Execution slip

The idea was right. The computation was not.

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

  4

  Why this is wrong, and why it looks right

  False — 4 would be the count for a product of two odd primes. 105 has three distinct prime factors (SymPy: 3 · 5 · 7), so the count is $2^3$, not $2^2$.

  From this question

  The number of solutions of the equation $x^2=1$ in the ring $\mathbb{Z}/105\mathbb{Z}$ is

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  Part B · june 2023 · free

• Option A

  3

  Why this is wrong, and why it looks right

  False. 3 is the difference in correct answers minus one, or what you get by taking 18 − 13 = 5 and subtracting 2 instead of 1. The quantity asked for is the difference in incorrect answers, which is 5 − (x − y) where x − y = 1.

  From this question

  In a test with multiple choice questions, candidates get 4 marks for a correct answer and lose 1 mark for an incorrect answer. Two candidates A and B attempting 18 and 13 questions, respectively, secure equal marks. How many more INCORRECT answers does A have compared to B?

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  Part A · september 2022

• Option C

  5

  Why this is wrong, and why it looks right

  False. 5 is the difference in attempted questions, 18 − 13, which would be the answer only if both candidates got the same number right. They do not — A gets exactly one more right than B, and that one shifts the incorrect gap down to 4.

  From this question

  In a test with multiple choice questions, candidates get 4 marks for a correct answer and lose 1 mark for an incorrect answer. Two candidates A and B attempting 18 and 13 questions, respectively, secure equal marks. How many more INCORRECT answers does A have compared to B?

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  Part A · september 2022

• Option D

  6

  Why this is wrong, and why it looks right

  False. 6 is 5 + 1 rather than 5 − 1, that is adding the correct-answer difference instead of subtracting it. More correct answers for A means fewer incorrect ones, so the adjustment reduces the gap.

  From this question

  In a test with multiple choice questions, candidates get 4 marks for a correct answer and lose 1 mark for an incorrect answer. Two candidates A and B attempting 18 and 13 questions, respectively, secure equal marks. How many more INCORRECT answers does A have compared to B?

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  Part A · september 2022

• Option A

  2.6

  Why this is wrong, and why it looks right

  False. 2.6 would follow from dividing by a total that omits the original solution — 144/5400 or similar. The new volume is 360 + 3600 = 3960, not 3600 alone.

  From this question

  A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?

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  Part A · september 2022

• Option C

  4.0

  Why this is wrong, and why it looks right

  False. 4.0 comes from dividing 144 by 3600 — the added water alone — giving exactly 4%. Forgetting that the original 360 ml is still in the beaker is the single most common slip here.

  From this question

  A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?

  See the full question

  Part A · september 2022

• Option D

  1.0

  Why this is wrong, and why it looks right

  False. 1.0 is roughly 40 divided by 40, that is treating the tenfold dilution as if it acted on the percentage rather than on the volume. Dilution by a factor of 11 takes 40% to 3.64%, not to 1%.

  From this question

  A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?

  See the full question

  Part A · september 2022

• Option A

  P(X > 9) ≥ 11/12

  Why this is wrong, and why it looks right

  False, and it is (B) with the inequality pointing the wrong way. A Poisson variable with mean 3/4 is concentrated near 0 — Markov's inequality gives P(X ≥ 9) ≤ (3/4)/9 = 1/12 (verified), so the upper tail is tiny, not overwhelming.

  From this question

  Suppose X ~ Poisson(3/4). Then which of the following statements is true?

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  Part B · december 2023

• Option C

  $E(X-3/4{)}^2\ge 11/12$

  Why this is wrong, and why it looks right

  False. $E(X-3/4{)}^2$ is the variance, and for a Poisson distribution the variance equals the mean, so it is 3/4 = 0.75. That is less than 11/12 ≈ 0.917 — SymPy gives the difference as −1/6, confirming the inequality fails.

  From this question

  Suppose X ~ Poisson(3/4). Then which of the following statements is true?

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  Part B · december 2023

• Option C

  Var(Y) = 3.04

  Why this is wrong, and why it looks right

  False, and the number offered is revealing. Variances do not mix with the weights. Correctly, $E[Y^2]=0.4(10+100)+0.6(4+9)=51.8$, so $Var(Y)=51.8-5.8^2=18.16($verified). SymPy also computes $0.4^2\cdot 10+0.6^2\cdot 4=3.04$ — precisely the option offered — which is the variance of the *linear combination$\ast 0.4X_1+0.6X_2$. This option is the right formula applied to the wrong object.

