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

(i) Working from the right-hand side:
\(\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{r^2 + 2r + 1 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}\).
This completes the proof.

(ii) Using the method of differences:
\(S_n = \sum_{r=1}^n \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right)\)
\(= \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\)
All intermediate terms cancel out, leaving:
\(S_n = 1 - \frac{1}{(n+1)^2}\).

(iii) The summation is:
\(\sum_{r=n+1}^{2n} \frac{2r+1}{r^2(r+1)^2} = S_{2n} - S_n\)
\(= \left(1 - \frac{1}{(2n+1)^2}\right) - \left(1 - \frac{1}{(n+1)^2}\right)\)
\(= \frac{1}{(n+1)^2} - \frac{1}{(2n+1)^2}\).

(iv) Taking the limit as \(n \to \infty\) of \(S_n\):
\(\sum_{r=1}^{\infty} \frac{2r+1}{r^2(r+1)^2} = \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)^2}\right) = 1\).

(i) M1: For attempting to combine the fractions on the RHS with a common denominator. A1: For obtaining the correct numerator and showing algebraic equivalence.
(ii) M1: For writing out the first few terms showing the cancellation of terms. A1: For correct intermediate form with cancellation shown. A1: For the final correct simplified expression \(1 - \frac{1}{(n+1)^2}\).
(iii) M1: For expressing the sum as \(S_{2n} - S_n\). M1: For substituting the expressions from part (ii). A1.7: For the correct simplified algebraic result \(\frac{1}{(n+1)^2} - \frac{1}{(2n+1)^2}\).
(iv) M1: For attempting to take the limit of \(S_n\) as \(n \to \infty\). A1: For the correct value 1.

(i) For \(n=1\):
LHS \(= (3(1)-1)(3(1)+2) = (2)(5) = 10\).
RHS \(= 1(3(1)^2 + 6(1) + 1) = 1(3 + 6 + 1) = 10\).
Since LHS = RHS, the statement is true for \(n=1\).

(ii) Assume the statement is true for \(n=k\), where \(k\) is a positive integer:
\(\sum_{r=1}^k (3r-1)(3r+2) = k(3k^2+6k+1)\).
Now we examine the case \(n=k+1\):
\(\sum_{r=1}^{k+1} (3r-1)(3r+2) = \left[ \sum_{r=1}^k (3r-1)(3r+2) \right] + [3(k+1)-1][3(k+1)+2]\)
\(= k(3k^2+6k+1) + (3k+2)(3k+5)\)
\(= 3k^3 + 6k^2 + k + (9k^2 + 21k + 10)\)
\(= 3k^3 + 15k^2 + 22k + 10\).
Now we factor this expression to see if it matches the form \((k+1)(3(k+1)^2 + 6(k+1) + 1)\):
\((k+1)(3(k^2+2k+1) + 6k+6 + 1) = (k+1)(3k^2+12k+10)\)
\(= 3k^3 + 12k^2 + 10k + 3k^2 + 12k + 10 = 3k^3 + 15k^2 + 22k + 10\).
Since the algebraic expressions are identical, the statement is true for \(n=k+1\) whenever it is true for \(n=k\).
Since the base case \(n=1\) has been shown to be true, by mathematical induction, the statement holds for all positive integers \(n\).

(i) B1.7: For evaluating both sides for \(n=1\) and explicitly stating they are equal to 10.
(ii) M1: For stating the induction hypothesis for \(n=k\). M1: For adding the \((k+1)\)-th term to the assumed sum. A1: For obtaining \(3k^3 + 15k^2 + 22k + 10\). M1: For attempting to expand or factor the target expression \((k+1)(3(k+1)^2 + 6(k+1) + 1)\). A1: For showing the expansion matches the sum. A1: For a complete and logically sound inductive argument concluding that the statement holds for all positive integers.

(i) From the coefficients of the given cubic equation:
\(\alpha + \beta + \gamma = 3\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = 5\)
\(\alpha\beta\gamma = 2\).

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

(iii) Let \(y = x^2\), so \(x = y^{1/2}\).
Substituting into the original equation:
\(y^{3/2} - 3y + 5y^{1/2} - 2 = 0\)
Rearranging terms with fractional powers onto one side:
\(y^{1/2}(y + 5) = 3y + 2\)
Squaring both sides:
\(y(y + 5)^2 = (3y + 2)^2\)
\(y(y^2 + 10y + 25) = 9y^2 + 12y + 4\)
\(y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\)
\(y^3 + y^2 + 13y - 4 = 0\).

(i) B1.7: For correctly stating all three relationships from Vieta's formulas.
(ii) M1: For using the correct algebraic identity for the sum of squares. A2: For calculating the correct value of \(-1\).
(iii) M1: For substituting \(y = x^2\) or attempting to group terms. M1: For squaring both sides to eliminate fractional powers. A1: For expanding the LHS and RHS terms correctly. A1: For collecting terms into standard cubic form. A2: For the final correct equation \(y^3 + y^2 + 13y - 4 = 0\).

(i) Expressing the equation of \(C\) by dividing the numerator by the denominator:
\(y = \frac{x(x - 2) - 3}{x - 2} = x - \frac{3}{x - 2}\).
As \(x \to 2\), \(y \to \pm \infty\), so the vertical asymptote is \(x = 2\).
As \(x \to \pm \infty\), \(\frac{3}{x - 2} \to 0\), so the oblique asymptote is \(y = x\).

(ii) For the \(y\)-intercept, let \(x = 0\):
\(y = \frac{0^2 - 2(0) - 3}{0 - 2} = \frac{-3}{-2} = 1.5\).
So the \(y\)-intercept is \((0, 1.5)\).
For the \(x\)-intercepts, let \(y = 0\):
\(x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0\).
So the \(x\)-intercepts are \((3, 0)\) and \((-1, 0)\).

(iii) The sketch must feature:
- A vertical asymptote at \(x = 2\).
- A diagonal oblique asymptote at \(y = x\).
- No stationary points, since \(\frac{dy}{dx} = 1 + \frac{3}{(x-2)^2} > 0\) for all \(x \neq 2\).
- Left branch in the upper-left area passing through \((-1, 0)\) and \((0, 1.5)\).
- Right branch in the lower-right area passing through \((3, 0)\).

(i) M1: For attempting algebraic division. A1.7: For finding \(y = x\). B1: For finding \(x = 2\).
(ii) B1: For finding the correct \(y\)-intercept \((0, 1.5)\). B2: For finding both \(x\)-intercepts \((3, 0)\) and \((-1, 0)\).
(iii) G1: For drawing the two asymptotes correctly with dashed lines. G2: For drawing two branches of the curve with correct asymptotic behavior. G1: For labeling the three intercepts correctly on the graph.

(i) The curve starts at the pole \((r=0\) when \(\theta=0)\) and increases to \(r=2a\) when \(\theta=\pi\). Since it's restricted to \(0 \le \theta \le \pi\), it represents the top half of a cardioid.

(ii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_0^{\pi} r^2 \mathrm{d}\theta = \frac{1}{2} a^2 \int_0^{\pi} (1 - \cos \theta)^2 \mathrm{d}\theta\)
\(= \frac{1}{2} a^2 \int_0^{\pi} (1 - 2\cos \theta + \cos^2 \theta) \mathrm{d}\theta\)
Using the identity \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\):
\(A = \frac{1}{2} a^2 \int_0^{\pi} \left(1.5 - 2\cos \theta + 0.5\cos 2\theta\right) \mathrm{d}\theta\)
\(= \frac{1}{2} a^2 \left[ 1.5\theta - 2\sin \theta + 0.25\sin 2\theta \right]_0^{\pi}\)
\(= \frac{1}{2} a^2 (1.5\pi - 0) = \frac{3}{4}\pi a^2\).

(iii) Tangent perpendicular to the initial line implies \(\frac{dx}{d\theta} = 0\).
\(x = r \cos \theta = a(1 - \cos \theta) \cos \theta = a(\cos \theta - \cos^2 \theta)\).
\(\frac{dx}{d\theta} = a(-\sin \theta + 2\cos \theta \sin \theta) = a\sin \theta(2\cos \theta - 1)\).
Setting this to 0 for \(0 < \theta < \pi\):
Since \(\sin \theta \neq 0\), we must have \(2\cos \theta - 1 = 0 \implies \cos \theta = \frac{1}{2}\), which gives \(\theta = \frac{\pi}{3}\).
At \(\theta = \frac{\pi}{3}\):
\(r = a\left(1 - \cos\frac{\pi}{3}\right) = \frac{1}{2}a\).
Converting to Cartesian coordinates:
\(x = r \cos \theta = \frac{1}{2}a \times \frac{1}{2} = \frac{1}{4}a\).
\(y = r \sin \theta = \frac{1}{2}a \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}a\).
Thus, the coordinates are \(\left(\frac{1}{4}a, \frac{\sqrt{3}}{4}a\right)\).

