Hypothesis dropped

A theorem applied without checking one of its preconditions.

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 D

  2/5

  Why this is wrong, and why it looks right

  False — 2/5 is the maximum likelihood estimate, 10/25 (verified), which is the answer with no prior at all. The Bayes estimate always lies strictly between the prior mean 1/2 and the MLE 2/5, and 3/7 ≈ 0.4286 does exactly that. Seeing your answer sit outside that interval is a quick sanity check.

  From this question

  The probability of a head in tossing a coin is p ∈ (0, 1). The coin is independently tossed 25 times and heads appear 10 times. The Bayes estimate of p with respect to the prior Beta(5, 5) and squared error loss is

  See the full question

  Part B · december 2023 · free

• Option D

  $L^{1}(\mathbb{R}) \subseteq L^{2}(\mathbb{R})$

  Why this is wrong, and why it looks right

  On $\mathbb{R}$ the measure is infinite, so neither Hölder direction gives a free inclusion. f(x)=$|x|^{-1/2}$ on (0,1) and 0 elsewhere has $\int$|f|$=2<\infty$ but $\int$|f|$^{2}=\int_{0}^{1}x^{-1}dx=\infty$, so $f \in L^{1}(\mathbb{R})$∖$L^{2}(\mathbb{R})$.

  From this question

  Which of the following are true?

  Part C · free

• Option A

  inf U = 5.

  Why this is wrong, and why it looks right

  False. 5 lies in [3, 6] but not in [3, 4], so it is not even an element of U. Choosing 5 almost always means solving $x^{2} - 9x + 18 \le 0$, reading off its interval, and never imposing the second constraint.

  From this question

  Consider the following subset of $\mathbb{R}: U =$ {$x \in \mathbb{R} : x^{2} - 9x + 18 \le 0, x^{2} - 7x + 12 \le 0$}. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option D

  12

  Why this is wrong, and why it looks right

  Twelve is just |$\mathbb{Z}/12\mathbb{Z}$| itself, which assumes every element generates the group. Only the elements coprime to 12 do; e.g. 2 generates only the order-6 subgroup {0,2,4,...,10}, not the whole group.

  From this question

  The number of generators of the cyclic group $\mathbb{Z}/12\mathbb{Z}$ is

  Part B · free

• Option D

  0

  Why this is wrong, and why it looks right

  0 is what 'multiply the diagonal entries' gives you, and that shortcut only works for triangular matrices. A = J − I has 0 on every diagonal entry (J has 1's on the diagonal too, so J − I zeroes them) and 1 everywhere off it — the matrix is nowhere near triangular, so multiplying the diagonal $(0^{5} = 0)$ throws away exactly the off-diagonal structure the real computation depends on.

  From this question

  Let A = J − I where J is the 5×5 all-ones matrix. Then det A equals

  Part B · free

• Option B

  The linear transformation $T : \mathbb{Q}^{3} \to \mathbb{Q}^{5}$ given by T(v) = Bᵗv is injective.

  Why this is wrong, and why it looks right

  False. Bᵀ is 5 × 3 with rank 2 — SymPy confirms rank 2 on a representative rank-2 matrix of this shape — so the map $\mathbb{Q}^{3} \to \mathbb{Q}^{5}$ has a one-dimensional kernel and is not injective. A tall matrix is injective only when its rank equals its number of columns, which is 3 here, not 2.

  From this question

  Let B be a 3×5 matrix with entries from $\mathbb{Q}$. Assume that {$v \in \mathbb{R}^{5}$ | Bv = 0} is a three-dimensional real vector space. Which of the following statements are true?

  See the full question

  Part C · june 2023 · free

• Option A

  $\sum Y_{i}x_{i} / \sum x_{i}^{2}$

  Why this is wrong, and why it looks right

  False here, though it is the familiar answer. $\sum{}Y_{i}x_{i}/\sum{}x_{i}^{2}$ is the ordinary least squares estimator, which is BLUE only when the errors have constant variance. Here $Var(\varepsilon_{i}) = x_{i}^{2}\sigma^{2}$ grows with $x_{i}$, so OLS over-weights the noisiest observations. It remains unbiased, just not best.