  From this question

  Let $X_1$ and $X_2$ be independent, $X_1$ gamma with mean 10 and variance 10, and $X_2$ ~ N(3, 4). Let $f_1,f_2$ be their densities. Define Y with density $f(y)=0.4f_1(y)+q{f}_2(y)$. Which of the following are true?

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  Part C · june 2023

• Option D

  $Y=0.4X_1+$ qX${}_2$

  Why this is wrong, and why it looks right

  False, and it is the conceptual error behind (C). A mixture is a random choice between two distributions — with probability 0.4 you observe a gamma variable, otherwise a normal one — not a weighted sum of both. The distinction shows up immediately in the variance (18.16 versus 3.04) and in the shape: a mixture here is bimodal, while a linear combination is not.

  From this question

  Let $X_1$ and $X_2$ be independent, $X_1$ gamma with mean 10 and variance 10, and $X_2$ ~ N(3, 4). Let $f_1,f_2$ be their densities. Define Y with density $f(y)=0.4f_1(y)+q{f}_2(y)$. Which of the following are true?

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  Part C · june 2023

• Option C

  $-(n-1)/\sum \ln X_i$

  Why this is wrong, and why it looks right

  False — this is the UMVUE of $\theta$ itself, with the (n − 1) correcting the bias of the MLE. It is the right estimator for the wrong quantity. Read which of $\theta$ and $1/\theta$ the question asks for before choosing.

  From this question

  For n ≥ 2, let $X_1,\dots ,X_n$ be a random sample with density f(x|$\theta )=$ $\theta x^{\theta -1}$ on $0<x<1,\theta >0$ unknown. Which of the following is the UMVUE for $1/\theta$?

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  Part B · december 2023

• Option D

  $-(2/n)\sum \ln X_i$

  Why this is wrong, and why it looks right

  False. This is twice the correct estimator, so its expectation is $2/\theta$ rather than $1/\theta$ — it is biased by a factor of 2 and cannot be a UMVUE.

  From this question

  For n ≥ 2, let $X_1,\dots ,X_n$ be a random sample with density f(x|$\theta )=$ $\theta x^{\theta -1}$ on $0<x<1,\theta >0$ unknown. Which of the following is the UMVUE for $1/\theta$?

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  Part B · december 2023

• Option A

  4.5

  Why this is wrong, and why it looks right

  False, and 4.5 is what intuition suggests: the water fraction fell by only one percentage point, so surely the melon barely changed. It did not — the solid matter went from 1% to 2% of the total, which means the total had to halve.

  From this question

  A 5 kg watermelon contains 99% water by weight. Some of the water evaporates and the melon now contains 98% water by weight. What is the weight (in kg) of watermelon now?

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  Part A · september 2022

• Option C

  4.8

  Why this is wrong, and why it looks right

  False. 4.8 comes from subtracting a small amount of water proportional to the one-point change, treating percentages as if they subtract directly. The relevant ratio is between the solid fractions, 1% and 2%, which differ by a factor of two rather than by one point.

  From this question

  A 5 kg watermelon contains 99% water by weight. Some of the water evaporates and the melon now contains 98% water by weight. What is the weight (in kg) of watermelon now?

  See the full question

  Part A · september 2022

• Option D

  4.9

  Why this is wrong, and why it looks right

  False. 4.9 is 5 minus 1% of 5, that is removing 0.05 kg — the weight of the solid rather than the weight of the evaporated water. The water lost is 4.95 − 2.45 = 2.5 kg, fifty times larger.

  From this question

  A 5 kg watermelon contains 99% water by weight. Some of the water evaporates and the melon now contains 98% water by weight. What is the weight (in kg) of watermelon now?

  See the full question

  Part A · september 2022

• Option A

  6

  Why this is wrong, and why it looks right

  False. 6 would be p − 2, or the treatment degrees of freedom for a 7 × 7 square. Solving (p − 1)(p − 2) = 42 gives p = 8, not 7 — at p = 7 the error degrees of freedom would be 6 × 5 = 30.

