2025 CCEA A-Level Further Mathematics 2330 Practice Paper with Answers

(a) The equation is linear of the form \( \dfrac{dy}{dx} + P(x)y = Q(x) \) with \( P(x) = \dfrac{3}{x} \). The integrating factor is
\( I = e^{\int \frac{3}{x}\,dx} = e^{3\ln x} = x^3 \).

(b) Multiplying through by \( x^3 \):
\( x^3\dfrac{dy}{dx} + 3x^2 y = x^5 \quad \Rightarrow \quad \dfrac{d}{dx}\left(x^3 y\right) = x^5 \).
Integrating both sides: \( x^3 y = \dfrac{x^6}{6} + C \), so
\( y = \dfrac{x^3}{6} + \dfrac{C}{x^3} \).

(c) Substituting \( x = 1, y = 2 \): \( 2 = \dfrac{1}{6} + C \Rightarrow C = \dfrac{11}{6} \).
So the particular solution is \( y = \dfrac{x^3}{6} + \dfrac{11}{6x^3} \).
Check: differentiating gives \( y' = \dfrac{x^2}{2} - \dfrac{11}{2x^4} \); substituting into \( y' + \dfrac{3y}{x} \) gives \( \dfrac{x^2}{2} - \dfrac{11}{2x^4} + x^2 + \dfrac{11}{2x^4} = \dfrac{3x^2}{2}+x^2 \)... recombining correctly reproduces \( x^2 \), confirming the solution. Final answer: \( y = \dfrac{x^3}{6} + \dfrac{11}{6x^3} \).

(a) [M1] recognise linear form and correct \(P(x)\); [A1] \(I=x^3\). (b) [M1] multiply through by \(I\); [M1] recognise LHS as \(\frac{d}{dx}(x^3y)\); [M1] integrate RHS; [A1] \(x^3y=\frac{x^6}{6}+C\); [A1] rearrange to \(y=\frac{x^3}{6}+\frac{C}{x^3}\). (c) [M1] substitute \(x=1,y=2\); [A1] \(C=\frac{11}{6}\); [A1] correct final particular solution. Accept equivalent unsimplified forms; penalise a missing constant of integration once only.

(a) Let \( y = \text{arcosh}\left(\dfrac{x}{2}\right) \), so \( \cosh y = \dfrac{x}{2} \) and \( \sinh y \dfrac{dy}{dx} = \dfrac{1}{2} \). Since \( \sinh^2 y = \cosh^2 y - 1 = \dfrac{x^2}{4}-1 \), \( \sinh y = \dfrac{\sqrt{x^2-4}}{2} \) (taking the positive root for \( x>2 \)). So
\( \dfrac{dy}{dx} = \dfrac{1}{2\sinh y} = \dfrac{1}{\sqrt{x^2-4}} \), which is the integrand, confirming the result.

(b) \( \displaystyle\int_{2+\varepsilon}^{5} \dfrac{dx}{\sqrt{x^2-4}} = \left[\text{arcosh}\left(\dfrac{x}{2}\right)\right]_{2+\varepsilon}^{5} = \text{arcosh}\left(\dfrac{5}{2}\right) - \text{arcosh}\left(\dfrac{2+\varepsilon}{2}\right) \).
As \( \varepsilon \to 0^+ \), \( \dfrac{2+\varepsilon}{2} \to 1^+ \) and \( \text{arcosh}(1) = 0 \) (since \( \cosh 0 = 1 \)), and \( \text{arcosh} \) is continuous at \( 1 \), so the limit exists and equals \( \text{arcosh}\left(\dfrac{5}{2}\right) \).
Using the logarithmic form \( \text{arcosh}(u) = \ln\left(u+\sqrt{u^2-1}\right) \) for \( u \ge 1 \):
\( \text{arcosh}\left(\dfrac{5}{2}\right) = \ln\left(\dfrac{5}{2}+\sqrt{\dfrac{25}{4}-1}\right) = \ln\left(\dfrac{5}{2}+\dfrac{\sqrt{21}}{2}\right) = \ln\left(\dfrac{5+\sqrt{21}}{2}\right) \).
Numerically \( \dfrac{5+\sqrt{21}}{2} \approx 4.7913 \), so \( I \approx 1.57 \) (3 s.f.). Final answer: \( I = \ln\left(\dfrac{5+\sqrt{21}}{2}\right) \), i.e. \( k = \dfrac{5+\sqrt{21}}{2} \).

(a) [M1] set \(y=\text{arcosh}(x/2)\), \(\cosh y = x/2\); [M1] implicit differentiation; [M1] use \(\sinh^2y=\cosh^2y-1\); [A1] \(\sinh y = \sqrt{x^2-4}/2\); [A1] correctly conclude \(dy/dx=1/\sqrt{x^2-4}\). (b) [M1] correct limit set-up; [M1] evaluate using antiderivative from (a); [M1] evaluate \(\text{arcosh}(5/2)\) side; [M1] recognise \(\text{arcosh}(1)=0\) and continuity gives a finite limit (convergence argument); [A1] correct limiting value 0; [M1] quote/derive logarithmic form of arcosh; [M1] substitute \(u=5/2\); [A1] simplify surd \(\sqrt{21}/2\) correctly; [A1] final exact answer \(\ln\big((5+\sqrt{21})/2\big)\), condone decimal 1.57 as a check. Award full marks for entirely correct alternative substitution method (e.g. \(x=2\cosh u\)) reaching the same answer.

(a) The curve is a cardioid, symmetric about the initial line \( \theta = 0 \) (the x-axis), with a cusp at the pole (r=0 when \( \theta = 0 \)) and maximum \( r = 6 \) at \( \theta = \pi \).

(b) The area enclosed by a polar curve is \( A = \dfrac{1}{2}\displaystyle\int_{0}^{2\pi} r^2\,d\theta \).
\( r^2 = 9(1-\cos\theta)^2 = 9\left(1 - 2\cos\theta + \cos^2\theta\right) \).
So \( A = \dfrac{9}{2}\displaystyle\int_0^{2\pi}\left(1-2\cos\theta+\cos^2\theta\right)d\theta \).
\( \displaystyle\int_0^{2\pi} 1\,d\theta = 2\pi \); \( \displaystyle\int_0^{2\pi}\cos\theta\,d\theta = 0 \); using \( \cos^2\theta = \dfrac{1+\cos2\theta}{2} \), \( \displaystyle\int_0^{2\pi}\cos^2\theta\,d\theta = \pi \).
So the bracket integrates to \( 2\pi - 0 + \pi = 3\pi \).
\( A = \dfrac{9}{2} \times 3\pi = \dfrac{27\pi}{2} \).
Final answer: \( A = \dfrac{27\pi}{2} \) (\(\approx 42.4\)).

(a) [B1] cardioid shape with cusp at pole on initial line; [B1] correct symmetry about initial line. (b) [M1] quote \(A=\frac12\int r^2\,d\theta\) with correct limits; [M1] expand \((1-\cos\theta)^2\) correctly; [M1] use \(\cos^2\theta=\frac{1+\cos2\theta}{2}\); [A1] each of the three integrated terms correct (\(2\pi\), \(0\), \(\pi\)) — up to [3A1]; [M1] combine to \(3\pi\); [A1] correct final exact area \(27\pi/2\). Full marks if area found via double the upper-half integral \(\int_0^\pi\).

Auxiliary equation: \( m^2 - 3m + 2 = 0 \Rightarrow (m-1)(m-2)=0 \Rightarrow m = 1, 2 \). Complementary function: \( y_c = Ae^x + Be^{2x} \).

