2025 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

We are given (u_1 = \_ln(3)) and (u_3 = \_ln(12)). Since this is an arithmetic sequence, we have (u_3 = (u_1 + 2d)). Substituting the given values: (\_ln(12) = \_ln(3) + 2d), which simplifies to (2d = \_ln(12) - \_ln(3) = \_ln(4)). This gives (2d = 2\_ln(2)), so (d = \_ln(2)). Now, we find the sum of the first 3 terms, (S_3): (S_3 = \_frac{3}{2}(u_1 + u_3) = \_frac{3}{2}(\_ln(3) + \_ln(12)) = \_frac{3}{2}\_ln(36)). Since (\_ln(36) = 2\_ln(6)), we get (S_3 = 3\_ln(6) = \_ln(216)). Alternatively, summing the terms directly: (u_1 = \_ln(3)), (u_2 = \_ln(6)), (u_3 = \_ln(12)), and (S_3 = \_ln(3) + \_ln(6) + \_ln(12) = \_ln(216)).

M1: Attempt to write an expression for (u_3) in terms of (u_1) and (d). A1: Correctly finding (d = \_ln(2)). M1: Attempt to use the sum formula or sum the terms directly. A1: Correct substitution. M1: Correct use of logarithm properties for addition. A1: Simplifying to (3\_ln(6)). A1: Correct final answer (\_ln(216)).

(a) The vertical asymptote is (x = 3) and the horizontal asymptote is (y = 2). (b) Since the graph passes through ((4, 11)), we have (11 = \_frac{2(4)+a}{4-3}) which simplifies to (11 = 8 + a), so (a = 3). (c) Let (y = \_frac{2x+3}{x-3}). Rearranging to make (x) the subject: (y(x-3) = 2x+3 \_Rightarrow xy - 3y = 2x+3 \_Rightarrow x(y-2) = 3y+3 \_Rightarrow x = \_frac{3y+3}{y-2}). Thus, (f^{-1}(x) = \_frac{3x+3}{x-2}).

(a) A1: Correct vertical asymptote (x = 3). A1: Correct horizontal asymptote (y = 2). (b) M1: Substituting ((4, 11)) into the function. A1: Showing (a = 3). (c) M1: Swapping (x) and (y) or rearranging. M1: Isolating (y). A1: Correct expression for (f^{-1}(x) = \_frac{3x+3}{x-2}).

Divide the entire equation by 2: (\_frac{\_sqrt{3}}{2} \_sin(2x) + \_frac{1}{2} \_cos(2x) = \_frac{1}{2}). Using the compound angle identity (\_sin(A+B) = \_sin A \_cos B + \_cos A \_sin B), we rewrite this as (\_sin(2x + \_frac{\_pi}{6}) = \_frac{1}{2}). Since (0 \_leq x \_leq \_pi), we have (\_frac{\_pi}{6} \_leq 2x + \_frac{\_pi}{6} \_leq \_frac{13\_pi}{6}). In this interval, (\_sin(\_theta) = \_frac{1}{2}) has solutions (\_theta = \_frac{\_pi}{6}, \_frac{5\_pi}{6}, \_frac{13\_pi}{6}). Thus, we solve: 1) (2x + \_frac{\_pi}{6} = \_frac{\_pi}{6} \_Rightarrow x = 0); 2) (2x + \_frac{\_pi}{6} = \_frac{5\_pi}{6} \_Rightarrow x = \_frac{\_pi}{3}); 3) (2x + \_frac{\_pi}{6} = \_frac{13\_pi}{6} \_Rightarrow x = \_pi). The solutions are (x = 0, \_frac{\_pi}{3}, \_pi).

M1: Dividing by 2 to prepare for compound angle identity. A1: Correctly rewriting as (\_sin(2x + \_frac{\_pi}{6}) = \_frac{1}{2}). M1: Determining the correct interval for the argument. A1: Finding one correct value of the angle. A1: Finding the other correct values of the angle. A1: Finding solutions (x = 0) and (x = \_pi). A1: Finding solution (x = \_frac{\_pi}{3}).

(a) The particle is at rest when (v(t) = 0), so (3(t^2 - 4t + 3) = 0 \_Rightarrow 3(t-1)(t-3) = 0), which gives (t = 1) or (t = 3). (b) The total distance is (\_int_{0}^{3} |v(t)| dt = \_int_{0}^{1} v(t) dt - \_int_{1}^{3} v(t) dt). Let the position function be (s(t) = t^3 - 6t^2 + 9t). We evaluate (s(t)) at the boundaries: (s(0) = 0), (s(1) = 4), (s(3) = 0). The distance from (t=0) to (t=1) is (|s(1) - s(0)| = 4). The distance from (t=1) to (t=3) is (|s(3) - s(1)| = 4). The total distance is (4 + 4 = 8) metres.

(a) M1: Setting (v(t) = 0). A1: Correct values (t = 1) and (t = 3). (b) M1: Recognizing that total distance requires splitting the interval. A1: Finding the correct antiderivative (s(t) = t^3 - 6t^2 + 9t). M1: Substituting boundaries (0, 1, 3). A1: Calculating (s(1) = 4) and (s(3) = 0). A1: Correct total distance of (8).

(a) Since the sum of probabilities is 1, (k + 2k + 3k + p = 1 \_Rightarrow p = 1 - 6k). The expectation is (\_text{E}(X) = 1(k) + 2(2k) + 3(3k) + 4(p) = 14k + 4p). Since (\_text{E}(X) = 3), we have (14k + 4(1-6k) = 3 \_Rightarrow -10k = -1 \_Rightarrow k = 0.1). Thus (p = 1 - 6(0.1) = 0.4). (b) To find the variance, first find (\_text{E}(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10). Then (\_text{Var}(X) = \_text{E}(X^2) - [\_text{E}(X)]^2 = 10 - 3^2 = 1).

(a) M1: Using sum of probabilities equals 1. M1: Finding expression for (\_text{E}(X)). A1: Equating to 3. M1: Solving the system of equations. A1: Correct values (k = 0.1, p = 0.4). (b) M1: Finding (\_text{E}(X^2)). A1: Correctly calculating (\_text{Var}(X) = 1).

Using substitution (u = e^x + x), we get (du = (e^x + 1) dx). Changing limits of integration: when (x = 0), (u = 1); when (x = 1), (u = e + 1). The integral becomes (\_int_{1}^{e+1} \_frac{1}{u} \_, du = [\_ln|u|]_{1}^{e+1} = \_ln(e+1) - \_ln(1) = \_ln(e+1)).

M1: Finding the differential (du = (e^x + 1)dx). A1: Correct substitution of integrand. M1: Changing limits of integration. A1: Writing definite integral in terms of (u). A1: Finding antiderivative (\_ln|u|). M1: Substituting limits. A1: Correct final answer (\_ln(e+1)).

