MATH40004 Revised Historical Coverage Practice Paper 2

Question 1: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. Three independent convergence tests, followed by a binomial-to-antiderivative method and an iterated chain rule. P P (a) n−3/2 converges by the p-test. For (log n)n−q, the integral test gives convergence for q > 1 and divergence for q = 1 (substitute u = log x). For the factorial series, e(n+1)1/3 /(n + 1)! e(n+1)1/3−n1/3 = → 0, en1/3 /n! n + 1 so it converges. (b) x2 3x4 5x6 (1 − x2) −1/2 = 1 + + + + · · · , |x| < 1. 2 8 16 Because (sin −1 x)′ = (1 − x2)−1/2 and sin −1 0 = 0, x3 3x5 5x7 sin −1 x = x + + + + · · · . 6 40 112 At x = 1/5, 1 1 3 5 1057129 S = + + + = = 0.2013579048 . . . . 5 6 53 40 55 112 57 5250000 The omitted positive terms are below 1.7 × 10−8 (for example by bounding the coefficient ratio and a geometric tail), so sin −1(1/5) = 0.201358 to six decimal places. (c) Repeated chain rule gives nY−1 ′ F (x) = cos(F (x)). n j j=0 Since |F (x)| ≤ 1 for j ≥ 1, those cosine factors cannot vanish. Thus stationary points are x = (2m + 1)π/2, from the first j factor cos x. For a = F (π/2), 0 < a = sin a < a ; hence a → L ≥ 0 and L = sin L, giving L = 0. n n n+1 n n n Marking points. Tests 6; binomial/arcsin 5; rational approximation/error 3; derivative product/stationary points/limit 6. Common errors. Using a derivative-heavy route for the binomial series; giving six decimals without controlling the positive tail; solving L = sin L before proving convergence. Self-check. The approximation must exceed 0.2 and be very close to the calculator value; F (π/2) must decrease. n Page 1 of 5

Question 2: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. Exploit evenness first; for the shifted function use the Fourier shift identity rather than re-integrating. (a) Since ϕ is even, X a ϕ(x) = 0 + a cos nx. 2 n n≥1 Direct integration gives Z 2 π/2 π a = (π/2 − x)dx = , 0 π 4 0 Z (cid:16) (cid:17) 2 π/2 2 nπ a = (π/2 − x) cos nx dx = 1 − cos . n π πn2 2 0 Hence π X∞ 2(1 − cos(nπ/2)) ϕ(x) = + cos nx. 8 πn2 n=1 Now X (cid:2) (cid:3) π ψ(x) = −ϕ(x − π/2) = − − a cos nx cos(nπ/2) + sin nx sin(nπ/2) . 8 n n≥1 (b) From F{e−|x|} = 2/(1 + ω2) and duality, (cid:26) (cid:27) 1 F (ω) = πe −|ω| . 1 + x2 The requested integral is the convolution cos(2·) ∗ e−|·|. In the transform domain, 2 2π gb(ω) = π[δ(ω − 2) + δ(ω + 2)] = [δ(ω − 2) + δ(ω + 2)]. 1 + ω2 5 Inverting, 2 g(x) = cos 2x. 5 Marking points. Sketch/evenness 2; coefficients 4; shift/reflection 4; duality 4; convolution/delta/inverse 6. Common errors. Writing a as the constant term instead of a /2; failing to expand cos(n(x − π/2)); losing the factor 2π in the 0 0 convolution theorem. R Self-check. At x = 0, the integral is positive and less than e−|u|du = 2; 2/5 is plausible. Page 2 of 5

Question 3: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. Euler–Cauchy means use z = log x; a repeated eigenvalue with one eigenvector means a Jordan chain. (a) Put Y (z) = y(ez). Then xy′ = Y ′ and x2y′′ = Y ′′ − Y ′, so Y ′′ − 4Y ′ + 4Y = 6z + e2z. The homogeneous operator is (D − 2)2, so Y = (c + c z)e2z, y = x2(c + c log x). c 1 2 c 1 2 The Wronskian of e2z, ze2z is e4z ̸= 0. For 6z, try az + b: substitution gives a = 3/2, b = 3/2. For the resonant e2z term, try cz2e2z; (D − 2)2[cz2e2z] = 2ce2z, so c = 1/2. Therefore 3 3 1 y = x2(c + c log x) + log x + + x2(log x)2. 1 2 2 2 2 (b) The characteristic polynomial is (λ − 1)2. An eigenvector is v = (1, −1)T . Choose w = (1/2, 0)T so (A − I)w = v. Then X(t) = et{c v + c (tv + w)}. 1 2 The origin is a defective unstable node; span{v} is invariant and all generic trajectories become parallel to it while moving away from the origin. Marking points. Change of variable 3; CF/Wronskian 3; two PI components 4; final form 1; Jordan data 5; phase portrait 4. Common errors. Missing the −Y ′ term in x2y′′; failing to multiply a resonant ansatz by z2; using a second eigenvector when none exists. Self-check. Applying (D − 2)2 to z2e2z/2 must return exactly e2z. Page 3 of 5

Question 4: complete classroom-method solution Difficulty budget for this question: A6, B4, C2, D3, Mastery5 = 20. Recognition signal. Factor the autonomous vector field. At branch intersections, compare stability on both sides before naming the bifurcation. (a) The branches are √ y = r/2, y = ± −r (r ≤ 0). 1 2,3 With F = (r − 2y)(r + y2), F = −2(−ry + r + 3y2). y √ √ √ √ On y = r/2, F = −r(r + 4)/2; on y = −r it is 2r( −r + 2); on y = − −r it is −2r( −r − 2). Stability follows y from F < 0. The line branch intersects a square-root branch at r = −4 and exchanges stability: transcritical. At r = 0 the y symmetric branches meet the line branch in a supercritical pitchfork (with the branch orientation read from decreasing r). (b)(i) (cid:18) (cid:19) (cid:18) (cid:19) 2u −2v 1 2u 2v J = , J −1 = . 2v 2u 4(u2 + v2) −2v 2u The off-diagonal entries prove (u ) = −(v ) . y x x y (ii) The zero contour is the circle x2 + y2 = 2 together with the line y = 1/2. The gradient equations give √ (0, 1), (0, −2/3), (± 7/2, 1/2). For (cid:18) (cid:19) 2y − 1 2x H = , 2x 6y − 1 (0, 1) is a minimum, (0, −2/3) a maximum, and the two points on y = 1/2 are saddles. The contour plot confirms closed curves around the extrema and sign changes across both zero components. Page 4 of 5 Marking points. Branches/stability/classification 9; Jacobian identity 5; zero set/stationary/Hessian/contours 6. Common errors. Treating every branch meeting as pitchfork; interchanging the coordinates of the saddle points; classifying by trace alone when the determinant is negative. Self-check. The two saddle points must lie at the intersections of the circle and the line y = 1/2. Page 5 of 5