(i) G3: For a correct sketch of the upper half of a cardioid, starting at the pole at angle 0 and ending on the negative initial line at distance \(2a\).
(ii) M1: For using the polar area formula \(\frac{1}{2} \int r^2 \mathrm{d}\theta\). M1: For expanding the integrand and using the double-angle identity. A1: For obtaining correct integrated terms. A1.7: For evaluating and showing the result \(\frac{3}{4}\pi a^2\).
(iii) M1: For writing \(x = r\cos\theta\) and differentiating with respect to \(\theta\). A1: For finding \(\theta = \frac{\pi}{3}\). A1: For calculating the correct Cartesian coordinates \(\left(\frac{1}{4}a, \frac{\sqrt{3}}{4}a\right)\).

(i) The matrix is singular if and only if its determinant is zero.
\(\det(M) = 2(2a - 2) - 1(2 - 3) + a(2 - 3a)\)
\(= 4a - 4 + 1 + 2a - 3a^2\)
\(= -3a^2 + 6a - 3\)
\(= -3(a - 1)^2\).
Setting \(\det(M) = 0\):
\(-3(a - 1)^2 = 0 \implies a = 1\).

(ii) For \(a = 0\), the matrix is \(M = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 1 \\ 3 & 2 & 2 \end{pmatrix}\).
From part (i), \(\det(M) = -3(0 - 1)^2 = -3\).
Now find the cofactor matrix \(C\):
\(C_{11} = +(0 - 2) = -2\)
\(C_{12} = -(2 - 3) = 1\)
\(C_{13} = +(2 - 0) = 2\)
\(C_{21} = -(2 - 0) = -2\)
\(C_{22} = +(4 - 0) = 4\)
\(C_{23} = -(4 - 3) = -1\)
\(C_{31} = +(1 - 0) = 1\)
\(C_{32} = -(2 - 0) = -2\)
\(C_{33} = +(0 - 1) = -1\)

So \(C = \begin{pmatrix} -2 & 1 & 2 \\ -2 & 4 & -1 \\ 1 & -2 & -1 \end{pmatrix}\).

The adjugate matrix is \(C^T = \begin{pmatrix} -2 & -2 & 1 \\ 1 & 4 & -2 \\ 2 & -1 & -1 \end{pmatrix}\).

Thus, \(M^{-1} = \frac{1}{\det(M)} C^T = -\frac{1}{3} \begin{pmatrix} -2 & -2 & 1 \\ 1 & 4 & -2 \\ 2 & -1 & -1 \end{pmatrix} = \begin{pmatrix} 2/3 & 2/3 & -1/3 \\ -1/3 & -4/3 & 2/3 \\ -2/3 & 1/3 & 1/3 \end{pmatrix}\).

(i) M1: For attempting to find the determinant of \(M\) in terms of \(a\). A2: For obtaining the simplified determinant \(-3a^2 + 6a - 3\). A1: For setting to 0 and solving to find \(a = 1\).
(ii) M1: For calculating the determinant when \(a = 0\) (value \(-3\)). M2: For finding the cofactors (at least 5 correct cofactors for M1, all correct for M2). A1.7: For the correct adjugate matrix. A2: For dividing the adjugate matrix by the determinant to obtain the correct inverse matrix.

(i) Let the direction vectors of \(l_1\) and \(l_2\) be \(\mathbf{u} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\).
A normal vector \(\mathbf{n}\) to both lines is:
\(\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{pmatrix} (-1)(-2) - (3)(1) \\ (3)(1) - (2)(-2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).
The magnitude of \(\mathbf{n}\) is:
\(|\mathbf{n}| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59}\).
Let \(\mathbf{p}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\mathbf{p}_2 = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\) be points on \(l_1\) and \(l_2\) respectively.
\(\mathbf{p}_2 - \mathbf{p}_1 = \begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix}\).
The shortest distance \(d\) is the projection of this vector onto \(\mathbf{n}\):
\(d = \frac{|(\mathbf{p}_2 - \mathbf{p}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(-1) + (-3)(7) + 5(3)|}{\sqrt{59}}\)
\(= \frac{|-1 - 21 + 15|}{\sqrt{59}} = \frac{|-7|}{\sqrt{59}} = \frac{7}{\sqrt{59}}\).

(ii) Since the plane contains \(l_1\) and is parallel to \(l_2\), its normal vector is parallel to \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).
Let the normal vector be \(\begin{pmatrix} 1 \\ -7 \\ -3 \end{pmatrix}\).
The equation of the plane containing \(\mathbf{p}_1\) is:
\(1(x - 1) - 7(y - 2) - 3(z + 1) = 0\)
\(x - 1 - 7y + 14 - 3z - 3 = 0\)
\(x - 7y - 3z = -10\).

(i) M1: For calculating the cross product of the direction vectors. A1: For obtaining the correct normal vector \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\). M1: For calculating a vector between any point on \(l_1\) and any point on \(l_2\). M1: For applying the projection formula. A1.7: For the correct exact distance \(\frac{7}{\sqrt{59}}\).
(ii) M1: For using the normal vector found in part (i). M1: For substituting the coordinates of a point on \(l_1\) into the plane equation. A2: For obtaining the correct plane equation \(x - 7y - 3z = -10\) (or any scalar multiple thereof).

(a) We have \( I_n = \int_0^1 x^n e^{2x} \, dx \).
Using integration by parts, let \( u = x^n \) and \( dv = e^{2x} \, dx \).
Then \( du = n x^{n-1} \, dx \) and \( v = \frac{1}{2} e^{2x} \).
Applying the integration by parts formula:
\[ I_n = \left[ \frac{1}{2} x^n e^{2x} \right]_0^1 - \int_0^1 \frac{n}{2} x^{n-1} e^{2x} \, dx \]
\[ I_n = \left( \frac{1}{2} e^2 - 0 \right) - \frac{n}{2} I_{n-1} \]
\[ I_n = \frac{1}{2} e^2 - \frac{n}{2} I_{n-1} \]
Multiplying both sides by 2 gives:
\[ 2I_n + nI_{n-1} = e^2 \]

(b) For \( n = 0 \):
\[ I_0 = \int_0^1 e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_0^1 = \frac{1}{2}e^2 - \frac{1}{2} \]
Using the reduction formula:
For \( n = 1 \):
\[ 2I_1 + I_0 = e^2 \implies 2I_1 + \left(\frac{1}{2}e^2 - \frac{1}{2}\right) = e^2 \implies 2I_1 = \frac{1}{2}e^2 + \frac{1}{2} \implies I_1 = \frac{1}{4}e^2 + \frac{1}{4} \]
For \( n = 2 \):
\[ 2I_2 + 2I_1 = e^2 \implies 2I_2 + 2\left(\frac{1}{4}e^2 + \frac{1}{4}\right) = e^2 \implies 2I_2 + \frac{1}{2}e^2 + \frac{1}{2} = e^2 \implies 2I_2 = \frac{1}{2}e^2 - \frac{1}{2} \implies I_2 = \frac{1}{4}e^2 - \frac{1}{4} \]
For \( n = 3 \):
\[ 2I_3 + 3I_2 = e^2 \implies 2I_3 + 3\left(\frac{1}{4}e^2 - \frac{1}{4}\right) = e^2 \implies 2I_3 + \frac{3}{4}e^2 - \frac{3}{4} = e^2 \implies 2I_3 = \frac{1}{4}e^2 + \frac{3}{4} \implies I_3 = \frac{1}{8}e^2 + \frac{3}{8} = \frac{e^2 + 3}{8} \]

(a)
- M1: Applies integration by parts to the definition of \( I_n \).
- A1: Obtains correct boundary terms and integral term.
- A1: Deduces reduction formula \( 2I_n + nI_{n-1} = e^2 \) with no errors shown.

