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To find the term independent of \( x \), we use the general term of the binomial expansion: \( T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{k}{x}\right)^r \). This simplifies to: \( T_{r+1} = \binom{6}{r} 2^{6-r} (-k)^r x^{12-3r} \). For the term to be independent of \( x \), the exponent of \( x \) must be 0: \( 12 - 3r = 0 \implies r = 4 \). Substituting \( r = 4 \) into the coefficient expression: \( \binom{6}{4} 2^{6-4} (-k)^4 = 15 \times 4 \times k^4 = 60k^4 \). We are given that this coefficient is 240: \( 60k^4 = 240 \implies k^4 = 4 \). Since \( k > 0 \), we take the positive real root: \( k = \sqrt{2} \).

M1: Attempt to write down the general term of the expansion with correct exponents of \( x \). A1: Correct simplification of the exponent of \( x \) to \( 12-3r \) and setting \( 12-3r = 0 \) to find \( r = 4 \). M1: Substituting \( r = 4 \) into the general term. A1: Correctly evaluating the coefficient as \( 60k^4 \). M1: Setting \( 60k^4 = 240 \) and attempting to solve for \( k \). A1: \( k^4 = 4 \). A1: \( k = \sqrt{2} \) (rejecting negative and non-real roots since \( k > 0 \)).

(a) Let \( u = e^x \). The equation \( f(x) = 0 \) becomes: \( u^2 - 3u - 4 = 0 \). Factoring the quadratic: \( (u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1 \). Since \( u = e^x \) and \( e^x > 0 \) for all real \( x \), we reject \( u = -1 \). Therefore, \( e^x = 4 \implies x = \ln 4 \). (b) For \( g(x) = \ln(f(x) + 4) \) to be defined, we require: \( f(x) + 4 > 0 \). Substituting \( f(x) \): \( (e^{2x} - 3e^x - 4) + 4 > 0 \implies e^{2x} - 3e^x > 0 \). Factoring the expression: \( e^x(e^x - 3) > 0 \). Since \( e^x > 0 \) for all \( x \), we must have: \( e^x - 3 > 0 \implies e^x > 3 \implies x > \ln 3 \).

(a) M1: Substitution of \( u = e^x \) (or equivalent) to form a quadratic equation. A1: Correct factors \( (e^x - 4)(e^x + 1) = 0 \). R1: Stating that \( e^x = -1 \) has no real solutions. A1: \( x = \ln 4 \). (b) M1: Setting the argument of the logarithm to be strictly positive: \( f(x) + 4 > 0 \). A1: Simplifying to \( e^x(e^x - 3) > 0 \). A1: Correct domain \( x > \ln 3 \).

First, we find the \( y \)-coordinate of the point of tangency by substituting \( x = 2 \) into the curve equation: \( y = 2 \ln(2(2) - 3) = 2 \ln(1) = 0 \). So, the point of tangency is \( (2, 0) \). Next, we find the derivative \( \frac{dy}{dx} \) using the product rule and chain rule: Let \( u = x \implies u' = 1 \), and \( v = \ln(2x-3) \implies v' = \frac{2}{2x-3} \). Then, \( \frac{dy}{dx} = 1 \cdot \ln(2x-3) + x \cdot \frac{2}{2x-3} = \ln(2x-3) + \frac{2x}{2x-3} \). Now, evaluate the gradient of the tangent at \( x = 2 \): \( m = \ln(2(2)-3) + \frac{2(2)}{2(2)-3} = \ln(1) + \frac{4}{1} = 4 \). Using the point-gradient form, the equation of the tangent line is: \( y - 0 = 4(x - 2) \implies y = 4x - 8 \). This is in the required form \( ay = bx + c \) where \( a = 1 \), \( b = 4 \), and \( c = -8 \).

A1: Correctly finding the \( y \)-coordinate as \( y = 0 \). M1: Attempting to use the product rule to find the derivative. A1: Correct derivative expression \( \frac{dy}{dx} = \ln(2x-3) + \frac{2x}{2x-3} \). M1: Substituting \( x = 2 \) into their derivative to find the gradient. A1: Correct gradient \( m = 4 \). M1: Attempting to use the point-gradient formula with their point and gradient. A1: Correct equation in the specified form \( y = 4x - 8 \) (or any integer multiple).

(a) Using the cosine rule on triangle \( ABC \): \( AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\widehat{ABC}) \). Substituting the given values: \( AC^2 = 5^2 + 7^2 - 2(5)(7)\left(\frac{1}{5}\right) = 25 + 49 - 14 = 60 \). Thus, \( AC = \sqrt{60} = 2\sqrt{15} \text{ cm} \). (b) Since the angle \( \widehat{ABC} \) is inside a triangle and its cosine is positive, it must be an acute angle. Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \( \sin(\widehat{ABC}) = \sqrt{1 - \cos^2(\widehat{ABC})} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5} \). (c) The area of the triangle is given by: \( \text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\widehat{ABC}) \). Substituting our values: \( \text{Area} = \frac{1}{2} \cdot 5 \cdot 7 \cdot \frac{2\sqrt{6}}{5} = 7\sqrt{6} \text{ cm}^2 \).

(a) M1: Attempt to use the cosine rule with correct substitution of given values. A1: Correct calculation of \( AC^2 = 60 \). A1: \( AC = 2\sqrt{15} \) (accept \( \sqrt{60} \)). (b) M1: Attempt to use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). A1: \( \sin(\widehat{ABC}) = \frac{2\sqrt{6}}{5} \) (accept \( \frac{\sqrt{24}}{5} \)). (c) M1: Attempt to use the area of a triangle formula \( \frac{1}{2}ab\sin C \). A1: Correct area \( 7\sqrt{6} \) (must be exact).

(a) Since the sum of all probabilities in a probability distribution must equal 1: \( a + b + 0.3 + 0.2 = 1 \implies a + b = 0.5 \) (Equation 1). The expectation \( \text{E}(X) \) is defined as \( \sum x \cdot P(X = x) \): \( \text{E}(X) = 1(a) + 2(b) + 3(0.3) + 4(0.2) = 2.4 \implies a + 2b + 0.9 + 0.8 = 2.4 \implies a + 2b = 0.7 \) (Equation 2). (b) We solve the system of equations by subtracting Equation 1 from Equation 2: \( (a + 2b) - (a + b) = 0.7 - 0.5 \implies b = 0.2 \). Substituting \( b = 0.2 \) back into Equation 1: \( a + 0.2 = 0.5 \implies a = 0.3 \).

(a) M1: Setting sum of probabilities to 1. A1: Correct simplified equation: \( a + b = 0.5 \). M1: Attempting to write the expression for expectation. A1: Correct simplified equation: \( a + 2b = 0.7 \). (b) M1: Attempting to solve the system of linear equations. A1: \( b = 0.2 \). A1: \( a = 0.3 \).

(a) Using the chain rule on \( g(x) = (x^2 + 1)^{1/2} \): \( g'(x) = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \). (b) We observe that \( f(x) = \frac{2x}{\sqrt{x^2+1}} = 2 g'(x) \). Therefore, the antiderivative of \( f(x) \) is \( 2g(x) = 2\sqrt{x^2+1} \). Using the fundamental theorem of calculus: \( \int_{0}^{\sqrt{3}} f(x) dx = \left[ 2\sqrt{x^2+1} \right]_{0}^{\sqrt{3}} = 2\sqrt{(\sqrt{3})^2 + 1} - 2\sqrt{0^2 + 1} = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2 \).

