2025 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

(i) We start from the right-hand side and combine into a single fraction:

\(\frac{1}{(r+1)^2} - \frac{1}{(r+2)^2} = \frac{(r+2)^2 - (r+1)^2}{(r+1)^2(r+2)^2}\)

\(= \frac{(r^2+4r+4) - (r^2+2r+1)}{(r+1)^2(r+2)^2}\)

\(= \frac{2r+3}{(r+1)^2(r+2)^2} = u_r\).

(ii) Using the result from (i), the sum is:

\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \left( \frac{1}{(r+1)^2} - \frac{1}{(r+2)^2} \right)\)

Writing out the terms:

\(r=1: \frac{1}{2^2} - \frac{1}{3^2}\)

\(r=2: \frac{1}{3^2} - \frac{1}{4^2}\)

\(\dots\)

\(r=n: \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2}\)

Summing these terms, the intermediate terms cancel out, leaving:

\(\sum_{r=1}^{n} u_r = \frac{1}{4} - \frac{1}{(n+2)^2}\).

(iii) We can express the sum from \(n+1\) to \(2n\) as:

\(\sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r\)

Using the formula from part (ii):

\(\sum_{r=n+1}^{2n} u_r = \left( \frac{1}{4} - \frac{1}{(2n+2)^2} \right) - \left( \frac{1}{4} - \frac{1}{(n+2)^2} \right)\)

\(= \frac{1}{(n+2)^2} - \frac{1}{(2n+2)^2}\)

\(= \frac{1}{(n+2)^2} - \frac{1}{4(n+1)^2}\).

(iv) As \(n \to \infty\), both \(\frac{1}{(n+2)^2} \to 0\) and \(\frac{1}{4(n+1)^2} \to 0\).

Using the result from (ii):

\(\sum_{r=1}^{\infty} u_r = \lim_{n \to \infty} \left( \frac{1}{4} - \frac{1}{(n+2)^2} \right) = \frac{1}{4}\).

(i) B1: For clear algebraic working showing the difference of the fractions yields the expression for \(u_r\).

(ii) M1: For writing out at least three terms of the sum to show the cancellation of intermediate terms.
A1: For obtaining the correct sum of \(\frac{1}{4} - \frac{1}{(n+2)^2}\) (or equivalent).

(iii) M1: For expressing the sum from \(n+1\) to \(2n\) as the difference between \(S_{2n}\) and \(S_n\).
A1: For obtaining the correct simplified expression \(\frac{1}{(n+2)^2} - \frac{1}{4(n+1)^2}\) (or equivalent).

(iv) B1: For stating the correct sum to infinity of \(\frac{1}{4}\).

Let \(f(n) = 7^{2n} + 16n - 1\). For \(n = 1\), \(f(1) = 7^{2(1)} + 16(1) - 1 = 49 + 16 - 1 = 64\), which is divisible by 64. Thus the statement is true for \(n = 1\). Assume the statement is true for \(n = k\), where \(k\) is a positive integer. That is, \(7^{2k} + 16k - 1 = 64M\) for some integer \(M\). We now consider \(n = k + 1\): \(f(k+1) = 7^{2(k+1)} + 16(k+1) - 1 = 7^{2k+2} + 16k + 16 - 1 = 49 \cdot 7^{2k} + 16k + 15\). Substituting \(7^{2k} = 64M - 16k + 1\), we get \(f(k+1) = 49(64M - 16k + 1) + 16k + 15 = 49(64M) - 784k + 49 + 16k + 15 = 49(64M) - 768k + 64 = 64(49M - 12k + 1)\). Since \(M\) and \(k\) are integers, \(49M - 12k + 1\) is an integer, so \(f(k+1)\) is divisible by 64. Therefore, since the statement is true for \(n = 1\), and if true for \(n = k\) it is also true for \(n = k + 1\), by mathematical induction the statement is true for all positive integers \(n\).

B1: Show that the statement holds for \(n = 1\). M1: State the inductive hypothesis (assume true for \(n = k\)). M1: Attempt to write \(f(k+1)\) and substitute the hypothesis. A1: Correctly expand and simplify to show the terms. A1: Factor out 64 to obtain \(64(49M - 12k + 1)\) or equivalent. A1: Complete logical argument concluding with the induction statement.

**(i) Shortest distance between \(l_1\) and \(l_2\):**

Let the direction vectors of \(l_1\) and \(l_2\) be \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\) respectively.

First, we find a vector normal to both lines, \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\):
\[\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 1 & -2 \end{vmatrix} = \mathbf{i}(2 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(2 - (-1)) = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\]

The magnitude of \(\mathbf{n}\) is:
\[|\mathbf{n}| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59}\]

Let \(A(1, 2, -1)\) be a point on \(l_1\) and \(B(2, -1, 1)\) be a point on \(l_2\). The displacement vector \(\vec{AB}\) is:
\[\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}\]

The shortest distance \(d\) between the skew lines is given by the projection of \(\vec{AB}\) onto the normal vector \(\mathbf{n}\):
\[d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(-1) + (-3)(7) + 2(3)|}{\sqrt{59}} = \frac{|-1 - 21 + 6|}{\sqrt{59}} = \frac{16}{\sqrt{59}} \approx 2.08\]

**(ii) Equation of the plane \(\Pi\) and perpendicular distance from \(C\):**

The plane \(\Pi\) contains \(l_1\) and is parallel to \(l_2\), so its normal vector is parallel to \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).
Using the normal vector \(\begin{pmatrix} 1 \\ -7 \\ -3 \end{pmatrix}\), the equation of the plane is:
\[x - 7y - 3z = D\]

Since \(\Pi\) contains the point \(A(1, 2, -1)\) on \(l_1\):
\[1 - 7(2) - 3(-1) = 1 - 14 + 3 = -10\]
Thus, \(D = -10\), giving the Cartesian equation:
\[x - 7y - 3z + 10 = 0\]

Now, we find the perpendicular distance from \(C(3, 4, 2)\) to \(\Pi\):
\[\text{Distance} = \frac{|1(3) - 7(4) - 3(2) + 10|}{\sqrt{1^2 + (-7)^2 + (-3)^2}} = \frac{|3 - 28 - 6 + 10|}{\sqrt{59}} = \frac{|-21|}{\sqrt{59}} = \frac{21}{\sqrt{59}} \approx 2.73\]

**(i)**
* **M1**: For finding the vector cross product of the two direction vectors.
* **A1**: For obtaining the correct normal vector \(\begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\) (or equivalent).
* **M1**: For using the formula \(\frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\) with their \(\vec{AB}\) and normal vector.
* **A1**: For the correct shortest distance \(\frac{16}{\sqrt{59}}\) or \(2.08\).

**(ii)**
* **M1**: For utilizing the normal vector found in (i) to set up the equation of the plane.
* **A1**: For substituting a point on \(l_1\) to determine the constant term.
* **A1**: For the correct Cartesian equation, e.g., \(x - 7y - 3z + 10 = 0\) (accept any equivalent form).
* **M1**: For using the formula for the perpendicular distance from a point to a plane.
* **A1**: For substituting the coordinates of \(C\) correctly into the formula.
* **A1**: For obtaining the correct distance \(\frac{21}{\sqrt{59}}\) or \(2.73\).