(b)
- B1: Obtains \( I_0 = \frac{1}{2}(e^2 - 1) \).
- M1: Uses reduction formula once to find \( I_1 \).
- A1: Obtains correct \( I_1 = \frac{1}{4}e^2 + \frac{1}{4} \).
- M1: Applies the reduction formula successively to obtain \( I_2 \) and \( I_3 \).
- A1: Obtains correct \( I_2 = \frac{1}{4}e^2 - \frac{1}{4} \).
- A1: Obtains correct exact value of \( I_3 = \frac{e^2 + 3}{8} \).

(a) Substituting the definitions of \( \cosh x \) and \( \sinh x \):
\[ 4\left(\frac{e^x + e^{-x}}{2}\right) - 3\left(\frac{e^x - e^{-x}}{2}\right) = 3 \]
Multiply both sides by 2:
\[ 4(e^x + e^{-x}) - 3(e^x - e^{-x}) = 6 \]
\[ e^x + 7e^{-x} = 6 \]
Multiply by \( e^x \):
\[ e^{2x} - 6e^x + 7 = 0 \]
Let \( u = e^x \), then:
\[ u^2 - 6u + 7 = 0 \]
Using the quadratic formula:
\[ u = \frac{6 \pm \sqrt{36 - 28}}{2} = 3 \pm \sqrt{2} \]
Since both values are positive, we have:
\[ x = \ln(3 \pm \sqrt{2}) \]

(b) Let \( I = \int_0^{\ln 2} \frac{1}{4\cosh x - 3\sinh x} \, dx \).
Using the simplification from (a), the denominator is \( \frac{e^x + 7e^{-x}}{2} \).
\[ I = \int_0^{\ln 2} \frac{2}{e^x + 7e^{-x}} \, dx \]
Using substitution \( t = e^x \), so \( dx = \frac{dt}{t} \).
When \( x = 0 \), \( t = 1 \).
When \( x = \ln 2 \), \( t = 2 \).
\[ I = \int_1^2 \frac{2}{t + 7/t} \frac{dt}{t} = \int_1^2 \frac{2}{t^2 + 7} \, dt \]
Using the standard form:
\[ I = \left[ \frac{2}{\sqrt{7}} \arctan\left(\frac{t}{\sqrt{7}}\right) \right]_1^2 \]
\[ I = \frac{2}{\sqrt{7}} \left( \arctan\left(\frac{2}{\sqrt{7}}\right) - \arctan\left(\frac{1}{\sqrt{7}}\right) \right) \]
Using the identity \( \arctan A - \arctan B = \arctan\left(\frac{A-B}{1+AB}\right) \):
\[ \frac{\frac{2}{\sqrt{7}} - \frac{1}{\sqrt{7}}}{1 + \frac{2}{7}} = \frac{\frac{1}{\sqrt{7}}}{\frac{9}{7}} = \frac{\sqrt{7}}{9} \]
Thus:
\[ I = \frac{2}{\sqrt{7}} \arctan\left(\frac{\sqrt{7}}{9}\right) = \frac{2\sqrt{7}}{7} \arctan\left(\frac{\sqrt{7}}{9}\right) \]

(a)
- M1: Uses exponential definitions of hyperbolic functions.
- A1: Obtains a correct quadratic equation in \( e^x \).
- M1: Solves the quadratic equation to find \( e^x \).
- A1: Obtains correct logarithmic forms for both solutions.

(b)
- M1: Integrates using the exponential form of the denominator.
- A1: Applies correct substitution \( t=e^x \) and finds correct limits.
- M1: Integrates to an arctan form.
- A1: Substitutes the limits correctly.
- A1: Uses the difference of arctangent identity to simplify to the final exact form.

(a) To find the eigenvalues, we solve the characteristic equation:
\[ \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \implies (3-\lambda)(-\lambda) - (-1)(2) = 0 \]
\[ \lambda^2 - 3\lambda + 2 = 0 \implies (\lambda-1)(\lambda-2) = 0 \]
So the eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).

For \( \lambda_1 = 1 \):
\[ \begin{pmatrix} 2 & -1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies y = 2x \]
An eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \).

For \( \lambda_2 = 2 \):
\[ \begin{pmatrix} 1 & -1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies y = x \]
An eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).

(b) Since \( \mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1} \), we have:
\[ \mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \]
The determinant of \( \mathbf{P} \) is \( 1(1) - 1(2) = -1 \).
\[ \mathbf{P}^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \]
Using \( \mathbf{A}^n = \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1} \):
\[ \mathbf{A}^n = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \]
\[ \mathbf{A}^n = \begin{pmatrix} 1 & 2^n \\ 2 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \]
\[ \mathbf{A}^n = \begin{pmatrix} -1 + 2^{n+1} & 1 - 2^n \\ -2 + 2^{n+1} & 2 - 2^n \end{pmatrix} \]

(a)
- M1: Sets up the characteristic equation and solves for eigenvalues.
- A1: Finds eigenvalues 1 and 2 correctly.
- M1: Substitutes eigenvalues back to obtain equations for eigenvectors.
- A1: Obtains correct eigenvector for eigenvalue 1.
- A1: Obtains correct eigenvector for eigenvalue 2.

(b)
- M1: Defines matrices \( \mathbf{P} \) and \( \mathbf{D} \) and finds \( \mathbf{P}^{-1} \).
- A1: Obtains correct inverse matrix \( \mathbf{P}^{-1} \).
- M1: Performs the matrix multiplication \( \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1} \).
- A1: Obtains the correct final expression for \( \mathbf{A}^n \).

(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 the real parts:
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \]
Substituting \( \sin^2\theta = 1 - \cos^2\theta \):
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 \]
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \]
\[ \cos(5\theta) = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta \]
\[ \cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \]

(b) The equation is \( 16x^5 - 20x^3 + 5x - \frac{1}{2} = 0 \).
Letting \( x = \cos\theta \) where \( 0 \le \theta \le \pi \):
\[ \cos(5\theta) = \frac{1}{2} \]
Solving in the interval \( 0 \le 5\theta \le 5\pi \):
\[ 5\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3} \]
\[ \theta = \frac{\pi}{15}, \frac{\pi}{3}, \frac{7\pi}{15}, \frac{11\pi}{15}, \frac{13\pi}{15} \]
Thus, the exact roots are:
\[ x = \cos\left(\frac{\pi}{15}\right), \frac{1}{2}, \cos\left(\frac{7\pi}{15}\right), \cos\left(\frac{11\pi}{15}\right), \cos\left(\frac{13\pi}{15}\right) \]

(a)
- M1: Uses de Moivre's theorem and expands binomially.
- A1: Extracts the correct real parts.
- M1: Substitutes \( \sin^2\theta = 1 - \cos^2\theta \) to obtain expression only in terms of cosines.
- A1: Show full working leading to the correct result.

(b)
- M1: Connects polynomial to the identity using \( x = \cos\theta \).
- A1: Obtains \( \cos(5\theta) = \frac{1}{2} \).
- M1: Solves to find at least three values of \( 5\theta \).
- A1: Finds all 5 values of \( \theta \) in the interval.
- A1: Lists the five exact solutions clearly.

(a) The auxiliary equation is:
\[ m^2 - 4m + 13 = 0 \implies m = \frac{4 \pm \sqrt{16-52}}{2} = 2 \pm 3i \]
So the complementary function is:
\[ y_{cf} = e^{2x}(A\cos 3x + B\sin 3x) \]

(b) Let the particular integral be \( y_{pi} = px^2 + qx + r \).
Then:
\[ y_{pi}' = 2px + q \]
\[ y_{pi}'' = 2p \]
Substituting into the differential equation:
\[ 2p - 4(2px + q) + 13(px^2 + qx + r) = 39x^2 + 15x - 19 \]
\[ 13px^2 + (13q - 8p)x + (2p - 4q + 13r) = 39x^2 + 15x - 19 \]
Equating coefficients:
- \( x^2 \): \( 13p = 39 \implies p = 3 \)
- \( x \): \( 13q - 8p = 15 \implies 13q - 24 = 15 \implies q = 3 \)
- Constant: \( 2p - 4q + 13r = -19 \implies 6 - 12 + 13r = -19 \implies r = -1 \)
So \( y_{pi} = 3x^2 + 3x - 1 \).