(a) M1: Applying chain rule to find \( g'(x) \). A1: Correct derivative \( \frac{x}{\sqrt{x^2+1}} \). (b) M1: Recognizing that \( f(x) = 2g'(x) \) or attempting integration by substitution with \( u = x^2+1 \). A1: Finding the antiderivative \( 2\sqrt{x^2+1} \). M1: Substituting limits \( 0 \) and \( \sqrt{3} \) into their antiderivative. A1: Correct calculation of upper limit value \( 4 \) and lower limit value \( 2 \). A1: Final exact answer of \( 2 \).

Using the power law of logarithms: \( 2\log_3(x-1) = \log_3(x-1)^2 \). Substitute this back into the equation: \( \log_3(x-1)^2 - \log_3(x+5) = 1 \). Apply the quotient law of logarithms: \( \log_3\left(\frac{(x-1)^2}{x+5}\right) = 1 \). Convert the logarithmic equation to its exponential form: \( \frac{(x-1)^2}{x+5} = 3^1 \implies (x-1)^2 = 3(x+5) \). Expand both sides: \( x^2 - 2x + 1 = 3x + 15 \). Rearrange into a standard quadratic form: \( x^2 - 5x - 14 = 0 \). Factoring the quadratic: \( (x-7)(x+2) = 0 \). This gives potential solutions \( x = 7 \) or \( x = -2 \). Since the original logarithmic terms require \( x - 1 > 0 \implies x > 1 \), we reject \( x = -2 \). Thus, the only valid solution is \( x = 7 \).

M1: Applying power rule of logarithms to get \( \log_3(x-1)^2 \). M1: Applying quotient rule to get \( \log_3\left(\frac{(x-1)^2}{x+5}\right) \). M1: Converting logarithmic equation to exponential form. A1: Correct quadratic equation in expanded form: \( x^2 - 5x - 14 = 0 \). M1: Attempting to solve the quadratic equation. A1: Finding \( x = 7 \) and \( x = -2 \). R1: Rejecting \( x = -2 \) because \( x > 1 \) and stating final answer \( x = 7 \).

(a) We complete the square for \( f(x) \): \( f(x) = x^2 - 4x + 4 - 4 + 7 = (x-2)^2 + 3 \). (b) From part (a), the vertex of the graph of \( f \) is at \( (2, 3) \). We apply the geometric transformations to this vertex: 1. Reflection in the \( y \)-axis changes the sign of the \( x \)-coordinate, so the vertex \( (2, 3) \) becomes \( (-2, 3) \). 2. Translation by the vector \( \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) adds 3 to the \( x \)-coordinate and subtracts 2 from the \( y \)-coordinate, so the new vertex coordinates are \( (-2+3, 3-2) = (1, 1) \).

(a) M1: Attempting to complete the square. A1: Correct form \( (x-2)^2 + 3 \). (b) A1: Correctly identifying the initial vertex of \( f \) as \( (2, 3) \). M1: Correctly applying the reflection in the \( y \)-axis to find intermediate coordinates \( (-2, 3) \). M1: Correctly applying translation vector to their intermediate vertex. A1: Vertex of \( g \) has \( x \)-coordinate 1. A1: Vertex of \( g \) has \( y \)-coordinate 1.

(a) To find the maximum point, we use the quotient rule to differentiate \( f(x) \):
\( f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \).
Setting \( f'(x) = 0 \) gives \( 1 - \ln x = 0 \implies x = e \).
Evaluating the function at \( x = e \):
\( f(e) = \frac{\ln e}{e} = \frac{1}{e} \).
Since \( f'(x) > 0 \) for \( 0 < x < e \) and \( f'(x) < 0 \) for \( x > e \), the point \( (e, \frac{1}{e}) \) is a local maximum.

(b) We differentiate \( f'(x) \) again using the quotient rule:
\( f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \ln x) \cdot 2x}{x^4} = \frac{-x - 2x + 2x\ln x}{x^4} = \frac{2\ln x - 3}{x^3} \).
Setting \( f''(x) = 0 \) gives \( 2\ln x - 3 = 0 \implies \ln x = \frac{3}{2} \implies x = e^{3/2} \).
Since \( f''(x) < 0 \) for \( 0 < x < e^{3/2} \) and \( f''(x) > 0 \) for \( x > e^{3/2} \), the concavity changes at this point, which confirms a point of inflection at \( x = e^{3/2} \).

(c) The region \( R \) is bounded by the curve, the x-axis (where \( y=0 \), intersecting at \( x=1 \) since \( \ln 1 = 0 \)), and the line \( x = e \).
Area \( A = \int_1^e \frac{\ln x}{x} dx \).
Let \( u = \ln x \), then \( du = \frac{1}{x} dx \).
When \( x = 1 \), \( u = 0 \). When \( x = e \), \( u = 1 \).
\( A = \int_0^1 u \, du = \left[ \frac{1}{2}u^2 \right]_0^1 = \frac{1}{2} \).

(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 for \( \int \frac{(\ln x)^2}{x^2} dx \).
Let \( u = (\ln x)^2 \) and \( dv = x^{-2} dx \).
Then \( du = \frac{2\ln x}{x} dx \) and \( v = -\frac{1}{x} \).
\( \int \frac{(\ln x)^2}{x^2} dx = -\frac{(\ln x)^2}{x} + 2 \int \frac{\ln x}{x^2} dx \).
Using integration by parts again on the second integral with \( U = \ln x \) and \( dV = x^{-2} dx \):
\( dU = \frac{1}{x} dx \) and \( V = -\frac{1}{x} \).
\( \int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} + \int x^{-2} dx = -\frac{\ln x}{x} - \frac{1}{x} \).
Thus, the general antiderivative is \( -\frac{(\ln x)^2}{x} - \frac{2\ln x}{x} - \frac{2}{x} \).
Evaluating this from \( 1 \) to \( e \):
At \( x = e \): \( -\frac{1}{e} - \frac{2}{e} - \frac{2}{e} = -\frac{5}{e} \).
At \( x = 1 \): \( 0 - 0 - 2 = -2 \).
So, \( V = \pi \left[ -\frac{5}{e} - (-2) \right] = \pi \left( 2 - \frac{5}{e} \right) \).

(a) [5 marks]
- M1 for attempting to use the quotient rule to find \( f'(x) \).
- A1 for obtaining the correct derivative: \( f'(x) = \frac{1 - \ln x}{x^2} \).
- M1 for setting \( f'(x) = 0 \) and solving for \( x \).
- A1 for \( x = e \).
- A1 for \( y = \frac{1}{e} \).

(b) [5 marks]
- M1 for attempting to differentiate \( f'(x) \) using the quotient rule.
- 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 obtaining \( x = e^{3/2} \).
- R1 for explaining that \( f''(x) \) changes sign across \( x = e^{3/2} \), confirming a point of inflection.

(c) [4 marks]
- M1 for identifying the lower limit \( x = 1 \) and setting up the integral: \( \int_1^e \frac{\ln x}{x} dx \).
- M1 for using substitution, e.g., \( u = \ln x \).
- A1 for correct antiderivative: \( \frac{1}{2}(\ln x)^2 \).
- A1 for the correct area of \( \frac{1}{2} \).

(d) [4 marks]
- M1 for setting up the volume integral: \( V = \pi \int_1^e \frac{(\ln x)^2}{x^2} dx \).
- M1 for using integration by parts.
- A1 for finding the correct antiderivative: \( -\frac{(\ln x)^2 + 2\ln x + 2}{x} \).
- A1 for substituting limits correctly to show the final result: \( \pi \left( 2 - \frac{5}{e} \right) \).

(a) Since \( f(x) \) is a probability density function, the total area under the curve must equal 1:
\( \int_0^2 k x^2 (2 - x) dx = 1 \)
\( k \int_0^2 (2x^2 - x^3) dx = 1 \)
\( k \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 = 1 \)
\( k \left( \frac{16}{3} - 4 \right) = 1 \)
\( k \left( \frac{4}{3} \right) = 1 \implies k = \frac{3}{4} \).

