2024 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

A loop of the curve is defined for values of \( \theta \) where \( r \ge 0 \). Since \( r = 3\sin(3\theta) \), a single loop begins at \( 3\theta = 0 \) and ends at \( 3\theta = \pi \), which gives the interval \( 0 \le \theta \le \frac{\pi}{3} \). The area \( A \) of this loop is given by the integral: \( A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} 9\sin^2(3\theta) \, d\theta \). Using the trigonometric identity \( \sin^2(3\theta) = \frac{1 - \cos(6\theta)}{2} \), we obtain: \( A = \frac{9}{4} \int_{0}^{\frac{\pi}{3}} (1 - \cos(6\theta)) \, d\theta = \frac{9}{4} \left[ \theta - \frac{1}{6}\sin(6\theta) \right]_{0}^{\frac{\pi}{3}} \). Evaluating this at the limits: \( A = \frac{9}{4} \left( \left( \frac{\pi}{3} - \frac{1}{6}\sin(2\pi) \right) - (0 - 0) \right) = \frac{9}{4} \left( \frac{\pi}{3} \right) = \frac{3\pi}{4} \).

M1: Set up the area integral with correct limits 0 and \( \pi/3 \). M1: Apply the correct double-angle trigonometric identity. A1: Correctly integrated expression. M1: Correct substitute of limits. A1.67: Obtain final exact area of \( \frac{3\pi}{4} \).

The matrix for a shear parallel to the \( y \)-axis has the general form \( \mathbf{T}_1 = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \). Under this transformation, the point \( (1, 0) \) maps to \( (1, k) \). Given that \( (1, 0) \) maps to \( (1, 4) \), we must have \( k = 4 \), so \( \mathbf{T}_1 = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \). The rotation matrix is given by \( \mathbf{T}_2 = \begin{pmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \). The combined transformation matrix is \( \mathbf{M} = \mathbf{T}_2 \mathbf{T}_1 = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \). Performing the matrix multiplication yields: \( \mathbf{M} = \begin{pmatrix} \frac{\sqrt{3}}{2}(1) + (-\frac{1}{2})(4) & \frac{\sqrt{3}}{2}(0) + (-\frac{1}{2})(1) \\ \frac{1}{2}(1) + \frac{\sqrt{3}}{2}(4) & \frac{1}{2}(0) + \frac{\sqrt{3}}{2}(1) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}-4}{2} & -\frac{1}{2} \\ \frac{1+4\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \).

B1: Obtain correct shear matrix \( \mathbf{T}_1 \). B1: Obtain correct rotation matrix \( \mathbf{T}_2 \). M1: Correctly apply order of multiplication \( \mathbf{T}_2 \mathbf{T}_1 \). A1: At least two correct entries in final matrix. A2.67: Obtain fully correct matrix \( \mathbf{M} \) in exact form.

Using partial fractions, we express the general term as: \( \frac{2}{(r+2)(r+4)} = \frac{A}{r+2} + \frac{B}{r+4} \). Solving for \( A \) and \( B \) gives \( 2 = A(r+4) + B(r+2) \). Substituting \( r = -2 \) yields \( 2 = 2A \implies A = 1 \), and substituting \( r = -4 \) yields \( 2 = -2B \implies B = -1 \). Thus, \( \frac{2}{(r+2)(r+4)} = \frac{1}{r+2} - \frac{1}{r+4} \). Writing out the terms of the sum: \( S_n = \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \dots + \left( \frac{1}{n+1} - \frac{1}{n+3} \right) + \left( \frac{1}{n+2} - \frac{1}{n+4} \right) \). After cancelling intermediate terms, we are left with: \( S_n = \frac{1}{3} + \frac{1}{4} - \frac{1}{n+3} - \frac{1}{n+4} = \frac{7}{12} - \frac{1}{n+3} - \frac{1}{n+4} \). To find the sum to infinity, we take the limit as \( n \to \infty \): \( S_{\infty} = \lim_{n \to \infty} S_n = \frac{7}{12} - 0 - 0 = \frac{7}{12} \).

M1: Decompose general term into partial fractions. A1: Correctly obtain \( \frac{1}{r+2} - \frac{1}{r+4} \). M1: Write down terms to show cancellation pattern. A1.67: Obtain correct finite sum expression \( S_n = \frac{7}{12} - \frac{1}{n+3} - \frac{1}{n+4} \). M1: Take limit of \( S_n \) as \( n \to \infty \). A1: Correctly state sum to infinity is \( \frac{7}{12} \).