**Part (a)** First, we state the relations between the roots and coefficients for the equation \(x^3 - 3x^2 + 2x - 5 = 0\): \(\alpha + \beta + \gamma = 3\), \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\), and \(\alpha\beta\gamma = 5\). To find \(\alpha^2 + \beta^2 + \gamma^2\), we use: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (3)^2 - 2(2) = 9 - 4 = 5\). To find \(\alpha^3 + \beta^3 + \gamma^3\), since \(\alpha\), \(\beta\), and \(\gamma\) are roots of the cubic equation, they satisfy \(\alpha^3 - 3\alpha^2 + 2\alpha - 5 = 0\), \(\beta^3 - 3\beta^2 + 2\beta - 5 = 0\), and \(\gamma^3 - 3\gamma^2 + 2\gamma - 5 = 0\). Summing these three equations gives \((\alpha^3 + \beta^3 + \gamma^3) - 3(\alpha^2 + \beta^2 + \gamma^2) + 2(\alpha + \beta + \gamma) - 15 = 0\). Let \(S_n = \alpha^n + \beta^n + \gamma^n\). Then \(S_3 - 3S_2 + 2S_1 - 15 = 0\). Substituting \(S_1 = 3\) and \(S_2 = 5\) gives \(S_3 - 3(5) + 2(3) - 15 = 0\), which simplifies to \(S_3 - 15 + 6 - 15 = 0\), so \(S_3 = 24\). Thus, \(\alpha^3 + \beta^3 + \gamma^3 = 24\). **Part (b)** We use the substitution \(y = x^2\), which means \(x = \sqrt{y}\). Substituting this into the cubic equation gives \((\sqrt{y})^3 - 3(\sqrt{y})^2 + 2\sqrt{y} - 5 = 0\), which simplifies to \(y\sqrt{y} - 3y + 2\sqrt{y} - 5 = 0\). Grouping the terms with \(\sqrt{y}\) yields \(\sqrt{y}(y + 2) = 3y + 5\). Squaring both sides gives \(y(y + 2)^2 = (3y + 5)^2\), which expands to \(y(y^2 + 4y + 4) = 9y^2 + 30y + 25\). This simplifies to the cubic equation \(y^3 - 5y^2 - 26y - 25 = 0\). **Part (c)** The roots of \(y^3 - 5y^2 - 26y - 25 = 0\) are \(y_1 = \alpha^2\), \(y_2 = \beta^2\), and \(y_3 = \gamma^2\). We want to find \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{1}{y_1} + \frac{1}{y_2} + \frac{1}{y_3} = \frac{y_1 y_2 + y_2 y_3 + y_3 y_1}{y_1 y_2 y_3}\). From the coefficients of the cubic in \(y\), the sum of products of roots pairwise is \(y_1 y_2 + y_2 y_3 + y_3 y_1 = -26\), and the product of roots is \(y_1 y_2 y_3 = 25\). Thus, \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{-26}{25} = -1.04\).

**Part (a)** * **B1**: State sum of roots \(\Sigma \alpha = 3\) and sum of pairwise products \(\Sigma \alpha\beta = 2\). * **M1**: Use the algebraic identity \(\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2\Sigma \alpha\beta\) to find \(\alpha^2 + \beta^2 + \gamma^2\). * **A1**: Obtain \(\alpha^2 + \beta^2 + \gamma^2 = 5\). * **A1**: Use the recurrence relation \(S_3 - 3S_2 + 2S_1 - 15 = 0\) (or other valid method) to show \(\alpha^3 + \beta^3 + \gamma^3 = 24\). **Part (b)** * **M1**: Rearrange the cubic equation to isolate terms for squaring, or substitute \(x = \sqrt{y}\). * **M1**: Square both sides to eliminate the fractional powers of \(y\). * **A1**: Obtain the correct cubic equation \(y^3 - 5y^2 - 26y - 25 = 0\) (or integer multiple). **Part (c)** * **M1**: Express \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}\) in terms of the sum of pairwise products and product of the roots of the equation in \(y\). * **A1ft**: Substitute the correct coefficients from their equation in part (b) (e.g., \(\Sigma y_1 y_2 = -26\) and \(y_1 y_2 y_3 = 25\)). * **A1**: Obtain the final answer \(-\frac{26}{25}\) or \(-1.04\).

(a) For \(\mathbf{A}\) to be singular, its determinant must be zero.
\(\det(\mathbf{A}) = 2((-1)(-1) - (1)(2)) - 1((1)(-1) - (1)(k+1)) + k((1)(2) - (-1)(k+1))\)
\(\det(\mathbf{A}) = 2(1 - 2) - (-1 - k - 1) + k(2 + k + 1)\)
\(\det(\mathbf{A}) = -2 + (k + 2) + k(k + 3) = k^2 + 4k\)
Setting \(\det(\mathbf{A}) = 0\):
\(k^2 + 4k = 0 \implies k(k+4) = 0\)
So, \(k = 0\) or \(k = -4\).

(b) When \(k = 1\), \(\det(\mathbf{A}) = 1^2 + 4(1) = 5\).
The matrix of cofactors \(\mathbf{C}\) is given by:
\(C_{11} = +((-1)(-1) - 2(1)) = -1\)
\(C_{12} = -((1)(-1) - (1)(2)) = 3\)
\(C_{13} = +((1)(2) - (-1)(2)) = 4\)
\(C_{21} = -((1)(-1) - (1)(2)) = 3\)
\(C_{22} = +((2)(-1) - (1)(2)) = -4\)
\(C_{23} = -((2)(2) - (1)(2)) = -2\)
\(C_{31} = +((1)(1) - (-1)(1)) = 2\)
\(C_{32} = -((2)(1) - (1)(1)) = -1\)
\(C_{33} = +((2)(-1) - (1)(1)) = -3\)

So the cofactor matrix is:
\(\mathbf{C} = \begin{pmatrix} -1 & 3 & 4 \\ 3 & -4 & -2 \\ 2 & -1 & -3 \end{pmatrix}\)

Taking the transpose of \(\mathbf{C}\) to find the adjugate matrix:
\(\mathbf{C}^T = \begin{pmatrix} -1 & 3 & 2 \\ 3 & -4 & -1 \\ 4 & -2 & -3 \end{pmatrix}\)

Thus, the inverse is:
\(\mathbf{A}^{-1} = \frac{1}{5} \begin{pmatrix} -1 & 3 & 2 \\ 3 & -4 & -1 \\ 4 & -2 & -3 \end{pmatrix}\).

(c) The system of equations can be written as:
\(\mathbf{A} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ -1 \\ 3 \end{pmatrix}\) where \(k = 1\).

Therefore,
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} 6 \\ -1 \\ 3 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -1 & 3 & 2 \\ 3 & -4 & -1 \\ 4 & -2 & -3 \end{pmatrix} \begin{pmatrix} 6 \\ -1 \\ 3 \end{pmatrix}\)

\(x = \frac{1}{5} ( -1(6) + 3(-1) + 2(3) ) = -0.6\)
\(y = \frac{1}{5} ( 3(6) - 4(-1) - 1(3) ) = 3.8\)
\(z = \frac{1}{5} ( 4(6) - 2(-1) - 3(3) ) = 3.4\)

So \(x = -0.6\), \(y = 3.8\), and \(z = 3.4\).

