MATH40002 2026 Practice Paper 3

MA TH40002 Analysis 1 Tony 2026 Summer Resit 1 Question 1: complete solution and marking guide Let rn = ∑∞ k=n ak, so rn → 0 and aj = rj − rj+1. Summation by parts gives n∑ j=1 jaj = n∑ j=1 rj − nrn+1. Hence the required expression is the Cesaro mean of rj minus rn+1, and both tend to 0. [5] For Sm = ∑m k=1 ank , non-negativity gives 0 ≤ Sm ≤ ∑nm n=1 an ≤ ∑∞ n=1 an. Thus (Sm) is increasing and bounded. [4] A direct calculation gives bn+1 − bn = 1 2n + 1 − 1 2n + 2 ≥ 0, so (bn) is increasing. Also bn ≤ n · 1/(n + 1) < 1. Hence it converges. [5] Since cn = rn − rn+1, cn√rn = (√rn − √rn+1) √rn + √rn+1 √rn ≤ 2(√rn − √rn+1). Comparison with the telescoping series proves convergence. [6] Tony proof and problem-solving template module 1. Template used in this question Series tails/summation by parts + positive-term subseries + monotone bounded sequences + telescoping comparison using square roots. 2. Recognising the method When ∑an converges and jaj appears, use tails or Abel transformation; for a positive-term subsequence, compare partial sums; when the tail rn appears, write cn = rn − rn+1。 3. Plain-language explanation These questions turn global convergence into controlled partial sums or tails, followed by monotone boundedness or telescoping. 4. General structure for use in an examination Weighted sum: define the tail rn and sum by parts; positive subseries: bound its partial sums by the original sum; sequence: prove monotonicity + boundedness; for a tail in the denominator, factor the difference into a difference of square roots. 5. Applying the template to this question The four parts correspond to four standard starting points; the square-root factorisation in (d) is especially central. 6. Variations on this template Weights can be changed to Cesaro averages and tails to other powers. For mixed signs, nonnegativity cannot be used directly to compare subseries. 7. Common errors and securing method marks Omitting −nrn+1 in summation by parts; ignoring positivity; failing to state the direction of telescoping comparison. To secure method marks, define the tail and state monotone boundedness. 2 MA TH40002 Analysis 1 Tony 2026 Summer Resit 8. Thirty-second recall card Weights in a convergent series → tails/summation by parts; positive subseries → upper bound for partial sums; differences of tails → telescoping square-root differences. 9. A minimal variation exercise If rn = ∑∞ j=n cj, express cn. Brief answer: cn = rn − rn+1. 3

MA TH40002 Analysis 1 Tony 2026 Summer Resit 2 Question 2: complete solution and marking guide The definition is: for every ε > 0 there is δ > 0 such that 0 < |x − a| < δ implies |f (x) − L| < ε. If the epsilon-delta limit holds and xn → a, choose N so that |xn − a| < δ for n ≥ N ; then f (xn) → L. Conversely, if the epsilon-delta statement fails, there is ε0 > 0 such that for every n one can choose xn with 0 < |xn − a| < 1/n and |f (xn) − L| ≥ ε0. Then xn → a but f (xn) ̸→ L, contradiction. [8] At any x, choose rational qn → x and irrational rn → x. The image sequences are constantly 1 and 0, so sequential continuity fails. [4] Let h(x) = f (x) − x. Then h(0) ≥ 0 and h(1) ≤ 0. IVT gives a zero. [4] Heine–Cantor: a continuous function on a compact set is uniformly continuous. Since [a, b] is compact, the con- clusion follows. [4] Tony proof and problem-solving template module 1. Template used in this question Epsilon-delta and the sequential criterion + two dense sequences + an IVT fixed point + Heine-Cantor. 2. Recognising the method For “iff every sequence”, prove both directions: use the definition forwards and select points within 1/n from failure of the condition backwards; for a fixed point set h = f − x. 3. Plain-language explanation The sequential criterion turns a function limit into sequence limits; the key reverse step is to choose one bad point for every n. 4. General structure for use in an examination Forward: given epsilon choose delta, then obtain N from xn → a. Reverse: assume epsilon-delta fails and choose 0 < |xn − a| < 1/n with an error bounded away from zero. For IVT, construct a difference function and examine its endpoints. 5. Applying the template to this question Use two dense sequences for the Dirichlet function; for the fixed point use h(0) ≥ 0, h(1) ≤ 0. 6. Variations on this template Variations include continuity, function limits and one-sided limits. The reverse construction must ensure xn ̸= a. 7. Common errors and securing method marks Reversing quantifiers; proving only one direction of the sequential criterion; omitting continuity for IVT. To secure method marks, write the bad-point construction and the difference function. 8. Thirty-second recall card Sequential criterion: use delta forwards and choose a bad point within 1/n backwards; fixed point: h = f − x has opposite endpoint signs; continuity on a compact interval: Heine-Cantor。 4 MA TH40002 Analysis 1 Tony 2026 Summer Resit 9. A minimal variation exercise Write the first step in constructing a sequence from “not continuous at a”. Brief answer: there is ε0 > 0 such that for every n one can choose xn with |xn − a| < 1/n and |f (xn) − f (a)| ≥ ε0. 5

