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

Using the geometric term formula \( u_n = u_1 r^{n-1} \), we have \( u_1 r = 6 \) and \( u_1 r^4 = \frac{16}{9} \). Dividing these two equations gives \( r^3 = \frac{16/9}{6} = \frac{8}{27} \). Taking the cube root, we find the common ratio \( r = \frac{2}{3} \). Substituting \( r \) back into the first equation, we get \( u_1 \left(\frac{2}{3}\right) = 6 \), which gives the first term \( u_1 = 9 \). Since \( |r| < 1 \), the sum to infinity exists and is calculated using \( S_{\infty} = \frac{u_1}{1 - r} \). Substituting the values, we obtain \( S_{\infty} = \frac{9}{1 - 2/3} = 27 \).

M1 for expressing terms in terms of \( u_1 \) and \( r \); A1 for finding \( r = \frac{2}{3} \); M1 for finding \( u_1 = 9 \); M1 for using the sum to infinity formula; A1 for the correct final calculation leading to 27.

(a) The composite function is found by substituting \( g(x) \) into \( f(x) \): \( (f \circ g)(x) = f(g(x)) = \ln(2(e^x + 1) - 3) = \ln(2e^x - 1) \). (b) Setting this expression equal to \( \ln(5) \) gives \( \ln(2e^x - 1) = \ln(5) \). This simplifies to \( 2e^x - 1 = 5 \), which gives \( 2e^x = 6 \). Dividing by 2, we get \( e^x = 3 \). Taking the natural logarithm of both sides yields \( x = \ln(3) \).

M1 for substituting \( g(x) \) into \( f(x) \); A1 for the correct simplified expression \( \ln(2e^x - 1) \); M1 for setting their expression equal to \( \ln(5) \) and removing logs; A1 for obtaining \( e^x = 3 \); A1 for the final answer \( x = \ln(3) \).

Using the identity \( \cos^2(x) = 1 - \sin^2(x) \), the equation becomes \( 2(1 - \sin^2(x)) + 3\sin(x) - 3 = 0 \). Expanding and simplifying gives \( 2 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \), which simplifies to \( 2\sin^2(x) - 3\sin(x) + 1 = 0 \). Factoring the quadratic in terms of \( \sin(x) \) gives \( (2\sin(x) - 1)(\sin(x) - 1) = 0 \). This yields two cases: \( \sin(x) = \frac{1}{2} \) or \( \sin(x) = 1 \). Within the interval \( 0 \le x \le 2\pi \), the solutions for \( \sin(x) = \frac{1}{2} \) are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). The solution for \( \sin(x) = 1 \) is \( x = \frac{\pi}{2} \). Combining these, the complete set of solutions is \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \).

M1 for using the Pythagorean identity; A1 for obtaining the simplified quadratic equation \( 2\sin^2(x) - 3\sin(x) + 1 = 0 \); M1 for factoring or solving the quadratic; A1 for finding \( \sin(x) = \frac{1}{2} \) and \( \sin(x) = 1 \); A2 for identifying all three correct angles.

(a) First, find the derivative of the curve: \( \frac{dy}{dx} = 3x^2 - 6x + k \). At \( x = 2 \), the gradient of the tangent is \( \frac{dy}{dx} = 3(2)^2 - 6(2) + k = k \). The given line is \( y = 4x + 3 \), which has a gradient of 4. Since the tangent is parallel to this line, their gradients are equal, so \( k = 4 \). (b) With \( k = 4 \), the y-coordinate of the point of tangency is \( y = 2^3 - 3(2)^2 + 4(2) - 5 = -1 \). Using the point-slope form with point \( (2, -1) \) and gradient 4: \( y - (-1) = 4(x - 2) \implies y = 4x - 9 \).

M1 for differentiating the equation of the curve; A1 for evaluating the gradient at \( x = 2 \) as \( k \); A1 for identifying the gradient of the line as 4 and setting \( k = 4 \); M1 for substituting \( x = 2 \) back into the curve equation to find the y-coordinate; A1 for finding the point \( (2, -1) \); A1 for writing the correct tangent equation.

Since the sum of probabilities must equal 1, we have \( a + b + 0.3 + 0.1 = 1 \implies a + b = 0.6 \). The expected value is given by \( \mathrm{E}(X) = \sum x P(X = x) \), so \( 1(a) + 2(b) + 3(0.3) + 4(0.1) = 2.1 \implies a + 2b + 1.3 = 2.1 \implies a + 2b = 0.8 \). We now have a system of two linear equations: (1) \( a + b = 0.6 \) and (2) \( a + 2b = 0.8 \). Subtracting the first equation from the second gives \( b = 0.2 \). Substituting \( b = 0.2 \) back into the first equation yields \( a = 0.4 \).

M1 for setting sum of probabilities to 1; A1 for the equation \( a + b = 0.6 \); M1 for using the expectation formula; A1 for the equation \( a + 2b = 0.8 \); M1 for solving the simultaneous equations; A1 for both correct values of \( a \) and \( b \).

Let \( u = -x^2 \). Then, the derivative is \( \frac{du}{dx} = -2x \), which gives \( x dx = -\frac{1}{2} du \). Changing the limits of integration: when \( x = 0 \), \( u = 0 \); when \( x = 1 \), \( u = -1 \). Substituting these into the integral, we get \( \int_{0}^{-1} e^u \left(-\frac{1}{2}\right) du = \frac{1}{2} \int_{-1}^{0} e^u du \). Integrating \( e^u \) gives \( \frac{1}{2} [e^u]_{-1}^{0} = \frac{1}{2} (e^0 - e^{-1}) = \frac{e-1}{2e} \).

M1 for choosing a suitable substitution (e.g., \( u = -x^2 \)); A1 for finding the correct differential relationship; A1 for updating the limits of integration correctly; M1 for integrating the expression; A1 for evaluating and expressing the result in exact form.

(a) Using the Cosine Rule: \( BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(A) \). Substituting the given values: \( BC^2 = 6^2 + 4^2 - 2(6)(4)\left(\frac{1}{4}\right) = 36 + 16 - 12 = 40 \). Thus, \( BC = \sqrt{40} = 2\sqrt{10} \text{ cm} \). (b) Since \( A \) is an angle in a triangle and \( \cos(A) = \frac{1}{4} > 0 \), \( A \) is acute. We find \( \sin(A) \) using \( \sin(A) = \sqrt{1 - \cos^2(A)} = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4} \). The area of the triangle is given by \( \text{Area} = \frac{1}{2} (AB)(AC)\sin(A) = \frac{1}{2} (6)(4)\left(\frac{\sqrt{15}}{4}\right) = 3\sqrt{15} \text{ cm}^2 \).

