Converse assumed

The theorem runs one way. You used it in the other.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option B

  $L^{1}[0,1] \subseteq L^{2}[0,1]$

  Why this is wrong, and why it looks right

  The reverse fails: f(x) = $x^{-1/2}$ has $\int_{0}^{1}$|f|dx = 2, so $f \in L^{1}[0,1]$, but $\int_{0}^{1}$|f|$^{2}$dx $= \int_{0}^{1}x^{-1}dx$ diverges, so $f \notin L^{2}[0,1]$. Finite measure only buys containment toward the higher exponent, not both ways.

  From this question

  Which of the following are true?

  Part C · free

• Option C

  The p-value is the probability that $H_{0}$ is true given the observed data.

  Why this is wrong, and why it looks right

  The p-value is P(data this extreme or more | $H_{0}$ true) — a probability about the data computed assuming $H_{0}$, not $P(H_{0}$ | data). Converting one to the other needs Bayes' theorem and a prior on $H_{0}$, which the p-value calculation never uses.

  From this question

  For testing $H_{0} : \theta = \theta_{0}$ against $H_{1} : \theta = \theta_{1}$ at level $\alpha$, which of the following are true?

  Part C · free

• Option B

  $-n/\sum \ln X_{i}$

  Why this is wrong, and why it looks right

  False — this estimates $\theta$, not $1/\theta. -n/\sum \ln X_{i}$ is the reciprocal of the correct estimator and is in fact the MLE of $\theta$. Reciprocals do not commute with expectation, so an unbiased estimator of $1/\theta$ cannot simply be inverted to give one for $\theta$.

  From this question

  For n ≥ 2, let $X_{1}, \dots, X_{n}$ be a random sample with density f(x|$\theta) =$ $\theta{}x^{\theta-1}$ on $0 < x < 1, \theta > 0$ unknown. Which of the following is the UMVUE for $1/\theta$?

  See the full question

  Part B · december 2023 · free

• Option B

  G is a cyclic group.

  Why this is wrong, and why it looks right

  False — 'need not be' is the point. P has order $17^{2}$ and may be $\mathbb{Z}_{17} \times \mathbb{Z}_{17}$ rather than $\mathbb{Z}_{289}$, in which case $G \cong \mathbb{Z}_{7} \times \mathbb{Z}_{17} \times \mathbb{Z}_{17}$ has no element of order 2023. Abelian does not imply cyclic once a prime appears squared; $\mathbb{Z}_{7} \times \mathbb{Z}_{289}$ is cyclic but it is not the only option.

  From this question

  Let G be a group of order 2023. Which of the following statements are true?

  See the full question

  Part C · june 2023 · free

• Option C

  f′ is continuous

  Why this is wrong, and why it looks right

  False: $f(x)=x^{2}\sin(1/x) (f(0)=0)$ is differentiable everywhere, but f′(x)=2x sin(1/x)−cos(1/x) for x≠0 oscillates with no limit as x→0, while f′(0)=0 — so f′ is discontinuous at 0 despite having the intermediate value property from option (0).

  From this question

  Let $f : \mathbb{R} \to \mathbb{R}$ be differentiable. Which of the following are true?

  Part C · free

• Option B

  $X_{n} \to X$ in probability ⟹ $X_{n} \to X$ almost surely.

  Why this is wrong, and why it looks right

  This assumes the converse of the true implication in option (a). The classic counterexample is a 'moving window' sequence on [0,1] — indicators of shrinking, cyclically sweeping intervals — which converges to 0 in probability, yet since some window always contains any fixed point, never converges there almost surely.

  From this question

  Let $X_{1}, X_{2}, \dots$ and X be random variables on a common probability space. Which of the following implications hold in general?

  Part C · free

• Option D

  $X_{n} \to X$ in distribution ⟹ $X_{n} \to X$ in probability.

  Why this is wrong, and why it looks right

  Convergence in distribution only constrains the marginal law of $X_{n}$, saying nothing about how $X_{n}$ and X are jointly related on the same probability space, so it generally can't force |$X_{n}-X$|→0 in probability. The implication does hold in the special case where X is almost surely a constant, but not otherwise.

