Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

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

  inf U = 4.

  Why this is wrong, and why it looks right

  False — 4 is the supremum of U, not the infimum. SymPy gives the roots as {3, 6} and {3, 4}, so U = [3, 4]: 3 is the left endpoint and 4 the right. This option is what correct algebra plus a misread of which endpoint was asked for produces, which is why it is the most-picked wrong answer.

  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 B

  The equation is hyperbolic everywhere.

  Why this is wrong, and why it looks right

  This would follow from misplacing the classification boundary itself — treating $B^{2}-4$AC≥0 as the hyperbolic condition instead of the strict $B^{2}-4$AC>0, which folds the boundary case into hyperbolic. But here $B^{2}-4$AC=16−16=0 exactly, landing precisely on the parabolic threshold; hyperbolic needs a strictly positive discriminant, which this equation never attains.

  From this question

  Consider the second-order PDE u_xx + 4u_xy + 4u_yy = 0. Which of the following are true?

  Part C · free

• Option D

  The equation has two distinct families of real characteristics.

  Why this is wrong, and why it looks right

  A zero discriminant gives the repeated root $(m-2)^{2}=0$, not two distinct roots, so there's exactly one family of real characteristics, not two. Parabolic is precisely the borderline case where the two hyperbolic families merge into a single one.

  From this question

  Consider the second-order PDE u_xx + 4u_xy + 4u_yy = 0. Which of the following are true?

  Part C · free

• Option C

  $\lambda = 0$ is an eigenvalue.

  Why this is wrong, and why it looks right

  $\lambda=0$ gives y″=0, so y=ax+b; y(0)=0 forces b=0 and $y(\pi)=0$ then forces a=0, leaving only the trivial solution. The pattern $\lambda_{n}=n^{2}$ tempts you to extend it down to n=0, but that boundary case has no nontrivial eigenfunction.

  From this question

  Consider $-y'' = \lambda{}y$ on $[0, \pi]$ with $y(0) = y(\pi) = 0$. Which of the following are true?

  Part C · free

• Option C

  |$I_{k}$| ≤ |$I_{k}$₊$_{1}$| for every integer k.

  Why this is wrong, and why it looks right

  False. The moduli are $2\pi/(k-1)$!, which decrease once the factorial starts biting. SymPy gives |$I_{2}$| $= 2\pi$ and |$I_{3}$| $= \pi$, so |$I_{2}$| > |$I_{3}$| and the claimed inequality fails at k = 2. The first two values being equal (|$I_{1}$| = |$I_{2}$| $= 2\pi$, since 0! = 1! = 1) is what makes the sequence look non-decreasing at a glance.

  From this question

  For an integer k, consider the contour integral $I_{k} = \int_{|z|=1}$ eᶻ/zᵏ dz. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option C

  There exists $\lambda$ such that $I_\lambda, K_\lambda$ both converge.

  Why this is wrong, and why it looks right

  False, and it is the option the question is built to catch. Both converging would need $\lambda < 1$ and $\lambda > 1$ at once. The two conditions are not merely different, they are complementary — there is no overlap window, and the single point $\lambda = 1$ they share is where both diverge.

  From this question

  For a real number $\lambda$, consider the improper integrals $I_\lambda = \int_{0}^{1}$ dx$/(1 - x)^\lambda$ and $K_\lambda = \int_{1}^\infty$ dx$/x^\lambda$. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option B

  The series is convergent.

  Why this is wrong, and why it looks right

  Not the intended answer, and it is worth being clear why: this statement is literally true — the series does converge. But 'conditionally convergent' in (C) is strictly more informative, and the examiner marks only the most precise true statement. When two options are nested like this, always choose the sharper one. If you found (B) attractive, your mathematics was right and your reading of the instruction was not.

  From this question

  Consider the series $\sum_{n=1}^\infty$ $a_{n}$, where $a_{n} =$ $(-1)^{n+1}(\sqrt{n+1} - \sqrt{n})$. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option C

  8

  Why this is wrong, and why it looks right

  False. 8 is p itself, not p − 1. Every source of variation in a Latin square — rows, columns, treatments — carries p − 1 degrees of freedom, one fewer than the number of levels.

  From this question

  In a Latin square design, the degrees of freedom for the error sum of squares is 42. Then the degrees of freedom for the sum of squares due to treatments is

  See the full question

  Part B · december 2023 · free

• Option A

  A_a is positive definite for all a < 3.