M1 for using the Cosine Rule; A1 for finding \( BC^2 = 40 \) and thus \( BC = 2\sqrt{10} \); M1 for using the Pythagorean identity to find \( \sin(A) \); A1 for \( \sin(A) = \frac{\sqrt{15}}{4} \); M1 for using the area formula; A1 for obtaining \( 3\sqrt{15} \).

Using the product rule of logarithms, we can combine the terms: \( \log_2(x(x-3)) = 2 \). Converting the logarithmic equation into exponential form gives \( x(x-3) = 2^2 \), which simplifies to \( x^2 - 3x = 4 \). Rearranging into standard quadratic form gives \( x^2 - 3x - 4 = 0 \). Factoring the quadratic, we get \( (x - 4)(x + 1) = 0 \), which gives the potential solutions \( x = 4 \) or \( x = -1 \). Since the original logarithmic expressions require the arguments to be positive, we must have \( x > 3 \). Therefore, we reject \( x = -1 \), leaving the single solution \( x = 4 \).

M1 for applying the product rule of logarithms; M1 for converting the logarithm to exponential form; A1 for the quadratic equation \( x^2 - 3x - 4 = 0 \); M1 for solving the quadratic equation; A1 for identifying the roots \( x = 4, -1 \); R1 for rejecting \( x = -1 \) and stating \( x = 4 \) as the only valid solution.

An arithmetic sequence \(u_n\) has \(u_1 = 4\) and common difference \(d\). Thus: \(u_4 = u_1 + 3d = 4 + 3d\) and \(u_{10} = u_1 + 9d = 4 + 9d\).

A geometric sequence \(v_n\) has \(v_1 = 4\) and common ratio \(r\). Thus: \(v_3 = v_1 r^2 = 4r^2\) and \(v_4 = v_1 r^3 = 4r^3\).

We are given \(u_4 = v_3\) and \(u_{10} = v_4\). This gives the system of equations:
1) \(4 + 3d = 4r^2 \implies 3d = 4r^2 - 4\)
2) \(4 + 9d = 4r^3 \implies 9d = 4r^3 - 4\)

Notice that \(9d = 3(3d)\). Substituting the first equation into the second:
\(3(4r^2 - 4) = 4r^3 - 4\)
\(12r^2 - 12 = 4r^3 - 4\)

Rearranging the terms:
\(4r^3 - 12r^2 + 8 = 0\)

Dividing the entire equation by 4:
\(r^3 - 3r^2 + 2 = 0\)

Since the sum of the coefficients is \(1 - 3 + 2 = 0\), \(r = 1\) is a root of this cubic. Using polynomial division or synthetic division, we can factor out \((r-1)\):
\((r-1)(r^2 - 2r - 2) = 0\)

Since we are given that \(r \neq 1\), we solve the quadratic equation:
\(r^2 - 2r - 2 = 0\)

Using the quadratic formula:
\(r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}\)
\(r = \frac{2 \pm \sqrt{4 + 8}}{2}\)
\(r = \frac{2 \pm \sqrt{12}}{2}\)
\(r = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}\)

Thus, the possible values of \(r\) are \(1 + \sqrt{3}\) and \(1 - \sqrt{3}\).

**M1** for expressing \(u_4\) and \(v_3\) in terms of \(d\) and \(r\) and equating them.
**A1** for expressing \(u_{10}\) and \(v_4\) in terms of \(d\) and \(r\) and equating them.
**M1** for attempting to eliminate \(d\) to obtain an equation in terms of \(r\) only.
**A1** for obtaining the simplified cubic equation \(r^3 - 3r^2 + 2 = 0\).
**M1** for identifying \(r=1\) as a root and factoring the cubic to get \((r - 1)(r^2 - 2r - 2) = 0\).
**A1** for solving the quadratic equation and stating the final correct values: \(r = 1 \pm \sqrt{3}\).

(a) To find the local maximum, we first find the derivative of \(f(x) = \frac{\ln x}{x}\) using the quotient rule: \(f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}\). Set \(f'(x) = 0\) to find critical points: \(1 - \ln x = 0 \implies \ln x = 1 \implies x = e\). The y-coordinate is \(f(e) = \frac{\ln e}{e} = \frac{1}{e}\). For \(0 < x < e\), \(f'(x) > 0\), and for \(x > e\), \(f'(x) < 0\), confirming \((e, e^{-1})\) is a local maximum. (b) To find the point of inflection, we find the second derivative using the quotient rule on \(f'(x)\): \(f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \ln x) \cdot 2x}{x^4} = \frac{-x - 2x + 2x \ln x}{x^4} = \frac{-3 + 2\ln x}{x^3}\). Set \(f''(x) = 0\): \(-3 + 2\ln x = 0 \implies \ln x = 1.5 \implies x = e^{3/2}\). The y-coordinate is \(f(e^{3/2}) = \frac{1.5}{e^{3/2}} = 1.5 e^{-1.5}\). Since the sign of \(f''(x)\) changes across \(x = e^{3/2}\), the point of inflection is \((e^{3/2}, 1.5 e^{-1.5})\). (c) The curve crosses the x-axis when \(f(x) = 0 \implies \ln x = 0 \implies x = 1\). The area is given by \(\int_{1}^{e^2} \frac{\ln x}{x} dx\). Let \(u = \ln x\), then \(du = \frac{1}{x} dx\). The limits of integration change: when \(x = 1\), \(u = 0\); when \(x = e^2\), \(u = 2\). The integral becomes \(\int_{0}^{2} u du = \left[ \frac{1}{2} u^2 \right]_{0}^{2} = \frac{1}{2}(4) - 0 = 2\). (d) The volume \(V\) is given by \(V = \pi \int_{1}^{e} [f(x)]^2 dx = \pi \int_{1}^{e} \frac{(\ln x)^2}{x^2} dx\). We use integration by parts: let \(u = (\ln x)^2\) and \(dv = x^{-2} dx\). Then \(du = \frac{2\ln x}{x} dx\) and \(v = -x^{-1}\). Thus, \(\int \frac{(\ln x)^2}{x^2} dx = -\frac{(\ln x)^2}{x} + 2 \int \frac{\ln x}{x^2} dx\). For the remaining integral, we use integration by parts again with \(u = \ln x\) and \(dv = x^{-2} dx\): \(\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} - \frac{1}{x}\). Combining these results gives: \(\int \frac{(\ln x)^2}{x^2} dx = -\frac{(\ln x)^2 + 2\ln x + 2}{x}\). Evaluating this from 1 to \(e\): at \(x = e\), the value is \(-\frac{1 + 2 + 2}{e} = -\frac{5}{e}\); at \(x = 1\), the value is \(-\frac{0 + 0 + 2}{1} = -2\). The definite integral is \(-\frac{5}{e} - (-2) = 2 - \frac{5}{e}\). Thus, the volume is \(V = \pi(2 - \frac{5}{e})\).

