2023 Edexcel IAL Further Mathematics (YFM01) Practice Paper with Answers

### (a)

\(\det(\mathbf{A}) = p(p+2) - (2)(-3) = p^2 + 2p + 6\)

### (b)

We complete the square for \(\det(\mathbf{A})\):
\[p^2 + 2p + 6 = (p+1)^2 + 5\]
Since \((p+1)^2 \ge 0\) for all real \(p\), we have:
\[\det(\mathbf{A}) \ge 5\]
Since \(\det(\mathbf{A}) \ne 0\) for all real \(p\), the matrix \(\mathbf{A}\) is non-singular for all real values of \(p\).

*Alternatively:*
Using the discriminant of \(p^2 + 2p + 6 = 0\):
\[\Delta = 2^2 - 4(1)(6) = -20\]
Since \(\Delta < 0\), the equation \(\det(\mathbf{A}) = 0\) has no real roots, so \(\det(\mathbf{A}) \ne 0\) for all real \(p\).

### (c)

The inverse of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by:
\[\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]

Applying this to \(\mathbf{A}\):
\[\mathbf{A}^{-1} = \frac{1}{p^2+2p+6} \begin{pmatrix} p+2 & -2 \\ 3 & p \end{pmatrix}\]

### (a)
* **B1:** Correct expression for the determinant: \(p^2 + 2p + 6\) (or equivalent).

### (b)
* **M1:** Attempt to show \(\det(\mathbf{A}) \ne 0\) by completing the square or calculating the discriminant of their quadratic expression.
* **A1:** Correct completion of square/discriminant and a clear conclusion that \(\det(\mathbf{A}) \ne 0\) (or \(\det(\mathbf{A}) \ge 5\)) for all real \(p\), hence \(\mathbf{A}\) is non-singular.

### (c)
* **M1:** Correct process for finding the inverse of a \(2 \times 2\) matrix, swapping the leading diagonal elements, changing the signs of the other two elements, and dividing by their determinant from (a).
* **A1:** Fully correct inverse matrix: \(\frac{1}{p^2+2p+6} \begin{pmatrix} p+2 & -2 \\ 3 & p \end{pmatrix}\) (or equivalent, e.g. written as a single matrix).

(a) We begin by expanding the summand: \(r(3r - 5) = 3r^2 - 5r\). Using the standard formulae for the sum of the first \(n\) integers and the sum of the first \(n\) squares, we have: \(\sum_{r=1}^n (3r^2 - 5r) = 3\sum_{r=1}^n r^2 - 5\sum_{r=1}^n r\) which equals \(3 \left[ \frac{1}{6}n(n+1)(2n+1) \right] - 5 \left[ \frac{1}{2}n(n+1) \right]\). Simplifying the coefficients gives: \(\frac{1}{2}n(n+1)(2n+1) - \frac{5}{2}n(n+1)\). Factorising out the common term \(\frac{1}{2}n(n+1)\): \(\frac{1}{2}n(n+1) [ (2n+1) - 5 ] = \frac{1}{2}n(n+1)(2n-4) = n(n+1)(n-2)\). This completes the proof. (b) To find the sum from \(r = 10\) to \(25\), we use: \(\sum_{r=10}^{25} r(3r - 5) = \sum_{r=1}^{25} r(3r - 5) - \sum_{r=1}^{9} r(3r - 5)\). Applying the formula from part (a): For \(n = 25\), the sum is \(25(26)(23) = 14950\). For \(n = 9\), the sum is \(9(10)(7) = 630\). Thus, the required value is \(14950 - 630 = 14320\).

Part (a): [M1] Attempts to expand and substitute standard formulae for the sum of \(r^2\) and the sum of \(r\). [A1] Obtains a correct unsimplified expression with correct formulae substituted. [M1] Shows a valid factorisation step by taking out at least \(\frac{1}{2}n(n+1)\) as a common factor. [A1] Fully simplifies to the given result with no errors in the working. Part (b): [M1] Applies the method \(S_{25} - S_9\) using the formula derived in part (a). [A1] Obtains the correct final answer of 14320.

(a) Since the coefficients of the polynomial are real, the complex roots must occur in conjugate pairs. Therefore, the conjugate of \( z_1 = 2 - 3\mathrm{i} \) is also a root: \( z_2 = 2 + 3\mathrm{i} \). (b) Method 1: The quadratic factor corresponding to the roots \( z_1 \) and \( z_2 \) is given by: \( (z - (2 - 3\mathrm{i}))(z - (2 + 3\mathrm{i})) = ((z - 2) + 3\mathrm{i})((z - 2) - 3\mathrm{i}) = (z - 2)^2 + 9 = z^2 - 4z + 13 \). Let the other quadratic factor be \( z^2 + Az + B \). We can write: \( (z^2 - 4z + 13)(z^2 + Az + B) = z^4 + p z^3 + q z^2 + 18z + 26 \). Expanding the left-hand side: \( z^4 + (A - 4)z^3 + (B - 4A + 13)z^2 + (13A - 4B)z + 13B = z^4 + p z^3 + q z^2 + 18z + 26 \). Comparing the constant terms: \( 13B = 26 \implies B = 2 \). Comparing the coefficients of \( z \): \( 13A - 4B = 18 \implies 13A - 4(2) = 18 \implies 13A = 26 \implies A = 2 \). Comparing the coefficients of \( z^3 \): \( p = A - 4 \implies p = 2 - 4 = -2 \). Comparing the coefficients of \( z^2 \): \( q = B - 4A + 13 \implies q = 2 - 4(2) + 13 = 7 \). Hence, \( p = -2 \) and \( q = 7 \). Method 2: Let the four roots be \( z_1, z_2, z_3, z_4 \). We have: \( z_1 + z_2 = 4 \) and \( z_1 z_2 = 13 \). The product of all four roots is: \( z_1 z_2 z_3 z_4 = 26 \implies 13(z_3 z_4) = 26 \implies z_3 z_4 = 2 \). The sum of products of the roots taken three at a time is: \( -(z_1 z_2 z_3 + z_1 z_2 z_4 + z_1 z_3 z_4 + z_2 z_3 z_4) = 18 \implies -[z_1 z_2(z_3 + z_4) + z_3 z_4(z_1 + z_2)] = 18 \implies -[13(z_3 + z_4) + 2(4)] = 18 \implies -13(z_3 + z_4) - 8 = 18 \implies z_3 + z_4 = -2 \). Now we can find \( p \) and \( q \): \( -p = z_1 + z_2 + z_3 + z_4 = 4 + (-2) = 2 \implies p = -2 \). \( q = z_1 z_2 + z_3 z_4 + (z_1 + z_2)(z_3 + z_4) = 13 + 2 + (4)(-2) = 7 \). (c) The remaining roots \( z_3 \) and \( z_4 \) are the roots of the quadratic factor \( z^2 + 2z + 2 = 0 \) (obtained from \( z^2 + Az + B \) with \( A = 2, B = 2 \)). Solving this: \( z = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm \mathrm{i} \). Thus, the remaining roots are \( z_3 = -1 + \mathrm{i} \) and \( z_4 = -1 - \mathrm{i} \). (d) On an Argand diagram, the four roots are represented by the points: \( (2, 3) \) for \( z_2 \), \( (2, -3) \) for \( z_1 \), \( (-1, 1) \) for \( z_3 \), and \( (-1, -1) \) for \( z_4 \). These points should show symmetry about the real axis.

(a) B1: Correctly states \( 2 + 3\mathrm{i} \) (or equivalent). (1 mark) (b) M1: Attempts to find the quadratic factor by expanding \( (z - (2 - 3\mathrm{i}))(z - (2 + 3\mathrm{i})) \) to obtain \( z^2 - 4z + 13 \). M1: Sets up a general quadratic factor \( z^2 + Az + B \) and equates coefficients, or uses relations between roots and coefficients to establish equations for the remaining roots. A1: Correctly finds \( A = 2 \) and \( B = 2 \) (or equivalent values, e.g., product of remaining roots is 2 and sum is -2). A1: Correct value of \( p = -2 \). A1: Correct value of \( q = 7 \). (5 marks) (c) M1: Solves their quadratic equation \( z^2 + 2z + 2 = 0 \) (or equivalent) using the quadratic formula or completing the square. A1: Correctly identifies the other two roots as \( -1 \pm \mathrm{i} \). (2 marks) (d) B1: Plots all 4 roots in the correct quadrants with reasonable symmetry about the real axis. B1: Correctly labels the coordinates or values on both axes (real axis values at 2 and -1, imaginary axis values at 3, 1, -1, and -3). (2 marks)

(a) Evaluating the function at the endpoints: \( \text{f}(1.3) = 1.3^3 - \cos(1.3) - 2 \approx 2.197 - 0.2674988... - 2 = -0.0704988... \approx -0.0705 \). \( \text{f}(1.4) = 1.4^3 - \cos(1.4) - 2 \approx 2.744 - 0.169967... - 2 = 0.574032... \approx 0.5740 \). Since there is a change of sign between \( \text{f}(1.3) < 0 \) and \( \text{f}(1.4) > 0 \), and \( \text{f}(x) \) is continuous on the interval, a root \( \alpha \) exists in \( [1.3, 1.4] \). (b) Using linear interpolation: \( \alpha \approx 1.3 + \frac{|\text{f}(1.3)|}{|\text{f}(1.3)| + |\text{f}(1.4)|} \times (1.4 - 1.3) \). \( \alpha \approx 1.3 + \frac{0.0704988}{0.0704988 + 0.5740328} \times 0.1 \approx 1.3 + \frac{0.0704988}{0.6445316} \times 0.1 \approx 1.3 + 0.010938... = 1.310938... \). To 3 decimal places, \( \alpha \approx 1.311 \). (c) First, find the derivative: \( \text{f}'(x) = 3x^2 + \sin(x) \). Evaluating \( \text{f}'(1.3) \): \( \text{f}'(1.3) = 3(1.3)^2 + \sin(1.3) \approx 5.07 + 0.963558... = 6.033558... \). Applying Newton-Raphson: \( x_1 = 1.3 - \frac{\text{f}(1.3)}{\text{f}'(1.3)} \approx 1.3 - \frac{-0.0704988}{6.033558} \approx 1.3 + 0.011684... = 1.311684... \). To 3 decimal places, \( x_1 \approx 1.312 \).

