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

The sum of the first \( n \) terms of an arithmetic sequence is given by:
\( S_n = \frac{n}{2} [2u_1 + (n - 1)d] \)

For \( n = 5 \), \( u_1 = a \), and \( d = 3 \):
\( S_5 = \frac{5}{2} [2a + 4(3)] = \frac{5}{2} [2a + 12] = 5a + 30 \)

The sum to infinity of a geometric sequence is given by:
\( S_{\infty} = \frac{v_1}{1 - r} \)

For \( v_1 = a \) and \( r = \frac{1}{2} \):
\( S_{\infty} = \frac{a}{1 - \frac{1}{2}} = 2a \)

Since the two sums are equal:
\( 5a + 30 = 2a \)
\( 3a = -30 \)
\( a = -10 \)

M1 for writing a correct expression for the sum of the first 5 terms of the arithmetic sequence.
A1 for simplifying it to \( 5a + 30 \).
M1 for writing a correct expression for the sum to infinity of the geometric sequence.
A1 for simplifying it to \( 2a \).
M1 for setting up the equation \( 5a + 30 = 2a \) and solving for \( a \).
A1 for obtaining the correct value \( a = -10 \).

First, find \( f^{-1}(5) \). Let \( y = f(x) = 2x + 1 \).
\( 5 = 2x + 1 \Rightarrow 2x = 4 \Rightarrow x = 2 \).
Therefore, \( f^{-1}(5) = 2 \).

Next, find the composite function \( (g \circ f)(x) \):
\( (g \circ f)(x) = g(f(x)) = g(2x + 1) \)

Substitute \( 2x+1 \) into the expression for \( g(x) \):
\( g(2x+1) = \frac{(2x+1) - 3}{(2x+1) + 1} = \frac{2x - 2}{2x + 2} = \frac{x-1}{x+1} \)

Now set \( (g \circ f)(x) = f^{-1}(5) \):
\( \frac{x-1}{x+1} = 2 \)
\( x - 1 = 2(x + 1) \)
\( x - 1 = 2x + 2 \)
\( x = -3 \)

M1 for a valid method to find \( f^{-1}(5) \).
A1 for \( f^{-1}(5) = 2 \).
M1 for substituting \( f(x) \) into \( g(x) \) to find the composite function \( (g \circ f)(x) \).
A1 for simplifying the composite function to \( \frac{x-1}{x+1} \).
M1 for setting up the equation \( \frac{x-1}{x+1} = 2 \) and attempting to solve for \( x \).
A1 for the final answer \( x = -3 \).

Using the trigonometric identity \( \cos^2(x) = 1 - \sin^2(x) \):
\( 2(1 - \sin^2(x)) - \sin(x) - 1 = 0 \)
\( 2 - 2\sin^2(x) - \sin(x) - 1 = 0 \)
\( 2\sin^2(x) + \sin(x) - 1 = 0 \)

Letting \( u = \sin(x) \), we factor the quadratic equation:
\( (2\sin(x) - 1)(\sin(x) + 1) = 0 \)

This yields two possible branches:
1) \( 2\sin(x) - 1 = 0 \Rightarrow \sin(x) = \frac{1}{2} \)
For \( 0 \le x \le 2\pi \), the solutions are:
\( x = \frac{\pi}{6}, \frac{5\pi}{6} \)

2) \( \sin(x) + 1 = 0 \Rightarrow \sin(x) = -1 \)
For \( 0 \le x \le 2\pi \), the solution is:
\( x = \frac{3\pi}{2} \)

Combining all solutions:
\( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} \)

M1 for using the identity \( \cos^2(x) = 1 - \sin^2(x) \) to form a quadratic in terms of \( \sin(x) \).
A1 for obtaining \( 2\sin^2(x) + \sin(x) - 1 = 0 \).
M1 for factoring the quadratic into \( (2\sin(x) - 1)(\sin(x) + 1) = 0 \).
A1 for identifying the key values \( \sin(x) = \frac{1}{2} \) and \( \sin(x) = -1 \).
A1 for finding \( x = \frac{\pi}{6}, \frac{5\pi}{6} \).
A1 for finding \( x = \frac{3\pi}{2} \).

The gradient of the tangent to the curve is given by the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = 3x^2 - 6x - 9 \)

A tangent is parallel to the \( x \)-axis when its gradient is zero, so we set \( \frac{dy}{dx} = 0 \):
\( 3x^2 - 6x - 9 = 0 \)

Dividing the equation by 3:
\( x^2 - 2x - 3 = 0 \)

Factoring the quadratic:
\( (x - 3)(x + 1) = 0 \)

This gives \( x = 3 \) and \( x = -1 \).

Now we find the corresponding \( y \)-coordinates by substituting these \( x \)-values back into the original curve equation:
For \( x = -1 \):
\( y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 \)
Point: \( (-1, 10) \)

For \( x = 3 \):
\( y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22 \)
Point: \( (3, -22) \)

Thus, the coordinates of the points are \( (-1, 10) \) and \( (3, -22) \).

M1 for differentiating to find \( \frac{dy}{dx} = 3x^2 - 6x - 9 \).
M1 for setting \( \frac{dy}{dx} = 0 \).
A1 for solving the quadratic to find \( x = 3 \) and \( x = -1 \).
M1 for substituting at least one of their \( x \)-values back into the curve equation to find a \( y \)-value.
A1 for finding the coordinate \( (-1, 10) \).
A1 for finding the coordinate \( (3, -22) \).

(a) Using the probability addition rule:
\( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) \)
\( 0.8 = 0.6 + 0.4 - \mathrm{P}(A \cap B) \)
\( \mathrm{P}(A \cap B) = 0.2 \)

(b) For \( A \) and \( B \) to be independent, \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) \).
\( \mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24 \)
Since \( 0.2 \neq 0.24 \), the events \( A \) and \( B \) are not independent.

(c) Using the conditional probability formula:
\( \mathrm{P}(A' \mid B) = \frac{\mathrm{P}(A' \cap B)}{\mathrm{P}(B)} \)
Since \( \mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.4 - 0.2 = 0.2 \):
\( \mathrm{P}(A' \mid B) = \frac{0.2}{0.4} = 0.5 \)

(a) M1 for using the addition rule of probability. A1 for \( \mathrm{P}(A \cap B) = 0.2 \).
(b) M1 for calculating the product \( \mathrm{P}(A) \times \mathrm{P}(B) = 0.24 \). R1 for comparing \( 0.2 \neq 0.24 \) and concluding they are not independent.
(c) M1 for finding \( \mathrm{P}(A' \cap B) = 0.2 \) or applying the conditional probability ratio. A1 for the answer \( 0.5 \).

Using the log product rule:
\( \log_2(x(x - 3)) = 2 \)

Converting the equation to exponential form:
\( x(x - 3) = 2^2 \)
\( x^2 - 3x = 4 \)
\( x^2 - 3x - 4 = 0 \)

Factoring the quadratic equation:
\( (x - 4)(x + 1) = 0 \)
This gives the roots \( x = 4 \) or \( x = -1 \).

Since the equation is only defined for \( x > 3 \), we reject \( x = -1 \).
Thus, the only valid solution is \( x = 4 \).

M1 for applying the log law \( \log_a(u) + \log_a(v) = \log_a(uv) \).
M1 for converting to exponential form to obtain \( x(x-3) = 4 \).
A1 for writing the correct quadratic equation \( x^2 - 3x - 4 = 0 \).
M1 for solving the quadratic equation to find potential roots.
A1 for finding both roots \( x = 4 \) and \( x = -1 \).
R1 for rejecting \( x = -1 \) and correctly identifying the final answer as \( x = 4 \).

Using integration by substitution, let \( u = x^2 \).
Then \( \mathrm{d}u = 2x \mathrm{d}x \), which means \( x \mathrm{d}x = \frac{1}{2} \mathrm{d}u \).

Change the limits of integration:
When \( x = 0 \), \( u = 0^2 = 0 \).
When \( x = 2 \), \( u = 2^2 = 4 \).

