2023 CCEA A-Level Further Mathematics 2330 Practice Paper with Answers

Basis (n=1): LHS \( =1(2)=2 \). RHS \( =\dfrac{1(2)(3)}{3}=2 \). So the statement holds for n=1.
Inductive step: Assume the statement is true for \( n=k \), i.e. \( \sum_{r=1}^{k} r(r+1) = \dfrac{k(k+1)(k+2)}{3} \). Then for \( n=k+1 \):
\( \sum_{r=1}^{k+1} r(r+1) = \sum_{r=1}^{k} r(r+1) + (k+1)(k+2) = \dfrac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)
\( = (k+1)(k+2)\left[\dfrac{k}{3}+1\right] = (k+1)(k+2)\cdot\dfrac{k+3}{3} = \dfrac{(k+1)(k+2)(k+3)}{3} \),
which is the formula with n replaced by \( k+1 \). So if the statement is true for n=k, it is true for n=k+1.
Since the statement is true for n=1, and true for n=k+1 whenever it is true for n=k, by the principle of mathematical induction, the statement is true for all positive integers n.

[6] B1: basis case n=1 verified (both sides =2); M1: assumes true for n=k and adds the (k+1)th term (k+1)(k+2); M1: correctly factorises out (k+1)(k+2); A1: correctly simplified to \( \frac{(k+1)(k+2)(k+3)}{3} \); B1: this is shown to match the given formula with n=k+1; B1: valid concluding statement referencing the induction principle (true for n=1, true for k+1 if true for k, therefore true for all positive integers n).

Using integration by parts with \( u=x, \; dv=e^{-2x}dx \), so \( du=dx, \; v=-\tfrac12e^{-2x}\):
\( \int x e^{-2x}\,dx = -\tfrac12xe^{-2x} + \tfrac12\int e^{-2x}\,dx = -\tfrac12xe^{-2x} - \tfrac14e^{-2x} + C \).
Consider \( \int_0^{N} xe^{-2x}\,dx = \left[-\tfrac12xe^{-2x}-\tfrac14e^{-2x}\right]_0^{N} = \left(-\tfrac12Ne^{-2N}-\tfrac14e^{-2N}\right) - \left(0-\tfrac14\right) \).
As \( N\to\infty \), \( Ne^{-2N}\to0 \) and \( e^{-2N}\to0 \) (exponential decay dominates the polynomial factor N), so the first bracket tends to 0. Hence \( \int_0^{\infty}xe^{-2x}\,dx = 0-\left(-\tfrac14\right) = \tfrac14 \), which is finite, so the integral converges. Final answer: \( \dfrac14 \).

[8] M1: correct choice u=x, dv=e^{-2x}dx; M1: correct du, v; M1: correct application of integration by parts formula; A1: correct antiderivative \( -\frac12xe^{-2x}-\frac14e^{-2x} \); M1: evaluates as a limit \( N\to\infty \) (not simply substituting ∞); A1: correctly argues \( Ne^{-2N}\to0 \) and \( e^{-2N}\to0 \); A1: correctly evaluates the lower limit (x=0) term; A1: final exact value 1/4, with explicit statement that the integral converges (finite limit).

The equation is linear, of the form \( \frac{dy}{dx}+P(x)y=Q(x) \) with \( P(x)=\frac{2}{x} \). Integrating factor: \( I(x) = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = x^2 \).
Multiplying through by \( x^2 \): \( x^2\dfrac{dy}{dx}+2xy = x^5 \), i.e. \( \dfrac{d}{dx}(x^2y) = x^5 \).
Integrating both sides: \( x^2y = \dfrac{x^6}{6}+C \).
Using \( y=2 \) when \( x=1 \): \( (1)^2(2) = \dfrac{1}{6}+C \Rightarrow C = 2-\dfrac16 = \dfrac{11}{6} \).
So \( x^2y = \dfrac{x^6}{6}+\dfrac{11}{6} \Rightarrow y = \dfrac{x^4}{6}+\dfrac{11}{6x^2} \).

[10] M1: recognises linear first-order form; M1: correct integrating factor set up \( e^{\int 2/x\,dx} \); A1: I(x)=x²; M1: multiplies through by x² correctly; A1: recognises LHS as \( \frac{d}{dx}(x^2y) \); M1: integrates RHS, \( \int x^5dx=\frac{x^6}{6} \); A1: \( x^2y=\frac{x^6}{6}+C \) (with +C); M1: substitutes x=1,y=2; A1: C=11/6; A1: final answer \( y=\frac{x^4}{6}+\frac{11}{6x^2} \).

Auxiliary equation (for the homogeneous part): \( m^2-5m+6=0 \Rightarrow (m-2)(m-3)=0 \Rightarrow m=2 \text{ or } m=3 \). Since the roots are real and distinct, the complementary function is \( y_{CF} = Ae^{2x}+Be^{3x} \).
For a particular integral, since \( e^{4x} \) is not part of the complementary function, try \( y_{PI}=Ce^{4x} \): \( y_{PI}'=4Ce^{4x} \), \( y_{PI}''=16Ce^{4x} \). Substituting: \( 16Ce^{4x}-5(4Ce^{4x})+6Ce^{4x} = e^{4x} \Rightarrow (16-20+6)Ce^{4x}=e^{4x} \Rightarrow 2Ce^{4x}=e^{4x} \Rightarrow C=\dfrac12 \).
General solution: \( y = Ae^{2x}+Be^{3x}+\dfrac12e^{4x} \).

[9] M1: forms auxiliary equation m²−5m+6=0; A1: correctly factorises/solves to m=2, m=3; A1: correct complementary function Ae^{2x}+Be^{3x}; M1: recognises need for particular integral of form Ce^{4x} (since 4 is not a root); M1: correctly differentiates y_PI twice; M1: correctly substitutes into the DE; A1: correctly forms 2C=1; A1: C=1/2; B1: full general solution correctly stated (CF + PI, with A, B as arbitrary constants).