(a) M1: Attempts to evaluate \( \text{f}(1.3) \) and \( \text{f}(1.4) \) with at least one correct to 1 s.f. (with calculator in radian mode). A1: Evaluates both correctly to at least 2 d.p. (\( \text{f}(1.3) \approx -0.07 \) and \( \text{f}(1.4) \approx 0.57 \)) and states that the change of sign combined with continuity implies a root in the interval. (b) M1: Sets up a correct linear interpolation equation or ratio, e.g., \( \alpha = 1.3 + 0.1 \times \frac{|\text{f}(1.3)|}{|\text{f}(1.3)| + |\text{f}(1.4)|} \). A1: Correct substitution of values, e.g., \( 1.3 + 0.1 \times \frac{0.0705}{0.0705 + 0.5740} \). A1: \( 1.311 \) (must be 3 decimal places, accept 1.311 only). (c) M1: Differentiates to find \( \text{f}'(x) = 3x^2 + \sin(x) \). M1: Applies the Newton-Raphson formula \( x_1 = 1.3 - \frac{\text{f}(1.3)}{\text{f}'(1.3)} \) using their evaluated values. A1: \( 1.312 \) (must be 3 decimal places, accept 1.312 only).

(a) From the given cubic equation \(x^3 - 3x^2 + 5x - 2 = 0\), we can identify the relationships between the roots and the coefficients:

\[\alpha + \beta + \gamma = 3\]
\[\alpha\beta + \beta\gamma + \gamma\alpha = 5\]
\[\alpha\beta\gamma = 2\]

We use the identity:

\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]

Substitute the values:

\[\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(5) = 9 - 10 = -1\]

(b) **Method 1: Using the recurrence relation**

Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of the cubic equation, they each satisfy it:

\[\alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0\]
\[\beta^3 - 3\beta^2 + 5\beta - 2 = 0\]
\[\gamma^3 - 3\gamma^2 + 5\gamma - 2 = 0\]

Summing these equations, we get:

\[(\alpha^3 + \beta^3 + \gamma^3) - 3(\alpha^2 + \beta^2 + \gamma^2) + 5(\alpha + \beta + \gamma) - 6 = 0\]

Letting \(S_n = \alpha^n + \beta^n + \gamma^n\), we have:

\[S_3 - 3S_2 + 5S_1 - 6 = 0\]

Substitute \(S_1 = 3\) and \(S_2 = -1\):

\[S_3 - 3(-1) + 5(3) - 6 = 0\]
\[S_3 + 3 + 15 - 6 = 0\]
\[S_3 + 12 = 0 \implies S_3 = -12\]

**Method 2: Using the algebraic identity**

\[\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))\]
\[\alpha^3 + \beta^3 + \gamma^3 - 3(2) = 3(-1 - 5)\]
\[\alpha^3 + \beta^3 + \gamma^3 - 6 = 3(-6) = -18\]
\[\alpha^3 + \beta^3 + \gamma^3 = -12\]

(c) **Method 1: Finding new coefficients**

Let the new roots be \(u = \alpha^2\), \(v = \beta^2\), and \(w = \gamma^2\).

The new cubic equation is of the form:

\[y^3 - (u+v+w)y^2 + (uv+vw+wu)y - uvw = 0\]

1. Sum of new roots:

\[u+v+w = \alpha^2 + \beta^2 + \gamma^2 = -1\]

2. Product of new roots:

\[uvw = (\alpha\beta\gamma)^2 = 2^2 = 4\]

3. Sum of products in pairs:

\[uv+vw+wu = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\]

Using the identity:

\[\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)\]
\[\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = 5^2 - 2(2)(3) = 25 - 12 = 13\]

Substituting these values back into the equation form:

\[y^3 - (-1)y^2 + 13y - 4 = 0\]
\[y^3 + y^2 + 13y - 4 = 0\]

**Method 2: Substitution**

Let \(y = x^2 \implies x = \sqrt{y}\).

Substitute into the original equation:

\[x^3 + 5x = 3x^2 + 2\]
\[x(x^2 + 5) = 3x^2 + 2\]
\[\sqrt{y}(y + 5) = 3y + 2\]

Square both sides:

\[y(y + 5)^2 = (3y + 2)^2\]
\[y(y^2 + 10y + 25) = 9y^2 + 12y + 4\]
\[y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\]
\[y^3 + y^2 + 13y - 4 = 0\]

(Any variable name like \(x\), \(y\), or \(w\) is acceptable.)

**Part (a)**
- **B1**: For correctly stating or using \(\alpha + \beta + \gamma = 3\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 5\).
- **M1**: For applying the correct identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\) with their values.
- **A1**: For obtaining the correct value of \(-1\).

**Part (b)**
- **M1**: For a complete and correct method to find \(\alpha^3 + \beta^3 + \gamma^3\), either by using the cubic identity or the recurrence relation \(S_3 - 3S_2 + 5S_1 - 6 = 0\) with their values.
- **A1ft**: For correct substitution of their values into their chosen formula.
- **A1**: For obtaining the correct value of \(-12\).

**Part (c)**
- **M1**: A complete method to obtain the new coefficients. Either by finding \(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\) or by substituting \(x = \sqrt{y}\) and isolating terms containing \(\sqrt{y}\) to square both sides.
- **A1**: Correct values for all three new coefficients: Sum \(= -1\), Sum of products in pairs \(= 13\), Product \(= 4\) (or correct expanded equation prior to final algebraic simplification).
- **A1**: For a correct cubic equation with integer coefficients, e.g., \(y^3 + y^2 + 13y - 4 = 0\) (must include "\(= 0\)").

(a) Differentiating \(y = c^2 x^{-1}\) with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -c^2 x^{-2} = -\frac{c^2}{x^2}\]
At \(P\left(ct, \frac{c}{t}\right)\), the gradient of the tangent is:
\[m_T = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}\]
Since the normal is perpendicular to the tangent, the gradient of the normal, \(m_N\), is:
\[m_N = -\frac{1}{m_T} = t^2\]
The equation of the normal at \(P\) is:
\[y - \frac{c}{t} = t^2(x - ct)\]
Multiplying both sides by \(t\):
\[ty - c = t^3(x - ct)\]
\[ty - c = t^3 x - ct^4\]
Rearranging gives:
\[t^3 x - ty = c(t^4 - 1)\]

(b) At \(A\) (on the \(x\)-axis, so \(y = 0\)):
\[t^3 x = c(t^4 - 1) \implies x = \frac{c(t^4 - 1)}{t^3}\]
So, \(A = \left(\frac{c(t^4 - 1)}{t^3}, 0\right)\).

At \(B\) (on the \(y\)-axis, so \(x = 0\)):
\[-ty = c(t^4 - 1) \implies y = -\frac{c(t^4 - 1)}{t}\]
So, \(B = \left(0, -\frac{c(t^4 - 1)}{t}\right)\).

(c) Since \(M(x, y)\) is the midpoint of \(AB\):
\[x = \frac{\frac{c(t^4 - 1)}{t^3} + 0}{2} = \frac{c(t^4 - 1)}{2t^3}\]
\[y = \frac{0 + \left(-\frac{c(t^4 - 1)}{t}\right)}{2} = -\frac{c(t^4 - 1)}{2t}\]
To eliminate \(t\), find the ratio of \(y\) to \(x\):
\[\frac{y}{x} = \frac{-\frac{c(t^4 - 1)}{2t}}{\frac{c(t^4 - 1)}{2t^3}} = -t^2 \implies t^2 = -\frac{y}{x}\]
We can write \(t^4\) as:
\[t^4 = (t^2)^2 = \left(-\frac{y}{x}\right)^2 = \frac{y^2}{x^2}\]
This gives:
\[t^4 - 1 = \frac{y^2}{x^2} - 1 = \frac{y^2 - x^2}{x^2}\]
Substitute this back into the expression for \(y\):
\[y = -\frac{c(t^4 - 1)}{2t} = -\frac{c}{2t} \left(\frac{y^2 - x^2}{x^2}\right) = \frac{c(x^2 - y^2)}{2tx^2}\]
Rearrange to express \(t\) in terms of \(x\) and \(y\):
\[2tx^2y = c(x^2 - y^2) \implies t = \frac{c(x^2 - y^2)}{2x^2y}\]
Squaring both sides:
\[t^2 = \frac{c^2(x^2 - y^2)^2}{4x^4y^2}\]
Substitute \(t^2 = -\frac{y}{x}\):
\[-\frac{y}{x} = \frac{c^2(x^2 - y^2)^2}{4x^4y^2}\]
Multiply both sides by \(4x^4y^2\):
\[-4x^3y^3 = c^2(x^2 - y^2)^2\]
Rearranging to the required form:
\[c^2(x^2 - y^2)^2 = -4x^3y^3\]

(a)
- M1: Attempts to differentiate \(y = \frac{c^2}{x}\) and substitute \(x = ct\) to find the gradient of the tangent.
- A1: Obtains the correct gradient of the normal, \(m_N = t^2\).
- A1*: Uses \(y - y_1 = m(x - x_1)\) with their normal gradient and \(P\left(ct, \frac{c}{t}\right)\), showing sufficient algebraic steps to reach the printed answer.