Rewrite the integral in terms of \( u \):
\( \int_{0}^{4} e^u \cdot \frac{1}{2} \mathrm{d}u = \frac{1}{2} \int_{0}^{4} e^u \mathrm{d}u \)

Evaluating the definite integral:
\( \frac{1}{2} [e^u]_0^4 = \frac{1}{2} (e^4 - e^0) = \frac{1}{2} (e^4 - 1) \)

M1 for setting up the substitution \( u = x^2 \) and finding \( \mathrm{d}u = 2x \mathrm{d}x \).
A1 for correctly expressing the integrand as \( \frac{1}{2}e^u \mathrm{d}u \).
M1 for changing the limits of integration to \( u = 0 \) and \( u = 4 \).
A1 for integrating to obtain \( \frac{1}{2}e^u \).
M1 for substituting the limits into the integrated expression.
A1 for the correct exact final answer \( \frac{1}{2}(e^4 - 1) \) (or equivalent form).

(a) The perimeter of a sector is given by:
\( P = 2r + r\theta \)
Given \( P = 24 \):
\( 2r + r\theta = 24 \)
\( r\theta = 24 - 2r \)
\( \theta = \frac{24 - 2r}{r} \)

(b) The area of a sector is given by:
\( A = \frac{1}{2} r^2 \theta \)
Substitute the expression for \( \theta \) from part (a) into the area formula:
\( 36 = \frac{1}{2} r^2 \left( \frac{24 - 2r}{r} \right) \)
\( 36 = \frac{1}{2} r (24 - 2r) \)
\( 36 = 12r - r^2 \)
\( r^2 - 12r + 36 = 0 \)

Factoring the quadratic:
\( (r - 6)^2 = 0 \)
This gives \( r = 6 \).

(a) M1 for stating the perimeter of a sector formula \( 2r + r\theta = 24 \).
A1 for writing \( \theta = \frac{24 - 2r}{r} \).
(b) M1 for substituting their expression for \( \theta \) into the area formula \( \frac{1}{2} r^2 \theta = 36 \).
A1 for simplifying the equation to a standard quadratic equation, e.g., \( r^2 - 12r + 36 = 0 \).
M1 for solving the quadratic equation.
A1 for obtaining the final correct value \( r = 6 \).

**(a)**
To find the local minimum, we find the first derivative of \( f(x) = x^2 \ln(x) \) using the product rule:
\( f'(x) = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x \)

Setting \( f'(x) = 0 \):
\( x(2\ln(x) + 1) = 0 \)

Since \( x > 0 \), we have:
\( 2\ln(x) + 1 = 0 \implies \ln(x) = -\frac{1}{2} \implies x = e^{-1/2} \)

The y-coordinate is:
\( f(e^{-1/2}) = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1} \left(-\frac{1}{2}\right) = -\frac{1}{2e} \)

We verify using the second derivative:
\( f''(x) = 2\ln(x) + 2 + 1 = 2\ln(x) + 3 \)
At \( x = e^{-1/2} \), \( f''(e^{-1/2}) = 2\left(-\frac{1}{2}\right) + 3 = 2 > 0 \), confirming it is a local minimum.

Thus, the local minimum coordinates are \( \left(e^{-1/2}, -\frac{1}{2e}\right) \).

**(b)**
To find the point of inflection, we set the second derivative to zero:
\( f''(x) = 2\ln(x) + 3 = 0 \implies \ln(x) = -\frac{3}{2} \implies x = e^{-3/2} \)

The y-coordinate is:
\( f(e^{-3/2}) = (e^{-3/2})^2 \ln(e^{-3/2}) = e^{-3} \left(-\frac{3}{2}\right) = -\frac{3}{2e^3} \)

Since \( f''(x) \) changes sign at \( x = e^{-3/2} \), this is a point of inflection.

Thus, the coordinates of the point of inflection are \( \left(e^{-3/2}, -\frac{3}{2e^3}\right) \).

**(c)**
Since \( f(x) \ge 0 \) for \( x \in [1, e] \), the area is given by:
\( \text{Area} = \int_1^e x^2 \ln(x) \, dx \)

Using integration by parts with \( u = \ln(x) \) and \( dv = x^2 \, dx \):
\( du = \frac{1}{x} \, dx \) and \( v = \frac{1}{3}x^3 \)

Then:
\( \int x^2 \ln(x) \, dx = \frac{1}{3}x^3 \ln(x) - \int \frac{1}{3}x^2 \, dx = \frac{1}{3}x^3 \ln(x) - \frac{1}{9}x^3 \)

Evaluating between the limits \( 1 \) and \( e \):
\( \text{Area} = \left[ \frac{1}{3}x^3 \ln(x) - \frac{1}{9}x^3 \right]_1^e \)
\( \text{Area} = \left( \frac{1}{3}e^3 \ln(e) - \frac{1}{9}e^3 \right) - \left( \frac{1}{3}(1)^3 \ln(1) - \frac{1}{9}(1)^3 \right) \)
\( \text{Area} = \left( \frac{1}{3}e^3 - \frac{1}{9}e^3 \right) - \left( 0 - \frac{1}{9} \right) \)
\( \text{Area} = \frac{2}{9}e^3 + \frac{1}{9} = \frac{2e^3 + 1}{9} \)

**(d)**
First, we find the boundaries of the region. The curve \( y = g(x) \) intersects the x-axis where \( g(x) = 0 \):
\( x \sqrt{\ln(x)} = 0 \implies \ln(x) = 0 \implies x = 1 \)
Thus, the region is bounded between \( x = 1 \) and \( x = e \).

The volume of revolution about the x-axis is given by:
\( V = \pi \int_1^e [g(x)]^2 \, dx = \pi \int_1^e \left(x \sqrt{\ln(x)}\right)^2 \, dx = \pi \int_1^e x^2 \ln(x) \, dx \)

Using the integration result from part (c):
\( V = \pi \left( \frac{2e^3 + 1}{9} \right) = \frac{\pi(2e^3 + 1)}{9} \)

**(a) [4 Marks]**
- **M1** for attempting to use the product rule to differentiate \( f(x) \).
- **A1** for the correct derivative: \( f'(x) = 2x\ln(x) + x \).
- **A1** for setting \( f'(x) = 0 \) and finding \( x = e^{-1/2} \).
- **A1** for the correct y-coordinate: \( y = -\frac{1}{2e} \).

**(b) [4 Marks]**
- **M1** for attempting to differentiate \( f'(x) \).
- **A1** for the correct second derivative: \( f''(x) = 2\ln(x) + 3 \).
- **A1** for setting \( f''(x) = 0 \) and finding \( x = e^{-3/2} \).
- **A1** for the correct y-coordinate: \( y = -\frac{3}{2e^3} \).

**(c) [6 Marks]**
- **M1** for setting up the correct definite integral: \( \int_1^e x^2\ln(x) \, dx \).
- **M1** for setting up integration by parts correctly: \( u = \ln(x) \), \( dv = x^2 \, dx \).
- **A1** for finding the correct first term \( \frac{1}{3}x^3\ln(x) \) and the simplified remaining integral \( \int \frac{1}{3}x^2 \, dx \).
- **A1** for obtaining the correct indefinite integral \( \frac{1}{3}x^3\ln(x) - \frac{1}{9}x^3 \).
- **M1** for correctly substituting limits \( e \) and \( 1 \).
- **A1** for the final simplified area: \( \frac{2e^3 + 1}{9} \).

**(d) [4 Marks]**
- **M1** for utilizing the formula for volume of revolution: \( V = \pi \int [g(x)]^2 \, dx \).
- **A1** for finding the lower bound \( x = 1 \) by solving \( g(x) = 0 \).
- **M1** for expressing \( V = \pi \int_1^e x^2\ln(x) \, dx \).
- **A1** for the correct final volume using the result of (c): \( \frac{\pi(2e^3 + 1)}{9} \).

