2024 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

(i) From the given equation \(x^3 - 3x^2 + 5x - 2 = 0\), we can identify the relations between the roots and the coefficients:
\(\sum \alpha = \alpha + \beta + \gamma = 3\)
\(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = 5\)

Using the identity for the sum of squares:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)
\(\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(5) = 9 - 10 = -1\).

(ii) We want to find a cubic equation with roots \(u = \alpha^2\), \(v = \beta^2\), and \(w = \gamma^2\). Let the new variable be \(y\).

Method 1: Substitution
Let \(y = x^2\), so \(x = \sqrt{y}\).
Substitute this into the original cubic equation:
\((\sqrt{y})^3 - 3(\sqrt{y})^2 + 5\sqrt{y} - 2 = 0\)
\(y\sqrt{y} - 3y + 5\sqrt{y} - 2 = 0\)
Group the square root terms on one side:
\(\sqrt{y}(y + 5) = 3y + 2\)
Square both sides to eliminate the square root:
\(y(y + 5)^2 = (3y + 2)^2\)
\(y(y^2 + 10y + 25) = 9y^2 + 12y + 4\)
\(y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\)
Rearrange the terms into standard cubic form:
\(y^3 + y^2 + 13y - 4 = 0\).

Method 2: Root Relations
Let the new cubic equation be \(y^3 - S_1 y^2 + S_2 y - S_3 = 0\).
\(S_1 = \alpha^2 + \beta^2 + \gamma^2 = -1\) (from part i)
\(S_2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\)
We know \(\alpha\beta\gamma = 2\).
So, \(S_2 = 5^2 - 2(2)(3) = 25 - 12 = 13\).
\(S_3 = \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 2^2 = 4\).
Substituting these values into the general form gives:
\(y^3 - (-1)y^2 + 13y - 4 = 0 \implies y^3 + y^2 + 13y - 4 = 0\).

(i)
M1: For using the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) with correct values of \(\sum \alpha = 3\) and \(\sum \alpha\beta = 5\).
A1: For obtaining \(-1\).

(ii)
Using Method 1 (Substitution):
M1: For substituting \(x = \sqrt{y}\) (or equivalent) into the given equation.
M1: For correctly separating the terms with \(\sqrt{y}\) from the other terms.
M1: For squaring both sides and expanding both expressions correctly.
A1: For the final correct cubic equation \(y^3 + y^2 + 13y - 4 = 0\) (or any non-zero constant multiple).
Using Method 2 (Root Relations):
M1: For stating the formula for \(S_2 = \sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\) or equivalent.
A1: For calculating \(S_2 = 13\).
B1: For calculating \(S_3 = 4\).
A1: For compiling these into the correct cubic equation \(y^3 + y^2 + 13y - 4 = 0\) (or any non-zero constant multiple).

Let \(\mathbf{M} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\). We wish to prove that \(\mathbf{M}^n = \begin{pmatrix} 2n+1 & -4n \\ n & 1-2n \end{pmatrix}\) for all \(n \in \mathbb{Z}^+\).

**Base Case:**
For \(n = 1\):
\[ \mathbf{M}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \]
Using the formula for \(n = 1\):
\[ \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \]
Since the left-hand side and right-hand side are equal, the base case holds.

**Inductive Hypothesis:**
Assume the result is true for \(n = k\) for some positive integer \(k\):
\[ \mathbf{M}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \]

**Inductive Step:**
We show that this implies the result is also true for \(n = k+1\):
\[ \mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \]
Perform the matrix multiplication:
- Row 1, Column 1: \((2k+1)(3) + (-4k)(1) = 6k + 3 - 4k = 2(k+1) + 1\)
- Row 1, Column 2: \((2k+1)(-4) + (-4k)(-1) = -8k - 4 + 4k = -4(k+1)\)
- Row 2, Column 1: \(k(3) + (1-2k)(1) = 3k + 1 - 2k = k+1\)
- Row 2, Column 2: \(k(-4) + (1-2k)(-1) = -4k - 1 + 2k = 1 - 2(k+1)\)

Thus,
\[ \mathbf{M}^{k+1} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix} \]
which is the required form for \(n = k+1\).

**Conclusion:**
Since the statement is true for \(n = 1\), and if it is true for \(n = k\) then it is true for \(n = k+1\), by mathematical induction the statement holds for all positive integers \(n\).

**B1**: Verifies the base case \(n = 1\) with clear, correct substitution on both sides.

**M1**: Formulates a clear inductive hypothesis stating \(\mathbf{M}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\) for some positive integer \(k\).

**M1**: Sets up the product for \(\mathbf{M}^{k+1}\) as \(\mathbf{M}^k \mathbf{M}\) (or \(\mathbf{M} \mathbf{M}^k\)) using the inductive hypothesis.

**A1**: Correctly performs matrix multiplication to find all elements of the resulting matrix.

**A1**: Algebraically manipulates and simplifies each element of the product matrix to show they explicitly match the form of the formula for \(n = k+1\).

**A1**: Provides a complete, mathematically coherent concluding statement referencing the induction steps.

**(i)** Starting with the right-hand side, we find a common denominator:
\[ \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} = \frac{(r+2)(r+3) - r(r+1)}{r(r+1)(r+2)(r+3)} \]
Expanding the numerator:
\[ (r+2)(r+3) - r(r+1) = (r^2 + 5r + 6) - (r^2 + r) = 4r + 6 \]
Thus,
\[ \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} = \frac{4r+6}{r(r+1)(r+2)(r+3)} \]
which is the required result.

**(ii)** We note that:
\[ \frac{2r+3}{r(r+1)(r+2)(r+3)} = \frac{1}{2} \left( \frac{4r+6}{r(r+1)(r+2)(r+3)} \right) \]
Using the result from part (i), we can write the sum as:
\[ S_n = \sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)(r+3)} = \frac{1}{2} \sum_{r=1}^{n} \left( \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} \right) \]
Listing the terms for the method of differences:
- For \( r = 1 \): \( \frac{1}{2} \left( \frac{1}{1\cdot 2} - \frac{1}{3\cdot 4} \right) \)
- For \( r = 2 \): \( \frac{1}{2} \left( \frac{1}{2\cdot 3} - \frac{1}{4\cdot 5} \right) \)
- For \( r = 3 \): \( \frac{1}{2} \left( \frac{1}{3\cdot 4} - \frac{1}{5\cdot 6} \right) \)
- For \( r = n-1 \): \( \frac{1}{2} \left( \frac{1}{(n-1)n} - \frac{1}{(n+1)(n+2)} \right) \)
- For \( r = n \): \( \frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+2)(n+3)} \right) \)

Summing these terms, we see that terms cancel diagonally (e.g., \( -\frac{1}{3\cdot 4} \) with \( +\frac{1}{3\cdot 4} \)), leaving only the first two positive terms and the last two negative terms:
\[ S_n = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{6} - \frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \right] \]
Simplifying the terms:
\[ \frac{1}{2} + \frac{1}{6} = \frac{2}{3} \]
\[ \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)} = \frac{(n+3) + (n+1)}{(n+1)(n+2)(n+3)} = \frac{2(n+2)}{(n+1)(n+2)(n+3)} = \frac{2}{(n+1)(n+3)} \]
Substituting these back into the expression for \( S_n \):
\[ S_n = \frac{1}{2} \left[ \frac{2}{3} - \frac{2}{(n+1)(n+3)} \right] = \frac{1}{3} - \frac{1}{(n+1)(n+3)} \]
Alternatively, combining into a single fraction:
\[ S_n = \frac{(n+1)(n+3) - 3}{3(n+1)(n+3)} = \frac{n^2 + 4n}{3(n+1)(n+3)} \]

**(iii)** As \( n \to \infty \), \( \frac{1}{(n+1)(n+3)} \to 0 \).
Thus, the sum to infinity is:
\[ \sum_{r=1}^{\infty} \frac{2r+3}{r(r+1)(r+2)(r+3)} = \frac{1}{3} \]

**(i)**
* **M1**: For putting the right-hand side over a common denominator of \( r(r+1)(r+2)(r+3) \).
* **A1**: For correct expansion and simplification to show that the numerator is indeed \( 4r + 6 \).

**(ii)**
* **M1**: For writing the general term of the sum as \( \frac{1}{2} \left( \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} \right) \).
* **M1**: For writing down a sufficient number of terms to demonstrate the canceling structure clearly.
* **A1**: For identifying the remaining uncancelled terms: \( \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{6} - \frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \right] \).
* **A1**: For simplifying to the final form \( \frac{1}{3} - \frac{1}{(n+1)(n+3)} \) or \( \frac{n^2 + 4n}{3(n+1)(n+3)} \).

**(iii)**
* **M1**: For evaluating the limit as \( n \to \infty \) by identifying that \( \frac{1}{(n+1)(n+3)} \to 0 \) (or equivalent justification).
* **A1**: For obtaining the correct sum to infinity: \( \frac{1}{3} \).

**(i) Find the matrix \(\mathbf{M}\)**

A shear parallel to the \(x\)-axis preserves the \(x\)-axis, so:

\(\mathbf{H} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)

Since it maps \((0,1)\) to \((3,1)\):

\(\mathbf{H} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\)

Thus, the matrix representing the shear is:

\(\mathbf{H} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\)

A reflection in the line \(y = -x\) maps:

\(\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 0 \\ -1 \end{pmatrix}\)

\(\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} -1 \\ 0 \end{pmatrix}\)

Thus, the matrix representing the reflection is:

\(\mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\)

The combined transformation of the shear followed by the reflection is represented by the matrix product \(\mathbf{M} = \mathbf{R}\mathbf{H}\):

\(\mathbf{M} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix}\)

**(ii) Show that the origin is the only invariant point**

An invariant point \(\begin{pmatrix} x \\ y \end{pmatrix}\) must satisfy \(\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\):

\(\begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives the system of linear equations:

1) \(-y = x \implies x + y = 0\)

2) \(-x - 3y = y \implies x + 4y = 0\)

Subtracting the first equation from the second equation:

\(3y = 0 \implies y = 0\)

Substituting \(y = 0\) back into \(x + y = 0\) gives \(x = 0\).

Thus, the only invariant point is the origin \((0,0)\).

**(iii) Find the equations of the invariant lines through the origin**

Let the equation of an invariant line through the origin be \(y = mx\).

Any point on this line can be represented as \(\begin{pmatrix} x \\ mx \end{pmatrix}\). Its image \(\begin{pmatrix} x' \\ y' \end{pmatrix}\) under \(\mathbf{M}\) is:

\(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} -mx \\ -x - 3mx \end{pmatrix}\)

For the line to be invariant, the image point must also lie on the line \(y' = mx'\):

\(-x - 3mx = m(-mx)\)

\(-x - 3mx = -m^2 x\)

Since this must hold for any point on the line other than the origin, we divide by \(x\) (where \(x \neq 0\)):

\(-1 - 3m = -m^2 \implies m^2 - 3m - 1 = 0\)

Using the quadratic formula to solve for \(m\):

\(m = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}\)

Thus, the equations of the invariant lines through the origin are:

\(y = \left(\frac{3 + \sqrt{13}}{2}\right)x\) and \(y = \left(\frac{3 - \sqrt{13}}{2}\right)x\)

**(i)**
* **M1**: For correctly finding the shear matrix \(\mathbf{H} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\).
* **M1**: For correctly finding the reflection matrix \(\mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\).
* **M1**: For writing the product in the correct order \(\mathbf{M} = \mathbf{R}\mathbf{H}\).
* **A1**: For obtaining the correct combined matrix \(\mathbf{M} = \begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix}\).

**(ii)**
* **M1**: For setting up the matrix equation \(\mathbf{M}\mathbf{x} = \mathbf{x}\) to form the simultaneous equations \(-y = x\) and \(-x - 3y = y\).
* **A1**: For solving the simultaneous equations showing clear algebraic steps leading to \(y=0\).
* **A1**: For obtaining \(x=0\) and correctly concluding that \((0,0)\) is the only invariant point.

**(iii)**
* **M1**: For setting up the equation of the line as \(y = mx\) and finding the transformed coordinates of a general point \((x, mx)\).
* **M1**: For applying the condition that the image point lies on the line to form the equation \(-x - 3mx = m(-mx)\).
* **M1**: For simplifying this equation to the quadratic form \(m^2 - 3m - 1 = 0\).
* **M1**: For solving the quadratic equation using a valid method.
* **A1**: For obtaining both values of the gradient \(m = \frac{3 \pm \sqrt{13}}{2}\).
* **A1**: For stating both final equations of the invariant lines: \(y = \left(\frac{3 + \sqrt{13}}{2}\right)x\) and \(y = \left(\frac{3 - \sqrt{13}}{2}\right)x\).