(i) By algebraic division, we can rewrite the equation of the curve:
\(y = \frac{x(x+2) - 3x - 2}{x+2} = x - \frac{3x+6-4}{x+2} = x - 3 + \frac{4}{x+2}\).
Thus, as \(x \to -2\), \(y \to \pm \infty\), so the vertical asymptote is \(x = -2\).
As \(x \to \pm \infty\), \(y \to x - 3\), so the oblique asymptote is \(y = x - 3\).

(ii) To find stationary points, we differentiate \(y = x - 3 + 4(x+2)^{-1}\) with respect to \(x\):
\(\frac{dy}{dx} = 1 - \frac{4}{(x+2)^2}\).
Setting \(\frac{dy}{dx} = 0\):
\(\frac{4}{(x+2)^2} = 1 \implies (x+2)^2 = 4 \implies x+2 = \pm 2\).
This yields \(x = 0\) or \(x = -4\).
- For \(x = 0\), \(y = \frac{-2}{2} = -1\).
- For \(x = -4\), \(y = \frac{16+4-2}{-2} = -9\).
Hence, the coordinates of the stationary points are \((0, -1)\) and \((-4, -9)\).

(iii) The y-intercept is at \((0, -1)\).
For the x-intercepts, set \(y = 0\):
\(x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = 2\) or \(x = -1\).
The coordinates are \((2, 0)\) and \((-1, 0)\).
The sketch should show two branches:
- A left branch with a local maximum at \((-4, -9)\) approaching the asymptotes \(x = -2\) and \(y = x - 3\).
- A right branch with a local minimum at \((0, -1)\) passing through \((-1, 0)\), \((0, -1)\), and \((2, 0)\).

(iv) The line \(y = k\) does not intersect the curve if \(k\) lies strictly between the local maximum value and the local minimum value.
Therefore, the set of values is \(-9 < k < -1\).

(i) M1: For attempting algebraic division to find the oblique asymptote.
A1: For correct vertical asymptote \(x = -2\).
A1: For correct oblique asymptote \(y = x - 3\).

(ii) M1: For differentiating and setting the derivative equal to 0.
A1: For finding \(x = 0\) and \(x = -4\).
A1: For both correct coordinates \((0, -1)\) and \((-4, -9)\).

(iii) B1: For drawing the asymptotes correctly and labeling them.
B1: For showing a correct two-branched shape.
B1: For labeling the stationary points.
B2: For all axes intercepts correctly labeled (deduct 1 mark if one is missing or incorrect).

(iv) M1: For identifying that the range of no real roots is determined by the y-coordinates of the stationary points.
A1.75: For the correct inequality \(-9 < k < -1\).

(i) The curve is a cardioid symmetric about the initial line, with its cusp at the pole. It passes through \((2a, 0)\), \((a, \pi/2)\), \((0, \pi)\), and \((a, 3\pi/2)\).

(ii) The y-coordinate is given by:
\(y = r\sin\theta = a(1+\cos\theta)\sin\theta = a\sin\theta + a\sin\theta\cos\theta = a\sin\theta + \frac{1}{2}a\sin 2\theta\).
For tangents parallel to the initial line, we set \(\frac{dy}{d\theta} = 0\):
\(\frac{dy}{d\theta} = a\cos\theta + a\cos 2\theta = a(\cos\theta + 2\cos^2\theta - 1) = a(2\cos^2\theta + \cos\theta - 1) = 0\).
Factorizing the quadratic expression:
\((2\cos\theta - 1)(\cos\theta + 1) = 0\).
So \\cos\theta = \frac{1}{2}\) or \(\cos\theta = -1\).
- If \(\cos\theta = \frac{1}{2}\), then \(\theta = \frac{\pi}{3}\) or \(\theta = \frac{5\pi}{3}\).
When \(\theta = \frac{\pi}{3}\), \(r = a(1 + \frac{1}{2}) = \frac{3}{2}a\).
When \(\theta = \frac{5\pi}{3}\), \(r = a(1 + \frac{1}{2}) = \frac{3}{2}a\).
- If \(\cos\theta = -1\), then \(\theta = \pi\), which gives \(r = 0\). This represents the pole where the tangent is along the initial line.
Thus, the required points are \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and \(\left(\frac{3}{2}a, \frac{5\pi}{3}\right)\).