  Why this is wrong, and why it looks right

  False. Sylvester's criterion needs all leading principal minors positive: they are 2, 3, and det A_a, which SymPy evaluates as 3a − 2. So A_a is positive definite exactly when a > 2/3. The range a < 3 includes a = 0, where the determinant is −2 (verified) — not positive definite.

  From this question

  For $a \in \mathbb{R}$, let A_a = [[2, −1, 0], [−1, 2, −1], [0, −1, a]]. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option C

  A_a is positive definite for all a ≥ −2.

  Why this is wrong, and why it looks right

  False. a ≥ −2 includes a = −2, where the determinant is 3(−2) − 2 = −8 (verified), so the matrix is indefinite there. The threshold is 2/3, well above −2.

  From this question

  For $a \in \mathbb{R}$, let A_a = [[2, −1, 0], [−1, 2, −1], [0, −1, a]]. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option A

  88

  Why this is wrong, and why it looks right

  False here, but it is the right answer to a neighbouring question — which is exactly why it is offered. The power sum $\sum{}k^m mod p$ equals −1 ≡ p − 1 precisely when (p − 1) | m, and 0 otherwise. SymPy computes both cases directly: $\sum{}k^{88} mod 89 = 88$, while $\sum{}k^{37} mod 89 = 0$. Since 88 ∤ 37 (verified: 37 mod 88 = 37), we are in the second case.

  From this question

  Which one of the following is equal to $1^{37} + 2^{37} + 3^{37} + \cdots + 88^{37}$ in $\mathbb{Z}/89\mathbb{Z}$?

  See the full question

  Part B · december 2023 · free

• Option D

  Some solutions spiral into the origin as $t \to \infty$.

  Why this is wrong, and why it looks right

  False. As shown above |x(t)| = |x(0)| for all t, so trajectories neither shrink nor grow — nothing spirals in. Spiralling (a stable or unstable focus) happens precisely when the eigenvalues are complex with nonzero real part; $Re(\lambda)=0$ is exactly the boundary between those two cases, and at that boundary you get a centre instead. If you marked this true, you applied 'complex eigenvalues means spiralling' without checking whether the real part was actually zero.

  From this question

  Consider the system x′ = Ax with A = [[0, 1], [−1, 0]]. Which of the following are true?

  Part C · free

• Option A

  $\sum 1/(n \log n)$

  Why this is wrong, and why it looks right

  Condensing: $\sum2^{k}\cdot1/(2^{k} \log 2^{k}) = \sum1/(k log2), a$ constant multiple of the harmonic series, which diverges. So $\sum1/(n \log n)$ diverges too — the single log power isn't enough.

  From this question

  Which of the following series converge?

  Part C · free

• Option D

  $\sum$ $1/n^{1 + 1/n}$

  Why this is wrong, and why it looks right

  $1/n^{1+1/n}$ = (1/n)·$n^{-1/n}$, and $n^{-1/n}$=$e^{-(\ln n)/n}$→e^0=1 since (ln n)/n→0. So $1/n^{1+1/n}$ is asymptotically 1/n, and by limit comparison with the divergent $\sum1/n, \sum1/n^{1+1/n}$ diverges too — the exponent only looks bigger than 1.

  From this question

  Which of the following series converge?

  Part C · free

• Option B

  f is non-constant and has no zeros in D.

  Why this is wrong, and why it looks right

  If f had no zeros in D, 1/f would also be holomorphic there. Maximum modulus on f gives |f| ≤ 3 in D; applied to 1/f (boundary value 1/3) gives |f| ≥ 3 in D. Together |f| ≡ 3 on the connected open set D forces f constant, contradicting non-constant.

  From this question

  Let f be holomorphic on the open unit disc D and continuous on its closure, with |f(z)| = 3 for every |z| = 1. Which of the following are possible?

  Part C · free

• Option C

  45

  Why this is wrong, and why it looks right

  False, and this is the engineered near-miss: 45 is exactly the answer if the first digit is forbidden from being 0, giving 9 + 8 + ⋯ + 1 = 45 (verified). The 10 extra PINs are those starting 0, from 000 up to 099. Everything about the calculation is right except the reading of 'PIN'.