(a) To find the normal vector \(\mathbf{n}\) of the plane \(\Pi\), we calculate the vector product of the direction vectors of \(l_1\) and \(l_2\): \(\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} (-1)(-2) - (1)(1) \\ (1)(1) - (2)(-2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\). Since the plane contains \(l_1\), it must pass through the point \((1, 2, -1)\). The Cartesian equation of the plane is in the form \(x + 5y + 3z = d\). Substituting the point: \(1(1) + 5(2) + 3(-1) = 8\). Thus, the equation of \(\Pi\) is \(x + 5y + 3z = 8\). (b) The shortest distance between two skew lines is equal to the perpendicular distance from any point on \(l_2\) to the plane \(\Pi\). A point on \(l_2\) is \((3, 0, 2)\). The perpendicular distance is given by: \(d = \frac{|1(3) + 5(0) + 3(2) - 8|}{\sqrt{1^2 + 5^2 + 3^2}} = \frac{|3 + 0 + 6 - 8|}{\sqrt{35}} = \frac{1}{\sqrt{35}} = \frac{\sqrt{35}}{35}\). (c) Since the line \(l_3\) is perpendicular to \(\Pi\), its direction vector is the normal vector of \(\Pi\), which is \(\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\). Since it passes through \(P(4, 5, -2)\), its vector equation is: \(\mathbf{r} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\). This gives parametric coordinates for any point on \(l_3\) as \(x = 4+t\), \(y = 5+5t\), \(z = -2+3t\). Substituting these into the plane equation \(x + 5y + 3z = 8\): \((4 + t) + 5(5 + 5t) + 3(-2 + 3t) = 8\) which simplifies to \(4 + t + 25 + 25t - 6 + 9t = 8 \implies 35t + 23 = 8 \implies 35t = -15 \implies t = -\frac{3}{7}\). Substituting \(t = -\frac{3}{7}\) back into the parametric coordinates gives: \(x = 4 - \frac{3}{7} = \frac{25}{7}\), \(y = 5 - \frac{15}{7} = \frac{20}{7}\), \(z = -2 - \frac{9}{7} = -\frac{23}{7}\). Thus, the coordinates of the point of intersection are \(\left(\frac{25}{7}, \frac{20}{7}, -\frac{23}{7}\right)\).

(a) M1: For calculating the cross product of the direction vectors of both lines. A1: For obtaining the correct normal vector \(\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\) (or any non-zero scalar multiple). M1: For substituting the point \((1, 2, -1)\) into the plane equation with their normal vector to find the constant. A1: For the correct Cartesian equation \(x + 5y + 3z = 8\) (or equivalent). (b) M1: For using the formula for the distance from a point on \(l_2\), e.g., \((3, 0, 2)\), to the plane \(\Pi\). A1: For substituting the correct values into the distance formula. A1: For simplifying to the exact form \(\frac{\sqrt{35}}{35}\). (c) B1: For writing a correct vector equation for the line \(l_3\). M1: For substituting the parametric coordinates of \(l_3\) into the Cartesian equation of \(\Pi\). A1: For solving for the parameter to get \(t = -\frac{3}{7}\) (or equivalent parameter value depending on the choice of line equation). M1: For substituting their parameter value back to find the coordinates of the point of intersection. A1: For obtaining the correct coordinates \(\left(\frac{25}{7}, \frac{20}{7}, -\frac{23}{7}\right)\).

**(a) Asymptotes** The vertical asymptote occurs where the denominator is zero and the numerator is non-zero. Thus, the vertical asymptote is: \[ x = 1 \] To find the oblique asymptote, we rewrite the expression for \( y \) using algebraic division: \[ y = \frac{x^2 + 2x + 1}{x-1} = x + 3 + \frac{4}{x-1} \] As \( x \to \pm\infty \), the term \( \frac{4}{x-1} \to 0 \). Therefore, the oblique asymptote is: \[ y = x + 3 \] **(b) Stationary points** Differentiating \( y = x + 3 + \frac{4}{x-1} \) with respect to \( x \): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-1)^2} \] For stationary points, set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \): \[ 1 - \frac{4}{(x-1)^2} = 0 \implies (x-1)^2 = 4 \implies x-1 = \pm 2 \] This gives: - For \( x - 1 = 2 \implies x = 3 \). The \( y \)-coordinate is \( y = 8 \). - For \( x - 1 = -2 \implies x = -1 \). The \( y \)-coordinate is \( y = 0 \). To determine the nature, we find the second derivative: \[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{8}{(x-1)^3} \] - At \( x = 3 \): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 1 > 0 \), which means \( (3, 8) \) is a **local minimum**. - At \( x = -1 \): \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -1 < 0 \), which means \( (-1, 0) \) is a **local maximum**. **(c) Sketch of \( C \)** - **Intercepts**: The \( x \)-intercept is at \( (-1, 0) \), which is also the local maximum (the curve touches the \( x \)-axis and turns back down). The \( y \)-intercept is at \( (0, -1) \). - **Asymptotes**: Drawn as dashed lines at \( x = 1 \) and \( y = x + 3 \). - **Left branch (\( x < 1 \))**: Approaches \( y = x + 3 \) from below, rises to touch the \( x \)-axis at \( (-1, 0) \), crosses the \( y \)-axis at \( (0, -1) \), and falls to \( -\infty \) as \( x \to 1^- \). - **Right branch (\( x > 1 \))**: Falls from \( +\infty \) near \( x = 1^+ \) to the local minimum at \( (3, 8) \), and then rises towards the oblique asymptote \( y = x+3 \) from above. **(d) Sketch of \( y^2 = \frac{(x+1)^2}{x-1} \)** - **Domain**: Since \( y^2 \ge 0 \), we must have \( \frac{(x+1)^2}{x-1} \ge 0 \). This holds for \( x > 1 \) and at the single isolated point \( x = -1 \). - **Isolated Point**: There is a single point at \( (-1, 0) \) because \( y^2 = 0 \implies y = 0 \) at this point, but no real points exist for other values in \( x < 1 \). - **Symmetrical Branches**: For \( x > 1 \), the curve is symmetrical about the \( x \)-axis. It approaches \( x = 1 \) as a vertical asymptote as \( x \to 1^+ \). - **Turning Points**: Symmetrical turning points exist at \( (3, \pm\sqrt{8}) = (3, \pm 2\sqrt{2}) \). The upper branch has a local minimum at \( (3, 2\sqrt{2}) \) and the lower branch has a local maximum at \( (3, -2\sqrt{2}) \).

**(a) [3 Marks]**
- **M1**: Identify vertical asymptote \( x = 1 \).
- **M1**: Use algebraic division or expansion to express \( y \) in the form \( x + A + \frac{B}{x-1} \).
- **A1**: Obtain the oblique asymptote \( y = x + 3 \) correctly.

**(b) [4 Marks]**
- **M1**: Differentiate to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \) (using quotient rule or simplified form).
- **A1**: Solve \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to find \( x = -1 \) and \( x = 3 \).
- **A1**: Provide coordinates of both stationary points: \( (-1, 0) \) and \( (3, 8) \).
- **B1**: Correctly justify the nature of both points using \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) or first derivative sign test.

**(c) [4 Marks]**
- **G1**: Draw both branches showing correct asymptotic behaviour.
- **G1**: Clearly show and label asymptotes \( x = 1 \) and \( y = x + 3 \).
- **G1**: Correctly position and label stationary points \( (-1, 0) \) and \( (3, 8) \).
- **G1**: Correctly label the intercepts at \( (-1, 0) \) and \( (0, -1) \).

**(d) [4 Marks]**
- **B1**: Clearly identify and plot the isolated point at \( (-1, 0) \).
- **G1**: Sketch two symmetrical branches for \( x > 1 \) that approach the vertical asymptote \( x = 1 \).
- **G1**: Label/mark the turning points of these branches at \( (3, 2\sqrt{2}) \) and \( (3, -2\sqrt{2}) \).
- **B1**: Ensure no other part of the curve is drawn in the region \( x < 1 \).

(a) \( C_1 \) is a circle of diameter 3, symmetric about the initial line, with its center at \( (1.5, 0) \) and passing through the pole. \( C_2 \) is a cardioid symmetric about the initial line, with its cusp at the pole pointing in the direction \( \theta = \pi \), and its furthest point at \( (2, 0) \). The curves intersect at \( \theta = \pm\frac{\pi}{3} \) and at the pole. The sketch should show the circle lying outside the cardioid for \( -\frac{\pi}{3} < \theta < \frac{\pi}{3} \), and inside the cardioid for \( \frac{\pi}{3} < |\theta| \le \frac{\pi}{2} \).

(b) To find the intersection points, set the two equations equal:
\( 3\cos\theta = 1 + \cos\theta \implies 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \).
Since \( -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} \) for \( C_1 \), we have \( \theta = \pm\frac{\pi}{3} \).
Substituting back into either equation gives \( r = 1.5 \).
So two points of intersection are \( \left(\frac{3}{2}, \frac{\pi}{3}\right) \) and \( \left(\frac{3}{2}, -\frac{\pi}{3}\right) \).
Additionally, at the pole, \( r = 0 \). For \( C_1 \), \( r = 0 \) when \( \theta = \pm\frac{\pi}{2} \). For \( C_2 \), \( r = 0 \) when \( \theta = \pi \). Since both curves pass through the pole, the pole is also a point of intersection.

(c) For \( C_2 \), the vertical distance from the initial line is \( y = r\sin\theta = (1 + \cos\theta)\sin\theta = \sin\theta + \frac{1}{2}\sin 2\theta \).
To find where the tangent is parallel to the initial line, we solve \( \frac{dy}{d\theta} = 0 \):
\( \frac{dy}{d\theta} = \cos\theta + \cos 2\theta = 2\cos^2\theta + \cos\theta - 1 = (2\cos\theta - 1)(\cos\theta + 1) = 0 \).
This gives \( \cos\theta = \frac{1}{2} \) or \( \cos\theta = -1 \).
In the interval \( -\pi < \theta \le \pi \), this yields \( \theta = \pm\frac{\pi}{3} \) or \( \theta = \pi \).

(d) By symmetry, the area \( A \) is given by:
\( A = 2 \left[ \frac{1}{2} \int_{0}^{\pi/3} (1 + \cos\theta)^2 \, d\theta + \frac{1}{2} \int_{\pi/3}^{\pi/2} (3\cos\theta)^2 \, d\theta \right] \)
\( = \int_{0}^{\pi/3} (1 + 2\cos\theta + \cos^2\theta) \, d\theta + 9 \int_{\pi/3}^{\pi/2} \cos^2\theta \, d\theta \)
Using \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \):
\( \int_{0}^{\pi/3} (1 + 2\cos\theta + \cos^2\theta) \, d\theta = \int_{0}^{\pi/3} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \, d\theta \)
\( = \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi/3} \)
\( = \frac{\pi}{2} + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \).

Now for the second integral:
\( 9 \int_{\pi/3}^{\pi/2} \cos^2\theta \, d\theta = \frac{9}{2} \int_{\pi/3}^{\pi/2} (1 + \cos 2\theta) \, d\theta \)
\( = \frac{9}{2} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\pi/3}^{\pi/2} \)
\( = \frac{9}{2} \left[ \left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) \right] \)
\( = \frac{9}{2} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \).

Adding the two parts together:
\( A = \left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right) + \left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8}\right) = \frac{5\pi}{4} \).

(a)
M1: For correctly drawing a circle passing through the pole with its center on the initial line.
M1: For correctly drawing a cardioid with its cusp at the pole pointing left and its vertex at (2,0) on the initial line.
A1: For a correct overall sketch showing both curves intersecting at two points in the first/fourth quadrants and at the pole, with correct relative sizes (circle outside cardioid near \( \theta = 0 \)).

(b)
M1: For equating \( 3\cos\theta = 1 + \cos\theta \) and solving for \( \cos\theta \).
A1: For finding \( \theta = \pm\frac{\pi}{3} \) and obtaining the coordinates \( \left(\frac{3}{2}, \pm\frac{\pi}{3}\right) \) (or equivalent).
A1: For correctly identifying the pole (or \( r=0 \)) as another intersection point.

(c)
M1: For writing \( y = (1 + \cos\theta)\sin\theta \) and differentiating to find \( \frac{dy}{d\theta} \).
A1: For setting \( \frac{dy}{d\theta} = 0 \) to obtain \( (2\cos\theta - 1)(\cos\theta + 1) = 0 \) or equivalent quadratic.
A1: For finding the correct values \( \theta = \pm\frac{\pi}{3} \) and \( \theta = \pi \).

(d)
M1: For expressing the area as a sum of two integrals with appropriate limits (e.g., using symmetry: from 0 to \( \pi/3 \) for the cardioid and from \( \pi/3 \) to \( \pi/2 \) for the circle).
M1: For using the identity \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \) to integrate both terms.
A1: For obtaining \( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \) for the first integral (including correct limits substitution).
A1: For obtaining \( \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \) for the second integral (including correct limits substitution).
M1: For adding the two areas together with the exact cancellations of the square root terms shown.
A1: For the final exact area of \( \frac{5\pi}{4} \).

We know that \(\omega^7 = 1\) and \(\omega \neq 1\). The sum of all seventh roots of unity is \(1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0\). This implies \(\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = -1\). Since \(\alpha + \beta = (\omega + \omega^2 + \omega^4) + (\omega^3 + \omega^5 + \omega^6) = \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6\), we have \(\alpha + \beta = -1\). Next, we find \(\alpha \beta\): \(\alpha \beta = (\omega + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6) = \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10}\). Using the relation \(\omega^7 = 1\), we simplify the powers: \(\omega^7 = 1\), \(\omega^8 = \omega\), \(\omega^9 = \omega^2\), and \(\omega^{10} = \omega^3\). Substituting these back in gives \(\alpha \beta = \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega + 1 + \omega^2 + \omega^3 = 3 + (\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6) = 3 - 1 = 2\). A quadratic equation with roots \(\alpha\) and \(\beta\) is given by \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\). Substituting our values, we obtain \(x^2 + x + 2 = 0\).