  From this question

  Consider the simple linear regression model $Y_{i} = \beta{}x_{i} + \varepsilon_{i}, i = 1, \dots, n$, where $E(\varepsilon_{i}) = 0, Cov(\varepsilon_{i}, \varepsilon_{k}) = 0$ for i ≠ k and $Var(\varepsilon_{i}) = x_{i}^{2}\sigma^{2}$. The best linear unbiased estimator of $\beta$ is

  See the full question

  Part B · june 2023 · free

• Option A

  The Dirichlet function $1_\mathbb{Q}$

  Why this is wrong, and why it looks right

  False. At every point x ∈ [0,1], every neighborhood contains both rationals and irrationals, so on every subinterval of every partition, sup f = 1 and inf f = 0 — the upper Darboux sum is always 1 and the lower Darboux sum is always 0, so upper integral minus lower integral equals 1 forever and never shrinks as the partition refines. Equivalently, by Lebesgue's criterion f is discontinuous at every point of [0,1], a set of measure 1, not measure 0. The trap: $1_\mathbb{Q}$ agrees with the (trivially Riemann integrable) zero function everywhere except on the measure-zero set $\mathbb{Q}$, and it's tempting to assume changing a function on a measure-zero set can't affect Riemann integrability — that a.e.-invariance is a fact about Lebesgue integration, not Riemann; here the change is dense enough to destroy continuity at every remaining point too, which is exactly what checking the actual discontinuity set, rather than just its 'exceptional' part, would catch.

  From this question

  Which of the following functions on [0, 1] are Riemann integrable?

  Part C · free

• Option B

  $\limsup(a_{n} + b_{n}) = \limsup a_{n} + \limsup b_{n}$

  Why this is wrong, and why it looks right

  False in general: with $a_{n}=(-1)^{n}, b_{n}=(-1)^{n}$⁺$^{1}$, every $a_{n}+b_{n}=0$, so $\limsup(a_{n}+b_{n})=0$, while $\limsup a_{n}+\limsup b_{n}=1+1=2$. Equality needs the extra hypothesis in option (2).

  From this question

  Let $(a_{n}), (b_{n})$ be bounded real sequences. Which of the following are true?

  Part C · free

• Option B

  (a) is not convergent, but (b) is convergent.

  Why this is wrong, and why it looks right

  False, and this is the engineered trap. $\sum1/\sqrt{n}$ diverges (p = 1/2 ≤ 1), so comparing term-by-term suggests (a) diverges too. But $\sin(n\pi/2)$ runs 1, 0, −1, 0 repeating, so the non-zero terms alternate: the series is $\sum(-1)^k/\sqrt{2k+1}$, which converges by Leibniz — the terms decrease to 0, and SymPy confirms that limit is 0. The mistake: the comparison and p-tests require non-negative terms. That hypothesis was dropped.

  From this question

  Consider the following infinite series: $(a) \sum_{n\ge1}$ $\sin(n\pi/2)/\sqrt{n}, (b) \sum_{n\ge1}$ $\log(1 + 1/n^{2})$. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option A

  0

  Why this is wrong, and why it looks right

  Treating the equation as homogeneous — solving y′+y=0 instead of y′+y=$e^{-x}$, dropping the forcing term — gives y=$Ce^{-x}$. y(0)=0 forces C=0, so y≡0 identically and y(1)=0 too.

  From this question

  Let y solve y′ + y = $e^{-x}$ with y(0) = 0. Then y(1) equals

  Part B · free

• Option D

  1 − 1/e

  Why this is wrong, and why it looks right

  Dropping the +y term and solving y′=$e^{-x}$ alone gives $y=\int_{0}$ˣ$e^{-t}dt$=$1-e^{-x}$ instead of the actual $xe^{-x}$. Evaluating at x=1 then gives 1−1/e, the answer to a different, simpler equation.

  From this question

  Let y solve y′ + y = $e^{-x}$ with y(0) = 0. Then y(1) equals

  Part B · free

• Option D

  The dominated convergence theorem applies

  Why this is wrong, and why it looks right

  For fixed $x > 0, f_{n}(x) = n$ whenever n < 1/x, so $sup_n f_{n}(x)$ is comparable to 1/x, which is not integrable near 0 since $\int_{0} (1/x)$dx diverges. No integrable g dominates the whole family, so the dominated convergence theorem's hypothesis fails.

  From this question

  Let $f_{n}(x) = n\cdot1_{(0, 1/n)}(x)$ on [0,1]. Which of the following are true?

