MATH40004 Revised Historical Coverage Practice Paper 3

Question 1: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. This question mixes global graph information, nonzero-centre Taylor expansion, a Riemann sum and a parametric arc-length calculation. (a)(i) h′ = −3x2e−x3 ≤ 0, so there is no local extremum; x = 0 is stationary. Also h′′ = 3xe−x3 (3x3 − 2), so inflection points occur at x = 0 and x = (2/3) 1/3. The graph decreases from +∞ to 0. (ii) Taylor’s theorem is as stated in Paper 1. For f (x) = x log x about 1, f (x) = ( x − 1) + 1 2 (x − 1)2 − (x − 1)3 6ξ2 , where ξ lies between 1 and x. Hence 1.1 log 1.1 ≈ 1 10 + 1 200 = 21 200 , |R| ≤ 10−3 6 = 1 6000 . (b) Since R π/4 0 tan x dx = 1 2 log 2 and the mesh width is π/(4n), π 4n n−1X k=0 tan kπ 4n → 1 2 log 2, which gives the stated limit. (c) In polar form r = e−t, θ = t: an inward logarithmic spiral. The speed is p (x′)2 + (y′)2 = √ 2e−t, so the total length is√ 2. Marking points. Graph analysis 5; Taylor/remainder/approximation 6; Riemann sum 4; spiral and length 5. Common errors. Calling the stationary inflection at zero a maximum; using width1/n instead of π/(4n); forgetting that an infinite number of turns can have finite length. Self-check. h must be monotone decreasing; the spiral starts at (1, 0) and tends to the origin. Page 1 of 4 MATH40004 Calculus and Applications 2026 Summer Resit Preparation

Question 2: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. V olumes use disc/shell formulas; the rod uses mass times centre = moment; Laplace inversion uses partial fractions. (a) Vx = π Z L 0 x4dx = πL5 5 , V y = 2π Z L 0 x x2dx = πL4 2 . They are equal at L = 5/2. For small L, the shell radius about the y-axis dominates; for large L, the fourth power of height in the disc volume makes Vx dominate. (b)(i) The mass is M = ρ0 R π/4 0 tan xdx = (ρ0/2) log 2. Therefore ¯x = ρ0J M = 2J log 2 . (ii) With tan x = x + x3/3 + · · · , J ≈ Z π/4 0 ( x2 + x4 3 ) dx = (π/4)3 3 + (π/4)5 15 . Multiplying by 2/ log 2 gives the required expression. (c) Integration by parts gives L{t} = 1 s2 , L{sin 3t} = 3 s2 + 9 . Since 1 s2(s2 + 9) = 1 9 ( 1 s2 − 1 s2 + 9 ) , g(t) = t 9 − 1 27 sin 3t. Marking points. Two volumes/threshold/explanation 8; moment formula 4; Taylor approximation 3; transforms and inver- sion 5. Common errors. Using the surface-area formula instead of volume; leaving ρ0 in the centre of mass; confusing 1/(s2 + 9) with the transform of sin 3t without the factor 3. Self-check. The centre must lie between 0 and π/4; the inverse transform is zero at t = 0. Page 2 of 4 MATH40004 Calculus and Applications 2026 Summer Resit Preparation

Question 3: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. Use oddness for the Fourier coefficients; use periodic boundary terms twice before applying Parseval. (a) The function is odd, so sin(x/3) = ∞X n=1 bn sin nx, b n = 2 π Z π 0 sin(x/3) sin nx dx. Product-to-sum gives bn = (−1)n+19 √ 3 n π(9n2 − 1) . At x = π, periodic extension has one-sided values ± √ 3/2, so the series converges to 0. At x = π/2, only odd n = 2m + 1 remain and simplification gives the required identity. (b) Periodicity yields E = −2 Z π −π (f ′)2dx + ε Z π −π (f ′′)2dx. If f = a0/2 + P(an cos nx + bn sin nx) and a0 = 0, Parseval gives E π = ∞X n=1 n2(εn2 − 2)(a2 n + b2 n). Thus E ≥ 0 for every such f exactly when ε ≥ 2. It is strictly positive for every nonzero zero-mean f if ε > 2. At ε = 2 , the first Fourier mode gives equality. Marking points. Coefficients 7; special-value identity 3; integrations by parts 4; Parseval expression 4; positivity threshold 2. Common errors. Using the function value at the periodic jump; retaining an a0 term despite zero mean; saying E > 0 at ε = 2 without checking the n = 1 mode. Self-check. The coefficient denominator must be 9n2 − 1 and the positivity threshold is set by the smallest mode n = 1. Page 3 of 4 MATH40004 Calculus and Applications 2026 Summer Resit Preparation

Question 4: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. For the system, eigenvectors give the separatrices. For the transform integral, compute the cosine trans- form first. For Euler–Cauchy, stay in z = log x. (a) The eigenpairs are λ = −2, vs = (1, 1)T and λ = 1, vu = (1, 4)T . Hence X = c1e−2t(1, 1)T + c2et(1, 4)T . The origin is a saddle: generic solutions diverge along y = 4x, while data on y = x decay to the origin. (b)(i) bfc(ω) = Z 1 0 (1 − x) cos ωx dx = 1 − cos ω ω2 . (ii) The full transform is twice this. Plancherel gives 1 2π Z ∞ −∞ [ 2(1 − cos ω) ω2 ]2 dω = Z 1 −1 (1 − |x|)2dx = 2 3 , which reduces to the stated π/6 integral. (c) In z = log x the equation is Y ′′ − 2Y ′ + Y = ez z . The CF is (c1 + c2z)ez. Put Yp = A(z)ez; then (D − 1)2(Aez) = A′′ez, so A′′ = 1/ z. Taking A = z log z modulo homogeneous terms, y = c1x + c2x log x + x log x log(log x). Marking points. Eigenanalysis/phase portrait 8; transform 4; energy identity 3; Euler change/CF/ansatz/final 5. Common errors. Reversing stable and unstable eigenlines; using the cosine transform directly in Plancherel without the factor two; choosing A = log z instead of integrating twice. Self-check. The long-time direction is the eigenvector with positive eigenvalue; differentiating z log z twice gives 1/z. Page 4 of 4