B1: Correctly shows \(\alpha + \beta = -1\) using the sum of roots of unity. M1: Expands \(\alpha \beta\) into nine terms. M1: Uses \(\omega^7 = 1\) to reduce powers of \(\omega\) to the range \(\omega^0\) to \(\omega^6\). A1: Obtains \(\alpha \beta = 2\) with clear working. A1: Writes down the final quadratic equation \(x^2 + x + 2 = 0\) (or any non-zero integer multiple of this equation).

Method 1: Using Differentiation
Let \(f(x) = \ln(1 + \tan x)\).
Then \(f(0) = \ln(1) = 0\).

Differentiating once with respect to \(x\):
\(f'(x) = \frac{\sec^2 x}{1 + \tan x}\)
At \(x = 0\):
\(f'(0) = \frac{1}{1} = 1\).

Differentiating implicitly by writing:
\(f'(x)(1 + \tan x) = \sec^2 x\)
\(f''(x)(1 + \tan x) + f'(x)\sec^2 x = 2\sec^2 x \tan x\)
At \(x = 0\):
\(f''(0)(1) + (1)(1) = 0 \implies f''(0) = -1\).

Differentiating again with respect to \(x\):
\(f'''(x)(1 + \tan x) + 2f''(x)\sec^2 x + 2f'(x)\sec^2 x \tan x = 4\sec^2 x \tan^2 x + 2\sec^4 x\)
At \(x = 0\):
\(f'''(0)(1) + 2(-1)(1) + 0 = 0 + 2 \implies f'''(0) - 2 = 2 \implies f'''(0) = 4\).

Substituting these values into Maclaurin's formula:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)
\(f(x) \approx x - \frac{1}{2}x^2 + \frac{2}{3}x^3\).

Method 2: Using Standard Series
From MF19, the series for \(\tan x\) is:
\(\tan x = x + \frac{1}{3}x^3 + \dots\)
We also have the series:
\
\ln(1 + u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \dots\)
Substituting \(u = \tan x\):
\(\ln(1 + \tan x) = \left(x + \frac{1}{3}x^3\right) - \frac{1}{2}\left(x + \frac{1}{3}x^3\right)^2 + \frac{1}{3}\left(x + \frac{1}{3}x^3\right)^3 - \dots\)
Expanding terms up to \(x^3\):
\(\ln(1 + \tan x) = x + \frac{1}{3}x^3 - \frac{1}{2}x^2 + \frac{1}{3}x^3 + \dots\)
\(\ln(1 + \tan x) \approx x - \frac{1}{2}x^2 + \frac{2}{3}x^3\)

M1: For a valid method of finding the expansion (either through successive differentiation or substitution of standard series).
A1: For finding correct derivatives \(f'(0) = 1\) and \(f''(0) = -1\) (or for correct expansion terms up to the quadratic level: \(x - \frac{1}{2}x^2\)).
M1: For finding the correct third derivative value \(f'''(0) = 4\) (or for correctly identifying and adding the cubic-level expansion terms: \(\frac{1}{3}x^3 + \frac{1}{3}x^3\)).
A1: For the correct final expansion \(x - \frac{1}{2}x^2 + \frac{2}{3}x^3\).

We first find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\) using the product rule: \( \frac{dx}{dt} = -e^{-t}\cos t - e^{-t}\sin t = -e^{-t}(\cos t + \sin t) \) and \( \frac{dy}{dt} = -e^{-t}\sin t + e^{-t}\cos t = e^{-t}(\cos t - \sin t) \). Using the parametric formula for the first derivative: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^{-t}(\cos t - \sin t)}{-e^{-t}(\cos t + \sin t)} = \frac{\sin t - \cos t}{\cos t + \sin t} \). To find the second derivative, we differentiate \( \frac{dy}{dx} \) with respect to \(t\). Using the quotient rule: \( \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(\cos t + \sin t)(\cos t + \sin t) - (\sin t - \cos t)(\cos t - \sin t)}{(\cos t + \sin t)^2} = \frac{(\cos^2 t + 2\sin t \cos t + \sin^2 t) + (\sin^2 t - 2\sin t \cos t + \cos^2 t)}{(\cos t + \sin t)^2} = \frac{2}{(\cos t + \sin t)^2} \). Alternatively, we can express \( \frac{dy}{dx} = \tan\left(t - \frac{\pi}{4}\right) \), which gives \( \frac{d}{dt}\left(\frac{dy}{dx}\right) = \sec^2\left(t - \frac{\pi}{4}\right) \). Now, we find the second derivative: \( \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{2}{(\cos t + \sin t)^2}}{-e^{-t}(\cos t + \sin t)} = \frac{-2e^t}{(\cos t + \sin t)^3} \). At the point where \( t = 0 \), we substitute to get: \( \frac{d^2y}{dx^2} = \frac{-2e^0}{(\cos 0 + \sin 0)^3} = -2 \).

M1: For attempting to differentiate \(x\) and \(y\) with respect to \(t\) using the product rule. A1: For obtaining both correct derivatives: \( \frac{dx}{dt} = -e^{-t}(\cos t + \sin t) \) and \( \frac{dy}{dt} = e^{-t}(\cos t - \sin t) \). A1: For simplifying to obtain \( \frac{dy}{dx} = \frac{\sin t - \cos t}{\cos t + \sin t} \) (or equivalent). M1: For applying the quotient rule (or chain rule on a trigonometric identity) to differentiate \( \frac{dy}{dx} \) with respect to \(t\). A1: For obtaining the correct derivative \( \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2}{(\cos t + \sin t)^2} \) (or equivalent). M1: For using the formula \( \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt} \) and substituting \( t = 0 \). A1: For obtaining the correct final exact value of \( -2 \).

Part (i):
We write \( I_n = \int_0^{\pi/4} \sec^{n-2} x \cdot \sec^2 x \, dx \).

Using integration by parts, let:
\( u = \sec^{n-2} x \implies \frac{du}{dx} = (n-2)\sec^{n-3} x \cdot \sec x \tan x = (n-2)\sec^{n-2} x \tan x \)
\( \frac{dv}{dx} = \sec^2 x \implies v = \tan x \).

Then:
\[ I_n = \left[ \sec^{n-2} x \tan x \right]_0^{\pi/4} - \int_0^{\pi/4} (n-2) \sec^{n-2} x \tan^2 x \, dx \]

At the upper limit \( x = \frac{\pi}{4} \), \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \) and \( \tan\left(\frac{\pi}{4}\right) = 1 \).
At the lower limit \( x = 0 \), \( \tan(0) = 0 \).
So, the boundary term is:
\[ \left[ \sec^{n-2} x \tan x \right]_0^{\pi/4} = (\sqrt{2})^{n-2} \cdot 1 - 0 = 2^{\frac{n-2}{2}} \]

Using the identity \( \tan^2 x = \sec^2 x - 1 \) in the integral, we obtain:
\[ I_n = 2^{\frac{n-2}{2}} - (n-2) \int_0^{\pi/4} \sec^{n-2} x (\sec^2 x - 1) \, dx \]
\[ I_n = 2^{\frac{n-2}{2}} - (n-2) \left( \int_0^{\pi/4} \sec^n x \, dx - \int_0^{\pi/4} \sec^{n-2} x \, dx \right) \]
\[ I_n = 2^{\frac{n-2}{2}} - (n-2) (I_n - I_{n-2}) \]
\[ I_n = 2^{\frac{n-2}{2}} - (n-2) I_n + (n-2) I_{n-2} \]
\[ (1 + n - 2) I_n = 2^{\frac{n-2}{2}} + (n-2) I_{n-2} \]
\[ (n-1) I_n = 2^{\frac{n-2}{2}} + (n-2) I_{n-2} \]

Part (ii):
Substituting \( n = 4 \) into the reduction formula:
\[ 3 I_4 = 2^1 + 2 I_2 = 2 + 2 I_2 \]

We calculate \( I_2 \):
\[ I_2 = \int_0^{\pi/4} \sec^2 x \, dx = \left[ \tan x \right]_0^{\pi/4} = 1 - 0 = 1 \]

Substituting \( I_2 = 1 \) back into the equation:
\[ 3 I_4 = 2 + 2(1) = 4 \implies I_4 = \frac{4}{3} \]

Part (i):
M1: For splitting \( \sec^n x \) into \( \sec^{n-2} x \sec^2 x \) and attempting integration by parts.
A1: For correct differentiation of \( \sec^{n-2} x \) to obtain \( (n-2)\sec^{n-2} x \tan x \).
M1: For evaluating the limits of the boundary term correctly to obtain \( 2^{\frac{n-2}{2}} \).
M1: For substituting \( \tan^2 x = \sec^2 x - 1 \) and writing the equation in terms of \( I_n \) and \( I_{n-2} \).
A1: For correct completion of the proof showing all intermediate steps clearly.

Part (ii):
M1: For substituting \( n = 4 \) into the reduction formula.
M1: For finding \( I_2 = 1 \) (either by integration or by reduction formula with \( n = 2 \)).
A1: For the exact value of \( I_4 = \frac{4}{3} \) (or equivalent single fraction).

(a) The interval \( [0, 1] \) is divided into \( n \) subintervals of width \( \Delta x = \frac{1}{n} \), with endpoints \( x_r = \frac{r}{n} \) for \( r = 0, 1, \dots, n \).
Since \( f(x) = e^{2x} \) is strictly increasing, the minimum value in each subinterval \( [x_r, x_{r+1}] \) occurs at the left endpoint \( x_r \).
Thus, the lower sum is:
\[ L_n = \sum_{r=0}^{n-1} f(x_r) \Delta x = \sum_{r=0}^{n-1} e^{2r/n} \cdot \frac{1}{n} = \frac{1}{n} \sum_{r=0}^{n-1} \left(e^{2/n}\right)^r \]
This is a geometric progression with first term \( a = 1 \), common ratio \( R = e^{2/n} \), and \( n \) terms.
Using the sum formula \( S_n = a \frac{R^n - 1}{R - 1} \):
\[ L_n = \frac{1}{n} \frac{(e^{2/n})^n - 1}{e^{2/n} - 1} = \frac{e^2 - 1}{n(e^{2/n} - 1)} \]
For the upper sum \( U_n \), the maximum value in each subinterval occurs at the right endpoint \( x_{r+1} \).
\[ U_n = \sum_{r=0}^{n-1} f(x_{r+1}) \Delta x = \sum_{r=1}^{n} e^{2r/n} \cdot \frac{1}{n} = e^{2/n} L_n = \frac{e^{2/n}(e^2 - 1)}{n(e^{2/n} - 1)} \]

(b) Let \( y = \frac{2}{n} \). As \( n \to \infty \), \( y \to 0 \).
Using the series expansion of \( e^y \):
\[ e^y = 1 + y + \frac{y^2}{2!} + \dots \]
Then:
\[ n(e^{2/n} - 1) = \frac{2}{y}(e^y - 1) = \frac{2}{y}\left( y + \frac{y^2}{2} + \dots \right) = 2 + y + \dots \]
Taking the limit as \( y \to 0 \):
\[ \lim_{n \to \infty} n(e^{2/n} - 1) = 2 \]
Since the function \( f(x) = e^{2x} \) is continuous and increasing, the area under the curve is bounded by \( L_n \) and \( U_n \):
\[ L_n \le \int_0^1 e^{2x} \, \mathrm{d}x \le U_n \]
As \( n \to \infty \):
\[ \lim_{n \to \infty} L_n = \frac{e^2 - 1}{2} \]
\[ \lim_{n \to \infty} U_n = \lim_{n \to \infty} e^{2/n} \cdot L_n = 1 \cdot \frac{e^2 - 1}{2} = \frac{e^2 - 1}{2} \]
By the squeeze theorem, the value of the integral is:
\[ \int_0^1 e^{2x} \, \mathrm{d}x = \frac{e^2 - 1}{2} \]

(c) The difference between the upper and lower sums is:
\[ U_n - L_n = \frac{e^{2/n}(e^2 - 1) - (e^2 - 1)}{n(e^{2/n} - 1)} = \frac{(e^2 - 1)(e^{2/n} - 1)}{n(e^{2/n} - 1)} = \frac{e^2 - 1}{n} \]
We require:
\[ \frac{e^2 - 1}{n} < 0.01 \implies n > 100(e^2 - 1) \]
Using \( e^2 \approx 7.389056 \):
\[ n > 100(6.389056) = 638.9056 \]
Thus, the smallest integer value of \( n \) is 639.

(a)
- M1: Write down the expression for the lower sum as a sum of rectangles: \( L_n = \sum_{r=0}^{n-1} \frac{1}{n} e^{2r/n} \).
- M1: Recognize the series as a geometric progression and state its first term and common ratio.
- A1: Correctly apply the sum formula for a geometric progression to obtain \( \frac{(e^{2/n})^n - 1}{e^{2/n} - 1} \).
- A1: Simplify correctly to show the given expression for \( L_n \).
- B1: Write down the correct expression for \( U_n \), e.g., \( U_n = \frac{e^{2/n}(e^2 - 1)}{n(e^{2/n} - 1)} \).

(b)
- M1: Use the substitution \( y = 2/n \) or substitute the Taylor series expansion for \( e^{2/n} \) directly.
- A1: Correctly expand to obtain \( n(e^{2/n}-1) = 2 + \frac{2}{n} + \mathcal{O}(n^{-2}) \) or equivalent.
- A1: Conclude the limit is 2 as \( n \to \infty \).
- B1: State that the integral is bounded by \( L_n \) and \( U_n \), and evaluate the limit of either to obtain \( \frac{1}{2}(e^2 - 1) \).