\( \sum_{r=1}^{n} r(2r-1) = \sum_{r=1}^{n}(2r^2-r) = 2\sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r \).
Using the standard results \( \sum r^2 = \dfrac{n(n+1)(2n+1)}{6} \) and \( \sum r = \dfrac{n(n+1)}{2} \):
\( = 2\cdot\dfrac{n(n+1)(2n+1)}{6} - \dfrac{n(n+1)}{2} = \dfrac{n(n+1)(2n+1)}{3} - \dfrac{n(n+1)}{2} \).
Writing both terms over a common denominator of 6:
\( = \dfrac{2n(n+1)(2n+1) - 3n(n+1)}{6} = \dfrac{n(n+1)\left[2(2n+1)-3\right]}{6} = \dfrac{n(n+1)(4n+2-3)}{6} = \dfrac{n(n+1)(4n-1)}{6} \), as required.

[10] M1: expands r(2r−1)=2r²−r; M1: splits sum into 2Σr²−Σr; A1: correct standard result for Σr² quoted; A1: correct standard result for Σr quoted; M1: correct substitution of both formulae; M1: forms common denominator 6; A1: correctly combines numerators; M1: factorises out n(n+1); A1: correctly simplifies bracket to (4n−1); B1: final result matches the given expression exactly, confirming the proof is complete.

(a) \( 3x^2+2 \equiv A(x^2+1) + (Bx+C)x = (A+B)x^2 + Cx + A \). Comparing coefficients: constant term: \( A=2 \); x term: \( C=0 \); x² term: \( A+B=3 \Rightarrow B=1 \). So \( \dfrac{3x^2+2}{x(x^2+1)} \equiv \dfrac{2}{x}+\dfrac{x}{x^2+1} \).
(b) \( \int\left(\dfrac{2}{x}+\dfrac{x}{x^2+1}\right)dx = 2\ln|x| + \tfrac12\ln(x^2+1) + K \) (the second term uses the reverse chain rule, since the numerator x is (up to a constant) the derivative of \( x^2+1 \)).

(a) [6] M1: correct partial fraction identity set up; M1: correct method (comparing coefficients or substituting values); A1: A=2; A1: C=0; A1: B=1; A1: fully correct partial fraction form stated. (b) [5] M1 ft: integrates A/x term to 2ln|x|; M1 ft: recognises Bx/(x²+1) term needs reverse chain rule; A1 ft: integrates to \( \frac12\ln(x^2+1) \); A1: Cx/(x²+1) term correctly handled (here C=0, no contribution — accept implicit); B1: +K included.

Let \( x=2\sin\theta \), so \( dx=2\cos\theta\,d\theta \), and \( 4-x^2 = 4-4\sin^2\theta = 4\cos^2\theta \), so \( (4-x^2)^{3/2} = (4\cos^2\theta)^{3/2} = 8\cos^3\theta \) (taking \( \cos\theta \ge 0 \)).
Substituting: \( \int \dfrac{1}{8\cos^3\theta}\cdot2\cos\theta\,d\theta = \int \dfrac{1}{4\cos^2\theta}\,d\theta = \dfrac14\int \sec^2\theta\,d\theta = \dfrac14\tan\theta + C \).
Since \( \sin\theta = \dfrac{x}{2} \), we have \( \cos\theta = \dfrac{\sqrt{4-x^2}}{2} \), so \( \tan\theta = \dfrac{x/2}{\sqrt{4-x^2}/2} = \dfrac{x}{\sqrt{4-x^2}} \). Substituting back: \( \dfrac14\cdot\dfrac{x}{\sqrt{4-x^2}} + C = \dfrac{x}{4\sqrt{4-x^2}} + C \).

[11] M1: correct substitution dx=2cosθ dθ; M1: correctly simplifies \( 4-x^2=4\cos^2\theta \); M1: correctly simplifies \( (4-x^2)^{3/2}=8\cos^3\theta \); M1: fully substitutes into the integral; A1: simplifies integrand to \( \frac14\sec^2\theta \); M1: integrates \( \sec^2\theta \) to \( \tan\theta \); A1: \( \frac14\tan\theta+C \); M1: constructs a right triangle / uses sinθ=x/2 to find cosθ; A1: correctly finds \( \cos\theta=\frac{\sqrt{4-x^2}}{2} \); A1: correctly finds \( \tan\theta=\frac{x}{\sqrt{4-x^2}} \); A1: final answer \( \frac{x}{4\sqrt{4-x^2}}+C \).

(a) Since \( -1\le\cos\theta\le1 \), r ranges between \( 2-1=1 \) and \( 2+1=3 \); r is never zero or negative, so the curve is a closed, convex (dimpled or oval) curve enclosing the pole, symmetric about the initial line (θ=0). Maximum r=3 occurs at θ=0 (where cosθ=1); minimum r=1 occurs at θ=π (where cosθ=−1).
(b) Area \( = \dfrac12\displaystyle\int_0^{2\pi} r^2\,d\theta = \dfrac12\int_0^{2\pi}(2+\cos\theta)^2\,d\theta = \dfrac12\int_0^{2\pi}\left(4+4\cos\theta+\cos^2\theta\right)d\theta \).
Using \( \cos^2\theta = \dfrac{1+\cos2\theta}{2} \): integrand \( = 4+4\cos\theta+\dfrac12+\dfrac12\cos2\theta = \dfrac{9}{2}+4\cos\theta+\dfrac12\cos2\theta \).
\( \int_0^{2\pi}\left(\dfrac92+4\cos\theta+\dfrac12\cos2\theta\right)d\theta = \left[\dfrac{9\theta}{2}+4\sin\theta+\dfrac14\sin2\theta\right]_0^{2\pi} = \left(9\pi+0+0\right)-0 = 9\pi \) (since \( \sin(2\pi)=\sin(4\pi)=0 \)).
Area \( = \dfrac12(9\pi) = \dfrac{9\pi}{2} \).

(a) [4] B1: max r=3; B1: at θ=0; B1: min r=1; B1: at θ=π (accept general shape description of a convex closed curve around the pole for partial credit if values correct). (b) [8] M1: correct area formula \( \frac12\int r^2\,d\theta \) with correct limits; M1: correctly expands \( (2+\cos\theta)^2 \); M1: correctly uses double-angle identity for cos²θ; A1: correctly simplified integrand; M1: correctly integrates each term; A1: correct antiderivative; M1: correctly evaluates at θ=2π and θ=0 (using sin2π=sin4π=0); A1: final exact answer \( \frac{9\pi}{2} \).