MA TH40002 Analysis 1 Tony 2026 Summer Resit 3 Question 3: complete solution and marking guide (i) FALSE. Take f (x) = sin(x2). It is bounded. Let xn = √ 2πn + π/2, y n = √ 2πn + 3π/2. Then |xn − yn| = π/(xn + yn) → 0, while |f (xn) − f (yn)| = 2. Hence f is not uniformly continuous. [3] (ii) TRUE: if |f ′| ≤ M , MVT gives |f (x) − f (y)| ≤ M |x − y|. [3] (iii) FALSE: define f (x) = 0 for −1 ≤ x < 1 and f (1) = 1 ; it is convex under the endpoint convention used in the course but discontinuous at 1. [3] (iv) TRUE: the limits bound both tails; continuity and EVT bound the remaining compact interval. [3] Non-existence: for every L ∈ R there is εL > 0 such that for every δ > 0 some x with 0 < |x − a| < δ satisfies |f (x) − L| ≥ εL. [2] f ′′(x0) exists if f is differentiable on a neighbourhood of x0 and f ′ is differentiable at x0. [3] MVT: if f is continuous on [a, b] and differentiable on (a, b), some c ∈ (a, b) satisfies f ′(c) = ( f (b) − f (a))/(b − a). [3] Tony proof and problem-solving template module 1. Template used in this question Global properties + a collection of counterexamples + precise definitions. 2. Recognising the method Boundedness/UC/convex endpoints/finite limits frequently occur in true-or-false questions. First check for missing hypotheses such as continuity, compactness or endpoint continuity in the statement. 3. Plain-language explanation One missing hypothesis often determines whether a statement is true. 4. General structure for use in an examination For a false statement, choose the simplest counterexample; for a true one, give a theorem; state definitions with precise quantifiers/neighbourhoods. 5. Applying the template to this question This question uses sin x2, MVT, an endpoint jump, and a tail + compact interval argument. 6. Variations on this template Variations include continuous but unbounded, uniformly continuous but unbounded, and the intermediate value property without continuity. A counterexample must meet every hypothesis. 7. Common errors and securing method marks A convex function is continuous on an open interval but may jump at endpoints of a closed interval: a common trap. To secure method marks, first state true/false and give the counterexample function. 8. Thirty-second recall card Look for missing hypotheses → for true statements state a theorem, and for false ones give a minimal counterexample; keep the quantifiers in order in definitions. 6 MA TH40002 Analysis 1 Tony 2026 Summer Resit 9. A minimal variation exercise Does “continuous and bounded” necessarily imply UC on R? Brief answer: no; for example, sin(x2). 7

MA TH40002 Analysis 1 Tony 2026 Summer Resit 4 Question 4: complete solution and marking guide The backward implication is immediate. Conversely, if f (x0) > 0, continuity gives an interval I around x0 on which f ≥ f (x0)/2. By monotonicity and additivity of the integral, ∫ b a f ≥ |I|f (x0)/2 > 0, contradiction. [5] Let A(x) and B(x) denote the two sides. Leibniz/FTC gives A′(x) = ∫ x 0 f (t)dt = B′(x), and A(0) = B(0) = 0 . Hence A = B. [5] If f has a positive maximum at an interior point x0, then f ′(x0) = 0 , f ′′(x0) ≤ 0, and the equation gives f (x0) = f ′′(x0) ≤ 0, contradiction. A negative minimum is similarly impossible. Since endpoint values are zero, f ≡ 0. [5] Let discontinuities be c1, . . . , cm and |f | ≤ M . Put the bad points in intervals of total length < ε /(4M ). On the compact complement, f is uniformly continuous; choose a partition so oscillation there contributes < ε /2. The bad intervals contribute at most 2M times their total length, also < ε/2. Thus U − L < ε . [5] Tony proof and problem-solving template module 1. Template used in this question Nonnegative continuous functions with zero integral + variable-upper-limit integral identities + the maximum principle + integrability with finitely many discontinuities. 2. Recognising the method For f ≥ 0 and integral 0, use a contradiction from local positivity; for two integral expressions, compare derivatives and initial values; for an ODE with zero endpoints, use a positive maximum/negative minimum; isolate finitely many bad points in small intervals. 3. Plain-language explanation All four proofs reduce a global conclusion to local information: near a positive point, derivatives, an extremum, or neighbourhoods of bad points. 4. General structure for use in an examination Zero integral: assume f (x0) > 0 for contradiction; identity: differentiate both sides and compare initial values; maximum principle: substitute an interior extremum; integrability: give the bad points small total interval length and use uniform continuity on the good set. 5. Applying the template to this question The opening line of each part almost determines half the marks. 6. Variations on this template Variations include integral test functions, different ODE signs and countably many discontinuities; the finite-point proof cannot simply be reused for countably many points. 7. Common errors and securing method marks Not stating FTC/Leibniz; overlooking that a positive maximum is interior because endpoints are zero. To secure method marks, give a locally positive interval, define A/B and differentiate, and state the derivative signs at extrema. 8. Thirty-second recall card Nonnegative function with zero integral → contradiction from local positivity; integral identity → equal derivatives and initial values; ODE with two zeros → positive maximum/negative minimum; finitely many discontinuities → small intervals around bad points. 8 MA TH40002 Analysis 1 Tony 2026 Summer Resit 9. A minimal variation exercise If continuous f ≥ 0 and f (x0) = c > 0 at some point, what bound should be established first? Brief answer: by continuity choose δ > 0 so that |x − x0| < δ implies f (x) > c/2. 9