v(t)\mathrm{d}t \approx 8.16\text{ m}\).
Therefore, the maximum displacement is \(8.16\text{ m}\), occurring at \(t = 6\text{ s}\).
(iii) Average velocity \(= \frac{s(6) - s(0)}{6 - 0} = \frac{1}{6} \int_0^6 v(t)\mathrm{d}t \approx \frac{8.163 - 3}{6} \approx 0.860\text{ m s}^{-1}\).

(a) M1 for substituting \(t=0\) into \(v(t)\). A1 for \(1\text{ m s}^{-1}\). [2 marks]

(b) M1 for finding the derivative expression \(v'(t) = -3\sin(t) + 1\). M1 for substituting \(t=2\). A1 for \(-1.73\text{ m s}^{-2}\). [3 marks]

(c) M1 for setting \(v(t) = 0\). A1 for \(t \approx 1.35\). A1 for \(t \approx 3.99\). [3 marks]

(d) M1 for integral of absolute value of \(v(t)\). M1 for correct limits \(0\) and \(6\). A1 for \(12.0\text{ m}\) (accept 11.98). [3 marks]

(e) (i) A1 for correct integral expression. [1 mark]
(ii) M1 for checking values of \(s(t)\) at boundary and/or critical points. A1 for correct displacements at \(t=1.35\) and \(t=6\). A1 for maximum displacement \(8.16\text{ m}\) (or 8.163). A1 for occurring at \(t = 6\text{ s}\). [4 marks]
(iii) M1 for formula for average velocity. A1 for \(0.860\text{ m s}^{-1}\) (or 0.861). [2 marks]

(a) Due West-East line represents a bearing of \(090^\circ\) from \(A\). Hence \(S\hat{A}B = 90^\circ - 55^\circ = 35^\circ\).
Due West line from \(B\) represents a bearing of \(270^\circ\). Hence \(S\hat{B}A = 315^\circ - 270^\circ = 45^\circ\).
In \(\triangle ASB\), the sum of angles is \(180^\circ\), so \(A\hat{S}B = 180^\circ - (35^\circ + 45^\circ) = 100^\circ\).

(b) Using the Sine Rule:
\(\frac{AS}{\sin(45^\circ)} = \frac{15}{\sin(100^\circ)} \implies AS = \frac{15\sin(45^\circ)}{\sin(100^\circ)}\).
Using GDC: \(AS \approx 10.77 \approx 10.8\text{ km}\).

(c) Using the Sine Rule:
\(\frac{BS}{\sin(35^\circ)} = \frac{15}{\sin(100^\circ)} \implies BS = \frac{15\sin(35^\circ)}{\sin(100^\circ)}\).
Using GDC: \(BS \approx 8.74\text{ km}\).

(d) (i) Distance \(= 120t\).
(ii) At time \(t\), the helicopter is at distance \(120t\) from \(A\) along the segment \(AS\). Its remaining distance to the original ship position \(S\) is \(10.77 - 120t\).
The ship has sailed directly towards \(B\) a distance of \(24t\) from \(S\).
Consider \(\triangle HSS'\) where \(H\) is the helicopter's position and \(S'\) is the ship's position. The side lengths are \(SH = 10.77 - 120t\), \(SS' = 24t\), and the included angle \(H\hat{S}S' = A\hat{S}B = 100^\circ\).
Applying the Cosine Rule:
\(d^2 = (10.77 - 120t)^2 + (24t)^2 - 2(10.77 - 120t)(24t)\cos(100^\circ)\)
\(d^2 = (10.77 - 120t)^2 + (24t)^2 - 48t(10.77 - 120t)\cos(100^\circ)\).

(e) Let \(f(t) = d^2\). To find the minimum distance, we find the minimum of the function \(d(t)\) or \(d(t)^2\) using GDC.
Plotting \(y = (10.77 - 120x)^2 + 576x^2 - 48x(10.77 - 120x)\cos(100^\circ)\) on GDC and finding the local minimum:
The minimum occurs at \(t \approx 0.08926 \approx 0.0893\text{ hours}\) (approximately \(5.36\text{ minutes}\)).
The minimum distance is \(d \approx \sqrt{4.636} \approx 2.15\text{ km}\).