(a) \( \cosh^2x-\sinh^2x = \left(\dfrac{e^x+e^{-x}}{2}\right)^2 - \left(\dfrac{e^x-e^{-x}}{2}\right)^2 = \dfrac{(e^{2x}+2+e^{-2x}) - (e^{2x}-2+e^{-2x})}{4} = \dfrac{4}{4}=1 \). ✓
(b) \( 2\sinh x\cosh x = 2\left(\dfrac{e^x-e^{-x}}{2}\right)\left(\dfrac{e^x+e^{-x}}{2}\right) = \dfrac{(e^x-e^{-x})(e^x+e^{-x})}{2} = \dfrac{e^{2x}-e^{-2x}}{2} = \sinh2x \). ✓
(c) \( 2\cosh x-\sinh x = 2\cdot\dfrac{e^x+e^{-x}}{2} - \dfrac{e^x-e^{-x}}{2} = \dfrac{2(e^x+e^{-x})-(e^x-e^{-x})}{2} = \dfrac{e^x+3e^{-x}}{2} \). Setting this equal to 2: \( e^x+3e^{-x}=4 \). Let \( u=e^x \) (u>0): \( u+\dfrac3u=4 \Rightarrow u^2-4u+3=0 \Rightarrow (u-1)(u-3)=0 \Rightarrow u=1 \text{ or } u=3 \). So \( e^x=1\Rightarrow x=0 \), or \( e^x=3 \Rightarrow x=\ln3 \).
(d) \( \sinh y = x \Rightarrow \dfrac{e^y-e^{-y}}{2}=x \Rightarrow e^y-e^{-y}=2x \). Let \( w=e^y \) (w>0): \( w-\dfrac1w=2x \Rightarrow w^2-2xw-1=0 \). By the quadratic formula: \( w = \dfrac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1} \). Since \( w=e^y>0 \) for all real y, and \( x-\sqrt{x^2+1}<0 \) for all real x (as \( \sqrt{x^2+1}>|x|\ge x \)), we must take the positive root: \( w = x+\sqrt{x^2+1} \). So \( e^y = x+\sqrt{x^2+1} \Rightarrow y = \ln\left(x+\sqrt{x^2+1}\right) \).

(a) [4] M1: correct exponential definitions substituted; M1: correctly squares both expressions; A1: correctly expands and combines numerators; A1: simplifies to 1 (identity fully shown, not assumed). (b) [4] M1: correct exponential definitions substituted; M1: correctly multiplies the two brackets; A1: correctly simplifies; A1: matches sinh2x definition. (c) [6] M1: correctly rewrites LHS in exponential form; A1: correctly simplified to \( \frac{e^x+3e^{-x}}{2} \); M1: forms equation \( e^x+3e^{-x}=4 \); M1: substitutes u=e^x to form quadratic \( u^2-4u+3=0 \); A1: correctly factorises/solves to u=1, u=3; A1: correctly gives x=0 and x=ln3 (both required). (d) [4] M1: correctly forms \( e^y-e^{-y}=2x \); M1: substitutes w=e^y to form quadratic \( w^2-2xw-1=0 \); M1: correctly solves via the quadratic formula; A1: correctly rejects the negative root (with valid reasoning) and states \( y=\ln(x+\sqrt{x^2+1}) \).

(a) \( e^x = 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\ldots \) and \( \cos x = 1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\ldots \). Multiplying and collecting terms up to \( x^4 \):
Constant: \( 1\times1=1 \).
x: \( 1\times x \) term: \( x \).
x²: \( \dfrac{x^2}{2}(1) + 1\times\left(-\dfrac{x^2}{2}\right) = \dfrac{x^2}{2}-\dfrac{x^2}{2}=0 \).
x³: \( \dfrac{x^3}{6}(1) + x\times\left(-\dfrac{x^2}{2}\right) = \dfrac{x^3}{6}-\dfrac{x^3}{2} = -\dfrac{x^3}{3} \).
x⁴: \( \dfrac{x^4}{24}(1) + \dfrac{x^2}{2}\times\left(-\dfrac{x^2}{2}\right) + 1\times\dfrac{x^4}{24} = \dfrac{x^4}{24}-\dfrac{x^4}{4}+\dfrac{x^4}{24} = -\dfrac{x^4}{6} \).
So \( f(x) \approx 1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6} \).
(b) \( g(x)=e^{2x}\cos2x = f(2x) \). Substituting \( 2x \) for x in (a): \( 1+(2x)-\dfrac{(2x)^3}{3}-\dfrac{(2x)^4}{6} = 1+2x-\dfrac{8x^3}{3}-\dfrac{16x^4}{6} = 1+2x-\dfrac{8x^3}{3}-\dfrac{8x^4}{3} \).
(c) \( e^{0.1}\cos(0.1) = f(0.1) \approx 1+0.1-\dfrac{(0.1)^3}{3}-\dfrac{(0.1)^4}{6} = 1.1 - 0.0003333 - 0.0000167 = 1.09965 \) (5 d.p.). (The true value is 1.0996497 to 7 d.p., so the series estimate agrees to 5 decimal places.)

(a) [8] B1: correct series for e^x up to x⁴; B1: correct series for cosx up to x⁴; M1: systematic multiplication/collection of terms up to x⁴; A1: constant + x terms correct; A1: x² coefficient =0 correctly found; A1: x³ coefficient correctly found (−1/3); A1: x⁴ coefficient correctly found (−1/6); A1: fully correct final series stated. (b) [4] M1: recognises g(x)=f(2x); M1: substitutes 2x for x in their (a); A1: correctly simplifies odd-power (x³) term; A1: correctly simplifies even-power (x⁴) term, fully correct series. (c) [4] M1 ft: substitutes x=0.1 into their (a); M1: correct arithmetic evaluation of each term; A1: 1.09965 (5 d.p., ft); B1: sensible comment on accuracy/agreement with true value.