(d) The transformation \(V\) maps the point \((1,0)\) to \((1,3)\).
The rotation matrix \(\mathbf{U}\) is given by:
\(\mathbf{U} = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\)

The image of the point \((1,0)\) under the combined transformation (\(V\) followed by \(U\)) is:
\(\mathbf{U} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} \cos \alpha - 3\sin \alpha \\ \sin \alpha + 3\cos \alpha \end{pmatrix}\)

We are given that this image is \((-1, 3)^T\), so:
1) \(\cos \alpha - 3\sin \alpha = -1\)
2) \(\sin \alpha + 3\cos \alpha = 3\)

From (1), we have \(\cos \alpha = 3\sin \alpha - 1\). Substituting this into (2):
\(\sin \alpha + 3(3\sin \alpha - 1) = 3 \implies 10\sin \alpha - 3 = 3 \implies \sin \alpha = 0.6\)

This gives \(\cos \alpha = 3(0.6) - 1 = 0.8\).
Since \(0 < \alpha < \frac{\pi}{2}\), we have:
\(\alpha = \arcsin(0.6) \approx 0.6435\) radians.

To 3 significant figures, \(\alpha = 0.644\) radians.

(a)
M1: Attempt to expand the determinant of \(\mathbf{A}\) in terms of \(k\).
A1: Obtain the correct quadratic expression \(k^2 + 4k\).
A1: Solve \(\det(\mathbf{A}) = 0\) to find \(k = 0\) and \(k = -4\).

(b)
M1: Find the determinant of \(\mathbf{A}\) for \(k=1\) (which is 5).
M1: Attempt to find the cofactors (award if at least 4 cofactors are correct).
A1: Correct cofactor matrix \(\mathbf{C}\) (or its transpose).
M1: Divide the transposed cofactor matrix by the determinant.
A1: Obtain the correct inverse matrix \(\mathbf{A}^{-1} = \frac{1}{5} \begin{pmatrix} -1 & 3 & 2 \\ 3 & -4 & -1 \\ 4 & -2 & -3 \end{pmatrix}\).

(c)
M1: Set up the matrix equation and express the solution as \(\mathbf{A}^{-1} \mathbf{b}\).
M1: Correctly perform the matrix multiplication.
A1: Obtain the correct solutions \(x = -0.6\), \(y = 3.8\), \(z = 3.4\) (accept exact fractional equivalents).

(d)
M1: Express the image of \((1,0)\) after \(V\) as \((1,3)^T\) and set up the multiplication \(\mathbf{U} \begin{pmatrix} 1 \\ 3 \end{pmatrix}\).
M1: Set up and solve the simultaneous equations for \(\sin \alpha\) and \(\cos \alpha\).
A1: Correctly deduce \(\alpha \approx 0.644\) radians (or \(36.9^\circ\)).

(i) Intercepts and Sketch:
For \( \theta = 0 \), \( r = a(3 + 2(1)) = 5a \). Point of intersection: \( (5a, 0) \).
For \( \theta = \pi \), \( r = a(3 + 2(-1)) = a \). Point of intersection: \( (a, \pi) \).
For \( \theta = \pm\frac{\pi}{2} \), \( r = a(3 + 0) = 3a \). Points of intersection: \( (3a, \frac{\pi}{2}) \) and \( (3a, -\frac{\pi}{2}) \).
The sketch is a closed curve symmetrical about the initial line, with a dimple at \( \theta = \pi \) and no loops, passing through the points calculated above.

(ii) Area calculation:
The area \( A \) is given by:
\[ A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, d\theta = a^2 \int_{0}^{\pi} (3 + 2\cos\theta)^2 \, d\theta \]
\[ A = a^2 \int_{0}^{\pi} (9 + 12\cos\theta + 4\cos^2\theta) \, d\theta \]
Using the double-angle identity \( \cos^2\theta = \frac{1}{2}(1 + \cos 2\theta) \):
\[ A = a^2 \int_{0}^{\pi} (9 + 12\cos\theta + 2(1 + \cos 2\theta)) \, d\theta = a^2 \int_{0}^{\pi} (11 + 12\cos\theta + 2\cos 2\theta) \, d\theta \]
Integrating term by term:
\[ A = a^2 \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_{0}^{\pi} \]
Evaluating the limits:
\[ A = a^2 \left( (11\pi + 0 + 0) - (0) \right) = 11\pi a^2 \]

(iii) Points where the tangent is perpendicular to the initial line:
These points occur where \( \frac{dx}{d\theta} = 0 \) and \( \frac{dy}{d\theta} \neq 0 \).
Using \( x = r\cos\theta \):
\[ x = a(3 + 2\cos\theta)\cos\theta = a(3\cos\theta + 2\cos^2\theta) \]
Differentiating with respect to \( \theta \):
\[ \frac{dx}{d\theta} = a(-3\sin\theta - 4\cos\theta\sin\theta) = -a\sin\theta(3 + 4\cos\theta) \]
Setting \( \frac{dx}{d\theta} = 0 \) (and since \( a > 0 \)):
\[ \sin\theta(3 + 4\cos\theta) = 0 \]
This gives two cases:
1) \( \sin\theta = 0 \implies \theta = 0 \) or \( \theta = \pi \) (for the interval \( -\pi < \theta \le \pi \)).
2) \( 3 + 4\cos\theta = 0 \implies \cos\theta = -\frac{3}{4} \implies \theta = \pm \arccos(-0.75) \approx \pm 2.41886 \approx \pm 2.42 \) rad (to 3 s.f.).

Finding the corresponding values of \( r \):
- For \( \theta = 0 \), \( r = 5a \).
- For \( \theta = \pi \), \( r = a \).
- For \( \theta = \pm 2.42 \), \( r = a(3 + 2(-0.75)) = 1.5a \).

We verify that \( \frac{dy}{d\theta} \neq 0 \) at these points, where \( y = r\sin\theta \):
\[ \frac{dy}{d\theta} = a(3\cos\theta + 2\cos 2\theta) \]
- For \( \theta = 0 \), \( \frac{dy}{d\theta} = 5a \neq 0 \).
- For \( \theta = \pi \), \( \frac{dy}{d\theta} = -a \neq 0 \).
- For \( \cos\theta = -0.75 \), \( \frac{dy}{d\theta} = a(-2.25 + 2(2(-0.75)^2 - 1)) = -2a \neq 0 \).

Thus, the required polar coordinates are:
\[ (5a, 0), \quad (a, \pi), \quad (1.5a, 2.42), \quad (1.5a, -2.42) \]

(i) Sketch [3 Marks]:
- B1: For a correct symmetrical, closed, dimpled egg-like shape with no loops, oriented primarily along the positive initial line.
- B1: For correctly identifying and labeling the intercepts on the initial line: \( (5a, 0) \) and \( (a, \pi) \).
- B1: For correctly identifying and labeling the intercepts on \( \theta = \pm\frac{\pi}{2} \): \( (3a, \frac{\pi}{2}) \) and \( (3a, -\frac{\pi}{2}) \).

(ii) Area [4 Marks]:
- M1: For using the area formula \( A = \frac{1}{2}\int r^2 \, d\theta \) with correct limits (e.g. from \( -\pi \) to \( \pi \), or using symmetry as \( 2 \times \frac{1}{2} \int_0^{\pi} \dots \)).
- A1: For expanding \( r^2 \) correctly and substituting the identity \( \cos^2\theta = \frac{1}{2}(1 + \cos 2\theta) \).
- M1: For integrating to a form \( p\theta + q\sin\theta + r\sin 2\theta \).
- A1: For obtaining the correct exact area \( 11\pi a^2 \) with no errors shown.