(a) M1 for finding angle \(S\hat{A}B = 35^\circ\). M1 for finding angle \(S\hat{B}A = 45^\circ\). A1 for showing/concluding \(A\hat{S}B = 100^\circ\). [3 marks]

(b) M1 for setting up the Sine Rule correctly. A1 for correct substitution of values. A1 for \(AS \approx 10.8\text{ km}\). [3 marks]

(c) M1 for setting up the Sine Rule for \(BS\). A1 for \(BS \approx 8.74\text{ km}\). [2 marks]

(d) (i) A1 for \(120t\). [1 mark]
(ii) M1 for identifying the sides of the triangle as \(10.77 - 120t\) and \(24t\). M1 for identifying the angle between these sides is \(100^\circ\). M1 for applying the Cosine Rule. A1 for algebraic completion to match the given expression. [4 marks]

(e) M1 for identifying the need to minimize \(d(t)\) or \(d^2(t)\). M2 for GDC graph setup and finding the minimum point. A1 for \(t \approx 0.0893\text{ hours}\). A1 for \(\text{minimum distance} \approx 2.15\text{ km}\). [5 marks]

**Part A**

(a) (i) For \(f_1(x) = \frac{\ln x}{x} = 0\), we have \(\ln x = 0 \implies x = 1\).
Thus, the intercept is \((1, 0)\).

(ii) Using the quotient rule on \(f_1(x)\):
\(f_1'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}\).
Setting \(f_1'(x) = 0 \implies 1 - \ln x = 0 \implies x = e\).
Substituting \(x = e\) back into \(f_1(x)\) gives \(y = \frac{\ln e}{e} = \frac{1}{e}\).
Since \(f_1'(x) > 0\) for \(0 < x < e\) and \(f_1'(x) < 0\) for \(x > e\), this point \((e, 1/e)\) is indeed a local maximum.

(iii) Differentiating \(f_1'(x)\) to find the second derivative:
\(f_1''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \ln x) \cdot 2x}{x^4} = \frac{-x - 2x + 2x \ln x}{x^4} = \frac{2 \ln x - 3}{x^3}\).
Setting \(f_1''(x) = 0 \implies 2 \ln x - 3 = 0 \implies \ln x = \frac{3}{2} \implies x = e^{3/2}\).
Since \(2 \ln x - 3\) changes sign from negative to positive as \(x\) increases through \(e^{3/2}\), there is a point of inflection at \(x = e^{3/2}\).

**Part B**

(b) (i) Using the quotient rule on \(f_k(x) = x^{-k}\ln x\):
\(f'_k(x) = \frac{\frac{1}{x} \cdot x^k - k x^{k-1} \ln x}{x^{2k}} = \frac{x^{k-1}(1 - k \ln x)}{x^{2k}} = \frac{1 - k \ln x}{x^{k+1}} = x^{-k-1}(1 - k \ln x)\).
Setting \(f'_k(x) = 0 \implies 1 - k \ln x = 0 \implies \ln x = \frac{1}{k} \implies x = e^{1/k}\).
Since \(1 - k \ln x > 0\) for \(x < e^{1/k}\) and \(1 - k \ln x < 0\) for \(x > e^{1/k}\), the maximum value occurs at \(x = e^{1/k}\).

(ii) Substitute \(x = e^{1/k}\) into \(f_k(x)\):
\(y = \frac{\ln(e^{1/k})}{(e^{1/k})^k} = \frac{1/k}{e} = \frac{1}{ke}\).
Thus, the maximum point is \((e^{1/k}, \frac{1}{ke})\).

(iii) Differentiating \(f'_k(x) = x^{-k-1}(1 - k \ln x)\) using the product rule:
\(f''_k(x) = -(k+1) x^{-k-2} (1 - k \ln x) + x^{-k-1} \left(-\frac{k}{x}\right)\)
\(= x^{-k-2} [ -(k+1)(1 - k \ln x) - k ]\)
\(= x^{-k-2} [ -k - 1 + k(k+1) \ln x - k ]\)
\(= x^{-k-2} [ k(k+1) \ln x - (2k + 1) ]\).
To find the point of inflection, we set \(f''_k(x) = 0 \implies k(k+1) \ln x - (2k+1) = 0 \implies \ln x = \frac{2k+1}{k(k+1)}\).
Thus, the \(x\)-coordinate of the inflection point is \(x = e^{\frac{2k+1}{k(k+1)}}\).

