Question 1
each question on a new sheet. Give complete reasons and state theorem hypotheses.
Question 1
(a) Prove that the set of functions f : Z2 → Z2 which are periodic in the first variable is
uncountable. [4 marks]
(b) Let A ⊂ (0, ∞) be non-empty and bounded above, and B ⊂ (1, ∞) non-empty. Prove
sup A
sup{a/b : a ∈ A, b ∈ B} = .
inf B
[5 marks]
(c) If a → a and b → b, prove from first principles that max(a , b ) → max(a, b). [5
n n n n
marks]
(d) Let a ≥ 0 and a ≤ a + 1/n2. Prove that (a ) converges. [6 marks]
n n+1 n n
1
Worked solution and marking guidance
Question 1 - Complete course-method solution
Historical official basis:2022 Q1(b); 2021 Q2(c); 2022 Q3(b/c).
Course-method evidence:official solutions and marker comments in the uploaded 2020–
2026 papers; method families: countability_diagonal; supremum_infimum; epsilon_N; mono-
tone_bounded_recurrence.
Recognition signal
Use an injection for uncountability of a function space; for the supremum of quotients first bound and then approximate; separate cases for a maximum limit; when upward errors are summable add the tail to construct a monotone
sequence.
First key step
Construct fg; choose sequences approaching sup/inf; assume a ≥ b without loss of generality; define tn as the tail of a p-series.
(a) For every functiong : Z ! Z, define fg(i, j) = (0, g(j)). Then fg(i + 1, j) = fg(i, j), and g 7! fg
is injective. The set of all g : Z ! Z is uncountable by diagonalisation.
(b) The displayed number is an upper bound becausea ≤ supA and b ≥ infB > 0. Choose
an ! supA from A and bn ! infB from B; then an/bn tends to the displayed number , proving
leastness.
(c) Assume a ≥ b. Given ε > 0, eventually both jan − aj < ε and jbn − bj < ε . If an ≥ bn, the
maximum isan. Otherwise bn > an > a − ε and bn < b + ε ≤ a + ε. Thus the maximum lies within
ε of a.
(d) Let tn = ∑∞
j=n 1/j2 and cn = an + tn. Then
cn+1 = an+1 + tn+1 ≤ an + 1/n2 + tn − 1/n2 = cn.
Thus (cn) is decreasing and bounded below by0, hence converges. Sincetn ! 0, an = cn − tn
converges.
Marking points
Uncountability: 4 marks; supremum of quotients: 5; maximum limit: 5; error-corrected sequence: 6.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Common errors
Reading the periodicity condition as simultaneous translation of both variables; failing to check whether inf B could be 0; using only composition with a continuous function for the maximum limit despite the requirement to work from first
principles; choosing the wrong sign in the corrected sequence.
Final self-check
cn must be nonincreasing and nonnegative; tn − tn+1 = 1/n2.
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 2
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 2
∑ ∑
(a) ∑Power series a
n
zn and b
n
zn have radii R
1
, R
2
∈ (0, ∞). Prove that the radius of
a b zn is at least R R . [6 marks]
n n 1 2
∑ ∑
(b) If a converges, prove that a /n2 converges. [3 marks]
n n
∑ ∑ ∑
(c) Let a > 0 and a < ∞. Put b = n −1 n a . Prove that b diverges. [4 marks]
n n n j=1 j n
(d) Suppose a → a > 0 and fix k ≥ 1.
n
( )
(i) Prove a · · · a 1/k → a. [3 marks]
n+1 n+k
(ii) Give a divergent positive sequence for which these geometric means converge
when k = 2. [2 marks]
(iii) State the alternating series test. [2 marks]
2
Worked solution and marking guidance
Question 2 - Complete course-method solution
Historical official basis:2021 Q1(e); 2022 Q1(c)(ii); 2021 Q3(d); 2020 Q3(a).
Course-method evidence:official solutions and marker comments in the uploaded 2020–
2026 papers; method families: power_series_radius; series_partial_sums; alternating_dirichlet.
Recognition signal
For the radius of the coefficientwise product, use two interior radii and geometric comparison; terms of a convergent series are bounded; a positive-term average includes at least the first-term harmonic lower bound;
squeeze the geometric mean of finitely many consecutive factors.
First key step
Choose x < R1, y < R2 and r < xy; find a uniform bound for janj; write bn ≥ a1/n.
(a) Fix r < R1R2. Choose x < R1, y < R2 with r < xy. Since the two series converge absolutely
at x, y, the sequencesjanjxn and jbnjyn are bounded byM, N. Thus
janbnjrn ≤ M N
( r
xy
) n
,
so the product-coefficient series converges absolutely atr.
(b) A convergent series has bounded terms, sayjanj ≤ M. Then janj/n2 ≤ M/n2.
(c) Since bn ≥ a1/n, comparison with the harmonic series gives divergence.
(d)(i)For anyε 2 (0, a), all k factors eventually lie in(a−ε, a+ε), so their geometric mean lies
in the same interval. (ii) Take an = 1 for even n and 4 for odd n; adjacent geometric means
are 2. (iii) If un # 0, then ∑(−1)n−1un converges, and the remainder has absolute value at
most the first omitted term.
Marking points
Hadamard-product radius: 6 marks; weighted absolute comparison: 3; divergence of the series of averages: 4; geometric means and the alternating test: 7.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Common errors
Proving only that the general term tends to 0 rather than convergence of the power series; treating af(n) as though it were a subsequence; omitting the monotonicity condition of the alternating test.
Final self-check
The comparison ratio r/(xy) must be strictly below 1.