**(a) (i)**
The tree diagram should have two initial branches for the die roll:
- Even (Bag A) with probability \( \frac{1}{2} \)
- Odd (Bag B) with probability \( \frac{1}{2} \)

From the "Bag A" branch, there are two sub-branches:
- Blue with probability \( \frac{2}{5} \)
- Red with probability \( \frac{3}{5} \)

From the "Bag B" branch, there are two sub-branches:
- Blue with probability \( \frac{4}{5} \)
- Red with probability \( \frac{1}{5} \)

**(ii)**
The probability of selecting a blue marble is:
\( P(\text{Blue}) = P(\text{Bag A} \cap \text{Blue}) + P(\text{Bag B} \cap \text{Blue}) \)
\( P(\text{Blue}) = \left(\frac{1}{2} \times \frac{2}{5}\right) + \left(\frac{1}{2} \times \frac{4}{5}\right) = \frac{1}{5} + \frac{2}{5} = \frac{3}{5} \)

**(b)**
Using conditional probability:
\( P(\text{Bag A} \mid \text{Blue}) = \frac{P(\text{Bag A} \cap \text{Blue})}{P(\text{Blue})} \)
\( P(\text{Bag A} \mid \text{Blue}) = \frac{\frac{1}{5}}{\frac{3}{5}} = \frac{1}{3} \)

**(c)**
Let \( Y \) be the number of times a blue marble is selected in 5 independent trials. \( Y \sim B\left(5, \frac{3}{5}\right) \).

**(i)**
\( P(Y = 3) = \binom{5}{3} \left(\frac{3}{5}\right)^3 \left(\frac{2}{5}\right)^2 \)
\( P(Y = 3) = 10 \times \frac{27}{125} \times \frac{4}{25} = 10 \times \frac{108}{3125} = \frac{1080}{3125} = \frac{216}{625} \) (or 0.3456)

**(ii)**
\( P(Y \ge 4) = P(Y = 4) + P(Y = 5) \)
\( P(Y = 4) = \binom{5}{4} \left(\frac{3}{5}\right)^4 \left(\frac{2}{5}\right)^1 = 5 \times \frac{81}{625} \times \frac{2}{5} = \frac{810}{3125} \)
\( P(Y = 5) = \left(\frac{3}{5}\right)^5 = \frac{243}{3125} \)
\( P(Y \ge 4) = \frac{810}{3125} + \frac{243}{3125} = \frac{1053}{3125} \) (or 0.33696)

**(d)**
Let \( W \) represent the player's winnings. We list the possible outcomes, their winnings, and their respective probabilities:
- Blue from Bag A: Winnings = $10, Probability = \( P(\text{Bag A} \cap \text{Blue}) = \frac{1}{2} \times \frac{2}{5} = \frac{1}{5} \)
- Blue from Bag B: Winnings = $5, Probability = \( P(\text{Bag B} \cap \text{Blue}) = \frac{1}{2} \times \frac{4}{5} = \frac{2}{5} \)
- Red marble (from either bag): Winnings = \( -x \), Probability = \( P(\text{Red}) = 1 - P(\text{Blue}) = 1 - \frac{3}{5} = \frac{2}{5} \)

For the expected winnings to be $0:
\( E(W) = 10 \cdot P(\text{Blue from A}) + 5 \cdot P(\text{Blue from B}) - x \cdot P(\text{Red}) = 0 \)
\( 10\left(\frac{1}{5}\right) + 5\left(\frac{2}{5}\right) - x\left(\frac{2}{5}\right) = 0 \)
\( 2 + 2 - \frac{2x}{5} = 0 \)
\( 4 = \frac{2x}{5} \implies 2x = 20 \implies x = 10 \)

**(a) [4 Marks]**
- **M1** for drawing a correct tree diagram structure.
- **A1** for correct branch probabilities on the die roll: \( 1/2 \) for both Bag A and Bag B.
- **A1** for correct branch probabilities for marbles: Bag A (Blue: \( 2/5 \), Red: \( 3/5 \)) and Bag B (Blue: \( 4/5 \), Red: \( 1/5 \)).
- **A1** for calculating \( P(\text{Blue}) = \frac{1}{2}\left(\frac{2}{5}\right) + \frac{1}{2}\left(\frac{4}{5}\right) = \frac{3}{5} \).

**(b) [3 Marks]**
- **M1** for writing down the correct conditional probability expression: \( P(\text{Bag A} \mid \text{Blue}) = \frac{P(\text{Bag A} \cap \text{Blue})}{P(\text{Blue})} \).
- **A1** for \( P(\text{Bag A} \cap \text{Blue}) = \frac{1}{5} \).
- **A1** for the final answer: \( \frac{1}{3} \).

**(c) [5 Marks]**
- **M1** for identifying the binomial distribution model \( Y \sim B\left(5, \frac{3}{5}\right) \).
- **A1** for calculating \( P(Y = 3) = \frac{216}{625} \).
- **M1** for setting up \( P(Y \ge 4) = P(Y = 4) + P(Y = 5) \).
- **A1** for correct value of \( P(Y = 4) = \frac{810}{3125} \) or \( P(Y = 5) = \frac{243}{3125} \).
- **A1** for final answer of \( \frac{1053}{3125} \).

**(d) [6 Marks]**
- **M1** for identifying that there are three distinct financial outcomes: winning $10, winning $5, and losing $x.
- **A1** for calculating \( P(\text{Blue from A}) = \frac{1}{5} \).
- **A1** for calculating \( P(\text{Blue from B}) = \frac{2}{5} \).
- **A1** for calculating \( P(\text{Red}) = \frac{2}{5} \).
- **M1** for formulating the equation for the expected value: \( 10\left(\frac{1}{5}\right) + 5\left(\frac{2}{5}\right) - x\left(\frac{2}{5}\right) = 0 \).
- **A1** for correctly solving to get \( x = 10 \).

**(a)**
For the arithmetic sequence:
\( u_2 = a + d \)
\( u_5 = a + 4d \)

For the geometric sequence:
\( v_2 = ar \)
\( v_3 = ar^2 \)

From the given information:
1) \( a + d = ar \implies d = a(r - 1) \)
2) \( a + 4d = ar^2 \)

Substitute the first equation into the second:
\( a + 4a(r - 1) = ar^2 \)

Since \( a > 0 \), we can divide the entire equation by \( a \):
\( 1 + 4(r - 1) = r^2 \)
\( r^2 - 4r + 3 = 0 \)

Factoring the quadratic equation:
\( (r - 3)(r - 1) = 0 \implies r = 3 \text{ or } r = 1 \)

Since the geometric sequence is non-constant, \( r \neq 1 \), hence \( r = 3 \).

Now, substitute \( r = 3 \) back into the equation for \( d \):
\( d = a(3 - 1) = 2a \).

**(b)**
The formula for the sum of the first 10 terms of an arithmetic sequence is:
\( S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) \)

Substituting \( d = 2a \) into the sum:
\( S_{10} = 5(2a + 9(2a)) = 5(20a) = 100a \)

We are given that \( S_{10} = 200 \):
\( 100a = 200 \implies a = 2 \).

**(c)**
Using \( a = 2 \) and \( r = 3 \), the sum of the first \( n \) terms of the geometric sequence, \( G_n \), is:
\( G_n = \frac{a(r^n - 1)}{r - 1} \)
\( G_n = \frac{2(3^n - 1)}{3 - 1} = \frac{2(3^n - 1)}{2} = 3^n - 1 \).

**(d)**
First, find the sum of the first \( 2n \) terms of the arithmetic sequence, \( S_{2n} \).
Since \( a = 2 \) and \( d = 2a = 4 \):
\( S_{2n} = \frac{2n}{2}(2a + (2n - 1)d) = n(2(2) + (2n - 1)4) \)
\( S_{2n} = n(4 + 8n - 4) = 8n^2 \).

We want to find the smallest integer \( n \) such that:
\( G_n > S_{2n} \implies 3^n - 1 > 8n^2 \)

We test positive integer values for \( n \):
- For \( n = 1 \): \( 3^1 - 1 = 2 \) and \( 8(1)^2 = 8 \). (False)
- For \( n = 2 \): \( 3^2 - 1 = 8 \) and \( 8(2)^2 = 32 \). (False)
- For \( n = 3 \): \( 3^3 - 1 = 26 \) and \( 8(3)^2 = 72 \). (False)
- For \( n = 4 \): \( 3^4 - 1 = 80 \) and \( 8(4)^2 = 128 \). (False)
- For \( n = 5 \): \( 3^5 - 1 = 242 \) and \( 8(5)^2 = 200 \). (True)

Thus, the smallest value of \( n \) is 5.