(iii) Perpendicular Tangents [6 Marks]:
- M1: For writing \( x = r\cos\theta \) and attempting to differentiate with respect to \( \theta \).
- A1: For finding the correct derivative \( \frac{dx}{d\theta} = -a\sin\theta(3 + 4\cos\theta) \) (or equivalent).
- M1: For setting \( \frac{dx}{d\theta} = 0 \) and solving for \( \theta \) in the interval \( -\pi < \theta \le \pi \).
- A1: For obtaining the four correct angles: \( \theta = 0 \), \( \theta = \pi \), and \( \theta \approx \pm 2.42 \) (or \( \pm 2.4188 \dots \) or exact equivalents).
- A1: For finding the correct corresponding values of \( r \): \( 5a \) for \( \theta = 0 \), \( a \) for \( \theta = \pi \), and \( 1.5a \) (or \( \frac{3}{2}a \)) for \( \theta \approx \pm 2.42 \).
- A1: For expressing the final answers as correct polar coordinates: \( (5a, 0) \), \( (a, \pi) \), \( (1.5a, 2.42) \), and \( (1.5a, -2.42) \) (accept exact form \( (1.5a, \pm \arccos(-0.75)) \)). (Candidate must show or state that \( \frac{dy}{d\theta} \neq 0 \) at these points, or imply it by checking).

**(i)**
By performing algebraic long division or rewriting the numerator, we can express the equation of the curve as:
\[y = \frac{2x(x+1) + 3(x+1) + 2}{x+1} = 2x + 3 + \frac{2}{x+1}\]
As \(x \to \pm \infty\), \(\frac{2}{x+1} \to 0\).
Therefore, the oblique asymptote is:
\[y = 2x + 3\]
The denominator is zero when \(x = -1\) and the numerator is non-zero.
Therefore, the vertical asymptote is:
\[x = -1\]

**(ii)**
Using the simplified form of the curve, we differentiate \(y\) with respect to \(x\):
\[\frac{dy}{dx} = 2 - \frac{2}{(x+1)^2}\]
For stationary points, we set \(\frac{dy}{dx} = 0\):
\[2 - \frac{2}{(x+1)^2} = 0 \implies (x+1)^2 = 1 \implies x+1 = \pm 1\]
This yields \(x = 0\) or \(x = -2\).
- At \(x = 0\), \(y = 5\), giving the stationary point \((0, 5)\).
- At \(x = -2\), \(y = -3\), giving the stationary point \((-2, -3)\).

To determine their nature, we find the second derivative:
\[\frac{d^2y}{dx^2} = \frac{4}{(x+1)^3}\]
- At \(x = 0\), \(\frac{d^2y}{dx^2} = 4 > 0\), so \((0, 5)\) is a local minimum.
- At \(x = -2\), \(\frac{d^2y}{dx^2} = -4 < 0\), so \((-2, -3)\) is a local maximum.

**(iii)**
The sketch of \(C\) should feature:
- The asymptotes \(x = -1\) and \(y = 2x + 3\) drawn as dashed lines.
- No intersections with the \(x\)-axis (since \(2x^2+5x+5=0\) has a negative discriminant \(\Delta = -15\)).
- The \(y\)-intercept at \((0,5)\), which is also the local minimum.
- Two branches:
- The right branch lies entirely in the region \(x > -1\) above \(y = 2x + 3\), with a minimum at \((0, 5)\).
- The left branch lies entirely in the region \(x < -1\) below \(y = 2x + 3\), with a maximum at \((-2, -3)\).

**(iv)**
The equation representing the intersections of \(C\) and the line \(y = kx + 5\) is:
\[\frac{2x^2 + 5x + 5}{x + 1} = kx + 5\]
Since the line passes through \((0,5)\), which lies on the curve, \(x = 0\) is always a solution. For \(x \neq -1\):
\[2x^2 + 5x + 5 = (kx + 5)(x + 1) = kx^2 + (k+5)x + 5\]
\[(2 - k)x^2 - kx = 0 \implies x[(2-k)x - k] = 0\]
The non-zero root is given by:
\[x = \frac{k}{2 - k}\]
For there to be two distinct real roots:
1. The equation must remain quadratic, meaning the second root must exist, so \(2 - k \neq 0 \implies k \neq 2\). (Geometrically, the line is parallel to the oblique asymptote \(y = 2x + 3\) when \(k = 2\), leaving only the intersection at \(x=0\)).
2. The second root must be distinct from the first, so \(x \neq 0 \implies k \neq 0\). (Geometrically, when \(k = 0\), the line is the tangent line \(y=5\) at the local minimum, yielding a repeated root).
3. The second root cannot be at the vertical asymptote \(x = -1\), which is always true since \(\frac{k}{2-k} = -1\) has no solutions.

Thus, the set of values is \(\{k \in \mathbb{R} : k \neq 0, k \neq 2\}\).

**(i)**
M1: For a valid attempt at algebraic long division or equivalent expansion to find the oblique asymptote.
A1: For correct oblique asymptote equation \(y = 2x + 3\).
A1: For correct vertical asymptote equation \(x = -1\).

**(ii)**
M1: For differentiating the expression for \(y\) (using quotient rule or from the simplified form).
A1: For obtaining a correct first derivative, e.g., \(\frac{dy}{dx} = 2 - \frac{2}{(x+1)^2}\).
M1: For setting the derivative to zero and solving for \(x\).
A1: For both stationary points correctly identified: \((0, 5)\) and \((-2, -3)\).
A1: For correct determination of nature for both stationary points with valid justification (e.g. second derivative test).

**(iii)**
G1: For both asymptotes correctly drawn (dashed) and labeled.
G1: For both branches of the curve correctly drawn in the appropriate asymptotic regions.
G1: For labeling the stationary points \((0, 5)\) and \((-2, -3)\) in their correct relative positions.
G1: For indicating the correct intercept on the \(y\)-axis at \((0,5)\) and showing no intersections with the \(x\)-axis.

**(iv)**
M1: For setting up the algebraic equation and multiplying through by \(x+1\).
A1: For simplifying to the quadratic form \(x[(2-k)x - k] = 0\) or equivalent.
M1: For identifying that the non-zero root must be distinct from \(0\) and must exist, leading to identifying \(k = 0\) and \(k = 2\) as critical boundary cases (either algebraically or through geometric reasoning about tangency and parallelism to the asymptote).
A1: For stating the correct set of values: \(k \neq 0\) and \(k \neq 2\) (accept equivalent interval notations).

(i) To find the eigenvalues of \(\mathbf{A}\), we solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). This gives \(\det\begin{pmatrix} 3 - \lambda & -1 \\ 2 & -\lambda \end{pmatrix} = 0\), which simplifies to \(\lambda^2 - 3\lambda + 2 = 0\). Factoring gives \((\lambda-1)(\lambda-2) = 0\), so the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 2\). (ii) For \(\lambda = 1\), the eigenvector equation is \((\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0}\), which is \(\begin{pmatrix} 2 & -1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\). This gives \(2x - y = 0\), so an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\). For \(\lambda = 2\), the eigenvector equation is \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}\), which is \(\begin{pmatrix} 1 & -1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\). This gives \(x - y = 0\), so an eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\). Therefore, we can write \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\). (iii) The inverse matrix is \(\mathbf{P}^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix}\). Then \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 2^n \\ 2 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2^{n+1}-1 & 1-2^n \\ 2^{n+1}-2 & 2-2^n \end{pmatrix}\).

(i) M1 for attempting to find eigenvalues via characteristic equation. A1 for correct eigenvalues 1 and 2. (ii) M1 for finding eigenvectors corresponding to both eigenvalues. A1 for correct P and D. (iii) M1 for calculating P^-1 and attempting to find P D^n P^-1. A1 for correct matrix expression for A^n.