(b) The expected value is given by:
\( E(X) = \int_0^2 x f(x) dx = \frac{3}{4} \int_0^2 x(2x^2 - x^3) dx = \frac{3}{4} \int_0^2 (2x^3 - x^4) dx \)
\( E(X) = \frac{3}{4} \left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_0^2 \)
\( E(X) = \frac{3}{4} \left( 8 - \frac{32}{5} \right) = \frac{3}{4} \left( \frac{8}{5} \right) = \frac{6}{5} \text{ or } 1.2 \).

(c) The mode is the value of \( x \) that maximizes \( f(x) \) in the interval \( [0, 2] \).
\( f'(x) = \frac{3}{4}(4x - 3x^2) = 0 \implies x(4 - 3x) = 0 \).
Since \( x \in [0, 2] \), we have critical points at \( x = 0 \) and \( x = \frac{4}{3} \).
Since \( f''(x) = \frac{3}{4}(4 - 6x) \), at \( x = \frac{4}{3} \) we have \( f''(\frac{4}{3}) = -3 < 0 \), which confirms a local maximum.
Thus, the mode is \( \frac{4}{3} \).

(d) \( P(1 \le X \le 2) = \int_1^2 \frac{3}{4} (2x^2 - x^3) dx \)
\( = \frac{3}{4} \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_1^2 \)
Evaluating at the limits:
At \( x = 2 \): \( \frac{3}{4} \left( \frac{16}{3} - 4 \right) = 1 \).
At \( x = 1 \): \( \frac{3}{4} \left( \frac{2}{3} - \frac{1}{4} \right) = \frac{3}{4} \left( \frac{5}{12} \right) = \frac{5}{16} \).
Therefore, \( P(1 \le X \le 2) = 1 - \frac{5}{16} = \frac{11}{16} \text{ or } 0.6875 \).

(e) First, we find \( E(X^2) \):
\( E(X^2) = \int_0^2 x^2 f(x) dx = \frac{3}{4} \int_0^2 (2x^4 - x^5) dx \)
\( E(X^2) = \frac{3}{4} \left[ \frac{2x^5}{5} - \frac{x^6}{6} \right]_0^2 = \frac{3}{4} \left( \frac{64}{5} - \frac{64}{6} \right) = \frac{3}{4} \cdot 64 \left( \frac{1}{30} \right) = \frac{8}{5} \).
Now we find the variance of \( X \):
\( \text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{8}{5} - \left( \frac{6}{5} \right)^2 = \frac{40}{25} - \frac{36}{25} = \frac{4}{25} \).
Using the properties of variance:
\( \text{Var}(Y) = \text{Var}(2X + 3) = 2^2 \text{Var}(X) = 4 \left( \frac{4}{25} \right) = \frac{16}{25} \text{ or } 0.64 \).

(a) [4 marks]
- M1 for setting up the integral equal to 1: \( \int_0^2 k x^2 (2 - x) dx = 1 \).
- A1 for correct integration of the polynomial: \( k [\frac{2x^3}{3} - \frac{x^4}{4}]_0^2 \).
- M1 for substituting limits correctly.
- A1 for showing clearly that \( k = 3/4 \).

(b) [4 marks]
- M1 for setting up \( E(X) = \int_0^2 x f(x) dx \).
- A1 for correct algebraic simplification of integrand: \( \frac{3}{4}(2x^3 - x^4) \).
- M1 for integration and limit substitution.
- A1 for \( 6/5 \) (or 1.2).

(c) [3 marks]
- M1 for finding the derivative of \( f(x) \) and setting to 0.
- A1 for solving to get \( x = 4/3 \).
- R1 for confirming it is a maximum (e.g. via second derivative or table of signs).

(d) [3 marks]
- M1 for setting up \( \int_1^2 f(x) dx \).
- M1 for substituting limits 1 and 2 into the antiderivative.
- A1 for \( 11/16 \) (or 0.6875).

(e) [4 marks]
- M1 for attempting to find \( E(X^2) = \int_0^2 x^2 f(x) dx \).
- A1 for \( E(X^2) = 8/5 \) (or 1.6).
- M1 for using \( \text{Var}(X) = E(X^2) - [E(X)]^2 \) to get \( 4/25 \).
- A1 for applying \( \text{Var}(2X+3) = 4\text{Var}(X) \) to find \( 16/25 \) (or 0.64).

(a) We use the double-angle identity for cosine: \( \cos(2x) = 1 - 2 \sin^2 x \).
Substituting this into the function:
\( f(x) = 2(1 - 2 \sin^2 x) - 4 \sin x + 1 \)
\( f(x) = 2 - 4 \sin^2 x - 4 \sin x + 1 \)
\( f(x) = -4 \sin^2 x - 4 \sin x + 3 \).
Thus, \( a = -4 \), \( b = -4 \), \( c = 3 \).

(b) To solve \( f(x) = 0 \):
\( -4 \sin^2 x - 4 \sin x + 3 = 0 \)
Multiplying by \(-1\):
\( 4 \sin^2 x + 4 \sin x - 3 = 0 \)
Letting \( u = \sin x \), we get the quadratic equation:
\( 4u^2 + 4u - 3 = 0 \)
Factorizing this quadratic:
\( (2u - 1)(2u + 3) = 0 \)
This yields:
\( \sin x = \frac{1}{2} \) or \( \sin x = -\frac{3}{2} \).
Since the range of the sine function is \( [-1, 1] \), the equation \( \sin x = -\frac{3}{2} \) has no real solutions.
For \( \sin x = \frac{1}{2} \) in the interval \( 0 \le x \le 2\pi \):
\( x = \frac{\pi}{6} \) or \( x = \frac{5\pi}{6} \).

(c) To find stationary points, we find \( f'(x) \):
\( f'(x) = -8 \sin x \cos x - 4 \cos x = -4 \cos x (2 \sin x + 1) \).
Setting \( f'(x) = 0 \):
Either \( \cos x = 0 \) or \( 2 \sin x + 1 = 0 \implies \sin x = -\frac{1}{2} \).
In the interval \( [0, 2\pi] \):
From \( \cos x = 0 \): \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).
From \( \sin x = -\frac{1}{2} \): \( x = \frac{7\pi}{6}, \frac{11\pi}{6} \).
We calculate the corresponding \( y \)-coordinates using \( f(x) = -4\sin^2 x - 4\sin x + 3 \):
- For \( x = \frac{\pi}{2} \): \( f(\pi/2) = -4(1)^2 - 4(1) + 3 = -5 \).
- For \( x = \frac{3\pi}{2} \): \( f(3\pi/2) = -4(-1)^2 - 4(-1) + 3 = 3 \).
- For \( x = \frac{7\pi}{6} \): \( f(7\pi/6) = -4(-1/2)^2 - 4(-1/2) + 3 = 4 \).
- For \( x = \frac{11\pi}{6} \): \( f(11\pi/6) = -4(-1/2)^2 - 4(-1/2) + 3 = 4 \).
Using the sign of the derivative or a table of values:
- Local minima occur at \( (\frac{\pi}{2}, -5) \) and \( (\frac{3\pi}{2}, 3) \).
- Local maxima occur at \( (\frac{7\pi}{6}, 4) \) and \( (\frac{11\pi}{6}, 4) \).

(d) To find the range of \( f(x) \) on the closed interval \( [0, 2\pi] \), we also check the boundary values:
\( f(0) = f(2\pi) = 3 \).
Comparing all y-values:
The global minimum is \(-5\) (at \( x = \pi/2 \)).
The global maximum is \(4\) (at \( x = 7\pi/6 \) and \( 11\pi/6 \)).
Therefore, the range of the function is \( [-5, 4] \) (or \( -5 \le f(x) \le 4 \)).