**(a) [5 Marks]**
- **M1** for writing expressions for \( u_2, u_5, v_2, \) and \( v_3 \).
- **M1** for setting up the system of equations: \( d = a(r - 1) \) and \( a + 4d = ar^2 \).
- **A1** for substituting \( d \) to obtain \( a + 4a(r - 1) = ar^2 \).
- **A1** for simplifying to the quadratic equation \( r^2 - 4r + 3 = 0 \).
- **A1** for solving to find \( r = 3 \) (including rejection of \( r = 1 \)) and stating \( d = 2a \).

**(b) [4 Marks]**
- **M1** for using the sum of an arithmetic sequence formula with \( n = 10 \).
- **A1** for substituting \( d = 2a \) into the sum formula to get \( 100a \).
- **M1** for setting \( 100a = 200 \).
- **A1** for finding \( a = 2 \).

**(c) [4 Marks]**
- **M1** for identifying the parameters for the geometric sequence: first term \( 2 \) and common ratio \( 3 \).
- **M1** for using the geometric sum formula.
- **A1** for substituting the values: \( \frac{2(3^n - 1)}{3 - 1} \).
- **A1** for simplifying to \( 3^n - 1 \).

**(d) [5 Marks]**
- **M1** for writing the sum of the first \( 2n \) terms of the arithmetic sequence: \( S_{2n} = \frac{2n}{2}(2a + (2n - 1)d) \).
- **A1** for substituting \( a = 2, d = 4 \) to obtain \( 8n^2 \).
- **M1** for setting up the inequality: \( 3^n - 1 > 8n^2 \).
- **M1** for systematic trial and error of integers \( n \).
- **A1** for identifying \( n = 5 \) as the smallest integer.

(a) Using the sum to infinity formula: \(S_{\infty} = \frac{u_1}{1 - r} = \frac{12}{1 - 0.85} = \frac{12}{0.15} = 80\). (b) We want to solve \(S_n > 0.98 S_{\infty}\). Substituting the formula for the sum of the first \(n\) terms: \(80(1 - 0.85^n) > 0.98 \times 80\), which simplifies to \(1 - 0.85^n > 0.98 \implies 0.85^n < 0.02\). Taking logarithms on both sides: \(n \ln(0.85) < \ln(0.02) \implies n > \frac{\ln(0.02)}{\ln(0.85)}\). Using a calculator: \(n > 24.08\). Since \(n\) must be an integer, the least value of \(n\) is \(25\).

M1 for substituting into the sum to infinity formula. A1 for S_infinity = 80. M1 for setting up the inequality 80(1-0.85^n) > 0.98 * 80. M1 for simplifying to 0.85^n < 0.02 or equivalent. A1 for n > 24.08. A1 for the final integer answer n = 25.

To find the points of intersection, we set \(f(x) = g(x)\), which gives \(e^{-0.2x} + 1 = x^2 - 3x\). Using a graphing display calculator (GDC), we can find the points of intersection by finding the roots of \(x^2 - 3x - e^{-0.2x} - 1 = 0\). The x-coordinates of the intersection points are \(x \approx -0.592\) and \(x \approx 3.44\). Substituting these values back into either function to find the corresponding y-coordinates: For \(x \approx -0.592\), \(y \approx (-0.592)^2 - 3(-0.592) \approx 2.13\). For \(x \approx 3.44\), \(y \approx (3.44)^2 - 3(3.44) \approx 1.50\). Thus, the points of intersection are \((-0.592, 2.13)\) and \((3.44, 1.50)\).

M1 for equating f(x) and g(x). A1, A1 for finding correct x-coordinates (x = -0.592 and x = 3.44). M1 for attempting to find corresponding y-coordinates. A1, A1 for correct y-coordinates (2.13 and 1.50).

(a) The particle is at rest when \(v(t) = 0\), so \(3t \sin(0.5t^2) - 2 = 0\). Using a GDC to find the roots in the interval \([0, 3]\), we obtain \(t \approx 1.13\) and \(t \approx 2.39\). (b) The total distance travelled is given by the integral of the speed: \(\int_{0}^{3} |v(t)| \, dt = \int_{0}^{3} |3t \sin(0.5t^2) - 2| \, dt\). Using the GDC's numerical integration feature, we find the total distance is approximately \(7.91\) meters.

M1 for setting v(t) = 0. A1, A1 for t = 1.13 and t = 2.39. M1 for setting up the integral of the absolute value of velocity. M1 for correct limits 0 and 3. A1 for 7.91.

(a) The area of a triangle is given by \(\text{Area} = \frac{1}{2} a c \sin B\). Substituting the given values: \(24 = \frac{1}{2} (9)(7) \sin B \implies 48 = 63 \sin B \implies \sin B = \frac{16}{21}\). The acute angle is \(B = \arcsin\left(\frac{16}{21}\right) \approx 49.63^{\circ}\). The obtuse angle is \(180^{\circ} - 49.63^{\circ} \approx 130.37^{\circ}\). Thus, the two possible values of \(\hat{B}\) are \(49.6^{\circ}\) and \(130^{\circ}\). (b) Since \(\hat{B}\) is obtuse, we use \(B \approx 130.37^{\circ}\). Using the cosine rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B\). Substituting the values: \(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(130.37^{\circ}) = 49 + 81 - 126(-0.6477) \approx 130 + 81.61 = 211.61\). Therefore, \(AC = \sqrt{211.61} \approx 14.5\text{ cm}\).

M1 for using the area formula 24 = 0.5 * 7 * 9 * sin B. A1 for sin B = 16/21. A1 for both angle solutions (49.6 and 130). M1 for using the cosine rule with the obtuse angle. A1 for substituting values into the cosine rule. A1 for AC = 14.5.

(a) The compound interest formula is \(FV = PV \left(1 + \frac{r}{100k}\right)^{kn}\), where \(k = 12\) (compounded monthly) and \(n = 5\) years. So, \(10200 = 8000 \left(1 + \frac{r}{1200}\right)^{60}\). Dividing both sides by 8000: \(1.275 = \left(1 + \frac{r}{1200}\right)^{60}\). Taking the 60th root: \(1 + \frac{r}{1200} = (1.275)^{1/60} \approx 1.004055\). Thus, \(\frac{r}{1200} \approx 0.004055 \implies r \approx 4.87\%\). (b) We now want to find the number of years \(t\) such that \(8000 \left(1 + \frac{4.87}{1200}\right)^{12t} \ge 15000\). Using the exact values from part (a): \(8000 (1.275)^{t/5} \ge 15000 \implies (1.275)^{t/5} \ge 1.875\). Taking logarithms: \(\frac{t}{5} \ln(1.275) \ge \ln(1.875) \implies t \ge 5 \times \frac{\ln(1.875)}{\ln(1.275)} \approx 12.92\). Since the question asks for the minimum number of complete years, she must keep the money in the account for \(13\) years.

M1 for substituting into the compound interest formula with monthly compounding. A1 for 1 + r/1200 = 1.004055 or equivalent. A1 for r = 4.87. M1 for setting up the inequality for 15000. M1 for solving for t using logs or GDC. A1 for t = 13.

(a) Let \(X\) be the mass of an apple. We are given \(X \sim N(150, \sigma^2)\) and \(P(X > 185) = 0.08\). Standardizing the variable: \(P\left(Z > \frac{185 - 150}{\sigma}\right) = 0.08\), which means \(P\left(Z < \frac{35}{\sigma}\right) = 0.92\). Using the inverse normal function on a GDC, we find the z-score corresponding to a cumulative probability of 0.92 is \(z \approx 1.405\). Thus, \(\frac{35}{\sigma} = 1.405 \implies \sigma = \frac{35}{1.405} \approx 24.9\text{ g}\). (b) We want to find \(P(130 < X < 160)\). Using the GDC's normal cumulative distribution function with mean \(\mu = 150\) and standard deviation \(\sigma = 24.909\), we get \(P(130 < X < 160) \approx 0.444\).