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 3
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 3
(a) Let f : R → R be non-decreasing. Prove that its left limit at every point exists and
equals sup{f (y) : y < x }. [5 marks]
0
(b) If f : [0, 1] → [0, 1] is continuous, prove it has a fixed point. [3 marks]
(c) If f : R → R is differentiable and even, prove f ′ (0) = 0. [3 marks]
′
(d) If lim x→∞ f (x) and lim x→∞ f (x) both exist and the first is finite, prove the second limit
is 0. [4 marks]
(e) Prove that x2 is not uniformly continuous on (0, ∞). [5 marks]
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Worked solution and marking guidance
Question 3 - Complete course-method solution
Historical official basis:2025 Q4(d); 2023 Q5(c); 2025 Q5(c/d); 2020 Q5(a)(iii).
Course-method evidence:official solutions and marker comments in the uploaded 2020–
2026 papers; method families: supremum_infimum; IVT_fixed_point; MVT_Rolle; deriva-
tive_infinity; uniform_continuity .
Recognition signal
Use sup for a monotone one-sided limit; a difference function for a fixed point; the chain rule for the derivative of an even function; if a function has a finite limit and its derivative has a limit,
argue by contradiction; use nearby large points to disprove uniform continuity of a polynomial.
First key step
Respectively write L = sup, h = f − x, f ′(x) = −f ′(−x), assume the derivative limit is nonzero, and choose n and n + 1/n.
(a) Let L be the stated supremum. Given ε > 0, choose y < x 0 with L − ε < f (y) ≤ L. For
y < x < x 0, monotonicity givesL − ε < f (x) ≤ L.
(b) Apply IVT to h(x) = f(x) − x, noting h(0) ≥ 0 and h(1) ≤ 0.
(c) Differentiatef(x) = f(−x) to obtain f ′(x) = −f ′(−x), then set x = 0.
(d) If f ′(x) ! ℓ > 0, then eventuallyf ′ > ℓ/2 and MVT/integration forcesf(x) ! 1. If ℓ < 0, it
forces f(x) ! −1. Hence ℓ = 0.
(e) Takexn = n and yn = n + 1/n. Their distance tends to 0, but y2
n − x2
n = 2 + 1/n2.
Marking points
One-sided limit: 5 marks; fixed point: 3; even function: 3; derivative limit: 4; failure of uniform continuity: 5.
Common errors
Assuming the one-sided limit already exists; using Rolle for an even function only gives a zero derivative somewhere, not necessarily at 0; assuming f ′ tends to 0 automatically by interchanging
limits.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Final self-check
For failure of uniform continuity, both points should tend to infinity while their distance tends to 0.
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 4
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 4
(a) Prove that a bounded function on [a, b] with finitely many discontinuities is Darboux
integrable. [5 marks]
∫
(b) If f : [−1, 1] → R is continuous, non-negative and 1 f = 0, prove f ≡ 0. [3 marks]
−1
(c) Let f ∈ C ∞ (−1, 1) and suppose |f (n)(x)| ≤ (n − 1)! for all n ≥ 1. Prove that its Taylor
polynomials at 0 converge uniformly to f . [4 marks]
(d) A C3 function meets the chord joining (a, f (a)) and (b, f (b)) at two distinct interior
points. Prove f (3)(c) = 0 for some c ∈ (a, b). [4 marks]
(e) Let f : [0, ∞) → R be continuous and f (x)/x → 1. Prove
∫
1 x 1
f (t) dt → .
x2 2
0
[4 marks]
Worked solution and marking guidance
Question 4 - Complete course-method solution
Historical official basis:2026 Q4(a); 2025 Q6(d); 2020 Q5(c); 2022 Q5(c)(ii); 2021 Q6(b)(ii).
Course-method evidence:official solutions and marker comments in the uploaded 2020–
2026 papers; method families: Darboux_sums; Taylor; MVT_Rolle; FTC_integral_identity .
Recognition signal
Darboux sums with finitely many bad points, nonnegative integrals, uniform Taylor remainders, subtracting a chord + Rolle, and asymptotics of integral averages are frequent methods from five different
years.
First key step
For integrability split into good and bad regions; for Taylor state a remainder bound; for the chord problem subtract the linear function; for integral asymptotics rewrite the limit as two-sided
inequalities.
(a) Enclose each discontinuity in intervals of arbitrarily small total length; on the remaining
compact set use uniform continuity to make oscillations small. Add the two contributions to
U − L.
(b) If f(x0) > 0, continuity gives f ≥ f(x0)/2 on a non-trivial interval, making the integral
positive.
(c) Taylor’s theorem gives, forjxj < 1,
jf(x) − Pn(x)j ≤ n!
(n + 1)!jxjn+1 ≤ 1
n + 1,
uniformly inx.
(d) Subtract the chord line to obtain a function with four zeros: the two endpoints and two
interior intersections. Repeated Rolle gives a zero of the third derivative; the chord has zero
third derivative.
(e) Given ε > 0, eventually (1 − ε)t ≤ f(t) ≤ (1 + ε)t. Integrate from a fixed N to x, absorb
the finite initial integral into a constant, divide byx2, and let x ! 1. The limit is squeezed
between(1 ± ε)/2.
Marking points
Integrability with finite discontinuities: 5 marks; nonnegative integral: 3; uniform Taylor convergence: 4; chord and Rolle: 4; integral asymptotics: 4.
4
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Common errors
An off-by-one error in (n − 1)! in the Taylor remainder; forgetting the endpoints are also chord intersections; ignoring the constant contribution from [0, N] in integral asymptotics.
Final self-check
The uniform Taylor error does not depend on x; after division by x2 the constant term tends to 0.
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
5