(c) The general solution is:
\[ y = e^{2x}(A\cos 3x + B\sin 3x) + 3x^2 + 3x - 1 \]
Using \( y(0) = 1 \):
\[ 1 = A - 1 \implies A = 2 \]
Differentiating the general solution:
\[ y' = 2e^{2x}(A\cos 3x + B\sin 3x) + e^{2x}(-3A\sin 3x + 3B\cos 3x) + 6x + 3 \]
Using \( y'(0) = 4 \):
\[ 4 = 2(A) + 3B + 3 \implies 4 = 4 + 3B + 3 \implies 3B = -3 \implies B = -1 \]
So the particular solution is:
\[ y = e^{2x}(2\cos 3x - \sin 3x) + 3x^2 + 3x - 1 \]

(a)
- M1: Forms the auxiliary quadratic equation and solves for complex roots.
- A1: Obtains the correct complementary function.

(b)
- M1: Proposes form \( px^2 + qx + r \) and computes its derivatives.
- M1: Equates coefficients to find \( p, q, r \).
- A1: Obtains correct values of \( p = 3, q = 3, r = -1 \).

(c)
- B1: Formulates general solution.
- M1: Uses condition \( y(0)=1 \) to find \( A \).
- M1: Differentiates general solution and uses condition \( y'(0)=4 \) to find \( B \).
- A1: Obtains the correct particular solution.

(a) Find the derivative of \( y \):
\[ \frac{dy}{dx} = x - \frac{1}{4x} \]
Then:
\[ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(x - \frac{1}{4x}\right)^2 = 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2} = x^2 + \frac{1}{2} + \frac{1}{16x^2} = \left(x + \frac{1}{4x}\right)^2 \]

(b) The arc length \( s \) is:
\[ s = \int_1^e \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_1^e \left(x + \frac{1}{4x}\right) \, dx \]
\[ s = \left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]_1^e = \left(\frac{1}{2}e^2 + \frac{1}{4}\right) - \left(\frac{1}{2} + 0\right) = \frac{1}{2}e^2 - \frac{1}{4} \]

(c) The surface area of revolution \( S \) about the \( y \)-axis is:
\[ S = 2\pi \int_1^e x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = 2\pi \int_1^e x \left(x + \frac{1}{4x}\right) \, dx \]
\[ S = 2\pi \int_1^e \left(x^2 + \frac{1}{4}\right) \, dx = 2\pi \left[ \frac{1}{3}x^3 + \frac{1}{4}x \right]_1^e \]
\[ S = 2\pi \left[ \left(\frac{1}{3}e^3 + \frac{1}{4}e\right) - \left(\frac{1}{3} + \frac{1}{4}\right) \right] = 2\pi \left(\frac{1}{3}e^3 + \frac{1}{4}e - \frac{7}{12}\right) = \pi \left(\frac{2}{3}e^3 + \frac{1}{2}e - \frac{7}{6}\right) \]

(a)
- M1: Differentiates \( y \) correctly.
- A1: Completes algebraic verification of the identity.

(b)
- M1: Writes down the arc length integral with the correct integrand and limits.
- A1: Integrates correctly.
- A1: Obtains exact value \( \frac{1}{2}e^2 - \frac{1}{4} \).

(c)
- M1: Uses correct formula for the surface area of revolution about the \( y \)-axis.
- A1: Obtains correct integrated expression.
- A1: Evaluates boundaries to obtain correct exact result.

(a) We find the modulus and argument of \( w \):
\[ r = |w| = \sqrt{(-8)^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = 16 \]
Since the real part is negative and imaginary part is positive, \( w \) lies in the second quadrant:
\[ \theta = \pi - \arctan\left(\frac{8\sqrt{3}}{8}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
So \( w = 16e^{i2\pi/3} \).

(b) To solve \( z^4 = 16e^{i(2\pi/3 + 2k\pi)} \):
\[ z = 2e^{i(\frac{\pi}{6} + \frac{k\pi}{2})} \]
Using \( k = 0, 1, -1, -2 \) to get principal arguments in \( (-\pi, \pi] \):
- For \( k = 0 \): \( z_1 = 2e^{i\pi/6} \)
- For \( k = 1 \): \( z_2 = 2e^{i2\pi/3} \)
- For \( k = -1 \): \( z_3 = 2e^{-i\pi/3} \)
- For \( k = -2 \): \( z_4 = 2e^{-i5\pi/6} \)

(c) The four roots lie on a circle centered at the origin of radius 2. Since their arguments differ by \( \frac{\pi}{2} \) (or 90 degrees), they form the vertices of a square centered at the origin on an Argand diagram.

(a)
- M1: Computes the modulus of \( w \).
- A1: Computes the argument of \( w \) in the second quadrant.
- A1: Expresses \( w \) in correct exponential form.

(b)
- M1: Applies de Moivre's theorem to find the fourth root.
- A1: Obtains the correct magnitude 2.
- A1: Correctly derives arguments for the four roots.
- A1: States all four roots with correct principal arguments.

(c)
- B1: Explains that the roots lie on a circle of radius 2.
- B1: Identifies the shape as a square centered at the origin.

(a) We have \( f(x) = e^{2x}\cos x \implies f(0) = 1 \).
Differentiating using the product rule:
\[ f'(x) = 2e^{2x}\cos x - e^{2x}\sin x = e^{2x}(2\cos x - \sin x) \implies f'(0) = 2 \]
\[ f''(x) = 2e^{2x}(2\cos x - \sin x) + e^{2x}(-2\sin x - \cos x) = e^{2x}(3\cos x - 4\sin x) \implies f''(0) = 3 \]
\[ f'''(x) = 2e^{2x}(3\cos x - 4\sin x) + e^{2x}(-3\sin x - 4\cos x) = e^{2x}(2\cos x - 11\sin x) \implies f'''(0) = 2 \]

(b) The Maclaurin's series formula is:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]
Substituting the derived values:
\[ f(x) = 1 + 2x + \frac{3}{2}x^2 + \frac{2}{6}x^3 + \dots = 1 + 2x + \frac{3}{2}x^2 + \frac{1}{3}x^3 + \dots \]

(c) The Maclaurin's expansions are:
\[ e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots \]
\[ \cos x = 1 - \frac{x^2}{2} + \dots \]
Multiplying the two expansions together and keeping terms up to \( x^3 \):
\[ f(x) \approx \left(1 + 2x + 2x^2 + \frac{4}{3}x^3\right)\left(1 - \frac{x^2}{2}\right) \]
\[ = 1 - \frac{x^2}{2} + 2x - x^3 + 2x^2 + \frac{4}{3}x^3 + \dots \]
\[ = 1 + 2x + \left(2 - \frac{1}{2}\right)x^2 + \left(\frac{4}{3} - 1\right)x^3 + \dots \]
\[ = 1 + 2x + \frac{3}{2}x^2 + \frac{1}{3}x^3 \]
This matches the result in part (b).

(a)
- M1: Differentiates using the product rule.
- A1: Obtains correct first derivative value \( f'(0) = 2 \).
- M1: Differentiates again to find \( f''(x) \).
- A1: Obtains correct second and third derivative values \( f''(0) = 3 \) and \( f'''(0) = 2 \).

(b)
- M1: Uses Maclaurin's series expansion formula with correct denominators.
- A1: Obtains correct series up to the term in \( x^3 \).

(c)
- B1: States correct Maclaurin's expansions for both \( e^{2x} \) and \( \cos x \).
- M1: Multiplies out the series neglecting terms of degree higher than 3.
- A1: Confirms match with part (b) cleanly.

(i) The horizontal and vertical displacement equations for a projectile are:
\(x = u t \cos\alpha \implies t = \frac{x}{u \cos\alpha}\)
\(y = u t \sin\alpha - \frac{1}{2}gt^2\)

Substituting \(t\) into the equation for \(y\):
\(y = x \tan\alpha - \frac{gx^2}{2u^2 \cos^2\alpha}\)

Since \(\frac{1}{\cos^2\alpha} = 1 + \tan^2\alpha\), we can write:
\(y = x \tan\alpha - \frac{gx^2}{2u^2}(1 + \tan^2\alpha)\)

Given that the projectile passes through \(P(d, h)\):
\(h = d \tan\alpha - \frac{gd^2}{2u^2}(1 + \tan^2\alpha)\)

Letting \(t = \tan\alpha\):
\(h = dt - \frac{gd^2}{2u^2}(1 + t^2)\)

Rearranging into a quadratic in \(t\):
\(\frac{gd^2}{2u^2}t^2 - dt + \left(h + \frac{gd^2}{2u^2}\right) = 0\) (as required).