(a) Let \( u=\sin^{n-1}x, \; dv=\sin x\,dx \), so \( du=(n-1)\sin^{n-2}x\cos x\,dx, \; v=-\cos x \).
\( I_n = \left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx \).
The boundary term vanishes at both limits (cos(π/2)=0 and sin(0)=0), so \( I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x(1-\sin^2x)\,dx = (n-1)\left(I_{n-2}-I_n\right) \).
So \( I_n = (n-1)I_{n-2} - (n-1)I_n \Rightarrow I_n + (n-1)I_n = (n-1)I_{n-2} \Rightarrow nI_n = (n-1)I_{n-2} \Rightarrow I_n = \dfrac{n-1}{n}I_{n-2} \), as required.
(b) \( I_0 = \int_0^{\pi/2}1\,dx = \dfrac{\pi}{2} \). \( I_1 = \int_0^{\pi/2}\sin x\,dx = \left[-\cos x\right]_0^{\pi/2} = 0-(-1)=1 \).
(c) \( I_5 = \dfrac{4}{5}I_3 = \dfrac{4}{5}\cdot\dfrac{2}{3}I_1 = \dfrac{4}{5}\cdot\dfrac{2}{3}\cdot1 = \dfrac{8}{15} \).
(d) \( I_6 = \dfrac{5}{6}I_4 = \dfrac{5}{6}\cdot\dfrac{3}{4}I_2 = \dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}I_0 = \dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{5\pi}{32} \).
(e) By the substitution \( x\to\dfrac{\pi}{2}-x \) (or the standard result that \( \int_0^{\pi/2}\cos^nx\,dx = \int_0^{\pi/2}\sin^nx\,dx = I_n \) for all n, since cos and sin are related by this symmetry over this interval), \( \int_0^{\pi/2}\cos^4x\,dx = I_4 = \dfrac{3}{4}I_2 = \dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{3\pi}{16} \).

(a) [9] M1: correct choice of u, dv; M1: correct du; A1: correct v; M1: correctly applies integration by parts formula; A1: boundary term shown to vanish (with reasoning); M1: correctly uses \( \cos^2x=1-\sin^2x \) to split integral into I_{n-2}-I_n; A1: correct equation \( I_n=(n-1)(I_{n-2}-I_n) \); M1: correctly collects I_n terms; A1: final result matches given formula. (b) [2] B1: I₀=π/2; B1: I₁=1. (c) [3] M1 ft: correct chain I₅=(4/5)I₃=(4/5)(2/3)I₁; A1: correct intermediate values; A1: I₅=8/15. (d) [3] M1 ft: correct chain I₆=(5/6)I₄=(5/6)(3/4)I₂=(5/6)(3/4)(1/2)I₀; A1: correct intermediate values; A1: I₆=5π/32. (e) [2] B1: recognises ∫cos⁴x dx over [0,π/2] equals I₄ (by symmetry); B1: correct value 3π/16 (ft from a correctly-found I₄).

(a) By De Moivre's theorem, \( \cos4\theta+i\sin4\theta = (\cos\theta+i\sin\theta)^4 \). Expanding the right-hand side using the binomial theorem, with \( c=\cos\theta, s=\sin\theta \):
\( (c+is)^4 = c^4+4ic^3s-6c^2s^2-4ics^3+s^4 \).
The real part of this expression equals \( \cos4\theta \): \( \cos4\theta = c^4-6c^2s^2+s^4 \).
Using \( s^2=1-c^2 \): \( s^4=(1-c^2)^2=1-2c^2+c^4 \), and \( -6c^2s^2=-6c^2(1-c^2)=-6c^2+6c^4 \).
So \( \cos4\theta = c^4 + (-6c^2+6c^4) + (1-2c^2+c^4) = 8c^4-8c^2+1 \), as required.
(b) By part (a), the equation \( 8c^4-8c^2+1=0 \) (with \( c=\cos\theta \)) is equivalent to \( \cos4\theta=0 \). For \( 0\le\theta\le90° \), \( 4\theta \) ranges over \( 0°\le4\theta\le360° \). \( \cos4\theta=0 \) when \( 4\theta=90° \) or \( 4\theta=270° \) (within this range), giving \( \theta=22.5° \) or \( \theta=67.5° \).
(c) Let \( \sqrt{-5+12i} = a+bi \), so \( (a+bi)^2 = -5+12i \Rightarrow a^2-b^2+2abi = -5+12i \). Equating real and imaginary parts: \( a^2-b^2=-5 \) and \( 2ab=12 \Rightarrow ab=6 \Rightarrow b=\dfrac{6}{a} \) (a≠0). Substituting: \( a^2-\left(\dfrac{6}{a}\right)^2=-5 \Rightarrow a^4+5a^2-36=0 \). Let \( u=a^2 \): \( u^2+5u-36=0 \Rightarrow (u+9)(u-4)=0 \Rightarrow u=4 \) (rejecting u=−9, since \( u=a^2\ge0 \)). So \( a^2=4\Rightarrow a=\pm2 \). If \( a=2 \), \( b=6/2=3 \); if \( a=-2 \), \( b=6/(-2)=-3 \). So the square roots are \( 2+3i \) and \( -2-3i \), i.e. \( \pm(2+3i) \). (Check: \( (2+3i)^2=4+12i-9=-5+12i \) ✓.)

(a) [6] M1: states De Moivre's theorem correctly; M1: expands (c+is)⁴ using the binomial theorem; A1: correctly identifies the real part c⁴−6c²s²+s⁴; M1: substitutes s²=1−c² into both s⁴ and the c²s² term; A1: correct intermediate expansion; A1: fully simplified to 8c⁴−8c²+1 (identity shown, not assumed). (b) [7] M1: recognises the equation is equivalent to cos4θ=0 using part (a); M1: correctly identifies range of 4θ as 0° to 360°; M1: correctly identifies cos4θ=0 at 4θ=90° and 4θ=270° (within this range, no other solutions); A1: θ=22.5°; A1: θ=67.5°; B1: recognises these are the only two solutions in the given range for θ; B1: (implicit/explicit) correct handling — no extra spurious solutions given. (c) [7] M1: sets up (a+bi)²=−5+12i and expands; M1: correctly equates real and imaginary parts (a²−b²=−5, 2ab=12); M1: correctly eliminates b (or a) to form a quartic in a; A1: correctly forms and factorises u²+5u−36=0 (or equivalent) to u=4 (rejecting u=−9 with reasoning); A1: a=±2; A1: correct corresponding b values found; A1: both square roots stated correctly as ±(2+3i), with a check shown.