(a) The general term in the binomial expansion is (\_binom{6}{r} (x^2)^{6-r} \_left(-\_frac{k}{x}\_right)^r = \_binom{6}{r} (-1)^r k^r x^{12-3r}). For the term to be independent of (x), we need (12 - 3r = 0 \_Rightarrow r = 4). The term is (\_binom{6}{4} (-1)^4 k^4 = 15k^4). (b) Setting this term equal to 240 gives (15k^4 = 240 \_Rightarrow k^4 = 16). Since (k \_gt 0), we have (k = 2).

(a) M1: Writing down the general term. A1: Simplifying powers of (x). M1: Solving for (r = 4). A1: Finding (15k^4). (b) M1: Equating term to 240. A1: Finding (k^4 = 16). A1: Solving for (k = 2).

(a) Using the chain rule, (f'(x) = \_frac{1}{x^2 - 3} \_cdot (2x) = \_frac{2x}{x^2 - 3}). (b) At (x = 2), the (y)-coordinate is (f(2) = \_ln(2^2 - 3) = \_ln(1) = 0). The gradient of the tangent is (f'(2) = \_frac{2(2)}{2^2 - 3} = 4). Using the tangent line equation, (y - 0 = 4(x - 2) \_Rightarrow y = 4x - 8).

(a) M1: Using chain rule for differentiation. A1: Correct derivative (f'(x) = \_frac{2x}{x^2 - 3}). (b) M1: Finding (y)-coordinate (y = 0). M1: Finding gradient (m = 4). M1: Setting up straight line equation. A1: Correct equation (y = 4x - 8).

(a) To find the y-intercept, set x = 0: f(0) = (0 - 0)e^0 = 0, so the y-intercept is (0,0). To find the x-intercepts, set f(x) = 0: (2x^2 - bx)e^{-x} = 0. Since e^{-x} > 0, we have x(2x - b) = 0, giving x = 0 or x = b/2. Thus, the points of intersection with the axes are (0,0) and (b/2, 0).

(b) Using the product rule with u = 2x^2 - bx and v = e^{-x}: u' = 4x - b and v' = -e^{-x}. Therefore, f'(x) = (4x - b)e^{-x} + (2x^2 - bx)(-e^{-x}) = (4x - b - 2x^2 + bx)e^{-x} = -(2x^2 - (b+4)x + b)e^{-x}.

(c) (i) A stationary point at x = 3 implies f'(3) = 0. Substituting x = 3 into f'(x) = 0: -(2(3)^2 - (b+4)(3) + b)e^{-3} = 0 => 18 - 3b - 12 + b = 0 => 6 - 2b = 0 => b = 3.

(ii) Substituting b = 3 into f'(x) gives f'(x) = -(2x^2 - 7x + 3)e^{-x}. Setting f'(x) = 0 gives 2x^2 - 7x + 3 = 0 => (2x - 1)(x - 3) = 0. Thus, the other stationary point is at x = 1/2 (or 0.5).

(iii) We can test the sign of f'(x) around the critical points x = 0.5 and x = 3. For x < 0.5, 2x - 1 < 0 and x - 3 < 0, so f'(x) < 0. For 0.5 < x < 3, 2x - 1 > 0 and x - 3 < 0, so f'(x) > 0. For x > 3, 2x - 1 > 0 and x - 3 > 0, so f'(x) < 0. Since f'(x) changes from negative to positive at x = 0.5, the point at x = 0.5 is a local minimum. Since f'(x) changes from positive to negative at x = 3, the point at x = 3 is a local maximum.

(d) To find the points of inflection, we differentiate f'(x) = -(2x^2 - 7x + 3)e^{-x} again using the product rule: f''(x) = -(4x - 7)e^{-x} - (-(2x^2 - 7x + 3))e^{-x} = (-4x + 7 + 2x^2 - 7x + 3)e^{-x} = (2x^2 - 11x + 10)e^{-x}. Setting f''(x) = 0 yields 2x^2 - 11x + 10 = 0.

(e) Since f(x) < 0 for 0 < x < 1.5, the area of R is given by Area = \int_{0}^{1.5} -f(x) dx = \int_{0}^{1.5} -(2x^2 - 3x)e^{-x} dx. Using integration by parts twice, the antiderivative of f(x) is -(2x^2 + x + 1)e^{-x}. Thus, the Area is [ (2x^2 + x + 1)e^{-x} ]_0^{1.5} = (2(1.5)^2 + 1.5 + 1)e^{-1.5} - (2(0)^2 + 0 + 1)e^0 = (4.5 + 1.5 + 1)e^{-1.5} - 1 = 7e^{-1.5} - 1.

(a) M1 for attempting to solve f(x) = 0, A1 for x = 0 and x = b/2, A1 for (0,0) and (b/2, 0) with y-intercept included.
(b) M1 for correct product rule application, A1 for correct derivatives u' and v', A1 for algebraic simplification leading to the given expression.
(c)(i) M1 for setting f'(3) = 0, A1 for showing b = 3.
(c)(ii) M1 for setting the quadratic to 0 with b = 3, A1 for x = 0.5.
(c)(iii) M1 for checking signs of f'(x) in intervals or using the second derivative test, A1 for identifying x = 0.5 as a local minimum, A1 for identifying x = 3 as a local maximum.
(d) M1 for product rule on f'(x), A1 for f''(x) = (2x^2 - 11x + 10)e^{-x}, A1 for setting f''(x) = 0 to obtain the given quadratic.
(e) M1 for expressing the area as the negative integral or using absolute value, A1 for the antiderivative (2x^2 + x + 1)e^{-x}, A1 for the final exact value 7e^{-1.5} - 1.

(a) (i) The vertical asymptote is found where the denominator is zero: x = 1. The horizontal asymptote is found by taking the limit as x approaches infinity: y = \lim_{x \to \pm\infty} \frac{3x - 2}{x - 1} = 3.

(ii) To find the inverse, set y = \frac{3x - 2}{x - 1} and solve for x: y(x - 1) = 3x - 2 => yx - y = 3x - 2 => x(y - 3) = y - 2 => x = \frac{y - 2}{y - 3}. Thus, g^{-1}(x) = \frac{x - 2}{x - 3}. The domain of g^{-1} is the range of g, which is x \in \mathbb{R}, x
eq 3.

(b) (i) For f(x) to be defined, the argument of the logarithm must be strictly positive: \frac{3x - 2}{x - 1} > 0. The critical values are x = 2/3 and x = 1. Testing the intervals: for x < 2/3, the expression is positive; for 2/3 < x < 1, the expression is negative; for x > 1, the expression is positive. Thus, the domain is x < 2/3 or x > 1 (which can be written as (-\infty, 2/3) \cup (1, \infty)).

(ii) To solve f(x) = 0: \ln\left(\frac{3x - 2}{x - 1}\right) = 0 => \frac{3x - 2}{x - 1} = 1 => 3x - 2 = x - 1 => 2x = 1 => x = 1/2. Since 1/2 < 2/3, this solution lies within the domain.