(iii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{0}^{2\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{2\pi} a^2(1+\cos\theta)^2 \, d\theta\)
\(= \frac{a^2}{2} \int_{0}^{2\pi} (1 + 2\cos\theta + \cos^2\theta) \, d\theta\)
\(= \frac{a^2}{2} \int_{0}^{2\pi} \left(1 + 2\cos\theta + \frac{1+\cos 2\theta}{2}\right) \, d\theta\)
\(= \frac{a^2}{2} \int_{0}^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \, d\theta\)
\(= \frac{a^2}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{2\pi}\)
\(= \frac{a^2}{2} \left( 3\pi \right) = \frac{3\pi a^2}{2}\).

(iv) The upper half-plane area is half of the total area: \(\frac{3\pi a^2}{4}\).
If the line \(\theta = \alpha\) divides this area in half, then:
\(\frac{1}{2} \int_{0}^{\alpha} a^2(1+\cos\theta)^2 \, d\theta = \frac{3\pi a^2}{8}\)
\(\int_{0}^{\alpha} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \, d\theta = \frac{3\pi}{4}\)
\(\left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\alpha} = \frac{3\pi}{4}\)
\(\frac{3}{2}\alpha + 2\sin\alpha + \frac{1}{4}\sin 2\alpha = \frac{3\pi}{4}\).
Multiplying by 2:
\(3\alpha + 4\sin\alpha + \sin 2\alpha = \frac{3\pi}{2}\).

(i) B1: For a correct cardioid shape.
B1: For symmetry about the initial line.
B1: For identifying and labeling the maximum distance at \(r = 2a\).

(ii) M1: For writing \(y = r\sin\theta\).
M1: For differentiating with respect to \(\theta\) and setting equal to 0.
A1: For solving to find \(\cos\theta = 1/2\).
A1: For correct polar coordinates \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and \(\left(\frac{3}{2}a, \frac{5\pi}{3}\right)\).

(iii) M1: For using \(A = \frac{1}{2} \int r^2 \, d\theta\).
M1: For expanding and using double-angle identity for \(\cos^2\theta\).
A1: For correct integration.
A1.75: For correct final area value \(\frac{3\pi a^2}{2}\).

(iv) M1: For setting up the integral equation with limits \(0\) to \(\alpha\) equated to half the upper area.
A1: For clear algebraic reduction to the given expression.

(i) The matrix is in the form \(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) with \(\cos\theta = \frac{1}{\sqrt{2}}\) and \(\sin\theta = \frac{1}{\sqrt{2}}\), which corresponds to \(\theta = 45^{\circ}\) (or \(\frac{\pi}{4}\) radians).
Thus, \(\mathbf{M}_1\) represents a rotation of \(45^{\circ}\) anticlockwise about the origin.

(ii) A shear parallel to the x-axis with the x-axis invariant has a matrix of the form \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\).
Applying this to \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\):
\(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 + k \\ 1 \end{pmatrix}\).
We are given that this is \(\begin{pmatrix} 4 \\ 1 \end{pmatrix}\), so \(1 + k = 4 \implies k = 3\).
Thus, \(\mathbf{M}_2 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\).

(iii) The matrix \(\mathbf{M}\) representing the combined transformation is \(\mathbf{M}_2 \mathbf{M}_1\):
\(\mathbf{M} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} & -\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \frac{4}{\sqrt{2}} & \frac{2}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 2\sqrt{2} & \sqrt{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}\).

(iv) Let the equation of the invariant line be \(y = mx + c\).
Applying the matrix \(\mathbf{W}\):
\(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} 3x + 2mx + 2c \\ 2x + 3mx + 3c \end{pmatrix}\).
For the line to be invariant, \(Y = mX + c\) must hold for all \(x\):
\(2x + 3mx + 3c = m(3x + 2mx + 2c) + c\)
\((2+3m)x + 3c = (3m+2m^2)x + (2mc + c)\).
Equating coefficients of \(x\):
\(2 + 3m = 3m + 2m^2 \implies 2m^2 = 2 \implies m = \pm 1\).
Equating the constant terms:
\(3c = 2mc + c \implies 2c(1 - m) = 0\).
- Case 1: If \(m = 1\), the equation \(2c(1-1) = 0\) is always true, so \(c\) can be any real number.
The invariant lines are \(y = x + c\) for all \(c \in \mathbb{R}\).
- Case 2: If \(m = -1\), we have \(2c(2) = 0 \implies 4c = 0 \implies c = 0\).
The invariant line is \(y = -x\).

(i) B1: For identifying rotation about the origin.
B1: For specifying the angle as \(45^{\circ}\) (or \(\pi/4\) radians) anticlockwise.