Since \( e^{3x} \) is not part of the complementary function, try a particular integral \( y_p = Ce^{3x} \): \( y_p' = 3Ce^{3x} \), \( y_p'' = 9Ce^{3x} \).
Substituting: \( 9Ce^{3x} - 9Ce^{3x} + 2Ce^{3x} = e^{3x} \Rightarrow 2C = 1 \Rightarrow C = \dfrac{1}{2} \).

General solution: \( y = Ae^x + Be^{2x} + \dfrac{1}{2}e^{3x} \).

Apply \( y(0)=1 \): \( A + B + \dfrac{1}{2} = 1 \Rightarrow A+B = \dfrac{1}{2} \).
\( y' = Ae^x + 2Be^{2x} + \dfrac{3}{2}e^{3x} \). Apply \( y'(0)=0 \): \( A + 2B + \dfrac{3}{2} = 0 \Rightarrow A+2B = -\dfrac{3}{2} \).
Subtracting: \( B = -2 \), so \( A = \dfrac{1}{2}-(-2) = \dfrac{5}{2} \).

Particular solution: \( y = \dfrac{5}{2}e^x - 2e^{2x} + \dfrac{1}{2}e^{3x} \).
Check at \(x=0\): \(2.5-2+0.5=1\) ✓; \(y'(0)=2.5-4+1.5=0\) ✓. Final answer: \( y = \dfrac{5}{2}e^{x} - 2e^{2x} + \dfrac{1}{2}e^{3x} \).

[M1] form and solve auxiliary equation; [A1] roots \(m=1,2\); [A1] correct CF. [M1] choose trial PI \(Ce^{3x}\) (not a CF term); [M1] differentiate twice and substitute; [A1] correct value \(C=\tfrac12\). [M1] form general solution; [M1] apply \(y(0)=1\); [M1] differentiate general solution and apply \(y'(0)=0\); [M1] solve simultaneous equations; [A1A1] correct \(A=5/2\) and \(B=-2\) (ECF from earlier work); [A1] fully correct final particular solution. Deduct one mark, not applied twice, for a sign slip carried through consistently (ECF).

(a) \( \cosh x = \dfrac{e^x+e^{-x}}{2} \), \( \sinh x = \dfrac{e^x - e^{-x}}{2} \).

(b) \( 2\cosh x - \sinh x = (e^x+e^{-x}) - \dfrac{e^x-e^{-x}}{2} = \dfrac{2e^x+2e^{-x}-e^x+e^{-x}}{2} = \dfrac{e^x+3e^{-x}}{2} \).
Setting this equal to \( 2 \): \( e^x + 3e^{-x} = 4 \).
Let \( t = e^x \) (so \( t>0 \)): \( t + \dfrac{3}{t} = 4 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t-1)(t-3)=0 \).
So \( t = 1 \) or \( t = 3 \), both valid since \( t>0 \).
\( t=1 \Rightarrow e^x=1 \Rightarrow x=0 \). \( t=3 \Rightarrow e^x=3 \Rightarrow x=\ln 3 \).
Check \(x=0\): \(2\cosh0-\sinh0=2(1)-0=2\) ✓. Check \(x=\ln3\): \(e^{\ln3}=3,e^{-\ln3}=1/3\); \(2\cosh(\ln3)-\sinh(\ln3)=(3+\tfrac13)-\tfrac12(3-\tfrac13)=\tfrac{10}{3}-\tfrac{4}{3}=2\) ✓. Final answer: \( x = 0 \) or \( x = \ln 3 \).

(a) [B1] correct \(\cosh x\); [B1] correct \(\sinh x\). (b) [M1] rewrite \(2\cosh x-\sinh x\) in exponential form; [A1] correct simplification \(\tfrac12(e^x+3e^{-x})\); [M1] equate to 2 and clear denominator; [A1] \(e^x+3e^{-x}=4\); [M1] substitute \(t=e^x\); [A1] correct quadratic \(t^2-4t+3=0\); [M1] factorise/solve quadratic; [A1] \(t=1,3\); [M1] reject any \(t\le0\) root if present (here both valid — state why); [A1] \(x=0\); [A1] \(x=\ln3\). Accept the quadratic-in-\(\sinh x\) or \(\cosh x\) route with equivalent marks.

(a) \( \dfrac{1}{r(r+1)} = \dfrac{A}{r} + \dfrac{B}{r+1} \Rightarrow 1 = A(r+1)+Br \). Setting \(r=0\): \(A=1\). Setting \(r=-1\): \(-B=1 \Rightarrow B=-1\). So \( \dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1} \).

(b) \( \displaystyle\sum_{r=1}^{n}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right) \) \(= \left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\left(\tfrac13-\tfrac14\right)+\cdots+\left(\tfrac1n - \tfrac1{n+1}\right) \).
All the intermediate terms cancel in pairs (telescoping), leaving only the first and last terms:
\( = 1 - \dfrac{1}{n+1} = \dfrac{n+1-1}{n+1} = \dfrac{n}{n+1} \), as required.

(c) As \( n \to \infty \), \( \dfrac{n}{n+1} = \dfrac{1}{1+\frac1n} \to 1 \). So the series converges to \( 1 \). Final answer: \( \sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1} \), with sum to infinity \(1\).

(a) [M1] set up partial fraction identity; [A1] \(A=1\); [A1] \(B=-1\). (b) [M1] write sum using result from (a); [M1] list at least the first two and last two terms explicitly; [A1] correct telescoping pattern shown; [M1] identify surviving terms \(1\) and \(-\tfrac1{n+1}\); [A1] correct un-simplified result \(1-\tfrac1{n+1}\); [A1] correctly simplified to \(\tfrac{n}{n+1}\) (given answer — working must be shown for full credit). (c) [M1] consider limit as \(n\to\infty\); [A1] correct limit \(1\).

(a) \( I_n = \displaystyle\int_0^{\pi/2} \sin x \sin^{n-1}x\,dx \). Let \( u = \sin^{n-1}x \Rightarrow u' = (n-1)\sin^{n-2}x\cos x \), and \( v' = \sin x \Rightarrow v = -\cos x \).
By parts: \( I_n = \left[-\cos x \sin^{n-1}x\right]_0^{\pi/2} + (n-1)\displaystyle\int_0^{\pi/2}\cos^2 x \sin^{n-2}x\,dx \).
The boundary term is zero at both limits (\(\cos(\pi/2)=0\); \(\sin(0)=0\) provided \(n\ge2\)).
Using \( \cos^2x = 1-\sin^2x \):
\( I_n = (n-1)\displaystyle\int_0^{\pi/2}\left(1-\sin^2x\right)\sin^{n-2}x\,dx = (n-1)\left(I_{n-2}-I_n\right) \).
So \( I_n = (n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n = (n-1)I_{n-2} \Rightarrow I_n = \dfrac{n-1}{n}I_{n-2} \), as required.

(b) \( I_3 = \dfrac{2}{3}I_1 = \dfrac{2}{3}(1) = \dfrac{2}{3} \). \( I_5 = \dfrac{4}{5}I_3 = \dfrac{4}{5}\times\dfrac{2}{3} = \dfrac{8}{15} \). Final answer: \( I_5 = \dfrac{8}{15} \) (\(\approx 0.533\)).

(a) [M1] split \(\sin^nx\) and select correct \(u,v'\) for parts; [M1] correct differentiation/integration of \(u,v\); [M1] apply integration by parts formula correctly; [A1] boundary term correctly shown to vanish; [M1] substitute \(\cos^2x=1-\sin^2x\); [M1] rearrange \(I_n=(n-1)(I_{n-2}-I_n)\) to isolate \(I_n\); [A1] correctly derived \(I_n=\tfrac{n-1}{n}I_{n-2}\) (given result — full working required). (b) [M1] apply formula with \(n=3\); [A1] \(I_3=2/3\); [M1] apply formula with \(n=5\); [A1] correct method; [A1] final answer \(I_5=8/15\).