(c) (i) Set g(x) = x => \frac{3x - 2}{x - 1} = x => 3x - 2 = x^2 - x => x^2 - 4x + 2 = 0. Using the quadratic formula: x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}.

(ii) Express g(x) in partial fractions / quotient form: g(x) = \frac{3(x-1) + 1}{x - 1} = 3 + \frac{1}{x-1}. This is of the form g(x) = \frac{1}{x - d} + c. Therefore, the sequence of transformations is: a translation of 1 unit to the right (along the x-axis) followed by a translation of 3 units upwards (along the y-axis).

(a)(i) A1 for x = 1, A1 for y = 3.
(a)(ii) M1 for interchanging x and y or attempting to make x the subject, A1 for correct intermediate step, A1 for g^{-1}(x) = \frac{x-2}{x-3}, A1 for stating domain x
eq 3.
(b)(i) M1 for setting the inequality \frac{3x-2}{x-1} > 0, A1 for finding critical values 2/3 and 1, A1 for testing intervals, A1 for correct final intervals x < 2/3 or x > 1.
(b)(ii) M1 for setting the fraction to 1, A1 for x = 1/2.
(c)(i) M1 for setting g(x) = x, A1 for forming the quadratic x^2 - 4x + 2 = 0, A1 for exact roots x = 2 \pm \sqrt{2}.
(c)(ii) M1 for rewriting g(x) as 3 + \frac{1}{x-1}, A1 for horizontal translation of 1 unit right, A1 for vertical translation of 3 units up.

(a) (i) The prime numbers in the support of X are 2 and 3. Therefore, P(X is prime) = P(X=2) + P(X=3) = b + b = 2b. We are given that P(X is prime) = 0.6, so 2b = 0.6 => b = 0.3.

(ii) We are given P(X \leq 2) = 0.6, which means P(X=1) + P(X=2) = a + b = 0.6. Substituting b = 0.3 gives a + 0.3 = 0.6 => a = 0.3. Since the sum of all probabilities must equal 1, we have a + 2b + c = 1 => 0.3 + 2(0.3) + c = 1 => 0.3 + 0.6 + c = 1 => c = 0.1.

(b) (i) E(X) = \sum x \cdot P(X=x) = 1(a) + 2(b) + 3(b) + 4(c) = 1(0.3) + 2(0.3) + 3(0.3) + 4(0.1) = 0.3 + 0.6 + 0.9 + 0.4 = 2.2.

(ii) First, compute E(X^2) = \sum x^2 \cdot P(X=x) = 1^2(0.3) + 2^2(0.3) + 3^2(0.3) + 4^2(0.1) = 1(0.3) + 4(0.3) + 9(0.3) + 16(0.1) = 0.3 + 1.2 + 2.7 + 1.6 = 5.8. Then, Var(X) = E(X^2) - (E(X))^2 = 5.8 - (2.2)^2 = 5.8 - 4.84 = 0.96.

(c) (i) The pairs (X_1, X_2) that sum to 5 are (1,4), (2,3), (3,2), and (4,1). Since the observations are independent, the probabilities are: P(1,4) = 0.3 \times 0.1 = 0.03; P(2,3) = 0.3 \times 0.3 = 0.09; P(3,2) = 0.3 \times 0.3 = 0.09; P(4,1) = 0.1 \times 0.3 = 0.03. Summing these probabilities gives P(X_1 + X_2 = 5) = 0.03 + 0.09 + 0.09 + 0.03 = 0.24.

(ii) We want to find the conditional probability P(at least one is 3 | X_1 + X_2 = 5). The favorable pairs that include at least one 3 and sum to 5 are (2,3) and (3,2). The sum of their probabilities is P(2,3) + P(3,2) = 0.09 + 0.09 = 0.18. Using the conditional probability formula: P = \frac{0.18}{0.24} = \frac{18}{24} = \frac{3}{4} = 0.75.

(a)(i) M1 for identifying prime outcomes 2 and 3, A1 for setting P(X=2) + P(X=3) = 2b, A1 for obtaining b = 0.3.
(a)(ii) M1 for setting a + b = 0.6, A1 for a = 0.3, A1 for using sum of probabilities = 1 to find c = 0.1.
(b)(i) M1 for attempting to evaluate E(X), A1 for E(X) = 2.2.
(b)(ii) M1 for attempting to evaluate E(X^2), A1 for E(X^2) = 5.8, M1 for using Var(X) = E(X^2) - [E(X)]^2, A1 for showing Var(X) = 0.96.
(c)(i) M1 for listing the four cases that sum to 5, A1 for calculating individual probabilities, A1 for final probability 0.24.
(c)(ii) M1 for identifying the relevant outcomes (2,3) and (3,2), A1 for their sum 0.18, A1 for dividing by 0.24 to get 0.75 (or 3/4).

(a) Let \(X\) be the mass of an apple. We are given \(X \sim N(150, \sigma^2)\) and \(P(X > 180) = 0.15\).
Standardizing this gives \(P\left(Z > \frac{180 - 150}{\sigma}\right) = 0.15\), which means \(P\left(Z < \frac{30}{\sigma}\right) = 0.85\).
Using the inverse normal function on a GDC, we find the \(z\)-score:
\(\frac{30}{\sigma} \approx 1.036433\)
\(\sigma = \frac{30}{1.036433} \approx 28.945 \approx 28.9\text{ g}\).

(b) We want to find \(P(140 < X < 160)\).
Using the normal cumulative distribution function on a GDC with \(\mu = 150\) and \(\sigma = 28.945\):
\(P(140 < X < 160) \approx 0.2699\).
To three significant figures, the probability is \(0.270\).

(a)
M1 for attempting to standardize or write a probability equation, e.g., \(P(Z < \frac{30}{\sigma}) = 0.85\).
A1 for finding the correct \(z\)-score of \(1.036\).
A1 for \(\sigma = 28.9\text{ g}\) (accept 28.945).

(b)
M1 for setting up the correct probability interval: \(P(140 < X < 160)\).
A1 for entering correct parameters into the GDC normalcdf function.
A1 for \(0.270\) (accept 0.27).

(a) To find the points of intersection, we set \(f(x) = g(x)\):
\(\ln(x^2 + 1) = \cos(x)\)
Using a GDC to find the roots of \(\ln(x^2 + 1) - \cos(x) = 0\) in the interval \([-2, 2]\), we obtain:
\(x \approx -0.916174\) and \(x \approx 0.916174\).
To three significant figures, the coordinates are \(x = -0.916\) and \(x = 0.916\).

(b) The area \(A\) of the enclosed region is given by the integral of the upper function minus the lower function between the intersection points:
\(A = \int_{-0.916174}^{0.916174} (\cos(x) - \ln(x^2 + 1)) \, dx\)
Evaluating this integral numerically using a GDC:
\(A \approx 1.0731\).
To three significant figures, the area is \(1.07\).

(a)
M1 for setting up the equation \(f(x) = g(x)\).
A1 for both correct \(x\)-coordinates: \(x \approx -0.916\) and \(x \approx 0.916\) (accept \(\pm 0.916\)).