(ii) M1: For identifying the general form \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\) and setting up an equation.
A1: For \(\mathbf{M}_2 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\).

(iii) M1: For performing matrix multiplication in the correct order \(\mathbf{M}_2 \mathbf{M}_1\).
A1.75: For the correct resulting matrix.

(iv) M1: For writing down the mapping of coordinates \((x, mx+c)\).
M1: For substituting into \(Y = mX + c\) and comparing coefficients.
A1: For setting up the two conditions \(2 = 2m^2\) and \(2c = 2mc\).
A1: For obtaining \(m = \pm 1\).
A1: For showing that if \(m=1\), then \(c\) can take any value.
A1: For showing that if \(m=-1\), then \(c=0\).
A1: For stating both sets of invariant lines clearly.

(i) First, we check if the lines are parallel. The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix}\). Since \(\mathbf{d}_1\) is not a scalar multiple of \(\mathbf{d}_2\), the lines are not parallel.

Next, we test for intersection by setting up a system of equations:
1) \(1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 2\)
2) \(2 - \lambda = -1 + \mu \implies \lambda + \mu = 3\)
3) \(-1 + \lambda = 2 + 3\mu \implies \lambda - 3\mu = 3\)

Adding equations (1) and (2):
\(3\lambda = 5 \implies \lambda = \frac{5}{3}\).
Substituting \(\lambda = \frac{5}{3}\) into equation (2):
\(\frac{5}{3} + \mu = 3 \implies \mu = \frac{4}{3}\).
Now test these values in equation (3):
LHS \(= \lambda - 3\mu = \frac{5}{3} - 3\left(\frac{4}{3}\right) = \frac{5}{3} - 4 = -\frac{7}{3}\).
RHS \(= 3\).
Since LHS \(\neq\) RHS, the lines do not intersect.
Since the lines are neither parallel nor intersecting, they are skew.

(ii) The normal vector \(\mathbf{n}\) to the plane \(\Pi\) must be perpendicular to both direction vectors:
\(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & 3 \end{vmatrix} = \mathbf{i}(-3-1) - \mathbf{j}(6-1) + \mathbf{k}(2-(-1)) = -4\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}\).
We can choose the normal vector to be \(\mathbf{n} = \begin{pmatrix} 4 \\ 5 \\ -3 \end{pmatrix}\).
Since \(\Pi\) contains \(l_1\), it passes through the point \((1, 2, -1)\).
The equation of the plane is:
\(4x + 5y - 3z = 4(1) + 5(2) - 3(-1) = 4 + 10 + 3 = 17\).
Thus, the equation is \(4x + 5y - 3z = 17\).

(iii) The shortest distance between the skew lines is the perpendicular distance from any point on \(l_2\) to the plane \(\Pi\).
A point on \(l_2\) is \((3, -1, 2)\).
The distance \(d\) is:
\(d = \frac{|4(3) + 5(-1) - 3(2) - 17|}{\sqrt{4^2 + 5^2 + (-3)^2}} = \frac{|12 - 5 - 6 - 17|}{\sqrt{16 + 25 + 9}} = \frac{|-16|}{\sqrt{50}} = \frac{16}{5\sqrt{2}} = \frac{8\sqrt{2}}{5}\).

(i) B1: For explaining why the lines are not parallel.
M1: For setting up the simultaneous equations for intersection.
A1: For finding values of \(\lambda\) and \(\mu\) using two of the equations.
A1: For showing the values do not satisfy the third equation, concluding that they are skew.

(ii) M1: For attempting the cross product of the direction vectors.
A1: For a correct normal vector (e.g., \(\begin{pmatrix} 4 \\ 5 \\ -3 \end{pmatrix}\) or \(\begin{pmatrix} -4 \\ -5 \\ 3 \end{pmatrix}\)).
M1: For substituting the point \((1, 2, -1)\) into the plane equation formula.
A1.75: For the final plane equation \(4x + 5y - 3z = 17\).

(iii) M1: For identifying a point on \(l_2\) (e.g., \((3, -1, 2)\)).
M1: For attempting to use the distance from a point to a plane formula (or projection of a connecting vector onto the unit normal).
A1: For evaluating the numerator of the distance formula.
A2: For simplifying to the exact value \(\frac{8\sqrt{2}}{5}\) (or \(\frac{16}{\sqrt{50}}\)).

