MATH40004 2026 Practice Paper 8

MATH40004 Calculus and Applications Original 2026 Resit Training Material Question 1 - Complete solution (Total: 20 marks) (a) If f has n + 1 derivatives on an interval containing a and x, then f(x) = n∑ k=0 f (k)(a) k! (x−a)k + f (n+1)(ξ) (n + 1)! (x−a)n+1, where ξlies between a and x. [2 marks] Suggested level: A (b) (i) From the exponential series, g(x) = 1 2 + x 3! + x2 4! +···= ∞∑ m=0 xm (m + 2)!. Thusg(0) = 1/2, and the radius of convergence is infinite. [4 marks] Suggested level: A/B (ii) As x→−∞, ex is negligible and g(x)∼(−x−1)/x2→0+. Also g(0) = 1/2, while g(x)→∞as x→∞. The strict convexity inequality ex > 1 +x for x̸= 0 shows g(x) > 0; at 0 the extension is 1 /2 > 0. [3 marks] Suggested level: B (iii) Use terms through degree 7: S7 = 7∑ k=0 (1/2)k k! = 354541 215040 = 1.648721168... The Lagrange remainder satisfies |R7|≤e1/2 (1/2)8 8! < 1.60×10−7 < 0.5×10−6. Hence e1/2 = 1.648721 to six decimal places. [4 marks] Suggested level: C (iv) Leibniz’ rule applied tox2g gives the stated identity because derivatives of x2 above order 2 vanish. Since x2g =ex−1−x is entire, the identity (or the power series) shows that g has derivatives of every order at 0, with g(n)(0) = n! (n + 2)! = 1 (n + 1)(n + 2) . [3 marks] Suggested level: C/D (c) Fora> 0, 0< 1−e−a <a . Thus (an) is positive and strictly decreasing, so it tends to L≥0. Passing to the limit gives L = 1−e−L. The function 1 −e−L−L is 0 at 0 and strictly decreasing for L >0, hence L = 0. [2 marks] Suggested level: D (d) Near 0, g(x)e−2x is bounded. As x→∞, g(x)e−2x∼e−x x2 , which is integrable. Therefore I exists. [2 marks] Suggested level: C 2

MATH40004 Calculus and Applications Original 2026 Resit Training Material Question 2 - Complete solution (Total: 20 marks) (a) (i) Since V =πdR2, 2πdRR′=−kπR2−2πqR, so, while R> 0, R′=−k 2dR−q d . Because k×area and q×length are volume rates, [k] =L/T, [q] =L2/T. [4 marks] Suggested level: A (ii) Put α=k/(2d) and β=q/d. Solving R′+αR=−βgives R(t) = ( R0 + β α ) e−αt−β α= ( R0 + 2q k ) e−kt/(2d)−2q k . The extinction time is te = 2d k log ( 1 +kR0 2q ) . [5 marks] Suggested level: B (iii) As k→0, R(t)→R0−(q/d)t, so ts = dR0 q . Since log(1 +s)<s for s> 0, te < 2d k kR0 2q =ts. The additional surface-loss mechanism shortens the lifetime. [3 marks] Suggested level: C (b) The function is odd and continuous under periodic extension. Therefore it has a sine series. For n≥1, bn = 2 π ∫ π 0 x(π−x) sin(nx)dx = 4(1−(−1)n) πn3 . Hence f(x) = 8 π ∞∑ m=0 sin((2m + 1)x) (2m + 1)3 . The series converges to f(x) at every real x because the periodic function is continuous. [6 marks] Suggested level: A/B Atx =π/2, f(π/2) =π2/4 and sin((2m + 1)π/2) = (−1)m, so π2 4 = 8 π ∞∑ m=0 (−1)m (2m + 1)3, which gives the required π3/32. [2 marks] Suggested level: B/C 3

MATH40004 Calculus and Applications Original 2026 Resit Training Material Question 3 - Complete solution (Total: 20 marks) (a) Let A = ( 2 1 −1 0 ) . (i) The characteristic polynomial is ( λ−1)2. An eigenvector is v = ( 1 −1 ) . Choose w = (1, 0)T , for which (A−I)w =v. Therefore X(t) =et [c1v +c2(tv +w)] . [5 marks] Suggested level: B/C (ii) On the y-axis, A (0 y ) = (y 0 ) , so the vector field is horizontal. The origin is a defective unstable node. Every non-zero trajectory moves away from the origin; as t→∞it becomes parallel to the eigenline y =−x (and if c2 = 0 it lies on that line exactly). [4 marks] Suggested level: A/B (iii) The forcing is etv. Since (D−A)(tetv) =etv, a particular integral is Xp =tetv. Thus X(t) =et [c1v +c2(tv +w) +tv] . [4 marks] Suggested level: D (b) Let M = 2xy + 2y2 +y and N =x + 2y. (i) My = 2x + 4y + 1̸= 1 =Nx, so the equation is not exact. [2 marks] Suggested level: A (ii) My−Nx N = 2x + 4y x + 2y = 2, so an integrating factor is µ(x) =e2x. A potential is obtained from Uy =e2x(x + 2y), namely U =e2x(xy +y2) +h(x). Comparison with Ux =e2xM givesh′= 0. Hence e2x(xy +y2) =C. [5 marks] Suggested level: B/C 4

MATH40004 Calculus and Applications Original 2026 Resit Training Material Question 4 - Complete solution (Total: 20 marks) (a) Write the fixed-point condition as r =R(x) :=x + 1 x−1, x ̸= 1. Equivalently, x±= r + 1± √ (r + 1)(r−3) 2 , which are real for r≤−1 or r≥3. The singular point is always x = 1. Since fx =−R′(x), R ′(x) = 1− 1 (x−1)2, a fixed point is stable for x< 0 or x> 2, and unstable for 0 <x< 1 or 1 <x< 2. Thus for r< −1 there is one stable point below 0 and one unstable point in (0 , 1); for−1<r< 3 there are none; and for r> 3 there is one unstable point in (1, 2) and one stable point above 2. At r =−1 and r = 3 the double points are half-stable. [5 marks] Suggested level: A/C (b) The turning points of R(x) are (r,x) = (−1, 0) and (3, 2), and both are saddle-node bifurcations. For r = 4, xu = 5− √ 5 2 , x s = 5 + √ 5 2 . The basin of the stable point xs is (5− √ 5 2 ,∞ ) . Initial data below xu move towards the singular barrier x = 1 (from the appropriate side). [4 marks] Suggested level: D (c) Fourier transformation gives (ω4 + 5ω2 + 4)ˆy = 1, ˆy = 1 (ω2 + 1)(ω2 + 4) = 1 3 ( 1 ω2 + 1− 1 ω2 + 4 ) . UsingF−1{(ω2 +a2)−1}=e−a|x|/(2a), y(x) = 1 6e−|x|−1 12e−2|x|. [5 marks] Suggested level: B/D (d) For ∆ = ( x,y )T , z =z(0, 0) +∇z(0, 0)·∆ + 1 2∆ THz(0, 0)∆ + o(∥∆∥2). Let F =ez +xz +x + 2y +xy−1. At the origin, Fz = 1, and implicit differentiation gives zx =−1, z y =−2. Differentiating once more, zxx = 1, z xy =−1, z yy =−4. Therefore z(x,y ) =−x−2y + 1 2x2−xy−2y2 +o(x2 +y2) . [6 marks] Suggested level: A/D 5