(a) \( \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \) rad s⁻¹.
(b) \( \ddot x = -\omega^2x = -\dfrac{\pi^2}{4}x \).
(c) Maximum speed \( = \omega A = \dfrac{\pi}{2}(0.5) = \dfrac{\pi}{4} = 0.785 \) m s⁻¹ (3 s.f.), occurring as P passes through O.
(d) Maximum acceleration \( = \omega^2A = \dfrac{\pi^2}{4}(0.5) = \dfrac{\pi^2}{8} = 1.23 \) m s⁻² (3 s.f.), occurring at the extremes \( x=\pm0.5 \).
(e) Using \( v^2=\omega^2(A^2-x^2) \): \( v^2 = \dfrac{\pi^2}{4}(0.5^2-0.3^2) = \dfrac{\pi^2}{4}(0.16) \Rightarrow v=\dfrac{\pi}{2}\sqrt{0.16} = \dfrac{\pi}{2}(0.4) = 0.628 \) m s⁻¹ (3 s.f.).
(f) With P starting at O moving in the positive direction, \( x=A\sin(\omega t) \). Setting \( x=0.3 \): \( 0.3 = 0.5\sin\left(\dfrac{\pi}{2}t\right) \Rightarrow \sin\left(\dfrac{\pi}{2}t\right)=0.6 \Rightarrow \dfrac{\pi}{2}t = \sin^{-1}(0.6) = 0.6435 \text{ rad (taking the first/least solution)} \Rightarrow t = \dfrac{0.6435}{\pi/2} = 0.410 \) s (3 s.f.).

(a) [2] M1: ω=2π/T; A1: ω=π/2. (b) [2] B1 ft: correct SHM equation form; B1 ft: ω² correctly evaluated (π²/4). (c) [3] M1 ft: v_max=ωA; M1: correct substitution; A1 ft: 0.785 m s⁻¹. (d) [2] M1 ft: a_max=ω²A; A1 ft: 1.23 m s⁻². (e) [3] M1 ft: correct formula v²=ω²(A²−x²); M1: correct substitution; A1 ft: 0.628 m s⁻¹. (f) [3] M1: correct model x=Asin(ωt) for these initial conditions; M1: correctly solves for t using arcsin; A1: 0.410 s (3 s.f.).

(a) By Newton's second law, taking the positive x-direction: \( 2\ddot x = -18x - 8\dot x \) (restoring force acts towards O, i.e. in the −x direction when x>0; resistive force opposes the motion, i.e. acts in the −ẋ direction). Dividing by 2: \( \ddot x = -9x-4\dot x \Rightarrow \ddot x+4\dot x+9x=0 \), as required.
(b) Auxiliary equation: \( m^2+4m+9=0 \). Discriminant \( = 4^2-4(1)(9) = 16-36=-20 <0 \), so the roots are complex: \( m = \dfrac{-4\pm\sqrt{-20}}{2} = -2\pm i\sqrt5 \). Since the discriminant of the auxiliary equation is negative (complex roots), the motion is lightly damped (underdamped) — the particle oscillates with an exponentially decaying amplitude.
(c) With complex roots \( -2\pm i\sqrt5 \), the general solution is \( x = e^{-2t}\left(P\cos\sqrt5t+Q\sin\sqrt5t\right) \). At \( t=0 \), \( x=1 \Rightarrow P=1 \). Differentiating: \( \dot x = -2e^{-2t}(P\cos\sqrt5t+Q\sin\sqrt5t) + e^{-2t}\left(-P\sqrt5\sin\sqrt5t+Q\sqrt5\cos\sqrt5t\right) \). At \( t=0 \), \( \dot x=0 \Rightarrow -2P+Q\sqrt5 = 0 \Rightarrow Q = \dfrac{2P}{\sqrt5} = \dfrac{2}{\sqrt5} = \dfrac{2\sqrt5}{5} \). So \( x = e^{-2t}\left(\cos\sqrt5t + \dfrac{2\sqrt5}{5}\sin\sqrt5t\right) \).

(a) [3] M1: correct application of Newton's second law with both forces (correct signs); M1: correctly divides by mass 2; A1: correctly rearranged to given form. (b) [5] M1: correct auxiliary equation; M1: correctly calculates discriminant; A1: correct complex roots −2±i√5; B1: correctly identifies light damping; B1: correct reason (discriminant negative / complex roots ⇒ oscillatory with decaying exponential envelope). (c) [7] M1 ft: correct general solution form using their roots; M1: applies x(0)=1 to find P=1; M1: differentiates x(t) correctly (product rule); M1: applies ẋ(0)=0; A1: correct equation −2P+Q√5=0; A1: correctly solves for Q; A1: fully correct final expression for x(t).

(a) Area of rectangle \( = 8\times6=48 \) cm². By symmetry, its centre of mass is at the centre: (4, 3).
(b) Area of semicircle \( = \dfrac12\pi r^2 = \dfrac12\pi(3)^2 = 4.5\pi \approx14.1 \) cm² (3 s.f.). The centre of mass of a semicircle lies a distance \( \dfrac{4r}{3\pi} \) from the centre of the flat (diameter) edge, along the axis of symmetry, towards the curved edge. Since the flat edge is at x=8 and the curved part bulges towards decreasing x, the centre of mass of the semicircular piece is at \( x = 8-\dfrac{4(3)}{3\pi} = 8-\dfrac{4}{\pi} \approx 6.73 \) (3 s.f.), \( y=3 \) (by symmetry). So centre of mass of the removed piece ≈ (6.73, 3).
(c) Treating the lamina as (rectangle) − (semicircle), using moments: let the composite area be \( A_c = 48-4.5\pi \approx 33.9 \) cm² (3 s.f.). Taking moments about the y-axis (for the x-coordinate): \( A_c\bar x = 48(4) - 4.5\pi(6.7268) \Rightarrow \bar x = \dfrac{192-95.1}{33.9} = \dfrac{96.9}{33.9} \approx 2.86 \) (3 s.f.). By symmetry (both the rectangle and the semicircle have centres of mass on the line y=3), \( \bar y = 3 \). So the centre of mass of the lamina is at approximately (2.86, 3).
(d) The lamina occupies \( 0\le x\le8, 0\le y\le6 \), minus the semicircular bite near x=8. The point (2.86, 3) is well within the rectangular region and far from the removed semicircular region (which only affects \( 5\le x\le8 \)), so the centre of mass does lie within the material of the lamina.