(b)
M1 for writing a correct integral expression for the area.
M1 for using correct limits (their values from part a).
M1 for correct integrand \(g(x) - f(x)\) (order must be correct for positive area).
A1 for \(1.07\) (accept 1.073).

(a) Using the Sine Rule:
\(\frac{\sin(B\hat{C}A)}{AB} = \frac{\sin(B\hat{A}C)}{BC}\)
\(\frac{\sin(B\hat{C}A)}{7.2} = \frac{\sin(42^\circ)}{5.4}\)
\(\sin(B\hat{C}A) = \frac{7.2 \sin(42^\circ)}{5.4} \approx 0.89218\)

Finding the acute angle case:
\(B\hat{C}A \approx \arcsin(0.89218) \approx 63.14^\circ \approx 63.1^\circ\)

Finding the obtuse angle case:
\(B\hat{C}A = 180^\circ - 63.14^\circ \approx 116.86^\circ \approx 117^\circ\).

(b) If \(B\hat{C}A\) is obtuse, \(B\hat{C}A \approx 116.86^\circ\).
The third angle, \(A\hat{B}C\), is:
\(A\hat{B}C = 180^\circ - 42^\circ - 116.86^\circ = 21.14^\circ\).

The area of the triangle is:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(A\hat{B}C)\)
\(\text{Area} = \frac{1}{2} \times 7.2 \times 5.4 \times \sin(21.14^\circ) \approx 7.011\text{ cm}^2\).
To three significant figures, the area is \(7.01\text{ cm}^2\).

(a)
M1 for substituting correct values into the Sine Rule formula.
A1 for finding the acute angle \(63.1^\circ\) (accept 63.14).
A1 for finding the obtuse angle \(117^\circ\) (accept 116.86).

(b)
M1 for calculating the correct angle \(A\hat{B}C = 21.14^\circ\).
M1 for substituting their values into the area formula \(\frac{1}{2}ac\sin(B)\).
A1 for \(7.01\text{ cm}^2\) (accept 7.011).

(a) Using the geometric series sum formula:
\(S_{10} = \frac{u_1(1 - r^{10})}{1 - r} = \frac{80(1 - 0.85^{10})}{1 - 0.85} \approx 428.33\).
To three significant figures, \(S_{10} \approx 428\).

(b) First find the sum to infinity:
\(S_\infty = \frac{u_1}{1 - r} = \frac{80}{1 - 0.85} = \frac{80}{0.15} \approx 533.33\).

We want to find the smallest integer \(k\) such that:
\(S_k > 0.95 \times S_\infty\)
\(\frac{80(1 - 0.85^k)}{0.15} > 0.95 \times \frac{80}{0.15}\)

Dividing both sides by \(\frac{80}{0.15}\):
\(1 - 0.85^k > 0.95\)
\(0.85^k < 0.05\)

Taking natural logarithms on both sides:
\(k \ln(0.85) < \ln(0.05)\)
Since \(\ln(0.85) < 0\), we reverse the inequality:
\(k > \frac{\ln(0.05)}{\ln(0.85)} \approx 18.43\).

Since \(k\) must be an integer, the least value is \(k = 19\).

(a)
M1 for correct substitution into the sum formula.
A1 for \(428\) (accept 428.33).

(b)
M1 for finding the sum to infinity \(S_\infty = 533.33\).
M1 for setting up the inequality \(1 - 0.85^k > 0.95\) (or equivalent with sums).
M1 for solving the inequality (accepting equations using logs or GDC table/graph tool).
A1 for \(k = 19\).

(a) The particle is at instantaneous rest when \(v(t) = 0\):
\(4\sin(t) - t = 0\)
Using a GDC to find the positive root in the interval \([0, 3]\):
\(t \approx 2.47458\text{ s}\).
To three significant figures, \(t \approx 2.47\text{ s}\).

(b) The total distance traveled is given by the integral of the speed:
\(\text{Distance} = \int_0^3 |v(t)| \, dt = \int_0^3 |4\sin(t) - t| \, dt\)

Since the particle changes direction at \(t \approx 2.47458\), we split the integral:
\(\text{Distance} = \int_0^{2.47458} (4\sin(t) - t) \, dt + \int_{2.47458}^3 (t - 4\sin(t)) \, dt\)
Evaluating numerically using GDC:
\(\text{Distance} \approx 4.0794 + 0.6194 = 4.6988\text{ m}\).
To three significant figures, the distance is \(4.70\text{ m}\).

(a)
M1 for setting \(v(t) = 0\).
A1 for \(t \approx 2.47\text{ s}\) (accept 2.475).

(b)
M1 for recognizing total distance is \(\int_0^3 |v(t)| \, dt\).
M1 for correct split of the integral based on their root from part (a) or using the absolute value function on GDC.
M1 for writing down correct numerical values for both parts of the distance (e.g., \(4.08\) and \(0.619\)).
A1 for \(4.70\text{ m}\) (accept 4.7).

(a) To find the point of intersection, we set \(f(x) = g(x)\):
\(e^{0.5x} - 2 = \frac{4}{x - 1} + 1\)
Using a GDC to find the point of intersection of the two graphs:
\(x \approx 3.159\)
\(y = f(3.159) = e^{0.5(3.159)} - 2 \approx 2.853\)
To three significant figures, the coordinates are \((3.16, 2.85)\).

(b) To solve \(f(x) > g(x)\), we look at where the graph of \(f\) lies above the graph of \(g\).
Since \(f\) is strictly increasing and \(g\) is strictly decreasing for \(x > 1\), the graph of \(f\) lies above \(g\) for all values of \(x\) to the right of the intersection point.
Therefore, the solution is \(x > 3.159\).
To three significant figures, \(x > 3.16\).

(a)
M1 for attempting to solve \(f(x) = g(x)\) (e.g., sketching graphs or setting up equation).
A1 for \(x \approx 3.16\).
A1 for \(y \approx 2.85\).

(b)
M1 for identifying that the inequality depends on the intersection point.
M1 for realizing \(f(x)\) is above \(g(x)\) to the right of the point.
A1 for \(x > 3.16\) (accept \(x > 3.159\)).

(a) Using the product rule with \(u = x^2\) and \(v = \ln(x)\):
\(u' = 2x\), \(v' = \frac{1}{x}\)
\(\frac{dy}{dx} = u'v + uv' = 2x\ln(x) + x^2 \left(\frac{1}{x}\right) = 2x\ln(x) + x\).

(b) At \(x = e\):
The \(y\)-coordinate is:
\(y = e^2 \ln(e) = e^2 \approx 7.389\).

The gradient of the tangent is:
\(m = \left.\frac{dy}{dx}\right|_{x=e} = 2e\ln(e) + e = 3e \approx 8.1548\).

Using the point-slope formula \(y - y_1 = m(x - x_1)\):
\(y - e^2 = 3e(x - e)\)
\(y = 3ex - 3e^2 + e^2\)
\(y = 3ex - 2e^2\).