  From this question

  The number of three digit PINs, in which the third digit is the sum of the first two digits, is

  See the full question

  Part A · september 2022 · free

• Option D

  parabolic in {(x, y) : xy ≠ 0}

  Why this is wrong, and why it looks right

  False. The region xy ≠ 0 is exactly where the discriminant is non-zero, so it is precisely where the equation is not parabolic. Parabolic behaviour occurs only on the degenerate set xy = 0, where all three second-order coefficients collapse.

  From this question

  The partial differential equation $x^{2} u_{xx} - 2$xy $u_{x}$ᵧ $- 3y^{2} u_yy + u_{x} - u_y = 0$ is

  See the full question

  Part B · december 2023 · free

• Option A

  F(x) has exactly one real root.

  Why this is wrong, and why it looks right

  False, because 'all antiderivatives' means the whole family F + C, and the root count changes with C. Take $f(x) = x^{3} + x$, whose only real root is 0 and is simple (SymPy: roots 0, ±i). Then $F = x^{4}/4 + x^{2}/2 + C$, and SymPy gives two real roots at C = −1, exactly one at C = 0, and none at C = +1. So 'exactly one' holds for a single value of C, not for all.

  From this question

  Let f(x) be a cubic polynomial with real coefficients. Suppose that f(x) has exactly one real root and that this root is simple. Which one of the following statements holds for ALL antiderivatives F(x) of f(x)?

  See the full question

  Part B · december 2023 · free

• Option A

  $\lim P(1/2 < Y_{n} < 3/2) = 1$

  Why this is wrong, and why it looks right

  False, and it is ruled out before any limit is taken. The distribution is supported on $[5, \infty)$ — SymPy confirms $F(5) = 1 - e^{0} = 0$ — so every $X_{i}$, and hence every $Y_{n}$, is at least 5. The event {$1/2 < Y_{n} < 3/2$} is impossible, with probability 0 for every n.

  From this question

  Let $X_{1}, X_{2}, \dots$ be i.i.d. with CDF F(x) = 0 for x < 5 and 1 − $e^{5-x}$ for x ≥ 5. Define $Y_{n} = \min${$X_{1}, \dots, X_{n}$} and $Z_{n} = \sqrt{n} (Y_{n} - 5)$, and let Z be standard normal. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option C

  There is a bijective function from X to Y.

  Why this is wrong, and why it looks right

  False. A bijection would need |Y| = |X|, but |Y| = 2^|X| and $2^{n} > n$ for every n ≥ 1 — even the smallest case fails, since |X| = 1 gives |Y| = 2 (SymPy: $2^{1} - 1 = 1$, so the gap is already positive). For |X| = 3 the sets have 3 and 8 elements.

  From this question

  Let X be a non-empty finite set and Y = {$f^{-1}(0)$ : f is a real-valued function on X}. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option D

  There is a surjective function from X to Y.

  Why this is wrong, and why it looks right

  False. A surjection X → Y needs |X| ≥ |Y|, that is $n \ge 2^{n}$, which holds for no n ≥ 1. The inequality runs the other way: the power set is always strictly larger.

  From this question

  Let X be a non-empty finite set and Y = {$f^{-1}(0)$ : f is a real-valued function on X}. Which one of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option A

  f(x) = [x] sin(1/x) for x ≠ 0, f(0) = 0 (where [x] is the integer part) has a discontinuity at 0 which is removable.

  Why this is wrong, and why it looks right

  False. Check both sides, because [x] jumps at 0. For 0 < x < 1 the integer part is 0, so f ≡ 0 on the right and the right limit is 0 — which is where most students stop. For −1 ≤ x < 0 the integer part is −1, so f(x) = −sin(1/x), which oscillates without limit as x → 0⁻. A removable discontinuity needs the two-sided limit to exist; here the left limit does not. The mistake: evaluating the side on which the floor function is harmless and never checking the other.

  From this question

  Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option B

  (0, 1] with the usual metric

  Why this is wrong, and why it looks right

  (0,1] is not complete: the sequence 1/n is Cauchy (it converges to 0 in $\mathbb{R})$ but 0 ∉ (0,1], so it has no limit inside the space. Removing just the single endpoint 0 is enough to destroy completeness.

  From this question

  Which of the following metric spaces are complete?

  Part C · free

• Option B

  only at interior points

  Why this is wrong, and why it looks right

  This is precisely what the maximum principle rules out: if u attained its max at an interior point (0<x<L, 0<t≤T) without u being constant, the equation u_t=u_xx would force a contradiction via the sign of u_xx there. The max is pinned to the boundary, not the interior.