  From this question

  Let $X_{1}, X_{2}, \dots$ and X be random variables on a common probability space. Which of the following implications hold in general?

  Part C · free

• Option B

  (x) is a maximal ideal.

  Why this is wrong, and why it looks right

  $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$, which is a domain but not a field, so (x) is prime but not maximal. It sits properly inside the strictly larger ideal (2,x), and primality only coincides with maximality in special rings like PIDs — not in $\mathbb{Z}[x]$.

  From this question

  Consider the ring $\mathbb{Z}[x]$. Which of the following are true?

  Part C · free

• Option C

  P(diseased | positive) > 1/2.

  Why this is wrong, and why it looks right

  P(D|+) = 1/6 ≈ 0.167, well below 1/2. The test's high sensitivity and specificity describe P(test | disease), not P(disease | test); with only 1% prevalence, the 4.95% of healthy people who false-positive outnumber the 0.99% of diseased people who true-positive.

  From this question

  A test for a disease has sensitivity 0.99 and specificity 0.95, and the disease affects 1% of the population. Which of the following are true?

  Part C · free

• Option B

  If P(A | B) = 1 then P(B | A) = 1.

  Why this is wrong, and why it looks right

  False, and here the symmetry genuinely breaks. P(A | B) = 1 means P(A ∩ B) = P(B), that is B is contained in A up to null sets. Then P(B | A) = P(B)/P(A), which is less than 1 whenever P(B) < P(A). Containment runs one way only.

  From this question

  Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?

  See the full question

  Part C · june 2023 · free

• Option D

  If P(A | B) > P(B) then P(B | A) > P(A).

  Why this is wrong, and why it looks right

  False. Written out, the hypothesis is $P(A \cap B) > P(B)^{2}$ and the conclusion $P(A \cap B) > P(A)^{2}$, which are different statements. Take B ⊂ A with P(A) = 9/10 and P(B) = 1/10: then P(A | B) = 1 > 1/10 = P(B), so the hypothesis holds, but P(B | A) = 1/9 (verified) and SymPy confirms 1/9 − 9/10 < 0, so the conclusion fails. Compare with (C): the same shape of statement, but comparing against the other event's probability destroys the symmetry.

  From this question

  Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?

  See the full question

  Part C · june 2023 · free

• Option D

  If $\nabla{}u \cdot \nabla{}v = 0$ at every point, then f is an entire function.

  Why this is wrong, and why it looks right

  False — it is (C) read backwards. Take f(z) = z̄, so u = x and v = −y: SymPy gives $\nabla{}u \cdot \nabla{}v = 1\cdot0 + 0\cdot(-1) = 0$, yet f is nowhere holomorphic. Orthogonal gradients are one consequence of Cauchy–Riemann, not equivalent to them.

  From this question

  Let $f : \mathbb{C} \to \mathbb{C}$ be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let $\nabla{}u = (u_{x}, u_y)$ denote the gradient. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023 · free

• Option D

  If $(a_{1} + \cdots + a_{n})/n$ converges then $(a_{n})$ converges

  Why this is wrong, and why it looks right

  False: $a_{n}=(-1)^{n}$ has Cesàro means $(a_{1}+\cdots+a_{n})/n \to 0$, but $(a_{n})$ itself never converges. Cesàro summability only follows FROM convergence, not the other way round.

  From this question

  Let $(a_{n})$ be a sequence of real numbers. Which of the following are true?

  Part C · free

• Option D

  If B(v, w) = 0 then either v = 0 or w = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  B(x, x) ≠ 0 for all nonzero $x \in \mathbb{R}^{4}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  deg h ≥ n for every nonzero $h \in \ker \varphi$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  nowhere

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  All Manuals are Booklets.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · june 2025

• Option B

  All Catalogues are Booklets.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · june 2025

• Option D

  All Catalogues are Manuals.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · june 2025

• Option A

  f is differentiable everywhere and f′ is continuous

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  f is differentiable everywhere and f′ is bounded on [−1, 1] but f′(0) ≠ 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  undefined, since the series diverges