(a) [3 marks]
- M1 for using the identity \( \cos(2x) = 1 - 2 \sin^2 x \).
- A1 for correct substitution: \( 2(1-2\sin^2 x) - 4\sin x + 1 \).
- A1 for simplifying to show \( -4 \sin^2 x - 4 \sin x + 3 \).

(b) [6 marks]
- M1 for expressing the equation in quadratic form: \( 4 \sin^2 x + 4 \sin x - 3 = 0 \).
- A1 for factorizing correctly: \( (2 \sin x - 1)(2 \sin x + 3) = 0 \).
- A1 for finding \( \sin x = 1/2 \).
- R1 for stating \( \sin x = -3/2 \) has no solution.
- A1 for \( x = \frac{\pi}{6} \).
- A1 for \( x = \frac{5\pi}{6} \).

(c) [5 marks]
- M1 for finding the derivative of \( f(x) \): \( f'(x) = -8 \sin x \cos x - 4 \cos x \).
- A1 for finding the critical values in the interval: \( x = \pi/2, 3\pi/2, 7\pi/6, 11\pi/6 \).
- M1 for finding the corresponding y-values: \( -5, 3, 4, 4 \).
- A1 for identifying local maxima at \( (7\pi/6, 4) \) and \( (11\pi/6, 4) \).
- A1 for identifying local minima at \( (\pi/2, -5) \) and \( (3\pi/2, 3) \).

(d) [4 marks]
- M1 for evaluating boundary points: \( f(0) = f(2\pi) = 3 \).
- M1 for comparing all stationary values and boundaries.
- A2 for the correct range: \( [-5, 4] \) (or \( -5 \le f(x) \le 4 \)).

(a) Let \(W\) be the weight of an apple. We require \(P(W > 165)\). Using the GDC with \(\mu = 150\) and \(\sigma = 12\), we find: \(P(W > 165) \approx 0.105649\). To 3 significant figures, this is \(0.106\). (b) We are given \(P(W < w) = 0.15\). Using the inverse normal function on the GDC: \(w \approx 137.563\). To 3 significant figures, \(w = 138\). (c) Let \(Y\) be the number of small apples in the sample of 5. Then \(Y \sim B(5, 0.15)\). We want to find \(P(Y \ge 2)\). Using the binomial cumulative distribution function on the GDC: \(P(Y \ge 2) = 1 - P(Y \le 1) \approx 1 - 0.83521 = 0.16479\). To 3 significant figures, this is \(0.165\).

(a) M1 for setting up the normal probability expression, e.g., \(P(W > 165)\). A1 for \(0.106\). (b) M1 for setting up the inverse normal equation, e.g., \(P(W < w) = 0.15\). A1 for \(138\). (c) M1 for identifying the binomial distribution, e.g., \(Y \sim B(5, 0.15)\), and the correct inequality \(P(Y \ge 2)\). A1 for \(0.165\).

(a) We find the intersection points by setting the two equations equal to each other: \(\mathrm{e}^{0.5x} = 4 - x^2\). Using a GDC to find the roots of \(\mathrm{e}^{0.5x} + x^2 - 4 = 0\), we get: \(x_1 \approx -1.9009\) and \(x_2 \approx 1.4069\). To 3 significant figures, the \(x\)-coordinates are \(-1.90\) and \(1.41\). (b) The upper curve is \(y = 4 - x^2\) and the lower curve is \(y = \mathrm{e}^{0.5x}\). The area \(A\) is given by: \(A = \int_{-1.9009}^{1.4069} (4 - x^2 - \mathrm{e}^{0.5x}) \mathrm{d}x\). Evaluating this integral on the GDC gives: \(A \approx 6.7473\). To 3 significant figures, the area is \(6.75\).

(a) M1 for setting up the equation \(\mathrm{e}^{0.5x} = 4 - x^2\). A1 for both correct values: \(-1.90\) and \(1.41\). (b) M1 for identifying the correct upper and lower curves. M1 for setting up the definite integral with their limits. A1 for correct integration process (or GDC syntax). A1 for \(6.75\).

(a) Using the formula for the \(n\)-th term of an arithmetic sequence: \(u_{20} = u_1 + 19d = 5 + 19(3) = 62\). (b) We want to find the smallest integer \(k\) such that \(v_k > u_k\). The terms are given by: \(u_k = 5 + (k-1)3 = 3k + 2\) and \(v_k = 2(1.2)^{k-1}\). We set up the inequality on the GDC: \(2(1.2)^{k-1} > 3k + 2\). Using the table of values or a graph: For \(k = 19\): \(u_{19} = 59\) and \(v_{19} = 2(1.2)^{18} \approx 53.24\) (since \(53.24 < 59\), this is false). For \(k = 20\): \(u_{20} = 62\) and \(v_{20} = 2(1.2)^{19} \approx 63.89\) (since \(63.89 > 62\), this is true). Thus, the smallest value of \(k\) is \(20\).

(a) M1 for correct substitution into the arithmetic term formula: \(5 + 19(3)\). A1 for \(62\). (b) M1 for setting up the inequality \(2(1.2)^{k-1} > 3k + 2\). M1 for showing evidence of evaluating or comparing terms around \(k=19\) and \(k=20\). A1 for showing \(v_{19} < u_{19}\) and \(v_{20} > u_{20}\). A1 for the final answer \(20\).

(a) Applying the Sine Rule: \(\frac{\sin(A\widehat{C}B)}{AB} = \frac{\sin(B\widehat{A}C)}{BC}\) which gives \(\frac{\sin(C)}{9} = \frac{\sin(48^\circ)}{7}\). This yields \(\sin(C) = \frac{9 \sin(48^\circ)}{7} \approx 0.95547\). The acute angle is \(C_1 = \arcsin(0.95547) \approx 72.83^\circ\). The obtuse angle is \(C_2 = 180^\circ - 72.83^\circ \approx 107.17^\circ\). Thus, the two possible values are \(72.8^\circ\) and \(107^\circ\) (to 3 significant figures). (b) Since \(A\widehat{C}B\) is acute, we use \(C \approx 72.83^\circ\). The third angle \(B\) is: \(B = 180^\circ - (48^\circ + 72.83^\circ) = 59.17^\circ\). The area of the triangle is: \(\text{Area} = \frac{1}{2} \cdot BC \cdot AB \cdot \sin(B) = \frac{1}{2} (7)(9) \sin(59.17^\circ) \approx 27.048\text{ cm}^2\). To 3 significant figures, the area is \(27.0\text{ cm}^2\).

(a) M1 for applying the sine rule. A1 for the acute angle \(72.8^\circ\). A1 for the obtuse angle \(107^\circ\). (b) M1 for finding the angle \(B = 59.2^\circ\) (or \(59.17^\circ\)). M1 for substituting into the area formula \(\frac{1}{2}ac\sin B\). A1 for \(27.0\text{ cm}^2\) (accept \(27\) or \(27.0\)).