(a) M1 for applying the quotient rule to find \(f'(x)\). A1 for correct derivative \(f'(x) = \frac{1 - \ln x}{x^2}\). M1 for setting \(f'(x) = 0\) and solving for \(x\). A1 for correct coordinates \((e, e^{-1})\). [4 marks]
(b) M1 for applying the quotient rule to find \(f''(x)\). A1 for correct second derivative \(f''(x) = \frac{2\ln x - 3}{x^3}\). M1 for setting \(f''(x) = 0\) and solving for \(x\). A1 for correct coordinates \((e^{3/2}, 1.5e^{-3/2})\). [4 marks]
(c) M1 for setting up the integral \(\int_{1}^{e^2} \frac{\ln x}{x} dx\). M1 for using the substitution \(u = \ln x\). A1 for changing limits and obtaining \(\int_{0}^{2} u du\). A1 for evaluating to 2. [4 marks]
(d) M1 for setting up the volume integral \(\pi \int_{1}^{e} \frac{(\ln x)^2}{x^2} dx\). M1 for first integration by parts. A1 for obtaining \(-\frac{(\ln x)^2}{x} + 2\int\frac{\ln x}{x^2}dx\). M1 for second integration by parts to obtain \(-\frac{(\ln x)^2 + 2\ln x + 2}{x}\). A1 for substituting limits of integration. A1 for the final exact answer \(\pi(2 - \frac{5}{e})\). [6 marks]

(a) From the definitions, \(u_2 = 4 + d\) and \(v_2 = 4r\). Since \(u_2 = v_2 + 2\), we have \(4 + d = 4r + 2 \implies d = 4r - 2\). Also, \(u_3 = 4 + 2d\) and \(v_3 = 4r^2\). Since \(u_3 = v_3\), we have \(4 + 2d = 4r^2\). Substituting the expression for \(d\) into this equation yields: \(4 + 2(4r - 2) = 4r^2 \implies 4 + 8r - 4 = 4r^2 \implies 4r^2 - 8r = 0\). Since \(d \neq 0\), \(r\) cannot be 1 (which would make \(d = 2\) but let's look at the roots: \(4r(r - 2) = 0\), so \(r = 0\) or \(r = 2\)). Since a geometric sequence has positive terms or non-zero ratio, and \(d \neq 0\), we take \(r = 2\). Substituting \(r = 2\) into the expression for \(d\) gives \(d = 4(2) - 2 = 6\). (b) The sum of the first \(k\) terms of the arithmetic sequence is \(S_k = \frac{k}{2}[2u_1 + (k-1)d] = \frac{k}{2}[2(4) + (k-1)6] = \frac{k}{2}[8 + 6k - 6] = \frac{k}{2}[6k + 2] = 3k^2 + k\). (c) The sum of the first \(k\) terms of the geometric sequence is \(G_k = \frac{v_1(r^k - 1)}{r-1} = \frac{4(2^k - 1)}{2-1} = 4(2^k - 1) = 2^{k+2} - 4\). (d) We require \(G_k - S_k > 2000 \implies 2^{k+2} - 4 - (3k^2 + k) > 2000 \implies 2^{k+2} - 3k^2 - k - 4 > 2000\). Testing values of \(k\): For \(k = 9\): \(2^{11} - 3(81) - 9 - 4 = 2048 - 243 - 13 = 1792 \le 2000\). For \(k = 10\): \(2^{12} - 3(100) - 10 - 4 = 4096 - 300 - 14 = 3782 > 2000\). Thus, the smallest value of \(k\) is 10. (e) \(v_n = v_1 r^{n-1} = 4 \cdot 2^{n-1} = 2^{n+1}\). Then, \(w_n = \log_2(2^{n+1}) = n+1\). The first term is \(w_1 = 1+1 = 2\). Consider the difference between consecutive terms: \(w_{n+1} - w_n = (n+2) - (n+1) = 1\). Since this difference is constant, \(w_n\) is an arithmetic sequence with first term 2 and common difference 1.

(a) M1 for writing expressions for \(u_2, v_2, u_3, v_3\). M1 for setting up the system of equations. M1 for substituting \(d = 4r - 2\) into the second equation. A1 for obtaining \(4r^2 - 8r = 0\). A1 for concluding \(r = 2\) and \(d = 6\). [5 marks]
(b) M1 for using the arithmetic sum formula. A1 for simplifying to \(3k^2 + k\). [2 marks]
(c) M1 for using the geometric sum formula. A1 for simplifying to \(2^{k+2} - 4\). [2 marks]
(d) M1 for setting up the inequality \(2^{k+2} - 3k^2 - k - 4 > 2000\). M1 for attempting to evaluate the expression for a specific value of \(k\). A1 for finding the value of the expression at \(k = 9\) (1792). A1 for finding the value at \(k = 10\) (3782). A1 for concluding that \(k = 10\). [5 marks]
(e) M1 for writing the general term \(v_n = 2^{n+1}\). A1 for substituting and simplifying \(w_n = n + 1\). M1 for showing \(w_{n+1} - w_n = 1\). A1 for stating first term is 2 and common difference is 1. [4 marks]

(a) Since \(AC\) is tangent to the circle at \(A\), the angle \(OAC\) is \(\frac{\pi}{2}\) (a right angle). In the right-angled triangle \(OAC\), we have \(\tan \theta = \frac{AC}{OA} = \frac{AC}{6}\), which gives \(AC = 6 \tan \theta\). The area of the triangle is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times AC = \frac{1}{2} \times 6 \times 6 \tan \theta = 18 \tan \theta\). (b) The area of sector \(OAB\) is \(\frac{1}{2} r^2 \theta = \frac{1}{2} (6^2) \theta = 18\theta\). The shaded region is the area of triangle \(OAC\) minus the area of sector \(OAB\). Therefore, the shaded area is \(18 \tan \theta - 18\theta = 18(\tan \theta - \theta)\). (c) We are given that the shaded area is \(18\sqrt{3} - 6\pi\). Set the expression equal to this value: \(18(\tan \theta - \theta) = 18\sqrt{3} - 6\pi \implies \tan \theta - \theta = \sqrt{3} - \frac{\pi}{3}\). Comparing the terms on both sides, we see by inspection that \(\theta = \frac{\pi}{3}\) (since \(\tan(\pi/3) = \sqrt{3}\)). (d) (i) \(f'(\theta) = \frac{d}{d\theta}(\tan \theta - \theta) = \sec^2 \theta - 1\). Using the trigonometric identity \(1 + \tan^2 \theta = \sec^2 \theta\), we obtain \(f'(\theta) = \tan^2 \theta\). (ii) Since \(\frac{d}{d\theta}(\tan \theta - \theta) = \tan^2 \theta\), the antiderivative is \(\int \tan^2 \theta \, d\theta = \tan \theta - \theta + C\). (e) Using the result from (d), we evaluate the definite integral: \(\int_{0}^{\pi/4} 3 \tan^2 \theta \, d\theta = 3 \left[ \tan \theta - \theta \right]_{0}^{\pi/4} = 3 \left( (\tan\frac{\pi}{4} - \frac{\pi}{4}) - (\tan 0 - 0) \right) = 3 \left( 1 - \frac{\pi}{4} \right) = 3 - \frac{3\pi}{4}\).