(ii) Substituting \(d = 12\), \(h = 5\), and \(g = 10\) into the quadratic equation:
\(\frac{10 \times 144}{2u^2}t^2 - 12t + \left(5 + \frac{10 \times 144}{2u^2}\right) = 0\)
\(\frac{720}{u^2}t^2 - 12t + \left(5 + \frac{720}{u^2}\right) = 0\)

For a valid trajectory to exist, \(\alpha\) (and hence \(t = \tan\alpha\)) must be real. Thus, the discriminant of this quadratic equation in \(t\) must be greater than or equal to zero:
\(\Delta = B^2 - 4AC \ge 0\)
\(12^2 - 4 \times \frac{720}{u^2} \left(5 + \frac{720}{u^2}\right) \ge 0\)
\(144 - \frac{14400}{u^2} - \frac{2073600}{u^4} \ge 0\)

Dividing the entire inequality by \(144\):
\(1 - \frac{100}{u^2} - \frac{14400}{u^4} \ge 0\)

Let \(y = \frac{1}{u^2}\). Then:
\(14400y^2 + 100y - 1 \le 0\)

Solving for the roots of the quadratic equation \(14400y^2 + 100y - 1 = 0\):
\(y = \frac{-100 \pm \sqrt{10000 - 4(14400)(-1)}}{28800} = \frac{-100 \pm \sqrt{67600}}{28800} = \frac{-100 \pm 260}{28800}\)

Since \(y > 0\), we take the positive root:
\(y \le \frac{160}{28800} = \frac{1}{180}\)

Therefore:
\(\frac{1}{u^2} \le \frac{1}{180} \implies u^2 \ge 180 \implies u \ge \sqrt{180} = 6\sqrt{5} \approx 13.4\text{ m/s}\).

(i)
- M1: Attempts to eliminate time \(t\) from horizontal and vertical motion equations.
- A1: Obtains the trajectory equation \(y = x\tan\alpha - \frac{gx^2}{2u^2}(1+\tan^2\alpha)\).
- A1: Substitutes \(x=d\), \(y=h\) and rearranges correctly to show the given quadratic in \(t\).

(ii)
- M1: Uses the discriminant condition \(B^2 - 4AC \ge 0\) for real roots of the quadratic.
- A1: Substitutes values and simplifies to obtain the inequality in terms of \(u^2\) or \(y = \frac{1}{u^2}\).
- M1: Solves the resulting quadratic inequality to find the range of \(u^2\).
- A1: Deduces the minimum speed \(u \ge 6\sqrt{5}\) (or accept \(13.4\text{ m/s}\)).

(i) Let the maximum extension of the string be \(x\) meters. At the point of maximum extension, the particle is instantaneously at rest, so its kinetic energy is zero.
By the conservation of energy, the loss in gravitational potential energy as the particle falls from \(O\) to its lowest point is equal to the gain in elastic potential energy of the string:
\(mg(l + x) = \frac{\lambda x^2}{2l}\)

Given \(m = 1\text{ kg}\), \(g = 10\text{ m/s}^2\), \(l = 0.8\text{ m}\), and \(\lambda = 40\text{ N}\):
\(10(0.8 + x) = \frac{40 x^2}{2(0.8)}\)
\(8 + 10x = 25x^2\)
\(25x^2 - 10x - 8 = 0\)

Factoring the quadratic equation:
\((5x - 4)(5x + 2) = 0\)

Since \(x > 0\), we have \(x = 0.8\text{ m}\) (as required).

(ii) The maximum speed of the particle occurs when its acceleration is zero, which is when the forces on the particle are in equilibrium.
This occurs when the tension in the string equals the weight of the particle:
\(T = mg \implies \frac{\lambda y}{l} = mg\)

where \(y\) is the extension at maximum speed:
\(\frac{40y}{0.8} = 10 \implies 50y = 10 \implies y = 0.2\text{ m}\).

By the conservation of energy between the release point \(O\) and the point of maximum speed:
\(mg(l + y) = \frac{1}{2}mv_{\text{max}}^2 + \frac{\lambda y^2}{2l}\)
\(10(0.8 + 0.2) = \frac{1}{2}(1)v_{\text{max}}^2 + \frac{40(0.2)^2}{2(0.8)}\)
\(10 = 0.5v_{\text{max}}^2 + 1\)
\(0.5v_{\text{max}}^2 = 9 \implies v_{\text{max}}^2 = 18 \implies v_{\text{max}} = 3\sqrt{2} \approx 4.24\text{ m/s}\).

(i)
- M1: Sets up the energy equation equating loss of GPE to gain in EPE.
- A1: Obtains the correct quadratic equation in terms of extension \(x\).
- A1: Solves the quadratic equation to show \(x = 0.8\text{ m}\).

(ii)
- M1: Identifies that maximum speed occurs at the equilibrium position where \(T = mg\).
- A1: Calculates the correct extension at equilibrium, \(y = 0.2\text{ m}\).
- M1: Sets up the conservation of energy equation including kinetic energy.
- A1: Obtains \(v_{\text{max}} = 3\sqrt{2}\text{ m/s}\) (or accept \(4.24\text{ m/s}\)).

(i) Let \(v\) be the speed of the particle when the radius \(OP\) makes an angle \(\theta\) with the downward vertical. By conservation of mechanical energy:
\(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mga(1 - \cos\theta)\)
\(v^2 = u^2 - 2ag(1 - \cos\theta)\)

Using Newton's second law along the radial direction towards the center of the sphere:
\(R - mg\cos\theta = \frac{mv^2}{a}\)

Substituting the expression for \(v^2\) into this equation:
\(R = mg\cos\theta + \frac{m}{a}\left(u^2 - 2ag(1 - \cos\theta)\right)\)
\(R = mg\cos\theta + \frac{mu^2}{a} - 2mg + 2mg\cos\theta\)
\(R = \frac{m}{a}(u^2 - 2ag + 3ag\cos\theta)\) (as required).

(ii) The particle leaves the surface of the sphere when the normal reaction force \(R = 0\).
Substituting \(u^2 = \frac{7ag}{2}\) into the expression for \(R\):
\(0 = \frac{m}{a}\left(\frac{7ag}{2} - 2ag + 3ag\cos\theta\right)\)
\(0 = mg\left(\frac{3}{2} + 3\cos\theta\right)\)

Since \(mg \neq 0\):
\(\frac{3}{2} + 3\cos\theta = 0 \implies \cos\theta = -\frac{1}{2}\).

The height \(h\) of the particle above the lowest point is given by:
\(h = a(1 - \cos\theta)\)

Substituting \(\cos\theta = -\frac{1}{2}\):
\(h = a\left(1 - \left(-\frac{1}{2}\right)\right) = 1.5a\).

(i)
- M1: Applies conservation of energy to find an expression for \(v^2\).
- A1: Correctly expresses \(v^2\) in terms of \(u\), \(a\), \(g\) and \(\theta\).
- M1: Resolves forces radially to form the equation of motion containing \(R\), \(mg\cos\theta\) and the centripetal acceleration term.
- A1: Successfully combines equations to derive the given expression for \(R\).

(ii)
- M1: Sets \(R = 0\) to find the condition for leaving the sphere.
- A1: Obtains \(\cos\theta = -0.5\).
- A1: Correctly calculates height as \(1.5a\).

(i) Using Newton's second law, the equation of motion is:
\(m a = -0.4 v^2\)

Since the mass \(m = 0.2\text{ kg}\) and acceleration \(a = v \frac{dv}{dx}\):
\(0.2 v \frac{dv}{dx} = -0.4 v^2\)

Since \(v \neq 0\) during the motion:
\(\frac{dv}{dx} = -2v\)

Separating variables and integrating:
\(\int \frac{1}{v} dv = \int -2 dx\)
\(\ln v = -2x + C\)

When \(x = 0\), the initial velocity is \(v = 5\):
\(\ln 5 = 0 + C \implies C = \ln 5\)

Thus:
\(\ln v - \ln 5 = -2x \implies \ln\left(\frac{v}{5}\right) = -2x\)
\(v = 5 e^{-2x}\) (as required).