(a) The amplitude is \(a = \frac{\text{Max} - \text{Min}}{2} = \frac{14.6 - 8.2}{2} = 3.2\). The vertical shift is \(c = \frac{\text{Max} + \text{Min}}{2} = \frac{14.6 + 8.2}{2} = 11.4\). The time from maximum to minimum is \(6.2\) hours, which is half the period. Thus, the period \(T = 12.4\) hours. This gives \(b = \frac{2\pi}{T} = \frac{2\pi}{12.4} \approx 0.506708\). To 3 significant figures, \(a = 3.2\), \(b = 0.507\), and \(c = 11.4\). (b) We set up the inequality \(3.2 \cos(0.5067 t) + 11.4 \ge 11.0\) for \(0 \le t \le 12\). Solving the boundary equation \(3.2 \cos(0.5067 t) + 11.4 = 11.0\) using GDC gives \(t_1 \approx 3.3473\) and \(t_2 \approx 9.0527\). For \(0 \le t \le 12\), the depth is at least \(11.0\text{ m}\) during the intervals \([0, 3.3473]\) and \([9.0527, 12]\). The total duration is \(3.3473 + (12 - 9.0527) = 6.2946\) hours. To 3 significant figures, the total time is \(6.29\) hours.

(a) A1 for \(a = 3.2\) and \(c = 11.4\). M1 for using half-period to find \(b\). A1 for \(b \approx 0.507\) (accept \(\frac{\pi}{6.2}\)). (b) M1 for setting up the equation \(d(t) = 11.0\) and finding the boundary times \(t_1 \approx 3.35\) and \(t_2 \approx 9.05\). M1 for identifying the correct intervals \([0, 3.35]\) and \([9.05, 12]\). A1 for \(6.29\) (accept \(6.29\) or \(6.30\) depending on rounding of intermediate values).

(a) Using the product rule and chain rule: \(f'(x) = 1 \cdot \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\). (b) At \(x = 2\): \(f(2) = 2 \ln(5) \approx 3.21888\), and \(f'(2) = \ln(5) + \frac{8}{5} \approx 3.20944\). The equation of the tangent is \(y - 2\ln(5) = (\ln(5) + 1.6)(x - 2)\). In slope-intercept form: \(y \approx 3.20944x - 3.2000\). To 3 significant figures, this is \(y = 3.21x - 3.20\). (c) To find the other intersection point, we solve: \(x \ln(x^2 + 1) = (\ln(5) + 1.6)x - 3.2000\). Using a numerical solver on the GDC, we find the non-tangent intersection point to be \(x \approx -6.3311\). To 3 significant figures, the \(x\)-coordinate is \(-6.33\).

(a) M1 for using the product rule. A1 for \(f'(x) = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\). (b) M1 for finding both \(f(2)\) and \(f'(2)\) numerically or analytically. A1 for the tangent equation, e.g., \(y = 3.21x - 3.20\) (or exact equivalent). (c) M1 for equating the function and the tangent equation. A1 for solving on GDC to get \(-6.33\).

(a) Entering the data into the GDC, we obtain Pearson's product-moment correlation coefficient: \(r \approx 0.99843\). To 3 significant figures, \(r = 0.998\). (b) From the GDC linear regression function, we find: \(a \approx 35.0151\) and \(b \approx -316.601\). Thus, the regression line equation is \(y = 35.0x - 317\) (to 3 significant figures). (c) Substituting \(x = 24\) into the regression equation: \(y = 35.0151(24) - 316.601 \approx 523.76\). To the nearest dollar, the estimate is \(524\) dollars. (d) The estimate is reliable because: 1. \(x = 24^\circ\text{C}\) lies within the range of the given data (interpolation), which is from \(16^\circ\text{C}\) to \(30^\circ\text{C}\). 2. The correlation coefficient \(r \approx 0.998\) is extremely strong, indicating a highly linear relationship.

(a) A1 for \(0.998\). (b) A1 for \(a = 35.0\), A1 for \(b = -317\) (accept \(y = 35.0x - 317\)). (c) A1 for \(524\) (accept \(523.76\)). (d) R1 for stating it is reliable because it is interpolation (or within range). R1 for stating it is reliable because the correlation is very strong (or \(r\) is close to 1).

(a) The composite function is defined as \((g \circ f)(x) = g(f(x)) = g(\ln(x-2)) = \mathrm{e}^{2\ln(x-2) - 1}\). For \(g(f(x))\) to be defined, \(x\) must be in the domain of \(f\), so the domain is \(x > 2\). Since \(x > 2\), the range of \(f\) is all real numbers. For any real input, \(g\) outputs positive values. Specifically, \((g \circ f)(x) = \frac{(x-2)^2}{\mathrm{e}} > 0\) since \(x \neq 2\). Thus, the range is \(y > 0\). (b) Setting \((g \circ f)(x) = 10\): \(\mathrm{e}^{2\ln(x-2) - 1} = 10\). Taking the natural logarithm of both sides: \(2\ln(x-2) - 1 = \ln(10)\). Simplifying: \(2\ln(x-2) = \ln(10) + 1 \Rightarrow \ln(x-2) = \frac{\ln(10) + 1}{2}\). Converting to exponential form: \(x - 2 = \mathrm{e}^{\frac{\ln(10)+1}{2}} = \sqrt{10\mathrm{e}}\). Thus, \(x = 2 + \sqrt{10\mathrm{e}} \approx 7.2137\). To 3 significant figures, \(x = 7.21\).

(a) A1 for the correct domain: \(x > 2\) (or equivalent). A1 for simplifying the composite expression. A1 for the correct range: \(y > 0\) (or equivalent). (b) M1 for equating the composite function to 10. M1 for correct algebraic steps to solve for \(x\). A1 for \(7.21\) (or exact value \(2 + \sqrt{10\mathrm{e}}\)).

Let \(Z\) be the standard normal variable, so \(Z \sim N(0, 1)\). Using the given probabilities, we can write: \(P(X < 45) = 0.15 \implies P\left(Z < \frac{45 - \mu}{\sigma}\right) = 0.15\). Using a GDC (inverse normal function): \(\frac{45 - \mu}{\sigma} \approx -1.03643...\) (Equation 1). We also know: \(P(X > 75) = 0.10 \implies P(X < 75) = 0.90\), which gives \(P\left(Z < \frac{75 - \mu}{\sigma}\right) = 0.90\). Using a GDC: \(\frac{75 - \mu}{\sigma} \approx 1.28155...\) (Equation 2). Now, we express both equations in terms of \(\mu\) and \(\sigma\): \(\mu - 1.03643\sigma = 45\) and \(\mu + 1.28155\sigma = 75\). Subtracting the first equation from the second equation to eliminate \(\mu\): \((1.28155 - (-1.03643))\sigma = 75 - 45 \implies 2.31798\sigma = 30 \implies \sigma = \frac{30}{2.31798} \approx 12.9423...\). Substituting \(\sigma\) back into the equation for \(\mu\): \(\mu = 75 - 1.28155(12.9423) \approx 58.4138...\). Thus, to 3 significant figures, \(\mu \approx 58.4\) and \(\sigma \approx 12.9\).

**(M1)** for attempting to standardize at least one probability using \(Z = \frac{X - \mu}{\sigma}\). **(A1)** for finding the first \(z\)-value: \(z_1 \approx -1.03643...\) (accept \(-1.04\)). **(A1)** for finding the second \(z\)-value: \(z_2 \approx 1.28155...\) (accept \(1.28\)). **(M1)** for setting up a correct system of two simultaneous equations in terms of \(\mu\) and \(\sigma\). **(A1)** for finding \(\sigma \approx 12.9\) (accept \(12.9423...\)). **(A1)** for finding \(\mu \approx 58.4\) (accept \(58.4138...\)).