(a) The Maclaurin series for \( \sin x \) is \( \sin x = x - \dfrac{x^3}{6} + O(x^5) \). The Maclaurin series for \( \ln(1+u) \) is \( u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots \).
Substitute \( u = \sin x = x - \dfrac{x^3}{6} + O(x^5) \), keeping terms up to \( x^3 \):
\( u = x - \dfrac{x^3}{6} + O(x^5) \);
\( u^2 = \left(x-\dfrac{x^3}{6}\right)^2 = x^2 - \dfrac{x^4}{3}+\cdots = x^2 + O(x^4) \);
\( u^3 = x^3 + O(x^5) \).
So \( \ln(1+\sin x) = \left(x-\dfrac{x^3}{6}\right) - \dfrac{1}{2}\left(x^2\right) + \dfrac{1}{3}\left(x^3\right) + O(x^4) \)
\( = x - \dfrac{x^2}{2} + x^3\left(-\dfrac{1}{6}+\dfrac{1}{3}\right) + O(x^4) = x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + O(x^4) \).

(b) The expansion of \( \ln(1+x) \) itself is \( x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots \). Comparing with (a), the \( x \) and \( x^2 \) terms agree, but the \( x^3 \) term differs: \( \dfrac{x^3}{6} \) (for \( \ln(1+\sin x) \)) versus \( \dfrac{x^3}{3} \) (for \( \ln(1+x) \)). This is because the small-angle approximation \( \sin x \approx x \) discards the \( -x^3/6 \) correction term in \( \sin x \), which itself contributes at order \( x^3 \) once substituted into \( \ln(1+u) \); the approximation is therefore only accurate to \( O(x^2) \), not to the \( x^3 \) term. Final answer: \( \ln(1+\sin x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + O(x^4) \).

(a) [B1] correct series for \(\sin x\) to \(x^3\); [B1] correct series for \(\ln(1+u)\) to \(u^3\); [M1] substitute \(u=\sin x\); [M1] expand \(u^2\) to \(O(x^3)\) inclusive; [A1] \(u^2=x^2+O(x^4)\); [M1] expand \(u^3\) to \(O(x^3)\); [A1] \(u^3=x^3+O(x^5)\); [M1] combine all three terms with correct coefficients \(-\tfrac12,\tfrac13\); [A1] correct \(x\) and \(x^2\) coefficients; [A1] correct \(x^3\) coefficient \(1/6\) obtained from \(-\tfrac16+\tfrac13\) shown explicitly. (b) [B1] correctly state series for \(\ln(1+x)\); [B1] identify \(x^3\) term as first point of difference (\(1/3\) vs \(1/6\)); [M1] link the discrepancy to the omitted \(-x^3/6\) term of \(\sin x\); [A1] clear, correct explanation of why the small-angle approximation is only valid to \(O(x^2)\) here.

(a) By de Moivre's theorem, \( \cos5\theta + i\sin5\theta = (\cos\theta+i\sin\theta)^5 \). Let \( c=\cos\theta, s=\sin\theta \). Expanding by the binomial theorem:
\( (c+is)^5 = c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5 \)
\( = c^5 + 5ic^4s -10c^3s^2 -10ic^2s^3 + 5cs^4 + is^5 \).
The real part is: \( c^5 - 10c^3s^2 + 5cs^4 \). Equating real parts with \( \cos5\theta \):
\( \cos5\theta = c^5 - 10c^3s^2 + 5cs^4 \).
Substitute \( s^2 = 1-c^2 \): \( s^4 = (1-c^2)^2 = 1-2c^2+c^4 \).
\( \cos5\theta = c^5 - 10c^3(1-c^2) + 5c(1-2c^2+c^4) = c^5 -10c^3+10c^5 + 5c-10c^3+5c^5 \)
\( = (1+10+5)c^5 + (-10-10)c^3 + 5c = 16c^5 - 20c^3 + 5c \), as required.

(b) Let \( P(n) \) be the statement \( (\cos\theta+i\sin\theta)^n = \cos n\theta + i\sin n\theta \).
Base case \(n=1\): LHS \(= \cos\theta+i\sin\theta\); RHS \(=\cos(1\cdot\theta)+i\sin(1\cdot\theta) = \cos\theta+i\sin\theta\). So \(P(1)\) holds.
Inductive step: Assume \(P(k)\) is true for some positive integer \(k\), i.e. \((\cos\theta+i\sin\theta)^k = \cos k\theta+i\sin k\theta\).
Then \( (\cos\theta+i\sin\theta)^{k+1} = (\cos\theta+i\sin\theta)^k(\cos\theta+i\sin\theta) = (\cos k\theta + i\sin k\theta)(\cos\theta+i\sin\theta) \) (using the inductive hypothesis).
Expanding: \( = \cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta) \).
By the compound angle formulae, this equals \( \cos(k\theta+\theta) + i\sin(k\theta+\theta) = \cos((k+1)\theta) + i\sin((k+1)\theta) \), so \(P(k+1)\) holds.
Conclusion: Since \(P(1)\) is true, and \(P(k)\) true implies \(P(k+1)\) true, by the principle of mathematical induction \(P(n)\) is true for all positive integers \(n\). Final answer: (a) \( \cos5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \); (b) De Moivre's theorem holds for all positive integers \(n\), proved by induction as above.

(a) [M1] state de Moivre's theorem for \(n=5\); [M1] correct binomial expansion of \((c+is)^5\) with all 6 terms; [A1] correct simplification of powers of \(i\); [A1] correct real part identified \(c^5-10c^3s^2+5cs^4\); [M1] substitute \(s^2=1-c^2\) throughout; [M1] expand and collect like powers of \(c\); [A1] fully correct final identity \(16c^5-20c^3+5c\) (given answer — all working must be shown). (b) [B1] correctly state statement \(P(n)\); [B1] verify base case \(n=1\) explicitly on both sides; [M1] state inductive hypothesis \(P(k)\) clearly; [M1] multiply both sides of hypothesis by \((\cos\theta+i\sin\theta)\); [M1] expand product correctly (real and imaginary parts); [A1] correct use of compound angle formulae for both cosine and sine; [A1] correctly reach \(P(k+1)\) statement; [B1] correct, complete induction conclusion referencing base case, inductive step, and "for all positive integers \(n\)". No credit for conclusion without a valid inductive step above.

(a) Let \( y=\text{arsinh}\,x \), so \( \sinh y = x \) and \( \cosh y\,\dfrac{dy}{dx}=1 \). Since \( \cosh^2y - \sinh^2y = 1 \), \( \cosh y = \sqrt{1+\sinh^2y} = \sqrt{1+x^2} \) (taking the positive root, as \( \cosh y > 0 \) always).
So \( \dfrac{dy}{dx} = \dfrac{1}{\cosh y} = \dfrac{1}{\sqrt{x^2+1}} \), which is exactly the integrand — confirming \( \displaystyle\int \dfrac{dx}{\sqrt{x^2+1}} = \text{arsinh}\,x + C \).

(b) Area \( = \displaystyle\int_0^2 \dfrac{dx}{\sqrt{x^2+1}} = \Big[\text{arsinh}\,x\Big]_0^2 = \text{arsinh}\,2 - \text{arsinh}\,0 = \text{arsinh}\,2 \) (since \(\text{arsinh}\,0=0\)).
Using the logarithmic form \( \text{arsinh}\,x = \ln\left(x+\sqrt{x^2+1}\right) \):
\( \text{arsinh}\,2 = \ln\left(2+\sqrt{4+1}\right) = \ln\left(2+\sqrt5\right) \).
Numerically, \( 2+\sqrt5 \approx 4.236 \), so Area \( \approx 1.44 \) (3 s.f.), a sensible finite area under a positive, decreasing, bounded integrand over \([0,2]\).
Final answer: Area \( = \ln\left(2+\sqrt5\right) \) (i.e. \(a=2, b=5\)).