(c)
- M1: Set up the inequality \( U_n - L_n < 0.01 \) and simplify \( U_n - L_n \) to \( \frac{e^2-1}{n} \).
- A1: Solve for \( n \) and state \( n = 639 \) (must be an integer).

(a) The auxiliary equation is:
\[m^2 + 4m + 4 = 0\implies (m+2)^2 = 0\]
This gives a repeated root of \(m = -2\).

Therefore, the complementary function (CF) is:
\[y_c = (A + Bx)e^{-2x}\]

For the particular integral (PI), since \(-2\) is a repeated root of multiplicity 2 of the auxiliary equation and the term on the right-hand side is of the form \(C e^{-2x}\), we try:
\[y_p = k x^2 e^{-2x}\]

Differentiating \(y_p\) using the product rule:
\[y_p' = k(2x - 2x^2)e^{-2x}\]
\[y_p'' = k(2 - 4x - 4x + 4x^2)e^{-2x} = k(2 - 8x + 4x^2)e^{-2x}\]

Substituting \(y_p\), \(y_p'\), and \(y_p''\) into the original differential equation:
\[k e^{-2x} \left[ (2 - 8x + 4x^2) + 4(2x - 2x^2) + 4x^2 \right] = 8e^{-2x}\]
\[k e^{-2x} \left[ 2 - 8x + 4x^2 + 8x - 8x^2 + 4x^2 \right] = 8e^{-2x}\]
\[2k e^{-2x} = 8e^{-2x} \implies k = 4\]

Thus, the general solution is:
\[y = (A + Bx)e^{-2x} + 4x^2e^{-2x}\]

(b) Applying the boundary condition \(y = 2\) at \(x = 0\):
\[2 = A e^0 + 0 \implies A = 2\]

Differentiating the general solution to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = B e^{-2x} - 2(A + Bx)e^{-2x} + 8x e^{-2x} - 8x^2 e^{-2x}\]

Applying the boundary condition \(\frac{dy}{dx} = 1\) at \(x = 0\):
\[1 = B - 2A \implies 1 = B - 4 \implies B = 5\]

Hence, the particular solution is:
\[y = (2 + 5x + 4x^2)e^{-2x}\]

(c) For stationary points, we set \(\frac{dy}{dx} = 0\):
\[\frac{dy}{dx} = (5 + 8x)e^{-2x} - 2(2 + 5x + 4x^2)e^{-2x} = 0\]
\[(1 - 2x - 8x^2)e^{-2x} = 0\]

Since \(e^{-2x} \neq 0\) for all real \(x\):
\[8x^2 + 2x - 1 = 0 \implies (4x - 1)(2x + 1) = 0\]

Since we are looking for a stationary point where \(x > 0\), we select \(x = \frac{1}{4}\).

Substituting \(x = \frac{1}{4}\) back into the particular solution to find the corresponding \(y\)-value:
\[y = \left(2 + 5\left(\frac{1}{4}\right) + 4\left(\frac{1}{16}\right)\right)e^{-2(1/4)}\]
\[y = \left(2 + \frac{5}{4} + \frac{1}{4}\right)e^{-1/2} = \frac{7}{2}e^{-1/2}\]

Thus, the exact coordinates of the stationary point are \(\left(\frac{1}{4}, \frac{7}{2}e^{-1/2}\right)\).

(a)
- M1: Solves auxiliary equation to find the repeated root \(m = -2\).
- A1: Obtains the correct complementary function \(y_c = (A + Bx)e^{-2x}\) (or equivalent).
- M1: For attempting to find a particular integral of the form \(y_p = k x^2 e^{-2x}\) and finding derivatives.
- M1: Substituting their derivatives into the differential equation and simplifying.
- A1: Obtains \(k = 4\).
- A1: States the correct general solution \(y = (A+Bx)e^{-2x} + 4x^2e^{-2x}\).

(b)
- M1: Uses the condition \(y(0) = 2\) to find \(A\).
- M1: Differentiates their general solution and uses \(y'(0) = 1\) to find \(B\).
- A1: Obtains the correct particular solution \(y = (2 + 5x + 4x^2)e^{-2x}\).

(c)
- M1: Sets \(\frac{dy}{dx} = 0\) and obtains a quadratic equation in \(x\).
- A1: Finds the correct positive \(x\)-coordinate, \(x = \frac{1}{4}\).
- A1: Finds the correct exact \(y\)-coordinate, \(y = \frac{7}{2}e^{-1/2}\) (accept \(\frac{7}{2\sqrt{e}}\)).

(i) Given \(u = y^{-2}\), we differentiate with respect to \(x\) to obtain \(\frac{\mathrm{d}u}{\mathrm{d}x} = -2y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\). Rearranging this expression gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x}\). Substituting this expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) into the original differential equation gives: \(-\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x} + y \cot x = y^3 \sin^2 x\). Dividing both sides by \(-\frac{1}{2}y^3\) (where \(y \neq 0\)), we get: \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2y^{-2} \cot x = -2 \sin^2 x\). Since \(u = y^{-2}\), this simplifies to: \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2u \cot x = -2 \sin^2 x\). Thus, \(P(x) = -2 \cot x\) and \(Q(x) = -2 \sin^2 x\). (ii) To solve the linear differential equation, we find the integrating factor \(I(x)\): \(I(x) = \mathrm{e}^{\int -2 \cot x \,\mathrm{d}x} = \mathrm{e}^{-2 \ln(\sin x)} = \mathrm{e}^{\ln(\sin^{-2} x)} = \csc^2 x = \frac{1}{\sin^2 x}\). Multiplying the differential equation by the integrating factor: \(\csc^2 x \frac{\mathrm{d}u}{\mathrm{d}x} - 2u \csc^2 x \cot x = -2 \sin^2 x \csc^2 x\), which is equivalent to: \(\frac{\mathrm{d}}{\mathrm{d}x}(u \csc^2 x) = -2\). Integrating both sides with respect to \(x\) gives: \(u \csc^2 x = \int -2 \,\mathrm{d}x = -2x + C\), where \(C\) is an arbitrary constant. Solving for \(u\) yields: \(u = (C - 2x) \sin^2 x\). Since \(u = y^{-2}\), we substitute this back to obtain: \(\frac{1}{y^2} = (C - 2x) \sin^2 x\), which gives the general solution: \(y^2 = \frac{1}{(C - 2x) \sin^2 x}\). (iii) We are given the boundary condition \(y = \sqrt{2}\) when \(x = \frac{\pi}{4}\). Since \(y > 0\) at this point, we take the positive square root for the explicit solution: \(y = \frac{1}{\sin x \sqrt{C - 2x}}\). Substituting the given values: \(\sqrt{2} = \frac{1}{\sin(\pi/4) \sqrt{C - 2(\pi/4)}}\). Using \(\sin(\pi/4) = \frac{1}{\sqrt{2}}\), we get: \(\sqrt{2} = \frac{1}{\frac{1}{\sqrt{2}} \sqrt{C - \frac{\pi}{2}}} = \frac{\sqrt{2}}{\sqrt{C - \frac{\pi}{2}}}\). This simplifies to \(\sqrt{C - \frac{\pi}{2}} = 1\), which yields \(C - \frac{\pi}{2} = 1\), so \(C = 1 + \frac{\pi}{2}\). Thus, the particular solution is: \(y = \frac{1}{\sin x \sqrt{1 + \frac{\pi}{2} - 2x}}\).

(i) M1: For differentiating \(u = y^{-2}\) using the chain rule to obtain \(\frac{\mathrm{d}u}{\mathrm{d}x} = k y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\) for some non-zero constant \(k\). A1: For the correct derivative relation \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x}\). M1: For substituting into the original differential equation and dividing the resulting equation by \(y^3\). A1: For obtaining the correct linear differential equation \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2u \cot x = -2 \sin^2 x\) with \(P(x)\) and \(Q(x)\) clearly identified or implied. (ii) M1: For attempting to find the integrating factor using \(\mathrm{e}^{\int P(x)\,\mathrm{d}x}\). A1: For obtaining the correct integrating factor \(\csc^2 x\) (or \(\frac{1}{\sin^2 x}\)). M1: For integrating the right-hand side, \(\int -2 \,\mathrm{d}x\), to get \(-2x\) (the constant of integration must be present in the solution). A1: For obtaining the correct expression for \(u\) in terms of \(x\), e.g., \(u = (C - 2x)\sin^2 x\). B1: For substituting back \(u = y^{-2}\) to get the correct general solution for \(y\), e.g., \(y^2 = \frac{1}{(C - 2x)\sin^2 x}\). (iii) M1: For substituting the initial conditions \(x = \frac{\pi}{4}\) and \(y = \sqrt{2}\) into their general solution and attempting to solve for \(C\). A1: For obtaining \(C = 1 + \frac{\pi}{2}\) (or equivalent). A1: For stating the final explicit particular solution \(y = \frac{1}{\sin x \sqrt{1 + \frac{\pi}{2} - 2x}}\) (or equivalent, e.g. \(y = \csc x (1 + \frac{\pi}{2} - 2x)^{-1/2}\)).

(i) Given \(u = y^{-2}\), we differentiate with respect to \(x\) to obtain \(\frac{\mathrm{d}u}{\mathrm{d}x} = -2y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\). Rearranging this expression gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x}\). Substituting this expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) into the original differential equation gives: \(-\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x} + y \cot x = y^3 \sin^2 x\). Dividing both sides by \(-\frac{1}{2}y^3\) (where \(y \neq 0\)), we get: \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2y^{-2} \cot x = -2 \sin^2 x\). Since \(u = y^{-2}\), this simplifies to: \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2u \cot x = -2 \sin^2 x\). Thus, \(P(x) = -2 \cot x\) and \(Q(x) = -2 \sin^2 x\). (ii) To solve the linear differential equation, we find the integrating factor \(I(x)\): \(I(x) = \mathrm{e}^{\int -2 \cot x \,\mathrm{d}x} = \mathrm{e}^{-2 \ln(\sin x)} = \mathrm{e}^{\ln(\sin^{-2} x)} = \csc^2 x = \frac{1}{\sin^2 x}\). Multiplying the differential equation by the integrating factor: \(\csc^2 x \frac{\mathrm{d}u}{\mathrm{d}x} - 2u \csc^2 x \cot x = -2 \sin^2 x \csc^2 x\), which is equivalent to: \(\frac{\mathrm{d}}{\mathrm{d}x}(u \csc^2 x) = -2\). Integrating both sides with respect to \(x\) gives: \(u \csc^2 x = \int -2 \,\mathrm{d}x = -2x + C\), where \(C\) is an arbitrary constant. Solving for \(u\) yields: \(u = (C - 2x) \sin^2 x\). Since \(u = y^{-2}\), we substitute this back to obtain: \(\frac{1}{y^2} = (C - 2x) \sin^2 x\), which gives the general solution: \(y^2 = \frac{1}{(C - 2x) \sin^2 x}\). (iii) We are given the boundary condition \(y = \sqrt{2}\) when \(x = \frac{\pi}{4}\). Since \(y > 0\) at this point, we take the positive square root for the explicit solution: \(y = \frac{1}{\sin x \sqrt{C - 2x}}\). Substituting the given values: \(\sqrt{2} = \frac{1}{\sin(\pi/4) \sqrt{C - 2(\pi/4)}}\). Using \(\sin(\pi/4) = \frac{1}{\sqrt{2}}\), we get: \(\sqrt{2} = \frac{1}{\frac{1}{\sqrt{2}} \sqrt{C - \frac{\pi}{2}}} = \frac{\sqrt{2}}{\sqrt{C - \frac{\pi}{2}}}\). This simplifies to \(\sqrt{C - \frac{\pi}{2}} = 1\), which yields \(C - \frac{\pi}{2} = 1\), so \(C = 1 + \frac{\pi}{2}\). Thus, the particular solution is: \(y = \frac{1}{\sin x \sqrt{1 + \frac{\pi}{2} - 2x}}\).

(i) M1: For differentiating \(u = y^{-2}\) using the chain rule to obtain \(\frac{\mathrm{d}u}{\mathrm{d}x} = k y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\) for some non-zero constant \(k\). A1: For the correct derivative relation \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x}\). M1: For substituting into the original differential equation and dividing the resulting equation by \(y^3\). A1: For obtaining the correct linear differential equation \(\frac{\mathrm{d}u}{\mathrm{d}x} - 2u \cot x = -2 \sin^2 x\) with \(P(x)\) and \(Q(x)\) clearly identified or implied. (ii) M1: For attempting to find the integrating factor using \(\mathrm{e}^{\int P(x)\,\mathrm{d}x}\). A1: For obtaining the correct integrating factor \(\csc^2 x\) (or \(\frac{1}{\sin^2 x}\)). M1: For integrating the right-hand side, \(\int -2 \,\mathrm{d}x\), to get \(-2x\) (the constant of integration must be present in the solution). A1: For obtaining the correct expression for \(u\) in terms of \(x\), e.g., \(u = (C - 2x)\sin^2 x\). B1: For substituting back \(u = y^{-2}\) to get the correct general solution for \(y\), e.g., \(y^2 = \frac{1}{(C - 2x)\sin^2 x}\). (iii) M1: For substituting the initial conditions \(x = \frac{\pi}{4}\) and \(y = \sqrt{2}\) into their general solution and attempting to solve for \(C\). A1: For obtaining \(C = 1 + \frac{\pi}{2}\) (or equivalent). A1: For stating the final explicit particular solution \(y = \frac{1}{\sin x \sqrt{1 + \frac{\pi}{2} - 2x}}\) (or equivalent, e.g. \(y = \csc x (1 + \frac{\pi}{2} - 2x)^{-1/2}\)).