(i) We use integration by parts for \( I_n = \int_0^1 1 \cdot (1-x^2)^n \, dx \).
Let \( u = (1-x^2)^n \) and \( \frac{dv}{dx} = 1 \).
Then \( \frac{du}{dx} = -2nx(1-x^2)^{n-1} \) and \( v = x \).
Integrating by parts gives:
\( I_n = \left[ x(1-x^2)^n \right]_0^1 - \int_0^1 x \left( -2nx(1-x^2)^{n-1} \right) \, dx \)
For \( n \ge 1 \), evaluating the boundary term yields:
\( \left[ x(1-x^2)^n \right]_0^1 = 1(0) - 0 = 0 \).
Thus,
\( I_n = 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \).
Writing \( x^2 = 1 - (1-x^2) \), we get:
\( I_n = 2n \int_0^1 \left( 1 - (1-x^2) \right) (1-x^2)^{n-1} \, dx \)
\( I_n = 2n \int_0^1 \left( (1-x^2)^{n-1} - (1-x^2)^n \right) \, dx \)
\( I_n = 2n (I_{n-1} - I_n) \)
\( I_n = 2n I_{n-1} - 2n I_n \)
\( (2n+1) I_n = 2n I_{n-1} \).

(ii) We apply the reduction formula successively:
For \( n = 3 \): \( I_3 = \frac{6}{7} I_2 \)
For \( n = 2 \): \( I_2 = \frac{4}{5} I_1 \)
For \( n = 1 \): \( I_1 = \frac{2}{3} I_0 \)
We calculate the base case \( I_0 \):
\( I_0 = \int_0^1 (1-x^2)^0 \, dx = \int_0^1 1 \, dx = [x]_0^1 = 1 \).
Therefore:
\( I_1 = \frac{2}{3} (1) = \frac{2}{3} \)
\( I_2 = \frac{4}{5} \left(\frac{2}{3}\right) = \frac{8}{15} \)
\( I_3 = \frac{6}{7} \left(\frac{8}{15}\right) = \frac{48}{105} = \frac{16}{35} \).

For part (i):
M1: For applying integration by parts to \( I_n \) with appropriate choices for \( u \) and \( v \).
A1: For obtaining \( I_n = 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx \) with limits evaluated correctly.
M1: For writing \( x^2 = 1 - (1-x^2) \) and splitting the integral to express it in terms of \( I_n \) and \( I_{n-1} \).
A1: For completing the proof to show the given relation \( (2n+1)I_n = 2n I_{n-1} \) with clear steps.

For part (ii):
M1: For evaluating \( I_0 = 1 \) (or calculating \( I_1 \) directly) and applying the reduction formula step-by-step to find \( I_3 \).
A1: For obtaining the correct exact value of \( \frac{16}{35} \).

**(i)**
Using the substitution \( x = e^t \), we have \( t = \ln x \) and \( \frac{dt}{dx} = \frac{1}{x} \).
By the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} \implies x \frac{dy}{dx} = \frac{dy}{dt} \]
Differentiating again with respect to \( x \):
\[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x} \frac{dy}{dt} \right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \frac{d}{dx} \left( \frac{dy}{dt} \right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \left( \frac{d^2 y}{dt^2} \frac{dt}{dx} \right) \]
\[ \frac{d^2 y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2 y}{dt^2} \implies x^2 \frac{d^2 y}{dx^2} = \frac{d^2 y}{dt^2} - \frac{dy}{dt} \]
Substituting these into the original differential equation:
\[ \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right) + 5 \left( \frac{dy}{dt} \right) + 4y = 9e^t \]
\[ \frac{d^2 y}{dt^2} + 4 \frac{dy}{dt} + 4y = 9e^t \]
as required.

**(ii)**
To solve the transformed equation, we first find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \implies m = -2 \text{ (repeated root)} \]
Thus, the CF is:
\[ y_c = (A + Bt)e^{-2t} \]
For the particular integral (PI), we try \( y_p = C e^t \):
\[ \frac{dy_p}{dt} = C e^t, \quad \frac{d^2 y_p}{dt^2} = C e^t \]
Substituting into the differential equation:
\[ C e^t + 4C e^t + 4C e^t = 9e^t \implies 9C = 9 \implies C = 1 \]
Thus, the PI is \( y_p = e^t \).
The general solution in terms of \( t \) is:
\[ y = (A + Bt)e^{-2t} + e^t \]
Using \( t = \ln x \) and \( e^t = x \), we obtain the general solution in terms of \( x \):
\[ y = \frac{A + B \ln x}{x^2} + x \]

**(iii)**
We are given the initial conditions: when \( x = 1 \), \( y = 2 \) and \( \frac{dy}{dx} = 1 \).
Substituting \( x = 1 \) and \( y = 2 \) into the general solution:
\[ 2 = \frac{A + B \ln(1)}{1^2} + 1 \implies 2 = A + 1 \implies A = 1 \]
Now, we differentiate \( y \) with respect to \( x \):
\[ y = x + (A + B \ln x)x^{-2} \]
\[ \frac{dy}{dx} = 1 - 2(A + B \ln x)x^{-3} + B x^{-1} x^{-2} = 1 + \frac{B - 2A - 2B \ln x}{x^3} \]
Substituting \( x = 1 \), \( \frac{dy}{dx} = 1 \), and \( A = 1 \):
\[ 1 = 1 + \frac{B - 2(1) - 2B(0)}{1} \implies 1 = 1 + B - 2 \implies B = 2 \]
Thus, the particular solution is:
\[ y = \frac{1 + 2 \ln x}{x^2} + x \]

**(i)**
- **M1**: Correct use of the chain rule to find \( \frac{dy}{dx} = \frac{1}{x} \frac{dy}{dt} \) or equivalent.
- **M1**: Differentiates again with respect to \( x \) to find \( \frac{d^2 y}{dx^2} \) using the product/chain rule.
- **A1**: Correctly shows the given differential equation in terms of \( y \) and \( t \) with all intermediate steps clear.

**(ii)**
- **M1**: Formulates and solves the auxiliary equation \( m^2 + 4m + 4 = 0 \).
- **A1**: Correct complementary function \( y_c = (A + Bt)e^{-2t} \) (or equivalent with different arbitrary constant letters).
- **M1**: Finds the correct form of the particular integral \( y_p = e^t \) by substituting \( y_p = C e^t \) and solving for \( C \).
- **A1**: Writes the correct general solution in terms of \( t \): \( y = (A + Bt)e^{-2t} + e^t \).
- **A1**: Substitutes back to find the general solution in terms of \( x \): \( y = \frac{A + B \ln x}{x^2} + x \).

**(iii)**
- **M1**: Applies boundary conditions \( y = 2 \) and \( \frac{dy}{dx} = 1 \) at \( x = 1 \) to find both constants \( A \) and \( B \).
- **A1**: Correct particular solution: \( y = \frac{1 + 2 \ln x}{x^2} + x \) (or equivalent standard form).