(a) Since \(f(t)\) is a probability density function, \(\int_{0}^{6} f(t) \, dt = 1\).
Using integration by parts on \(\int t e^{-0.5t} \, dt\):
Let \(u = t \implies du = dt\)
Let \(dv = e^{-0.5t} \, dt \implies v = -2e^{-0.5t}\)
\(\int t e^{-0.5t} \, dt = -2t e^{-0.5t} - \int -2e^{-0.5t} \, dt = -2t e^{-0.5t} - 4e^{-0.5t} = -2(t+2)e^{-0.5t}\)

Evaluating this from \(0\) to \(6\):
\(\left[ -2(t+2)e^{-0.5t} \right]_{0}^{6} = -16e^{-3} - (-4e^0) = 4 - 16e^{-3}\)

Thus, \(k(4 - 16e^{-3}) = 1 \implies k = \frac{1}{4 - 16e^{-3}} = \frac{e^3}{4e^3 - 16}\). [AG]

(b) The probability is given by \(P(T > 4) = \int_{4}^{6} k t e^{-0.5t} \, dt\).
Using the antiderivative found in part (a):
\(P(T > 4) = k \left[ -2(t+2)e^{-0.5t} \right]_{4}^{6} = k \left( -16e^{-3} - (-12e^{-2}) \right) = k(12e^{-2} - 16e^{-3})\)
Using \(k \approx 0.312165\):
\(P(T > 4) \approx 0.312165 \times (1.62402 - 0.79659) \approx 0.258\) (to 3 s.f.).

(c) The mean lifetime is \(E(T) = \int_{0}^{6} t f(t) \, dt = k \int_{0}^{6} t^2 e^{-0.5t} \, dt\).
Using integration by parts on \(\int t^2 e^{-0.5t} \, dt\):
Let \(u = t^2 \implies du = 2t \, dt\)
Let \(dv = e^{-0.5t} \, dt \implies v = -2e^{-0.5t}\)
\(\int t^2 e^{-0.5t} \, dt = -2t^2 e^{-0.5t} + 4 \int t e^{-0.5t} \, dt = -2t^2 e^{-0.5t} - 8(t+2)e^{-0.5t} = -2e^{-0.5t}(t^2 + 4t + 8)\)

Evaluating this from \(0\) to \(6\):
\(\left[ -2e^{-0.5t}(t^2 + 4t + 8) \right]_{0}^{6} = -2e^{-3}(36 + 24 + 8) - (-16) = 16 - 136e^{-3}\)

So \(E(T) = k (16 - 136e^{-3})\).
Using \(k \approx 0.312165\):
\(E(T) \approx 0.312165 \times (16 - 6.77104) \approx 2.88\) thousand hours.

(d) (i) Let \(X\) be the number of components that last more than 4 thousand hours. \(X \sim B(500, p)\) where \(p \approx 0.258296\).
Expected value \(E(X) = n p = 500 \times 0.258296 \approx 129\) (or \(129.15\)).

(ii) We require \(P(X \ge 140) = 1 - P(X \le 139)\).
Using GDC Binomial Cumulative Distribution:
\(P(X \le 139) \approx 0.85764\)
\(P(X \ge 140) \approx 1 - 0.85764 = 0.142\) (to 3 s.f.).
*(Note: If normal approximation is used with continuity correction, \(\mu = 129.15, \sigma \approx 9.787 \implies P(X \ge 139.5) \approx 0.145\). Without continuity correction, \(P(X \ge 140) \approx 0.134\). Accept these with appropriate working.)*

(a)
- M1: Attempt to integrate \(f(t)\) and set to 1.
- A1: Correctly applying integration by parts once.
- A1: Correct limits evaluation to obtain \(4 - 16e^{-3}\).
- AG: Clearly showing the algebraic steps to reach the final expression.

(b)
- M1: Attempt to integrate \(f(t)\) from 4 to 6 (either analytically or on GDC).
- A1: Correct substitution of limits or setup.
- A1: Correct numerical answer to 3 s.f. (accept 0.258).

(c)
- M1: Attempt to write the integral expression for the expected value.
- M1: Correct application of integration by parts to obtain the correct antiderivative \(-2e^{-0.5t}(t^2 + 4t + 8)\).
- A1: Correct evaluation of the definite integral.
- A1: Correct numerical answer (accept 2.88).

(d)(i)
- M1: Attempt to find \(n p\).
- A1: Correct answer (accept 129 or 129.15).

(d)(ii)
- M1: Identifying the binomial distribution parameters \(n=500, p=0.2583\).
- M1: Recognising the need to compute \(P(X \ge 140)\) as \(1 - P(X \le 139)\).
- A2: Correct answer using exact binomial method (accept 0.142).
- (Award up to A2 for normal approximation method with correct steps: M1 for finding mean/variance, A1 for correct probability: accept 0.145 or 0.134).

(a) Initial velocity is when \(t = 0\):
\(v(0) = 3(0) \cos(0) - e^0 + 2 = -1 + 2 = 1 \text{ m s}^{-1}\).

(b) Acceleration is the derivative of velocity, \(a(t) = v'(t)\).
Using GDC numerical derivative at \(t = 3\), or analytically:
\(v'(t) = 3\cos(0.4t) - 1.2t\sin(0.4t) + 0.1e^{-0.1t}\)
\(a(3) = v'(3) = 3\cos(1.2) - 3.6\sin(1.2) + 0.1e^{-0.3} \approx -2.19 \text{ m s}^{-2}\).

(c) The particle is at rest when \(v(t) = 0\).
Using GDC to solve \(3t \cos(0.4t) - e^{-0.1t} + 2 = 0\) in the interval \(0 \le t \le 10\):
\(t \approx 4.19\) seconds.

(d) Total distance is given by \(\int_{0}^{10} |v(t)| \, dt\).
Using the GDC to evaluate this definite integral directly:
Total distance \(\approx 105\) m (or more precisely \(105.1\) m).

(e) Displacement is given by \(\int_{0}^{10} v(t) \, dt\).
Using the GDC to evaluate this definite integral directly:
Displacement \(\approx -74.1\) m.

(f) Displacement of particle B at \(t = 10\) is:
\(\int_{0}^{10} (-2t + k) \, dt = \left[ -t^2 + kt \right]_{0}^{10} = -100 + 10k\).
Set B's displacement equal to the first particle's displacement:
\(-100 + 10k = -74.086\)
\(10k = 25.914 \implies k \approx 2.59\).

(a)
- M1: Substituting \(t = 0\) into the velocity equation.
- A1: Correct value \(1\).

(b)
- M1: Recognising acceleration as \(v'(t)\).
- A1: Correct derivative formula or clear setup on GDC.
- A1: Correct answer \(-2.19\).

(c)
- M1: Setting \(v(t) = 0\).
- A2: Plotting/sketching or using solver on GDC to identify root.
- A1: Correct value of \(t = 4.19\) s.

(d)
- M1: Recognising total distance is \(\int |v(t)| \, dt\).
- A1: Correct limits from \(0\) to \(10\).
- A2: Correct evaluated distance \(105\) (accept 105.1).

(e)
- M1: Recognising displacement is \(\int v(t) \, dt\).
- A1: Correct evaluated displacement \(-74.1\) (accept \(-74.1\) or \(-74.08\)).

(f)
- M1: Correct integration of \(v_B(t)\) to find displacement in terms of \(k\).
- M1: Equating their expression to their answer from (e).
- A1: Correct value of \(k = 2.59\).

(a) (i) The vertical range of motion is \(84 - 4 = 80\) m.
The amplitude is half the range: \(\frac{80}{2} = 40\).
Since the cabin starts at its minimum height at \(t = 0\), the cosine curve is reflected: \(a = -40\). [AG]

(ii) The vertical shift is the midline of the motion:
\(c = \frac{84 + 4}{2} = 44\). [AG]

(iii) The time to go from minimum to maximum height is 8 minutes, which represents half a full period.
Therefore, the full period \(P = 16\) minutes.
\(b = \frac{2\pi}{P} = \frac{2\pi}{16} = \frac{\pi}{8}\). [AG]

(b) Substituting \(t = 10\) into the model:
\(h(10) = -40 \cos\left(\frac{10\pi}{8}\right) + 44 = -40 \cos\left(\frac{5\pi}{4}\right) + 44\)
\(h(10) = -40 \left(-\frac{\sqrt{2}}{2}\right) + 44 = 20\sqrt{2} + 44 \approx 72.3\) meters.