(a) M1 for recognizing \(\angle OAC = \frac{\pi}{2}\). M1 for finding \(AC = 6 \tan \theta\). A1 for showing the area is \(18 \tan \theta\). [3 marks]
(b) M1 for finding the area of sector \(OAB = 18\theta\). M1 for subtracting sector area from triangle area. A1 for \(18(\tan \theta - \theta)\). [3 marks]
(c) M1 for setting up the equation \(18(\tan \theta - \theta) = 18\sqrt{3} - 6\pi\). M1 for simplifying to \(\tan \theta - \theta = \sqrt{3} - \frac{\pi}{3}\). R2 for identifying and justifying \(\theta = \frac{\pi}{3}\). [4 marks]
(d) (i) M1 for differentiating \(\tan \theta - \theta\) to get \(\sec^2 \theta - 1\). M2 for applying the identity \(\sec^2 \theta - 1 = \tan^2 \theta\). (ii) A2 for \(\tan \theta - \theta + C\). [5 marks]
(e) M1 for using the fundamental theorem of calculus with the antiderivative. A1 for substituting limits \(\frac{\pi}{4}\) and 0. A1 for final exact answer \(3 - \frac{3\pi}{4}\). [3 marks]

The sum of the first \( n \) terms of the arithmetic sequence is: \( S_n^{(A)} = \frac{n}{2} [2(12) + (n-1)4.5] = 2.25n^2 + 9.75n \). The sum of the first \( n \) terms of the geometric sequence is: \( S_n^{(G)} = \frac{3(1.2^n - 1)}{1.2 - 1} = 15(1.2^n - 1) \). We solve the inequality: \( 15(1.2^n - 1) > 2.25n^2 + 9.75n \). Using a GDC to find values of both sums for different values of \( n \): For \( n = 26 \): \( S_{26}^{(A)} = 2.25(26)^2 + 9.75(26) = 1774.5 \) and \( S_{26}^{(G)} = 15(1.2^{26} - 1) \approx 1702.13 \) (here \( S_n^{(G)} < S_n^{(A)} \)). For \( n = 27 \): \( S_{27}^{(A)} = 2.25(27)^2 + 9.75(27) = 1903.5 \) and \( S_{27}^{(G)} = 15(1.2^{27} - 1) \approx 2045.56 \) (here \( S_n^{(G)} > S_n^{(A)} \)). Thus, the smallest value of \( n \) is \( 27 \).

(M1) for a correct expression for the sum of the arithmetic sequence.
(M1) for a correct expression for the sum of the geometric sequence.
(M1) for setting up the inequality or equation \( S_n^{(G)} > S_n^{(A)} \).
(A1) for calculating the values at \( n = 26 \) and \( n = 27 \) (or indicating graphical intersection).
(A2) for the final correct integer answer \( n = 27 \).

(a) To find the points of intersection, set \( e^{0.5x} = 3x - 1 \). Using a GDC solver for the equation \( e^{0.5x} - 3x + 1 = 0 \), we obtain \( x \approx 0.841 \) and \( x \approx 5.47 \).
(b) The inequality \( f(x) < g(x) \) is satisfied on the interval between the two intersection points since the exponential curve lies below the straight line in this region. This gives \( 0.841 < x < 5.47 \).

(a)
(M1) for equating the two functions: \( e^{0.5x} = 3x - 1 \).
(A1) for \( x \approx 0.841 \).
(A1) for \( x \approx 5.47 \).
(b)
(M1) for recognizing that the solution lies between the two intersection points (e.g., from a sketch).
(A2) for the correct interval \( 0.841 < x < 5.47 \) (accept interval notation).

(a) The particle is at rest when \( v(t) = 0 \), which gives \( 3t^2 e^{-0.6t} - t = 0 \). Since \( t > 0 \), we can simplify this to \( 3t e^{-0.6t} - 1 = 0 \). Using a GDC to solve this equation yields \( t \approx 0.432 \) and \( t \approx 4.24 \) seconds.
(b) The total distance travelled is the integral of the speed: \( \int_{0}^{6} |v(t)| \, dt = \int_{0}^{6} |3t^2 e^{-0.6t} - t| \, dt \). Evaluating this using the numerical integration function on a GDC gives \( \text{Distance} \approx 6.67 \text{ m} \).

(a)
(M1) for setting \( v(t) = 0 \).
(A1) for \( t \approx 0.432 \).
(A1) for \( t \approx 4.24 \).
(b)
(M1) for writing the correct absolute value integral expression.
(M1) for an attempt to evaluate the integral numerically using GDC.
(A1) for the correct distance \( 6.67 \).

(a) Using the Sine Rule: \( \frac{\sin(A\widehat{C}B)}{72} = \frac{\sin 40^\circ}{55} \implies \sin(A\widehat{C}B) = \frac{72 \sin 40^\circ}{55} \approx 0.8415 \). This yields two possible values: \( A\widehat{C}B_1 = \arcsin(0.8415) \approx 57.3^\circ \) and \( A\widehat{C}B_2 = 180^\circ - 57.29^\circ \approx 123^\circ \). Both are valid because the sum of angles in each case is less than \( 180^\circ \).
(b) The area is \( \text{Area} = \frac{1}{2}(72)(55)\sin(A\widehat{B}C) \). To minimize the area, we must minimize \( \sin(A\widehat{B}C) \). Since \( A\widehat{B}C = 180^\circ - (40^\circ + A\widehat{C}B) \): Case 1: \( A\widehat{C}B = 57.29^\circ \implies A\widehat{B}C = 82.71^\circ \). Case 2: \( A\widehat{C}B = 122.71^\circ \implies A\widehat{B}C = 17.29^\circ \). Since \( \sin(17.29^\circ) < \sin(82.71^\circ) \), the area is minimized when \( A\widehat{B}C \approx 17.29^\circ \). Then, by the Sine Rule, \( AC = \frac{55 \sin 17.29^\circ}{\sin 40^\circ} \approx 25.4\text{ m} \).