Converting to three significant figures:
\(m = 3e \approx 8.15\)
\(c = -2e^2 \approx -14.8\)

Thus, the equation is \(y = 8.15x - 14.8\).

(a)
M1 for attempting product rule (at least one term correct).
A1 for \(2x\ln(x) + x\).

(b)
M1 for finding the gradient of the tangent at \(x = e\) (accept \(3e\) or \(8.15\)).
M1 for finding the \(y\)-coordinate (accept \(e^2\) or \(7.39\)).
M1 for attempting to find the equation of a line using their point and gradient.
A1 for \(y = 8.15x - 14.8\).

(a) The volume of a cylinder is \(V =
\pi r^2 h = 500\).
Thus, the height is \(h = \frac{500}{\pi r^2}\).

The total surface area of a closed cylinder is:
\(A = 2\pi r^2 + 2\pi r h\)

Substituting \(h\) into the surface area formula:
\(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right)\)
\(A = 2\pi r^2 + \frac{1000}{r}\). [Show]

(b) To minimize \(A(r)\), we find the derivative \(A'(r)\):
\(A'(r) = 4\pi r - \frac{1000}{r^2}\)
Setting \(A'(r) = 0\):
\(4\pi r = \frac{1000}{r^2}\)
\(r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}\)
\(r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30127\text{ cm}\).

Substituting this radius back into the surface area equation:
\(A(4.30127) = 2\pi (4.30127)^2 + \frac{1000}{4.30127} \approx 348.73\text{ cm}^2\).

To three significant figures, the radius is \(4.30\text{ cm}\) and the minimum surface area is \(349\text{ cm}^2\).

(a)
M1 for writing \(h\) in terms of \(r\) using the volume formula.
A1 for algebraic substitution and simplification leading to the shown result.

(b)
M1 for differentiating to get \(A'(r) = 4\pi r - \frac{1000}{r^2}\) (or using GDC solver/graph minimum feature).
M1 for setting \(A'(r) = 0\) (or equivalent).
A1 for \(r \approx 4.30\text{ cm}\).
A1 for \(\text{Minimum Area} \approx 349\text{ cm}^2\).

Part (a): Let \( W \sim N(150, \sigma^2) \). We are given that \( P(W > 175) = 0.15 \), which implies \( P(W \le 175) = 0.85 \). Standardizing the variable gives \( P(Z \le \frac{175 - 150}{\sigma}) = 0.85 \). Using a GDC to find the inverse normal of \( 0.85 \), we get \( \frac{25}{\sigma} \approx 1.03643 \). Solving for \( \sigma \), we find \( \sigma \approx \frac{25}{1.03643} \approx 24.1202 \). Thus, \( \sigma \approx 24.1\text{ g} \) to 3 significant figures. Part (b): Let \( X \) be the number of apples in the sample of 8 that weigh more than \( 175\text{ g} \). Since each apple has a probability of \( 0.15 \) of weighing more than \( 175\text{ g} \), \( X \) follows a binomial distribution \( X \sim B(8, 0.15) \). We need to find \( P(X \ge 2) \), which is equal to \( 1 - P(X \le 1) \). Using a GDC for cumulative binomial probability, we find \( P(X \le 1) \approx 0.657181 \). Therefore, \( P(X \ge 2) \approx 1 - 0.657181 = 0.342819 \), which is \( 0.343 \) to 3 significant figures.

Part (a): \( P(W \le 175) = 0.85 \) or \( z = 1.03643... \) (M1). For standardizing the equation: \( \frac{175 - 150}{\sigma} = 1.03643... \) (A1). \( \sigma = 24.1 \) (accept \( 24.1202... \)) (A1). Part (b): Recognizing binomial distribution, \( X \sim B(8, 0.15) \) (M1). For identifying the correct probability expression: \( P(X \ge 2) = 1 - P(X \le 1) \) (M1). \( = 0.343 \) (accept \( 0.342819... \)) (A1).

(a) Let \(X \sim N(\mu, \sigma^2)\). We are given \(P(X < 120) = 0.08\) and \(P(X > 180) = 0.15\). Using a standard normal distribution table or GDC, the z-scores corresponding to these probabilities are: \(z_{0.08} \approx -1.40507\) and \(z_{0.85} \approx 1.03643\). This gives two simultaneous equations: \(\frac{120 - \mu}{\sigma} = -1.40507\) and \(\frac{180 - \mu}{\sigma} = 1.03643\). Rearranging gives: \(\mu - 1.40507\sigma = 120\) and \(\mu + 1.03643\sigma = 180\). Subtracting the first equation from the second gives \(2.4415\sigma = 60\), hence \(\sigma \approx 24.575\) (or \(24.6\) g to 3 s.f.). Substituting back gives \(\mu = 180 - 1.03643(24.575) \approx 154.53\) (or \(155\) g to 3 s.f.). (b) To find \(P(135 < X < 165)\) with \(\mu = 154.53\) and \(\sigma = 24.575\), we use the normal cumulative distribution function on a GDC: \(P(135 < X < 165) \approx 0.45157\), which is \(0.452\) to 3 s.f. (c) Let \(Y\) be the number of Premium apples in a sample of 10. Then \(Y \sim B(10, 0.45157)\). We want to find \(P(Y \ge 4) = 1 - P(Y \le 3)\). Using a binomial cumulative distribution on a GDC: \(P(Y \le 3) \approx 0.25950\). Thus, \(P(Y \ge 4) \approx 1 - 0.25950 = 0.7405\), which is \(0.741\) to 3 s.f. (d) Let \(F\) be the number of flawless apples in a bag of 6 Premium apples. Then \(F \sim B(6, 0.9)\). The probability that all 6 apples in a bag are flawless is \(P(F = 6) = 0.9^6 \approx 0.531441\). The probability that a bag is not completely flawless is \(1 - 0.531441 = 0.468559\). The expected selling price per bag is \(E[S] = 7.50 \times 0.531441 + 5.00 \times 0.468559 = 3.9858 + 2.3428 = 6.3286\). Since the cost to produce each bag is $2.00, the expected profit per bag is \(6.3286 - 2.00 = 4.3286\). For 100 bags, the expected profit is \(100 \times 4.3286 = \$432.86\), which is \(\$433\) to 3 s.f.

(a) M1 for attempting to find z-scores. A1 for \(z_1 = -1.405\) and A1 for \(z_2 = 1.036\). M1 for setting up the two linear equations. A1 for finding \(\sigma = 24.6\). A1 for finding \(\mu = 155\). (b) M1 for setting up standard normal range. A2 for \(0.452\). (c) M1 for identifying Binomial distribution model. A1 for correct parameters \(n=10, p=0.452\). A1 for \(0.741\). (d) M1 for setting up flawless binomial probability. A1 for \(P(F=6) = 0.531\). M1 for identifying alternative outcome probability. M1 for expected revenue calculation. A1 for \(E[S] = 6.33\). M1 for subtracting cost. A1 for final answer \(\$433\) (or \(\$432.86\)).