  Part C · free

• Option D

  f(x) = eˣ on $[0, \infty)$

  Why this is wrong, and why it looks right

  False: for any fixed $\delta>0$, $e^{x+\delta}-e$ˣ=eˣ$(e^\delta-1)\to\infty$ as $x\to\infty$, so no single $\delta$ controls the output gap for $\varepsilon=1$ across all of $[0,\infty). [0,\infty)$ is unbounded, so Heine–Cantor's compactness hypothesis simply isn't available here.

  From this question

  Which of the following functions are uniformly continuous on the given domain?

  Part C · free

• Option A

  For $c_{1}, c_{2} \in \mathbb{C}$, the level curves $u = c_{1}$ and $v = c_{2}$ are orthogonal wherever they intersect.

  Why this is wrong, and why it looks right

  False. f is only assumed real-differentiable, so the Cauchy–Riemann equations need not hold and there is no reason for the level curves to meet at right angles. Even for entire f the statement is delicate at points where f′ = 0, since the level sets there need not be curves at all. The hypothesis granted is much weaker than the one the claim needs.

  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 B

  $\nabla{}u \cdot \nabla{}v = 0$ at every point.

  Why this is wrong, and why it looks right

  False for exactly the same reason: without holomorphy, u and v are unrelated real functions and their gradients can point anywhere. Take f(x + iy) = x + ix, so u = v = x and $\nabla{}u \cdot \nabla{}v = 1 \ne 0$. Option (C) is this same claim with the missing hypothesis restored.

  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 C

  $K(x,t) = \sqrt{t}(1 - x)$ for $t < x; \sqrt{x}(1 - t)$ for t > x

  Why this is wrong, and why it looks right

  False. $\sqrt{t}$ is not a solution of the homogeneous equation y″ = 0, so this kernel cannot reproduce the differential operator. Each branch of a Green's function must itself solve the homogeneous problem on its side of the diagonal.

  From this question

  For the unknown $y : [0,1] \to \mathbb{R}$, consider the boundary value problem y″(x) + 2y(x) = 0 for x ∈ (0,1), y(0) = y(1) = 0. It is given that it corresponds to the integral equation $y(x) = 2\int_{0}^{1} K(x, t) y(t)$ dt. Which of the following is the kernel K(x, t)?

  See the full question

  Part B · june 2023 · free

• Option C

  A compact subspace of any topological space is closed in it.

  Why this is wrong, and why it looks right

  False. This needs the ambient space to be at least T1 (Hausdorff suffices); without it, compact subsets need not be closed. Take X = {a,b} with the indiscrete topology (only ∅ and X open), so the only closed sets are ∅ and X and the singleton {a} is not closed — yet {a} is finite, and every finite set is compact regardless of the topology. So {a} is a compact subspace that isn't closed in X. If you marked this true, you applied 'compact ⟹ closed' as if it needed no hypothesis on the ambient space, when it needs Hausdorff — exactly what makes option 0 work.

  From this question

  Which of the following statements about compact spaces are true?

  Part C · free

• Option C

  $A^{3} = 0$.

  Why this is wrong, and why it looks right

  False unless c = 0, which is not given. With the same example A = [[0,1,0],[0,0,0],[0,0,1]], SymPy shows $A^{3}$ has rank 1 — non-zero. A is nilpotent only when the whole characteristic polynomial is $T^{3}$, and $T^{2}$ dividing p permits a non-zero third eigenvalue.

  From this question

  Let A be a 3×3 real matrix whose characteristic polynomial p(T) is divisible by $T^{2}$. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option B

  $\int f_{n} \to 0$.

  Why this is wrong, and why it looks right

  $\int_{0}^{1} f_{n} = n\cdot($length of (0,1/n)) = n·(1/n) = 1 for every n, so $\int{}f_{n}\to1$, not 0. The mass doesn't vanish; it escapes into an ever-thinner, ever-taller spike.

  From this question

  Let $f_{n} = n\cdot1_{(0, 1/n)}$ on [0, 1] with Lebesgue measure. Which of the following are true?

  Part C · free

• Option C

  The dominated convergence theorem applies to {$f_{n}$}.

  Why this is wrong, and why it looks right

  If DCT applied we'd need $\int{}f_{n}\to\int(\lim f_{n})=\int0=0$, but $\int{}f_{n}\equiv1$ contradicts that. By contrapositive, no integrable g can dominate all $f_{n}$ — indeed $sup_n f_{n}(x)$ grows like 1/x near 0, which isn't integrable on (0,1).