(a) To find the eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \). This gives \( \det \begin{pmatrix} -3-\lambda & 1 & 3 \\ -5 & 3-\lambda & 3 \\ -7 & 1 & 7-\lambda \end{pmatrix} = 0 \). Expanding along the first row: \( (-3-\lambda) [ (3-\lambda)(7-\lambda) - 3 ] - 1 [ -5(7-\lambda) - (-21) ] + 3 [ -5 - (-7(3-\lambda)) ] = 0 \). Simplifying this expression yields \( (-3-\lambda) ( \lambda^2 - 10\lambda + 18 ) - (5\lambda - 14) + 3(16 - 7\lambda) = 0 \). Expanding and combining like terms gives the characteristic equation \( \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \). Factorising the cubic equation gives \( (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0 \). Thus, the eigenvalues are \( \lambda = 1, 2, 4 \). (b) Next, we find the eigenvectors. For \( \lambda = 1 \): we solve \( (A - I)\mathbf{v} = \mathbf{0} \), which leads to the system \( -4x + y + 3z = 0 \), \( -5x + 2y + 3z = 0 \), and \( -7x + y + 6z = 0 \). Subtracting the first equation from the second gives \( -x + y = 0 \), so \( x = y \). Substituting this into the first equation yields \( -3x + 3z = 0 \), so \( x = z \). Thus, a corresponding eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \). For \( \lambda = 2 \): we solve \( (A - 2I)\mathbf{v} = \mathbf{0} \), which leads to the system \( -5x + y + 3z = 0 \), \( -5x + y + 3z = 0 \), and \( -7x + y + 5z = 0 \). Subtracting the first equation from the third gives \( -2x + 2z = 0 \), so \( x = z \). Substituting this into the first equation gives \( y = 2x \). Thus, a corresponding eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \). For \( \lambda = 4 \): we solve \( (A - 4I)\mathbf{v} = \mathbf{0} \), which leads to the system \( -7x + y + 3z = 0 \), \( -5x - y + 3z = 0 \), and \( -7x + y + 3z = 0 \). Adding the first two equations gives \( -12x + 6z = 0 \), so \( z = 2x \). Substituting this into the first equation gives \( -x + y = 0 \), so \( y = x \). Thus, a corresponding eigenvector is \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \). (c) We can construct \( P \) and \( D \) using the eigenvectors and eigenvalues: \( P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \) and \( D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \). To find \( P^{-1} \), we calculate the determinant of \( P \): \( \det(P) = 1(4-1) - 1(2-1) + 1(1-2) = 1 \). Finding the matrix of cofactors and transposing gives \( P^{-1} = \begin{pmatrix} 3 & -1 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix} \). Now, we compute \( A^n = P D^n P^{-1} \). First, find \( P D^n = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} = \begin{pmatrix} 1 & 2^n & 4^n \\ 1 & 2^{n+1} & 4^n \\ 1 & 2^n & 2 \cdot 4^n \end{pmatrix} \). Multiplying this by \( P^{-1} \) gives \( A^n = \begin{pmatrix} 1 & 2^n & 4^n \\ 1 & 2^{n+1} & 4^n \\ 1 & 2^n & 2 \cdot 4^n \end{pmatrix} \begin{pmatrix} 3 & -1 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 - 2^n - 4^n & -1 + 2^n & -1 + 4^n \\ 3 - 2^{n+1} - 4^n & -1 + 2^{n+1} & -1 + 4^n \\ 3 - 2^n - 2 \cdot 4^n & -1 + 2^n & -1 + 2 \cdot 4^n \end{pmatrix} \).

Part (a): M1 for setting up det(A - lambda I) = 0. A1 for the correct cubic equation lambda^3 - 7*lambda^2 + 14*lambda - 8 = 0. M1 for attempting to factorise the cubic equation. A2 for obtaining lambda = 1, 2, 4 (A1 for any two correct, A2 for all three). [5 marks total] Part (b): M1 for attempting to solve (A - lambda I)v = 0 for at least one eigenvalue. A1 for a correct eigenvector corresponding to lambda = 1. A1 for a correct eigenvector corresponding to lambda = 2. A1 for a correct eigenvector corresponding to lambda = 4. A1 for expressing the final set of eigenvectors clearly. [5 marks total] Part (c): B1 for correct matrices P and D. M1 for a valid method to find P^-1. A1 for the correct inverse matrix P^-1. M1 for attempting the multiplication P * D^n * P^-1. A2 for the correct final matrix A^n (A1 for at least 5 correct elements, A2 for all correct elements). [6 marks total]

Let the line of centers of the spheres at the moment of impact be the \(x\)-axis, and the perpendicular direction in the plane of motion be the \(y\)-axis.

Before collision, the velocity components of sphere \(A\) are:
\(u_{Ax} = u \cos \alpha = 0.8u\)
\(u_{Ay} = u \sin \alpha = \sqrt{1 - 0.8^2} u = 0.6u\)

Sphere \(B\) is initially at rest, so:
\(u_{Bx} = 0\)
\(u_{By} = 0\)

Since the spheres are smooth, there is no impulse parallel to the tangent plane (along the \(y\)-axis). Thus, the \(y\)-components of the velocities remain unchanged after the collision:
\(v_{Ay} = 0.6u\)
\(v_{By} = 0\)

Let \(v_{Ax}\) and \(v_{Bx}\) be the velocity components along the line of centers after the collision.

Applying conservation of linear momentum parallel to the line of centers:
\(m u_{Ax} + 5m u_{Bx} = m v_{Ax} + 5m v_{Bx}\)
\(m(0.8u) + 0 = m v_{Ax} + 5m v_{Bx}\)
\(v_{Ax} + 5v_{Bx} = 0.8u\) --- (Equation 1)

Applying Newton's law of restitution parallel to the line of centers:
\(v_{Bx} - v_{Ax} = e(u_{Ax} - u_{Bx})\)
\(v_{Bx} - v_{Ax} = \frac{7}{8}(0.8u - 0) = 0.7u\) --- (Equation 2)

From Equation 2, we have \(v_{Bx} = v_{Ax} + 0.7u\). Substituting this into Equation 1:
\(v_{Ax} + 5(v_{Ax} + 0.7u) = 0.8u\)
\(6v_{Ax} + 3.5u = 0.8u\)
\(6v_{Ax} = -2.7u \implies v_{Ax} = -0.45u\)

(i) The speed \(V_A\) of sphere \(A\) immediately after the collision is given by:
\(V_A = \sqrt{v_{Ax}^2 + v_{Ay}^2} = \sqrt{(-0.45u)^2 + (0.6u)^2}\)
\(V_A = \sqrt{0.2025u^2 + 0.36u^2} = \sqrt{0.5625u^2} = 0.75u\) (or \(\frac{3}{4}u\)).

(ii) To find the deflection of the path of \(A\), we compare the velocity vectors of \(A\) before and after the collision:
Before collision, the direction vector is \(\mathbf{u}_A = \begin{pmatrix} 0.8u \\ 0.6u \end{pmatrix}\).
After collision, the direction vector is \(\mathbf{v}_A = \begin{pmatrix} -0.45u \\ 0.6u \end{pmatrix}\).

Taking the scalar (dot) product of \(\mathbf{u}_A\) and \(\mathbf{v}_A\):
\(\mathbf{u}_A \cdot \mathbf{v}_A = (0.8u)(-0.45u) + (0.6u)(0.6u)\)
\(\mathbf{u}_A \cdot \mathbf{v}_A = -0.36u^2 + 0.36u^2 = 0\)

Since the scalar product is zero, the initial and final velocity vectors are perpendicular. Hence, the direction of motion of \(A\) is deflected through exactly \(90^\circ\).

M1: For writing a correct conservation of momentum equation along the line of centers.
M1: For writing a correct restitution equation along the line of centers.
A1: For obtaining \(v_{Ax} = -0.45u\) (or equivalent).
A1: For obtaining the correct speed of \(A\) after collision, \(V_A = 0.75u\) (or \(\frac{3}{4}u\)).
M1: For an appropriate method to show perpendicularity (e.g., using the scalar product of initial and final velocity vectors, or trigonometric angle subtraction).
A1: For showing \(\mathbf{u}_A \cdot \mathbf{v}_A = 0\) or equivalent clear proof to conclude the angle of deflection is indeed \(90^\circ\).

Part (i): Let \(x\) be the extension of the string when the particle is at its lowest point. The total distance fallen from \(O\) is \(0.8 + x\). By conservation of energy, the loss in gravitational potential energy equals the gain in elastic potential energy: \(mg(L + x) = \frac{\lambda x^2}{2L}\). Substituting the given values: \(0.4 \times 10(0.8 + x) = \frac{16x^2}{2 \times 0.8}\), which simplifies to \(4(0.8 + x) = 10x^2\). Rearranging gives \(10x^2 - 4x - 3.2 = 0\), or \(25x^2 - 10x - 8 = 0\). Factoring the quadratic: \((5x - 4)(5x + 2) = 0\). Since the extension \(x\) must be positive, we have \(x = 0.8\text{ m}\). Thus, the maximum distance below \(O\) is \(0.8 + 0.8 = 1.6\text{ m}\). Part (ii): The maximum speed occurs when the acceleration is zero, which is at the equilibrium position where tension equals weight: \(T = mg \implies \frac{\lambda x}{L} = mg\). Substituting values gives \(\frac{16x}{0.8} = 4 \implies 20x = 4 \implies x = 0.2\text{ m}\). At this point, the distance fallen is \(0.8 + 0.2 = 1.0\text{ m}\). Using conservation of energy between the release point and this position: \(mg(L + x) = \frac{1}{2}mv^2 + \frac{\lambda x^2}{2L}\). Substituting the values: \(0.4 \times 10 \times 1.0 = \frac{1}{2}(0.4)v^2 + \frac{16(0.2)^2}{1.6}\), which simplifies to \(4 = 0.2v^2 + 0.4 \implies 3.6 = 0.2v^2 \implies v^2 = 18\). Thus, the maximum speed is \(v = \sqrt{18} = 3\sqrt{2}\text{ m s}^{-1}\) (or approximately \(4.24\text{ m s}^{-1}\)).

Part (i): M1: Set up a conservation of energy equation equating loss in GPE to gain in EPE. A1: Correct quadratic equation in \(x\), e.g., \(25x^2 - 10x - 8 = 0\). A1: Correct maximum distance of \(1.6\text{ m}\). Part (ii): M1: Attempt to find the extension at equilibrium by setting \(T = mg\). A1: Correct extension \(x = 0.2\text{ m}\). M1: Set up conservation of energy equation at this point to find \(v\). A1: Correct maximum speed of \(3\sqrt{2}\text{ m s}^{-1}\) or \(4.24\text{ m s}^{-1}\).

(i) Let the horizontal and vertical unit vectors be \(\mathbf{i}\) and \(\mathbf{j}\) respectively.
Given \(\sin\alpha = 0.8\), we have \(\cos\alpha = 0.6\).
The initial velocity vector is:
\(\mathbf{u} = 20\cos\alpha\mathbf{i} + 20\sin\alpha\mathbf{j} = 12\mathbf{i} + 16\mathbf{j}\)

At time \(t\), the velocity vector \(\mathbf{v}\) is:
\(\mathbf{v} = 12\mathbf{i} + (16 - gt)\mathbf{j} = 12\mathbf{i} + (16 - 10t)\mathbf{j}\)

The direction of motion is perpendicular to the initial direction when \(\mathbf{u} \cdot \mathbf{v} = 0\):
\((12\mathbf{i} + 16\mathbf{j}) \cdot (12\mathbf{i} + (16 - 10t)\mathbf{j}) = 0\)
\(12(12) + 16(16 - 10t) = 0\)
\(144 + 256 - 160t = 0\)
\(400 - 160t = 0 \implies t = 2.5\text{ s}\)

Alternative trigonometric method:
Let \(\theta\) be the angle of the velocity vector with the horizontal at time \(t\).
\(\tan\theta = \frac{16 - 10t}{12}\)
Since \(\mathbf{v}\) is perpendicular to \(\mathbf{u}\):
\(\tan\theta = -\frac{1}{\tan\alpha} = -\frac{\cos\alpha}{\sin\alpha} = -\frac{0.6}{0.8} = -0.75\)
\(\frac{16 - 10t}{12} = -0.75 \implies 16 - 10t = -9 \implies t = 2.5\text{ s}\)

(ii) Substituting \(t = 2.5\) into the velocity vector:
\(\mathbf{v} = 12\mathbf{i} + (16 - 10(2.5))\mathbf{j} = 12\mathbf{i} - 9\mathbf{j}\)

The speed is the magnitude of the velocity vector:
\(v = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ m s}^{-1}\)

Alternative geometric method:
Using the right-angled vector triangle, the speed is:
\(v = u\cot\alpha = 20 \times \frac{0.6}{0.8} = 15\text{ m s}^{-1}\)

(i)
M1: For expressing the velocity vector in terms of \(t\) (or equivalent trigonometric expression for \(\tan\theta\)).
M1: For setting up the perpendicularity condition, e.g., \(\mathbf{u} \cdot \mathbf{v} = 0\) or \(\tan\theta = -\cot\alpha\).
A1: For obtaining \(t = 2.5\text{ s}\) (or equivalent fraction).

(ii)
M1: For substituting their value of \(t\) to find the vertical component of velocity (or using the geometric relation \(v = u\cot\alpha\)).
A1: For obtaining speed \(= 15\text{ m s}^{-1}\).