(a) The vertical translation \(D\) is the midpoint of the range: \(D = \frac{26+14}{2} = 20\). The amplitude \(A\) is half the range: \(A = \frac{26-14}{2} = 6\). The period of the temperature cycle is 24 hours, so \(B = \frac{2\pi}{24} = \frac{\pi}{12}\). Since the maximum occurs at \(t = 15\), and the peak of cosine occurs when the argument is 0, we have \(C = 15\). Hence \(T(t) = 6\cos\left(\frac{\pi}{12}(t-15)\right) + 20\). (b) At 08:00, \(t = 8\). Substituting into the function: \(T(8) = 6\cos\left(\frac{\pi}{12}(8-15)\right) + 20 = 6\cos\left(-\frac{7\pi}{12}\right) + 20 \approx 18.447\). So the temperature is \(18.4^\circ\text{C}\). (c) The cooling system is active when \(T(t) \ge 23 \implies 6\cos\left(\frac{\pi}{12}(t-15)\right) + 20 \ge 23 \implies \cos\left(\frac{\pi}{12}(t-15)\right) \ge 0.5\). This occurs when \(-\frac{\pi}{3} \le \frac{\pi}{12}(t-15) \le \frac{\pi}{3} \implies -4 \le t - 15 \le 4 \implies 11 \le t \le 19\). The total active hours is \(19 - 11 = 8\) hours. (d) The active interval is \([11, 19]\). The total energy consumed is the integral of \(P(t)\) over this interval: \(E = \int_{11}^{19} 0.5\left(6\cos\left(\frac{\pi}{12}(t-15)\right) - 3\right)^2 dt\). Evaluating this integral on a GDC yields \(E \approx 18.683\) kWh (or \(18.7\) kWh to 3 s.f.). (e) \(\frac{S(24)}{24} = \frac{1}{24}\int_0^{24} T(t) dt\). Evaluating this integral: \(\int_0^{24} \left(6\cos\left(\frac{\pi}{12}(t-15)\right) + 20\right) dt = [20t + \frac{72}{\pi}\sin(\frac{\pi}{12}(t-15))]_0^{24} = 480\) because the sine terms cancel over a full period. Thus, \(\frac{S(24)}{24} = \frac{480}{24} = 20^\circ\text{C}\). This represents the average temperature of the server room over the 24-hour period.

(a) M1 for calculating midpoint for \(D\). A1 for showing \(A=6\). M1 for calculating period for \(B\). A1 for phase shift argument. A1 for complete verification. (b) M1 for substituting \(t=8\) into formula. A1 for \(18.4^\circ\text{C}\). (c) M1 for setting up inequality \(T(t) \ge 23\). A1 for finding boundaries 11 and 19. A2 for 8 hours. (d) M1 for establishing integral boundaries. M1 for integrating \(P(t)\). A3 for correct GDC evaluation to \(18.7\). (e) M1 for identifying definition of average value of a function. A1 for finding the value of 20. A1 for correct interpretation.

(a) Direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}\). Since \(\mathbf{d}_1 \neq k \mathbf{d}_2\) for any scalar \(k\), the lines are not parallel. To show they do not intersect, we set up simultaneous equations: \(2 + 3\lambda = 1 - \mu\) (1), \(-1 + \lambda = 4 + 2\mu\) (2), \(4 - 2\lambda = 1 + 2\mu\) (3). Solving (1) and (2) gives \(\lambda = \frac{3}{7}\) and \(\mu = -\frac{16}{7}\). Substituting these values into (3) yields: LHS: \(4 - 2(\frac{3}{7}) = \frac{22}{7}\), RHS: \(1 + 2(-\frac{16}{7}) = -\frac{25}{7}\). Since LHS \(\neq\) RHS, there is no intersection. Thus, the lines are skew. (b) Let \(P\) be on \(L_1\) and \(Q\) be on \(L_2\). \(\vec{PQ} = \mathbf{r}_2(\mu) - \mathbf{r}_1(\lambda) = \begin{pmatrix} -1 - 3\lambda - \mu \\ 5 - \lambda + 2\mu \\ -3 + 2\lambda + 2\mu \end{pmatrix}\). Since \(\vec{PQ}\) is perpendicular to both lines, we have \(\vec{PQ} \cdot \mathbf{d}_1 = 0 \implies 14\lambda + 5\mu = 8\) and \(\vec{PQ} \cdot \mathbf{d}_2 = 0 \implies 5\lambda + 9\mu = -5\). Solving these simultaneous equations yields \(\lambda \approx 0.9604\) and \(\mu \approx -1.0891\). Substituting these back into the line equations gives \(P(4.88, -0.0396, 2.08)\) and \(Q(2.09, 1.82, -1.18)\). (c) Let \(X\) be a point on \(L_1\) with parameter \(\lambda\). \(\vec{BX} = \begin{pmatrix} 3\lambda - 8 \\ \lambda - 9 \\ -2\lambda - 8 \end{pmatrix}\). For minimum distance, \(\vec{BX} \cdot \mathbf{d}_1 = 0 \implies 3(3\lambda - 8) + 1(\lambda - 9) - 2(-2\lambda - 8) = 0 \implies 14\lambda - 17 = 0 \implies \lambda = \frac{17}{14} \approx 1.214\). Substituting \(\lambda\) back into \(\vec{BX}\) gives \(\vec{BX} \approx \begin{pmatrix} -4.357 \\ -7.786 \\ -10.429 \end{pmatrix}\). The minimum distance is the magnitude \(|\vec{BX}| \approx \sqrt{(-4.357)^2 + (-7.786)^2 + (-10.429)^2} \approx 13.7\).

(a) M1 for comparing direction vectors. A1 for stating they are not parallel. M1 for setting up system of equations for intersection. A1 for finding solutions to a pair. A1 for showing inconsistency in third equation and concluding skew. (b) M1 for setting up vector \(\vec{PQ}\). M1 for dot products with direction vectors. A1 for system of equations. A1 for solving for parameters. A3 for finding coordinates to 3 s.f. (c) M1 for finding vector \(\vec{BX}\). M1 for finding parameter where dot product is 0. A1 for finding \(\lambda = 1.21\). A1 for vector coordinates. A2 for final magnitude of \(13.7\).

Part A
(a) Differentiating using the product rule:
\( f_n'(x) = n x^{n-1} e^{-x} - x^n e^{-x} = x^{n-1} e^{-x} (n - x) \).
Since \( x > 0 \) and \( e^{-x} > 0 \), \( f_n'(x) = 0 \implies x = n \).
For \( 0 < x < n \), \( f_n'(x) > 0 \), and for \( x > n \), \( f_n'(x) < 0 \). Thus, \( x = n \) is a unique local maximum.