  From this question

  Let $f_{n} = n\cdot1_{(0, 1/n)}$ on [0, 1] with Lebesgue measure. Which of the following are true?

  Part C · free

• Option D

  $\pi$

  Why this is wrong, and why it looks right

  Using only the pole at $z=i, 2\pi{}i\cdot{}Res_{z=i}$ $= 2\pi{}i\cdot(1/(2i)) = \pi, a$ real nonzero value. But z=−i is also inside |z|=2 with residue −1/(2i), and the residue theorem requires summing over every enclosed singularity, not just one.

  From this question

  The value of $\oint_{|z|=2}$ dz$/(z^{2} + 1)$, taken counterclockwise, is

  Part B · free

• Option C

  Every function $f : X \to \mathbb{R}$ is continuous

  Why this is wrong, and why it looks right

  This would hold if X were discrete, but 0 is a limit point of {1/n}, not an isolated point of X. Taking f(1/n)=0 for all n and f(0)=1 gives a function continuous at every isolated 1/n yet discontinuous at 0, since f(1/n)=0 ↛ 1=f(0).

  From this question

  Let X = {$1/n : n \in \mathbb{N}$} ∪ {0} with the usual metric from $\mathbb{R}$. Which of the following are true?

  Part C · free

• Option C

  a centre

  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

  a saddle

  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

  {$1/n : n \in \mathbb{N}$}

  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

• Option B

  It always converges from any starting point

  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

  Compact subsets of any infinite topological space are closed.

  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

  $\beta_{n}$ is rational for infinitely many 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 D

  g has a removable singularity at 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 2024

• Option D

  If $q_{1}(T)v = 0$ for some nonzero $v \in \mathbb{R}^{4}$, then {v, Tv, $T^{2}v, T^{3}v$} is a basis of $\mathbb{R}^{4}$.

  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 C · december 2024

• Option A

  For every $x \in \mathbb{R}, \lim(t\to\infty) u(x, t) = 0$

  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 A

  $(1/\sigma^{2}) Y^{T}X\beta$̂ has a central $\chi^{2}_{p+1}$ distribution

  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

  $(1/\sigma^{2}) \sum(Y_{i} -$ Ȳ$)^{2}$ has a central $\chi^{2}_{n-1}$ distribution

  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

  5/4

  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 B

  $2\pi{}i$

  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

  $\pi{}i$

  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

  $-2\pi{}i$

  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

  $f(z) = z^{2}$ is possible

  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

  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 B

• Option B

  Let $A_{2} =$ {$x^{2} + y^{2} \le 1$}. For each $\gamma \in p(A_{2})$ there exists $\varepsilon > 0$ with $(\gamma - \varepsilon, \gamma + \varepsilon) \subseteq p(A_{2})$.

  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 C

  1, inconclusive

  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

  The iteration converges for every choice of initial guess.

  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

  (n/(n − 1))p(1 − p)

  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

  There exists a non-trivial solution $\varphi$ of ODE, and $\alpha,\beta\in\mathbb{R}, \alpha<\beta$ such that $\varphi$ has infinitely many zeros in $[\alpha,\beta]$.

  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

  There exists a non-trivial solution of ODE which vanishes at most at one point in $(0,\infty)$.

  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 2025

• Option A

  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 D

  exactly on the real axis

  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

  The origin is asymptotically stable.

  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

  {$(x, y) \in \mathbb{Q}^{2}$ | $x^{2} + y^{2} = 1$} is compact in the subspace topology.

  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 C

  F is left-continuous on $\mathbb{R}$

  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 C · december 2025

• Option A

  $(e/(1 + e^{2}))(e$ˣ + e⁻ˣ$) + x^{2} - 1$

  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 2024

• Option B

  $(e/(1 + e^{2}))(e$ˣ + e⁻ˣ$) + 1 - x^{2}$

  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 · june 2024

• Option D

  $(e/(1 + e^{2}))(e$ˣ + e⁻ˣ$) + \sin(2\pi{}x)$

  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 C · june 2024

• Option C

  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 A · december 2024

• Option D

  9

  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 2024

• Option C

  det A > 0 alone

  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

  If f has no zeros then f is constant

  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

  F is differentiable on (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 D

  Every subgroup of prime index in G is normal.