Let \(W = mg\) be the weight of the ladder, acting at its midpoint (distance \(a\) from \(A\)). Let \(2W = 2mg\) be the weight of the man, at distance \(x\) from \(A\). Let \(R\) and \(F_1\) be the normal reaction and the frictional force at \(A\), where \(F_1 = 0.5 R\) at the point of slipping. Let \(S\) and \(F_2\) be the normal reaction and the frictional force at \(B\), where \(F_2 = 0.5 S\) at the point of slipping. Resolving horizontally: \(F_1 = S\), which gives \(0.5 R = S\), so \(R = 2S\). Resolving vertically: \(R + F_2 = 3W\). Substituting \(R = 2S\) and \(F_2 = 0.5 S\) gives \(2S + 0.5 S = 3W\), which simplifies to \(2.5 S = 3W\), so \(S = 1.2 W\) and \(R = 2.4 W\). It follows that \(F_2 = 0.6 W\). Taking moments about \(A\): \(W a \cos 45^\circ + 2W x \cos 45^\circ = S (2a \sin 45^\circ) + F_2 (2a \cos 45^\circ)\). Since \(\sin 45^\circ = \cos 45^\circ\), we can divide the equation by \(W \cos 45^\circ\) to get: \(a + 2x = 2a \left(\frac{S}{W}\right) + 2a \left(\frac{F_2}{W}\right)\). Substituting \(\frac{S}{W} = 1.2\) and \(\frac{F_2}{W} = 0.6\) gives: \(a + 2x = 2a(1.2) + 2a(0.6) = 2.4a + 1.2a = 3.6a\). Solving for \(x\) gives: \(2x = 2.6a\), which yields \(x = 1.3a\).

M1: Set up horizontal and vertical equations of equilibrium. M1: Use the limiting friction condition at both contacts to express \(S\) and \(R\) in terms of \(W\). A1: Obtain correct values \(S = 1.2W\) and \(R = 2.4W\) (or equivalent). M1: Take moments about a point, with correct number of terms and trigonometric factors. A1: Obtain a correct moments equation, e.g., \(W a \cos 45^\circ + 2W x \cos 45^\circ = S 2a \sin 45^\circ + F_2 2a \cos 45^\circ\). M1: Substitute \(S\) and \(F_2\) and solve for \(x\). A1: Obtain final answer \(1.3a\) (or \(\frac{13}{10}a\)).

(i) Let \(\theta\) be the angle the string \(AP\) makes with the vertical axis.
The distance from the axis of rotation is \(OP = 0.3\text{ m}\) and the vertical height is \(OA = 0.4\text{ m}\).
Thus, the length of the string is:
\(AP = \sqrt{OA^2 + OP^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m}\)

From the geometry of the triangle, we have:
\(\sin\theta = \frac{OP}{AP} = \frac{0.3}{0.5} = 0.6\)
\(\cos\theta = \frac{OA}{AP} = \frac{0.4}{0.5} = 0.8\)

Using Hooke's law, the tension \(T\) in the elastic string is:
\(T = \frac{\lambda x}{L} = \frac{3.75 \times (0.5 - 0.3)}{0.3} = \frac{3.75 \times 0.2}{0.3} = 2.5\text{ N}\)

Resolving forces vertically for the particle in contact with the turntable:
\(R + T\cos\theta = mg\)
where \(R\) is the normal reaction, \(m = 0.5\text{ kg}\), and \(g = 10\text{ m s}^{-2}\).
\(R + 2.5 \times 0.8 = 0.5 \times 10\)
\(R + 2.0 = 5.0 \implies R = 3\text{ N}\) (as required).

(ii) The maximum frictional force is:
\(F_{\text{max}} = \mu R = 0.3 \times 3.0 = 0.9\text{ N}\)

Let \(F\) be the frictional force acting radially inwards towards \(O\). The equation of motion for the circular path of radius \(r = 0.3\text{ m}\) is:
\(T\sin\theta + F = mr\omega^2\)
\(2.5 \times 0.6 + F = 0.5 \times 0.3 \times \omega^2\)
\(1.5 + F = 0.15\omega^2 \implies F = 0.15\omega^2 - 1.5\)

For no slipping, the frictional force must satisfy \(-F_{\text{max}} \le F \le F_{\text{max}}\):
\(-0.9 \le 0.15\omega^2 - 1.5 \le 0.9\)

Adding \(1.5\) to all parts of the inequality:
\(0.6 \le 0.15\omega^2 \le 2.4\)

Dividing by \(0.15\):
\(4 \le \omega^2 \le 16\)

Since \(\omega > 0\), taking the square root gives:
\(2 \le \omega \le 4\)

(i)
M1: For attempting to find the length of the string and the value of \(\cos\theta\) or \(\sin\theta\).
A1: For finding the tension \(T = 2.5\text{ N}\) using Hooke's law.
A1: For resolving vertically and showing that \(R = 3\text{ N}\).

(ii)
B1: For finding the maximum friction force \(F_{\text{max}} = 0.9\text{ N}\).
M1: For setting up the horizontal equation of motion involving tension component, friction, and centripetal acceleration.
M1: For formulating the inequality \(-0.9 \le F \le 0.9\) and attempting to solve for \(\omega\).
A1: For obtaining the correct range \(2 \le \omega \le 4\) (accept equivalent notations).

(i) The equation of motion is: \(m \frac{\mathrm{d}v}{\mathrm{d}t} = -R\) which gives \(0.2 \frac{\mathrm{d}v}{\mathrm{d}t} = -0.1(v + v^2)\), simplifying to \(\frac{\mathrm{d}v}{\mathrm{d}t} = -0.5 v(1+v)\). Separating the variables: \(\int \frac{1}{v(1+v)} \mathrm{d}v = -0.5 \int \mathrm{d}t\). Using partial fractions: \(\int \left( \frac{1}{v} - \frac{1}{1+v} \right) \mathrm{d}v = -0.5 \int \mathrm{d}t\), which integrates to \(\ln\left(\frac{v}{1+v}\right) = -0.5t + C\). Given that \(v = 10\) when \(t = 0\), we find \(C = \ln\left(\frac{10}{11}\right)\). Substituting \(v = 2\) gives \(\ln\left(\frac{2}{3}\right) = -0.5t + \ln\left(\frac{10}{11}\right)\), which simplifies to \(0.5t = \ln\left(\frac{10}{11} \times \frac{3}{2}\right) = \ln\left(\frac{15}{11}\right)\). Therefore, \(t = 2 \ln\left(\frac{15}{11}\right)\). (ii) Using \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\), we have \(0.2 v \frac{\mathrm{d}v}{\mathrm{d}x} = -0.1(v + v^2)\). Since \(v > 0\) during the motion, dividing by \(v\) yields \(2 \frac{\mathrm{d}v}{\mathrm{d}x} = -(1+v)\). Separating variables gives \(\int \frac{2}{1+v} \mathrm{d}v = -\int \mathrm{d}x\), which integrates to \(2 \ln(1+v) = -x + K\). Using the initial condition \(x = 0, v = 10\), we get \(2 \ln(11) = K\). Thus, \(x = 2 \ln(11) - 2 \ln(1+v) = 2 \ln \left( \frac{11}{1+v} \right)\). When \(v = 2\), \(x = 2 \ln \left( \frac{11}{3} \right) \approx 2.5986\text{ m}\), which is \(2.60\text{ m}\) to 3 significant figures.

(i) M1: For setting up the Newton's second law equation with the correct resistive force. M1: For separating variables and integrating with partial fractions. A1: For finding the correct constant of integration. A1: For obtaining \(t = 2 \ln(15/11)\) or equivalent. (ii) M1: For substituting the variable-force acceleration formula and separating variables. A1: For obtaining the correct integrated form with constant of integration. M1: For using the initial condition to solve for \(K\) and deriving the given equation for \(x\). A1: For substituting \(v=2\). A1: For calculating the final distance to 3 s.f. as \(2.60\text{ m}\) (accept 2.60, reject 2.6).

(a) Let \(\theta\) be the angle that the radius from \(O\) to the particle makes with the upward vertical.

By conservation of mechanical energy, taking the level of \(O\) as the reference for potential energy:
\[ E_i = m g a \]
\[ E_f = \frac{1}{2} m v^2 + m g a \cos \theta \]

Equating these gives:
\[ \frac{1}{2} m v^2 = m g a (1 - \cos \theta) \implies v^2 = 2 g a (1 - \cos \theta) \]

The equation of motion for the particle along the radial direction towards the center \(O\) is:
\[ m g \cos \theta - R = \frac{m v^2}{a} \]
where \(R\) is the normal reaction force.

The particle leaves the surface of the sphere when \(R = 0\):
\[ g \cos \theta = \frac{v^2}{a} \]

Substituting \(v^2\):
\[ g \cos \theta = 2 g (1 - \cos \theta) \implies \cos \theta = 2 - 2 \cos \theta \implies 3 \cos \theta = 2 \implies \cos \theta = \frac{2}{3} \]

(b) At the point of departure \(P\), \(\cos \theta = \frac{2}{3}\), which gives:
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \frac{\sqrt{5}}{3} \]

The coordinates of \(P\) relative to \(O\) (with the positive \(x\)-axis horizontal and the positive \(y\)-axis vertically upward) are:
\[ x_P = a \sin \theta = \frac{\sqrt{5}}{3} a \]
\[ y_P = a \cos \theta = \frac{2}{3} a \]

The speed at departure is:
\[ v = \sqrt{2 g a \left(1 - \frac{2}{3}\right)} = \sqrt{\frac{2}{3} g a} \]

The velocity vector at departure makes an angle \(\theta\) below the horizontal, so its horizontal and vertical components of velocity are:
\[ u_x = v \cos \theta = \frac{2}{3} \sqrt{\frac{2}{3} g a} = \frac{2\sqrt{6}}{9}\sqrt{g a} \]
\[ u_y = -v \sin \theta = -\frac{\sqrt{5}}{3} \sqrt{\frac{2}{3} g a} = -\frac{\sqrt{30}}{9}\sqrt{g a} \]

Since the sphere of radius \(a\) rests on the table, the table is at a vertical height of \(-a\) relative to \(O\). The vertical displacement from \(P\) to the table is:
\[ y = -a - y_P = -a - \frac{2}{3} a = -\frac{5}{3} a \]

Using the vertical motion equation for a projectile:
\[ y = u_y t - \frac{1}{2} g t^2 \]
\[ -\frac{5}{3} a = -\frac{\sqrt{30}}{9}\sqrt{g a} t - \frac{1}{2} g t^2 \]
\[ \frac{1}{2} g t^2 + \frac{\sqrt{30}}{9}\sqrt{g a} t - \frac{5}{3} a = 0 \]

Multiplying by \(\frac{18}{a}\):
\[ 9 \left(\sqrt{\frac{g}{a}} t\right)^2 + 2\sqrt{30} \left(\sqrt{\frac{g}{a}} t\right) - 30 = 0 \]

Let \(z = \sqrt{\frac{g}{a}} t\). Then:
\[ 9 z^2 + 2\sqrt{30} z - 30 = 0 \]

Solving for the positive root of this quadratic equation:
\[ z = \frac{-2\sqrt{30} + \sqrt{120 - 4(9)(-30)}}{18} = \frac{-2\sqrt{30} + \sqrt{1200}}{18} = \frac{10\sqrt{3} - \sqrt{30}}{9} \]

Thus, the time of flight is:
\[ t = \left(\frac{10\sqrt{3} - \sqrt{30}}{9}\right) \sqrt{\frac{a}{g}} \]

The horizontal distance traveled as a projectile is:
\[ x_f = u_x t = \left(\frac{2\sqrt{6}}{9}\sqrt{g a}\right) \left(\frac{10\sqrt{3} - \sqrt{30}}{9}\right) \sqrt{\frac{a}{g}} = \frac{2\sqrt{6}(10\sqrt{3} - \sqrt{30})}{81} a \]
\[ x_f = \frac{2(30\sqrt{2} - 6\sqrt{5})}{81} a = \frac{20\sqrt{2} - 4\sqrt{5}}{27} a \]

The total horizontal distance from the center of the sphere to where the particle strikes the table is:
\[ d = x_P + x_f = \frac{\sqrt{5}}{3} a + \frac{20\sqrt{2} - 4\sqrt{5}}{27} a = \frac{9\sqrt{5} + 20\sqrt{2} - 4\sqrt{5}}{27} a = \frac{20\sqrt{2} + 5\sqrt{5}}{27} a = \frac{5}{27}(4\sqrt{2} + \sqrt{5})a \]

**Part (a)**
* **M1**: Use conservation of energy to find \(v^2\) in terms of \(g, a, \theta\).
* **M1**: Write radial equation of motion including \(R\) and setting \(R = 0\).
* **M1**: Equate \(g \cos \theta\) and \(v^2 / a\).
* **A1**: Correctly obtain \(\cos \theta = 2/3\).

**Part (b)**
* **B1**: State coordinates of departure point \(P\) or components of velocity \(u_x\) and \(u_y\) correctly.
* **M1**: Formulate the quadratic equation for the time of flight \(t\) using the vertical drop of \(\frac{5}{3} a\).
* **A1**: Solve the quadratic equation to obtain \(t = \frac{10\sqrt{3} - \sqrt{30}}{9} \sqrt{\frac{a}{g}}\) or equivalent.
* **M1**: Use \(x_f = u_x t\) to find the horizontal projectile displacement.
* **A1**: Correctly sum the horizontal displacements to find \(d = \frac{5}{27}(4\sqrt{2} + \sqrt{5})a\) (allow equivalent exact simplified forms).