  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

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  Part B · december 2023

• Option D

  inf U = 2.

  Why this is wrong, and why it looks right

  False. 2 satisfies neither inequality. This is the factorisation-slip option: reading $x^2-7x+12$ as (x − 2)(x − 5) — which is actually $x^2-7x+10$ — plants a spurious root at 2.

  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 A

  (0, 5)ᵀ, [[5, −3], [−3, 40]]

  Why this is wrong, and why it looks right

  False on the off-diagonal. The variance 40 is right, but the covariance must also be rescaled: $Cov(X_1,5-2X_2)=-2\cdot Cov(X_1,X_2)=-2(-3)=6($verified), not −3. Scaling one coordinate scales every entry of the covariance matrix that involves it, not just the diagonal one.

  From this question

  Let $X=(X_1,X_2{)}^T$ be bivariate normal with mean (0, 0)ᵀ and covariance $\sum =[[5,-3],[-3,10]]$. The mean vector and covariance matrix of $Y=(X_1,5-2X_2{)}^T$ are

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  Part B · june 2023

• Option B

  (0, 5)ᵀ, [[5, −6], [−6, 20]]

  Why this is wrong, and why it looks right

  False on both entries. The variance should be $(-2{)}^2\cdot 10=40$, not 20 — the factor is squared. And the covariance should be +6, not −6: multiplying by −2 flips the sign of the original −3 to give a positive value.

  From this question

  Let $X=(X_1,X_2{)}^T$ be bivariate normal with mean (0, 0)ᵀ and covariance $\sum =[[5,-3],[-3,10]]$. The mean vector and covariance matrix of $Y=(X_1,5-2X_2{)}^T$ are

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  Part B · june 2023

• Option C

  n(n+1)/2

  Why this is wrong, and why it looks right

  False, and it is the engineered near-miss: n(n+1)/2 is $\sum i$, whereas the eigenvalue is $\sum i^2$. At n = 3, SymPy gives $\sum i=6$ but the actual eigenvalue is 14. Writing A = vvᵀ makes the non-zero eigenvalue $v^{T}v=\sum i^2$, so a moment's inattention between 'sum' and 'sum of squares' lands exactly here.

  From this question

  Let $A=(a_{ij})$ be the n × n real matrix with $a_{ij}=$ ij for all 1 ≤ i, j ≤ n. If n ≥ 3, which one of the following is an eigenvalue of A?

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  Part B · december 2023

• Option D

  $li{m}_{k\to \infty }$ |$I_k$| $=\infty$.

  Why this is wrong, and why it looks right

  False — the moduli tend to 0, not infinity. SymPy confirms $\lim 2\pi /\Gamma (k)=0$ as $k\to \infty$: the factorial in the denominator dominates everything. The values run $2\pi ,2\pi ,\pi ,\pi /3$, and shrink from there.

  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?

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  Part C · december 2023

• Option A

  elliptic in {(x, y) : y > 0}

  Why this is wrong, and why it looks right

  False. With $A=x^2,B=-2$xy, $C=-3y^2$, SymPy expands the discriminant $B^2-4$AC to $16x^2{y}^2$, which is a square and therefore never negative. Elliptic requires $B^2-4$AC < 0, so the equation is elliptic nowhere — least of all on a half-plane where x is unrestricted.

  From this question

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

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  Part B · december 2023

• Option B

  parabolic in {(x, y) : x > 0, y > 0}

  Why this is wrong, and why it looks right

  False. Parabolic requires $B^2-4$AC = 0, but on {x > 0, y > 0} the discriminant $16x^2{y}^2$ is strictly positive. The sign error that leads here is dropping the minus in $C=-3y^2$: with $C=+3y^2$ the discriminant would be $4x^2{y}^2-12x^2{y}^2=-8x^2{y}^2<0$, and the classification would flip entirely.

  From this question

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

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  Part B · december 2023

• Option C

  1/2

  Why this is wrong, and why it looks right

  False — 1/2 is the prior mean, 5/(5 + 5) (verified). Choosing it means never updating: the data pull the estimate down from 1/2 towards the observed proportion, landing at 3/7. A Bayes estimate equal to the prior mean would require the data to say nothing at all.