  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 D

  There exists $x \in \mathbb{R}$ such that f(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 B · december 2024

• Option A

  ȳ

  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 2025

• Option C

  (4ȳ_S + 5ȳ_B)/9

  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 2025

• Option D

  f is differentiable at (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 2025

• Option B

  If |g′(x*)| > 1, the iteration cannot converge to x* from any starting point whatsoever.

  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

  Y has a Double Exponential distribution with location parameter ln(ln 2) and scale parameter 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 2025

• Option C

  (P) has infinitely many solutions.

  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 A

  Rahman-Violet, George-Red, Vedant-Green

  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

  Rahman-Green, George-Red, Vedant-Violet

  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

  Rahman-Red, George-Violet, Vedant-Green

  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

  If p divides |G|, then E(p) is a subgroup of G.

  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

  For all p, E(p) is a subgroup of G.

  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

  $\sum{}a_{n}^{2}$ converges

  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

  $({\bar{X}} - (\sigma/\sqrt{n})$ $z_{0.05}$, ${\bar{X}} + (\sigma/\sqrt{n})$ $z_{0.05})$

  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

  $S_{1}$ and $S_{2}$ are both infinite.

  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

  diagonalisable 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

• Option D

  is 6 feet

  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

  {$(x, y) : x^{2} + y^{2} < 1$}

  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

• Option C

  {(x, y) : x ≥ 0, y ≥ 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 D

  {(x, y) : xy = 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

• Option D

  (X+Y) and (X−Y) are independent

  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

  converges conditionally

  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

  Chennai

  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

  (6, 3/4)

  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 2025

• Option D

  (6, 1/4)

  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 2025

• Option C

  $(1/2\pi{}i) \int_C ($zf)′/(zf) dz $= n_{0} - n_p$.

  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 B

  B is more probable than A

  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

  Both have the same probability

  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

  Whether A or B is more probable depends upon how many faces are painted green

  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 B

  $(\sum{}t X_{t}, \sum{}t$ tX$_{t}, \sum{}t t^{2}X_{t}^{2})$ is a jointly minimal sufficient statistic for $(\alpha,\sigma)$

  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

  $X_{4}-X_{3}-X_{2}+X_{1}$ is an ancillary statistic

  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

  1.5 second

  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 A

  no painter is a professional

  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 B

  all artists are professionals

  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

  $(1/\sum{}n_{i}) \sum_{i}\sum_{j} Y_{ij}$ is an unbiased estimator of $\mu$

  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 · december 2023

• Option A

  $\lim (x\to\infty) f(x) = \infty$

  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 2025

• Option B

  $\lim (x\to\infty) f(x) = 0$

  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 B · december 2025

• Option D

  For all x, y ∈ X, S_G(x) is isomorphic to S_G(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 B · december 2025

• Option A

  If X × Y is Hausdorff, then the statement S 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 B

  If Y is connected, then the statement S 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 C

  If X × Y is regular, then the statement S 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 C

  ⟨u, v⟩ ≠ ⟨u, w⟩ for all non-zero vectors u, v, w with v ≠ w

  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 A

  There is no stationary distribution

  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

  Stationary distribution exists and is unique

  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 C

  There are exactly two stationary distributions

  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

  For each x ∈ (0,7), u(x,t) → 0 as $t \to \infty$.

  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 B · december 2025

• Option A

  S is empty.

  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

  $\mathbb{Z}[x]/(x - 101)$

  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 D

  The series converges for all $\alpha \in (0,\infty)$ and for all $\beta \in (0,\infty)$.

  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 2025

• Option C

  $2/\sqrt{3}$

  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 2025

• Option D

  $1/\sqrt{3}$

  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 2025

• Option A

  $\int_{\gamma1}$ tan(z)dz $= 2\pi{}i$

  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

  $\int_{\gamma1}$ tan(z)dz $= -2\pi{}i$

  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

  For each $x_{0} \in \mathbb{R}, e^t\cdot{}u(x_{0},t) \to 0$ as $t \to \infty$.

  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 B · december 2025

• Option B

  For each $x_{0} \in \mathbb{R}$, |$u(x_{0},t)$| → 0 as $t \to \infty$.