**Part C**

(c) (i) We have \(x_M = e^{1/k}\) and \(y_M = \frac{1}{ke}\).
From \(x_M = e^{1/k}\), taking the natural logarithm of both sides gives \(\ln x_M = \frac{1}{k} \implies k = \frac{1}{\ln x_M}\).
Substituting this expression for \(k\) into \(y_M\):
\(y_M = \frac{1}{\left(\frac{1}{\ln x_M}\right) e} = \frac{\ln x_M}{e}\).
Since \(k > 0\), we must have \(x_M = e^{1/k} > 1\).
Therefore, the locus of maximum points is given by \(y = \frac{\ln x}{e}\) for \(x > 1\).

(ii) The sketch should show a standard logarithmic shape starting at \((1, 0)\) (as an open circle or starting boundary) and increasing slowly with downwards concavity. As \(x_M \to 1^+\), \(y_M \to 0^+\), and as \(x_M \to \infty\), \(y_M \to \infty\).

**Part D**

(d) (i) Let \(u = \ln x \implies \mathrm{d}u = \frac{1}{x}\mathrm{d}x\) and \(\mathrm{d}v = x^{-k}\mathrm{d}x \implies v = \frac{x^{1-k}}{1-k}\) for \(k \neq 1\).
Using integration by parts \(\int u \mathrm{d}v = u v - \int v \mathrm{d}u\):
\(\int \frac{\ln x}{x^k} \mathrm{d}x = \ln x \left( \frac{x^{1-k}}{1-k} \right) - \int \frac{x^{1-k}}{1-k} \cdot \frac{1}{x} \mathrm{d}x\)
\(= \frac{x^{1-k}}{1-k} \ln x - \frac{1}{1-k} \int x^{-k} \mathrm{d}x\)
\(= \frac{x^{1-k}}{1-k} \ln x - \frac{1}{1-k} \left( \frac{x^{1-k}}{1-k} \right) + C\)
\(= \frac{x^{1-k}}{1-k} \left( \ln x - \frac{1}{1-k} \right) + C\).

(ii) For the improper integral \(A_k = \lim_{b \to \infty} \left[ \frac{x^{1-k}}{1-k} \left( \ln x - \frac{1}{1-k} \right) \right]_{e^{1/k}}^{b}\):
If \(k < 1\), then \(1-k > 0\). As \(b \to \infty\), \(b^{1-k} \to \infty\) and \(\ln b \to \infty\), so the limit diverges.
If \(k = 1\), the integral is \(\int_e^{\infty} \frac{\ln x}{x} \mathrm{d}x = \lim_{b \to \infty} \left[ \frac{1}{2}(\ln x)^2 \right]_e^b = \infty\), which diverges.
If \(k > 1\), then \(1-k < 0\). Let \(p = k-1 > 0\). The upper limit becomes:
\(\lim_{b \to \infty} \frac{\ln b - \frac{1}{1-k}}{(1-k)b^p} = 0\) since a power function grows faster than a logarithmic function.
Evaluating at the lower limit \(x = e^{1/k}\):
\(\frac{(e^{1/k})^{1-k}}{1-k} \left( \ln(e^{1/k}) - \frac{1}{1-k} \right) = \frac{e^{\frac{1}{k}-1}}{1-k} \left( \frac{1}{k} - \frac{1}{1-k} \right) = \frac{e^{\frac{1}{k}-1}}{1-k} \left( \frac{1-k-k}{k(1-k)} \right) = \frac{(1-2k) e^{\frac{1}{k}-1}}{k(1-k)^2}\).
Thus, subtracting the lower limit yields:
\(A_k = 0 - \frac{(1-2k) e^{\frac{1}{k}-1}}{k(1-k)^2} = \frac{(2k-1) e^{\frac{1}{k}-1}}{k(1-k)^2}\).
This shows that the integral converges if and only if \(k > 1\).