  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 C

  $Z_n\to Z$ in distribution as $n\to \infty$

  Why this is wrong, and why it looks right

  False. The scaling is wrong: $Y_n-5$ ~ Exp(n) has size of order 1/n, so the natural normaliser is n, not $\sqrt{n}$. With $\sqrt{n}$ the product has mean $\sqrt{n}/n=1/\sqrt{n}\to 0($verified), so $Z_n\to 0$ in probability, a degenerate limit. And even correctly scaled, $n(Y_n-5)$ ~ Exp(1) — extreme-value limits are exponential or Weibull, never normal. $\sqrt{n}$ is the central-limit scaling for sums, not for minima.

  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?

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  Part B · december 2023

• Option D

  $\lim P(1<Z_n<2)=\Phi (2)-\Phi (1)$

  Why this is wrong, and why it looks right

  False, for the same reason as (C). Since $Z_n\to 0$ in probability, $P(1<Z_n<2)\to 0$, not $\Phi (2)-\Phi (1)\approx 0.136$. Any normal-looking answer here signals that a central limit theorem has been applied to a minimum.

  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?

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  Part B · december 2023

• Option B

  $E(M^2)=17$

  Why this is wrong, and why it looks right

  False. $E(M^2)=(9+16+25)/3=50/3\approx 16.67($verified), not 17. The near-miss is what makes this tempting — 50/3 rounds to 16.67, and 17 is what you get by rounding up or by mis-adding the squares to 51.

  From this question

  Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is drawn. Let M be the sample mean. Which of the following statements are true?

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  Part C · december 2023

• Option D

  Var(M) = 1

  Why this is wrong, and why it looks right

  False. $Var(M)=E(M^2)-(EM{)}^2=50/3-16=2/3($verified), not 1. Note that Var(M) = 2/3 is much smaller than the population variance 8/3, reflecting both the averaging and the finite population correction.

  From this question

  Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is drawn. Let M be the sample mean. Which of the following statements are true?

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  Part C · december 2023

• Option D

  The series is absolutely convergent.

  Why this is wrong, and why it looks right

  False. Absolute convergence would require $\sum 1/(\sqrt{n+1}+\sqrt{n})$ to converge, but that series is comparable to $\sum 1/(2\sqrt{n}),a$ divergent p-series with p = ½ ≤ 1. The rationalisation is what exposes the true size of the terms — without it, $\sqrt{n+1}-\sqrt{n}$ looks deceptively small.

  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?

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  Part B · june 2023

• Option A

  x(t) is a bounded solution for some $x_0\ne 0$.

  Why this is wrong, and why it looks right

  False. SymPy gives the eigenvalues of A as 6 and 1 — both strictly positive (the characteristic polynomial ${\lambda }^2-7\lambda +6$ has roots 1 and 6, verified). Every non-zero solution is a combination of $e^{6t}$ and $e^t$ modes, so all of them grow without bound. Bounded solutions would need an eigenvalue with non-positive real part.

  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?

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  Part B · june 2023

• Option D

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

  Why this is wrong, and why it looks right

  False. rank(BBᵀ) = rank(B) = 2 for a real (or rational) matrix — SymPy confirms rank 2 for the corresponding 3 × 3 product. A map ${\mathbb{Q}}^3\to {\mathbb{Q}}^3$ of rank 2 has a non-trivial kernel, so it is not injective. Multiplying by the transpose cannot raise the rank.

  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 B

  $K(x,t)=t^2(1-x)$ for $t<x;x^2(1-t)$ for t > x

  Why this is wrong, and why it looks right

  False. The branches $t^2(1-x)$ and $x^2(1-t)$ do not match at t = x — they give $x^2(1-x)$ and $x^2(1-x)$, which happen to agree, but the derivative jump is then 2x·… rather than the required −1, and the function no longer solves y″ = 0 away from t = x. A Green's function for a second-order operator is piecewise linear here, not quadratic.

  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 A

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

  Why this is wrong, and why it looks right

  Half right. (a) does converge, but so does $(b):\log (1+1/n^2)$ ~ $1/n^2$, and SymPy evaluates the sum numerically to ≈ 1.3018 — finite. Dismissing log(1 + u) as 'too slow to decay' overlooks that $u=1/n^2$ is already summable, and log(1 + u) ≤ u.

  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 B

  n ≡ 1 (mod 35).