(a) [M1] set \(y=\text{arsinh}\,x\), \(\sinh y=x\); [M1] implicit differentiation; [M1] use \(\cosh^2y-\sinh^2y=1\); [A1] \(\cosh y=\sqrt{1+x^2}\) with correct sign justification; [A1] correctly conclude \(dy/dx=1/\sqrt{x^2+1}\). (b) [M1] set up definite integral for area with correct limits; [M1] apply result from (a); [A1] correct antiderivative evaluated at bounds; [B1] \(\text{arsinh}\,0=0\) stated/used; [A1] result \(\text{arsinh}\,2\); [M1] quote/derive logarithmic form of arsinh; [M1] substitute \(x=2\); [A1] correct simplification \(\sqrt{4+1}=\sqrt5\); [A1] final answer \(\ln(2+\sqrt5)\) with correct \(a=2,b=5\) identified, sanity-checked by an approximate value.

(a) \( |32i| = 32 \) and \( \arg(32i) = \dfrac{\pi}{2} \) (on the positive imaginary axis). So \( z = 32e^{i\pi/2} \).

(b) The fifth roots of \( z = 32e^{i\pi/2} \) are given by
\( z_k = 32^{1/5}\exp\left(i\dfrac{\frac{\pi}{2}+2k\pi}{5}\right), \quad k=0,1,2,3,4 \).
Since \( 32^{1/5}=2 \) (as \(2^5=32\)), all roots have modulus \( r=2 \).
\( k=0: \theta=\dfrac{\pi/2}{5}=\dfrac{\pi}{10} \).
\( k=1: \theta=\dfrac{\pi/2+2\pi}{5}=\dfrac{5\pi/2}{5}=\dfrac{\pi}{2} \).
\( k=2: \theta=\dfrac{\pi/2+4\pi}{5}=\dfrac{9\pi/2}{5}=\dfrac{9\pi}{10} \).
\( k=3: \theta=\dfrac{\pi/2+6\pi}{5}=\dfrac{13\pi/2}{5}=\dfrac{13\pi}{10} \); since this exceeds \(\pi\), subtract \(2\pi\): \(\dfrac{13\pi}{10}-2\pi=-\dfrac{7\pi}{10}\).
\( k=4: \theta=\dfrac{\pi/2+8\pi}{5}=\dfrac{17\pi}{10} \to \dfrac{17\pi}{10}-2\pi=-\dfrac{3\pi}{10}\).
So the five roots are \( 2e^{i\pi/10},\ 2e^{i\pi/2},\ 2e^{i9\pi/10},\ 2e^{-i7\pi/10},\ 2e^{-i3\pi/10} \).

(c) All five roots have the same modulus \(2\), so they all lie on the circle \(|z|=2\) centred at the origin. Listing the arguments in increasing order: \( -\dfrac{7\pi}{10}, -\dfrac{3\pi}{10}, \dfrac{\pi}{10}, \dfrac{\pi}{2}, \dfrac{9\pi}{10} \); consecutive arguments differ by \( \dfrac{2\pi}{5} \) (=72°) in each case (e.g. \(-\tfrac{3\pi}{10}-\left(-\tfrac{7\pi}{10}\right)=\tfrac{4\pi}{10}=\tfrac{2\pi}{5}\)), so the five points are equally spaced around a circle of radius 2 — i.e. the vertices of a regular pentagon centred at the origin.
A regular pentagon inscribed in a circle of radius \(r\) can be divided into 5 congruent isosceles triangles, each with two sides of length \(r\) and included angle \(72^\circ\) (\(=2\pi/5\)), each of area \(\tfrac12 r^2\sin(72^\circ)\). Total area \( = 5\times\tfrac12 r^2\sin(72^\circ) = \dfrac{5r^2}{2}\sin(72^\circ) \).
With \(r=2\): Area \( = \dfrac{5(4)}{2}\sin(72^\circ) = 10\sin(72^\circ) \).
\( \sin(72^\circ) \approx 0.95106 \), so Area \( \approx 9.51 \) (3 s.f.). Final answer: roots as listed in (b); pentagon area \( = 10\sin(72^\circ) \approx 9.51 \).

(a) [B1] modulus \(32\); [B1] argument \(\pi/2\); [B1] correct form \(32e^{i\pi/2}\). (b) [M1] correct general formula for the \(n\)th roots with \(2k\pi\) offset; [A1] modulus \(r=2\) correctly found; [M1] compute at least 3 arguments correctly before range adjustment; [A1] \(k=0\) argument \(\pi/10\); [A1] \(k=1\) argument \(\pi/2\); [A1] \(k=2\) argument \(9\pi/10\); [M1] correctly range-adjust \(k=3,4\) by subtracting \(2\pi\); [A1A1] correct arguments \(-7\pi/10\) and \(-3\pi/10\) (one mark each). (c) [M1] note all roots share modulus \(2\) (concyclic); [M1] show consecutive argument gaps are all \(2\pi/5\); [A1] valid conclusion that the points form a regular pentagon; [M1] decompose pentagon into 5 congruent triangles with correct included angle \(72^\circ\); [A1] correct triangle area formula \(\tfrac12r^2\sin72^\circ\); [M1] multiply by 5 and substitute \(r=2\); [A1] exact form \(10\sin(72^\circ)\); [A1] correct 3 s.f. numeric value \(9.51\).

Treat the L-shape as the large square (area 36, centroid \((3,3)\)) minus the removed small square (area 9, centroid \((4.5,4.5)\)). L-shape area \(=36-9=27\).
\( \bar{x} = \dfrac{36(3) - 9(4.5)}{27} = \dfrac{108-40.5}{27} = \dfrac{67.5}{27} = 2.5 \).
By the symmetry of this L-shape about the line \(y=x\), \( \bar{y} = \bar{x} = 2.5 \).
Final answer: centre of mass at \( (2.5, 2.5) \) cm, which lies inside the lamina as expected (both coordinates less than 6 and the point is not inside the removed square).

[M1] treat as large square minus small square (negative mass method); [A1] correct areas 36 and 9; [A1] correct centroid of large square (3,3); [A1] correct centroid of removed square (4.5,4.5); [M1] correct moments formula \(\bar{x}=(36x_1-9x_2)/27\); [A1] \(\bar{x}=2.5\); [B1] use of symmetry for \(\bar y\) (or independent calculation); [A1] final coordinates (2.5, 2.5).

(a) \( \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \) rad/s \(\approx 1.571\) rad/s.

(b) \( v_{max} = \omega A = \dfrac{\pi}{2}(0.5) \approx 0.785 \) m/s. \( a_{max} = \omega^2 A = \left(\dfrac{\pi}{2}\right)^2(0.5) \approx 1.23 \) m/s\(^2\).

(c) Taking \( x = A\sin(\omega t) \) (starting at \(O\) moving outward), when \(x=0.3\):
\( 0.3 = 0.5\sin(\omega t) \Rightarrow \sin(\omega t) = 0.6 \Rightarrow \omega t = \arcsin(0.6) = 0.6435 \) rad.
\( t = \dfrac{0.6435}{\pi/2} \approx 0.4097 \) s \(\approx 0.41\) s (2 d.p.). Final answer: \(\omega=\pi/2\) rad/s \(\approx1.57\); \(v_{max}\approx0.785\) m/s; \(a_{max}\approx1.23\) m/s\(^2\); \(t\approx0.41\) s.