Using the identity \(\cosh^2 x = 1 + \sinh^2 x\), the equation becomes \(3(1 + \sinh^2 x) - 5\sinh x = 5\). Simplifying this gives the quadratic equation \(3\sinh^2 x - 5\sinh x - 2 = 0\). Factoring this quadratic yields \((3\sinh x + 1)(\sinh x - 2) = 0\), which gives the solutions \(\sinh x = 2\) or \(\sinh x = -\frac{1}{3}\). Using the logarithmic definition of the inverse hyperbolic sine, \(\sinh^{-1} y = \ln(y + \sqrt{y^2 + 1})\). For \(\sinh x = 2\), we find \(x = \ln(2 + \sqrt{5})\). For \(\sinh x = -\frac{1}{3}\), we find \(x = \ln\left(-\frac{1}{3} + \sqrt{\frac{1}{9} + 1}\right) = \ln\left(\frac{\sqrt{10}-1}{3}\right)\).

M1: Substitute \(\cosh^2 x = 1 + \sinh^2 x\) to form a quadratic in \(\sinh x\). A1: Correctly solve the quadratic equation to obtain \(\sinh x = 2\) and \(\sinh x = -1/3\). M1: Apply the logarithmic form of the inverse hyperbolic sine function to at least one value. A1: Correctly obtain \(x = \ln(2 + \sqrt{5})\). A1: Correctly obtain \(x = \ln\left(\frac{\sqrt{10}-1}{3}\right)\).

We use the substitution \(u = e^x\), which means \(dx = \frac{du}{u}\). When \(x = 0\), \(u = 1\), and when \(x = \ln 3\), \(u = 3\). Expressing \(\cosh x\) in terms of \(u\) gives \(\cosh x = \frac{u + u^{-1}}{2}\). The integrand's denominator is \(4\left(\frac{u + u^{-1}}{2}\right) + 5 = 2u + 2u^{-1} + 5 = \frac{2u^2 + 5u + 2}{u}\). Substituting these into the integral gives \(\int_1^3 \frac{1}{\frac{2u^2 + 5u + 2}{u}} \frac{du}{u} = \int_1^3 \frac{1}{2u^2 + 5u + 2} \, du\). Factoring the denominator, we get \(2u^2 + 5u + 2 = (2u + 1)(u + 2)\). Expressing this as partial fractions, we have \(\frac{1}{(2u+1)(u+2)} = \frac{2/3}{2u+1} - \frac{1/3}{u+2}\). Integrating this yields \(\left[ \frac{1}{3} \ln(2u+1) - \frac{1}{3} \ln(u+2) \right]_1^3 = \left[ \frac{1}{3} \ln\left(\frac{2u+1}{u+2}\right) \right]_1^3\). Evaluating at the upper limit \(u = 3\) gives \(\frac{1}{3} \ln\left(\frac{7}{5}\right)\), and at the lower limit \(u = 1\) gives \(\frac{1}{3} \ln(1) = 0\). Thus, the exact value is \(\frac{1}{3} \ln\left(\frac{7}{5}\right)\).

M1: Apply the substitution \(u = e^x\) and write \(\cosh x\) in terms of \(u\). A1: Obtain the correct simplified integral \(\int_1^3 \frac{1}{2u^2+5u+2} \, du\) with correct limits. M1: Decompose the integrand using partial fractions. A1: Find the correct antiderivative \(\frac{1}{3} \ln\left(\frac{2u+1}{u+2}\right)\). A1: Correctly substitute limits to obtain the final exact value \(\frac{1}{3} \ln\left(\frac{7}{5}\right)\).

Let \(z = \cos\theta + \text{i}\sin\theta\). Then \(z^n - z^{-n} = 2\text{i}\sin(n\theta)\). Consider \((z - z^{-1})^5 = (2\text{i}\sin\theta)^5 = 32\text{i}\sin^5\theta\). Expanding using the binomial theorem, we get \(32\text{i}\sin^5\theta = (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1})\). Substituting the sine expressions, we obtain \(32\text{i}\sin^5\theta = 2\text{i}\sin(5\theta) - 10\text{i}\sin(3\theta) + 20\text{i}\sin\theta\). Dividing by \(2\text{i}\) gives \(16\sin^5\theta = \sin(5\theta) - 5\sin(3\theta) + 10\sin\theta\), which simplifies to \(\sin(5\theta) - 5\sin(3\theta) = 16\sin^5\theta - 10\sin\theta\). Substituting this into the given equation, we obtain \(16\sin^5\theta - 10\sin\theta + 6\sin\theta = 0\), which reduces to \(16\sin^5\theta - 4\sin\theta = 0\). Factoring gives \(4\sin\theta(4\sin^4\theta - 1) = 0\). In the interval \(0 < \theta < \pi\), \(\sin\theta > 0\), so we solve \(4\sin^4\theta - 1 = 0\), which gives \(\sin^2\theta = \frac{1}{2}\). Since \(\sin\theta > 0\) in the given interval, we have \(\sin\theta = \frac{1}{\sqrt{2}}\). This gives the solutions \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\).