(a) Let \(\theta\) be the angle that the radius from \(O\) to the particle makes with the upward vertical.

By conservation of mechanical energy, taking the level of \(O\) as the reference for potential energy:
\[ E_i = m g a \]
\[ E_f = \frac{1}{2} m v^2 + m g a \cos \theta \]

Equating these gives:
\[ \frac{1}{2} m v^2 = m g a (1 - \cos \theta) \implies v^2 = 2 g a (1 - \cos \theta) \]

The equation of motion for the particle along the radial direction towards the center \(O\) is:
\[ m g \cos \theta - R = \frac{m v^2}{a} \]
where \(R\) is the normal reaction force.

The particle leaves the surface of the sphere when \(R = 0\):
\[ g \cos \theta = \frac{v^2}{a} \]

Substituting \(v^2\):
\[ g \cos \theta = 2 g (1 - \cos \theta) \implies \cos \theta = 2 - 2 \cos \theta \implies 3 \cos \theta = 2 \implies \cos \theta = \frac{2}{3} \]

(b) At the point of departure \(P\), \(\cos \theta = \frac{2}{3}\), which gives:
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \frac{\sqrt{5}}{3} \]

The coordinates of \(P\) relative to \(O\) (with the positive \(x\)-axis horizontal and the positive \(y\)-axis vertically upward) are:
\[ x_P = a \sin \theta = \frac{\sqrt{5}}{3} a \]
\[ y_P = a \cos \theta = \frac{2}{3} a \]

The speed at departure is:
\[ v = \sqrt{2 g a \left(1 - \frac{2}{3}\right)} = \sqrt{\frac{2}{3} g a} \]

The velocity vector at departure makes an angle \(\theta\) below the horizontal, so its horizontal and vertical components of velocity are:
\[ u_x = v \cos \theta = \frac{2}{3} \sqrt{\frac{2}{3} g a} = \frac{2\sqrt{6}}{9}\sqrt{g a} \]
\[ u_y = -v \sin \theta = -\frac{\sqrt{5}}{3} \sqrt{\frac{2}{3} g a} = -\frac{\sqrt{30}}{9}\sqrt{g a} \]

Since the sphere of radius \(a\) rests on the table, the table is at a vertical height of \(-a\) relative to \(O\). The vertical displacement from \(P\) to the table is:
\[ y = -a - y_P = -a - \frac{2}{3} a = -\frac{5}{3} a \]

Using the vertical motion equation for a projectile:
\[ y = u_y t - \frac{1}{2} g t^2 \]
\[ -\frac{5}{3} a = -\frac{\sqrt{30}}{9}\sqrt{g a} t - \frac{1}{2} g t^2 \]
\[ \frac{1}{2} g t^2 + \frac{\sqrt{30}}{9}\sqrt{g a} t - \frac{5}{3} a = 0 \]

Multiplying by \(\frac{18}{a}\):
\[ 9 \left(\sqrt{\frac{g}{a}} t\right)^2 + 2\sqrt{30} \left(\sqrt{\frac{g}{a}} t\right) - 30 = 0 \]

Let \(z = \sqrt{\frac{g}{a}} t\). Then:
\[ 9 z^2 + 2\sqrt{30} z - 30 = 0 \]

Solving for the positive root of this quadratic equation:
\[ z = \frac{-2\sqrt{30} + \sqrt{120 - 4(9)(-30)}}{18} = \frac{-2\sqrt{30} + \sqrt{1200}}{18} = \frac{10\sqrt{3} - \sqrt{30}}{9} \]

Thus, the time of flight is:
\[ t = \left(\frac{10\sqrt{3} - \sqrt{30}}{9}\right) \sqrt{\frac{a}{g}} \]

The horizontal distance traveled as a projectile is:
\[ x_f = u_x t = \left(\frac{2\sqrt{6}}{9}\sqrt{g a}\right) \left(\frac{10\sqrt{3} - \sqrt{30}}{9}\right) \sqrt{\frac{a}{g}} = \frac{2\sqrt{6}(10\sqrt{3} - \sqrt{30})}{81} a \]
\[ x_f = \frac{2(30\sqrt{2} - 6\sqrt{5})}{81} a = \frac{20\sqrt{2} - 4\sqrt{5}}{27} a \]

The total horizontal distance from the center of the sphere to where the particle strikes the table is:
\[ d = x_P + x_f = \frac{\sqrt{5}}{3} a + \frac{20\sqrt{2} - 4\sqrt{5}}{27} a = \frac{9\sqrt{5} + 20\sqrt{2} - 4\sqrt{5}}{27} a = \frac{20\sqrt{2} + 5\sqrt{5}}{27} a = \frac{5}{27}(4\sqrt{2} + \sqrt{5})a \]

**Part (a)**
* **M1**: Use conservation of energy to find \(v^2\) in terms of \(g, a, \theta\).
* **M1**: Write radial equation of motion including \(R\) and setting \(R = 0\).
* **M1**: Equate \(g \cos \theta\) and \(v^2 / a\).
* **A1**: Correctly obtain \(\cos \theta = 2/3\).

**Part (b)**
* **B1**: State coordinates of departure point \(P\) or components of velocity \(u_x\) and \(u_y\) correctly.
* **M1**: Formulate the quadratic equation for the time of flight \(t\) using the vertical drop of \(\frac{5}{3} a\).
* **A1**: Solve the quadratic equation to obtain \(t = \frac{10\sqrt{3} - \sqrt{30}}{9} \sqrt{\frac{a}{g}}\) or equivalent.
* **M1**: Use \(x_f = u_x t\) to find the horizontal projectile displacement.
* **A1**: Correctly sum the horizontal displacements to find \(d = \frac{5}{27}(4\sqrt{2} + \sqrt{5})a\) (allow equivalent exact simplified forms).

First, we calculate the sample mean (\bar{x}): [\bar{x} = \frac{\sum x}{n} = \frac{124.0}{8} = 15.5] Next, we calculate the unbiased estimate of the population variance (s^2): [s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{7} \left( 1934.5 - \frac{124.0^2}{8} \right) = \frac{1}{7}(1934.5 - 1922.0) = \frac{12.5}{7} \approx 1.7857] Since the population is normally distributed with unknown variance and the sample size is small (n = 8), we use the (t)-distribution with (v = n - 1 = 7) degrees of freedom. For a 95% confidence interval, the critical value is (t_{7}(0.975) = 2.365). The confidence interval is given by: [\bar{x} \pm t \sqrt{\frac{s^2}{n}} = 15.5 \pm 2.365 \sqrt{\frac{1.7857}{8}} = 15.5 \pm 2.365 \times 0.47246 = 15.5 \pm 1.117] This yields: Lower limit = (15.5 - 1.117 = 14.38) (to 2 d.p.) Upper limit = (15.5 + 1.117 = 16.62) (to 2 d.p.) Therefore, the 95% confidence interval is [14.38, 16.62].

M1: For calculating the sample mean (\bar{x} = 15.5) and attempting to calculate the unbiased estimate of the population variance (s^2). A1: For finding the correct value of (s^2 = \frac{12.5}{7} \approx 1.786) (or (s \approx 1.336)). B1: For stating or using the correct critical value (t_7(0.975) = 2.365). A1: For obtaining the correct interval [14.38, 16.62] (accept alternative bracket notations or (14.38) to (16.62)).

Let \(m\) be the population median fuel efficiency.

State the hypotheses:
\(H_0: m = 35.0\)
\(H_1: m > 35.0\)

Under \(H_0\), the number of cars with a fuel efficiency greater than 35.0, denoted by \(X\), follows a binomial distribution:
\(X \sim B(12, 0.5)\)

We determine the signs of the differences (value \(- 35.0\)) for the 12 observations:
- \(36.2 > 35.0\) (+)
- \(34.8 < 35.0\) (-)
- \(38.1 > 35.0\) (+)
- \(35.5 > 35.0\) (+)
- \(37.0 > 35.0\) (+)
- \(34.9 < 35.0\) (-)
- \(36.6 > 35.0\) (+)
- \(35.2 > 35.0\) (+)
- \(39.4 > 35.0\) (+)
- \(34.1 < 35.0\) (-)
- \(37.3 > 35.0\) (+)
- \(36.0 > 35.0\) (+)

This gives 9 plus signs (+) and 3 minus signs (-).

Since we are testing for an increase, the test statistic is the number of positive signs, \(x = 9\).

We find the probability of obtaining at least 9 positive signs under \(H_0\):
\(P(X \ge 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)\)
\(P(X \ge 9) = \left[ \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12} \right] (0.5)^{12}\)
\(P(X \ge 9) = \frac{220 + 66 + 12 + 1}{4096} = \frac{299}{4096} \approx 0.0730\)

Compare the p-value with the significance level \(\alpha = 0.10\):
Since \(0.0730 < 0.10\), we reject \(H_0\).

There is sufficient evidence at the 10% level of significance to support the claim that the fuel additive increases the median fuel efficiency.

B1: State both null and alternative hypotheses correctly, defining the parameter as the median.
M1: Use binomial distribution \(B(12, 0.5)\) and state the correct number of positive signs (9).
A1: Calculate the correct probability of \(P(X \ge 9) = \frac{299}{4096} \approx 0.0730\) (or find critical region \(X \ge 9\)).
M1: Compare their calculated probability with 0.10 (or compare the test statistic 9 to the critical value 9).
A1: Reject \(H_0\) and state a correct conclusion in context.

Let \( D = \text{Score After} - \text{Score Before} \).

State the null and alternative hypotheses:
- \( H_0 \): The median of the differences is 0 (the training course does not improve manual dexterity scores).
- \( H_1 \): The median of the differences is greater than 0 (the training course improves manual dexterity scores).

Calculate the differences \( D_i \) for each participant:
- \( A \): \( 48 - 42 = +6 \)
- \( B \): \( 53 - 55 = -2 \)
- \( C \): \( 49 - 38 = +11 \)
- \( D \): \( 69 - 62 = +7 \)
- \( E \): \( 52 - 47 = +5 \)
- \( F \): \( 47 - 51 = -4 \)
- \( G \): \( 52 - 39 = +13 \)
- \( H \): \( 65 - 56 = +9 \)

Rank the absolute differences \( |D_i| \) in ascending order:
- \( |D_i| = 2 \) (from B): Rank 1
- \( |D_i| = 4 \) (from F): Rank 2
- \( |D_i| = 5 \) (from E): Rank 3
- \( |D_i| = 6 \) (from A): Rank 4
- \( |D_i| = 7 \) (from D): Rank 5
- \( |D_i| = 9 \) (from H): Rank 6
- \( |D_i| = 11 \) (from C): Rank 7
- \( |D_i| = 13 \) (from G): Rank 8

Calculate the sum of the ranks of the positive and negative differences:
- Sum of negative ranks, \( T^- = 1 + 2 = 3 \)
- Sum of positive ranks, \( T^+ = 3 + 4 + 5 + 6 + 7 + 8 = 33 \)

The test statistic is \( T = \min(T^+, T^-) = 3 \).

For a one-tailed test at the 5% significance level with \( n = 8 \), the critical value from the Wilcoxon signed-rank tables is 5.

Since \( T = 3 \le 5 \), the test statistic lies in the critical region.

Therefore, we reject the null hypothesis \( H_0 \). There is sufficient evidence at the 5% significance level to conclude that the training course improves manual dexterity scores.

- **B1**: For both hypotheses stated correctly in terms of the median difference.
- **M1**: For calculating the differences \( D = \text{After} - \text{Before} \) (or vice versa).
- **A1**: For obtaining the correct set of differences: \( +6, -2, +11, +7, +5, -4, +13, +9 \).
- **M1**: For ranking the absolute values of the differences correctly.
- **A1**: For obtaining the correct sum of negative ranks \( T^- = 3 \) (or positive ranks \( T^+ = 33 \)) and identifying the test statistic as \( T = 3 \).
- **B1**: For stating the correct critical value of 5 (or stating the critical region \( T \le 5 \)).
- **M1**: For comparing their test statistic with their critical value.
- **A1**: For a correct conclusion in context, rejecting \( H_0 \) and stating there is evidence that the training course improves scores.