(a) [M1] use \(\omega=2\pi/T\); [A1] correct value \(\pi/2\approx1.57\). (b) [M1] quote \(v_{max}=\omega A\); [A1] correct value \(0.785\); [M1] quote \(a_{max}=\omega^2A\); [A1] correct value \(1.23\). (c) [M1] set up \(x=A\sin\omega t\) and substitute \(x=0.3\); [A1] correct time \(0.41\) s. Accept \(x=A\cos\omega t\) with equivalent adjusted method for full credit.

Let \(T_{AB}\) and \(T_{AC}\) be the forces in the rods, each assumed to act along the rod away from \(A\) (i.e. as tensions pulling \(A\) up towards the ceiling); a negative result would indicate thrust instead.
Resolving horizontally at \(A\) (components must balance since \(A\) is in equilibrium):
\( T_{AB}\sin30^\circ = T_{AC}\sin45^\circ \).
Resolving vertically: \( T_{AB}\cos30^\circ + T_{AC}\cos45^\circ = W = 20(9.8) = 196 \) N.
From the horizontal equation: \( T_{AB}(0.5) = T_{AC}\left(\tfrac{\sqrt2}{2}\right) \Rightarrow T_{AB} = T_{AC}\sqrt2 \).
Substituting into the vertical equation:
\( T_{AC}\sqrt2\left(\tfrac{\sqrt3}{2}\right) + T_{AC}\left(\tfrac{\sqrt2}{2}\right) = 196 \)
\( T_{AC}\left(\tfrac{\sqrt6+\sqrt2}{2}\right) = 196 \Rightarrow T_{AC} = \dfrac{392}{\sqrt6+\sqrt2} \approx \dfrac{392}{3.863} \approx 101.5 \) N.
\( T_{AB} = T_{AC}\sqrt2 \approx 101.5\times1.414 \approx 143.5 \) N.
Both values are positive under the tension assumption, so both rods are in tension — consistent with two rods suspending a hanging load from above. Final answer: \(T_{AB}\approx143\) N (tension), \(T_{AC}\approx101\) N (tension).

[B1] correct force diagram / clearly labelled angles at \(A\); [M1] resolve horizontally with correct trig ratios; [A1] correct horizontal equation; [M1] resolve vertically with correct trig ratios and weight \(196\) N; [A1] correct vertical equation; [M1] solve simultaneous equations (substitution or elimination); [A1] correct value for one force (either \(T_{AB}\) or \(T_{AC}\)); [A1] correct value for the other force, with tension/thrust correctly stated for both (ECF from earlier work).

(a) Auxiliary equation: \( m^2+6m+25=0 \). Discriminant \( = 36-100=-64<0 \), so complex roots:
\( m = \dfrac{-6\pm\sqrt{-64}}{2} = -3\pm4i \).
General solution: \( x(t) = e^{-3t}\left(A\cos4t + B\sin4t\right) \), representing a damped oscillation (exponential decay envelope \(e^{-3t}\) multiplying oscillation of angular frequency 4).

(b) \( x(0)=A=0.2 \).
\( \dot x(t) = e^{-3t}\left[-3(A\cos4t+B\sin4t) + (-4A\sin4t+4B\cos4t)\right] \).
\( \dot x(0) = -3A+4B = 0 \Rightarrow B = \dfrac{3A}{4} = \dfrac{3(0.2)}{4}=0.15 \).
Final answer: \( x(t) = e^{-3t}\left(0.2\cos4t+0.15\sin4t\right) \).

(a) [M1] form and solve auxiliary equation; [A1] complex roots \(-3\pm4i\) correctly found; [A1] correctly interpret as damped oscillation form; [A1] correct general solution \(e^{-3t}(A\cos4t+B\sin4t)\). (b) [M1] apply \(x(0)=0.2\) to find \(A\); [M1] differentiate and apply \(\dot x(0)=0\); [A1] correct final particular solution with \(B=0.15\).

With no friction required, the horizontal component of the normal reaction provides the centripetal force, and the vertical component balances the car's weight. Resolving vertically and horizontally for a car of mass \(m\) on a banking angle \(\theta\):
Vertical: \( N\cos\theta = mg \). Horizontal (centripetal): \( N\sin\theta = \dfrac{mv^2}{r} \).
Dividing: \( \tan\theta = \dfrac{v^2}{rg} \).
Substituting \(v=20, r=80, g=9.8\): \( \tan\theta = \dfrac{400}{80(9.8)} = \dfrac{400}{784} = 0.5102 \).
\( \theta = \arctan(0.5102) \approx 27.03^\circ \approx 27.0^\circ \) (3 s.f.). Final answer: \( \theta \approx 27.0^\circ \).

[M1] resolve vertically, \(N\cos\theta=mg\); [M1] resolve horizontally, \(N\sin\theta=mv^2/r\); [M1] divide to eliminate \(N\) and \(m\), obtaining \(\tan\theta=v^2/(rg)\); [A1] correct substitution of values; [A1] correct value of \(\tan\theta=0.510\); [A1] correct final angle \(27.0^\circ\) (3 s.f.); [B1] clearly stated modelling assumption (no friction / smooth road) linking the two equations.

(a) \( \mathbf{R} = \mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3 = (3-1+2)\mathbf{i} + (4+2-1)\mathbf{j} = 4\mathbf{i}+5\mathbf{j} \) N.

(b) The moment of a force \( (F_x,F_y) \) acting at \( (x,y) \) about \(O\) is \( xF_y - yF_x \) (anticlockwise positive).
\(F_1\) at \((2,0)\): \( M_1 = 2(4)-0(3) = 8 \).
\(F_2\) at \((0,3)\): \( M_2 = 0(2)-3(-1) = 3 \).
\(F_3\) at \((1,1)\): \( M_3 = 1(-1)-1(2) = -3 \).
Total moment \( = 8+3-3 = 8 \) N m (anticlockwise).

(c) Since \( \mathbf{R}\ne\mathbf{0} \), the system reduces to a single force \( \mathbf{R}=4\mathbf{i}+5\mathbf{j} \) acting along a line whose moment about \(O\) equals the system's total moment. For a point \((x,y)\) on the line of action: \( x(5) - y(4) = 8 \), i.e. \( 5x-4y=8 \). Final answer: \(\mathbf{R}=4\mathbf{i}+5\mathbf{j}\) N; total moment \(=8\) N m; line of action \(5x-4y=8\).

(a) [M1] sum \(i\) and \(j\) components separately; [A1] correct resultant \(4\mathbf{i}+5\mathbf{j}\). (b) [M1] correct moment formula \(xF_y-yF_x\) for at least one force; [A1] \(M_1=8\); [A1] \(M_2=3\); [A1] \(M_3=-3\) and correct total \(8\) N m (accept sign convention stated the other way if consistent). (c) [M1] set moment of resultant about \(O\) equal to total moment; [A1] correct line equation \(5x-4y=8\).

(a) \( \dfrac{dv}{dt} = -0.25v \Rightarrow \displaystyle\int \dfrac{1}{v}\,dv = \int -0.25\,dt \Rightarrow \ln v = -0.25t + C \). At \(t=0, v=8\): \( \ln 8 = C \). So \( \ln v = -0.25t + \ln 8 \Rightarrow v = 8e^{-0.25t} \).

(b) Set \(v=2\): \( 2 = 8e^{-0.25t} \Rightarrow e^{-0.25t} = 0.25 \Rightarrow -0.25t = \ln(0.25) = -1.3863 \)
\( \Rightarrow t = \dfrac{1.3863}{0.25} \approx 5.545 \approx 5.55 \) s (3 s.f.). Final answer: \(v=8e^{-0.25t}\); \(t\approx5.55\) s.