(ii) Since \(v = \frac{dx}{dt}\), we have:
\(\frac{dx}{dt} = 5 e^{-2x}\)

Separating variables to find \(t\) in terms of \(x\):
\(e^{2x} dx = 5 dt\)

Integrating both sides:
\(\int_0^1 e^{2x} dx = \int_0^T 5 dt\)
\(\left[ \frac{1}{2} e^{2x} \right]_0^1 = 5T\)
\(\frac{1}{2}(e^2 - 1) = 5T\)
\(T = \frac{e^2 - 1}{10} \approx 0.639\text{ seconds}\).

(i)
- M1: Formulates Newton's second law using \(a = v \frac{dv}{dx}\).
- A1: Obtains the correct first-order differential equation \(\frac{dv}{dx} = -2v\).
- M1: Integrates using the initial conditions \(v=5\) when \(x=0\).
- A1: Obtains the given relation \(v = 5 e^{-2x}\).

(ii)
- M1: Replaces \(v\) with \(\frac{dx}{dt}\) and separates variables.
- A1: Performs integration correctly with limits or constant of integration.
- A1: Reaches the correct value of \(t = \frac{e^2 - 1}{10}\) (or accept \(0.639\text{ s}\)).

(i) Let \(N_1\) and \(F_1\) be the normal and frictional forces at the ground \(A\). Let \(N_2\) and \(F_2\) be the normal and frictional forces at the wall \(B\).
At the point of slipping (limiting equilibrium):
\(F_1 = \mu_1 N_1\) (pointing towards the wall)
\(F_2 = \mu_2 N_2\) (pointing vertically upwards)

Resolving forces horizontally:
\(F_1 = N_2 \implies N_2 = \mu_1 N_1\)

Resolving forces vertically:
\(N_1 + F_2 = W \implies N_1 + \mu_2 N_2 = W\)

Substituting \(N_2 = \mu_1 N_1\) into the vertical equation:
\(N_1(1 + \mu_1 \mu_2) = W \implies N_1 = \frac{W}{1 + \mu_1 \mu_2}\)
\(N_2 = \frac{\mu_1 W}{1 + \mu_1 \mu_2}\)
\(F_2 = \frac{\mu_1 \mu_2 W}{1 + \mu_1 \mu_2}\)

Taking moments about \(A\) for the rod of length \(2a\):
\(W \cos\theta \times a = N_2 \sin\theta \times 2a + F_2 \cos\theta \times 2a\)
\(W \cos\theta = 2 N_2 \sin\theta + 2 F_2 \cos\theta\)

Substitute the expressions for \(N_2\) and \(F_2\):
\(W \cos\theta = 2 \left(\frac{\mu_1 W}{1 + \mu_1 \mu_2}\right) \sin\theta + 2 \left(\frac{\mu_1 \mu_2 W}{1 + \mu_1 \mu_2}\right) \cos\theta\)

Dividing through by \(W\):
\(\cos\theta = \frac{2\mu_1}{1 + \mu_1 \mu_2} \sin\theta + \frac{2\mu_1 \mu_2}{1 + \mu_1 \mu_2} \cos\theta\)
\(\cos\theta \left(1 - \frac{2\mu_1 \mu_2}{1 + \mu_1 \mu_2}\right) = \frac{2\mu_1}{1 + \mu_1 \mu_2} \sin\theta\)
\(\cos\theta \left(\frac{1 - \mu_1 \mu_2}{1 + \mu_1 \mu_2}\right) = \frac{2\mu_1}{1 + \mu_1 \mu_2} \sin\theta\)

Dividing both sides by \(\cos\theta\) and multiplying by \(1 + \mu_1 \mu_2\):
\(1 - \mu_1 \mu_2 = 2\mu_1 \tan\theta \implies \tan\theta = \frac{1 - \mu_1 \mu_2}{2\mu_1}\) (as required).

(ii) Given \(\mu_1 = 0.5\) and \(\mu_2 = 0.4\):
\(\tan\theta = \frac{1 - (0.5)(0.4)}{2(0.5)} = \frac{1 - 0.2}{1} = 0.8\)
\(\theta = \arctan(0.8) \approx 38.6598^\circ\)

To the nearest \(0.1^\circ\), \(\theta = 38.7^\circ\).

(i)
- M1: Resolves forces horizontally and vertically using friction laws \(F_1 = \mu_1 N_1\) and \(F_2 = \mu_2 N_2\).
- A1: Obtains \(N_1\) and \(N_2\) in terms of \(W\), \(\mu_1\), and \(\mu_2\).
- M1: Takes moments about point \(A\) (or any other point) and substitutes force components.
- A1: Rearranges equations systematically to arrive at the given expression for \(\tan\theta\).

(ii)
- M1: Evaluates \(\tan\theta\) using the given values of \(\mu_1\) and \(\mu_2\).
- A1: Obtains \(\theta = 38.7^\circ\).

(i) Let the direction of the initial velocity of \(A\) be positive.
Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) after the collision, respectively.
By conservation of linear momentum:
\(3m(u) + m(-2u) = 3m v_A + m v_B\)
\(3v_A + v_B = u\) --- (Equation 1)

By Newton's experimental law of impact:
\(v_B - v_A = e(u - (-2u)) = 3eu \implies v_B = v_A + 3eu\) --- (Equation 2)

Substitute Equation 2 into Equation 1:
\(3v_A + (v_A + 3eu) = u\)
\(4v_A = u(1 - 3e) \implies v_A = \frac{u(1 - 3e)}{4}\) (as required).

Using this to find \(v_B\):
\(v_B = \frac{u(1 - 3e)}{4} + 3eu = \frac{u(1 + 9e)}{4}\).

(ii) If \(e = \frac{2}{3}\):
\(v_A = \frac{u(1 - 2)}{4} = -\frac{u}{4}\)
\(v_B = \frac{u(1 + 6)}{4} = \frac{7u}{4}\)

The initial kinetic energy is:
\(K_i = \frac{1}{2}(3m)u^2 + \frac{1}{2}(m)(-2u)^2 = 1.5 m u^2 + 2 m u^2 = 3.5 m u^2 = \frac{7}{2} m u^2\)

The final kinetic energy is:
\(K_f = \frac{1}{2}(3m)\left(-\frac{u}{4}\right)^2 + \frac{1}{2}(m)\left(\frac{7u}{4}\right)^2\)
\(K_f = \frac{3}{32} m u^2 + \frac{49}{32} m u^2 = \frac{52}{32} m u^2 = \frac{13}{8} m u^2\)

The loss of kinetic energy is:
\(K_{\text{loss}} = K_i - K_f = \frac{7}{2} m u^2 - \frac{13}{8} m u^2 = \frac{15}{8} m u^2 = 1.875 m u^2\).

(i)
- M1: Sets up conservation of linear momentum equation.
- M1: Sets up restitution law equation.
- A1: Solves the simultaneous equations to find \(v_A = \frac{u(1-3e)}{4}\).
- A1: Finds the correct final velocity of \(B\) as \(v_B = \frac{u(1+9e)}{4}\).

(ii)
- M1: Calculates the initial kinetic energy.
- M1: Substitutes \(e = \frac{2}{3}\) to find numerical velocities and calculates final kinetic energy.
- A1: Correctly evaluates the loss of kinetic energy as \(\frac{15}{8} m u^2\) (or accept \(1.875 mu^2\)).

(i) Since the lengths of the triangle formed by \(AP\), \(BP\), and \(AB\) are \(3a\), \(4a\), and \(5a\) respectively, and \(3^2 + 4^2 = 5^2\), the triangle \(APB\) is a right-angled triangle at \(P\).
Let \(\alpha\) be the angle \(BAP\) and \(\beta\) be the angle \(ABP\).
\(\sin\alpha = \frac{4}{5}\), \(\cos\alpha = \frac{3}{5}\)
\(\sin\beta = \frac{3}{5}\), \(\cos\beta = \frac{4}{5}\)

The radius \(r\) of the horizontal circle is given by:
\(r = AP \sin\alpha = 3a \times \frac{4}{5} = 2.4a\)

Let \(T_1\) be the tension in the upper string \(AP\) and \(T_2\) be the tension in the lower string \(BP\).
Resolving vertically for equilibrium in the vertical direction:
\(T_1 \cos\alpha - T_2 \cos\beta - mg = 0\)
\(\frac{3}{5} T_1 - \frac{4}{5} T_2 = mg \implies 3 T_1 - 4 T_2 = 5mg\) --- (Equation 1)

Resolving horizontally (towards the center of the circular path):
\(T_1 \sin\alpha + T_2 \sin\beta = m \omega^2 r\)
\(\frac{4}{5} T_1 + \frac{3}{5} T_2 = m \omega^2 (2.4a)\)
\(4 T_1 + 3 T_2 = 12 m \omega^2 a\) --- (Equation 2)

From Equation 1, express \(T_1\) in terms of \(T_2\):
\(T_1 = \frac{5mg + 4T_2}{3}\)

Substitute this into Equation 2:
\(4\left(\frac{5mg + 4T_2}{3}\right) + 3 T_2 = 12 m \omega^2 a\)
\(20mg + 16 T_2 + 9 T_2 = 36 m \omega^2 a\)
\(25 T_2 = 36 m \omega^2 a - 20mg\)
\(T_2 = \frac{m}{25}(36 \omega^2 a - 20 g)\) (as required).