For part (a): Differentiating \( y = e^{\arctan x} \) with respect to \( x \) using the chain rule gives \( \frac{dy}{dx} = e^{\arctan x} \cdot \frac{1}{1+x^2} \). Since \( y = e^{\arctan x} \), we substitute to get \( \frac{dy}{dx} = \frac{y}{1+x^2} \). Multiplying both sides by \( 1+x^2 \) yields \( (1+x^2)\frac{dy}{dx} = y \). For part (b): Differentiating \( (1+x^2)\frac{dy}{dx} = y \) once with respect to \( x \) using the product rule gives \( (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = \frac{dy}{dx} \), which simplifies to \( (1+x^2)\frac{d^2y}{dx^2} + (2x-1)\frac{dy}{dx} = 0 \). Differentiating a second time with respect to \( x \) using the product rule gives \( (1+x^2)\frac{d^3y}{dx^3} + 2x\frac{d^2y}{dx^2} + (2x-1)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0 \). Grouping the \( \frac{d^2y}{dx^2} \) terms gives the required result: \( (1+x^2)\frac{d^3y}{dx^3} + (4x-1)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0 \). For part (c): Evaluate \( y \) and its derivatives at \( x = 0 \). We have \( y(0) = e^{\arctan 0} = e^0 = 1 \). From part (a), at \( x = 0 \), \( (1)\cdot y'(0) = y(0) \implies y'(0) = 1 \). From the first derivative relation \( (1+x^2)y'' + (2x-1)y' = 0 \) at \( x = 0 \), we get \( (1)y''(0) + (-1)y'(0) = 0 \implies y''(0) = y'(0) = 1 \). From the result in part (b), at \( x = 0 \), \( (1)y'''(0) + (-1)y''(0) + 2y'(0) = 0 \implies y'''(0) - 1 + 2(1) = 0 \implies y'''(0) = -1 \). Differentiating the second derivative relation once more to find the fourth derivative: \( (1+x^2)y^{(4)} + 2xy''' + (4x-1)y''' + 4y'' + 2y'' = 0 \), which simplifies to \( (1+x^2)y^{(4)} + (6x-1)y''' + 6y'' = 0 \). At \( x = 0 \), this gives \( (1)y^{(4)}(0) + (-1)y'''(0) + 6y''(0) = 0 \implies y^{(4)}(0) - (-1) + 6(1) = 0 \implies y^{(4)}(0) = -7 \). Substituting these values into Maclaurin's series formula: \( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \dots \) yields \( y(x) = 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{7}{24}x^4 \).

Part (a): M1 for differentiating \( y = e^{\arctan x} \) correctly using the chain rule. A1 for obtaining the printed equation \( (1+x^2)\frac{dy}{dx} = y \) showing clear working. Part (b): M1 for differentiating the relation once using the product rule. A1 for the correct first derivative relation \( (1+x^2)y'' + (2x-1)y' = 0 \). M1 for differentiating a second time using the product rule. A1 for obtaining the printed third derivative relation correctly. Part (c): M1 for evaluating at least three of the values \( y(0) = 1 \), \( y'(0) = 1 \), \( y''(0) = 1 \), and \( y'''(0) = -1 \) correctly. M1 for differentiating the relation in (b) to obtain a correct expression for the fourth derivative and substituting \( x=0 \) to find \( y^{(4)}(0) = -7 \). M1 for substituting their values into Maclaurin's general expansion formula. A1 for the correct final series: \( 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{7}{24}x^4 \).

For part (a): Differentiating \( y = e^{\arctan x} \) with respect to \( x \) using the chain rule gives \( \frac{dy}{dx} = e^{\arctan x} \cdot \frac{1}{1+x^2} \). Since \( y = e^{\arctan x} \), we substitute to get \( \frac{dy}{dx} = \frac{y}{1+x^2} \). Multiplying both sides by \( 1+x^2 \) yields \( (1+x^2)\frac{dy}{dx} = y \). For part (b): Differentiating \( (1+x^2)\frac{dy}{dx} = y \) once with respect to \( x \) using the product rule gives \( (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = \frac{dy}{dx} \), which simplifies to \( (1+x^2)\frac{d^2y}{dx^2} + (2x-1)\frac{dy}{dx} = 0 \). Differentiating a second time with respect to \( x \) using the product rule gives \( (1+x^2)\frac{d^3y}{dx^3} + 2x\frac{d^2y}{dx^2} + (2x-1)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0 \). Grouping the \( \frac{d^2y}{dx^2} \) terms gives the required result: \( (1+x^2)\frac{d^3y}{dx^3} + (4x-1)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0 \). For part (c): Evaluate \( y \) and its derivatives at \( x = 0 \). We have \( y(0) = e^{\arctan 0} = e^0 = 1 \). From part (a), at \( x = 0 \), \( (1)\cdot y'(0) = y(0) \implies y'(0) = 1 \). From the first derivative relation \( (1+x^2)y'' + (2x-1)y' = 0 \) at \( x = 0 \), we get \( (1)y''(0) + (-1)y'(0) = 0 \implies y''(0) = y'(0) = 1 \). From the result in part (b), at \( x = 0 \), \( (1)y'''(0) + (-1)y''(0) + 2y'(0) = 0 \implies y'''(0) - 1 + 2(1) = 0 \implies y'''(0) = -1 \). Differentiating the second derivative relation once more to find the fourth derivative: \( (1+x^2)y^{(4)} + 2xy''' + (4x-1)y''' + 4y'' + 2y'' = 0 \), which simplifies to \( (1+x^2)y^{(4)} + (6x-1)y''' + 6y'' = 0 \). At \( x = 0 \), this gives \( (1)y^{(4)}(0) + (-1)y'''(0) + 6y''(0) = 0 \implies y^{(4)}(0) - (-1) + 6(1) = 0 \implies y^{(4)}(0) = -7 \). Substituting these values into Maclaurin's series formula: \( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \dots \) yields \( y(x) = 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{7}{24}x^4 \).

Part (a): M1 for differentiating \( y = e^{\arctan x} \) correctly using the chain rule. A1 for obtaining the printed equation \( (1+x^2)\frac{dy}{dx} = y \) showing clear working. Part (b): M1 for differentiating the relation once using the product rule. A1 for the correct first derivative relation \( (1+x^2)y'' + (2x-1)y' = 0 \). M1 for differentiating a second time using the product rule. A1 for obtaining the printed third derivative relation correctly. Part (c): M1 for evaluating at least three of the values \( y(0) = 1 \), \( y'(0) = 1 \), \( y''(0) = 1 \), and \( y'''(0) = -1 \) correctly. M1 for differentiating the relation in (b) to obtain a correct expression for the fourth derivative and substituting \( x=0 \) to find \( y^{(4)}(0) = -7 \). M1 for substituting their values into Maclaurin's general expansion formula. A1 for the correct final series: \( 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{7}{24}x^4 \).

(a) By de Moivre's theorem, we have:
\[ \cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5 \]
Using the binomial expansion on the right-hand side:
\[ (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \]
Equating the imaginary parts of both sides gives:
\[ \sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \]
To express this purely in terms of \(\sin\theta\), we substitute \ \cos^2\theta = 1 - \sin^2\theta\):
\[ \sin 5\theta = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta \]
\[ \sin 5\theta = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta \]
\[ \sin 5\theta = (5\sin\theta - 10\sin^3\theta + 5\sin^5\theta) - 10\sin^3\theta + 11\sin^5\theta \]
\[ \sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \]

(b) Consider the equation \(\sin 5\theta = 0\):
\[ 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 0 \]
Factoring out \(\sin\theta\):
\[ \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0 \]
For \(\sin\theta \neq 0\), the roots of the equation \(16x^4 - 20x^2 + 5 = 0\) are given by \(x = \sin\theta\), where \(\sin 5\theta = 0\) and \(\sin\theta \neq 0\).