(b)
- B1: Correct coordinates of \(A\) (or equivalent form, e.g., \(x = c(t - t^{-3}), y = 0\)).
- B1: Correct coordinates of \(B\) (or equivalent form, e.g., \(x = 0, y = c(t^{-1} - t^3)\)).

(c)
- M1: Uses the midpoint formula to obtain expressions for the coordinates of \(M\) in terms of \(c\) and \(t\).
- M1: Considers \(y/x\) or similar to express \(t^2\) in terms of \(x\) and \(y\).
- M1: Substitute \(t^4-1\) or \(t^2\) to form an equation for \(t\) in terms of \(x\) and \(y\).
- A1*: Completes algebraic steps correctly with no errors to show the printed Cartesian equation \(c^2(x^2 - y^2)^2 = -4x^3y^3\).

(a) Comparing \(\mathbf{P}\) with the general rotation matrix \(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\):
\(\cos\theta = -\frac{1}{\sqrt{2}}\) and \(\sin\theta = \frac{1}{\sqrt{2}}\).
This corresponds to \(\theta = 135^\circ\) (or \(\frac{3\pi}{4}\) radians).
So \(\mathbf{P}\) represents a rotation about the origin through \(135^\circ\) anticlockwise.

(b) The matrix representing the stretch \(V\) is:
\[\mathbf{Q} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\]
Since \(W\) is \(U\) followed by \(V\), we have:
\[\mathbf{R} = \mathbf{Q}\mathbf{P} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{k}{\sqrt{2}} & -\frac{k}{\sqrt{2}} \end{pmatrix}\]

(c) Triangle \(T\) has vertices at \(A(1, 2)\), \(B(4, 2)\) and \(C(1, 6)\).
The base \(AB = 4 - 1 = 3\) and the height \(AC = 6 - 2 = 4\).
\[\text{Area}(T) = \frac{1}{2} \times 3 \times 4 = 6\]

(d) The determinant of \(\mathbf{R}\) is:
\[\det \mathbf{R} = \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{k}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{k}{\sqrt{2}}\right) = \frac{k}{2} + \frac{k}{2} = k\]
Since \(k > 0\), the area scale factor is \(k\).
\[\text{Area}(T') = |\det \mathbf{R}| \times \text{Area}(T) \implies 12\sqrt{2} = k \times 6 \implies k = 2\sqrt{2}\]

(e) Substituting \(k = 2\sqrt{2}\) into \(\mathbf{R}\):
\[\mathbf{R} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{2\sqrt{2}}{\sqrt{2}} & -\frac{2\sqrt{2}}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix}\]
Applying \(\mathbf{R}\) to the vertices of \(T\):
\[A' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{3}{\sqrt{2}} \\ -2 \end{pmatrix} = \begin{pmatrix} -\frac{3\sqrt{2}}{2} \\ -2 \end{pmatrix}\]
\[B' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{6}{\sqrt{2}} \\ 4 \end{pmatrix} = \begin{pmatrix} -3\sqrt{2} \\ 4 \end{pmatrix}\]
\[C' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 6 \end{pmatrix} = \begin{pmatrix} -\frac{7}{\sqrt{2}} \\ -10 \end{pmatrix} = \begin{pmatrix} -\frac{7\sqrt{2}}{2} \\ -10 \end{pmatrix}\]
So the exact coordinates are \(A'\left(-\frac{3\sqrt{2}}{2}, -2\right)\), \(B'\left(-3\sqrt{2}, 4\right)\), and \(C'\left(-\frac{7\sqrt{2}}{2}, -10\right)\).

(a)
- M1: Identifies the transformation as a rotation.
- A1: Fully describes the rotation: Rotation about the origin \((0,0)\) through \(135^\circ\) (or \(\frac{3\pi}{4}\) radians) anticlockwise.

(b)
- M1: For representing \(V\) by the matrix \(\mathbf{Q} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) and writing the multiplication in the correct order \(\mathbf{R} = \mathbf{Q}\mathbf{P}\).
- A1: Fully correct matrix \(\mathbf{R} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{k}{\sqrt{2}} & -\frac{k}{\sqrt{2}} \end{pmatrix}\) (o.e.).

(c)
- B1: Correct area of triangle \(T\) is 6.

(d)
- M1: Identifies that the area scale factor is given by \(\det \mathbf{R}\) and sets up the equation \(12\sqrt{2} = 6 \times \det \mathbf{R}\) (or equivalent).
- A1: For \(k = 2\sqrt{2}\).

(e)
- B1: Correct matrix \(\mathbf{R}\) with \(k = 2\sqrt{2}\) substituted, i.e. \(\begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix}\) (o.e.).
- M1: Correctly attempts to multiply their matrix \(\mathbf{R}\) by at least one of the position vectors of \(A\), \(B\) or \(C\).
- A1: Any two of the image coordinates correct (in exact form).
- A1: All three exact coordinates correct: \(A'\left(-\frac{3\sqrt{2}}{2}, -2\right)\), \(B'\left(-3\sqrt{2}, 4\right)\), and \(C'\left(-\frac{7\sqrt{2}}{2}, -10\right)\) (o.e. e.g. using \(\frac{1}{\sqrt{2}}\) instead of rationalised denominator).

(a) Comparing \(\mathbf{P}\) with the general rotation matrix \(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\):
\(\cos\theta = -\frac{1}{\sqrt{2}}\) and \(\sin\theta = \frac{1}{\sqrt{2}}\).
This corresponds to \(\theta = 135^\circ\) (or \(\frac{3\pi}{4}\) radians).
So \(\mathbf{P}\) represents a rotation about the origin through \(135^\circ\) anticlockwise.

(b) The matrix representing the stretch \(V\) is:
\[\mathbf{Q} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\]
Since \(W\) is \(U\) followed by \(V\), we have:
\[\mathbf{R} = \mathbf{Q}\mathbf{P} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{k}{\sqrt{2}} & -\frac{k}{\sqrt{2}} \end{pmatrix}\]

(c) Triangle \(T\) has vertices at \(A(1, 2)\), \(B(4, 2)\) and \(C(1, 6)\).
The base \(AB = 4 - 1 = 3\) and the height \(AC = 6 - 2 = 4\).
\[\text{Area}(T) = \frac{1}{2} \times 3 \times 4 = 6\]

(d) The determinant of \(\mathbf{R}\) is:
\[\det \mathbf{R} = \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{k}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{k}{\sqrt{2}}\right) = \frac{k}{2} + \frac{k}{2} = k\]
Since \(k > 0\), the area scale factor is \(k\).
\[\text{Area}(T') = |\det \mathbf{R}| \times \text{Area}(T) \implies 12\sqrt{2} = k \times 6 \implies k = 2\sqrt{2}\]

(e) Substituting \(k = 2\sqrt{2}\) into \(\mathbf{R}\):
\[\mathbf{R} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{2\sqrt{2}}{\sqrt{2}} & -\frac{2\sqrt{2}}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix}\]
Applying \(\mathbf{R}\) to the vertices of \(T\):
\[A' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{3}{\sqrt{2}} \\ -2 \end{pmatrix} = \begin{pmatrix} -\frac{3\sqrt{2}}{2} \\ -2 \end{pmatrix}\]
\[B' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{6}{\sqrt{2}} \\ 4 \end{pmatrix} = \begin{pmatrix} -3\sqrt{2} \\ 4 \end{pmatrix}\]
\[C' = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 6 \end{pmatrix} = \begin{pmatrix} -\frac{7}{\sqrt{2}} \\ -10 \end{pmatrix} = \begin{pmatrix} -\frac{7\sqrt{2}}{2} \\ -10 \end{pmatrix}\]
So the exact coordinates are \(A'\left(-\frac{3\sqrt{2}}{2}, -2\right)\), \(B'\left(-3\sqrt{2}, 4\right)\), and \(C'\left(-\frac{7\sqrt{2}}{2}, -10\right)\).

(a)
- M1: Identifies the transformation as a rotation.
- A1: Fully describes the rotation: Rotation about the origin \((0,0)\) through \(135^\circ\) (or \(\frac{3\pi}{4}\) radians) anticlockwise.

(b)
- M1: For representing \(V\) by the matrix \(\mathbf{Q} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) and writing the multiplication in the correct order \(\mathbf{R} = \mathbf{Q}\mathbf{P}\).
- A1: Fully correct matrix \(\mathbf{R} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{k}{\sqrt{2}} & -\frac{k}{\sqrt{2}} \end{pmatrix}\) (o.e.).

(c)
- B1: Correct area of triangle \(T\) is 6.

(d)
- M1: Identifies that the area scale factor is given by \(\det \mathbf{R}\) and sets up the equation \(12\sqrt{2} = 6 \times \det \mathbf{R}\) (or equivalent).
- A1: For \(k = 2\sqrt{2}\).