(b) The \( y \)-coordinate is \( f_n(n) = n^n e^{-n} = \left(\frac{n}{e}\right)^n \).
Since \( \frac{n}{e} > 1 \) for all \( n \geq 3 \), the base of the exponent grows beyond 1, so \( \lim_{n \to \infty} \left(\frac{n}{e}\right)^n = \infty \).

(c) Finding the second derivative:
\( f_n''(x) = (n-1)x^{n-2}e^{-x}(n-x) - x^{n-1}e^{-x} \)
\( f_n''(x) = x^{n-2}e^{-x} [ (n-1)(n-x) - x ] = x^{n-2}e^{-x} [ x^2 - 2nx + n(n-1) ] \).
Setting \( f_n''(x) = 0 \) for \( x > 0 \) gives the quadratic equation:
\( x^2 - 2nx + n(n-1) = 0 \).
Using the quadratic formula:
\( x = \frac{2n \pm \sqrt{4n^2 - 4n(n-1)}}{2} = n \pm \sqrt{n} \).
Thus, the points of inflection have coordinates \( (n \pm \sqrt{n}, f_n(n \pm \sqrt{n})) \).

Part B
(d) \( I_0 = \int_0^\infty e^{-x} \, dx = [-e^{-x}]_0^\infty = 0 - (-1) = 1 \).

(e) Using integration by parts with \( u = x^n \implies du = n x^{n-1} dx \) and \( dv = e^{-x} dx \implies v = -e^{-x} \):
\( I_n = [-x^n e^{-x}]_0^\infty + n \int_0^\infty x^{n-1} e^{-x} \, dx \).
Since \( \lim_{x \to \infty} x^n e^{-x} = 0 \) and \( 0^n e^0 = 0 \), the boundary term vanishes.
Thus, \( I_n = n I_{n-1} \).

(f) Since \( I_0 = 1 \) and \( I_n = n I_{n-1} \), it follows that \( I_n = n! \).

(g) For \( n = 0 \):
LHS: \( J_0(t) = \int_0^t e^{-x} \, dx = 1 - e^{-t} \).
RHS: \( 0! \left( 1 - e^{-t} \frac{t^0}{0!} \right) = 1 - e^{-t} \). Thus, true for \( n = 0 \).
Assume the formula holds for \( n = k \): \( J_k(t) = k! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \right) \).
For \( n = k + 1 \):
\( J_{k+1}(t) = \int_0^t x^{k+1} e^{-x} \, dx \). Using integration by parts with \( u = x^{k+1} \) and \( dv = e^{-x} dx \):
\( J_{k+1}(t) = [-x^{k+1} e^{-x}]_0^t + (k+1) \int_0^t x^k e^{-x} \, dx = -t^{k+1} e^{-t} + (k+1) J_k(t) \).
Substituting the induction hypothesis:
\( J_{k+1}(t) = -t^{k+1} e^{-t} + (k+1) k! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \right) \)
\( = -t^{k+1} e^{-t} + (k+1)! - (k+1)! e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \)
\( = (k+1)! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} - e^{-t} \frac{t^{k+1}}{(k+1)!} \right) \)
\( = (k+1)! \left( 1 - e^{-t} \sum_{r=0}^{k+1} \frac{t^r}{r!} \right) \).
Thus, by the principle of mathematical induction, the statement is true for all \( n \in \mathbb{N} \).

Part C
(h) \( \text{P}(X_t \leq n) = \sum_{k=0}^n \frac{e^{-t} t^k}{k!} \).

(i) From Part B, \( J_n(t) = n! - n! e^{-t} \sum_{k=0}^n \frac{t^k}{k!} \).
Since \( I_n = n! \), we can rewrite this as:
\( J_n(t) = I_n - n! \text{P}(X_t \leq n) \).
Thus, \( n! \text{P}(X_t \leq n) = I_n - J_n(t) = \int_0^\infty x^n e^{-x} \, dx - \int_0^t x^n e^{-x} \, dx = \int_t^\infty x^n e^{-x} \, dx \).
Dividing by \( n! \) gives: \( \text{P}(X_t \leq n) = \frac{1}{n!} \int_t^\infty x^n e^{-x} \, dx \).

(j) For large \( n \), a Poisson distribution with parameter \( t = n \) has mean \( \mu = n \) and variance \( \sigma^2 = n \).
By the Central Limit Theorem, \( X_n \) can be approximated by a normal distribution \( Y \sim \text{N}(n, n) \).
Therefore, \( \text{P}(X_n \leq n) \approx \text{P}(Y \leq n) = \text{P}\left(Z \leq \frac{n-n}{\sqrt{n}}\right) = \text{P}(Z \leq 0) = 0.5 \).
Thus, \( \lim_{n \to \infty} \text{P}(X_n \leq n) = 0.5 \).

(k) Since \( \frac{1}{n!} \int_n^\infty x^n e^{-x} \, dx = \text{P}(X_n \leq n) \), we have:
\( \lim_{n \to \infty} \frac{1}{n!} \int_n^\infty x^n e^{-x} \, dx = \lim_{n \to \infty} \text{P}(X_n \leq n) = 0.5 \).

Part A
(a) [2 marks]
M1: Attempt to differentiate \( f_n(x) \) using the product rule.
A1: Correct derivative set to 0 and obtaining \( x = n \) with valid reasoning for local maximum.

(b) [2 marks]
A1: Correct maximum coordinate \( \left(\frac{n}{e}\right)^n \).
A1: Correct justification of limit tending to \( \infty \).

(c) [2 marks]
M1: Setting second derivative to 0.
A1: Correct coordinates \( n \pm \sqrt{n} \).

Part B
(d) [1 mark]
A1: Correct calculation of \( I_0 = 1 \).

(e) [4 marks]
M1: Correct choice of parts \( u = x^n \) and \( dv = e^{-x} dx \).
A1: Correct application of integration by parts formula.
R1: Clear explanation of why the boundary term vanishes as \( x \to \infty \).
A1: Correctly obtaining the recurrence relation \( I_n = n I_{n-1} \).

(f) [1 mark]
A1: Deducing \( I_n = n! \).

(g) [5 marks]
A1: Correctly showing the base case \( n = 0 \).
M1: Stating assumption for \( n = k \) and attempting \( n = k+1 \) using integration by parts.
A1: Correct expression for \( J_{k+1}(t) \) in terms of \( J_k(t) \).
M1: Correct algebraic expansion and regrouping with the summation.
R1: Clear concluding statement indicating induction is complete.

Part C
(h) [1 mark]
A1: Correct sum expression for Poisson cumulative probability.

(i) [3 marks]
M1: Attempting to link \( J_n(t) \) and \( I_n \).
M1: Factoring \( n! \) and identifying the cumulative Poisson sum.
A1: Clearly showing the final integration identity.

(j) [3 marks]
M1: Identifying mean and variance of the Poisson distribution.
M1: Standardizing to the standard normal distribution.
A1: Evaluating to the limit value of 0.5.