(iii) As \(k \to 1^+\), the numerator \((2k-1)e^{\frac{1}{k}-1} \to (2(1)-1)e^{0} = 1\), while the denominator \(k(1-k)^2 \to 0^+\).
Therefore, \(A_k \to \infty\).

**Part A (6 Marks)**
- **(a)(i)** [1 Mark]
- **A1** for correct coordinates \((1, 0)\).
- **(a)(ii)** [2 Marks]
- **M1** for setting \(f_1'(x) = 0 \implies 1 - \ln x = 0 \implies x = e\).
- **A1** for correct coordinates \((e, 1/e)\).
- **(a)(iii)** [3 Marks]
- **M1** for correct derivative \(f_1''(x) = \frac{2 \ln x - 3}{x^3}\).
- **A1** for finding \(x = e^{3/2}\).
- **R1** for explaining that \(f_1''(x)\) changes sign at \(x = e^{3/2}\), establishing it is a point of inflection.

**Part B (9 Marks)**
- **(b)(i)** [3 Marks]
- **M1** for applying quotient or product rule on \(f_k(x)\).
- **A1** for obtaining \(f'_k(x) = x^{-k-1}(1 - k\ln x)\).
- **R1** for demonstrating that the derivative changes sign from positive to negative at \(x = e^{1/k}\), confirming a maximum.
- **(b)(ii)** [2 Marks]
- **A1** for \(x\)-coordinate \(e^{1/k}\).
- **A1** for \(y\)-coordinate \(\frac{1}{ke}\).
- **(b)(iii)** [4 Marks]
- **M1** for applying product rule to differentiate \(x^{-k-1}(1-k\ln x)\).
- **A1** for showing algebraic steps leading to \(f''_k(x) = x^{-k-2} [ k(k+1) \ln x - (2k + 1) ]\).
- **M1** for setting \(f''_k(x) = 0\) and solving for \(x\).
- **A1** for \(x = e^{\frac{2k+1}{k(k+1)}}\).

**Part C (4 Marks)**
- **(c)(i)** [2 Marks]
- **M1** for expressing \(k = \frac{1}{\ln x_M}\) and substituting into \(y_M\).
- **A1** for \(y = \frac{\ln x}{e}\) with specified domain \(x > 1\).
- **(c)(ii)** [2 Marks]
- **A1** for correct basic logarithmic shape with downward concavity.
- **A1** for starting point labeled/indicated at \((1,0)\) and correct asymptotic behavior.

**Part D (8 Marks)**
- **(d)(i)** [3 Marks]
- **M1** for correct choice of parts (\(u = \ln x\), \(\mathrm{d}v = x^{-k}\)).
- **A1** for intermediate step \(\frac{x^{1-k}}{1-k}\ln x - \int \frac{x^{-k}}{1-k} \mathrm{d}x\).
- **A1** for final simplified expression \(\frac{x^{1-k}}{1-k} \left( \ln x - \frac{1}{1-k} \right) + C\).
- **(d)(ii)** [4 Marks]
- **R1** for identifying divergence when \(k \le 1\).
- **M1** for applying limits to the integrated expression when \(k > 1\).
- **A1** for evaluating the limit at infinity as \(0\).
- **A1** for correct final evaluation at the lower limit leading to \(A_k = \frac{(2k-1) e^{\frac{1}{k}-1}}{k(1-k)^2}\).
- **(d)(iii)** [1 Mark]
- **A1** for stating \(A_k \to \infty\).

**Part A**

(a) (i) For \(n = 3\), the roots of unity are given by \(z_k = e^{\mathrm{i}\frac{2k\pi}{3}}\) for \(k = 1, 2\).
Using Euler's formula:
\(z_1 = \cos\left(\frac{2\pi}{3}\right) + \mathrm{i}\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\).
\(z_2 = \cos\left(\frac{4\pi}{3}\right) + \mathrm{i}\sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - \mathrm{i}\frac{\sqrt{3}}{2}\).

(ii) Distances from \(z_0 = 1\):
\(|z_1 - 1| = \left|-\frac{3}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right| = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3}\).
Since \(z_2\) is the complex conjugate of \(z_1\), its distance to \(1\) is the same:
\(|z_2 - 1| = \sqrt{3}\).

(iii) The product of the distances is:
\(D_3 = |z_1 - 1||z_2 - 1| = \sqrt{3} \times \sqrt{3} = 3\).