(e)
- B1: Correct matrix \(\mathbf{R}\) with \(k = 2\sqrt{2}\) substituted, i.e. \(\begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 2 & -2 \end{pmatrix}\) (o.e.).
- M1: Correctly attempts to multiply their matrix \(\mathbf{R}\) by at least one of the position vectors of \(A\), \(B\) or \(C\).
- A1: Any two of the image coordinates correct (in exact form).
- A1: All three exact coordinates correct: \(A'\left(-\frac{3\sqrt{2}}{2}, -2\right)\), \(B'\left(-3\sqrt{2}, 4\right)\), and \(C'\left(-\frac{7\sqrt{2}}{2}, -10\right)\) (o.e. e.g. using \(\frac{1}{\sqrt{2}}\) instead of rationalised denominator).

(a) For a parabola with equation \(y^2 = 4ax\), the focus is \(S(a, 0)\) and the directrix is \(x = -a\).
Comparing \(y^2 = 16x\) with \(y^2 = 4ax\), we have:

\(4a = 16 \implies a = 4\)

Therefore, the focus \(S\) has coordinates \(S(4, 0)\) and the directrix \(L\) has equation \(x = -4\).

(b) The focus-directrix property of a parabola states that the distance from any point on the parabola to the focus is equal to the perpendicular distance from that point to the directrix.
Thus, \(SP = PQ\).

The point \(P\) has coordinates \((4t^2, 8t)\). Since \(t > 0\), the x-coordinate of \(P\) is positive.
The perpendicular distance from \(P(4t^2, 8t)\) to the directrix \(x = -4\) is:

\(PQ = 4t^2 - (-4) = 4t^2 + 4 = 4(t^2 + 1)\)

Since \(SP = PQ\), we have:

\(SP = 4(t^2 + 1)\)

(c) The coordinates of \(S\) are \((4, 0)\) and the coordinates of \(P\) are \((4t^2, 8t)\).
Let the coordinates of the midpoint \(M\) be \((x, y)\).
Using the midpoint formula:

\(x = \frac{4 + 4t^2}{2} = 2 + 2t^2\)

\(y = \frac{0 + 8t}{2} = 4t\)

We can rearrange the equation for \(y\) to express \(t\) in terms of \(y\):

\(t = \frac{y}{4}\)

Substituting this into the equation for \(x\):

\(x = 2 + 2\left(\frac{y}{4}\right)^2 = 2 + 2\left(\frac{y^2}{16}\right) = 2 + \frac{y^2}{8}\)

Rearranging to the required form:

\(x - 2 = \frac{y^2}{8} \implies y^2 = 8(x - 2)\)

This is a parabola of the form \(y^2 = a(x - b)\), where \(a = 8\) and \(b = 2\).

(d) Given that \(SP = 20\):

\(4(t^2 + 1) = 20 \implies t^2 + 1 = 5 \implies t^2 = 4\)

Since \(t > 0\), we have \(t = 2\).
Substituting \(t = 2\) into the coordinates of \(P\):

\(P = (4(2^2), 8(2)) = (16, 16)\)

The point \(Q\) is the projection of \(P\) onto the directrix \(x = -4\), so:

\(Q = (-4, 16)\)

The trapezium \(OSPQ\) has vertices \(O(0, 0)\), \(S(4, 0)\), \(P(16, 16)\), and \(Q(-4, 16)\).
The side \(OS\) lies along the x-axis with length:

\(OS = 4 - 0 = 4\)

The side \(QP\) is horizontal and lies along the line \(y = 16\) with length:

\(QP = 16 - (-4) = 20\)

Since \(OS\) and \(QP\) are parallel, \(OSPQ\) is a trapezium with parallel sides of lengths 4 and 20, and height \(h = 16\).
The area of the trapezium is:

\(\text{Area} = \frac{1}{2}(OS + QP) \times h = \frac{1}{2}(4 + 20) \times 16 = 12 \times 16 = 192\)

**(a)**
* **B1:** Correct coordinates for the focus \(S(4, 0)\).
* **B1:** Correct equation for the directrix \(L: x = -4\).

**(b)**
* **M1:** Identifies that \(SP = PQ\) (focus-directrix property) and writes down an expression for the distance \(PQ\) from \(P(4t^2, 8t)\) to the line \(x = -4\).
* **A1*:** Correctly simplifies to show \(SP = 4(t^2 + 1)\) with no errors and a clear explanation referencing the focus-directrix property.

**(c)**
* **M1:** Attempts to find the coordinates of \(M\) using the midpoint formula with their \(S\) and \(P\). Must see at least one coordinate correct.
* **A1:** Correct parametric equations for \(M\): \(x = 2 + 2t^2\) and \(y = 4t\).
* **M1:** Attempts to eliminate the parameter \(t\) to find a Cartesian equation, e.g., by substituting \(t = y/4\) into their expression for \(x\).
* **A1:** Obtains \(y^2 = 8(x - 2)\) or clearly states \(a = 8\) and \(b = 2\) with a correct equation.

**(d)**
* **M1:** Sets their expression for \(SP\) equal to 20 and solves for \(t\). Must select the positive root \(t = 2\).
* **A1:** Finds the correct coordinates of \(P(16, 16)\) and \(Q(-4, 16)\).
* **A1:** Finds the correct area of the trapezium \(OSPQ\) as 192.

(a) For a parabola with equation \(y^2 = 4ax\), the focus is \(S(a, 0)\) and the directrix is \(x = -a\).
Comparing \(y^2 = 16x\) with \(y^2 = 4ax\), we have:

\(4a = 16 \implies a = 4\)

Therefore, the focus \(S\) has coordinates \(S(4, 0)\) and the directrix \(L\) has equation \(x = -4\).

(b) The focus-directrix property of a parabola states that the distance from any point on the parabola to the focus is equal to the perpendicular distance from that point to the directrix.
Thus, \(SP = PQ\).

The point \(P\) has coordinates \((4t^2, 8t)\). Since \(t > 0\), the x-coordinate of \(P\) is positive.
The perpendicular distance from \(P(4t^2, 8t)\) to the directrix \(x = -4\) is:

\(PQ = 4t^2 - (-4) = 4t^2 + 4 = 4(t^2 + 1)\)

Since \(SP = PQ\), we have:

\(SP = 4(t^2 + 1)\)

(c) The coordinates of \(S\) are \((4, 0)\) and the coordinates of \(P\) are \((4t^2, 8t)\).
Let the coordinates of the midpoint \(M\) be \((x, y)\).
Using the midpoint formula:

\(x = \frac{4 + 4t^2}{2} = 2 + 2t^2\)

\(y = \frac{0 + 8t}{2} = 4t\)

We can rearrange the equation for \(y\) to express \(t\) in terms of \(y\):

\(t = \frac{y}{4}\)

Substituting this into the equation for \(x\):

\(x = 2 + 2\left(\frac{y}{4}\right)^2 = 2 + 2\left(\frac{y^2}{16}\right) = 2 + \frac{y^2}{8}\)

Rearranging to the required form:

\(x - 2 = \frac{y^2}{8} \implies y^2 = 8(x - 2)\)

This is a parabola of the form \(y^2 = a(x - b)\), where \(a = 8\) and \(b = 2\).

(d) Given that \(SP = 20\):

\(4(t^2 + 1) = 20 \implies t^2 + 1 = 5 \implies t^2 = 4\)

Since \(t > 0\), we have \(t = 2\).
Substituting \(t = 2\) into the coordinates of \(P\):

\(P = (4(2^2), 8(2)) = (16, 16)\)

The point \(Q\) is the projection of \(P\) onto the directrix \(x = -4\), so:

\(Q = (-4, 16)\)

The trapezium \(OSPQ\) has vertices \(O(0, 0)\), \(S(4, 0)\), \(P(16, 16)\), and \(Q(-4, 16)\).
The side \(OS\) lies along the x-axis with length:

\(OS = 4 - 0 = 4\)

The side \(QP\) is horizontal and lies along the line \(y = 16\) with length:

\(QP = 16 - (-4) = 20\)

Since \(OS\) and \(QP\) are parallel, \(OSPQ\) is a trapezium with parallel sides of lengths 4 and 20, and height \(h = 16\).
The area of the trapezium is:

\(\text{Area} = \frac{1}{2}(OS + QP) \times h = \frac{1}{2}(4 + 20) \times 16 = 12 \times 16 = 192\)

**(a)**
* **B1:** Correct coordinates for the focus \(S(4, 0)\).
* **B1:** Correct equation for the directrix \(L: x = -4\).

**(b)**
* **M1:** Identifies that \(SP = PQ\) (focus-directrix property) and writes down an expression for the distance \(PQ\) from \(P(4t^2, 8t)\) to the line \(x = -4\).
* **A1*:** Correctly simplifies to show \(SP = 4(t^2 + 1)\) with no errors and a clear explanation referencing the focus-directrix property.

**(c)**
* **M1:** Attempts to find the coordinates of \(M\) using the midpoint formula with their \(S\) and \(P\). Must see at least one coordinate correct.
* **A1:** Correct parametric equations for \(M\): \(x = 2 + 2t^2\) and \(y = 4t\).
* **M1:** Attempts to eliminate the parameter \(t\) to find a Cartesian equation, e.g., by substituting \(t = y/4\) into their expression for \(x\).
* **A1:** Obtains \(y^2 = 8(x - 2)\) or clearly states \(a = 8\) and \(b = 2\) with a correct equation.

**(d)**
* **M1:** Sets their expression for \(SP\) equal to 20 and solves for \(t\). Must select the positive root \(t = 2\).
* **A1:** Finds the correct coordinates of \(P(16, 16)\) and \(Q(-4, 16)\).
* **A1:** Finds the correct area of the trapezium \(OSPQ\) as 192.