M1 for standardizing the normal distribution equation. M1 for finding the z-score 1.405 using GDC. A1 for standard deviation 24.9. M1 for setting up the probability P(130 < X < 160). A1 for correct inputs in GDC. A1 for probability 0.444.

(a) The rate of change of temperature is given by the derivative \(T'(t)\). Using the product rule: \(T'(t) = 15 \ln(t + 1) + 15t \cdot \frac{1}{t + 1} - 8t\). Substituting \(t = 2\): \(T'(2) = 15 \ln(3) + 15(2) \cdot \frac{1}{3} - 8(2) = 15 \ln(3) + 10 - 16 = 15 \ln(3) - 6 \approx 10.5\,^{\circ}\text{C/min}\). (b) The maximum temperature occurs when \(T'(t) = 0\). Using a GDC to solve \(15 \ln(t + 1) + \frac{15t}{t+1} - 8t = 0\) in the interval \([0, 5]\), we find the critical point at \(t \approx 4.88\) minutes. Substituting this value back into the temperature function: \(T(4.88) = 20 + 15(4.88) \ln(5.88) - 4(4.88)^2 \approx 54.4\,^{\circ}\text{C}\).

M1 for attempting to find the derivative T'(t) (using product rule). A1 for correct T'(t). A1 for rate of change = 10.5. M1 for setting T'(t) = 0. A1 for finding t = 4.88. A1 for maximum temperature = 54.4.

(a) Since the sum of probabilities must equal 1: \(k + 0.25 + k^2 + 0.3 = 1 \implies k^2 + k + 0.55 = 1 \implies k^2 + k - 0.45 = 0\). Solving this quadratic equation for \(k\) using the quadratic formula: \(k = \frac{-1 \pm \sqrt{1 - 4(1)(-0.45)}}{2} = \frac{-1 \pm \sqrt{2.8}}{2}\). Since \(k\) must be positive, we choose the positive root: \(k \approx 0.33665 \approx 0.337\) (to 3 s.f.). (b) The expected value is given by \(\text{E}(X) = \sum x \cdot P(X=x) = 1(k) + 2(0.25) + 3(k^2) + 4(0.3)\). Substituting \(k \approx 0.33665\): \(\text{E}(X) = 0.33665 + 0.5 + 3(0.33665)^2 + 1.2 \approx 2.38\) (to 3 s.f.).

M1 for setting the sum of probabilities to 1. A1 for the quadratic equation k^2 + k - 0.45 = 0. A1 for k = 0.337. M1 for setting up the expected value formula. M1 for substituting the value of k. A1 for E(X) = 2.38.

(a) Let \(V \sim N(\mu, 8^2)\).

We are given \(P(V < 490) = 0.15\).

Using the standard normal variable \(Z = \frac{V - \mu}{\sigma}\):

\(P\left(Z < \frac{490 - \mu}{8}\right) = 0.15\)

Using a GDC to find the inverse normal value:

\(\frac{490 - \mu}{8} \approx -1.036433\)

\(490 - \mu \approx -8.29146\)

\(\mu \approx 498.291\)

So \(\mu = 498\) mL (to 3 significant figures).

(b) We want to find \(P(495 < V < 510)\).

Using a GDC with lower limit \(495\), upper limit \(510\), \(\mu = 498.291\), and \(\sigma = 8\):

\(P(495 < V < 510) \approx 0.588\) (to 3 significant figures).

If the rounded value of \(\mu = 498\) is used, the calculation gives \(P(495 < V < 510) \approx 0.579\).

**(a)**

**M1** for attempting to set up a standardizing equation or using inverse normal on a GDC, e.g., \(P(V < 490) = 0.15\) or \(\frac{490 - \mu}{8} = z\).

**A1** for finding the correct z-score of \(-1.0364...\) (or \(-1.04\)).

**A1** for \(\mu = 498\) (accept \(498.291...\)).

**(b)**

**M1** for setting up the required probability statement, e.g., \(P(495 < V < 510)\).

**A1** for substituting their mean \(\mu\) and standard deviation \(\sigma = 8\) into the normal cumulative distribution function.

**A1** for \(0.588\) (accept \(0.579\) if \(\mu = 498\) is used).

(a) Let \(V \sim N(\mu, 8^2)\).

We are given \(P(V < 490) = 0.15\).

Using the standard normal variable \(Z = \frac{V - \mu}{\sigma}\):

\(P\left(Z < \frac{490 - \mu}{8}\right) = 0.15\)

Using a GDC to find the inverse normal value:

\(\frac{490 - \mu}{8} \approx -1.036433\)

\(490 - \mu \approx -8.29146\)

\(\mu \approx 498.291\)

So \(\mu = 498\) mL (to 3 significant figures).

(b) We want to find \(P(495 < V < 510)\).

Using a GDC with lower limit \(495\), upper limit \(510\), \(\mu = 498.291\), and \(\sigma = 8\):

\(P(495 < V < 510) \approx 0.588\) (to 3 significant figures).

If the rounded value of \(\mu = 498\) is used, the calculation gives \(P(495 < V < 510) \approx 0.579\).

**(a)**

**M1** for attempting to set up a standardizing equation or using inverse normal on a GDC, e.g., \(P(V < 490) = 0.15\) or \(\frac{490 - \mu}{8} = z\).

**A1** for finding the correct z-score of \(-1.0364...\) (or \(-1.04\)).

**A1** for \(\mu = 498\) (accept \(498.291...\)).

**(b)**

**M1** for setting up the required probability statement, e.g., \(P(495 < V < 510)\).

**A1** for substituting their mean \(\mu\) and standard deviation \(\sigma = 8\) into the normal cumulative distribution function.

**A1** for \(0.588\) (accept \(0.579\) if \(\mu = 498\) is used).

(a) \(L \sim N(800, 50^2)\). Using a GDC: \(P(780 < L < 830) \approx 0.381\).

(b) We require \(P(L > h) = 0.10\), which is equivalent to \(P(L \le h) = 0.90\). Using the inverse normal function on a GDC: \(h \approx 864\) hours.

(c) Let \(p\) be the probability that a bulb lasts longer than 830 hours: \(p = P(L > 830) \approx 0.274253\).
(i) Let \(X\) be the number of lightbulbs that last longer than 830 hours out of a sample of 12. Then \(X \sim B(12, 0.274253)\). Using a GDC binomial probability distribution function: \(P(X = 3) \approx 0.252\).
(ii) We want to find \(P(X \ge 2) = 1 - P(X \le 1)\). Using the binomial cumulative distribution function on a GDC: \(P(X \le 1) \approx 0.11718\). Thus, \(P(X \ge 2) = 1 - 0.11718 \approx 0.883\).

(d) Under the new process, \(L_{new} \sim N(\mu_{new}, 40^2)\). We are given \(P(L_{new} > 850) = 0.15\). Standardizing this gives: \(P\left(Z > \frac{850 - \mu_{new}}{40}\right) = 0.15\). From standard normal tables or GDC: \(\frac{850 - \mu_{new}}{40} \approx 1.03643\). Solving for \(\mu_{new}\): \(850 - \mu_{new} \approx 41.457 \implies \mu_{new} \approx 808.543 \approx 809\) hours.

(e) Under the new distribution parameters, \(L_{new} \sim N(808.543, 40^2)\). The probability of a bulb lasting less than 750 hours is: \(p_{750} = P(L_{new} < 750) \approx 0.07163\). For a batch of 150 bulbs, the expected number is: \(E(Y) = 150 \times 0.07163 \approx 10.7\).

(a) [3 marks]
M1 for setting up the correct normal probability expression \(P(780 < L < 830)\).
A1 for inputting correct parameters \(\mu = 800\) and \(\sigma = 50\) into the GDC.
A1 for \(0.381\) (or \(38.1\%\)).

(b) [3 marks]
M1 for writing the probability statement \(P(L > h) = 0.10\) or \(P(L \le h) = 0.90\).
M1 for demonstrating a correct method to find \(h\) using the inverse normal feature on GDC.
A1 for \(864\) (accept 864.08).