  Why this is wrong, and why it looks right

  False. 35 = 5 × 7, and although n ≡ 1 (mod 7) holds, the mod-5 part does not: n ≡ 4 (mod 5), not 1. The witness settles it — 64 mod 35 = 29, not 1. This option is what you get by combining the modulus 7 with the modulus 15 without splitting off the 5 and checking it.

  From this question

  Let $n\in \mathbb{Z}$ be such that n ≡ 1 (mod 7) and n ≡ 4 (mod 15). Which of the following statements are true?

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  Part C · december 2023

• Option D

  n ≡ 1 (mod 5).

  Why this is wrong, and why it looks right

  False. n ≡ 4 (mod 15) reduces to n ≡ 4 (mod 5), not 1 — the witness gives 64 mod 5 = 4. Reducing '4 mod 15' to '1 mod 5' is the slip; 4 and 1 are only congruent modulo 3.

  From this question

  Let $n\in \mathbb{Z}$ be such that n ≡ 1 (mod 7) and n ≡ 4 (mod 15). Which of the following statements are true?

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  Part C · december 2023

• Option A

  100

  Why this is wrong, and why it looks right

  False. 100 is the total number of roots in $\mathbb{C}$, by the fundamental theorem of algebra — not the number inside the unit disc. Rouché's theorem localises them, and it selects the dominant term on the circle, not the leading term of the polynomial.

  From this question

  How many roots does the polynomial $z^{100}-50z^{30}+40z^{10}+6z+1$ have in the open disc {$z\in \mathbb{C}$ : |z| < 1}?

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  Part B · december 2023

• Option B

  50

  Why this is wrong, and why it looks right

  False. 50 is the coefficient of the dominant term, not its degree. Rouché counts zeros of the dominant term $-50z^{30}$, and a constant multiple contributes nothing to the zero count — only the exponent 30 does.

  From this question

  How many roots does the polynomial $z^{100}-50z^{30}+40z^{10}+6z+1$ have in the open disc {$z\in \mathbb{C}$ : |z| < 1}?

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  Part B · december 2023

• Option B

  1.65

  Why this is wrong, and why it looks right

  False. 1.65 is roughly what a single step of size 0.1 would give if the slope were mistakenly evaluated as $\sqrt{6}\approx 2.45$ and then multiplied by a larger step, or what four steps would give. The problem specifies h = 0.05 from x = 1 to x = 1.1, which is exactly two steps.

  From this question

  Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx $=\sqrt{3x+2y+1},y(1)=1$, at x = 1.1 (rounded to two decimal places) is

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  Part B · december 2023

• Option D

  1.15

  Why this is wrong, and why it looks right

  False. 1.15 is close to the result of taking only the first step and stopping (1.1225 rounds to 1.12), or of using a step size half the one specified. The second step contributes another 0.126, which is what separates this from the right answer.

  From this question

  Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx $=\sqrt{3x+2y+1},y(1)=1$, at x = 1.1 (rounded to two decimal places) is

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  Part B · december 2023

• Option B

  $\phi (2)=2$

  Why this is wrong, and why it looks right

  False: $\phi (2)=2^2=4$, not 2. This option (and (D)) is what you get by assuming $\phi (x)=x$, which fits $\phi (1)=1$ but solves nothing — SymPy substitutes y = x and obtains 2x(x − 1), which is not identically zero. Matching the single condition $\phi (1)=1$ is not the same as solving the equation.

  From this question

  Consider the initial value problem $x^2{y}^{\prime \prime }-2x^2{y}^{\prime }+(4x-2)y=0,y(0)=0$. Suppose $y=\phi (x)$ is a polynomial solution satisfying $\phi (1)=1$. Which of the following statements are true?

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  Part C · december 2023

• Option D

  $\phi (3)=3$

  Why this is wrong, and why it looks right

  False: $\phi (3)=3^2=9$, not 3. Same underlying error as (B) — the linear guess $\phi (x)=x$.

  From this question

  Consider the initial value problem $x^2{y}^{\prime \prime }-2x^2{y}^{\prime }+(4x-2)y=0,y(0)=0$. Suppose $y=\phi (x)$ is a polynomial solution satisfying $\phi (1)=1$. Which of the following statements are true?

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  Part C · december 2023

• Option A

  2

  Why this is wrong, and why it looks right

  False — 2 is the x-coordinate, that is the initial value u(2, 0). The solution changes as y increases; reading the answer straight off the initial data ignores the evolution entirely.