The solutions to \(\sin 5\theta = 0\) are:
\[ 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \quad \text{for } k \in \mathbb{Z} \]
Since \(\sin\theta \neq 0\), \(k\) is not a multiple of 5. To obtain 4 distinct values for \(x = \sin\theta\) corresponding to the quartic equation, we can choose:
\[ \theta = \pm\frac{\pi}{5}, \pm\frac{2\pi}{5} \]
This gives the roots as:
\[ x = \sin\left(\pm\frac{\pi}{5}\right) = \pm\sin\left(\frac{\pi}{5}\right) \]
\[ x = \sin\left(\pm\frac{2\pi}{5}\right) = \pm\sin\left(\frac{2\pi}{5}\right) \]

These values for \(\alpha\) (namely \(\frac{\pi}{5}\) and \(\frac{2\pi}{5}\)) are indeed in the interval \(0 < \alpha < \frac{\pi}{2}\).

Thus, the roots are:
\[ x = \pm\sin\left(\frac{\pi}{5}\right) \quad \text{and} \quad x = \pm\sin\left(\frac{2\pi}{5}\right) \]

Part (a):
- M1: For stating de Moivre's theorem \((\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta\) and attempting binomial expansion.
- A1: For identifying and correctly writing down the expression for the imaginary part: \(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\).
- M1: For substituting \(\cos^2\theta = 1 - \sin^2\theta\) to express the entire equation in terms of \(\sin\theta\).
- A1: For correct algebraic simplification to yield the given identity: \(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\).

Part (b):
- M1: For establishing the link between the roots of \(16x^4 - 20x^2 + 5 = 0\) and the equation \(\sin 5\theta = 0\) with \(\sin\theta \neq 0\).
- M1: For solving \(\sin 5\theta = 0\) to obtain \(\theta = \frac{k\pi}{5}\).
- A1: For identifying appropriate values of \(\theta\) (e.g., \(\pm\frac{\pi}{5}, \pm\frac{2\pi}{5}\) or equivalent set of 4 values).
- A1: For writing the roots correctly in the requested form: \(x = \pm\sin\left(\frac{\pi}{5}\right)\) and \(x = \pm\sin\left(\frac{2\pi}{5}\right)\).

(i) Let \( u = y^2 \). Differentiating both sides with respect to \( x \) using the chain rule gives:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 2y \frac{\mathrm{d}y}{\mathrm{d}x} \]

Multiply the original differential equation by \( y \):
\[ 2xy \frac{\mathrm{d}y}{\mathrm{d}x} + y^2 = x^3 \]

Substitute \( 2y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x} \) and \( y^2 = u \):
\[ x \frac{\mathrm{d}u}{\mathrm{d}x} + u = x^3 \]

Divide through by \( x \) (since \( x > 0 \)):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{1}{x} u = x^2 \]

(ii) The equation
\[ \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{1}{x} u = x^2 \]
is a first-order linear differential equation. Find the integrating factor, \( I(x) \):
\[ I(x) = e^{\int \frac{1}{x} \mathrm{d}x} = e^{\ln x} = x \]

Multiply the differential equation by the integrating factor \( x \):
\[ x \frac{\mathrm{d}u}{\mathrm{d}x} + u = x^3 \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}(ux) = x^3 \]

Integrate both sides with respect to \( x \):
\[ ux = \int x^3 \mathrm{d}x \]
\[ ux = \frac{1}{4}x^4 + C \]

Divide by \( x \) to solve for \( u \):
\[ u = \frac{1}{4}x^3 + \frac{C}{x} \]

Since \( u = y^2 \), the general solution is:
\[ y^2 = \frac{1}{4}x^3 + \frac{C}{x} \]

(iii) Substitute the initial conditions \( x = 1 \) and \( y = 2 \) into the general solution:
\[ 2^2 = \frac{1}{4}(1)^3 + \frac{C}{1} \]
\[ 4 = \frac{1}{4} + C \implies C = \frac{15}{4} \]

Thus, the particular solution is:
\[ y^2 = \frac{1}{4}x^3 + \frac{15}{4x} \] (or equivalent, such as \( y^2 = \frac{x^4+15}{4x} \))

(i)
* M1: For differentiating \( u = y^2 \) with respect to \( x \) to get \( \frac{\mathrm{d}u}{\mathrm{d}x} = 2y \frac{\mathrm{d}y}{\mathrm{d}x} \) or equivalent.
* A1: For multiplying the given differential equation by \( y \) and substituting correctly to obtain the target equation.

(ii)
* M1: For calculating the integrating factor \( e^{\int \frac{1}{x} \mathrm{d}x} = x \).
* M1: For multiplying the differential equation by their integrating factor to write \( \frac{\mathrm{d}}{\mathrm{d}x}(ux) = x^3 \) or equivalent.
* A1: For integrating to obtain \( ux = \frac{1}{4}x^4 + C \).
* M1: For dividing by \( x \) and replacing \( u \) with \( y^2 \).
* A1: For obtaining the correct general solution \( y^2 = \frac{1}{4}x^3 + \frac{C}{x} \) (or equivalent form).

(iii)
* M1: For substituting \( x = 1, y = 2 \) into their general solution to find \( C \).
* A1: For obtaining the correct particular solution \( y^2 = \frac{1}{4}x^3 + \frac{15}{4x} \) (or \( y^2 = \frac{x^4 + 15}{4x} \)).

(a) To find the eigenvalues, we solve the characteristic equation \(\det(A - \lambda I) = 0\):
\[ \begin{vmatrix} 2-\lambda & 1 & 0 \\ 0 & 3-\lambda & 0 \\ 1 & 1 & 1-\triangle \end{vmatrix} = 0 \]
Expanding along the third column:
\[ (1-\lambda) \begin{vmatrix} 2-\lambda & 1 \\ 0 & 3-\lambda \end{vmatrix} = (1-\lambda)(2-\lambda)(3-\lambda) = 0 \]
Thus, the eigenvalues of \(A\) are \(\lambda = 1, 2, 3\).

(b) To find the eigenvectors:
For \(\lambda = 1\):
\[ (A - I)\mathbf{v} = 0 \implies \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x = 0 \]
Thus, an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda = 2\):
\[ (A - 2I)\mathbf{v} = 0 \implies \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x = z \]
Thus, an eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda = 3\):
\[ (A - 3I)\mathbf{v} = 0 \implies \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = y, x = z \]
Thus, an eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

(c) The characteristic equation of \(A\) is \(\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0\). By the Cayley-Hamilton theorem:
\[ A^3 - 6A^2 + 11A - 6I = 0 \]
Multiplying through by \(A^{-1}\):
\[ A^2 - 6A + 11I - 6A^{-1} = 0 \implies A^{-1} = \frac{1}{6}(A^2 - 6A + 11I) \]
Next, we calculate \(A^2\):
\[ A^2 = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 0 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 0 \\ 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 5 & 0 \\ 0 & 9 & 0 \\ 3 & 5 & 1 \end{pmatrix} \]
Now, substitute \(A^2\) and \(A\) into the expression for \(A^{-1}\):
\[ A^2 - 6A + 11I = \begin{pmatrix} 4 & 5 & 0 \\ 0 & 9 & 0 \\ 3 & 5 & 1 \end{pmatrix} - \begin{pmatrix} 12 & 6 & 0 \\ 0 & 18 & 0 \\ 6 & 6 & 6 \end{pmatrix} + \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} = \begin{pmatrix} 3 & -1 & 0 \\ 0 & 2 & 0 \\ -3 & -1 & 6 \end{pmatrix} \]
Therefore,
\[ A^{-1} = \frac{1}{6} \begin{pmatrix} 3 & -1 & 0 \\ 0 & 2 & 0 \\ -3 & -1 & 6 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{6} & 0 \\ 0 & \frac{1}{3} & 0 \\ -\frac{1}{2} & -\frac{1}{6} & 1 \end{pmatrix} \]

(a)
M1: For attempts to evaluate \(\det(A - \lambda I) = 0\).
A1: For obtaining the correct characteristic equation \((1-\lambda)(2-\lambda)(3-\lambda) = 0\) (or equivalent cubic).
A1: For listing the eigenvalues \(1, 2, 3\).