(b) (i) For \(n = 4\), the roots of unity are \(z_k = e^{\mathrm{i}\frac{2k\pi}{4}} = e^{\mathrm{i}\frac{k\pi}{2}}\) for \(k = 1, 2, 3\).
Thus:
\(z_1 = \mathrm{i}, \quad z_2 = -1, \quad z_3 = -\mathrm{i}\).

(ii) Distances to \(z_0 = 1\):
\(|z_1 - 1| = |\mathrm{i} - 1| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\).
\(|z_2 - 1| = |-1 - 1| = |-2| = 2\).
\(|z_3 - 1| = |-\mathrm{i} - 1| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}\).
Hence, the product of the distances is:
\(D_4 = |z_1 - 1||z_2 - 1||z_3 - 1| = \sqrt{2} \times 2 \times \sqrt{2} = 4\).

**Part B**

(c) (i) Expanding the right-hand side:
\((z-1)(z^{n-1} + z^{n-2} + \dots + z + 1) = z(z^{n-1} + z^{n-2} + \dots + 1) - (z^{n-1} + z^{n-2} + \dots + 1)\)
\(= (z^n + z^{n-1} + \dots + z) - (z^{n-1} + z^{n-2} + \dots + 1)\)
\(= z^n - 1\).
Hence, \(z^n - 1 = (z-1)(z^{n-1} + z^{n-2} + \dots + z + 1)\).

(ii) The polynomial equation \(z^n - 1 = 0\) has exactly \(n\) distinct roots: \(1, z_1, z_2, \dots, z_{n-1}\).
By part (c)(i), the roots of \(z^n - 1 = 0\) must satisfy either \(z - 1 = 0\) or \(P(z) = z^{n-1} + z^{n-2} + \dots + z + 1 = 0\).
Since the root \(1\) satisfies \(z - 1 = 0\), and all roots are distinct, the remaining \(n-1\) roots (which are \(z_1, z_2, \dots, z_{n-1}\)) must be the roots of \(P(z) = 0\).

(iii) Since \(P(z)\) is a monic polynomial of degree \(n-1\) with roots \(z_1, z_2, \dots, z_{n-1}\), it can be factored as:
\(P(z) = (z - z_1)(z - z_2) \dots (z - z_{n-1}) = \prod_{k=1}^{n-1} (z - z_k)\).

(iv) Evaluating \(P(z)\) at \(z = 1\) using its algebraic definition:
\(P(1) = 1^{n-1} + 1^{n-2} + \dots + 1 + 1 = n\) (since there are exactly \(n\) terms).
Using the factored form of \(P(z)\) evaluated at \(z = 1\):
\(P(1) = \prod_{k=1}^{n-1} (1 - z_k)\).
Therefore, \(\prod_{k=1}^{n-1} (1 - z_k) = n\).

(v) Since \(D_n = \prod_{k=1}^{n-1} |z_k - 1| = \prod_{k=1}^{n-1} |1 - z_k| = \left| \prod_{k=1}^{n-1} (1 - z_k) \right|\),
substituting the result from (c)(iv) yields:
\(D_n = |n| = n\) (since \(n\) is a positive integer).

**Part C**

(d) (i) Express \(e^{\mathrm{i}\theta} - 1\) using trigonometric identities:
\(e^{\mathrm{i}\theta} - 1 = \cos\theta - 1 + \mathrm{i}\sin\theta\).
Using the half-angle identities \(\cos\theta - 1 = -2\sin^2\left(\frac{\theta}{2}\right)\) and \(\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\):
\(e^{\mathrm{i}\theta} - 1 = -2\sin^2\left(\frac{\theta}{2}\right) + 2\mathrm{i}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) = 2\sin\left(\frac{\theta}{2}\right)\left(-\sin\left(\frac{\theta}{2}\right) + \mathrm{i}\cos\left(\frac{\theta}{2}\right)\right)\).
Taking the modulus of both sides:
\(|e^{\mathrm{i}\theta} - 1| = \left|2\sin\left(\frac{\theta}{2}\right)\right| \sqrt{\left(-\sin\left(\frac{\theta}{2}\right)\right)^2 + \left(\cos\left(\frac{\theta}{2}\right)\right)^2} = 2\left|\sin\left(\frac{\theta}{2}\right)\right|\).