M1: Relate \(\sin(n\theta)\) to complex numbers and expand \((z-z^{-1})^5\). A1: Establish the correct identity relating multiple angles and powers: \(\sin(5\theta) - 5\sin(3\theta) = 16\sin^5\theta - 10\sin\theta\). M1: Substitute into the given equation to obtain a polynomial equation in terms of \(\sin\theta\). A1: Solve for the non-zero root to obtain \(\sin\theta = \frac{1}{\sqrt{2}}\). A1: State both correct solutions \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\) with no extra values in the range.

(a) We write \( I_n = \int_0^1 x^{n-1} \cdot x\sqrt{1-x^2} \, dx \). Let \( u = x^{n-1} \) and \( dv = x\sqrt{1-x^2} \, dx \). Then \( du = (n-1)x^{n-2} \, dx \) and \( v = -\frac{1}{3}(1-x^2)^{3/2} \). Using integration by parts: \( I_n = \left[ -\frac{1}{3}x^{n-1}(1-x^2)^{3/2} \right]_0^1 + \frac{n-1}{3} \int_0^1 x^{n-2}(1-x^2)^{3/2} \, dx \). Since \( n \ge 2 \), the boundary term is 0. Thus, \( I_n = \frac{n-1}{3} \int_0^1 x^{n-2}(1-x^2)\sqrt{1-x^2} \, dx = \frac{n-1}{3} \int_0^1 \left( x^{n-2}\sqrt{1-x^2} - x^n\sqrt{1-x^2} \right) dx = \frac{n-1}{3} (I_{n-2} - I_n) \). Multiplying both sides by 3 gives: \( 3I_n = (n-1)I_{n-2} - (n-1)I_n \implies (n+2)I_n = (n-1)I_{n-2} \implies I_n = \frac{n-1}{n+2} I_{n-2} \). (b) Using the reduction formula: \( I_5 = \frac{4}{7} I_3 \) and \( I_3 = \frac{2}{5} I_1 \). We evaluate \( I_1 = \int_0^1 x\sqrt{1-x^2} \, dx = \left[ -\frac{1}{3}(1-x^2)^{3/2} \right]_0^1 = \frac{1}{3} \). Therefore, \( I_5 = \frac{4}{7} \times \frac{2}{5} \times \frac{1}{3} = \frac{8}{105} \).

(a) M1: For applying integration by parts with correct choice of u and dv. A1: For obtaining correct du and v. M1: For showing that the boundary term vanishes for n >= 2. M1: For writing (1-x^2)^{3/2} as (1-x^2) * sqrt(1-x^2) and expressing the integral in terms of I_{n-2} and I_n. A1: For correct intermediate algebraic simplification. A1: For completing the proof of the reduction formula. [6.8 marks] (b) M1: For expressing I_5 in terms of I_1 using the reduction formula twice. M1: For evaluating I_1 by direct integration. A1: For obtaining I_1 = 1/3. A1: For correct final evaluation of I_5 as 8/105. [5 marks]

First, we write the right-hand side in exponential form: \( 9\sinh x = 9 \left( \frac{e^x - e^{-x}}{2} \right) = \frac{9}{2}e^x - \frac{9}{2}e^{-x} \). The auxiliary equation for the homogeneous equation is \( m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \), which gives a repeated root of \( m = -2 \). Thus, the complementary function (CF) is \( y_c = (A + Bx)e^{-2x} \). For the particular integral (PI), we try \( y_p = C e^x + D e^{-x} \). Differentiating gives: \( y_p' = C e^x - D e^{-x} \) and \( y_p'' = C e^x + D e^{-x} \). Substituting these into the original differential equation: \( (C e^x + D e^{-x}) + 4(C e^x - D e^{-x}) + 4(C e^x + D e^{-x}) = \frac{9}{2}e^x - \frac{9}{2}e^{-x} \). Grouping terms: \( 9C e^x + D e^{-x} = \frac{9}{2}e^x - \frac{9}{2}e^{-x} \). Equating coefficients: \( 9C = \frac{9}{2} \implies C = \frac{1}{2} \) and \( D = -\frac{9}{2} \). Thus, the particular integral is \( y_p = \frac{1}{2}e^x - \frac{9}{2}e^{-x} \). The general solution is \( y = y_c + y_p = (A + Bx)e^{-2x} + \frac{1}{2}e^x - \frac{9}{2}e^{-x} \).