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  path-connected but not connected

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  $\lim(y\to+\infty)$ |f′(iy)| $= +\infty$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  If F(x) = 0 for all x ∈ [0, 1], then f(x) = 0 for all x ∈ [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  lim(x→0) f(x,0) = lim(y→0) f(0,y).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  8/25

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  $p_{0} > p_{3}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option B

  $X_{n}$ converges to X in probability as $n \to \infty$ implies that $X_{n}$ converges to X almost surely as $n \to \infty$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  There exists $(a, b) \in \mathbb{R}^{2}$ such that h_(a,b) is not differentiable at t = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  The function f is differentiable at the point (0, 0).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  no professionals are teachers

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · december 2025

• Option C

  f must be uniformly continuous.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  If |X| divides n, then G is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  population of young people is going to increase with a faster rate in the long run if the same status continues

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · december 2025

• Option D

  proportion of elderly people is going to decrease in the long run if the same status continues

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · december 2025

• Option A

  Anwar was the most senior

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · december 2025

• Option C

  S ~ $W_{12}(19, 19\sum)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  G has a subgroup of order 6.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  $A_{4}$ has a subgroup of order 6

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  (P) is FALSE, (Q) is true.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  Both (P) and (Q) are FALSE.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  If $a_{n}^{1/n}$ → L then $a_{n}$₊$_{1}/a_{n} \to L$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  (0.95X, X)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option A

  1/64

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  rank A = 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  rank A = 1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option A

  $Var(\beta$̃$_{1}) > Var(\beta$̃$_{2}), E(\beta$̃$_{1}-\beta)^{2} > E(\beta$̃$_{2}-\beta)^{2}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  The maximum likelihood estimator of $\theta$ is min{$-2X_{(1)}/3, 2X_{(n)}$}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  $X_{n} \to X$ almost surely

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  f maps an open neighbourhood of (0, 0) in $\mathbb{R}^{2}$ onto an open subset of $\mathbb{R}^{3}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  If A is one-to-one but not onto, then m > n

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option B

  If A is onto but not one-to-one, then m < n

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  If A is one-to-one, then m = n

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option C

  P(Y<1/2)={S(1/2)}$^{4}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  A is symmetric.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  If A is a real 3×2 matrix and B a real 2×3 matrix with BA $= I_{2}$, then AB $= I_{3}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option A

  There is no continuous real-valued function g defined on any interval of $\mathbb{R}$ containing 0 such that f(x, g(x)) = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option D

  $u(\pi) - u'(\pi) = e^\pi$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  S has exactly one element

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  There exists a holomorphic surjective map $f : \Omega_{2} \to \Omega_{1}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  There exists a holomorphic injective map $f : \Omega_{2} \to \Omega_{1}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  $y(x) = (1/2)(x^{2} - x + 2)$ is an extremal for $J_{1}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option B

  Ramesh

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · june 2025

• Option C

  a is not divisible by 48.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  f ∘ g (when defined)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  A is symmetric

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  A is bounded above but not bounded below.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  There exists a real invertible matrix P such that PAP⁻$^{1}$ is a diagonal matrix and Pᵗ$P = I_{3}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  A has only real eigenvalues and it is not diagonalizable over $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  Statement 1

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · september 2022

• Option B

  H is normal only if G is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  If |1 − āz| < |z − a|, then |z| < 1

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  If A is diagonalisable then A is invertible

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  There does not exist $(s_{1}, s_{2}) \in \mathbb{R}^{2}$ such that $q_{2}(s_{1}, s_{2}) = -5$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  Suppose that there exists a subspace W of $\mathbb{C}^{n}$ such that $T(W) \subseteq W, \mathbb{C}^{n} = V + W$ and V ∩ W = {0}. Then there exists a positive integer k such that Tᵏ is the identity map.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  $L^{1}(\mu) \subseteq L^{2}(\mu)$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  The preimage of a connected set under p is a connected set.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Closed and bounded ⇒ compact
• Bounded ⇒ totally bounded
• Connected ⇒ path-connected
• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Differentiable ⇒ continuously differentiable

Other mistakes that cost marks