  From this question

  Let u solve the heat equation u_t = u_xx on 0 < x < L, 0 < t ≤ T, with continuous data. By the maximum principle, the maximum of u over the closed rectangle is attained

  Part B · free

• Option C

  only on the top edge t = T

  Why this is wrong, and why it looks right

  This reverses the direction of diffusion: heat spreads and evens out values forward in time, so the final time t=T is the least likely place for a fresh maximum, not the most likely. The initial edge t=0, which does carry the max candidates, is part of the parabolic boundary, not the top.

  From this question

  Let u solve the heat equation u_t = u_xx on 0 < x < L, 0 < t ≤ T, with continuous data. By the maximum principle, the maximum of u over the closed rectangle is attained

  Part B · free

• Option D

  g is Lipschitz on [0, 1].

  Why this is wrong, and why it looks right

  $g'(x)=1/(2\sqrt{x})\to\infty$ as x→0⁺, so |g(x)−g(0)|/|x−0| $= 1/\sqrt{x}$ is unbounded near 0. No finite L can satisfy |g(x)−g(0)|≤Lx for all small x>0.

  From this question

  Let $f(x) = x^{2}$ on $\mathbb{R}$ and $g(x) = \sqrt{x}$ on $[0, \infty)$. Which of the following are true?

  Part C · free

• Option D

  For $y' = \lambda{}y$ with real $\lambda < 0$, explicit Euler is stable for every step size h > 0.

  Why this is wrong, and why it looks right

  False. Applying explicit Euler to $y'=\lambda{}y$ gives $y_{n+1}=(1+h\lambda)y_n$, so $y_n=(1+h\lambda)^{n}y_0$. For the numerical solution to stay bounded like the true solution $e^{\lambda{}t}$→0 does $(\lambda<0$ real), you need |$1+h\lambda$|$\le1, i.e. -2\le{}h\lambda\le0, i.e. h\le2/$|$\lambda$|. Past that threshold |$1+h\lambda$|>1 and the iterates grow without bound even though the exact solution decays — explicit Euler is only conditionally stable here, not stable for every h>0. If you marked this true, check whether you tested one small h and generalized, rather than checking the stability condition itself.

  From this question

  Consider solving y′ = f(t, y) numerically with step size h. Which of the following are true?

  Part C · free

• Option B

  $e^{-6t}$|x(t)| → 0 as $t \to \infty$, for all $x_{0} \ne 0$.

  Why this is wrong, and why it looks right

  False for the initial conditions lying along the dominant eigendirection. If $x_{0}$ is an eigenvector for $\lambda = 6$, then |x(t)| = |$x_{0}|e^{6t}$ exactly, so $e^{-6t}$|x(t)| = |$x_{0}$|, a non-zero constant. The claim is 'for all $x_{0} \ne 0'$, and this one starting point defeats it — damping at exactly the top growth rate leaves a constant, not zero.

  From this question

  Suppose x(t) is the solution of the initial value problem in $\mathbb{R}^{2}$: ẋ = Ax, $x(0) = x_{0}$, where A = [[5, 4], [1, 2]]. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option C

  $e^{-t}$|x(t)| $\to \infty$ as $t \to \infty$, for all $x_{0} \ne 0$.

  Why this is wrong, and why it looks right

  False, defeated by the other eigendirection. Along the $\lambda = 1$ eigenvector, |x(t)| = |$x_{0}|e^{t}$, so $e^{-t}$|x(t)| = |$x_{0}$| — constant, not tending to infinity. The slowest mode is what breaks this one, just as the fastest broke (B).

  From this question

  Suppose x(t) is the solution of the initial value problem in $\mathbb{R}^{2}$: ẋ = Ax, $x(0) = x_{0}$, where A = [[5, 4], [1, 2]]. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option C

  There exists an integer n ≥ 1 such that n $e^{-n}$ ≥ x.

  Why this is wrong, and why it looks right

  False — and the whole question is built around this option. n·e^(−n) is bounded: the function t·e^(−t) peaks at t = 1 with value 1/e ≈ 0.3679 and decays thereafter. Sampled across [0.5, 40], the largest value SymPy finds is exactly that peak. So for any x > 1/e there is no such n. The mistake: reading $'n \to \infty'$ off the leading factor and assuming the product must grow. e^(−n) decays faster than n grows, and comparing growth rates — not merely spotting an n — is what is being tested.