(a)
(M1) for correct substitution into the Sine Rule equation.
(A1) for \( 57.3^\circ \).
(A1) for \( 123^\circ \).
(b)
(M1) for identifying that the area is minimized when the angle \( A\widehat{B}C = 17.29^\circ \) is chosen.
(M1) for applying the Sine Rule (or Cosine Rule) to find side \( AC \).
(A1) for \( AC \approx 25.4\text{ m} \).

(a) Let \( X \sim N(150, 18^2) \). We want to find \( P(130 < X < 170) \). Using a GDC (normalcdf), we obtain \( P(130 < X < 170) \approx 0.7335 \approx 0.734 \).
(b) The probability that an apple is small is \( p = P(X < 125) \). Using a GDC, we find \( p \approx 0.082449 \). Let \( Y \) be the number of small apples in a sample of 10. \( Y \sim B(10, 0.082449) \). We need to find \( P(Y \ge 3) = 1 - P(Y \le 2) \). Using binomcdf on a GDC, we find \( P(Y \le 2) \approx 0.95853 \). Thus, \( P(Y \ge 3) = 1 - 0.95853 \approx 0.0415 \).

(a)
(M1) for setting up the normal probability equation.
(A1) for \( 0.734 \).
(b)
(M1) for finding \( p = P(X < 125) \approx 0.0824 \).
(M1) for identifying the binomial distribution \( Y \sim B(10, p) \).
(M1) for expressing the required probability as \( 1 - P(Y \le 2) \).
(A1) for \( 0.0415 \).

(a) The general term in the expansion is: \( T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left( \frac{k}{x} \right)^r = \binom{9}{r} 2^{9-r} k^r x^{18-3r} \). For the term independent of \( x \), set the exponent to 0: \( 18 - 3r = 0 \implies r = 6 \). Substituting \( r = 6 \) gives: \( T_7 = \binom{9}{6} 2^3 k^6 = 84 \times 8 \times k^6 = 672 k^6 \).
(b) Setting the coefficient equal to \( 145152 \): \( 672 k^6 = 145152 \implies k^6 = 216 \). Since \( k > 0 \), \( k = 216^{1/6} = (6^3)^{1/6} = 6^{1/2} = \sqrt{6} \).

(a)
(M1) for obtaining the general term in terms of \( r \).
(A1) for setting \( 18-3r=0 \) and finding \( r = 6 \).
(A1) for finding the coefficient \( 84 \times 8 = 672 \) and showing the term is \( 672 k^6 \).
(b)
(M1) for setting \( 672 k^6 = 145152 \).
(A1) for simplifying to \( k^6 = 216 \).
(A1) for the correct exact value of \( k = \sqrt{6} \).

(a) Entering the data into a GDC and performing linear regression analysis, we obtain the correlation coefficient: \( r \approx -0.996 \).
(b) From the GDC, the equation of the regression line is: \( y = ax + b \), where \( a \approx -4.22 \) and \( b \approx 117 \) (specifically, \( y = -4.218x + 116.9 \)).
(c) Substituting \( x = 15 \) into the equation: \( y = -4.218(15) + 116.89 \approx 53.62 \). The estimate is \( 53.6 \) cups (or rounded to the nearest integer, \( 54 \) cups).

(a)
(M1) for entering the data into GDC.
(A1) for \( r \approx -0.996 \).
(b)
(M1) for identifying the coefficients \( a \approx -4.22 \) and \( b \approx 117 \).
(A1) for the equation \( y = -4.22x + 117 \) (or equivalent with higher precision).
(c)
(M1) for substituting \( x = 15 \) into their regression line.
(A1) for \( 53.6 \) (accept \( 54 \) or answers derived from rounded coefficients like \( 53.7 \)).

(a) Using a GDC to find the maximum point of the function on \( [0, \pi/2] \), we get \( x \approx 0.615 \) (or analytically, \( x = \arctan(1/\sqrt{2}) \approx 0.61548 \)).
(b) The volume is given by \( V = \pi \int_{0}^{\pi/2} [f(x)]^2 \, dx = \pi \int_{0}^{\pi/2} 4 \cos^2(x) \sin(x) \, dx \). Evaluating this using a GDC (or analytically using substitution \( u = \cos(x) \)): \( \int_{0}^{\pi/2} 4 \cos^2(x) \sin(x) \, dx = \left[ -\frac{4}{3}\cos^3(x) \right]_{0}^{\pi/2} = \frac{4}{3} \). Thus, \( V = \frac{4\pi}{3} \approx 4.19 \).

(a)
(M1) for attempting to find the maximum on the GDC (or setting \( f'(x) = 0 \)).
(A2) for \( x \approx 0.615 \).
(b)
(M1) for setting up the correct integral expression for volume of revolution.
(M1) for evaluating the integral using a GDC or analytically.
(A1) for \( 4.19 \) (or \( \frac{4\pi}{3} \)).

(a) We are given that \(W \sim N(\mu, 12^2)\) and \(P(W < 980) = 0.08\). Standardizing the variable: \(P(Z < \frac{980 - \mu}{12}) = 0.08\). Using the inverse normal function on a calculator: \(z \approx -1.40507\). Setting up the equation: \(\frac{980 - \mu}{12} = -1.40507\), which gives \(980 - \mu = -16.8609\), hence \(\mu = 996.8609\). Therefore, \(\mu = 996.86\) grams (correct to two decimal places). (b) We want to find \(P(990 < W < 1010)\) where \(W \sim N(996.86..., 12^2)\). Using a graphic display calculator with lower bound 990, upper bound 1010, \(\mu = 996.86...\), and \(\sigma = 12\): \(P(990 < W < 1010) \approx 0.579\). (If using the rounded value of \(\mu = 997\), then \(P(990 < W < 1010) \approx 0.580\).)

(a) \(P(W < 980) = 0.08\). Attempt to standardize or use GDC inverse normal: \(z = -1.40507...\) (M1). Correct equation: \(\frac{980 - \mu}{12} = -1.40507...\) (A1). \(\mu = 996.86\) (A1). (b) Attempt to find the interval probability \(P(990 < W < 1010)\) (M1). Correct GDC inputs and setup (A1). 0.579 (or 0.580 if using \(\mu = 997\)) (A1).

(a) Let \(X \sim N(142, 8.5^2)\).
Using a GDC to find \(P(X > 150)\):
\(P(X > 150) \approx 0.17327 \approx 0.173\).

(b) Let \(Y\) be the number of apples weighing more than 150g, so \(Y \sim B(10, 0.17327)\).
We require \(P(Y \ge 3) = 1 - P(Y \le 2)\).
Using a GDC for binomial cumulative probability:
\(P(Y \le 2) \approx 0.74837\).
Thus, \(P(Y \ge 3) \approx 1 - 0.74837 \approx 0.25163 \approx 0.252\).