  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 B · december 2025

• Option C

  (1, 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

• Option B

  infinitely many solutions on the interval (−2, 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 B · december 2024

• Option C

  a unique solution and its maximal interval of existence is $(-\infty, 1)$

  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 2024

• Option D

  2X̄ is the UMVUE of $\theta$

  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

  A does not have any eigenvalue on the imaginary axis.

  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

  $A^{2}$ is the identity matrix.

  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

  f is bounded

  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

  The ideal (X, 5) is a maximal ideal in 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 · june 2024

• Option C

  10

  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

  infant mortality rate has increased, reducing the net fertility ratio

  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

  $2x^{2} + 3$xy $+ y^{2} + 3x + 2y + 1$ is irreducible in $\mathbb{Q}[x,y]$.

  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 D

  $x^{4} + 4x^{2} + 3$ is irreducible in $\mathbb{Z}[x]$.

  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 C · december 2025

• Option C

  Sam was the most junior

  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

  Prakash was the most junior

  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

  all real p

  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

  $E(t) \ge \pi$ for all 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 2025

• Option C

  $E(t) \ge \pi/2$ for all 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 2025

• Option B

  There exist $\varepsilon>0, \delta>0$ and a continuously differentiable function $h:\Omega_\varepsilon\to(-\delta,\delta)$ such that h(0,0)=0 and F(x,h(x,z),z)=−1 for all $(x,z)\in\Omega_\varepsilon$.

  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

  If AAᵀ = I, then DF$_A : M_{2}(\mathbb{R}) \to S$ need not be surjective.

  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 D

  If A is not invertible, then DF$_A : M_{2}(\mathbb{R}) \to S$ is surjective.

  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 D

  The characteristic polynomial of A splits into linear factors over $\mathbb{C}$.

  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

  $f(v_{2}, v_{1}) = -s$.

  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

  If f is positive definite, then r ≠ s, s ≠ t and r ≠ t.

  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 D

  If f is positive definite, then $f(v_{1}, v_{2}) = 0$.

  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 B

  Posterior mean = Maximum likelihood estimate of $\theta$, Posterior variance < Var(X|$\theta)$

  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 2025

• Option C

  Posterior mean > Maximum likelihood estimate of $\theta$, Posterior variance = Var(X|$\theta)$

  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 2025

• Option D

  Posterior mean > Maximum likelihood estimate of $\theta$, Posterior variance < Var(X|$\theta)$

  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 2025

• Option A

  YᵀAY$/\sigma^{2}$ follows a central $\chi^{2}$ distribution with (n−1) degrees of freedom.

  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

  YᵀBY/YᵀCY follows a central F distribution with (p,n−p) degrees of freedom.

  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

  YᵀBY and YᵀCY are independently distributed if and only if $\beta_{1}=\beta_{2}=\cdots=\beta_{p}=0$.

  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 2025

• Option B

  The BVP does NOT have a solution.

  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

  indiscrete

  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

  necessarily infinite

  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

  For all z ∈ 𝔻{0}, the sequence {$f_{n}(z)$}$_{n}\ge1$ converges.

  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 C

  The sequence {$f_{n}$}$_{n}\ge1$ converges pointwise to a holomorphic function on 𝔻[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 B · december 2025

• Option B

  (P) is true, (Q) is 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 AB = BA then A and B are simultaneously diagonalisable

  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

  f(x) = 1/x for x ∈ (0, 1], 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 C

• Option A

  If v and w are eigenvectors of B for distinct eigenvalues, then ⟨v,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 2025

• Option C

  For every subspace $W \subseteq \mathbb{R}^{4}, W + W\perp = \mathbb{R}^{4}$, where W⊥ = {$v \in \mathbb{R}^{4}$ ∣ ⟨v,w⟩ = 0 for all w ∈ W}.

  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

  If $W \subseteq \mathbb{R}^{4}$ is an eigenspace of B, then W has an orthonormal basis.

  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

  $\limsup x_{n} = \sqrt{2}$

  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 B · december 2025

• Option D

  $\limsup x_{n} = 1 + \sqrt{2}$

  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 B · december 2025

• Option A

  f∘g is uniformly continuous, but g∘f is not 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 B · december 2025

• Option B

  g∘f is uniformly continuous, but f∘g is not 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 B · december 2025

• Option C

  Neither f∘g nor g∘f is 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 B · december 2025

• Option A

  f is bounded

  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

  All three of the above hold for i.i.d. samples from any distribution with finite variance.