(c) [6 marks]
(i) M1 for finding the success probability \(p \approx 0.274\) (or better).
M1 for setting up the binomial distribution model \(B(12, 0.274)\).
A1 for \(0.252\).
(ii) M1 for expressing the cumulative probability as \(1 - P(X \le 1)\) or writing the sum \(P(X=2) + \dots + P(X=12)\).
A1 for correct cumulative probability subtraction step.
A1 for \(0.883\).

(d) [4 marks]
M1 for standardizing the boundary: \(\frac{850 - \mu_{new}}{40}\).
A1 for finding the correct z-score \(z \approx 1.036\).
M1 for setting up the equation \(\frac{850 - \mu_{new}}{40} = 1.036\).
A1 for \(\mu_{new} = 809\) (accept 808.54).

(e) [2 marks]
M1 for finding \(P(L_{new} < 750) \approx 0.0716\) using the updated mean.
A1 for \(10.7\).

(a) The particle is at rest when \(v(t) = 0\).
\((t - 3) e^{-0.2t} \sin(t) = 0\).
Since \(e^{-0.2t} > 0\) for all \(t\), we have:
\(t - 3 = 0 \implies t = 3\), or \(\sin(t) = 0 \implies t = 0, \pi, 2\pi, 3\pi\).
For \(0 \le t \le 10\), the values are \(t = 0\), \(t = 3\), \(t \approx 3.14\), \(t \approx 6.28\), and \(t \approx 9.42\).

(b) Acceleration is given by \(a(t) = v'(t)\). Using a GDC to find the numerical derivative at \(t = 4\):
\(a(4) = v'(4) \approx -0.566 \text{ m s}^{-2}\).

(c) Total distance travelled in the first 6 seconds is given by the integral of speed:
\(d = \int_{0}^{6} |v(t)| dt = \int_{0}^{6} |(t-3)e^{-0.2t}\sin(t)| dt\).
Using a GDC to compute this definite integral yields:
\(d \approx 2.76 \text{ m}\).

(d) Speed is the magnitude of velocity, \(|v(t)|\). Evaluating the local maxima of \(y = |(t-3)e^{-0.2t}\sin(t)|\) on the GDC over \(0 \le t \le 10\):
- Peak speed in \([0, 3]\) occurs at \(t \approx 1.02\) with speed \(1.37 \text{ m s}^{-1}\).
- Peak speed in \([\pi, 2\pi]\) occurs at \(t \approx 4.80\) with speed \(0.690 \text{ m s}^{-1}\).
- Peak speed in \([2\pi, 3\pi]\) occurs at \(t \approx 7.50\) with speed \(0.941 \text{ m s}^{-1}\).
Comparing these, the absolute maximum speed is \(1.37 \text{ m s}^{-1}\) at \(t \approx 1.02\) seconds.

(e) Displacement is given by \(s(t) = s(0) + \int_{0}^{t} v(x) dx\).
We are given \(s(0) = -2\).
\(s(8) = -2 + \int_{0}^{8} (t-3)e^{-0.2t}\sin(t) dt\).
Using a GDC to evaluate the integral:
\(\int_{0}^{8} v(t) dt \approx -2.04\).
Therefore, \(s(8) \approx -2 - 2.04 = -4.04 \text{ m}\).

(a) [3 marks]
M1 for equating the velocity function to zero: \(v(t) = 0\).
A1 for \(t = 3\).
A1 for finding \(t = 3.14, 6.28, 9.42\) (accept in terms of \(\pi\) and accept inclusion of boundary \(t=0\)).

(b) [3 marks]
M1 for recognizing that \(a(t) = v'(t)\).
M1 for attempting to compute the derivative at \(t = 4\) using GDC or product/chain rules.
A1 for \(-0.566\) (accept \(-0.566 \text{ m s}^{-2}\)).

(c) [4 marks]
M1 for recognizing that total distance is the integral of the absolute value of velocity.
A1 for setting up the integral expression \(\int_{0}^{6} |v(t)| dt\) with correct limits.
M1 for attempting to evaluate using GDC numeric integration.
A1 for \(2.76\).

(d) [4 marks]
M1 for considering the graph of the speed function \(|v(t)|\) or examining critical points of \(v(t)\).
M1 for locating the primary peak at \(t \approx 1.02\).
A1 for maximum speed of \(1.37\) (accept \(1.37 \text{ m s}^{-1}\)).
A1 for \(t \approx 1.02\) (accept \(1.02 \text{ s}\)).

(e) [4 marks]
M1 for expressing displacement as the sum of initial displacement and the integral of velocity.
A1 for setting up the equation: \(-2 + \int_{0}^{8} v(t) dt\).
M1 for calculating the definite integral value \(\approx -2.04\) on GDC.
A1 for \(-4.04\) (accept \(-4.04 \text{ m}\)).

(a)
(i) The amplitude is given by: \(a = \frac{\text{maximum depth} - \text{minimum depth}}{2} = \frac{14.6 - 8.2}{2} = \frac{6.4}{2} = 3.2\).
(ii) The vertical shift is given by the average line: \(d = \frac{\text{maximum depth} + \text{minimum depth}}{2} = \frac{14.6 + 8.2}{2} = 11.4\).

(b) The time elapsed between high tide (03:00) and low tide (09:15) is 6 hours and 15 minutes, which is 6.25 hours. This represents half of a full cycle (period).
Therefore, Period = \(2 \times 6.25 = 12.5\) hours.
Using \(\text{Period} = \frac{2\pi}{b}\):
\(b = \frac{2\pi}{12.5} = \frac{4\pi}{25} \approx 0.503\).
Since a maximum occurs at \(t = 3\), we can align the cosine function without phase shift by choosing \(c = 3\).

(c) At 12:00, \(t = 12\). Substituting the parameters into the function:
\(D(12) = 3.2 \cos\left(\frac{4\pi}{25}(12 - 3)\right) + 11.4 = 3.2 \cos\left(\frac{36\pi}{25}\right) + 11.4 \approx 10.8\) m.

(d) We solve \(D(t) \ge 10.5\) for \(0 \le t \le 24\).
Using a GDC to find intersections of \(y_1 = 3.2 \cos(0.50265(t-3)) + 11.4\) and \(y_2 = 10.5\):
The boundary times are \(t_1 \approx 6.690\), \(t_2 \approx 11.810\), and \(t_3 \approx 19.190\).
By inspecting the graph:
- \(D(t) \ge 10.5\) for \(t \in [0, 6.690]\) (since at \(t=0\), \(D(0) \approx 11.6 > 10.5\)).
- \(D(t) \ge 10.5\) for \(t \in [11.810, 19.190]\).
Total duration is \((6.690 - 0) + (19.190 - 11.810) = 6.690 + 7.380 = 14.1\) hours (to 3 s.f.).

(e) The rate of change of depth is given by \(D'(t)\). At 15:00, \(t = 15\).
Using the GDC to find the derivative of \(D(t)\) at \(t = 15\):
\(D'(15) \approx 0.400 \text{ m h}^{-1}\).

(a) [4 marks]
(i) M1 for attempting to use the formula \(\frac{\text{max} - \text{min}}{2}\).
A1 for showing \(a = 3.2\).
(ii) M1 for attempting to use the formula \(\frac{\text{max} + \text{min}}{2}\).
A1 for showing \(d = 11.4\).

(b) [4 marks]
M1 for calculating the time interval between max and min as 6.25 hours.
M1 for calculating the period as 12.5 hours.
A1 for \(b = \frac{4\pi}{25}\) (or \(0.503\)).
A1 for \(c = 3\).

(c) [2 marks]
M1 for substituting \(t = 12\) into their model equation.
A1 for \(10.8\) (or \(10.79\)).

(d) [4 marks]
M1 for setting up the inequality \(D(t) \ge 10.5\) or the equation \(D(t) = 10.5\).
A1 for finding the intersection times \(t \approx 6.69\), \(11.8\), \(19.2\).
M1 for identifying the valid intervals and setting up a sum of durations: \(6.69 + (19.19 - 11.81)\).
A1 for \(14.1\) hours (accept \(14.07\)).

(e) [4 marks]
M1 for recognizing that rate of change is the derivative \(D'(t)\).
M1 for calculating the symbolic derivative \(D'(t) = -3.2 b \sin(b(t-3))\) (or writing the GDC derivative notation).
M1 for substituting \(t = 15\) into the derivative expression.
A1 for \(0.400\) (or \(0.4\)).