(c) (i) "Standard" apples weigh between 130g and 150g:
\(P(130 < X < 150) \approx 0.74895 \approx 0.749\).

(ii) Composted apples weigh less than 130g:
\(P(X < 130) \approx 0.07778 \approx 0.0778\).

(iii) The expected revenue is:
\(E(R) = 0.80 \times P(\text{Premium}) + 0.50 \times P(\text{Standard}) + 0.05 \times P(\text{Composted})\)
\(E(R) = 0.80(0.17327) + 0.50(0.74895) + 0.05(0.07778)\)
\(E(R) = 0.138616 + 0.374475 + 0.003889 \approx 0.51698\).
Expected profit = \(E(R) - \text{cost} = 0.51698 - 0.20 = 0.31698 \approx \$0.317\).

(d) Let \(p_1 = P(X > w)\). The probability of an apple being composted remains \(0.07778\), so the probability of being "Standard" is \(1 - 0.07778 - p_1 = 0.92222 - p_1\).
The expected profit is now $0.33, meaning the expected revenue must be:
\(E(R) = 0.33 + 0.20 = 0.53\).
Setting up the equation for \(E(R)\):
\(0.80 p_1 + 0.50(0.92222 - p_1) + 0.05(0.07778) = 0.53\)
\(0.30 p_1 + 0.46111 + 0.00389 = 0.53\)
\(0.30 p_1 + 0.4650 = 0.53\)
\(0.30 p_1 = 0.0650 \implies p_1 \approx 0.21667\).
Thus, \(P(X > w) = 0.21667 \implies P(X \le w) = 0.78333\).
Using inverse normal on the GDC:
\(w = \text{invNorm}(0.78333, 142, 8.5) \approx 148.66 \approx 149\) g.

(a)
M1 for attempting to use normal distribution (e.g., writing \(P(X > 150)\) or showing standardisation).
A1 for standardising or writing correct parameters.
A1 for 0.173.

(b)
M1 for identifying binomial distribution with parameters \(n=10\) and \(p \approx 0.173\).
M1 for setting up the calculation \(1 - P(Y \le 2)\).
A1 for 0.252.

(c)
(i) A1 for 0.749.
(ii) A1 for 0.0778.
(iii) M1 for attempting to find expected revenue: \(0.80 \times P(\text{Premium}) + 0.50 \times P(\text{Standard}) + 0.05 \times P(\text{Composted})\).
A1 for correct substitution of their probabilities.
M1 for subtracting the production cost of $0.20.
A1 for $0.317 (accept $0.32 from correct rounding).

(d)
M1 for relating the expected profit to expected revenue: \(E(R) = 0.53\).
M1 for setting up an algebraic equation in terms of \(p_1 = P(X > w)\): \(0.30 p_1 + 0.465 = 0.53\).
A1 for solving \(p_1 \approx 0.217\) (or \(P(X \le w) \approx 0.783\)).
A1 for finding \(w \approx 149\) (accept 148.66).

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

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

(iii) The time between consecutive maximums is \(15 - 3 = 12\) hours, which represents the period \(T\).
\(b = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6}\).

(iv) Since the maximum of a standard cosine function occurs when the argument is \(0\), and the first maximum occurs at \(t = 3\), we have \(t - c = 0 \implies c = 3\).

(b) We require \(d(t) \ge 11.5\) for \(0 \le t \le 24\).
Using GDC to solve \(3.2 \cos\left(\frac{\pi}{6}(t - 3)\right) + 11.4 = 11.5\):
Let's find the intersection points:
\(t_1 \approx 0.0597\)
\(t_2 \approx 5.94\)
\(t_3 \approx 12.1\) (specifically \(12.0597\))
\(t_4 \approx 17.9\) (specifically \(17.9403\))
Since \(d(3) = 14.6 \ge 11.5\), the ship can enter safely between \(t_1\) and \(t_2\), and between \(t_3\) and \(t_4\).
Thus, the intervals of time are:
\(0.0597 \le t \le 5.94\) and \(12.1 \le t \le 17.9\).

(c) We solve \(d(t) = 2s(t)\):
\(3.2 \cos\left(\frac{\pi}{6}(t - 3)\right) + 11.4 = 2 \left[1.8 \sin\left(\frac{\pi}{6}(t - 1)\right) + 6.5\right]\)
Using a GDC to find the intersection points of \(y_1 = d(t)\) and \(y_2 = 2s(t)\) for \(0 \le t \le 24\):
We find the following intersection values:
\(t_1 = 1.00\)
\(t_2 \approx 11.2\)
\(t_3 = 13.0\)
\(t_4 \approx 23.2\)

These are the times when the water depth in the harbor is exactly twice the depth in the bay.

(a) (i)
M1 for substituting max and min values into amplitude formula.
A1 for showing \(a = 3.2\).

(ii)
M1 for attempting to find the midline.
A1 for \(k = 11.4\).

(iii)
M1 for finding the period \(T = 12\) hours.
A1 for obtaining \(b = \frac{\pi}{6}\).

(iv)
A1 for \(c = 3\) (accept other correct phase shifts like \(15\)).

(b)
M1 for setting up the inequality \(d(t) \ge 11.5\).
M1 for attempting to solve the equation \(d(t) = 11.5\) (e.g., finding at least two boundaries).
A1 for boundaries: \(0.0597\) (or \(0.060\)) and \(5.94\).
A1 for boundaries: \(12.1\) (or \(12.06\)) and \(17.9\) (or \(17.94\)).
A1 for expressing the intervals correctly (using inequality or interval notation).

(c)
M1 for setting up the equation \(d(t) = 2s(t)\).
M1 for correct substitution of both functions.
M1 for a valid method to solve the equation on GDC (e.g., graphing both sides and looking for intersections).
A1 for \(t_1 = 1.00\) and \(t_3 = 13.0\).
A2 for \(t_2 \approx 11.2\) and \(t_4 \approx 23.2\) (award A1 if only one of these decimal roots is found).

(a) (i) The particle is at rest when \(v(t) = 0\).
Using GDC to find the roots of \(3 t \cos(0.4 t) - 2 = 0\) for \(0 \le t \le 10\):
\(t \approx 0.694\) s and \(t \approx 3.43\) s.

(ii) Acceleration \(a(t) = v'(t)\). We require \(v'(t) < 0\).
Using the GDC to find the derivative of \(v(t)\) and setting it to \(0\):
\(v'(t) = 3 \cos(0.4 t) - 1.2 t \sin(0.4 t) = 0\).
The roots of \(v'(t) = 0\) in \(0 \le t \le 10\) are:
\(t \approx 2.15\) s and \(t \approx 8.51\) s.
Since \(v'(0) = 3 > 0\), the acceleration is negative in the interval:
\(2.15 < t < 8.51\) s.