Let \(f(n) = 4^{n+1} + 5^{2n-1}\). Step 1: Base Case. For \(n = 1\): \(f(1) = 4^{1+1} + 5^{2(1)-1} = 4^2 + 5^1 = 16 + 5 = 21\). Since 21 is divisible by 21, the statement is true for \(n = 1\). Step 2: Inductive Hypothesis. Assume that the statement is true for \(n = k\), where \(k \in \mathbb{Z}^+\). That is, assume \(f(k) = 4^{k+1} + 5^{2k-1} = 21m\) for some integer \(m\). Step 3: Inductive Step. Show that the statement is true for \(n = k+1\). \(f(k+1) = 4^{(k+1)+1} + 5^{2(k+1)-1} = 4^{k+2} + 5^{2k+1} = 4 \cdot 4^{k+1} + 5^2 \cdot 5^{2k-1} = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1}\). We can rewrite this expression to use our inductive hypothesis: \(f(k+1) = 4(4^{k+1} + 5^{2k-1}) + (25 - 4) \cdot 5^{2k-1} = 4f(k) + 21 \cdot 5^{2k-1}\). Substituting \(f(k) = 21m\), we obtain: \(f(k+1) = 4(21m) + 21 \cdot 5^{2k-1} = 21(4m + 5^{2k-1})\). Since \(m\) and \(k\) are positive integers, \((4m + 5^{2k-1})\) is an integer. Therefore, \(f(k+1)\) is divisible by 21. Step 4: Conclusion. The statement is true for \(n=1\). If the statement is true for \(n=k\), then it has been shown to be true for \(n=k+1\). Hence, by the principle of mathematical induction, the statement is true for all positive integers \(n\).

- **B1**: Evaluates \(f(1) = 21\) and states that it is divisible by 21. - **M1**: Writes down the expression for \(f(k+1)\) and attempts to split the index of at least one term, e.g., writing \(4^{k+2}\) as \(4 \cdot 4^{k+1}\) or \(5^{2k+1}\) as \(25 \cdot 5^{2k-1}\). - **M1**: Uses the induction hypothesis by expressing \(f(k+1)\) in terms of \(f(k)\), e.g., \(4f(k) + \dots\) or \(25f(k) - \dots\). - **A1**: Obtains a correct simplified expression in terms of \(f(k)\), such as \(4f(k) + 21 \cdot 5^{2k-1}\) or \(25f(k) - 21 \cdot 4^{k+1}\). - **A1**: Correctly factors out 21 to show \(f(k+1) = 21(4m + 5^{2k-1})\) or equivalent. - **A1 (cso)**: Fully correct proof with a complete concluding explanation that mentions: 1. True for \(n=1\), 2. If true for \(n=k\) then true for \(n=k+1\), and 3. Therefore true for all positive integers \(n\) (by mathematical induction).

**Part (a)**

Let \( y = \ln(\cos x + \sin x) \).

Using the chain rule to differentiate with respect to \( x \):
\[ \frac{\text{d}y}{\text{d}x} = \frac{1}{\cos x + \sin x} \cdot (-\sin x + \cos x) = \frac{\cos x - \sin x}{\cos x + \sin x} \]

Now, differentiate again with respect to \( x \) using the quotient rule:
\[ \frac{\text{d}^2y}{\text{d}x^2} = \frac{(-\sin x - \cos x)(\cos x + \sin x) - \cos x - \sin x)(\cos x - \sin x)}{(\cos x + \sin x)^2} \]
\[ \frac{\text{d}^2y}{\text{d}x^2} = \frac{-(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x + \sin x)^2} \]
\[ \frac{\text{d}^2y}{\text{d}x^2} = \frac{-(\cos^2 x + 2\sin x \cos x + \sin^2 x) - (\cos^2 x - 2\sin x \cos x + \sin^2 x)}{(\cos x + \sin x)^2} \]

Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ \frac{\text{d}^2y}{\text{d}x^2} = \frac{-1 - 1}{(\cos x + \sin x)^2} = \frac{-2}{(\cos x + \sin x)^2} \]

Since \( y = \ln(\cos x + \sin x) \), we have:
\[ \text{e}^y = \cos x + \sin x \implies \text{e}^{2y} = (\cos x + \sin x)^2 \]

Therefore,
\[ \frac{\text{d}^2y}{\text{d}x^2} = \frac{-2}{\text{e}^{2y}} = -2\text{e}^{-2y} \quad \text{(as required)} \]

**Part (b)**

At \( x = 0 \):
\[ y(0) = \ln(\cos 0 + \sin 0) = \ln(1) = 0 \]

From the first derivative expression:
\[ y'(0) = \frac{\cos 0 - \sin 0}{\cos 0 + \sin 0} = \frac{1}{1} = 1 \]

From the result of part (a):
\[ y''(0) = -2\text{e}^{-2(0)} = -2 \]

To find the third derivative, differentiate \( \frac{\text{d}^2y}{\text{d}x^2} = -2\text{e}^{-2y} \) with respect to \( x \):
\[ \frac{\text{d}^3y}{\text{d}x^3} = \frac{\text{d}}{\text{d}x}\left(-2\text{e}^{-2y}\right) = 4\text{e}^{-2y}\frac{\text{d}y}{\text{d}x} \]

At \( x = 0 \):
\[ y'''(0) = 4\text{e}^0 \cdot y'(0) = 4(1)(1) = 4 \]

The Maclaurin series expansion is:
\[ y = y(0) + x y'(0) + \frac{x^2}{2!} y''(0) + \frac{x^3}{3!} y'''(0) + \dots \]

Substituting the evaluated derivatives:
\[ y = 0 + x(1) + \frac{x^2}{2}(-2) + \frac{x^3}{6}(4) + \dots \]
\[ y = x - x^2 + \frac{2}{3}x^3 \]

**Part (a)**

* **M1**: Differentiates \( y \) once using the chain rule to obtain \( \frac{\text{d}y}{\text{d}x} = \frac{\cos x - \sin x}{\cos x + \sin x} \) (or equivalent implicit step such as \( \text{e}^y \frac{\text{d}y}{\text{d}x} = \cos x - \sin x \)).
* **M1**: Differentiates a second time using the quotient, product, or implicit rule to obtain a correct form of the second derivative.
* **A1\***: Completes the algebraic proof with no errors or omitted steps to establish \( \frac{\text{d}^2y}{\text{d}x^2} = -2\text{e}^{-2y} \).

**Part (b)**

* **B1**: Identifies \( y(0) = 0 \) and \( y'(0) = 1 \).
* **B1**: Uses the result from (a) to obtain \( y''(0) = -2 \).
* **M1**: Differentiates the equation in (a) to find an expression for \( y''' \) in terms of \( y \) and \( y' \), and evaluates it at \( x = 0 \) to find \( y'''(0) = 4 \).
* **M1**: Correctly applies the Maclaurin series formula with their evaluated derivatives at \( x = 0 \).
* **A1**: Correct simplified expression: \( x - x^2 + \frac{2}{3}x^3 \) (or equivalent equation \( y = x - x^2 + \frac{2}{3}x^3 \)).

(a) Let \(\frac{4r + 4}{r^2(r+2)^2} = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r+2} + \frac{D}{(r+2)^2}\). Multiplying both sides by the denominator: \(4r + 4 = Ar(r+2)^2 + B(r+2)^2 + Cr^2(r+2) + Dr^2\). Substituting \(r = 0\) gives \(4 = 4B \implies B = 1\). Substituting \(r = -2\) gives \(-4 = 4D \implies D = -1\). Equating the coefficient of \(r^3\) gives \(A + C = 0\). Equating the coefficient of \(r\) gives \(4A + 4B = 4 \implies 4A + 4 = 4 \implies A = 0\), which also means \(C = 0\). Therefore, \(\frac{4r+4}{r^2(r+2)^2} = \frac{1}{r^2} - \frac{1}{(r+2)^2}\). (b) Using the partial fractions from part (a): \(\sum_{r=1}^{n} \frac{4r+4}{r^2(r+2)^2} = \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+2)^2} \right)\). Listing the terms of the sum: for \(r = 1\), \(1 - \frac{1}{9}\); for \(r = 2\), \(\frac{1}{4} - \frac{1}{16}\); for \(r = 3\), \(\frac{1}{9} - \frac{1}{25}\); ...; for \(r = n-1\), \(\frac{1}{(n-1)^2} - \frac{1}{(n+1)^2}\); for \(r = n\), \(\frac{1}{n^2} - \frac{1}{(n+2)^2}\). Summing these terms, all intermediate terms cancel out, leaving: \(1 + \frac{1}{4} - \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2} = \frac{5}{4} - \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2}\). Combining over a common denominator: \(\frac{5(n+1)^2(n+2)^2 - 4(n+2)^2 - 4(n+1)^2}{4(n+1)^2(n+2)^2}\). Expanding the numerator: \(5(n^2+3n+2)^2 - 4(n^2+4n+4) - 4(n^2+2n+1) = 5(n^4+6n^3+13n^2+12n+4) - 8n^2 - 24n - 20 = 5n^4 + 30n^3 + 57n^2 + 36n = n(5n^3 + 30n^2 + 57n + 36)\). Since \(n = -3\) is a root of the cubic, we can factorise: \(5n^3 + 30n^2 + 57n + 36 = (n+3)(5n^2 + 15n + 12)\). Thus, the numerator is \(n(n+3)(5n^2 + 15n + 12)\), and the sum is \(\frac{n(n+3)(5n^2 + 15n + 12)}{4(n+1)^2(n+2)^2}\) as required.