**Part a**
The derivative of the parabola \(y = x^2\) is \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\). At \(P(p, p^2)\), the slope of the tangent is \(2p\).
For \(p \neq 0\), the slope of the normal is \(-\frac{1}{2p}\).
The equation of the normal line is therefore:
\(y - p^2 = -\frac{1}{2p}(x - p)\)
Multiplying by \(2p\) gives:
\(2p(y - p^2) = -(x - p) \implies 2py - 2p^3 = -x + p \implies 2py + x = 2p^3 + p\).
If \(p = 0\), the normal at \((0,0)\) is the y-axis, which has equation \(x = 0\). Substituting \(p = 0\) into \(2py + x = 2p^3 + p\) yields \(x = 0\). Thus, the equation also holds for \(p = 0\).

**Part b**
We have the system of normal line equations:
(1) \(x + 2py = 2p^3 + p\)
(2) \(x + 2qy = 2q^3 + q\)
Subtracting (2) from (1):
\(2(p-q)y = 2(p^3 - q^3) + (p - q)\)
Since \(p \neq q\), we can divide by \(p - q\). Recall that \(p^3 - q^3 = (p - q)(p^2 + pq + q^2)\):
\(2y = 2(p^2 + pq + q^2) + 1 \implies y = p^2 + pq + q^2 + \frac{1}{2}\).
Substituting this back into (1) to solve for \(x\):
\(x = 2p^3 + p - 2p\left(p^2 + pq + q^2 + \frac{1}{2}\right)\)
\(x = 2p^3 + p - 2p^3 - 2p^2 q - 2pq^2 - p = -2pq(p + q)\).

**Part c**
Taking the limit of \(I(p, q)\) as \(q \to p\):
\(x = \lim_{q \to p} [-2pq(p + q)] = -2p(p)(p + p) = -4p^3\)
\(y = \lim_{q \to p} \left[p^2 + pq + q^2 + \frac{1}{2}\right] = p^2 + p^2 + p^2 + \frac{1}{2} = 3p^2 + \frac{1}{2}\).

**Part d**
From the y-coordinate of \(C(p)\):
\(y - \frac{1}{2} = 3p^2 \implies \left(y - \frac{1}{2}\right)^3 = 27p^6\).
From the x-coordinate of \(C(p)\):
\(x = -4p^3 \implies x^2 = 16p^6 \implies p^6 = \frac{x^2}{16}\).
Substituting this into the first equation:
\[\left(y - \frac{1}{2}\right)^3 = 27\left(\frac{x^2}{16}\right) \implies 16\left(y - \frac{1}{2}\right)^3 = 27x^2\]

**Part e**
The normal line at \(P(p, p^2)\) passes through \(A(u, v)\) if and only if \(x = u\) and \(y = v\) satisfy the normal line equation:
\(2pv + u = 2p^3 + p \implies 2p^3 + (1-2v)p - u = 0\).

**Part f**
(i) Let \(g(p) = 2p^3 + (1-2v)p - u\). Then \(g'(p) = 6p^2 + (1-2v)\).
If \(v \le \frac{1}{2}\), then \(1-2v \ge 0\). Since \(6p^2 \ge 0\) for all \(p\), \(g'(p) \ge 0\) for all \(p\) (with at most one single point where \(g'(p) = 0\) if \(v=1/2\)).
Hence, \(g(p)\) is strictly increasing. Since \(g(p)\) is a cubic polynomial, its range is \(\mathbb{R}\), and by the Intermediate Value Theorem combined with strict monotonicity, it has exactly one real root \(p\). Thus, exactly one normal can be drawn.

(ii) If \(v > \frac{1}{2}\), then \(1-2v < 0\). Setting \(g'(p) = 0\) gives:
\(6p^2 = 2v - 1 \implies p = \pm \sqrt{\frac{2v-1}{6}}\).
The local maximum is at \(p_1 = -\sqrt{\frac{2v-1}{6}}\) and the local minimum is at \(p_2 = \sqrt{\frac{2v-1}{6}}\).
To find the y-coordinates, we can simplify \(g(p)\) at these points using \(2p^3 = \frac{2v-1}{3}p\):
\(g(p) = p(2p^2 + 1 - 2v) - u = p\left(\frac{2v-1}{3} - (2v-1)\right) - u = -\frac{2}{3}(2v-1)p - u\).
For the local maximum:
\(y_{\max} = \frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\).
For the local minimum:
\(y_{\min} = -\frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\).

**Part g**
A cubic equation has three distinct real roots if and only if it has a local maximum and a local minimum with opposite signs, meaning \(y_{\max} > 0\) and \(y_{\min} < 0\).
This is equivalent to:
\(y_{\max} \cdot y_{\min} < 0 \implies \left(\frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\right)\left(-\frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\right) < 0\)
\(u^2 - \frac{4}{9}(2v-1)^2\left(\frac{2v-1}{6}\right) < 0 \implies u^2 < \frac{2}{27}(2v-1)^3\)
\(27u^2 < 2(2v-1)^3 = 2\left(2\left(v - \frac{1}{2}\right)\right)^3 = 16\left(v - \frac{1}{2}\right)^3\).

**Part a** [3 marks]
- **M1** for finding \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\) and finding the tangent slope \(2p\).
- **A1** for obtaining the normal slope \(-\frac{1}{2p}\), writing the equation \(y - p^2 = -\frac{1}{2p}(x - p)\), and rearranging to the given form.
- **R1** for explaining that for \(p = 0\), the normal is \(x = 0\), which matches the equation when \(p = 0\) is substituted.

**Part b** [4 marks]
- **M1** for setting up the system of equations and subtracting them: \(2(p-q)y = 2(p^3-q^3) + (p-q)\).
- **M1** for factoring \(p^3 - q^3 = (p-q)(p^2 + pq + q^2)\).
- **A1** for dividing by \(p-q\) (stating \(p \neq q\)) to obtain \(y = p^2 + pq + q^2 + \frac{1}{2}\).
- **A1** for substituting \(y\) back and simplifying to show \(x = -2pq(p+q)\).

**Part c** [4 marks]
- **M1** for stating that the coordinates of \(C(p)\) are found by taking the limits of the coordinates of \(I(p, q)\) as \(q \to p\).
- **A1.5** for evaluating \(\lim_{q \to p} x = -4p^3\).
- **A1.5** for evaluating \(\lim_{q \to p} y = 3p^2 + \frac{1}{2}\).

**Part d** [4 marks]
- **M1** for writing \(y - \frac{1}{2} = 3p^2\).
- **A1** for cubing both sides to get \(\left(y - \frac{1}{2}\right)^3 = 27p^6\).
- **A1** for squaring \(x\) to get \(x^2 = 16p^6\).
- **A1** for substituting \(p^6 = \frac{x^2}{16}\) to show \(27x^2 = 16\left(y - \frac{1}{2}\right)^3\).

**Part e** [3 marks]
- **M1** for noting that the normal line at \(P(p, p^2)\) passes through \(A(u, v)\) if the coordinates satisfy the normal's equation.
- **A1** for substituting \(x = u\) and \(y = v\) to get \(2pv + u = 2p^3 + p\).
- **A1** for rearranging to obtain the given cubic polynomial equation.

**Part f** [5 marks]
- **A1** for finding the derivative \(g'(p) = 6p^2 + (1-2v)\).
- **R1** for explaining that if \(v \le \frac{1}{2}\), \(g'(p) \ge 0\) for all \(p\), making \(g(p)\) strictly increasing.
- **R1** for concluding that there is exactly one real root, hence exactly one normal.
- **A1** for setting \(g'(p) = 0\) and finding critical points \(p = \pm \sqrt{\frac{2v-1}{6}}\).
- **A1** for finding correct coordinates of the local minimum and maximum.