(b) (i) Since \(s(0) = 0\):
\(s(t) = \int_{0}^{t} v(x) \, dx = \int_{0}^{t} (3 x \cos(0.4 x) - 2) \, dx\).

(ii) Using GDC to integrate \(v(t)\) from \(0\) to \(6\):
\(s(6) = \int_{0}^{6} (3 t \cos(0.4 t) - 2) \, dt \approx -14.2\) m.

(iii) Total distance traveled is
\(d = \int_{0}^{6} |v(t)| \, dt\).
Using GDC to compute the integral of the absolute value of the velocity:
\(d \approx 22.2\) m.

(c) (i) The velocity of particle \(Q\) is given by:
\(v_Q(t) = s'_Q(t) = 2t - 6\).
We want \(v(t) = v_Q(t) \implies 3 t \cos(0.4 t) - 2 = 2t - 6\).
Using GDC to find the intersection of the two velocity curves on the interval \(0 \le t \le 10\):
\(t \approx 3.27\) s.

(ii) At \(t \approx 3.271\):
Displacement of \(P\) is \(s(3.271) = \int_0^{3.271} v(t) \, dt \approx 3.26\) m.
Displacement of \(Q\) is \(s_Q(3.271) = (3.271)^2 - 6(3.271) + 8 \approx -0.928\) m.
The distance between the two particles is:
\(|s_P(3.27) - s_Q(3.27)| \approx |3.26 - (-0.928)| \approx 4.19\) m.

(a) (i)
M1 for setting \(v(t) = 0\).
A1 for \(t \approx 0.694\).
A1 for \(t \approx 3.43\).

(ii)
M1 for identifying that \(a(t) = v'(t) < 0\).
M1 for finding critical values \(t \approx 2.15\) and \(t \approx 8.51\).
A1 for correct interval \(2.15 < t < 8.51\) (or \([2.15, 8.51]\)).

(b) (i)
A1 for \(s(t) = \int_0^t (3x \cos(0.4x) - 2) dx\) (accept dummy variable \(t\)).

(ii)
M1 for setting up the definite integral \(\int_0^6 v(t) dt\).
A1 for \(-14.2\).

(iii)
M1 for setting up the integral of absolute velocity: \(\int_0^6 |v(t)| dt\) (or split integrals).
M1 for evaluating sections using critical points from (a)(i).
A1 for \(22.2\).

(c) (i)
M1 for finding \(v_Q(t) = 2t - 6\).
M1 for setting \(v(t) = v_Q(t)\).
A1 for \(t \approx 3.27\).

(ii)
M1 for attempting to find the displacement of \(P\) at \(t \approx 3.27\) (e.g., \(\int_0^{3.27} v(t) dt\)).
M1 for substituting \(t \approx 3.27\) into \(s_Q(t)\).
A1 for \(4.19\) m.

Part A: (a) The gradient of the tangent at \(x = t\) is \(y' = 2t\). Since \(t \neq 0\), the gradient of the normal line is \(m = -1/(2t)\). The equation of the normal line is \(y - t^2 = -1/(2t)(x - t)\). Multiplying by \(2t\): \(2ty - 2t^3 = -x + t\), which rearranges to \(x + 2ty - (2t^3 + t) = 0\). (b) For \(t = 0\), the point is the origin \((0,0)\). The tangent is the horizontal line \(y = 0\) and the normal is the vertical line \(x = 0\). Substituting \(t = 0\) into the normal formula gives \(x = 0\), so the formula holds. (c) Substituting the coordinates of \(Q(x_0, y_0)\) into the normal equation yields \(x_0 + 2ty_0 - 2t^3 - t = 0\). Rearranging gives \(2t^3 + (1 - 2y_0)t - x_0 = 0\). Part B: (a) Let \(g(t) = 2t^3 + pt + q\). A multiple root occurs when \(g(t) = 0\) and \(g'(t) = 6t^2 + p = 0\). From the derivative, \(t^2 = -p/6\). Substituting this into the cubic equation: \(2t(-p/6) + pt + q = 0 \implies (2p/3)t = -q \implies t = -3q/(2p)\) (for \(p \neq 0\)). Squaring this gives \(t^2 = 9q^2/(4p^2)\). Equating the two expressions for \(t^2\): \(9q^2/(4p^2) = -p/6 \implies 54q^2 = -4p^3 \implies 8p^3 + 27q^2 = 0\). If \(p=0\), then \(t=0\) and \(q=0\), which also satisfies the relation. (b) Substituting \(p = 1 - 2y_0\) and \(q = -x_0\) into \(8p^3 + 27q^2 = 0\) yields \(8(1 - 2y_0)^3 + 27(-x_0)^2 = 0 \implies 27x^2 = 8(2y-1)^3\). (c) The curve \(27x^2 = 8(2y-1)^3\) is only defined for \(y \ge 1/2\) and has a cusp at \((0, 1/2)\). Testing a point inside the cusp, such as \((0, 1)\): \(2t^3 - t = 0 \implies t(2t^2 - 1) = 0 \implies t = 0, \pm 1/\sqrt{2}\) (3 distinct real roots/normals). Thus, the region with 3 distinct normals is \(27x^2 < 8(2y-1)^3\) and \(y > 1/2\). Testing a point outside, such as \((0, 0)\): \(2t^3 + t = 0 \implies t(2t^2 + 1) = 0 \implies t = 0\) (only 1 real root/normal). Thus, the region with 1 normal is \(27x^2 > 8(2y-1)^3\). Part C: (a) Given \(F(x,y,t) = x + 2ty - 2t^3 - t = 0\). Setting \(\partial F / \partial t = 2y - 6t^2 - 1 = 0\) yields \(y = 3t^2 + 1/2\). Substituting this back into \(F(x,y,t) = 0\) gives \(x + 2t(3t^2 + 1/2) - 2t^3 - t = 0 \implies x + 4t^3 = 0 \implies x = -4t^3\). (b) From \(y = 3t^2 + 1/2\), we have \(2y-1 = 6t^2 \implies t^2 = (2y-1)/6\). From \(x = -4t^3\), we have \(x^2 = 16t^6 = 16(t^2)^3 = 16((2y-1)/6)^3 = 16(2y-1)^3 / 216 = 8(2y-1)^3 / 27\), which simplifies to \(27x^2 = 8(2y-1)^3\).