(k) [3 marks]
M1: Linking the integral limit to the probability limit.
A2: Stating the final limit value is 0.5.

Part A
(a) Expanding the right-hand side:
\( (z-1)(z^{n-1} + z^{n-2} + \dots + 1) = (z^n + z^{n-1} + \dots + z) - (z^{n-1} + z^{n-2} + \dots + 1) = z^n - 1 \).

(b) (i) The equation \( z^n - 1 = 0 \) has exactly \( n \) distinct complex roots, which are \( z_k \) for \( k = 0, 1, \dots, n-1 \). By the Factor Theorem, \( z^n - 1 \) can be written as \( a \prod_{k=0}^{n-1} (z - z_k) \). Since the coefficient of \( z^n \) on both sides is 1, \( a = 1 \). Hence, \( z^n - 1 = \prod_{k=0}^{n-1} (z - z_k) \).
(ii) Since \( z_0 = 1 \), we can write the product as:
\( z^n - 1 = (z - 1) \prod_{k=1}^{n-1} (z - z_k) \).
Using the result from (a), we have:
\( (z-1)\sum_{j=0}^{n-1} z^j = (z - 1) \prod_{k=1}^{n-1} (z - z_k) \).
Since \( z \neq 1 \), we can divide both sides by \( z-1 \) to get:
\( \prod_{k=1}^{n-1} (z - z_k) = \sum_{j=0}^{n-1} z^j \).

(c) Taking the limit of both sides as \( z \to 1 \):
LHS: \( \lim_{z \to 1} \prod_{k=1}^{n-1} (z - z_k) = \prod_{k=1}^{n-1} (1 - z_k) \).
RHS: \( \lim_{z \to 1} \sum_{j=0}^{n-1} z^j = 1 + 1 + \dots + 1 = n \) (since there are \( n \) terms in the sum).
Therefore, \( \prod_{k=1}^{n-1} (1 - z_k) = n \).

Part B
(d) In the complex plane, the distance between the points representing complex numbers \( z_a \) and \( z_b \) is given by the modulus \( |z_a - z_b| \). Since \( A_0 \) is represented by \( 1 \) and \( A_k \) is represented by \( z_k \), the distance is \( |1 - z_k| \).

(e) The product of the lengths of the chords is \( \prod_{k=1}^{n-1} |1 - z_k| = \left| \prod_{k=1}^{n-1} (1 - z_k) \right| \).
Using the result from (c), this is equal to \( |n| = n \).

(f) \( \prod_{k=0}^{n-1} |w - z_k|^2 = \left| \prod_{k=0}^{n-1} (w - z_k) \right|^2 \).
From Part B(i), this is \( |w^n - 1|^2 \).
Since \( w = e^{i\theta} \), we have \( w^n = e^{in\theta} \).
\( |e^{in\theta} - 1|^2 = (\cos(n\theta) - 1)^2 + \sin^2(n\theta) = \cos^2(n\theta) - 2\cos(n\theta) + 1 + \sin^2(n\theta) = 2 - 2\cos(n\theta) \).

(g) The product of the distances is \( \sqrt{2 - 2\cos(n\theta)} \).
The maximum value occurs when \( \cos(n\theta) = -1 \). This maximum value is \( \sqrt{2 - 2(-1)} = \sqrt{4} = 2 \).
Since \( 0 < \theta < \frac{2\pi}{n} \), this maximum occurs when \( n\theta = \pi \implies \theta = \frac{\pi}{n} \).

Part C
(h) \( |u - z_k|^2 = (u - z_k)\overline{(u - z_k)} = (u - z_k)(\bar{u} - \bar{z}_k) = u\bar{u} - u\bar{z}_k - \bar{u}z_k + z_k\bar{z}_k \).
Since \( z_k \) is on the unit circle, \( z_k\bar{z}_k = |z_k|^2 = 1 \).
Also, \( u\bar{u} = |u|^2 \) and \( u\bar{z}_k + \bar{u}z_k = u\bar{z}_k + \overline{u\bar{z}_k} = 2\text{Re}(u\bar{z}_k) \).
Thus, \( |u - z_k|^2 = |u|^2 - 2\text{Re}(u\bar{z}_k) + 1 \).

(i) Summing over all \( k \):
\( \sum_{k=0}^{n-1} |u - z_k|^2 = \sum_{k=0}^{n-1} (|u|^2 + 1) - 2\text{Re}\left( u \sum_{k=0}^{n-1} \bar{z}_k \right) \).
Since \( \sum_{k=0}^{n-1} z_k = 0 \), its complex conjugate is also zero, so \( \sum_{k=0}^{n-1} \bar{z}_k = 0 \).
Therefore, the second term vanishes.
The first term is a constant summed \( n \) times, yielding \( n(|u|^2 + 1) \).
This depends only on \( |u| \), the distance of \( Q \) from the origin.

(j) If \( Q \) lies on the circumcircle, then \( |u| = 1 \).
The sum of the squares of the distances is \( n(1^2 + 1) = 2n \).

Part A
(a) [1 mark]
A1: Correct algebraic expansion showing LHS equals RHS.

(b) [4 marks]
(i) M1: Applying the Factor Theorem with roots \( z_k \).
A1: Justifying why the leading coefficient is 1.
(ii) M1: Correctly separating the term with \( z_0 = 1 \).
A1: Dividing by \( z-1 \) under the condition \( z \neq 1 \).

(c) [3 marks]
M1: Stating the limit approach as \( z \to 1 \).
A1: Evaluating LHS limit.
A1: Evaluating RHS limit to show the result \( n \).

Part B
(d) [1 mark]
A1: Clear explanation relating geometric distance to the modulus of the difference of complex numbers.

(e) [2 marks]
M1: Recognizing that the chord product is the modulus of the polynomial product.
A1: Obtaining the value \( n \).

(f) [4 marks]
M1: Expressing the product of squared moduli as the modulus of the product squared.
A1: Recognizing this equals \( |w^n - 1|^2 \).
M1: Substituting \( w = e^{i\theta} \).
A1: Clear trigonometric expansion leading to \( 2 - 2\cos(n\theta) \).

(g) [3 marks]
M1: Taking the square root to get the distance product.
A1: Finding the maximum value of 2.
A1: Correctly solving for \( \theta = \frac{\pi}{n} \).

Part C
(h) [2 marks]
M1: Using \( |z|^2 = z\bar{z} \) to expand \( |u - z_k|^2 \).
A1: Correctly simplifying terms using \( |z_k|^2 = 1 \) and the real part identity.

(i) [5 marks]
M1: Setting up the summation over all \( k \).
A1: Splitting the sum and factoring out \( u \).
M1: Explaining why the conjugate sum is 0.
A1: Evaluating the constant term sum to \( n(|u|^2 + 1) \).
R1: Concluding clearly that the sum depends only on the distance from the origin \( |u| \).

(j) [3 marks]
M1: Stating that \( |u| = 1 \) on the circumcircle.
A2: Correctly evaluating the final sum to \( 2n \).