(a) [2] B1: area 48 cm²; B1: centre of mass (4,3). (b) [5] M1: correct semicircle area formula; A1: area ≈14.1 cm²; M1: correct standard result 4r/(3π) for the centroid distance from the flat edge; A1: correctly applies this to find x-coordinate ≈6.73; B1: y-coordinate =3 (symmetry). (c) [6] M1: correct 'rectangle minus semicircle' (subtraction of moments) method set up; M1: correct composite area 48−4.5π; A1: composite area ≈33.9 cm² (3 s.f.); M1: correct moment equation for x̄; A1: x̄≈2.86 (3 s.f., ft); B1: ȳ=3 stated (by symmetry, ft). (d) [2] B1: correct conclusion (lies within the material); B1: valid supporting reason referencing the location of the removed region relative to the found centre of mass.

Let θ=20°, r=80 m, μ=0.2, g=9.8 m s⁻².
(a) With no friction required, resolving vertically and horizontally for the normal reaction N: \( N\cos\theta = mg \) and \( N\sin\theta = \dfrac{mv_0^2}{r} \). Dividing: \( \tan\theta = \dfrac{v_0^2}{rg} \Rightarrow v_0 = \sqrt{rg\tan\theta} = \sqrt{80(9.8)\tan20°} = \sqrt{80(9.8)(0.3640)} = 16.9 \) m s⁻¹ (3 s.f.).
(b) At maximum speed, the car is on the point of sliding up/outward across the banking, so friction acts down the slope (towards the centre-line, opposing the outward tendency), with magnitude \( \mu N \). Resolving perpendicular and along the incline (or vertically/horizontally) gives the standard result: \( v_{max}^2 = \dfrac{rg(\tan\theta+\mu)}{1-\mu\tan\theta} = \dfrac{80(9.8)(0.3640+0.2)}{1-0.2(0.3640)} = \dfrac{80(9.8)(0.5640)}{0.9272} \Rightarrow v_{max} = 21.8 \) m s⁻¹ (3 s.f.).
(c) At minimum speed, the car is on the point of sliding down/inward across the banking, so friction acts up the slope, with magnitude \( \mu N \), opposing this tendency: \( v_{min}^2 = \dfrac{rg(\tan\theta-\mu)}{1+\mu\tan\theta} = \dfrac{80(9.8)(0.3640-0.2)}{1+0.2(0.3640)} = \dfrac{80(9.8)(0.1640)}{1.0728} \Rightarrow v_{min} = 10.9 \) m s⁻¹ (3 s.f.).

(a) [4] M1: correctly resolves vertically (Ncosθ=mg) and horizontally (Nsinθ=mv₀²/r); M1: correctly eliminates N (division); A1: v₀²=rg tanθ; A1: v₀=16.9 m s⁻¹ (3 s.f.). (b) [6] M1: correctly identifies friction acts down the slope at maximum speed; M1: correct resolution including friction (perpendicular and along the incline, or vertical/horizontal, consistently); M1: correctly eliminates N; A1: correct formula for v²_max; M1: correct substitution; A1: v_max=21.8 m s⁻¹ (3 s.f.). (c) [5] M1: correctly identifies friction acts up the slope at minimum speed; M1: correct resolution including friction, with correct sign change from (b); A1: correct formula for v²_min; M1: correct substitution; A1: v_min=10.9 m s⁻¹ (3 s.f.).

(a) \( \mathbf{v} = \int\mathbf{a}\,dt = (3t^2+C_1)\mathbf{i}+(-4t+C_2)\mathbf{j}+(t^2+C_3)\mathbf{k} \). Using \( \mathbf{v}(0)=2\mathbf{i}+3\mathbf{j}-\mathbf{k} \): \( C_1=2, C_2=3, C_3=-1 \). So \( \mathbf{v} = (3t^2+2)\mathbf{i}+(3-4t)\mathbf{j}+(t^2-1)\mathbf{k} \).
(b) \( \mathbf{r} = \int\mathbf{v}\,dt = (t^3+2t+D_1)\mathbf{i}+(3t-2t^2+D_2)\mathbf{j}+\left(\dfrac{t^3}{3}-t+D_3\right)\mathbf{k} \). Using \( \mathbf{r}(0)=\mathbf{i}+2\mathbf{j} \): \( D_1=1, D_2=2, D_3=0 \). So \( \mathbf{r} = (t^3+2t+1)\mathbf{i}+(-2t^2+3t+2)\mathbf{j}+\left(\dfrac{t^3}{3}-t\right)\mathbf{k} \).
(c) At \( t=2 \): \( \mathbf{v} = (3(4)+2)\mathbf{i}+(3-8)\mathbf{j}+(4-1)\mathbf{k} = 14\mathbf{i}-5\mathbf{j}+3\mathbf{k} \). Speed \( = \sqrt{14^2+(-5)^2+3^2} = \sqrt{196+25+9} = \sqrt{230} = 15.2 \) m s⁻¹ (3 s.f.).
(d) j-component of velocity: \( 3-4t=0 \Rightarrow t = \dfrac34 = 0.75 \) s.

(a) [4] M1: integrates each component correctly (unsimplified, with constants); M1: applies initial velocity condition correctly; A1: correct i and j components; A1: correct k component (full vector). (b) [5] M1 ft: integrates their v correctly (with constants); M1: applies initial position condition correctly; A1: correct i component; A1: correct j component; A1: correct k component. (c) [4] M1 ft: finds v at t=2 using their (a); M1: correct speed formula \( \sqrt{v_i^2+v_j^2+v_k^2} \); A1: correct value under the root (230); A1: 15.2 m s⁻¹ (3 s.f.). (d) [2] M1 ft: sets j-component of their (a) to zero; A1: t=0.75 s.