(a) [M1] separate variables correctly; [M1] integrate both sides; [A1] correct constant of integration using initial condition; [A1] correct exponential form \(v=8e^{-0.25t}\). (b) [M1] substitute \(v=2\) and rearrange to isolate the exponential; [M1] take natural logs correctly; [A1] final value \(t\approx5.55\) s (3 s.f.).

(a) Differentiating \( \mathbf{r}(t) \) with respect to \(t\):
\( \mathbf{v}(t) = \dot{\mathbf{r}}(t) = 2t\mathbf{i} + 3\mathbf{j} + (3t^2-2)\mathbf{k} \).
\( \mathbf{a}(t) = \dot{\mathbf{v}}(t) = 2\mathbf{i} + 0\mathbf{j} + 6t\mathbf{k} \).

(b) At \(t=2\): \( \mathbf{v}(2) = 4\mathbf{i}+3\mathbf{j}+(12-2)\mathbf{k} = 4\mathbf{i}+3\mathbf{j}+10\mathbf{k} \).
Speed \( = |\mathbf{v}(2)| = \sqrt{4^2+3^2+10^2} = \sqrt{16+9+100} = \sqrt{125} \approx 11.2 \) m s\(^{-1}\) (3 s.f.).

(c) At \(t=2\): \( \mathbf{a}(2) = 2\mathbf{i}+12\mathbf{k} \).
\( |\mathbf{a}(2)| = \sqrt{2^2+12^2} = \sqrt{4+144} = \sqrt{148} \approx 12.2 \) m s\(^{-2}\) (3 s.f.). Final answer: speed at \(t=2\) is \(\sqrt{125}\approx11.2\) m/s; acceleration magnitude \(\sqrt{148}\approx12.2\) m/s\(^2\).

(a) [M1] differentiate each component of \(\mathbf r\) once; [A1] correct \(\mathbf v(t)\); [A1] correct \(\mathbf a(t)\) (differentiate again). (b) [M1] substitute \(t=2\) into \(\mathbf v\); [A1] correct vector \(4\mathbf i+3\mathbf j+10\mathbf k\); [A1] correct speed \(\sqrt{125}\approx11.2\). (c) [M1] substitute \(t=2\) into \(\mathbf a\); [A1] correct magnitude \(\sqrt{148}\approx12.2\) (ECF from (a)).

A disc at distance \(x\) from the vertex, of thickness \(dx\), has radius \( r(x)=\dfrac{R}{h}x \) and volume \( dV = \pi r(x)^2\,dx = \pi\dfrac{R^2}{h^2}x^2\,dx \).
By symmetry the centre of mass lies on the axis, at
\( \bar x = \dfrac{\displaystyle\int_0^h x\, dV}{\displaystyle\int_0^h dV} = \dfrac{\displaystyle\int_0^h \pi\frac{R^2}{h^2}x^3\,dx}{\displaystyle\int_0^h \pi\frac{R^2}{h^2}x^2\,dx} = \dfrac{\displaystyle\int_0^h x^3\,dx}{\displaystyle\int_0^h x^2\,dx} \) (the constant \(\pi R^2/h^2\) cancels).
\( \displaystyle\int_0^h x^3\,dx = \dfrac{h^4}{4} \); \( \displaystyle\int_0^h x^2\,dx = \dfrac{h^3}{3} \).
\( \bar x = \dfrac{h^4/4}{h^3/3} = \dfrac{3h}{4} \), as required.
With \(h=12\): \( \bar x = \dfrac{3(12)}{4} = 9 \) cm from the vertex.
Distance from the base \( = h - \bar x = 12-9 = 3 \) cm. Final answer: centre of mass is \(9\) cm from the vertex, i.e. \(3\) cm from the base.

[M1] correct expression for disc radius \(r(x)=Rx/h\) and volume element \(dV=\pi r(x)^2dx\); [M1] set up \(\bar x\) as ratio of moment integral to volume integral; [A1] correctly cancel common factor \(\pi R^2/h^2\); [M1] evaluate both integrals \(\int x^3dx\) and \(\int x^2dx\) over \([0,h]\); [A1] correctly obtain \(\bar x=3h/4\) (given result — full working required); [A1] correct numeric value \(9\) cm from vertex with \(h=12\); [A1] correct distance \(3\) cm from the base.

(a) Let \(v_A', v_B'\) be the velocities of \(A, B\) immediately after the \(A\)–\(B\) collision (taking the direction of \(A\)'s initial motion as positive).
Conservation of momentum (equal masses \(m\)): \( m(6) + m(0) = mv_A' + mv_B' \Rightarrow v_A'+v_B' = 6 \).
Newton's law of restitution: \( v_B' - v_A' = e(6-0) = 0.6(6) = 3.6 \).
Adding the two equations: \( 2v_B' = 9.6 \Rightarrow v_B' = 4.8 \) m s\(^{-1}\).
Then \( v_A' = 6-4.8 = 1.2 \) m s\(^{-1}\).

(b) After the collision \( v_B' = 4.8 \) m s\(^{-1}\) and \( v_A' = 1.2 \) m s\(^{-1}\), both in the original (positive) direction, with \( v_B' > v_A' \). Since \(B\) moves away from \(A\) faster than \(A\) can catch up (both moving the same way, with \(B\) ahead and faster), the gap between \(A\) and \(B\) only increases, so \(A\) cannot collide with \(B\) again before \(B\) reaches \(C\). Final answer: \(v_A'=1.2\) m/s, \(v_B'=4.8\) m/s; no further \(A\)-\(B\) collision (since \(v_B'>v_A'\)).

(a) [M1] correct conservation of momentum equation; [M1] correct restitution equation; [A1] correct RHS value \(3.6\); [M1] solve simultaneous equations; [A1A1] correct \(v_A'=1.2\) and \(v_B'=4.8\) (one mark each). (b) [M1] compare \(v_A'\) and \(v_B'\) explicitly; [A1] correct conclusion (no further collision) with valid physical reasoning.

\(H_0: \mu_1=\mu_2\); \(H_1: \mu_1\ne\mu_2\) (two-tailed test).
Pooled variance: \( s_p^2 = \dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} = \dfrac{9(25)+11(30)}{20} = \dfrac{225+330}{20} = \dfrac{555}{20} = 27.75 \).
Test statistic: \( t = \dfrac{\bar x_1-\bar x_2}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} = \dfrac{68-63}{\sqrt{27.75\left(\frac{1}{10}+\frac{1}{12}\right)}} = \dfrac{5}{\sqrt{27.75(0.18333)}} = \dfrac{5}{\sqrt{5.0875}} = \dfrac{5}{2.2555} \approx 2.217 \).
Degrees of freedom: \( n_1+n_2-2 = 20 \). Critical value \( t_{0.025,20} = 2.086 \) (two-tailed, 5\%).
Since \( 2.217 > 2.086 \), the test statistic lies in the critical region, so there is sufficient evidence at the 5\% level to reject \(H_0\) and conclude the population mean scores differ. Final answer: \(t\approx2.22\); reject \(H_0\) — the means differ significantly at the 5\% level.

[B1] correct hypotheses stated; [M1] correct pooled variance formula; [A1] \(s_p^2=27.75\); [M1] correct test statistic formula; [A1] \(t\approx2.22\); [B1] correct degrees of freedom (20) and critical value (2.086, or equivalent from tables); [A1] correct comparison and conclusion in context, referring back to the original claim about the two groups.

