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 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?

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  Part B · june 2023 · 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?

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  Part B · december 2023

• 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

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  Part B · december 2023

• 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?

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  Part C · june 2023

• 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)?

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  Part B · june 2023

• 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. $\sum 1/\sqrt{n}$ 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: $(a){\sum }_{n\ge 1}$ $\sin (n\pi /2)/\sqrt{n},(b){\sum }_{n\ge 1}$ $\log (1+1/n^2)$. Which one of the following statements is true?

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  Part B · december 2023

• 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 $Var({\epsilon }_i)=x_i^2{\sigma }^2$ 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 $Y_i=\beta x_i+{\epsilon }_i,i=1,\dots ,n$, where $E({\epsilon }_i)=0,Cov({\epsilon }_i,{\epsilon }_k)=0$ for i ≠ k and $Var({\epsilon }_i)=x_i^2{\sigma }^2$. The best linear unbiased estimator of $\beta$ is

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  Part B · june 2023

• 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?

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  Part B · december 2023

• 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?

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  Part B · december 2023

• Option A

  Both (a) and (b) are true.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option A

  ⟨x, y⟩ $={\sum }_{i,j=1}^n$ $x_i{y}_j$

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option D

  ⟨x, y⟩ $={\sum }_{j=1}^n$ $x_j{y}_n$₋${}_j$₊${}_1$

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

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

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option C

  1.5 second

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• 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.

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  Part C · june 2023

• Option D

  $S_1$ and $S_2$ are both infinite.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  f(x) = eˣ

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  f(x) = x − 1

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  M is the most popular English magazine in the country.

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option C

  M is the most popular magazine in the country.

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option D

  f need not be onto.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  f has infinitely many zeros in [0, 1].

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  $(1/\sum n_i){\sum }_i{\sum }_j{Y}_{ij}$ is an unbiased estimator of $\mu$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

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

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $(1/2\pi i){\int }_C($zf)′/(zf) dz $=n_0-n_p$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  (P) has infinitely many solutions.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option B

  Let $A_2=$ {$x^2+y^2\le 1$}. For each $\gamma \in p(A_2)$ there exists $\epsilon >0$ with $(\gamma -\epsilon ,\gamma +\epsilon )\subseteq p(A_2)$.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $(\sum$ iY${}_i,\sum i^2{Y}_i^2)$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

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

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

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

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

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

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

Other mistakes that cost marks