M1: For writing the auxiliary equation and solving for m. A1: For finding m = -2 as a repeated root. A1: For writing the correct CF: (A + Bx)e^{-2x}. B1: For expressing 9 sinh x in exponential form. M1: For proposing a PI of the form C e^x + D e^{-x}. M1: For differentiating the PI and substituting into the differential equation. A1: For obtaining the correct equation in C and D: 9C e^x + D e^{-x} = 4.5 e^x - 4.5 e^{-x}. A1: For finding C = 1/2. A1: For finding D = -9/2. A1: For combining CF and PI to write the general solution. [11.8 marks]

(a) We have \( y = a \cosh\left(\frac{x}{a}\right) \), so the derivative is \( \frac{dy}{dx} = \sinh\left(\frac{x}{a}\right) \). The arc length is given by \( s = \int_0^{x_1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_0^{x_1} \sqrt{1 + \sinh^2\left(\frac{x}{a}\right)} dx \). Using the hyperbolic identity \( 1 + \sinh^2 u = \cosh^2 u \), we have \( \sqrt{1 + \sinh^2\left(\frac{x}{a}\right)} = \cosh\left(\frac{x}{a}\right) \). Thus, \( s = \int_0^{x_1} \cosh\left(\frac{x}{a}\right) dx = \left[ a \sinh\left(\frac{x}{a}\right) \right]_0^{x_1} = a \sinh\left(\frac{x_1}{a}\right) - a \sinh(0) = a \sinh\left(\frac{x_1}{a}\right) \). (b) The surface area \( S \) generated by rotating the arc about the \( x \)-axis is given by \( S = 2\pi \int_0^{x_1} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \). Substituting \( y \) and \( \sqrt{1 + (dy/dx)^2} \) gives: \( S = 2\pi \int_0^{x_1} a \cosh\left(\frac{x}{a}\right) \cosh\left(\frac{x}{a}\right) dx = 2\pi a \int_0^{x_1} \cosh^2\left(\frac{x}{a}\right) dx \). Using the identity \( \cosh^2 u = \frac{1 + \cosh(2u)}{2} \): \( S = 2\pi a \int_0^{x_1} \frac{1}{2} \left( 1 + \cosh\left(\frac{2x}{a}\right) \right) dx = \pi a \left[ x + \frac{a}{2} \sinh\left(\frac{2x}{a}\right) \right]_0^{x_1} = \pi a \left[ x_1 + \frac{a}{2} \sinh\left(\frac{2x_1}{a}\right) \right] \) as required.

(a) B1: For correctly finding dy/dx. M1: For substituting into the arc length formula. A1: For using the identity 1 + sinh^2(u) = cosh^2(u). A1: For correct integration. A1: For obtaining the final expression for s. [5 marks] (b) M1: For using the correct formula for surface area of revolution. A1: For setting up the integral 2\pi a \int \cosh^2(x/a) dx. M1: For using the double angle identity for cosh^2(u). A1: For performing the integration correctly. A1: For substituting the limits 0 and x_1 and obtaining the final simplified expression. [6.8 marks]

(a) To find the eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \): \( \det \begin{pmatrix} 2-\lambda & 2 & -1 \\ 0 & 3-\lambda & 0 \\ -1 & 2 & 2-\lambda \end{pmatrix} = 0 \). Expanding along the second row: \( (3-\lambda) \det \begin{pmatrix} 2-\lambda & -1 \\ -1 & 2-\lambda \end{pmatrix} = 0 \implies (3-\lambda) [ (2-\lambda)^2 - 1 ] = 0 \implies (3-\lambda)(2-\lambda-1)(2-\lambda+1) = 0 \implies (3-\lambda)(1-\lambda)(3-\lambda) = 0 \). Thus, the eigenvalues are \( \lambda = 1 \) and \( \lambda = 3 \) (repeated twice). (b) For \( \lambda = 1 \), we solve \( (A-I)\mathbf{v} = \mathbf{0} \): \( \begin{pmatrix} 1 & 2 & -1 \\ 0 & 2 & 0 \\ -1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0 \) and \( x - z = 0 \). An eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \). For \( \lambda = 3 \), we solve \( (A-3I)\mathbf{v} = \mathbf{0} \): \( \begin{pmatrix} -1 & 2 & -1 \\ 0 & 0 & 0 \\ -1 & 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies -x + 2y - z = 0 \). Since there are two free variables, we can choose two linearly independent vectors: Setting \( y = 1, x = 0 \implies z = 2 \), we get \( \mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \). Setting \( y = 0, x = 1 \implies z = -1 \), we get \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \). (c) Since we have three linearly independent eigenvectors, the matrix is diagonalisable. We can write: \( D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \) and \( P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 2 & -1 \end{pmatrix} \).