(a) Since \(\mathrm{G}_X(t)\) is a probability generating function, we must have \(\mathrm{G}_X(1) = 1\).
Using the given expression:
\[ \mathrm{G}_X(1) = \frac{1}{(a - b)^2} = 1 \implies (a - b)^2 = 1 \]
Since \(a\) and \(b\) are positive constants with \(a > b\), we have \(a - b > 0\). Taking the positive square root gives:
\[ a - b = 1 \]

(b) Differentiating \(\mathrm{G}_X(t) = (a - bt)^{-2}\) with respect to \(t\):
\[ \mathrm{G}_X'(t) = -2(a - bt)^{-3} \cdot (-b) = 2b(a - bt)^{-3} \]
Differentiating a second time:
\[ \mathrm{G}_X''(t) = -6b(a - bt)^{-4} \cdot (-b) = 6b^2(a - bt)^{-4}
\]
When \(t = 1\), the term \(a - bt = a - b = 1\). Therefore:
\[ \mathrm{G}_X'(1) = 2b(1)^{-3} = 2b \]
\[ \mathrm{G}_X''(1) = 6b^2(1)^{-4} = 6b^2 \]
The variance of \(X\) is given by the formula:
\[ \operatorname{Var}(X) = \mathrm{G}_X''(1) + \mathrm{G}_X'(1) - [ \mathrm{G}_X'(1) ]^2 \]
Substituting our results:
\[ \operatorname{Var}(X) = 6b^2 + 2b - (2b)^2 = 2b^2 + 2b \]
We are given that \(\operatorname{Var}(X) = 24\):
\[ 2b^2 + 2b = 24 \implies b^2 + b - 12 = 0 \]
Factorising this quadratic equation:
\[ (b + 4)(b - 3) = 0 \]
Since \(b\) must be positive, we have \(b = 3\).
Substituting \(b = 3\) into \(a - b = 1\):
\[ a = 3 + 1 = 4 \]

(c) With \(a = 4\) and \(b = 3\), the probability generating function is:
\[ \mathrm{G}_X(t) = (4 - 3t)^{-2} \]
We can find \(\mathrm{P}(X = 2)\) as the coefficient of \(t^2\) in the binomial expansion of \(\mathrm{G}_X(t)\), or by using the derivative:
\[ \mathrm{P}(X = 2) = \frac{\mathrm{G}_X''(0)}{2!} \]
Using \(\mathrm{G}_X''(t) = 6b^2(a-bt)^{-4} = 54(4-3t)^{-4}\):
\[ \mathrm{G}_X''(0) = 54(4)^{-4} = \frac{54}{256} = \frac{27}{128} \]
Thus:
\[ \mathrm{P}(X = 2) = \frac{27}{256} \]

(a)
B1: For stating \(\mathrm{G}_X(1) = 1\) and showing clearly that \(a - b = 1\) since \(a > b\).

(b)
M1: For differentiating \(\mathrm{G}_X(t)\) to find \(\mathrm{G}_X'(t)\) using the chain rule.
A1: For correct first and second derivatives:
\(\mathrm{G}_X'(t) = 2b(a - bt)^{-3}\) and \(\mathrm{G}_X''(t) = 6b^2(a - bt)^{-4}\).
M1: For substituting \(t = 1\) and using the identity \(\operatorname{Var}(X) = \mathrm{G}_X''(1) + \mathrm{G}_X'(1) - [\mathrm{G}_X'(1)]^2 = 24\).
A1: For setting up and solving the correct quadratic equation \(b^2 + b - 12 = 0\).
A1: For obtaining \(b = 3\) and \(a = 4\) (must reject \(b = -4\)).

(c)
B1: For finding the correct probability \(\mathrm{P}(X=2) = \frac{27}{256}\) (or exact equivalent fraction/decimal).

First, we state the null and alternative hypotheses:
\(H_0\): Preferred book genre is independent of age group.
\(H_1\): Preferred book genre is not independent of age group.

Next, calculate the row and column totals:
- Row totals: Under 30: \(48 + 32 + 20 = 100\); 30 and over: \(32 + 48 + 40 = 120\). Total \(N = 220\).
- Column totals: Sci-Fi: \(48 + 32 = 80\); Mystery: \(32 + 48 = 80\); Biography: \(20 + 40 = 60\).

Calculate the expected frequencies using \(E = \frac{\text{Row Total} \times \text{Column Total}}{N}\):
- Under 30 and Sci-Fi: \(\frac{100 \times 80}{220} = 36.364\)
- Under 30 and Mystery: \(\frac{100 \times 80}{220} = 36.364\)
- Under 30 and Biography: \(\frac{100 \times 60}{220} = 27.273\)
- 30 and over and Sci-Fi: \(\frac{120 \times 80}{220} = 43.636\)
- 30 and over and Mystery: \(\frac{120 \times 80}{220} = 43.636\)
- 30 and over and Biography: \(\frac{120 \times 60}{220} = 32.727\)

Calculate the \(\chi^2\) test statistic using \(\sum \frac{(O - E)^2}{E}\):
- For (Under 30, Sci-Fi): \(\frac{(48 - 36.364)^2}{36.364} = 3.7236\)
- For (Under 30, Mystery): \(\frac{(32 - 36.364)^2}{36.364} = 0.5236\)
- For (Under 30, Biography): \(\frac{(20 - 27.273)^2}{27.273} = 1.9394\)
- For (30 and over, Sci-Fi): \(\frac{(32 - 43.636)^2}{43.636} = 3.1030\)
- For (30 and over, Mystery): \(\frac{(48 - 43.636)^2}{43.636} = 0.4364\)
- For (30 and over, Biography): \(\frac{(40 - 32.727)^2}{32.727} = 1.6162\)

Summing these values gives:
\(\chi^2 = 3.7236 + 0.5236 + 1.9394 + 3.1030 + 0.4364 + 1.6162 = 11.342\)

Determine the degrees of freedom:
\(d.f. = (r - 1)(c - 1) = (2 - 1)(3 - 1) = 2\).

At the 1% significance level, the critical value of \(\chi^2\) with 2 degrees of freedom is \(9.210\).

Since our calculated test statistic \(11.34 > 9.210\), we reject the null hypothesis \(H_0\).

There is significant evidence at the 1% level to conclude that preferred book genre is not independent of age group.

B1: For stating both hypotheses correctly in terms of independence/association.
M1: For a correct method of calculating the expected values.
A1: For all expected values correct (accept 3 significant figures: 36.4, 36.4, 27.3, 43.6, 43.6, 32.7).
M1: For a correct method of calculating the test statistic \(\chi^2\).
A1: For obtaining a test statistic of \(\chi^2 \approx 11.3\) or \(11.34\).
B1: For stating 2 degrees of freedom and finding the correct critical value of 9.210 (or 9.21).
M1: For comparing their test statistic with their critical value.
A1: For a correct conclusion in context, consistent with their comparison.

Let \(d = y - x\) be the difference in jump heights (After \(-\) Before).

**Step 1: State the hypotheses**
\(H_0: \mu_d = 0\)
\(H_1: \mu_d > 0\)
where \(\mu_d\) is the population mean difference in jump height.

**Step 2: Calculate the differences \(d_i = y_i - x_i\)**
* Athlete A: \(65.1 - 62.4 = 2.7\)
* Athlete B: \(59.9 - 58.1 = 1.8\)
* Athlete C: \(64.2 - 65.0 = -0.8\)
* Athlete D: \(58.0 - 55.3 = 2.7\)
* Athlete E: \(72.1 - 70.2 = 1.9\)
* Athlete F: \(63.3 - 59.8 = 3.5\)
* Athlete G: \(63.5 - 64.1 = -0.6\)
* Athlete H: \(64.7 - 61.5 = 3.2\)

**Step 3: Calculate the mean and variance of the differences**
* \(\sum d = 2.7 + 1.8 - 0.8 + 2.7 + 1.9 + 3.5 - 0.6 + 3.2 = 14.4\)
* \(\bar{d} = \frac{14.4}{8} = 1.8\)
* \(\sum d^2 = 2.7^2 + 1.8^2 + (-0.8)^2 + 2.7^2 + 1.9^2 + 3.5^2 + (-0.6)^2 + 3.2^2\)
* \(\sum d^2 = 7.29 + 3.24 + 0.64 + 7.29 + 3.61 + 12.25 + 0.36 + 10.24 = 44.92\)

Unbiased estimate of population variance \(s^2\):
\(s^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{7} \left( 44.92 - \frac{14.4^2}{8} \right) = \frac{1}{7} (44.92 - 25.92) = \frac{19}{7} \approx 2.7143\)
Standard deviation \(s = \sqrt{\frac{19}{7}} \approx 1.6475\)

**Step 4: Calculate the test statistic \(t\)**
\(t = \frac{\bar{d}}{s / \sqrt{n}} = \frac{1.8}{\sqrt{2.7143 / 8}} = \frac{1.8}{0.5825} \approx 3.090\) (or \(3.09\) to 3 s.f.)

**Step 5: Determine the critical value**
For a one-tailed test at the 5% level of significance with degrees of freedom \(\nu = n - 1 = 7\):
Critical value \(t_{7}(0.95) = 1.895\).

**Step 6: Conclusion**
Since the calculated test statistic \(t = 3.09 > 1.895\), we reject \(H_0\).
There is significant evidence at the 5% level to suggest that the training program has increased the mean jump height of the athletes.

* **B1**: Formulate correct hypotheses, defining \(\mu_d\) (or equivalent defining the difference direction).
* **M1**: Calculate the differences \(d_i\) for all 8 athletes.
* **A1**: Obtain correct mean difference \(\bar{d} = 1.8\) (or \(-1.8\)).
* **A1**: Obtain correct unbiased variance estimate \(s^2 = \frac{19}{7} \approx 2.71\) (or \(s \approx 1.65\)).
* **M1**: Use the correct formula for the \(t\)-test statistic.
* **A1**: Obtain \(t = 3.09\) (or \(-3.09\)).
* **B1**: State correct critical value of \(1.895\) (or \(-1.895\)) from \(t_7\) distribution.
* **A1**: Compare test statistic with critical value correctly and state a correct contextual conclusion (reject \(H_0\), training program has increased the mean jump height).

(a) Since \(f(x)\) is a probability density function, the total area under the curve must be equal to 1:
\[ \int_{0}^{4} f(x) \, dx = 1 \]
We split this integral into the two non-zero intervals:
\[ \int_{0}^{1} k(1+x) \, dx + \int_{1}^{4} \frac{k}{x^2} \, dx = 1 \]
Evaluating the first integral:
\[ k \left[ x + \frac{x^2}{2} \right]_{0}^{1} = k \left(1 + \frac{1}{2}\right) = \frac{3}{2}k \]
Evaluating the second integral:
\[ k \left[ -\frac{1}{x} \right]_{1}^{4} = k \left(-\frac{1}{4} - (-1)\right) = \frac{3}{4}k \]
Adding the two parts together:
\[ \frac{3}{2}k + \frac{3}{4}k = 1 \implies \frac{9}{4}k = 1 \implies k = \frac{4}{9} \quad \text{[AG]} \]

(b) Let \(m\) be the median of \(X\). First, we determine which interval the median lies in. The probability that \(X < 1\) is:
\[ P(X < 1) = \int_{0}^{1} \frac{4}{9}(1+x) \, dx = \frac{4}{9} \left( \frac{3}{2} \right) = \frac{2}{3} \approx 0.667 \]
Since \(0.667 > 0.5\), the median \(m\) must lie in the interval \(0 \le m < 1\).
Thus, we solve:
\[ \int_{0}^{m} \frac{4}{9}(1+x) \, dx = 0.5 \]
\[ \frac{4}{9} \left[ x + \frac{x^2}{2} \right]_{0}^{m} = \frac{1}{2} \]
\[ m + \frac{m^2}{2} = \frac{9}{8} \]
Multiplying through by 8 to obtain a quadratic equation:
\[ 4m^2 + 8m - 9 = 0 \]
Using the quadratic formula for \(m > 0\):
\[ m = \frac{-8 + \sqrt{8^2 - 4(4)(-9)}}{2(4)} = \frac{-8 + \sqrt{208}}{8} = \frac{-8 + 4\sqrt{13}}{8} = \frac{\sqrt{13}-2}{2} \]
Evaluating this to 3 significant figures:
\[ m \approx 0.803 \]

(c) The expectation \(E(X)\) is defined by:
\[ E(X) = \int_{0}^{4} x f(x) \, dx \]
Splitting the integral into the two intervals:
\[ E(X) = \int_{0}^{1} x \left[\frac{4}{9}(1+x)\right] \, dx + \int_{1}^{4} x \left[\frac{4}{9x^2}\right] \, dx \]
Integrating the first term:
\[ \int_{0}^{1} \frac{4}{9}(x+x^2) \, dx = \frac{4}{9} \left[ \frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{1} = \frac{4}{9} \left( \frac{1}{2} + \frac{1}{3} \right) = \frac{4}{9} \left( \frac{5}{6} \right) = \frac{10}{27} \]
Integrating the second term:
\[ \int_{1}^{4} \frac{4}{9x} \, dx = \frac{4}{9} [ \ln x ]_{1}^{4} = \frac{4}{9} \ln 4 - 0 = \frac{4}{9} \ln(2^2) = \frac{8}{9} \ln 2 \]
Combining both parts gives:
\[ E(X) = \frac{10}{27} + \frac{8}{9} \ln 2 \]
Thus, \(A = \frac{10}{27}\) and \(B = \frac{8}{9}\).

(a)
- M1: For an attempt to integrate both sections of the piecewise function, sum the results, and set equal to 1.
- A1: For obtaining correct integrated terms in terms of \(k\) (i.e. \(\frac{3}{2}k\) and \(\frac{3}{4}k\)).
- A1: For showing that \(k = \frac{4}{9}\) with no errors.

(b)
- M1: For identifying that the median lies in the interval \(0 \le x < 1\) (by finding \(P(X < 1) = \frac{2}{3}\) or similar) and setting up the integral equation.
- A1: For obtaining a correct quadratic equation in \(m\), e.g., \(4m^2 + 8m - 9 = 0\) (or equivalent).
- A1: For the correct value \(0.803\) (accept exact \(\frac{\sqrt{13}-2}{2}\)).

(c)
- M1: For expressing \(E(X)\) as the sum of two integrals on the appropriate domains.
- A1: For correct integration of the first part to get \(\frac{10}{27}\).
- A1: For correct integration of the second part to get \(\frac{4}{9} \ln 4\) or \(\frac{8}{9} \ln 2\).
- A1: For correct final answer in the requested form: \(\frac{10}{27} + \frac{8}{9} \ln 2\) (or clearly identifying \(A = \frac{10}{27}\) and \(B = \frac{8}{9}\)).