**Part g** [4 marks]
- **M1** for identifying that three distinct roots require \(y_{\max} > 0\) and \(y_{\min} < 0\) (or \(y_{\max} \cdot y_{\min} < 0\)).
- **A1** for setting up the inequality \(\left(\frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\right)\left(-\frac{2}{3}(2v-1)\sqrt{\frac{2v-1}{6}} - u\right) < 0\).
- **A1** for simplifying this to \(u^2 < \frac{2}{27}(2v-1)^3\).
- **A1** for showing this is algebraically equivalent to \(27u^2 < 16\left(v - \frac{1}{2}\right)^3\).

**Part a**
(i) \(z_0 = 1\), \(z_1 = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\), \(z_2 = e^{i\frac{4\pi}{3}} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).
(ii) \(1 - z_1 = 1 - \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{3}{2} - i\frac{\sqrt{3}}{2}\).
Thus, \(|1 - z_1|^2 = \left(\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{9}{4} + \frac{3}{4} = 3 \implies |1 - z_1| = \sqrt{3}\).
By symmetry, \(|1 - z_2| = |1 - z_1| = \sqrt{3}\).
So, \(|1 - z_1| \cdot |1 - z_2| = \sqrt{3} \cdot \sqrt{3} = 3\).

**Part b**
For \(n = 4\), the roots are \(z_0 = 1, z_1 = i, z_2 = -1, z_3 = -i\).
The distances are:
\(|1 - z_1| = |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
\(|1 - z_2| = |1 - (-1)| = 2\)
\(|1 - z_3| = |1 + i| = \sqrt{2}\)
Multiply the distances:
\(|1 - z_1| \cdot |1 - z_2| \cdot |1 - z_3| = \sqrt{2} \cdot 2 \cdot \sqrt{2} = 4\).

**Part c**
Conjecture: \(P_n = n\) for all \(n \ge 2\).
For \(n = 2\), the roots are \(z_0 = 1\) and \(z_1 = -1\). The product of distances is \(|1 - z_1| = |1 - (-1)| = 2\), which matches \(n = 2\).

**Part d**
(i) Expanding the right-hand side:
\((z - 1)(z^{n-1} + z^{n-2} + \dots + z + 1)\)
\(= z(z^{n-1} + z^{n-2} + \dots + 1) - (z^{n-1} + z^{n-2} + \dots + 1)\)
\(= (z^n + z^{n-1} + \dots + z) - (z^{n-1} + z^{n-2} + \dots + 1)\)
\(= z^n - 1\).
(ii) The roots of \(z^n - 1 = 0\) are \(z_0, z_1, \dots, z_{n-1}\). Since \(z_0 = 1\), we can factorize:
\(z^n - 1 = (z - 1)(z - z_1)(z - z_2)\dots(z - z_{n-1}) = (z - 1)\prod_{k=1}^{n-1} (z - z_k)\).
Equating the two expressions for \(z^n - 1\):
\((z - 1)\prod_{k=1}^{n-1} (z - z_k) = (z - 1)(z^{n-1} + z^{n-2} + \dots + z + 1)\).
Since \(z \neq 1\), we can divide both sides by \(z - 1\) to get:
\(\prod_{k=1}^{n-1} (z - z_k) = z^{n-1} + z^{n-2} + \dots + z + 1\).

**Part e**
Since both sides of the identity in Part d are continuous functions of \(z\), we can take the limit as \(z \to 1\) (or substitute \(z = 1\)):
\(\prod_{k=1}^{n-1} (1 - z_k) = 1^{n-1} + 1^{n-2} + \dots + 1 + 1\).
The right-hand side has \(n\) terms, each equal to 1, so their sum is \(n\).
Taking the modulus of both sides:
\(\left| \prod_{k=1}^{n-1} (1 - z_k) \right| = |n| = n\).
Using the property that the modulus of a product is the product of the moduli, we get:
\(\prod_{k=1}^{n-1} |1 - z_k| = n\). This proves the conjecture.

**Part f**
The factored form of \(w^n - 1\) evaluated at \(w\) on the unit circle is:
\(w^n - 1 = \prod_{k=0}^{n-1} (w - z_k)\).
Taking the modulus of both sides:
\(|w^n - 1| = \left| \prod_{k=0}^{n-1} (w - z_k) \right| = \prod_{k=0}^{n-1} |w - z_k| = Q_n(\theta)\).
Substituting \(w = e^{i\theta}\):
\(Q_n(\theta) = |e^{in\theta} - 1|\).
We can factor out \(e^{i\frac{n\theta}{2}}\):
\(|e^{in\theta} - 1| = \left| e^{i\frac{n\theta}{2}} \left( e^{i\frac{n\theta}{2}} - e^{-i\frac{n\theta}{2}} \right) \right|\).
Since \(|e^{i\frac{n\theta}{2}}| = 1\) and \(e^{i\frac{n\theta}{2}} - e^{-i\frac{n\theta}{2}} = 2i\sin\left(\frac{n\theta}{2}\right)\), we have:
\(Q_n(\theta) = \left| 2i\sin\left(\frac{n\theta}{2}\right) \right| = 2\left|\sin\left(\frac{n\theta}{2}\right)\right|\).

**Part g**
The maximum possible value of \(Q_n(\theta) = 2\left|\sin\left(\frac{n\theta}{2}\right)\right|\) is 2, since the maximum of the sine function is 1.
This maximum occurs when:
\(\left|\sin\left(\frac{n\theta}{2}\right)\right| = 1 \implies \frac{n\theta}{2} = \frac{\pi}{2} + k\pi\) for \(k \in \mathbb{Z}\)
\(\theta = \frac{\pi}{n} + \frac{2k\pi}{n} = \frac{(2k+1)\pi}{n}\).
For \( \theta \in [0, 2\pi) \), the values are \(\theta = \frac{(2k+1)\pi}{n}\) for \(k = 0, 1, \dots, n-1\).

**Part a** [3 marks]
- **A1** for writing down the roots in Cartesian form: \(z_0 = 1, z_1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, z_2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).
- **A1** for calculating \(|1 - z_1| = \sqrt{3}\).
- **A1** for showing \(|1 - z_1| \cdot |1 - z_2| = 3\).

**Part b** [4 marks]
- **A1** for identifying the other three roots as \(i\), \(-1\), \(-i\).
- **A1** for calculating \(|1 - i| = \sqrt{2}\) and \(|1 + i| = \sqrt{2}\).
- **A1** for calculating \(|1 - (-1)| = 2\).
- **A1** for multiplying the values together to obtain \(4\).

**Part c** [4 marks]
- **A2** for stating the conjecture: \(P_n = n\).
- **A1** for stating that for \(n = 2\), the roots are \(1\) and \(-1\).
- **A1** for showing that \(|1 - (-1)| = 2\) which matches the conjecture.

**Part d** [5 marks]
- **M1** for expanding \((z-1)(z^{n-1} + z^{n-2} + \dots + z + 1)\).
- **A1** for showing algebraic cancellations leading to \(z^n - 1\).
- **M1** for expressing \(z^n - 1\) in factored form \((z-1)\prod_{k=1}^{n-1} (z-z_k)\).
- **A1** for equating the two expressions.
- **R1** for explaining that dividing by \(z - 1\) (valid since \(z \neq 1\)) yields the result.

**Part e** [4 marks]
- **M1** for substituting \(z = 1\) (or taking the limit as \(z \to 1\)) into the identity.
- **A1** for evaluating the RHS to get \(n\).
- **A1** for taking the modulus of both sides.
- **A1** for using the property of moduli to conclude \(\prod_{k=1}^{n-1} |1 - z_k| = n\).

**Part f** [5 marks]
- **M1** for writing the product as \(|w^n - 1|\).
- **A1** for substituting \(w = e^{i\theta}\) to get \(|e^{in\theta} - 1|\).
- **M1** for factoring out \(e^{i\frac{n\theta}{2}}\).
- **A1** for using \(|e^{i\frac{n\theta}{2}}| = 1\).
- **A1** for applying Euler's relation to get the final form \(2\left|\sin\left(\frac{n\theta}{2}\right)\right|\).

**Part g** [3 marks]
- **A1** for stating the maximum value is 2.
- **M1** for setting \(\frac{n\theta}{2} = \frac{\pi}{2} + k\pi\).
- **A1** for finding \(\theta = \frac{(2k+1)\pi}{n}\) for \(k = 0, 1, \dots, n-1\).