Taking the common initial direction of motion as positive: \( u_A=6, u_B=2 \).
(a) Conservation of momentum: \( m_Au_A+m_Bu_B = m_Av_A+m_Bv_B \Rightarrow 3(6)+5(2) = 3v_A+5v_B \Rightarrow 28 = 3v_A+5v_B \). Newton's law of restitution: \( v_B-v_A = e(u_A-u_B) = 0.6(6-2) = 2.4 \Rightarrow v_B = v_A+2.4 \). Substituting: \( 3v_A+5(v_A+2.4)=28 \Rightarrow 8v_A+12=28 \Rightarrow v_A=2 \) m s⁻¹. Then \( v_B=2+2.4=4.4 \) m s⁻¹. Both velocities are positive, so both spheres continue moving in their original direction, A at 2 m s⁻¹ and B at 4.4 m s⁻¹.
(b) Since \( v_B(4.4) > v_A(2) \), sphere B is now moving away from A faster than A is moving, so the separation between A (behind) and B (ahead) continues to increase. No second collision occurs (assuming no other obstacles).
(c) Impulse on B \( = m_B(v_B-u_B) = 5(4.4-2) = 5(2.4) = 12 \) N s. By Newton's third law, the impulse exerted by A on B has the same magnitude as the impulse exerted on B (by A), so the magnitude of the impulse exerted by A on B is 12 N s.

(a) [8] M1: correct momentum equation with correct signs; M1: correct restitution equation; A1: correctly forms v_B=v_A+2.4; M1: correctly substitutes to eliminate v_B (or v_A); A1: v_A=2 m s⁻¹; A1: v_B=4.4 m s⁻¹; B1: both directions correctly stated as the original direction (positive). (b) [3] M1: compares v_A and v_B correctly; A1: correct conclusion (v_B>v_A, so no second collision); B1: valid physical reasoning (separation increasing / B moving away from A faster than A follows). (c) [4] M1 ft: impulse = m_B(v_B−u_B); M1: correct substitution using their v_B; A1 ft: 12 N s; B1: correctly notes this equals the impulse exerted by A on B (Newton's third law/definition of impulse on B).

(a) Since X and Y are independent, \( E(X+Y)=E(X)+E(Y)=50+30=80 \), and \( \mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)=16+9=25 \) (variances add for independent variables, whether the combination is a sum or difference). The sum of two independent Normal variables is itself Normal, so \( X+Y\sim N(80,25) \).
(b) \( P(X+Y>85) = P\left(Z>\dfrac{85-80}{\sqrt{25}}\right) = P(Z>1) = 0.159 \) (3 s.f.).
(c) \( E(D)=E(X)-2E(Y)=50-2(30)=-10 \). \( \mathrm{Var}(D)=\mathrm{Var}(X)+2^2\mathrm{Var}(Y)=16+4(9)=16+36=52 \) (variance of −2Y is \( (-2)^2\mathrm{Var}(Y) \); variances still add for independent X and Y even though we are subtracting). So \( D\sim N(-10,52) \).
(d) \( T\sim N(4(30),4(9)) = N(120,36) \) (mean and variance both scale by n=4 for the sum of n independent observations from the same distribution). \( P(T>125) = P\left(Z>\dfrac{125-120}{\sqrt{36}}\right) = P(Z>0.833) = 0.202 \) (3 s.f.).
(e) \( W=2Y_1 \sim N(2(30), 2^2(9)) = N(60,36) \), since \( E(2Y_1)=2E(Y_1)=60 \) and \( \mathrm{Var}(2Y_1)=2^2\mathrm{Var}(Y_1)=36 \). In contrast, \( Y_1+Y_2 \sim N(60, 9+9)=N(60,18) \). Although both have the same mean (60), \( W=2Y_1 \) doubles a single random deviation from the mean (so the variance is scaled by \( 2^2=4 \)), whereas \( Y_1+Y_2 \) sums two independent random deviations, which do not simply double — some of the randomness in \( Y_1 \) and \( Y_2 \) is independent and can partly offset, so the variance only doubles (adds), giving a smaller variance (18) than doubling a single observation (36).

(a) [3] B1: mean=80 with correct justification (sum of means); B1: variance=25 with correct justification (sum of variances, independence used); B1: correctly states X+Y is Normal. (b) [3] M1: correct standardisation using their (a); A1: z=1; A1: 0.159 (3 s.f.). (c) [3] M1: correct mean −10; M1: correct variance calculation with (−2)²=4 scaling; A1: D~N(−10,52) fully stated. (d) [3] M1: correct mean and variance for sum of 4 independent observations (120, 36); M1: correct standardisation; A1: 0.202 (3 s.f.). (e) [3] B1: correct distribution of W, N(60,36); B1: correct variance of Y₁+Y₂ (18) for comparison; B1: valid explanation referencing the difference between scaling a single variable (×4 variance) versus summing two independent variables (variances simply add, not quadruple).

(a) The sample size (n=10) is small, and the population variance is unknown — it must be estimated from the sample using s. When the population variance is estimated from a small sample (rather than being known), the extra uncertainty this introduces means the sampling distribution of the standardised statistic follows a t-distribution (with n−1 degrees of freedom), not a standard Normal distribution; the t-distribution has heavier tails to reflect this additional uncertainty.
(b) Standard error \( = \dfrac{s}{\sqrt n} = \dfrac{3.2}{\sqrt{10}} = 1.012 \) (4 s.f.). 95% CI: \( \bar x \pm t_{9,0.025}\times\text{SE} = 24.6 \pm 2.262(1.012) = 24.6\pm2.289 \). So the confidence interval is (22.3, 26.9) (3 s.f.).
(c) Let μ be the population mean. \( H_0: \mu=26 \); \( H_1: \mu\ne26 \) (two-tailed test). Test statistic: \( t = \dfrac{\bar x-\mu_0}{s/\sqrt n} = \dfrac{24.6-26}{1.012} = \dfrac{-1.4}{1.012} = -1.38 \) (3 s.f.). Since \( |t|=1.38 < 2.262 \) (the critical value), the result is not significant at the 5% level: we do not reject \( H_0 \). There is insufficient evidence, at the 5% significance level, to reject the researcher's claim that the population mean is 26. (This is consistent with part (b): the value 26 lies within the 95% confidence interval found.)