  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

  W is constant on I.

  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

  Every solution on I is a linear combination of $y_{1}$ and $y_{2}$.

  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

• Option D

  $S_{n}({\bar{X}}_{n})$⁻$^{1}$ is a consistent estimator of $\theta$

  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

  State 3 is aperiodic and ergodic

  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

  |f(−1/5)| ≤ 1/5

  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 · december 2025

• Option C

  |f′(−1/2)| ≤ 1/2

  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 B

  (CP) has a unique solution defined on some neighbourhood of the origin

  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 C

  (CP) has a unique solution defined on some neighbourhood of the point (0, 1) in the xy-plane

  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

  States 0 and 2 are recurrent.

  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

  $\partial^{2}w/\partial\xi^{2} + 4(\partial^{2}w/\partial\xi\partial\eta) + \partial^{2}w/\partial\eta^{2} = 0$

  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 B

  (P) is 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 C · december 2025

• Option A

  (X−1)/|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 B · december 2025

• Option C

  (Z−1)/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 B · december 2025

• Option D

  (X−Y)/(X+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 B · december 2025

• Option D

  Cov(X+Y,X−Y) = 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 2025

• Option B

  $\int_\gamma1$ dz$/(z \sin z) = 2\pi{}i$

  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

  $\int_\gamma2$ dz$/(z \sin z) = 2\pi{}i$

  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

  f need not be onto.

  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

  open but not closed

  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

  f must be a polynomial of positive degree.

  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

  There exist $a \in \mathbb{Z}$ and a prime p > 11 such that p ∤ aᵖ − a.

  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 2025

• Option B

  $\mathbb{Q} \cap [0, 1]$

  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

• Option D

  7

  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

  Every unbiased estimator attains the Cramér–Rao lower bound.

  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

  Figure B

  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

  Figure D

  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

  X × Y is compact with respect to the product topology

  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

  120

  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 2025

• Option A

  not calculable due to insufficient data

  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 2024

• Option B

  1 or 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

• Option C

  1 or 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 B

• Option D

  4 only

  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

  (q^(2n) − q^n)(q^(2n) − q^(n+1))/q^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 · december 2025

• Option C

  (q^(2n) − 1)(q^(2n) − q)⋯(q^(2n) − q^(n−1))/q^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 · december 2025

• Option B

  Volume of the rectangular container is larger than that of the square container.

  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 2024

• Option C

  Volume of the triangular container is smaller than that of the square container.

  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 2024

• Option D

  Volume of the square container is larger than that of the circular container.

  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 2024

• Option A

  11

  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

  13

  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

  (0.5, 0.5)

  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 B

• Option C

  There is exactly one differentiable real-valued function g defined on an 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 B

  y ≡ 0 is the only stationary function (extremal) for (P).

  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 2023

• Option B

  non-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 A

  Both (a) and (b) are 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 · june 2023

• Option B

  Every $\beta_{i}, 1 \le i \le n$, has infinitely many linear unbiased estimators, but a unique best linear unbiased estimator.

  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

  $u(\pi) = 2e^\pi$.

  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

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

  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 is an orthonormal basis of W.

  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 D

  S contains an orthonormal basis of W.

  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

  S is the empty 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 2024

• Option A

  The sequence of functions {$f_{n}$}$_{n}\ge1$ is uniformly bounded.

  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 sequence of functions {$f_{n}$}$_{n}\ge1$ converges uniformly to the constant function 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 2025

• Option D

  The set {$f_{n} : n\in\mathbb{N}$} is compact in C[0,1], where C[0,1] denotes the space of all real valued continuous functions on [0,1] equipped with the supremum norm.

  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 = constant or y = constant

  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 B

  $y/x^{2} =$ constant

  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

  yx = constant

  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 has infinitely many zeros in [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 B · december 2023

• Option C

  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

• Option D

  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 B

• Option B

  There exists $D \in M_{3}(\mathbb{R})$ and a polynomial $f \in \mathbb{R}[X]$ satisfying f(A) = f(B) = 0, f(C) = I and f(D) = I.