(ii) For the lower string to remain taut, we must have \(T_2 > 0\):
\(\frac{m}{25}(36 \omega^2 a - 20 g) > 0\)
\(36 \omega^2 a > 20 g\)
\(\omega^2 > \frac{20g}{36a} = \frac{5g}{9a}\)
\(\omega > \frac{1}{3}\sqrt{\frac{5g}{a}}\).

(i)
- M1: Recognizes the geometry of the 3-4-5 right-angled triangle to find \(\sin\alpha\), \(\cos\alpha\), \(\sin\beta\), and \(\cos\beta\).
- A1: Sets up the vertical equilibrium equation correctly.
- M1: Sets up the horizontal motion equation with centripetal force.
- A1: Solves the system of equations correctly to find the given expression for \(T_2\).

(ii)
- M1: Uses the condition \(T_2 > 0\) for the string to remain taut.
- A1: Solves the inequality to obtain the correct range for \(\omega\).
- A1: Expresses the minimum angular speed as \(\frac{1}{3}\sqrt{\frac{5g}{a}}\).

(i) Since \(f(x)\) is a probability density function, the total area under the curve is 1:
\[ \int_0^4 k x^2 (4 - x) \, dx = 1 \]
\[ k \int_0^4 (4x^2 - x^3) \, dx = 1 \]
\[ k \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_0^4 = 1 \]
\[ k \left( \frac{256}{3} - 64 \right) = 1 \]
\[ k \left( \frac{64}{3} \right) = 1 \implies k = \frac{3}{64}. \]

(ii) For \(0 \le x \le 4\), the cumulative distribution function is:
\[ \text{F}(x) = \int_0^x \frac{3}{64} (4t^2 - t^3) \, dt = \frac{3}{64} \left[ \frac{4t^3}{3} - \frac{t^4}{4} \right]_0^x = \frac{16x^3 - 3x^4}{256}. \]

(iii) First, we find the mean of \(X\):
\[ \text{E}(X) = \int_0^4 x f(x) \, dx = \int_0^4 \frac{3}{64} (4x^3 - x^4) \, dx = \frac{3}{64} \left[ x^4 - \frac{x^5}{5} \right]_0^4 \]
\[ \text{E}(X) = \frac{3}{64} \left( 256 - \frac{1024}{5} \right) = \frac{3}{64} \left( \frac{256}{5} \right) = 2.4. \]
We want to find \(\text{P}(X > 2.4) = 1 - \text{F}(2.4)\):
\[ \text{F}(2.4) = \frac{16(2.4)^3 - 3(2.4)^4}{256} = \frac{221.184 - 99.5328}{256} = 0.4752. \]
\[ \text{P}(X > 2.4) = 1 - 0.4752 = 0.5248 \approx 0.525 \text{ (3 s.f.)}. \]

M1: Set up the integral \(\int_0^4 k(4x^2 - x^3) dx = 1\) and attempt to integrate.
A1: Obtain \(k = \frac{3}{64}\) correctly.
M1: Integrate the PDF with limits 0 to \(x\) to find \(\text{F}(x)\).
A1: Correct expression for \(\text{F}(x)\) for \(0 \le x \le 4\).
M1: Find \[E(X)\) by integrating \(x f(x)\).
A1: Obtain \(\text{E}(X) = 2.4\).
M1: Calculate \(1 - \text{F}(2.4)\).
A1: Correct probability \(0.525\) (or \(0.5248\)).

Let \(d\) represent the difference in scores (After - Before).
\(H_0\): The median of the differences is zero (the new method has no effect).
\(H_1\): The median of the differences is greater than zero (the new method improves scores).
This is a one-tailed test.

Calculate the differences \(d_i = \text{After} - \text{Before}\) for each student:
- A: \(70 - 65 = +5\)
- B: \(71 - 72 = -1\)
- C: \(65 - 58 = +7\)
- D: \(88 - 80 = +8\)
- E: \(74 - 69 = +5\)
- F: \(73 - 74 = -1\)
- G: \(68 - 60 = +8\)
- H: \(82 - 85 = -3\)
- I: \(77 - 70 = +7\)
- J: \(73 - 66 = +7\)

Now find the absolute differences and their ranks:
Absolute differences: \(5, 1, 7, 8, 5, 1, 8, 3, 7, 7\).
Order and rank them, taking averages for ties:
- Value 1 (B, F): ranks 1 and 2. Average rank = 1.5
- Value 3 (H): rank 3
- Value 5 (A, E): ranks 4 and 5. Average rank = 4.5
- Value 7 (C, I, J): ranks 6, 7, and 8. Average rank = 7
- Value 8 (D, G): ranks 9 and 10. Average rank = 9.5

Sum the ranks for positive and negative differences:
Negative differences (B, F, H):
\(W_- = 1.5 \text{ (for B)} + 1.5 \text{ (for F)} + 3 \text{ (for H)} = 6\).
Positive differences (A, C, D, E, G, I, J):
\(W_+ = 4.5 + 7 + 9.5 + 4.5 + 9.5 + 7 + 7 = 49\).

The test statistic is \(T = \min(W_-, W_+) = 6\).

For a one-tailed test at the 5% level of significance with \(n = 10\), the critical value is 10.
Since \(T = 6 \le 10\), we reject \(H_0\).
There is sufficient evidence at the 5% level of significance to conclude that the new teaching method increases memory retention scores.

B1: Correct hypotheses stated in words or symbols (one-tailed test).
M1: Compute correct differences for all 10 students.
M1: Assign ranks to the absolute values of the differences, appropriately handling ties.
A1: Correct rank values for both positive and negative differences.
M1: Calculate the test statistic \(T = 6\).
B1: Identify the correct critical value of 10 from the tables.
M1: Compare \(T\) with the critical value and make a decision to reject \(H_0\).
A1: State final conclusion in context.

(i) For any valid probability generating function, we must have \(G_X(1) = 1\):
\[ G_X(1) = k \ln(1 - 0.8(1)) = k \ln(0.2) = 1 \]
Since \(0.2 = \frac{1}{5} = 5^{-1}\), we have:
\[ k \ln(5^{-1}) = -k \ln 5 = 1 \implies k = -\frac{1}{\ln 5}. \]

(ii) The mean of \(X\) is given by \(\text{E}(X) = G_X'(1)\). First, find the derivative:
\[ G_X'(t) = -\frac{1}{\ln 5} \cdot \frac{-0.8}{1 - 0.8t} = \frac{0.8}{\ln 5 (1 - 0.8t)}. \]
Evaluating at \(t = 1\):
\[ \text{E}(X) = G_X'(1) = \frac{0.8}{\ln 5 (1 - 0.8)} = \frac{0.8}{0.2 \ln 5} = \frac{4}{\ln 5} \approx 2.49. \]

(iii) To find the variance, we need \(G_X''(1)\):
\[ G_X''(t) = \frac{d}{dt} \left[ \frac{0.8}{\ln 5} (1 - 0.8t)^{-1} \right] = \frac{0.8}{\ln 5} (-1)(1 - 0.8t)^{-2} (-0.8) = \frac{0.64}{\ln 5 (1 - 0.8t)^2}. \]
Evaluating at \(t = 1\):
\[ G_X''(1) = \frac{0.64}{\ln 5 (1 - 0.8)^2} = \frac{0.64}{0.04 \ln 5} = \frac{16}{\ln 5}. \]
Now use the variance formula:
\[ \text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 \]
\[ \text{Var}(X) = \frac{16}{\ln 5} + \frac{4}{\ln 5} - \left( \frac{4}{\ln 5} \right)^2 = \frac{20}{\ln 5} - \frac{16}{(\ln 5)^2} \approx 12.427 - 6.177 = 6.25 \text{ (3 s.f.)}. \]

M1: Set \(G_X(1) = 1\) and write an equation for \(k\).
A1: Show clearly that \(k = -\frac{1}{\ln 5}\).
M1: Differentiate \(G_X(t)\) to find \(G_X'(t)\).
A1: Obtain the correct derivative \(G_X'(t)\).
A1: Calculate \(\text{E}(X) = \frac{4}{\ln 5}\) or \(2.49\).
M1: Differentiate \(G_X'(t)\) to find \(G_X''(t)\).
A1: Correct value \(G_X''(1) = \frac{16}{\ln 5}\).
M1: Use the correct formula for \(\text{Var}(X)\).
A1: Obtain the correct variance of \(\frac{20}{\ln 5} - \frac{16}{(\ln 5)^2}\) or \(6.25\).