(c) The rate of change of height is \(h'(t)\):
\(h'(t) = 40 \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{8} t\right) = 5\pi \sin\left(\frac{\pi}{8} t\right)\)
At \(t = 3\):
\(h'(3) = 5\pi \sin\left(\frac{3\pi}{8}\right) \approx 14.5 \text{ m min}^{-1}\).

(d) We solve \(-40 \cos\left(\frac{\pi}{8} t\right) + 44 \ge 60\).
Using GDC to find the intersections of \(y = -40 \cos\left(\frac{\pi}{8} t\right) + 44\) and \(y = 60\) on the interval \([0, 20]\):
Critical points are at \(t_1 \approx 5.048\) and \(t_2 \approx 10.952\).
(The next cycle's points are beyond \(20\)).
The time interval is \(10.952 - 5.048 = 5.90\) minutes.

(e) Points of inflection occur where \(h''(t) = 0\) and changes sign:
\(h''(t) = \frac{5\pi^2}{8} \cos\left(\frac{\pi}{8} t\right) = 0 \implies \cos\left(\frac{\pi}{8} t\right) = 0\)
For \(0 < t < 8\), \(\frac{\pi}{8} t = \frac{\pi}{2} \implies t = 4\).
At \(t = 4\), \(h(4) = 44\).
So the coordinates are \((4, 44)\).
Physical significance: At \(t = 4\), the vertical speed is at its maximum (or the vertical acceleration is zero).

(a)(i)
- M1: Attempt to relate amplitude to max/min values.
- A1: Clearly justifying why \(a = -40\).

(a)(ii)
- M1: Attempt to find the average height of max and min.
- A1: Correctly showing that \(c = 44\).

(a)(iii)
- M1: Finding the period \(P = 16\).
- A1: Applying \(b = \frac{2\pi}{P}\) to show \(b = \frac{\pi}{8}\).

(b)
- M1: Correct substitution of \(t = 10\) in the formula.
- A1: Correct height \(72.3\).

(c)
- M1: Finding the derivative \(h'(t)\).
- A1: Substituting \(t = 3\).
- A1: Correct rate of change \(14.5\).

(d)
- M1: Setting up inequality or equation \(h(t) = 60\).
- A1, A1: Finding both intersection values \(5.048\) and \(10.952\).
- A1: Correct duration \(5.90\) (3 s.f.).

(e)
- M1: Setting \(h''(t) = 0\) and solving.
- A1: Correct coordinates \((4, 44)\).
- R1: Explaining that this is when the vertical velocity is maximized or vertical acceleration is zero.

**(a)**
Taking the modulus of both sides:
\(|z^n| = |(z-1)^n| \implies |z|^n = |z-1|^n\).
Since moduli are non-negative, \(|z| = |z-1|\).
This represents the set of points in the complex plane equidistant from \(0\) and \(1\).
The locus is the perpendicular bisector of the segment joining \(0\) and \(1\), which is the vertical line \(\text{Re}(z) = \frac{1}{2}\).
*(Alternatively, let \(z = x + iy\). then \(x^2 + y^2 = (x-1)^2 + y^2 \implies 2x = 1 \implies x = \frac{1}{2}\).)*

**(b)**
For \(n = 3\), the equation is \(z^3 = (z-1)^3\).
Since \(z \neq z-1\), we can divide: \(\left(\frac{z}{z-1}\right)^3 = 1\).
Let \(u = \frac{z}{z-1}\). Then \(u^3 = 1\) (where \(u \neq 1\)).
Thus, \(u_1 = e^{i 2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(u_2 = e^{-i 2\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).
Solving for \(z\):
\(u(z-1) = z \implies z(u-1) = u \implies z = \frac{u}{u-1}\).
For \(u_1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\):
\(z_1 = \frac{-1/2 + i\sqrt{3}/2}{-3/2 + i\sqrt{3}/2} = \frac{-1 + i\sqrt{3}}{-3 + i\sqrt{3}} = \frac{(-1 + i\sqrt{3})(-3 - i\sqrt{3})}{12} = \frac{6 - 2i\sqrt{3}}{12} = \frac{1}{2} - i\frac{\sqrt{3}}{6}\).
By conjugate roots symmetry:
\(z_2 = \frac{1}{2} + i\frac{\sqrt{3}}{6}\).

**(c)**
Let \(\frac{z}{z-1} = e^{i\theta_k}\) with \(\theta_k = \frac{2k\pi}{n}\) for \(k = 1, 2, \dots, n-1\).
Solving for \(z\):
\(z_k = \frac{e^{i\theta_k}}{e^{i\theta_k} - 1} = \frac{e^{i\theta_k/2}}{e^{i\theta_k/2} - e^{-i\theta_k/2}}\).
Using Euler's formula:
\(z_k = \frac{\cos(\theta_k/2) + i\sin(\theta_k/2)}{2i\sin(\theta_k/2)} = \frac{1}{2} - \frac{i}{2}\cot(\theta_k/2)\).
Since \(\theta_k/2 = \frac{k\pi}{n}\), we obtain:
\(z_k = \frac{1}{2}\left(1 - i\cot\left(\frac{k\pi}{n}\right)\right)\).

**(d)**
\(|z_k|^2 = \frac{1}{4}\left(1 + \cot^2\left(\frac{k\pi}{n}\right)\right) = \frac{1}{4\sin^2\left(\frac{k\pi}{n}\right)}
\)\prod_{k=1}^{n-1} |z_k|^2 = \frac{1}{4^{n-1}} \frac{1}{\prod_{k=1}^{n-1} \sin^2\left(\frac{k\pi}{n}\right)}\).
Since \(\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}\), the squared product is \(\frac{n^2}{4^{n-1}}\).
Thus, \(\prod_{k=1}^{n-1} |z_k|^2 = \frac{1}{4^{n-1}} \cdot \frac{4^{n-1}}{n^2} = \frac{1}{n^2}\).

**(e)**
Take the modulus: \(|z| = \rho|z-1|\).
Let \(z = x + iy\):
\(x^2 + y^2 = \rho^2((x-1)^2+y^2) \implies (1-\rho^2)x^2 + 2\rho^2 x + (1-\rho^2)y^2 = \rho^2\).
Divide by \(1-\rho^2\):
\(x^2 + \frac{2\rho^2}{1-\rho^2}x + y^2 = \frac{\rho^2}{1-\rho^2}\).
Complete the square:
\(\left(x + \frac{\rho^2}{1-\rho^2}\right)^2 + y^2 = \frac{\rho^2}{(1-\rho^2)^2}\).
Thus, the center is \(\left(\frac{\rho^2}{\rho^2-1}, 0\right)\) and the radius is \(\frac{\rho}{|\rho^2-1|}\).

**(f)**
For \(n=2, w=4 \implies \rho = 2\):
Center: \(\left(\frac{4}{3}, 0\right)\), Radius: \(\frac{2}{3}\).

**(g)**
The roots of \(z^2 = 4(z-1)^2\) are \(z = 2\) and \(z = 2/3\).
The sketch shows a circle with center \((4/3, 0)\) and radius \(2/3\) passing through the roots on the real axis.

**(a)** [3.5 marks]
M1: For taking the modulus of both sides.
A1: For obtaining \(|z| = |z-1|\).
R1: For explaining that this represents the perpendicular bisector between 0 and 1.
A0.5: For concluding \(\text{Re}(z) = 1/2\).