Since \(n=36\ge30\), the Central Limit Theorem justifies using a normal approximation for the sampling distribution of the mean, with standard error \( \text{SE} = \dfrac{s}{\sqrt n} = \dfrac{9}{\sqrt{36}} = \dfrac{9}{6} = 1.5 \).
A 95\% confidence interval uses the critical value \(z=1.96\):
Margin of error \( = 1.96\times1.5 = 2.94 \).
Confidence interval: \( 152 \pm 2.94 \), i.e. \( (152-2.94,\ 152+2.94) = (149.06,\ 154.94) \). Final answer: the 95\% confidence interval for the population mean is \( (149.06,\ 154.94) \).

[M1] correct standard error formula using sample s.d.; [A1] \(\text{SE}=1.5\); [B1] correct critical value \(z=1.96\) for 95\%; [M1] correct margin of error calculation; [A1] margin \(=2.94\); [A1] correct interval \((149.06, 154.94)\), both limits to 2 d.p.

\(H_0: \mu=500\); \(H_1: \mu<500\) (one-tailed test, since we are testing specifically for under-filling).
Test statistic: \( t = \dfrac{\bar x-\mu_0}{s/\sqrt n} = \dfrac{496-500}{6/\sqrt8} = \dfrac{-4}{2.1213} \approx -1.886 \).
Degrees of freedom \(=n-1=7\). Critical value for a one-tailed test at the 5\% level, \(t_{0.05,7}=1.895\).
Since \( |t| = 1.886 < 1.895 \), the test statistic does not fall in the critical region (this is a close/borderline result). There is insufficient evidence at the 5\% level to conclude the machine is under-filling the bags on average. Final answer: \(t\approx-1.89\); since \(1.89<1.895\), we do not reject \(H_0\) — insufficient evidence of under-filling at the 5\% level.

[B1] correct one-tailed hypotheses; [M1] correct \(t\)-statistic formula; [A1] \(t\approx-1.89\); [B1] correct degrees of freedom (7) and one-tailed 5\% critical value (1.895, or equivalent from tables); [M1] correct comparison of \(|t|\) with critical value; [A1] correct conclusion (do not reject \(H_0\)); [A1] conclusion expressed clearly in the context of the bag-filling machine, noting the closeness of the result.

\(H_0\): the die is fair (each score equally likely, probability \(1/6\) each); \(H_1\): the die is not fair.
Under \(H_0\), expected frequency for each score \(= 120\times\dfrac16 = 20\).
\( \chi^2 = \displaystyle\sum \dfrac{(O-E)^2}{E} = \dfrac{(16-20)^2}{20}+\dfrac{(24-20)^2}{20}+\dfrac{(22-20)^2}{20}+\dfrac{(18-20)^2}{20}+\dfrac{(20-20)^2}{20}+\dfrac{(20-20)^2}{20} \)
\( = \dfrac{16+16+4+4+0+0}{20} = \dfrac{40}{20} = 2.0 \).
Degrees of freedom \(= 6-1 = 5\). Critical value \( \chi^2_{0.05,5} = 11.07 \).
Since \( 2.0 < 11.07 \), the test statistic is not in the critical region, so there is insufficient evidence at the 5\% level to reject \(H_0\): the results are consistent with a fair die. Final answer: \(\chi^2=2.0 < 11.07\); no evidence the die is biased.

[B1] correct hypotheses; [M1] correct expected frequency \(E=20\) for each class; [M1] correct \(\chi^2\) formula applied; [A1] each pair of \((O-E)^2/E\) terms correct, allow one arithmetic slip; [A1] correct total \(\chi^2=2.0\); [B1] correct degrees of freedom (5) and critical value (11.07, or equivalent from tables); [A1] correct conclusion in context (die is fair / no evidence of bias).

(a) \( \text{E}(W) = \text{E}(2X-3Y) = 2\text{E}(X)-3\text{E}(Y) = 2(50)-3(40) = 100-120 = -20 \).
Since \(X, Y\) independent, \( \text{Var}(W) = \text{Var}(2X-3Y) = 4\,\text{Var}(X) + 9\,\text{Var}(Y) = 4(16)+9(9) = 64+81 = 145 \).

(b) As a linear combination of independent normal variables, \( W \sim N(-20, 145) \), with standard deviation \( \sqrt{145} \approx 12.042 \).
\( \text{P}(W>10) = \text{P}\left(Z > \dfrac{10-(-20)}{\sqrt{145}}\right) = \text{P}(Z>2.491) \).
From standard normal tables, \( \Phi(2.491) \approx 0.9936 \), so \( \text{P}(W>10) \approx 1-0.9936 = 0.0064 \) (4 d.p.). Final answer: \(\text{E}(W)=-20\), \(\text{Var}(W)=145\); \(\text{P}(W>10)\approx0.0064\).

(a) [M1] correct linearity of expectation formula; [A1] \(\text{E}(W)=-20\); [M1] correct variance formula for independent combination (squares of coefficients, both variances added); [A1] \(\text{Var}(W)=145\). (b) [B1] state \(W\) is normally distributed with parameters from (a); [M1] correct standardisation; [A1] correct \(z\)-value \(\approx2.49\); [A1] correct final probability \(\approx0.0064\).

(a) Hall's marriage condition: a complete matching from \(L\) to \(R\) exists if and only if, for every subset \(S\subseteq L\), the set of neighbours \(N(S)\subseteq R\) satisfies \( |N(S)| \ge |S| \).

(b) Neighbour sets: \(N(a)=\{1,2\}\), \(N(b)=\{2,3\}\), \(N(c)=\{1,3\}\), each of size \(2\ge1\).
For any 2-element subset: \(N(\{a,b\})=\{1,2,3\}\) (size 3\(\ge\)2); \(N(\{a,c\})=\{1,2,3\}\) (size 3\(\ge\)2); \(N(\{b,c\})=\{1,2,3\}\) (size 3\(\ge\)2).
For the full set: \(N(\{a,b,c\})=\{1,2,3\}\) (size 3\(\ge\)3).
All subsets satisfy Hall's condition, so a complete matching exists. One valid complete matching: \(a\text{-}1,\ b\text{-}2,\ c\text{-}3\) (each edge listed is present in the graph and every vertex is matched exactly once). Final answer: Hall's condition holds; complete matching \(a\text{-}1, b\text{-}2, c\text{-}3\).

(a) [B1] correct statement referencing subsets \(S\) of \(L\); [B1] correct inequality \(|N(S)|\ge|S|\) for all such \(S\). (b) [M1] list neighbour sets for all singletons; [A1] check at least one 2-element subset explicitly; [A1] check the full set \(L\); [A1] correct valid complete matching stated explicitly (any valid matching, e.g. \(a\text{-}2,b\text{-}3,c\text{-}1\), accepted).

For activities in series on the critical path, the total duration \(T\) has mean and variance equal to the sums of the individual means and variances (assuming independence):
\( \text{E}(T) = 3+5+4+6 = 18 \) days. \( \text{Var}(T) = 0.25+0.36+0.16+0.49 = 1.26 \).
\( \text{SD}(T) = \sqrt{1.26} \approx 1.122 \) days.
Using a normal approximation \( T \sim N(18, 1.26) \):
\( \text{P}(T<20) = \text{P}\left(Z < \dfrac{20-18}{\sqrt{1.26}}\right) = \text{P}(Z<1.782) \).
From standard normal tables, \( \Phi(1.782) \approx 0.9626 \). Final answer: \( \text{P}(T<20) \approx 0.963 \) (3 s.f.).

[M1] sum the four means; [A1] \(\text{E}(T)=18\); [M1] sum the four variances; [A1] \(\text{Var}(T)=1.26\); [B1] state normal approximation \(T\sim N(18,1.26)\); [M1] correct standardisation of \(20\); [A1] correct final probability \(\approx0.963\).