Hypotheses:
\(H_0\): The phenotypes of the offspring follow the genetic ratio \(9:4:3\).
\(H_1\): The phenotypes of the offspring do not follow the genetic ratio \(9:4:3\).

First, calculate the expected frequencies. The total number of observed flowers is \(142 + 53 + 45 = 240\).
The parts of the ratio sum to \(9 + 4 + 3 = 16\).
- Expected Red: \(240 \times \frac{9}{16} = 135\)
- Expected Pink: \(240 \times \frac{4}{16} = 60\)
- Expected White: \(240 \times \frac{3}{16} = 45\)

Now calculate the test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
- For Red: \\frac{(142 - 135)^2}{135} = \frac{49}{135} \approx 0.3630
- For Pink: \\frac{(53 - 60)^2}{60} = \frac{49}{60} \approx 0.8167
- For White: \\frac{(45 - 45)^2}{45} = 0

Summing these values:
\[ \chi^2 = 0.3630 + 0.8167 + 0 = 1.1797 \approx 1.18. \]

The number of degrees of freedom is \(v = c - 1 = 3 - 1 = 2\).
At the 5% significance level, the critical value \(\chi^2_{0.05}(2) = 5.991\).

Since our calculated value \(1.18 < 5.991\), we fail to reject \(H_0\).
There is insufficient evidence at the 5% significance level to reject the genetic model. The biologist's sample supports the \(9:4:3\) genetic ratio.

B1: State correct hypotheses \(H_0\) and \(H_1\) in context.
M1: Calculate the expected frequencies based on the \(9:4:3\) ratio.
A1: Correct expected frequencies (135, 60, 45).
M1: Calculate individual terms \(\frac{(O - E)^2}{E}\) and sum them.
A1: Correct test statistic \(\chi^2 \approx 1.18\).
B1: State correct degrees of freedom \(v = 2\) and critical value \(5.991\).
M1: Compare \(\chi^2\) statistic with critical value.
A1: State correct conclusion in context (fail to reject \(H_0\), sample supports model).

Let \(d = \text{After} - \text{Before}\) be the differences:
- Car 1: \(13.1 - 12.4 = 0.7\)
- Car 2: \(14.0 - 14.1 = -0.1\)
- Car 3: \(12.6 - 11.8 = 0.8\)
- Car 4: \(13.9 - 13.5 = 0.4\)
- Car 5: \(15.4 - 15.0 = 0.4\)
- Car 6: \(13.5 - 12.9 = 0.6\)
- Car 7: \(13.1 - 13.2 = -0.1\)
- Car 8: \(14.9 - 14.3 = 0.6\)

The sample of differences is: \(0.7, -0.1, 0.8, 0.4, 0.4, 0.6, -0.1, 0.6\).
Calculate the sample mean of differences \(\bar{d}\):
\[ \bar{d} = \frac{0.7 - 0.1 + 0.8 + 0.4 + 0.4 + 0.6 - 0.1 + 0.6}{8} = \frac{3.3}{8} = 0.4125. \]

Calculate the unbiased sample variance \(s_d^2\):
\[ \sum d^2 = 0.7^2 + (-0.1)^2 + 0.8^2 + 0.4^2 + 0.4^2 + 0.6^2 + (-0.1)^2 + 0.6^2 = 2.19. \]
\[ s_d^2 = \frac{1}{7} \left( 2.19 - \frac{3.3^2}{8} \right) = \frac{1}{7} \left( 2.19 - 1.36125 \right) = 0.11839. \]
\[ s_d = \sqrt{0.11839} \approx 0.3441. \]

We use a t-distribution with \(v = n - 1 = 7\) degrees of freedom.
For a 95% confidence interval, the critical value \(t_{0.025, 7} = 2.365\).

The margin of error (ME) is:
\[ \text{ME} = t \times \frac{s_d}{\sqrt{n}} = 2.365 \times \frac{0.3441}{\sqrt{8}} = 2.365 \times 0.12165 \approx 0.2877. \]

Therefore, the 95% confidence interval is:
\[ 0.4125 \pm 0.2877 \implies (0.1248, 0.7002) \approx [0.125, 0.700] \text{ km/l}. \]

Since the entire confidence interval is strictly positive (greater than 0), it supports the company's claim that the treatment increases fuel efficiency.

M1: Compute correct paired differences for the 8 cars.
A1: Correct sample mean difference \(\bar{d} = 0.4125\).
M1: Use formula for unbiased sample standard deviation \(s_d\).
A1: Correct sample standard deviation \(s_d \approx 0.344\).
B1: Identify the correct critical value \(t = 2.365\) for \(v = 7\).
M1: Compute the confidence interval bounds using \(\bar{d} \pm t \frac{s_d}{\sqrt{n}}\).
A1: Correct interval limits \([0.125, 0.700]\).
A1: State that because the interval is completely above zero, it supports the claim.

(i) First, we find the cumulative distribution function of \(X\), \(\text{F}_X(x)\), for \(x \ge 0\):
\[ \text{F}_X(x) = \int_0^x \frac{1}{2} e^{-t/2} \, dt = \left[ -e^{-t/2} \right]_0^x = 1 - e^{-x/2}. \]
Since \(X \ge 0\), we have \(Y = e^{X/4} \ge e^0 = 1\).
For \(y \ge 1\), the cumulative distribution function of \(Y\) is:
\[ \text{F}_Y(y) = \text{P}(Y \le y) = \text{P}(e^{X/4} \le y). \]
Taking the natural logarithm on both sides:
\[ \text{P}\left(\frac{X}{4} \le \ln y\right) = \text{P}(X \le 4 \ln y). \]
Since \(y \ge 1\), \(\ln y \ge 0\), so \(4 \ln y \ge 0\). We can substitute this into \(\text{F}_X\):
\[ \text{F}_Y(y) = \text{F}_X(4 \ln y) = 1 - e^{-(4 \ln y)/2} = 1 - e^{-2 \ln y} = 1 - y^{-2}. \]

(ii) The probability density function of \(Y\), \(f_Y(y)\), is the derivative of \(\text{F}_Y(y)\) with respect to \(y\):
\[ f_Y(y) = \frac{d}{dy} \left( 1 - y^{-2} \right) = 2 y^{-3} \quad \text{for } y \ge 1. \]
Therefore, the probability density function is:
\[ f_Y(y) = \begin{cases} \frac{2}{y^3} & \text{for } y \ge 1, \\ 0 & \text{otherwise.} \end{cases} \]

M1: Find the cumulative distribution function of \(X\) by integration.
A1: Correct CDF \(\text{F}_X(x) = 1 - e^{-x/2}\) for \(x \ge 0\).
M1: Express \(\text{P}(Y \le y)\) in terms of \(X\) as \(\text{P}(X \le 4 \ln y)\).
M1: Substitute \(4 \ln y\) into \(\text{F}_X\) and simplify using properties of logs/exponentials.
A1: Correctly obtain \(\text{F}_Y(y) = 1 - y^{-2}\) for \(y \ge 1\).
M1: Differentiate \(\text{F}_Y(y)\) to find the PDF \(f_Y(y)\).
A1: Obtain \(2y^{-3}\).
A1: Specify the correct domain \(y \ge 1\) and state the full piecewise function.