(a) M1: Attempts to use a valid method to find at least one constant, such as substituting values of \(r\) or equating coefficients. A1: Correct partial fraction representation, \(\frac{1}{r^2} - \frac{1}{(r+2)^2}\) (or equivalent values of constants \(A=0, B=1, C=0, D=-1\)). (b) M1: Writes down at least the first three terms and the last two terms of the series to show the cancellation pattern. A1: Identifies the correct remaining terms of the sum, \(1 + \frac{1}{4} - \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2}\). M1: Expresses the sum over a common denominator of \(4(n+1)^2(n+2)^2\) and attempts to expand the numerator. A1*: Achieves the given result through clear, correct, and complete algebraic manipulation (cso).

**(a)**

Rewrite the given differential equation in standard form by dividing through by \( x+1 \) (since \( x > -1 \)):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x+1} y = \frac{\ln(x+1)}{(x+1)^2} \]

This is a first order linear differential equation of the form \( \frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x) \), where \( P(x) = \frac{2}{x+1} \).

Find the integrating factor, \( I(x) \):
\[ I(x) = \mathrm{e}^{\int \frac{2}{x+1} \mathrm{d}x} = \mathrm{e}^{2\ln(x+1)} = (x+1)^2 \]

Multiply the differential equation in standard form by the integrating factor:
\[ (x+1)^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2(x+1)y = \ln(x+1) \]
\[ \frac{\mathrm{d}}{\mathrm{d}x} \left[ y(x+1)^2 \right] = \ln(x+1) \]

Integrate both sides with respect to \( x \):
\[ y(x+1)^2 = \int \ln(x+1) \mathrm{d}x \]

To evaluate the integral, use the substitution \( u = x+1 \), so \( \mathrm{d}u = \mathrm{d}x \):
\[ \int \ln(x+1) \mathrm{d}x = \int \ln(u) \mathrm{d}u = u\ln(u) - u + C = (x+1)\ln(x+1) - (x+1) + C \]

Thus:
\[ y(x+1)^2 = (x+1)\ln(x+1) - (x+1) + C \]

Divide both sides by \( (x+1)^2 \) to find the general solution in the form \( y = \mathrm{f}(x) \):
\[ y = \frac{(x+1)\ln(x+1) - (x+1) + C}{(x+1)^2} = \frac{\ln(x+1) - 1}{x+1} + \frac{C}{(x+1)^2} \]

**(b)**

We are given the initial condition \( y = 1 \) when \( x = 0 \).

Substitute these values into the general solution:
\[ 1 = \frac{\ln(0+1) - 1}{0+1} + \frac{C}{(0+1)^2} \]
\[ 1 = \frac{0 - 1}{1} + \frac{C}{1} \]
\[ 1 = -1 + C \implies C = 2 \]

Substitute \( C = 2 \) back into the general solution to obtain the particular solution:
\[ y = \frac{\ln(x+1) - 1}{x+1} + \frac{2}{(x+1)^2} \]

Alternatively, combining the fractions yields:
\[ y = \frac{(x+1)\ln(x+1) - (x+1) + 2}{(x+1)^2} = \frac{(x+1)\ln(x+1) - x + 1}{(x+1)^2} \]

**(a)**
- **M1**: Divides the differential equation by \( (x+1) \) to obtain the standard form: \( \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x+1} y = \frac{\ln(x+1)}{(x+1)^2} \).
- **M1**: Integrates \( P(x) = \frac{2}{x+1} \) to obtain an integrating factor of the form \( \mathrm{e}^{k\ln(x+1)} \).
- **A1**: Correct integrating factor of \( (x+1)^2 \) or equivalent.
- **M1**: Multiplies by their integrating factor to write the LHS as a derivative of a product: \( \frac{\mathrm{d}}{\mathrm{d}x} [y \times \text{I.F.}] = \text{I.F.} \times Q(x) \).
- **M1**: Attempts integration of \( \ln(x+1) \) using substitution or integration by parts, including an arbitrary constant \( C \).
- **A1**: A fully correct general solution in the form \( y = \mathrm{f}(x) \), such as \( y = \frac{\ln(x+1) - 1}{x+1} + \frac{C}{(x+1)^2} \) or any equivalent form.

**(b)**
- **M1**: Substitutes \( x = 0 \) and \( y = 1 \) into their general solution to find the value of the constant of integration.
- **A1**: Correctly identifies the constant of integration (e.g., \( C = 2 \) for the standard form, or \( C = 1 \) if the integral of \( \ln(x+1) \) was written as \( (x+1)\ln(x+1) - x + C \)).
- **A1**: Correct particular solution in the form \( y = \mathrm{g}(x) \), such as \( y = \frac{(x+1)\ln(x+1) - x + 1}{(x+1)^2} \) or \( y = \frac{\ln(x+1) - 1}{x+1} + \frac{2}{(x+1)^2} \).

To find the set of values of \( x \) for which \( \frac{x + 2}{|x| - 1} > 2 \), we must analyze the inequality by considering the definition of \( |x| \).

**Case 1: \( x \ge 0 \)**
The inequality becomes:
\[ \frac{x + 2}{x - 1} > 2 \]
The expression is undefined at \( x = 1 \).

* **Subcase 1.1: \( x > 1 \)**
Since \( x - 1 > 0 \), we can multiply both sides of the inequality by \( x - 1 \) without changing the inequality sign:
\[ x + 2 > 2(x - 1) \]
\[ x + 2 > 2x - 2 \implies x < 4 \]
This gives the interval \( 1 < x < 4 \).

* **Subcase 1.2: \( 0 \le x < 1 \)**
Since \( x - 1 < 0 \), multiplying both sides of the inequality by \( x - 1 \) reverses the inequality sign:
\[ x + 2 < 2(x - 1) \]
\[ x + 2 < 2x - 2 \implies x > 4 \]
There is no overlap between \( 0 \le x < 1 \) and \( x > 4 \), so there are no solutions in this interval.

**Case 2: \( x < 0 \)**
The inequality becomes:
\[ \frac{x + 2}{-x - 1} > 2 \implies \frac{x + 2}{x + 1} < -2 \]
The expression is undefined at \( x = -1 \).

* **Subcase 2.1: \( -1 < x < 0 \)**
Since \( x + 1 > 0 \), multiplying both sides by \( x + 1 \) preserves the inequality sign:
\[ x + 2 < -2(x + 1) \]
\[ x + 2 < -2x - 2 \implies 3x < -4 \implies x < -\frac{4}{3} \]
There is no overlap between \( -1 < x < 0 \) and \( x < -\frac{4}{3} \), so there are no solutions in this interval.

* **Subcase 2.2: \( x < -1 \)**
Since \( x + 1 < 0 \), multiplying both sides by \( x + 1 \) reverses the inequality sign:
\[ x + 2 > -2(x + 1) \]
\[ x + 2 > -2x - 2 \implies 3x > -4 \implies x > -\frac{4}{3} \]
Combining this with \( x < -1 \) gives the interval \( -\frac{4}{3} < x < -1 \).

**Conclusion:**
Combining the valid intervals from both cases, the complete solution set is:
\[ -\frac{4}{3} < x < -1 \quad \text{or} \quad 1 < x < 4 \]
Or in interval notation:
\[ \left(-\frac{4}{3}, -1\right) \cup (1, 4) \]

- **M1**: For a valid attempt to split the inequality into cases based on the definition of \( |x| \) (e.g., considering \( x > 0 \) and \( x < 0 \) separately) or equivalent algebraic squaring/graphical methods.
- **A1**: For finding both non-asymptotic critical values: \( x = 4 \) and \( x = -\frac{4}{3} \).
- **B1**: For identifying the vertical asymptotes/domain restrictions: \( x = 1 \) and \( x = -1 \).
- **M1**: For testing intervals or using inequality algebra correctly to determine which regions satisfy the original inequality.
- **A1**: For obtaining either \( 1 < x < 4 \) or \( -\frac{4}{3} < x < -1 \).
- **A1**: For the complete and correct solution: \( -\frac{4}{3} < x < -1 \text{ or } 1 < x < 4 \) (accept equivalent interval/set notation).

(a) Given $w = \frac{z - \text{i}}{z + 2}$, we rearrange to make $z$ the subject:
$w(z+2) = z - \text{i} \implies wz + 2w = z - \text{i}$
$wz - z = -2w - \text{i} \implies z(w - 1) = -(2w + \text{i})$
$z = \frac{2w + \text{i}}{1 - w}$

Substitute $w = u + \text{i}v$ into the expression for $z$:
$z = \frac{2(u + \text{i}v) + \text{i}}{1 - (u + \text{i}v)} = \frac{2u + \text{i}(2v+1)}{(1-u) - \text{i}v}$

Multiply numerator and denominator by the conjugate of the denominator, $(1-u) + \text{i}v$:
$z = \frac{[2u + \text{i}(2v+1)][(1-u) + \text{i}v]}{[(1-u) - \text{i}v][(1-u) + \text{i}v]}$
$z = \frac{2u(1-u) - v(2v+1) + \text{i}[2uv + (2v+1)(1-u)]}{(1-u)^2 + v^2}$

Since $z$ lies on the imaginary axis, its real part is zero:
$\text{Re}(z) = 0 \implies 2u(1-u) - v(2v+1) = 0$
$2u - 2u^2 - 2v^2 - v = 0$
$2u^2 + 2v^2 - 2u + v = 0$
$u^2 + v^2 - u + \frac{1}{2}v = 0$
So $a = -1$ and $b = \frac{1}{2}$.