  From this question

  Let x be a real number. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option B

  2 and −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 A

  (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

• Option B

  The first principal component can only explain 32% of the total variation for some $\rho$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The second principal component can explain more than 32% of the total variation for any $\rho$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 variance of the first principal component is $1 + 2\rho$ for any $\rho$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  [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

• 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 · june 2023

• Option C

  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 2023

• Option A

  f(z) = 2024 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 · june 2024

• Option D

  There exists a positive integer N such that for all n≥N, $b_{n,2024}$ $= a_{2024} - 1$.

  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 D

  For every $t \in (0, \infty), \lim(x\to-\infty) u(x, t) = \infty$

  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

  P(|X − 4/5| ≥ 1) = 19/25

  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 C · june 2025

• Option D

  The value of the moment generating function, M_X(t), at t = 1 is $(e^{2} - 1)/5$

  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 · june 2025

• Option B

  If $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite, then f is Lebesgue integrable 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 · june 2025

• Option D

  If $\sum_{k\ge0}$ $a_k \mu(A_k)$ is finite and f is bounded, then f is Lebesgue integrable 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 · june 2025

• Option C

  C > 180/19

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  [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 2025

• Option B

  (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 2025

• Option D

  (−10⁻$^{2025},10$⁻$^{2025}]$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 series converges at x = 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 A

  f is monotone.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  f is continuous but not strictly monotone.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  f is continuous but 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 2024

• Option D

  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

  $X_{(4)}$ is a degenerate random variable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 C

  The MLE is unbiased.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  A is 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 B

• Option D

  A has a non-zero eigenvalue

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The series converges at 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

• Option D

  The ideal generated by $\psi(X)$ is a prime 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 C · december 2023

• Option A

  $99 C_{0}$

  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 A · september 2022

• Option B

  (1,2;3,4) ∈ 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 2025

• Option C

  The distribution of $2\sqrt{2}(X_{1} - 5)/\surd(\sum_{i}$₌$_{3}^{10}(X_{i} - 5)^{2})$ is t-distribution with degrees of freedom 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 C · june 2025

• Option C

  zf(z) → 0 as |z| $\to \infty$.

  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 · june 2025

• Option D

  f(z) → 0 as |z| $\to \infty$.

  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 · june 2025

• Option A

  become acidic.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  Var(X) > 3Var(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 2024

• Option B

  2Var(Y) < Var(X) < 3Var(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 2024

• Option C

  Var(Y) < Var(X) < 2Var(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 2024

• Option D

  Quadratic form (c) is positive definite.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The BVP has a non-zero solution if $\lambda = 2$

  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 2024

• Option A

  $H_{0}$ is rejected

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  k = −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 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 · june 2024

• Option D

  max{min{$X_{1}, X_{2}, \dots, X_{n}$}, 1 − max{$X_{1}, X_{2}, \dots, X_{n}$}}

  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 2024

• Option C

  $f_{n}(x) = x^{n}$ is equicontinuous 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

• Option C

  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

• Option A

  u changes sign in the interior of 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 · june 2023

• Option D

  Var(X) > 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 2023

• Option A

  If f ∈ B, then dim ker f = 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 2024

• Option D

  If f ∈ A, then the image of f is a one-dimensional $\mathbb{Q}-$vector space.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  Q is a positive definite quadratic form on V.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

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

  f′ extends continuously to 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

• Option A

  P{$Y_{n} \in [0, 2n]$}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 C

  P{$Y_{n} \in [2n, 4n]$}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  {$-\sqrt{2}, \sqrt{2}$}

  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 2024

• Option D

  f(ℍ) is a convex 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 · december 2023

• Option C

  $[{\bar{X}} - (S/\sqrt{n})$ $t_{n-1,0.9}$, $\infty)$

  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 2023

• Option C

  $S_{1}$ has exactly one element and $S_{2}$ has exactly two elements.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  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 A · june 2025

• Option D

  32

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  $\sum_{n\ge2}$ 1/(n log 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 D

  K = [−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 C · december 2023

• Option B

  P(X ≤ 4) = P(X ≥ 22)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  Sunday

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

• Option A

  det(B) > 0 for every nonzero $(a, b, c, d) \in \mathbb{R}^{4}$.