Part A: (a) M1 for attempting to find the derivative \(y' = 2t\), A1 for finding normal gradient \(m = -1/(2t)\), A1 for clear algebraic steps to reach the given form. (b) R1 for explaining that at \(t=0\), the normal is \(x=0\) and showing that substituting \(t=0\) into the formula gives \(x=0\). (c) M1 for substituting \(x_0, y_0\) for \(x, y\), A1 for rearranging terms correctly. Part B: (a) M1 for setting \(g'(t) = 0\), A1 for \(t^2 = -p/6\), M1 for substituting into \(g(t) = 0\) to express \(t\) in terms of \(p\) and \(q\), M1 for squaring \(t\) and equating to \(t^2\), A2 for completing the algebraic steps to show \(8p^3 + 27q^2 = 0\). (b) M1 for substituting \(p = 1 - 2y\) and \(q = -x\) into the condition, A2 for the correct Cartesian equation, A1 for description/sketch showing a cusp at \((0, 0.5)\). (c) M1 for testing a point in the 'inside' region, A1 for showing it yields 3 real roots, M1 for testing a point in the 'outside' region, A1 for showing it yields 1 real root, A1 for correctly identifying the regions in terms of inequalities. Part C: (a) M1 for finding \(\partial F/\partial t = 2y - 6t^2 - 1\), A1 for \(y = 3t^2 + 1/2\), M1 for substituting into \(F = 0\), A1 for \(x = -4t^3\). (b) M1 for expressing \(t^2\) in terms of \(y\) and substituting into \(x^2 = 16(t^2)^3\), A1 for showing equivalence to \(27x^2 = 8(2y-1)^3\).

Part A: (a) The roots of \(z^n - 1 = 0\) are the \(n\)-th roots of unity, given by \(z_k = e^{i \frac{2k\pi}{n}}\) for \(k = 0, 1, \dots, n-1\). By the Factor Theorem, since the leading coefficient is 1, we have \(z^n - 1 = \prod_{k=0}^{n-1} (z - e^{i \frac{2k\pi}{n}})\). (b) Since \(z_0 = 1\), we can write \(z^n - 1 = (z - 1) \prod_{k=1}^{n-1} (z - e^{i \frac{2k\pi}{n}})\). For \(z \neq 1\), dividing by \(z - 1\) yields \(\prod_{k=1}^{n-1} (z - e^{i \frac{2k\pi}{n}}) = \frac{z^n - 1}{z - 1} = \sum_{j=0}^{n-1} z^j\). Since both sides are polynomial expressions, they are continuous, so we can substitute \(z = 1\) to get \(\prod_{k=1}^{n-1} (1 - e^{i \frac{2k\pi}{n}}) = \sum_{j=0}^{n-1} 1^j = n\). (c) The distance from \(A_0\) (represented by 1) to \(A_k\) (represented by \(z_k\)) is \(A_0 A_k = |1 - z_k|\). Thus, the product of the chord lengths is \(\prod_{k=1}^{n-1} A_0 A_k = \prod_{k=1}^{n-1} |1 - e^{i \frac{2k\pi}{n}}| = |\prod_{k=1}^{n-1} (1 - e^{i \frac{2k\pi}{n}})| = |n| = n\). Part B: (a) The product is \(\prod_{k=0}^{n-1} |w - z_k| = |\prod_{k=0}^{n-1} (w - z_k)| = |w^n - 1|\). Substituting \(w = e^{i\theta}\): \(|w^n - 1| = |e^{in\theta} - 1| = |e^{in\theta/2}(e^{in\theta/2} - e^{-in\theta/2})| = |e^{in\theta/2}| \cdot |2i \sin(n\theta/2)| = 2|\sin(n\theta/2)|\). (b) Since the maximum value of \(|\sin(n\theta/2)|\) is 1, the maximum value of the product is 2. This occurs when \(n\theta/2 = \pi/2 + m\pi \implies \theta = \pi/n + 2m\pi/n\), which geometrically corresponds to the midpoints of the arcs between adjacent vertices. Part C: (a) The product of squared distances is \(|w^n - 1|^2\). Since \(w = d e^{i\theta}\), \(w^n = d^n e^{in\theta}\). Thus, \(|w^n - 1|^2 = |d^n e^{in\theta} - 1|^2 = (d^n \cos(n\theta) - 1)^2 + (d^n \sin(n\theta))^2 = d^{2n}\cos^2(n\theta) - 2d^n\cos(n\theta) + 1 + d^{2n}\sin^2(n\theta) = d^{2n} - 2d^n\cos(n\theta) + 1\). (b) Let \(D\) be the product of distances. Then \(D^2 = d^{2n} - 2d^n\cos(n\theta) + 1\). The maximum of \(D^2\) occurs when \(\cos(n\theta) = -1\), giving \(D^2 = d^{2n} + 2d^n + 1 = (d^n + 1)^2 \implies D = d^n + 1\). The minimum of \(D^2\) occurs when \(\cos(n\theta) = 1\), giving \(D^2 = d^{2n} - 2d^n + 1 = (d^n - 1)^2 \implies D = |d^n - 1|\). (c) For \(n = 6\) and \(d = 2\), the maximum product of distances is \(d^n + 1 = 2^6 + 1 = 64 + 1 = 65\). The minimum product of distances is \(|d^n - 1| = |2^6 - 1| = 63\).

Part A: (a) M1 for recognizing that \(z_k\) are roots of \(z^n-1=0\), A1 for factoring correctly using the Factor Theorem. (b) M1 for isolating the \((z-1)\) term, M1 for dividing by \((z-1)\) to get the geometric series sum, M1 for substituting \(z=1\), A1 for concluding the result is \(n\). (c) M1 for expressing the chord lengths as \(|1 - z_k|\), M1 for expressing the product of chord lengths as the modulus of the product from (b), A1 for obtaining \(n\). Part B: (a) M1 for expressing the product of distances as \(|w^n - 1|\), M1 for substituting \(w = e^{i\theta}\), M1 for factoring out \(e^{in\theta/2}\) or using Euler's formula, A1 for showing the final trigonometric form. (b) A1 for maximum value of 2, M1 for solving \(\sin(n\theta/2) = \pm 1\), A1 for identifying that this corresponds to the midpoints of the arcs between adjacent vertices. Part C: (a) M1 for expressing the product as \(|d^n e^{in\theta} - 1|^2\), M1 for writing this in Cartesian form as \((d^n\cos(n\theta) - 1)^2 + (d^n\sin(n\theta))^2\), A1 for expanding, M1 for using \(\cos^2(n\theta) + \sin^2(n\theta) = 1\), A1 for obtaining the result. (b) M1 for finding the maximum of the quadratic-like form when \(\cos(n\theta) = -1\), A1 for showing maximum is \(d^n + 1\), A1 for showing minimum is \(|d^n - 1|\) when \(\cos(n\theta) = 1\). (c) M1 for identifying \(n=6\) and \(d=2\), A1 for \(2^6 = 64\), A1 for maximum value 65, A1 for minimum value 63.