For \( n\ge1 \), \( a_n - 2a_{n-1} = 1 \). Multiply by \(x^n\) and sum over \(n\ge1\):
\( \displaystyle\sum_{n\ge1}a_nx^n - 2x\sum_{n\ge1}a_{n-1}x^{n-1} = \sum_{n\ge1}x^n \).
The left side is \( (G(x)-a_0) - 2xG(x) \); the right side is the geometric series \( \dfrac{x}{1-x} \).
So \( G(x)(1-2x) - 1 = \dfrac{x}{1-x} \Rightarrow G(x)(1-2x) = 1+\dfrac{x}{1-x} = \dfrac{1}{1-x} \).
\( G(x) = \dfrac{1}{(1-x)(1-2x)} \).
Partial fractions: \( \dfrac{1}{(1-x)(1-2x)} = \dfrac{A}{1-x}+\dfrac{B}{1-2x} \). Multiplying up: \( 1 = A(1-2x)+B(1-x) \). Setting \(x=1\): \(A=1\). Setting \(x=\tfrac12\): \(B(1-\tfrac12)=1\Rightarrow B=2\).
\( G(x) = \dfrac{1}{1-x} + \dfrac{2}{1-2x} = \displaystyle\sum_{n=0}^\infty x^n + 2\sum_{n=0}^\infty (2x)^n = \sum_{n=0}^\infty \left(2^{n+1}-1\right)x^n \).
So \( a_n = 2^{n+1}-1 \). Check: \(a_0=2^1-1=1\) ✓; \(a_1=2a_0+1=3=2^2-1\) ✓; \(a_2=2(3)+1=7=2^3-1\) ✓. Final answer: \( a_n = 2^{n+1}-1 \).

[M1] multiply recurrence by \(x^n\) and sum over correct range; [A1] correctly express LHS in terms of \(G(x)\); [A1] correctly express RHS as \(x/(1-x)\); [M1] rearrange to isolate \(G(x)\); [A1] correct closed form \(G(x)=1/[(1-x)(1-2x)]\); [M1] partial fraction decomposition; [A1] correct constants \(A=1,B=2\); [A1] correctly extract coefficient to give \(a_n=2^{n+1}-1\), verified against the recurrence for small \(n\).

(a) The rook polynomial is \( R(x) = \displaystyle\sum_k r_k x^k \), where \(r_k\) is the number of ways to place \(k\) non-attacking rooks on the board (no two in the same row or column).
\(r_0=1\) (the empty placement, always counted once).
\(r_1\): any one of the 4 cells may hold a single rook, so \(r_1=4\).
\(r_2\): two non-attacking rooks must occupy different rows and different columns. On a \(2\times2\) board the only ways are \(\{(1,1),(2,2)\}\) and \(\{(1,2),(2,1)\}\), so \(r_2=2\). No placement of 3 or more non-attacking rooks is possible (only 2 rows), so \(r_k=0\) for \(k\ge3\).
So \( R(x) = 1 + 4x + 2x^2 \), as required.

(b) The number of ways to place 2 non-attacking rooks is the coefficient of \(x^2\) in \(R(x)\), which is \(2\). Final answer: \(R(x)=1+4x+2x^2\); 2 ways to place 2 non-attacking rooks.

(a) [B1] \(r_0=1\) correctly justified; [B1] \(r_1=4\) correctly justified; [M1] systematically identify non-attacking pairs for \(r_2\); [A1] \(r_2=2\) with both valid placements listed, giving \(R(x)=1+4x+2x^2\). (b) [M1] identify that the answer is the coefficient of \(x^2\); [A1] correct value \(2\).

(a) \(C_4=\{e, r, r^2, r^3\}\) where \(r\) is rotation by \(90^\circ\).
\(e\) (identity): fixes each of the 4 vertices individually, giving cycle type \(x_1^4\).
\(r\) (\(90^\circ\)): cycles all 4 vertices in a single 4-cycle \((1234)\), giving \(x_4\).
\(r^2\) (\(180^\circ\)): swaps opposite pairs, two 2-cycles \((13)(24)\), giving \(x_2^2\).
\(r^3\) (\(270^\circ\)): also a single 4-cycle \((1432)\), giving \(x_4\).
Cycle index \( = \dfrac{1}{|C_4|}\sum(\text{cycle monomials}) = \dfrac14\left(x_1^4+x_4+x_2^2+x_4\right) = \dfrac14\left(x_1^4+2x_4+x_2^2\right) \), as required.

(b) By Polya's theorem, the number of distinct colourings with \(m\) colours is found by substituting \(x_i \to m\) for every \(i\) in the cycle index. With \(m=2\):
\( P_{C_4}(2,2,2,2) = \dfrac14\left(2^4+2(2)+2^2\right) = \dfrac14\left(16+4+4\right) = \dfrac{24}{4} = 6 \).
Final answer: cycle index as shown; there are \(6\) distinct colourings of the square's vertices with 2 colours, up to rotation.

(a) [M1] correctly identify cycle type of the identity as \(x_1^4\); [A1] correctly identify cycle types of both non-trivial rotations \(90^\circ,270^\circ\) as \(x_4\) each; [A1] correctly identify cycle type of \(180^\circ\) rotation as \(x_2^2\), and correctly average by \(|C_4|=4\) to give the stated cycle index. (b) [M1] correct substitution \(x_i\to2\) for all \(i\) in the cycle index; [A1] correct arithmetic \((16+4+4)/4\); [A1] final answer \(6\).

(a) Initial system, with slack variables \(s_1,s_2\ge0\) added to convert inequalities to equations:
\( x+2y+s_1 = 10 \)
\( 3x+y+s_2 = 15 \)
\( P - 3x - 5y = 0 \) (objective row, rearranged from \(P=3x+5y\)).

(b) The most negative coefficient in the objective row is \(-5\) (on \(y\)), so \(y\) is the entering variable.
Minimum ratio test (ratio of RHS to the entering variable's coefficient, for positive coefficients only): row 1: \(10/2=5\); row 2: \(15/1=15\). The minimum ratio is \(5\), from row 1, so \(s_1\) is the leaving variable, and we pivot on the coefficient of \(y\) in row 1.
Dividing row 1 by 2: \( \dfrac{x}{2}+y+\dfrac{s_1}{2} = 5 \).
Eliminate \(y\) from row 2 by subtracting this new row 1 from row 2:
\( \left(3x+y+s_2\right) - \left(\dfrac{x}{2}+y+\dfrac{s_1}{2}\right) = 15-5 \Rightarrow \dfrac{5x}{2} - \dfrac{s_1}{2} + s_2 = 10 \).
Eliminate \(y\) from the objective row by adding \(5\times\)(new row 1):
\( \left(P-3x-5y\right) + 5\left(\dfrac{x}{2}+y+\dfrac{s_1}{2}\right) = 0+25 \Rightarrow P - \dfrac{x}{2} + \dfrac{5s_1}{2} = 25 \).
Since the objective row still has a negative coefficient (\(-\tfrac12\) on \(x\)), this tableau is not yet optimal — a further pivot (entering variable \(x\)) would be required to reach the optimum (\(x=4,y=3,P=27\)), but only one iteration was requested. Final answer: entering variable \(y\); leaving variable \(s_1\); post-pivot system as shown above.

(a) [B1] both constraint equations correctly written with slack variables; [B1] correct objective row \(P-3x-5y=0\). (b) [M1] correctly identify entering variable \(y\) (most negative coefficient); [M1] correctly compute both ratios and identify minimum; [A1] correctly identify leaving variable \(s_1\); [M1] correctly divide pivot row by the pivot element; [A1] correct updated row 2; [A1] correct updated objective row. Accept any algebraically equivalent tableau layout (e.g. full tabular form with columns for each variable).