(a) M1: For setting up det(A - \lambda I) = 0. A1: For expanding determinant correctly. A1: For obtaining the characteristic cubic equation. A1: For finding the correct eigenvalues 1, 3, 3. [4 marks] (b) M1: For using (A - I)v = 0 to find eigenvector for \lambda = 1. A1: For obtaining a correct eigenvector v_1. M1: For using (A - 3I)v = 0 to find equations for eigenvectors for \lambda = 3. A1: For choosing two linearly independent eigenvectors for \lambda = 3. A1: For verifying independence and correctness. [5 marks] (c) B1: For writing down a correct diagonal matrix D. B1: For writing down a non-singular matrix P with columns matching the eigenvectors of D in the correct order. [2.8 marks]

(a) By de Moivre's theorem, \( \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \). Expanding using the Binomial Theorem: \( (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \). Equating real parts: \( \cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \). Substituting \( \sin^2\theta = 1 - \cos^2\theta \): \( \cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \). (b) To solve \( \cos(5\theta) = 0 \), we have \( 5\theta = \frac{\pi}{2} + k\pi \implies \theta = \frac{\pi}{10} + \frac{k\pi}{5} \) for \( k = 0, 1, 2, 3, 4 \). This gives five roots: \( \cos\left(\frac{\pi}{10}\right), \cos\left(\frac{3\pi}{10}\right), \cos\left(\frac{5\pi}{10}\right), \cos\left(\frac{7\pi}{10}\right), \cos\left(\frac{9\pi}{10}\right) \). Since \( \cos\left(\frac{5\pi}{10}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \), the remaining four values are roots of the quartic equation obtained by setting \( \cos(5\theta)/\cos\theta = 0 \): \( 16x^4 - 20x^2 + 5 = 0 \), where \( x = \cos\theta \). Let \( y = x^2 \), then \( 16y^2 - 20y + 5 = 0 \). Using the quadratic formula: \( y = \frac{20 \pm \sqrt{400 - 320}}{32} = \frac{20 \pm \sqrt{80}}{32} = \frac{5 \pm \sqrt{5}}{8} \). Thus, \( x^2 = \frac{5 \pm \sqrt{5}}{8} \implies x = \pm \sqrt{\frac{5 \pm \sqrt{5}}{8}} \). (c) The four roots correspond to \( \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10} \). Since \( \frac{\pi}{10} \) is in the first quadrant, \( \cos\left(\frac{\pi}{10}\right) > 0 \). Also, \( \cos\left(\frac{\pi}{10}\right) > \cos\left(\frac{3\pi}{10}\right) \) since the cosine function is decreasing on \( [0, \pi/2] \). Comparing the two positive roots, \( \sqrt{\frac{5+\sqrt{5}}{8}} > \sqrt{\frac{5-\sqrt{5}}{8}} \). Therefore, \( \cos\left(\frac{\pi}{10}\right) \) must be the larger positive root, which is \( \sqrt{\frac{5+\sqrt{5}}{8}} \).

(a) M1: For using de Moivre's theorem to express cos(5\theta) + i sin(5\theta). A1: For binomial expansion and equating real parts. M1: For substituting sin^2(\theta) = 1 - cos^2(\theta) into the expression. A1: For algebraic simplification. A1: For completing the proof correctly. [5 marks] (b) M1: For solving cos(5\theta) = 0 to obtain the angles. A1: For showing that cos(\pi/2) = 0 factorizes out, leaving the quartic. M1: For solving the quadratic in x^2. A1: For obtaining the four roots x = \pm \sqrt{(5 \pm \sqrt{5})/8}. [4 marks] (c) M1: For identifying that cos(\pi/10) must be a positive root. M1: For justifying why cos(\pi/10) is the larger of the two positive roots. A1: For concluding the proof. [2.8 marks]