  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 C

  There exists $D \in M_{3}(\mathbb{R})$ and a polynomial $f \in \mathbb{R}[X]$ satisfying f(A) = 0, f(B) = f(C) = I and f(D) ≠ 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 2025

• Option D

  There exists $D \in M_{3}(\mathbb{R})$ and a polynomial $f \in \mathbb{R}[X]$ satisfying f(A) = f(B) = 0, f(C) = I and f(D) = 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 2025

• Option A

  X is uncountable.

  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

  X is infinite and countable.

  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 C

  Every element of X is an open 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 B · december 2025

• Option A

  The function f(1/z) has a removable singularity at 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 2025

• Option B

  The function f(1/z) has an essential singularity at 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 2025

• Option D

  If S is not a singleton set, then $\int_{\gamma{}R}$ f(z)dz = 0 for some R > 0 such that S ∩ {$z \in \mathbb{C}$ : |z| = 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 2025

• Option C

  The function h with h(x) = 1 if $f(x) \in \mathbb{Q}$ and 0 otherwise

  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 restricted to $\mathbb{Q} \cap [0,1]$, extended by 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 A

  X is metrizable.

  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

  R(1/2,1/2,1/3) = 2/3

  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 C

  $(\sum$ iY$_{i}, \sum i^{2}Y_{i}^{2})$

  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 B · december 2023

• Option A

  Let f(X) ∈ 𝔽_p[X] and $\alpha$ be a root of f in 𝔽̄_p. Then 𝔽$_p(\alpha)$ is the splitting field of f in 𝔽̄_p.

  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

  P(Z ≥ 3/5) = 0.6, P(Z = 4/25) = 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 2025

• Option C

  P(Z ≥ 3/5) = 0.7, P(Z = 4/25) = 0.16

  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

  P(Z ≥ 3/5) = 0.8, P(Z = 4/25) = 0.16

  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

  There exists a finite group G such that |H_G| is even.

  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

  f is injective, but f⁻$^{1} : f(K) \to K$ need not be 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 · june 2024

• Option A

  f is a real valued function.

  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

  f(z) = 1 for all $z\in\mathbb{C}$.

  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

  If f∈F is a continuous function and f(n)=0 for all positive integers n, then $\lim(x\to\infty)\int_{0}$ˣf(t)dt exists in $\mathbb{R}$.

  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 C · december 2025

• Option C

  If each $f_{n}$ is bounded and continuous, $(f_{n})_{n}\ge_{1}$ converges pointwise to a bounded and continuous function f, then the convergence is uniform.

  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 C

  p = 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 D

  A

  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

  The sequence $(y_{n})$ does NOT converge

  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 A

  T is a contraction and has a unique fixed point

  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

  In every calendar month of 2024, at least one person was troublesome.

  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

  f has either poles or essential singularities at −1 and 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 D

  $\mathbb{Z}/8 \cong \mathbb{Z}/4 \times \mathbb{Z}/2$.

  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

• Option B

  There exists a subspace W of $\mathbb{C}^{n}$ such that $T(W) \subseteq W, \mathbb{C}^{n} = V + W$ and V ∩ 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 C · december 2024

• Option B

  f(x) = eˣ

  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

  f(x) = x − 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 C

  There exists $x\in(0,\infty)$ such that f(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 · december 2025

• Option D

  On $\mathbb{R}$ with Lebesgue measure, $L^{2} \subseteq L^{1}$ also holds.

  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

  Every point $x \in \mathbb{R}$ has an open neighbourhood U_x such that the restriction $p|_{U_x}$ is a homeomorphism onto an open set in $\mathbb{R}$.

  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 C · june 2025

• Option A

  ⟨x, y⟩ $= \sum_{i,j=1}^n$ $x_{i}y_{j}$

  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 2023

• Option D

  ⟨x, y⟩ $= \sum_{j=1}^n$ $x_{j} y_{n}$₋$_{j}$₊$_{1}$

  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 B · june 2023

• Option A

  $f_{n} \to 0$ uniformly and $f_{n}' \to 0$ uniformly

  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 bounded subset of $\mathbb{C}$

  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

  The ideal $(\sqrt{2} - 1)\mathbb{Z}[\sqrt{2}]$ is maximal in $\mathbb{Z}[\sqrt{2}]$.

  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 B · june 2025

• Option B

  M is the most popular English magazine in the country.

  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 C

  M is the most popular magazine in the country.

  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

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
• Completeness is a topological property
• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Differentiable ⇒ continuously differentiable

Other mistakes that cost marks