  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 C · december 2024

• Option B

  det(B) > 0 for infinitely many $(a, b, c, d) \in \mathbb{R}^{4}$.

  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 C · december 2024

• Option A

  elliptic

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  hyperbolic

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  Bayes estimate of $\beta$ tends to ȳ/x̄ as $\tau^{2} \to 0$

  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 C

  $\mu + \alpha_{1} + \alpha_{2}$ is an estimable linear parametric 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 2023

• Option A

  The series converges for all $\alpha \in (0,1)$ and for all $\beta \in (1,\infty)$.

  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

  The series converges for $\alpha = 1$ and $\beta = 1$.

  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 D

  For each $x_{0} \in \mathbb{R}$, there exists $a t_{0} > 0$ such that $u(x_{0},t_{0}) \le -4$.

  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 B

  $\Delta{}u =$ ue$^(u^{2}) + f$ in B(0,1), u = 0 on $\partial{}B(0,1)$

  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 · june 2024

• Option A

  S is an unbounded subset of $\mathbb{R}$.

  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 2024

• Option C

  There exists u ∈ X such that ‖u‖ = 1 and T(u) = 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

  S is a closed subset of $\mathbb{R}$.

  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 2024

• Option A

  12

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  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 2025

• 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 A · december 2025

• Option D

  ${\bar{R}}^{2} > 0$

  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 · 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 2025

• Option D

  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 A · december 2025

• Option D

  $\int_{0}^{1} 1/x$ dx

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  For $\alpha = 2$, there exists an unbounded solution of the ODE 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 2024

• Option B

  p > 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

  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 B

• Option D

  $\lim(x\to0+) x^{2}y_{1}(x)/y_{2}(x)$ exists.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  $u(1) = -5/(5 - 32\lambda)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  conj(f(z̄)) + f(z̄) is an entire 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 · june 2023

• Option B

  $dim_\mathbb{R}(W) = 2$.

  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

  There exists a solution u to BVP such that $u(x,y)=(1+e^{3})/2$ for some $(x,y)\in\Omega\cup\partial\Omega$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  20

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  For all z with |z| ≥ 1, 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 A

  There exists a unique solution u(x,t) of IBVP such that u(1,t) = e − sin t, for all t < 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 C

  There exists a solution u(x,t) of IBVP such that u(x,t) = e^(x−t), for all t > 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 D

  There exists a solution u(x,t) of IBVP such that u(x,t) = e − sin(t−x), for all t < 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 D

  If a = 0, then the sequence {$x_{n}$} 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 2023

• Option D

  The type of the equation is the same at every point of the 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

• Option B

  37/64

  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 B

  There exists a solution of the ODE which is unbounded on $(0, \infty)$

  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 2024

• Option C

  Every solution of the ODE is unbounded on $(0, \infty)$

  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 2024

• Option C

  If the observed value of X̄ is 1.3, then $\varphi$ rejects $H_{0}$

  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

  $\chi^{2}_{np}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  $u(x_{0}, y_{0}) = \sqrt{2} e^\pi$

  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 C · december 2024

• Option D

  $u(x_{0}, y_{0}) = -e^\pi$

  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 C · december 2024

• Option C

  u(−x, t) = u(x, t), whenever x > 0, 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 B · december 2024

• Option A

  There exists $x \in \mathbb{R}$ such that f(x) is not 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 · june 2024

• Option B

  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 D

  The power series converges for every z ∈ {$1/5^{n} : n \in \mathbb{N}$}

  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 C · june 2024

• Option A

  f has a pole at z = 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 D

  f has an essential singularity at z = 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

  a = 1 and b = 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

  a = 2 and b = 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

  a = 1 and b = 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 A

  $\sum x^{n}$ on (−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 C

• Option B

  The integral I is not convergent

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The integral $I_{a}$ converges for all 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 C · june 2024

• Option B

  $(x^{n}g(x))_{n}\ge_{1}$ does not converge uniformly 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 C

  $(x^{n}h(x))_{n}\ge_{1}$ converges uniformly 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 C

  If $T^{2}$ has only non-negative eigenvalues, then T is 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 C · december 2025

• Option A

  The sum of squares residuals is strictly positive with probability 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 B

  For every $\beta_{i}, 1 \le i \le n$, there are infinitely many linear unbiased estimators.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 need not be one-one.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  $limsup_{n} \exp(\cos((n\pi + (-1)^{n}\cdot2e)/(2n))) > 1$.