(b)
M1: For an attempt to solve \((A-\lambda I)\mathbf{v} = 0\) for any one eigenvalue.
A1: For a correct eigenvector corresponding to \(\lambda = 1\), e.g., \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\) (or any non-zero scalar multiple).
M1: For attempting to find the remaining eigenvectors.
A1: For correct eigenvectors corresponding to \(\lambda = 2\) and \(\lambda = 3\), e.g., \(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) (or any non-zero scalar multiples).

(c)
M1: For stating the Cayley-Hamilton equation and multiplying by \(A^{-1}\) to express \(A^{-1}\) in terms of \(A^2\), \(A\), and \(I\).
M1: For calculating \(A^2\) correctly as \(\begin{pmatrix} 4 & 5 & 0 \\ 0 & 9 & 0 \\ 3 & 5 & 1 \end{pmatrix}\).
A1: For evaluating \(A^2 - 6A + 11I\) to get \(\begin{pmatrix} 3 & -1 & 0 \\ 0 & 2 & 0 \\ -3 & -1 & 6 \end{pmatrix}\).
A1: For the final correct inverse matrix \(A^{-1} = \begin{pmatrix} 1/2 & -1/6 & 0 \\ 0 & 1/3 & 0 \\ -1/2 & -1/6 & 1 \end{pmatrix}\) (or fractional equivalent).

**(a)**
Substitute the exponential definitions of \(\cosh x\) and \(\sinh x\):
\(5\left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right) = 7\)

Multiply the entire equation by 2:
\(5(e^x + e^{-x}) - (e^x - e^{-x}) = 14\)
\(4e^x + 6e^{-x} = 14\)

Divide by 2 and multiply by \(e^x\):
\(2e^{2x} - 7e^x + 3 = 0\)

Factor the quadratic equation:
\((2e^x - 1)(e^x - 3) = 0\)

This yields:
\(e^x = \frac{1}{2}\) or \(e^x = 3\)

Taking the natural logarithm of both sides:
\(x = -\ln 2\) and \(x = \ln 3\).

**(b)**
Using the identity \(\cosh^2 x = 1 + \sinh^2 x\), write the integral as:
\(\int_{0}^{\ln 3} \cosh x (1 + \sinh^2 x) \, \mathrm{d}x\)

Use the substitution \(u = \sinh x\), which gives \(\mathrm{d}u = \cosh x \, \mathrm{d}x\).

Find the new limits of integration:
- When \(x = 0\), \(u = \sinh 0 = 0\).
- When \(x = \ln 3\), \(u = \sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - 1/3}{2} = \frac{4}{3}\).

Substitute into the integral:
\(\int_{0}^{4/3} (1 + u^2) \, \mathrm{d}u = \left[ u + \frac{1}{3}u^3 \right]_{0}^{4/3}\)

Evaluate at the limits:
\(\left(\frac{4}{3} + \frac{1}{3}\left(\frac{4}{3}\right)^3\right) - 0 = \frac{4}{3} + \frac{64}{81} = \frac{108}{81} + \frac{64}{81} = \frac{172}{81}\).

**(c) (i)**
The arc length \(s\) is given by:
\(s = \int_{0}^{\ln 2} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \, \mathrm{d}x\)

Since \(y = \cosh x\), we have \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sinh x\).

Substitute into the formula:
\(s = \int_{0}^{\ln 2} \sqrt{1 + \sinh^2 x} \, \mathrm{d}x = \int_{0}^{\ln 2} \cosh x \, \mathrm{d}x\)

Evaluate the integral:
\(s = \left[ \sinh x \right]_{0}^{\ln 2} = \sinh(\ln 2) - \sinh(0)\)

Since \(\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - 1/2}{2} = \frac{3}{4}\), we have:
\(s = \frac{3}{4}\).

**(c) (ii)**
The surface area of revolution \(S\) about the \(x\)-axis is given by:
\(S = 2\pi \int_{0}^{\ln 2} y \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \, \mathrm{d}x = 2\pi \int_{0}^{\ln 2} \cosh x \cdot \cosh x \, \mathrm{d}x = 2\pi \int_{0}^{\ln 2} \cosh^2 x \, \mathrm{d}x\)

Use the double-angle identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\):
\(S = 2\pi \int_{0}^{\ln 2} \frac{1}{2}(\cosh 2x + 1) \, \mathrm{d}x = \pi \left[ \frac{1}{2}\sinh 2x + x \right]_{0}^{\ln 2}\)

Evaluate at the limits:
- At \(x = \ln 2\):
\(\sinh(2\ln 2) = \sinh(\ln 4) = \frac{4 - 1/4}{2} = \frac{15}{8}\)

So:
\(\pi \left( \frac{1}{2} \left(\frac{15}{8}\right) + \ln 2 \right) = \pi \left( \frac{15}{16} + \ln 2 \right)\)

- At \(x = 0\), the term is 0.

Thus, the exact surface area is \(\pi \left( \frac{15}{16} + \ln 2 \right)\).

**(a)**
- **M1**: Substitute exponential definitions for \(\cosh x\) and \(\sinh x\) into the equation.
- **A1**: Obtain a correct simplified quadratic in \(e^x\), e.g., \(2e^{2x} - 7e^x + 3 = 0\).
- **M1**: Solve the quadratic equation for \(e^x\).
- **A1**: Both answers correct and in exact logarithmic form: \(x = -\ln 2\) (or equivalent) and \(x = \ln 3\).

**(b)**
- **M1**: Express \(\cosh^3 x\) as \(\cosh x (1 + \sinh^2 x)\) and use substitution \(u = \sinh x\) (or equivalent method, e.g., triple-angle identity).
- **A1**: Correctly transform integration limits to \(0\) and \(\frac{4}{3}\) (or find the correct antiderivative \(\sinh x + \frac{1}{3}\sinh^3 x\)).
- **M1**: Integrate and substitute the limits correctly.
- **A1**: Correct final simplified fraction \(\frac{172}{81}\).

**(c)(i)**
- **M1**: Use the formula for arc length \(s = \int \sqrt{1 + y'^2} \, \mathrm{d}x\).
- **A1**: Use identity \(1 + \sinh^2 x = \cosh^2 x\) to simplify the integrand to \(\cosh x\), and integrate to \(\sinh x\).
- **A1**: Substitute the limits and show that \(s = \frac{3}{4}\) (answer given).

**(c)(ii)**
- **M1**: Use the correct formula for surface area of revolution \(S = 2\pi \int y \sqrt{1 + y'^2} \, \mathrm{d}x\).
- **A1**: Obtain \(2\pi \int_{0}^{\ln 2} \cosh^2 x \, \mathrm{d}x\) and apply the identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\).
- **M1**: Integrate to obtain \(\pi \left[ \frac{1}{2}\sinh 2x + x \right]\) and substitute the limits.
- **A1**: Correct exact surface area \(\pi \left( \frac{15}{16} + \ln 2 \right)\) (or any equivalent exact form).