Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

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 A

  f(x) = sin(1/x)

  Why this is wrong, and why it looks right

  False. sin(1/x) has no limit as x → 0⁺ — SymPy samples it on (0.001, 0.05] and finds values ranging down to −0.99998 and up near +1, oscillating without settling. A continuous function on a bounded interval is uniformly continuous exactly when it extends continuously to the closure, and this one cannot.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023 · free

• Option C

  f(x) = eˣ cos(1/x)

  Why this is wrong, and why it looks right

  False. The factor eˣ tends to 1 at x = 0 and is harmless, but cos(1/x) oscillates exactly as sin(1/x) does. Multiplying an oscillating factor by a well-behaved one leaves the oscillation intact.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023

• Option D

  $f(x)=\cos x\cdot \cos (\pi /x)$

  Why this is wrong, and why it looks right

  False, for the same reason. cos x → 1 as x → 0, while $\cos (\pi /x)$ oscillates without limit — the $\pi$ in the numerator changes the frequency but not the behaviour. Three of the four options oscillate at 0; the exponential in (B) is the only one that damps.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023

• Option A

  G is a union of proper subgroups.

  Why this is wrong, and why it looks right

  False. Any cyclic group refutes it: a generator lies in no proper subgroup, so the proper subgroups cannot cover G. For $\mathbb{Z}/6$, SymPy gives the divisors 1, 2, 3, 6 — so the proper subgroups have orders 1, 2, 3, and their union misses the totient(6) = 2 generators, namely 1 and 5.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023

• Option B

  G is a union of proper subgroups if |G| has at least two distinct prime divisors.

  Why this is wrong, and why it looks right

  False, and $\mathbb{Z}/6$ is the counterexample the option seems designed to survive. |$\mathbb{Z}/6$| = 6 has two distinct prime divisors, yet $\mathbb{Z}/6$ is cyclic and so is not a union of proper subgroups. Having several prime factors says nothing about being cyclic.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023

• Option C

  If G is abelian, then G is a union of proper subgroups.

  Why this is wrong, and why it looks right

  False. $\mathbb{Z}/p$ is abelian and cyclic, so it is not a union of proper subgroups — in fact it has none besides the trivial one. Abelian is much weaker than non-cyclic.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023

• Option A

  S is in bijection with the set of rational numbers.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  S is in bijection with the set of real numbers.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option A

  The product of the numbers is zero

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option B

  At most two of these numbers are positive

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option C

  There cannot be exactly one zero

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option B

  X + I is invertible.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  Y is invertible.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  K is an algebraic extension of $\mathbb{Q}$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  $\mathbb{C}$ is an algebraic extension of K.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  $\int$₋${}_1^{1}f(x)x^n$ dx = 0 for all n ≥ 0 odd.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $\int$₋${}_1^{1}f(x)x^n$ dx = 0 for all n ≥ 0 even.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  f = 0.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  f is a constant function.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  There exists a compact subset K of $\mathbb{C}$ such that $f^{-1}(K)$ is not compact.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  The subsequences {$x_{kn}$}${}_n$ are convergent for every k ≥ 2.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option D

  lim |$x_n$₊${}_1-x_n$| = 0.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option A

  f never vanishes in [0, 1].

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  f is increasing.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  $A_n$ is finite for every integer n ≥ 1.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  $A_n$ is finite for some integer n ≥ 1.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $A_n$ is infinite for some integer n ≥ 1.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  f(z̄) is an entire function.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option B

  conj(f(z)) is an entire function.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option B

  V is the union of 3 proper subspaces.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  ${\mathbb{Q}}^2$ is connected.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

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