  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 D

  $lim_{n} \exp(\tan((n\pi^{2} + (-1)^{n}e^{2})/(7n)))$ does not exist.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  f is differentiable at 0 when $\alpha = 1/2$.

  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 2024

• Option C

  f is continuous at 0 when $\alpha = 0$.

  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

  P(Y < −1) < P(Y > 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 B

  The function 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 B · june 2025

• Option C

  c = 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 · june 2025

• Option B

  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 B · june 2025

• Option D

  no such f exists.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

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

  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 C

• Option C

  There are infinitely many $\alpha \in \mathbb{R}$ such that $q(\alpha{}v + w) = 0$.

  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 · june 2025

• Option A

  7/36

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  11/36

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  13/36

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  700

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 A

  taller than 21, but shorter than 18 others

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 C

  taller than 20, but shorter than 19 others

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 D

  taller than 19, but shorter than 19 others

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 A

  The optimal solution is 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 B

  An optimal point is (4, 1, 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 optimal solution is 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 · december 2024

• Option A

  $\tfrac{1}{2}$

  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 A · june 2024

• Option A

  For each $p < 100, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ 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 2024

• Option B

  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 A

  $T_{n}$ is an unbiased 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 2024

• Option D

  max{$X_{1}, X_{2}, \dots, X_{n}$} 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 2024

• Option D

  If A is a real non-zero 3×3 diagonal matrix, then there is a real matrix B such that 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 B

  For all $n \ge 1, p_{n}$ has a zero in the set {$z \in \mathbb{C}$ : |z| ≤ 2023}.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  If u and v satisfy the Cauchy–Riemann equations at a single point, then f is differentiable at that 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 A

  There exists $\alpha \in (0, 1)$ such that for all f, g ∈ C[0, 1], ‖T(f) − T(g)‖ $\le \alpha$‖f − 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 · june 2025

• Option A

  The first principal component based on $\sum$ explains exactly 90% of the total variability

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The second principal component based on $\sum$ explains exactly 10% of the total variability

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  0.4625

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  0.5425

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  0.6225

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  720

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  840

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  $T_{n}$ 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 2024

• Option B

  $T_{n}$ is an unbiased 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 2024

• Option D

  8/15

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  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 B

• Option A

  $W_{0} =$ {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 A

  Maximum likelihood estimator of $\theta$ is min{$X_{(1)}, X_{(n)}/2$}

  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 B

  parabolic

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  no 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 B

• Option C

  f has a Taylor series expansion $f(z) = \sum a_{n}z^{n}$, where $a_{0} = 1, a_{1} = 0$ and $a_{n}$₊$_{2} = a_{n} + a_{n}$₊$_{1}$ for n ≥ 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 · june 2023

• Option C

  145

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  S ≠ ∅ and S is a proper subset of $(1, \infty)$

  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 2023

• Option C

  $(2, \infty)$ is a proper subset of 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 B · december 2023

• Option D

  S ∩ (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 A

  $(Y + (1/n)\ln(1 - \alpha/2), Y - (1/n)\ln(1 - \alpha/2))$

  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 2025

• Option D

  They are conformal at every point 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 C

• Option D

  T is not 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 A

  There exist non-negative integers p, q such that 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 2025

• Option C

  For all non-negative integers p and q, (A) or (B) 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 · june 2025

• Option D

  There exists p ≠ q such that both (A) and (B) 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 · june 2025

• Option B

  If g(x) ≠ 0 for each x ∈ [a, b], then f/g is Riemann integrable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  There is a positive $y_{0}$ such that the solution of the IVP is unbounded

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  A′ ⊆ 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 C

• Option A

  For M = 0, the maximum likelihood estimate of $\theta$ is equal to 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

  16th from the top and 26th from the bottom

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  15th from the top and 27th from the bottom

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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 D

  16th from the top and 27th from the bottom

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  There are infinitely many extremals

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question 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

  The mean and median of the data will be different if and only if r > 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 · june 2025

• Option B

  Variance of the data is positive if and only if r > 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 · june 2025

• Option C

  Mean and mode of the data will be same if and only if r = 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 · june 2025

• Option D

  $\inf A = 3 + 2\sqrt{2}$

  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 2024

• 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

  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

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