(b) To find the centre and radius, complete the square for $u$ and $v$:
$\left(u - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(v + \frac{1}{4}\right)^2 - \frac{1}{16} = 0$
$\left(u - \frac{1}{2}\right)^2 + \left(v + \frac{1}{4}\right)^2 = \frac{5}{16}$

The centre of $C$ has coordinates $\left(\frac{1}{2}, -\frac{1}{4}\right)$.
The radius of $C$ is $\sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$.

(c) The point excluded from the circle corresponds to the limit as $z \to \infty$ along the imaginary axis:
$w = \lim_{z \to \infty} \frac{z - \text{i}}{z + 2} = \lim_{z \to \infty} \frac{1 - \text{i}/z}{1 + 2/z} = 1$
Thus, the excluded point in the $w$-plane is $1 + 0\text{i}$, which has coordinates $(1, 0)$.

Part (a):
M1: Correctly attempts to make $z$ the subject of the formula. Shows at least one intermediate step of grouping $z$ terms.
A1: Correct expression $z = \frac{2w + \text{i}}{1 - w}$ (or any equivalent form).
M1: Substitutes $w = u + \text{i}v$ and multiplies the numerator and denominator by the complex conjugate of the denominator, $(1-u) + \text{i}v$, to find the real part.
M1: Sets the real part of $z$ equal to 0 to form an equation in $u$ and $v$.
A1: Obtains the correct equation $u^2 + v^2 - u + \frac{1}{2}v = 0$ (or equivalent with $a = -1, b = 0.5$).

Part (b):
M1: Completes the square for $u$ and $v$ in their equation of the circle.
A1: Correct coordinates of the centre $\left(\frac{1}{2}, -\frac{1}{4}\right)$ and correct exact radius $\frac{\sqrt{5}}{4}$.

Part (c):
B1: Correct coordinates $(1, 0)$ (accept $w = 1$ value).

(a) By de Moivre's theorem, \(\cos(6\theta) + i\sin(6\theta) = (\cos\theta + i\sin\theta)^6\). Using the binomial expansion: \((\cos\theta + i\sin\theta)^6 = \cos^6\theta + 6i\cos^5\theta\sin\theta - 15\cos^4\theta\sin^2\theta - 20i\cos^3\theta\sin^3\theta + 15\cos^2\theta\sin^4\theta + 6i\cos\theta\sin^5\theta - \sin^6\theta\). Equating the real parts gives: \(\cos(6\theta) = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\). Since \(\sin^2\theta = 1 - \cos^2\theta\), we can write this as: \(\cos(6\theta) = \cos^6\theta - 15\cos^4\theta(1-\cos^2\theta) + 15\cos^2\theta(1-\cos^2\theta)^2 - (1-\cos^2\theta)^3\). Expanding the terms: \((1-\cos^2\theta)^2 = 1 - 2\cos^2\theta + \cos^4\theta\) and \((1-\cos^2\theta)^3 = 1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta\). Substituting these back into the expression: \(\cos(6\theta) = \cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta(1 - 2\cos^2\theta + \cos^4\theta) - (1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta) = \cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta - 30\cos^4\theta + 15\cos^6\theta - 1 + 3\cos^2\theta - 3\cos^4\theta + \cos^6\theta\). Grouping like terms: \(\cos(6\theta) = (1 + 15 + 15 + 1)\cos^6\theta - (15 + 30 + 3)\cos^4\theta + (15 + 3)\cos^2\theta - 1 = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\). (b) The equation to solve is \(64x^6 - 96x^4 + 36x^2 - 3 = 0\). We can rewrite this as: \(2(32x^6 - 48x^4 + 18x^2 - 1) - 1 = 0\). Let \(x = \cos\theta\). Then: \(2\cos(6\theta) - 1 = 0\), which simplifies to \(\cos(6\theta) = \frac{1}{2}\). Since we require the solutions to have \(0 < \alpha < \pi\), we find the solutions for \(6\theta\) in the interval \(0 < 6\theta < 6\pi\): \(6\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}\). Dividing by 6, we get: \(\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\). Since \(x = \cos\theta\), the six distinct roots of the equation are: \(x = \cos\left(\frac{\pi}{18}\right)\), \(x = \cos\left(\frac{5\pi}{18}\right)\), \(x = \cos\left(\frac{7\pi}{18}\right)\), \(x = \cos\left(\frac{11\pi}{18}\right)\), \(x = \cos\left(\frac{13\pi}{18}\right)\), and \(x = \cos\left(\frac{17\pi}{18}\right)\).

(a) M1: Applies de Moivre's theorem to write \(\cos(6\theta) + i\sin(6\theta) = (\cos\theta + i\sin\theta)^6\) and attempts binomial expansion to identify the real part. A1: Obtains a correct expression for \(\cos(6\theta)\) in terms of powers of \(\cos\theta\) and \(\sin\theta\): \(\cos(6\theta) = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\). M1: Uses the identity \(\sin^2\theta = 1 - \cos^2\theta\) to express \(\cos(6\theta)\) entirely in terms of \(\cos\theta\). M1: Expands the powers \((1 - \cos^2\theta)^2\) and \((1 - \cos^2\theta)^3\) correctly. A1*: Fully correct algebraic simplification leading to the given expression \(\cos(6\theta) = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\) with no errors seen. (b) M1: Recognises the link between the equation and part (a) by substituting \(x = \cos\theta\) to obtain \(2\cos(6\theta) - 1 = 0\) or equivalent. A1: Identifies the correct values for \(6\theta\) in the range \((0, 6\pi)\), at least three correct values or a general form. A1: Obtains all six correct values for \(x\) in the form \(x = \cos\alpha\), with \(\alpha = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\).

(a)
The area \(A\) enclosed by the polar curve \(C\) is given by:
\[A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, \mathrm{d}\theta\]
Using the symmetry of the curve about the initial line:
\[A = \int_{0}^{\pi} r^2 \, \mathrm{d}\theta = \int_{0}^{\pi} a^2(2 + \cos\theta)^2 \, \mathrm{d}\theta\]
\[A = a^2 \int_{0}^{\pi} (4 + 4\cos\theta + \cos^2\theta) \, \mathrm{d}\theta\]
Using the double-angle identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[A = a^2 \int_{0}^{\pi} \left(4 + 4\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
\[A = a^2 \int_{0}^{\pi} \left(\frac{9}{2} + 4\cos\theta + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
Integrating term-by-term:
\[A = a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\]
Evaluating at the limits:
\[A = a^2 \left( \left(\frac{9}{2}\pi + 0 + 0\right) - (0 + 0 + 0) \right) = \frac{9}{2}\pi a^2\)\n\n(b)\nThe distance of a point on the curve from the initial line is given by \(y = r\sin\theta\).\nTangents to \(C\) are parallel to the initial line when \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\).\n\[y = a(2 + \cos\theta)\sin\theta = a(2\sin\theta + \sin\theta\cos\theta) = a\left(2\sin\theta + \frac{1}{2}\sin 2\theta\right)\]
Differentiating with respect to \(\theta\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + \cos 2\theta)\]
Using the double-angle identity \(\cos 2\theta = 2\cos^2\theta - 1\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + 2\cos^2\theta - 1) = a(2\cos^2\theta + 2\cos\theta - 1)\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
\[2\cos^2\theta + 2\cos\theta - 1 = 0\]
Using the quadratic formula for \(\cos\theta\):
\[\cos\theta = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}\]
Since \(-1 \le \cos\theta \le 1\):
\[\frac{-1 - \sqrt{3}}{2} \approx -1.366 \quad (\text{reject since } < -1)\]
\[\frac{-1 + \sqrt{3}}{2} \approx 0.366 \quad (\text{accept})\]
Thus, we have:
\[\cos\theta = \frac{\sqrt{3}-1}{2}\]
Since \(-\pi < \theta \le \pi\), this gives two solutions:
\[\theta = \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\]
Substituting back to find the corresponding values of \(r\):
\[r = a(2 + \cos\theta) = a\left(2 + \frac{\sqrt{3}-1}{2}\right) = a\left(\frac{4 + \sqrt{3} - 1}{2}\right) = \frac{a(3+\sqrt{3})}{2}\]
Therefore, the polar coordinates of the points are:
\[\left(\frac{a(3+\sqrt{3})}{2}, \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\right)\]

(a)
- M1: For attempting to use the polar area formula \(\frac{1}{2}\int r^2 \, \mathrm{d}\theta\) with appropriate limits.
- A1: For expanding correctly and substituting the double-angle identity \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\) to get a fully integrable form.
- M1: For integrating to obtain an expression of the form \(P\theta + Q\sin\theta + R\sin 2\theta\).
- A1: For the correct final exact area of \(\frac{9}{2}\pi a^2\) (or equivalent form).

(b)
- M1: For using \(y = r\sin\theta\) and establishing the condition \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\).
- A1: For correctly differentiating \(y\) to find \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + \cos 2\theta)\) or equivalent.
- M1: For converting to a quadratic equation in \(\cos\theta\) using the double-angle identity \(\cos 2\theta = 2\cos^2\theta - 1\).
- A1: For solving the quadratic equation to obtain \(\cos\theta = \frac{\sqrt{3}-1}{2}\) (and identifying/rejecting the invalid root).
- M1: For substituting their valid value of \(\cos\theta\) into the polar equation to obtain the value of \(r\) in terms of \(a\).
- A1: For providing both correct coordinate pairs: \(\left(\frac{a(3+\sqrt{3})}{2}, \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\right)\) (or equivalent).