(a) [2] B1: population variance unknown/estimated from a small sample; B1: correct consequence — t-distribution used to account for the extra uncertainty (heavier tails), valid because the population itself is Normal. (b) [6] M1: correct standard error s/√n; A1: SE=1.012 (or equivalent); M1: correct CI formula x̄±t×SE; M1: correct substitution; A1: correct margin of error (±2.29, 3 s.f.); A1: correct interval (22.3, 26.9) (3 s.f.). (c) [7] B1: correct hypotheses (μ=26 vs μ≠26); M1: correct test statistic formula; M1: correct substitution; A1: t=−1.38 (3 s.f.); M1 ft: correctly compares |t| to 2.262; A1 ft: correct decision (do not reject H0); A1: correct conclusion stated in context, ideally noting consistency with part (b).

(a) \( H_0 \): the die is fair (each score 1–6 is equally likely to occur, i.e. P(score)=1/6 for each score). \( H_1 \): the die is not fair (the scores are not all equally likely).
(b) If the die is fair, each score is equally likely, so the expected frequency for each score, out of 120 rolls, is \( 120\times\dfrac16 = 20 \).
(c) \( \chi^2 = \sum\dfrac{(O-E)^2}{E} \):
Score 1: \( (15-20)^2/20 = 25/20 = 1.25 \).
Score 2: \( (22-20)^2/20 = 4/20 = 0.20 \).
Score 3: \( (18-20)^2/20 = 4/20 = 0.20 \).
Score 4: \( (25-20)^2/20 = 25/20 = 1.25 \).
Score 5: \( (17-20)^2/20 = 9/20 = 0.45 \).
Score 6: \( (23-20)^2/20 = 9/20 = 0.45 \).
Total: \( \chi^2 = 1.25+0.20+0.20+1.25+0.45+0.45 = 3.80 \).
(d) Degrees of freedom \( = 6-1=5 \). The calculated \( \chi^2 \) (3.80) is less than the critical value (11.070) at the 5% significance level with 5 degrees of freedom. So the result is not significant: we do not reject \( H_0 \). There is no evidence, at the 5% significance level, that the die is biased — the observed results are consistent with a fair die.
(e) Each expected frequency should be at least 5 (here each expected frequency is 20, so this condition is satisfied).

(a) [2] B1: H0 correctly stated (die fair/equal probabilities); B1: H1 correctly stated (die not fair). (b) [2] M1: 120×(1/6); A1: 20 for each score. (c) [6] M1: correct (O−E)² for at least two scores; M1: correct division by E throughout; A1: individual values correct (allow one slip, ft); A1: correctly totals to 3.80; B1: table/working clearly laid out; B1: correct formula Σ(O−E)²/E used explicitly. (d) [4] B1: degrees of freedom = 5; M1 ft: correctly compares their χ² to 11.070; A1 ft: correct decision (do not reject H0); A1: correct conclusion in context (no evidence of bias). (e) [1] B1: valid condition (each expected frequency ≥5).

(a) Prim's algorithm, starting at A:
1. Tree = {A}. Cheapest edge from the tree to a new vertex: AB (weight 4). Add AB. Tree = {A,B}.
2. Cheapest edge from {A,B} to a new vertex: candidates AC(7), AD(9), BC(3), BE(6) — cheapest is BC (weight 3). Add BC. Tree = {A,B,C}.
3. Cheapest edge from {A,B,C}: candidates AD(9), BE(6), CD(5), CE(8), CF(11) — cheapest is CD (weight 5). Add CD. Tree = {A,B,C,D}.
4. Cheapest edge from {A,B,C,D}: candidates BE(6), CE(8), CF(11), DF(12) — cheapest is BE (weight 6). Add BE. Tree = {A,B,C,D,E}.
5. Cheapest edge from {A,B,C,D,E}: candidates CE(8) [both endpoints already in tree, ignore], CF(11), DF(12), EF(2) — cheapest is EF (weight 2). Add EF. Tree = {A,B,C,D,E,F} — all vertices included.
Edges added, in order: AB(4), BC(3), CD(5), BE(6), EF(2).
(b) Total weight \( = 4+3+5+6+2 = 20 \) km.
(c) A minimal spanning tree for a network with n vertices must contain exactly \( n-1 \) edges. Here n=6 vertices (A–F), so the tree must have \( 6-1=5 \) edges — which matches the 5 edges found in part (a). ✓
(d) At every stage of Prim's algorithm above, there was a single, strictly cheapest available edge (no ties in weight between competing edges at any selection step), so the minimal spanning tree found is unique.
(e) By the time C is being considered for further connections, C is already connected to the tree (via BC), and by the completion of the algorithm, F is connected to the tree via the cheaper route B→E→F (total 6+2=8, i.e. only the edge EF, weight 2, was needed once E was in the tree) rather than via CF (weight 11); since Prim's algorithm always selects the cheapest available edge to extend the tree at each stage, and cheaper edges connecting F (namely EF) were available, CF (the most expensive edge in the whole network) was never the cheapest option and so is excluded.

(a) [8] M1: correctly starts at A and identifies AB as cheapest first edge; A1: AB(4) added correctly; M1: correctly identifies next cheapest edge from the growing tree; A1: BC(3) added correctly; M1: correctly identifies next cheapest edge; A1: CD(5) added correctly; A1: BE(6) then EF(2) added correctly, completing the tree (in this order or any correct valid order given ties do not occur); B1: clearly shows the algorithm process (tree growing, candidate edges considered at each stage), not just a final list. (b) [1] B1 ft: 20 km (sum of their 5 edges). (c) [2] B1: correctly states n−1 edges; B1: correctly verifies 6−1=5 matches their answer to (a). (d) [2] B1: correctly states the tree is unique; B1: valid reason (no tied/equal weights among the edges actually competing at each stage of the algorithm). (e) [2] B1: correctly identifies that a cheaper route to F already exists via the tree (through E); B1: correct link to Prim's algorithm always choosing the cheapest available edge at each stage, so the more expensive CF is never selected.