**(b)** [4 marks]
M1: For rewriting as \(\left(\frac{z}{z-1}\right)^3 = 1\).
A1: For identifying the non-trivial roots of unity.
M1: For solving \(z = \frac{u}{u-1}\) algebraically.
A1: For finding the correct roots \(z = \frac{1}{2} \pm i\frac{\sqrt{3}}{6}\).

**(c)** [5 marks]
M1: For setting \(\frac{z}{z-1} = e^{i\theta_k}\).
M1: For solving for \(z_k = \frac{e^{i\theta_k}}{e^{i\theta_k} - 1}\).
M1: For dividing numerator/denominator by \(e^{i\theta_k/2}\).
A1: For showing \(z_k = \frac{1}{2} - \frac{i}{2}\cot(\theta_k/2)\).
A1: For concluding the given formula.

**(d)** [4 marks]
M1: For calculating \(|z_k|^2 = \frac{1}{4}(1+\cot^2(k\pi/n))\).
A1: For simplifying to \(\frac{1}{4\sin^2(k\pi/n)}\).
M1: For writing the product of the sequence.
A1: For obtaining the final product \(\frac{1}{n^2}\).

**(e)** [6 marks]
M1: For taking the modulus and obtaining \(|z| = \rho|z-1|\).
M1: For substituting \(z = x + iy\).
A1: For expanding and rearranging.
M1: For completing the square.
A1: For the correct center.
A1: For the correct radius.

**(f)** [3 marks]
A1: For finding \(\rho = 2\).
A1: For center \((4/3, 0)\).
A1: For radius \(2/3\).

**(g)** [2 marks]
A1: For drawing a circle with the correct center and radius.
A1: For marking the roots \(2\) and \(2/3\) on the real axis.

**(a)**
A circle with center at \((a, 0)\) passing through \((0,0)\) has radius \(R = |a|\).
Equation: \((x-a)^2 + y^2 = a^2 \implies x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0\).

**(b)**
Differentiating with respect to \(x\):
\(2x + 2y \frac{dy}{dx} - 2a = 0 \implies a = x + y \frac{dy}{dx}\).
Substitute \(a\) back into the original equation:
\(x^2 + y^2 - 2x\left(x + y \frac{dy}{dx}\right) = 0 \implies y^2 - x^2 - 2xy \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\).

**(c)**
Orthogonal curves have perpendicular tangents. Thus, the new gradient is \(-\frac{1}{m_1}\):
\(\frac{dy}{dx} = -\frac{2xy}{y^2 - x^2} = \frac{2xy}{x^2 - y^2}\).

**(d)**
Let \(y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}\).
\(v + x\frac{dv}{dx} = \frac{2v}{1-v^2} \implies x\frac{dv}{dx} = \frac{v(1+v^2)}{1-v^2}\).
Separating variables:
\(\frac{1-v^2}{v(1+v^2)} dv = \frac{1}{x} dx\).
Using partial fractions:
\(\frac{1-v^2}{v(1+v^2)} = \frac{1}{v} - \frac{2v}{1+v^2}\).
Integrating:
\(\ln|v| - \ln(1+v^2) = \ln|x| + C \implies \frac{v}{1+v^2} = kx\).
Since \(v = y/x\):
\(\frac{y/x}{1+y^2/x^2} = kx \implies \frac{xy}{x^2+y^2} = kx \implies x^2+y^2 = \frac{1}{k}y\).
Let \(2b = 1/k\):
\(x^2+y^2-2by = 0\).
This represents a family of circles passing through the origin with centers on the \(y\)-axis.

**(e)**
For \(\lambda > -b^2\), since \(a > b\), we have \(a^2 + \lambda > 0\) and \(b^2 + \lambda > 0\). Both denominators are positive, representing an ellipse.
For \(-a^2 < \lambda < -b^2\), \(a^2 + \lambda > 0\) but \(b^2 + \lambda < 0\). One positive and one negative denominator represents a hyperbola.

**(f)**
For the ellipse: \(c^2 = A - B = (a^2+\lambda) - (b^2+\lambda) = a^2 - b^2\).
For the hyperbola: \(c^2 = A + |B| = (a^2+\lambda) - (b^2+\lambda) = a^2 - b^2\).
Thus, the foci are always at \(\left(\pm\sqrt{a^2-b^2}, 0\right)\).

**(g)**
Implicit differentiation:
\(\frac{2x}{a^2+\lambda} + \frac{2yy'}{b^2+\lambda} = 0 \implies \frac{b^2+\lambda}{a^2+\lambda} = -\frac{yy'}{x}\).
Let \(u = a^2+\lambda\) and \(v = b^2+\lambda \implies u-v = a^2-b^2\).
We have \(v = -\frac{yy'}{x}u\).
Substituting into \(u-v = a^2-b^2\):
\(u\left(1 + \frac{yy'}{x}\right) = a^2-b^2 \implies u = \frac{(a^2-b^2)x}{x+yy'}\), which yields \(v = -\frac{(a^2-b^2)yy'}{x+yy'}\).
Substitute \(u\) and \(v\) back into the conic equation:
\(\frac{x(x+yy')}{a^2-b^2} - \frac{y(x+yy')}{(a^2-b^2)y'} = 1\).
Multiply by \((a^2-b^2)y'\):
\((x+yy')(xy'-y) = (a^2-b^2)y'\).
Expanding and rearranging yields:
\(xy(y')^2 + (x^2 - y^2 - (a^2-b^2))y' - xy = 0\).

**(h)**
Let \(y' = p\). Replacing \(p\) with \(-1/p\):
\(xy\left(-\frac{1}{p}\right)^2 + (x^2 - y^2 - (a^2-b^2))\left(-\frac{1}{p}\right) - xy = 0\).
Multiply by \(-p^2\):
\(-xy + (x^2-y^2-(a^2-b^2))p + xyp^2 = 0\),
which is identical to the original equation.

**(i)**
Since the differential equation describes the entire family of confocal conics and is self-orthogonal, the orthogonal trajectories of the ellipses are the hyperbolas. Thus, the ellipses and hyperbolas in this family intersect each other orthogonally at every point of intersection.

**(a)** [2 marks]
M1: For writing the standard equation of a circle.
A1: For substituting \(R = |a|\) and simplifying.

**(b)** [4 marks]
M1: For implicit differentiation.
A1: For expressing \(a\) in terms of \(x\), \(y\), and \(y'\).
M1: For substituting \(a\) back.
A1: For simplifying to the given differential equation.

**(c)** [1.5 marks]
M1: For using \(m_2 = -1/m_1\).
A0.5: For writing down the correct orthogonal ODE.

**(d)** [5 marks]
M1: For substitution \(y = vx\) and differentiating.
A1: For getting the separable form.
M1: For integration with partial fractions.
A1: For integrating to logs.
A1: For obtaining the final cartesian circle equation.

**(e)** [2 marks]
A1: For explaining the ellipse condition.
A1: For explaining the hyperbola condition.

**(f)** [3 marks]
M1: For using focal distance formulas.
A1: For showing ellipse foci.
A1: For showing hyperbola foci are identical.

**(g)** [5 marks]
M1: For differentiating the family equation.
A1: For obtaining the ratio of denominators.
M1: For algebraically solving for the denominators.
A1: For substitution back into the conic equation.
A1: For algebraic simplification to the final form.

**(h)** [3 marks]
M1: For replacing \(y'\) with \(-1/y'\).
M1: For algebraic manipulation (multiplying by \(-y'^2\)).
A1: For obtaining the identical equation.

**(i)** [2 marks]
R1: Identifying that the family consists of both conics.
R1: Concluding that they intersect orthogonally.