(a)
The area \(A\) enclosed by the polar curve \(C\) is given by:
\[A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, \mathrm{d}\theta\]
Using the symmetry of the curve about the initial line:
\[A = \int_{0}^{\pi} r^2 \, \mathrm{d}\theta = \int_{0}^{\pi} a^2(2 + \cos\theta)^2 \, \mathrm{d}\theta\]
\[A = a^2 \int_{0}^{\pi} (4 + 4\cos\theta + \cos^2\theta) \, \mathrm{d}\theta\]
Using the double-angle identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[A = a^2 \int_{0}^{\pi} \left(4 + 4\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
\[A = a^2 \int_{0}^{\pi} \left(\frac{9}{2} + 4\cos\theta + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
Integrating term-by-term:
\[A = a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\]
Evaluating at the limits:
\[A = a^2 \left( \left(\frac{9}{2}\pi + 0 + 0\right) - (0 + 0 + 0) \right) = \frac{9}{2}\pi a^2\]

(b)
The distance of a point on the curve from the initial line is given by \(y = r\sin\theta\).
Tangents to \(C\) are parallel to the initial line when \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\).
\[y = a(2 + \cos\theta)\sin\theta = a(2\sin\theta + \sin\theta\cos\theta) = a\left(2\sin\theta + \frac{1}{2}\sin 2\theta\right)\]
Differentiating with respect to \(\theta\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + \cos 2\theta)\]
Using the double-angle identity \(\cos 2\theta = 2\cos^2\theta - 1\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + 2\cos^2\theta - 1) = a(2\cos^2\theta + 2\cos\theta - 1)\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
\[2\cos^2\theta + 2\cos\theta - 1 = 0\]
Using the quadratic formula for \(\cos\theta\):
\[\cos\theta = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}\]
Since \(-1 \le \cos\theta \le 1\):
\[\frac{-1 - \sqrt{3}}{2} \approx -1.366 \quad (\text{reject since } < -1)\]
\[\frac{-1 + \sqrt{3}}{2} \approx 0.366 \quad (\text{accept})\]
Thus, we have:
\[\cos\theta = \frac{\sqrt{3}-1}{2}\]
Since \(-\pi < \theta \le \pi\), this gives two solutions:
\[\theta = \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\]
Substituting back to find the corresponding values of \(r\):
\[r = a(2 + \cos\theta) = a\left(2 + \frac{\sqrt{3}-1}{2}\right) = a\left(\frac{4 + \sqrt{3} - 1}{2}\right) = \frac{a(3+\sqrt{3})}{2}\]
Therefore, the polar coordinates of the points are:
\[\left(\frac{a(3+\sqrt{3})}{2}, \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\right)\]

(a)
- M1: For attempting to use the polar area formula \(\frac{1}{2}\int r^2 \, \mathrm{d}\theta\) with appropriate limits.
- A1: For expanding correctly and substituting the double-angle identity \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\) to get a fully integrable form.
- M1: For integrating to obtain an expression of the form \(P\theta + Q\sin\theta + R\sin 2\theta\).
- A1: For the correct final exact area of \(\frac{9}{2}\pi a^2\) (or equivalent form).

(b)
- M1: For using \(y = r\sin\theta\) and establishing the condition \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\).
- A1: For correctly differentiating \(y\) to find \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(2\cos\theta + \cos 2\theta)\) or equivalent.
- M1: For converting to a quadratic equation in \(\cos\theta\) using the double-angle identity \(\cos 2\theta = 2\cos^2\theta - 1\).
- A1: For solving the quadratic equation to obtain \(\cos\theta = \frac{\sqrt{3}-1}{2}\) (and identifying/rejecting the invalid root).
- M1: For substituting their valid value of \(\cos\theta\) into the polar equation to obtain the value of \(r\) in terms of \(a\).
- A1: For providing both correct coordinate pairs: \(\left(\frac{a(3+\sqrt{3})}{2}, \pm \arccos\left(\frac{\sqrt{3}-1}{2}\right)\right)\) (or equivalent).

**(a)**
Given the substitution \( x = e^t \), we have \( t = \ln x \).

Using the chain rule to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} \implies x \frac{dy}{dx} = \frac{dy}{dt} \]

Differentiating again with respect to \( x \) to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x} \frac{dy}{dt}\right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \frac{d}{dx}\left(\frac{dy}{dt}\right) \]

Since \( \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d}{dt}\left(\frac{dy}{dt}\right) \frac{dt}{dx} = \frac{1}{x} \frac{d^2y}{dt^2} \), we obtain:
\[ \frac{d^2y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2y}{dt^2} = \frac{1}{x^2} \left( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right) \implies x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt} \]

Substituting these expressions into the original differential equation:
\[ \left( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right) + 5 \left( \frac{dy}{dt} \right) + 4y = 12 (e^t)^2 \]

Simplifying terms:
\[ \frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4y = 12 e^{2t} \]

**(b)**
First, solve the homogeneous equation:
\[ \frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4y = 0 \]

The auxiliary equation is:
\[ m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \implies m = -2 \text{ (repeated root)} \]

Thus, the complementary function (CF) is:
\[ y_{CF} = (A + Bt)e^{-2t} \]

To find the particular integral (PI), try a solution of the form:
\[ y_{PI} = C e^{2t} \]
\[ \frac{dy_{PI}}{dt} = 2C e^{2t}, \quad \frac{d^2y_{PI}}{dt^2} = 4C e^{2t} \]

Substitute these into the non-homogeneous equation:
\[ 4C e^{2t} + 4(2C e^{2t}) + 4(C e^{2t}) = 12 e^{2t} \]
\[ (4 + 8 + 4)C e^{2t} = 12 e^{2t} \implies 16C = 12 \implies C = \frac{3}{4} \]

So, the general solution in terms of \( y \) and \( t \) is:
\[ y = (A + Bt)e^{-2t} + \frac{3}{4} e^{2t} \]

Convert back to the variable \( x \) using \( e^t = x \) and \( t = \ln x \):
\[ y = (A + B\ln x)x^{-2} + \frac{3}{4} x^2 \]
\[ y = \frac{A + B\ln x}{x^2} + \frac{3}{4} x^2 \]

**(c)**
Apply the boundary conditions. When \( x = 1 \), \( y = 2 \):
\[ 2 = A + \frac{3}{4} \implies A = \frac{5}{4} \]

Differentiate the general solution with respect to \( x \):
\[ y = (A + B\ln x)x^{-2} + \frac{3}{4} x^2 \]
\[ \frac{dy}{dx} = -2(A + B\ln x)x^{-3} + B x^{-3} + \frac{3}{2} x \]

When \( x = 1 \), \( \frac{dy}{dx} = \frac{1}{2} \):
\[ \frac{1}{2} = -2A + B + \frac{3}{2} \]

Substitute \( A = \frac{5}{4} \):
\[ \frac{1}{2} = -2\left(\frac{5}{4}\right) + B + \frac{3}{2} \implies \frac{1}{2} = -\frac{5}{2} + B + \frac{3}{2} \implies \frac{1}{2} = B - 1 \implies B = \frac{3}{2} \]

Therefore, the particular solution is:
\[ y = \frac{\frac{5}{4} + \frac{3}{2}\ln x}{x^2} + \frac{3}{4} x^2 \implies y = \frac{5 + 6\ln x}{4x^2} + \frac{3}{4} x^2 \]

**(a)**
- **M1**: Attempts to find \( \frac{dy}{dx} \) using the chain rule with the substitution \( x = e^t \) or \( t =
\ln x \).
- **A1**: Correct expression for \( \frac{dy}{dx} \) (e.g., \( x \frac{dy}{dx} = \frac{dy}{dt} \)).
- **M1**: Attempts to find \( \frac{d^2y}{dx^2} \) using the product rule and chain rule.
- **A1**: Correct expression for \( \frac{d^2y}{dx^2} \) (e.g., \( x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt} \)).
- **A1* (cso)**: Fully correct substitution into the original equation leading to the given answer with no errors.

**(b)**
- **M1**: Sets up and solves the auxiliary equation \( m^2 + 4m + 4 = 0 \).
- **A1**: Correct complementary function \( y_{CF} = (A + Bt)e^{-2t} \) (accept other constant labels).
- **M1**: Proposes a particular integral of the form \( y_{PI} = C e^{2t} \), differentiates twice, and substitutes into the transformed equation.
- **A1**: Obtains \( C = \frac{3}{4} \).
- **A1**: Converts back to \( x \) to find the correct general solution \( y = \frac{A + B\ln x}{x^2} + \frac{3}{4} x^2 \).

**(c)**
- **M1**: Uses \( y = 2 \) at \( x = 1 \) to find the value of \( A \).
- **M1**: Differentiates their general solution and uses \( \frac{dy}{dx} = \frac{1}{2} \) at \( x = 1 \) to find the value of \( B \).
- **A1**: Obtains the correct particular solution: \( y = \frac{5 + 6\ln x}{4x^2} + \frac{3}{4} x^2 \) (or equivalent form, e.g., \( y = \frac{5}{4}x^{-2} + \frac{3}{2}x^{-2}\ln x + \frac{3}{4}x^2 \)).