(ii) Setting \(\theta = \frac{2k\pi}{n}\) for \(k = 1, \dots, n-1\):
\(|z_k - 1| = \left|e^{\mathrm{i}\frac{2k\pi}{n}} - 1\right| = 2\left|\sin\left(\frac{k\pi}{n}\right)\right|\).
Since \(1 \le k \le n-1\), the angle \(\frac{k\pi}{n}\) lies in \((0, \pi)\), so \(\sin\left(\frac{k\pi}{n}\right) > 0\). Thus, \(|z_k - 1| = 2\sin\left(\frac{k\pi}{n}\right)\).
Multiplying these terms together:
\(D_n = \prod_{k=1}^{n-1} |z_k - 1| = \prod_{k=1}^{n-1} \left(2\sin\left(\frac{k\pi}{n}\right)\right) = 2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)\).

(iii) From Part B, we have \(D_n = n\).
From Part C (d)(ii), we have \(D_n = 2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)\).
Equating these two expressions:
\(2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = n\).
Dividing by \(2^{n-1}\) gives the desired identity:
\(\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}\).

**Part A (9 Marks)**
- **(a)(i)** [2 Marks]
- **A1** for \(z_1 = -\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\).
- **A1** for \(z_2 = -\frac{1}{2} - \mathrm{i}\frac{\sqrt{3}}{2}\).
- **(a)(ii)** [2 Marks]
- **A1** for showing \(|z_1-1| = \sqrt{3}\) with working.
- **A1** for showing \(|z_2-1| = \sqrt{3}\).
- **(a)(iii)** [1 Mark]
- **A1** for \(D_3 = 3\).
- **(b)(i)** [1 Mark]
- **A1** for \(z_1 = \mathrm{i}, z_2 = -1, z_3 = -\mathrm{i}\).
- **(b)(ii)** [3 Marks]
- **M1** for calculating the three distances: \(\sqrt{2}, 2, \sqrt{2}\).
- **A1** for writing the product product expression.
- **A1** for verifying that the product is indeed \(4\).

**Part B (11 Marks)**
- **(c)(i)** [2 Marks]
- **M1** for expanding the RHS: \(z(z^{n-1} + \dots + 1) - (z^{n-1} + \dots + 1)\).
- **A1** for showing complete cancellation of intermediate terms leading to \(z^n-1\).
- **(c)(ii)** [2 Marks]
- **R1** for explaining that \(1\) is the root corresponding to \(z-1=0\).
- **R1** for explaining that the other \(n-1\) distinct roots must satisfy \(P(z) = 0\) due to the fundamental theorem of algebra.
- **(c)(iii)** [2 Marks]
- **M1** for representing as a product of linear factors.
- **A1** for \(P(z) = \prod_{k=1}^{n-1} (z - z_k)\).
- **(c)(iv)** [3 Marks]
- **A1** for showing \(P(1) = n\) by summing the terms of the geometric series.
- **M1** for substituting \(z=1\) into the factored form.
- **A1** for equating them to get \(\prod_{k=1}^{n-1} (1 - z_k) = n\).
- **(c)(v)** [2 Marks]
- **M1** for taking modulus on both sides: \(|P(1)| = n\).
- **A1** for concluding \(D_n = n\).

**Part C (8 Marks)**
- **(d)(i)** [3 Marks]
- **M1** for writing \(e^{\mathrm{i}\theta} - 1 = \cos\theta - 1 + \mathrm{i}\sin\theta\).
- **M1** for applying double-angle identities to factor out \(2\sin(\theta/2)\).
- **A1** for showing the modulus is indeed \(2|\sin(\theta/2)|\).
- **(d)(ii)** [2 Marks]
- **M1** for substituting \(\theta = \frac{2k\pi}{n}\) and justifying that \(\sin\left(\frac{k\pi}{n}\right) > 0\).
- **A1** for writing \(D_n = 2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)\).
- **(d)(iii)** [3 Marks]
- **R1** for recalling \(D_n = n\).
- **M1** for equating \(2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = n\).
- **A1** for obtaining \(\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}\).