  From this question

  Let u(x, y) be the solution of the Cauchy problem $u\cdot u_x+u_y=0$ for $x\in \mathbb{R},y>0$, with u(x, 0) = x for $x\in \mathbb{R}$. Which of the following is the value of u(2, 3)?

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  Part B · june 2023

• Option D

  1/3

  Why this is wrong, and why it looks right

  False — 1/3 is what you get from x/(1 + y) with the roles of x and y swapped, or from 1/y. Note that this solution never forms a shock for y > 0, since the characteristics fan out rather than converge: the denominator 1 + y stays positive.

  From this question

  Let u(x, y) be the solution of the Cauchy problem $u\cdot u_x+u_y=0$ for $x\in \mathbb{R},y>0$, with u(x, 0) = x for $x\in \mathbb{R}$. Which of the following is the value of u(2, 3)?

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  Part B · june 2023

• Option A

  ${\sum }_{l=0}^{\infty }$ 1/(l+1)!

  Why this is wrong, and why it looks right

  False — only one factorial appears, so this is the series for a single exponential rather than a product of two. SymPy's partial sums separate them clearly: the first four terms of this series give 41/24 ≈ 1.708, while the correct series gives 229/144 ≈ 1.590.

  From this question

  Let $f(z)=\exp (z+1/z),z\in \mathbb{C}$∖{0}. The residue of f at z = 0 is

  See the full question

  Part B · june 2023

• Option D

  ${\sum }_{l=0}^{\infty }$ $1/((l^2+l)!)$

  Why this is wrong, and why it looks right

  False. $(l^2+l)$! is a single factorial of a product-like expression, which never arises from multiplying two series — the two factorials stay separate. It is the right shape only if you mistakenly combine m! and (m+1)! into one factorial of m(m+1).

  From this question

  Let $f(z)=\exp (z+1/z),z\in \mathbb{C}$∖{0}. The residue of f at z = 0 is

  See the full question

  Part B · june 2023

• Option A

  $4/(1+e^2)$

  Why this is wrong, and why it looks right

  False. The denominator should be $e^2-1$, not $e^2+1$. That sign comes from evaluating ${\int }_0^1{t}^2{e}^{2t}$ dt: SymPy gives $e^2/4-1/4,a$ difference. Adding the boundary terms instead of subtracting produces this option.

  From this question

  The value of $\lambda$ for which the integral equation $y(x)=\lambda {\int }_0^1{x}^2$ $e^{x+t}$ y(t) dt has a non-zero solution is

  See the full question

  Part B · december 2023

• Option D

  $2/(e^2-1)$

  Why this is wrong, and why it looks right

  False. The denominator is right but the numerator is halved. ${\int }_0^1{t}^2{e}^{2t}$ dt $=(e^2-1)/4$, and inverting a quarter gives a factor 4, not 2. Mis-integrating by parts once rather than twice is the usual source of the missing factor.

  From this question

  The value of $\lambda$ for which the integral equation $y(x)=\lambda {\int }_0^1{x}^2$ $e^{x+t}$ y(t) dt has a non-zero solution is

  See the full question

  Part B · december 2023

• Option B

  1/4

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option D

  1/15

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option C

  (a) is true and (b) is false.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  ⟨x, y⟩ $={\sum }_{i,j=1}^n$ $(x_i^2+y_j^2)$

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option D

  (P) has infinitely many stationary functions (extremals).

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  1.0 second

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option B

  3/4

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option C

  4/3

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option B

  $li{m}_{t\to \infty }$ u(5, t) = 2

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  $li{m}_{t\to \infty }$ u(5, t) = 1/2

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  $li{m}_{t\to \infty }$ u(5, t) = 0

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  525

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option B

  625

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option C

  550

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option D

  There are two distinct differentiable real-valued functions g on an interval of $\mathbb{R}$ containing 0 such that f(x, g(x)) = 0.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option C

  $91C_0$

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option D

  $81C_0$

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

• Option D

  If $X_1$ and $Y_1$ are independent, then $S_{\rho }=1/2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $\mathrm{limsup}{a}_{2n}=1/2$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  $\mathrm{limsup}{a}_{3n}=0$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  $(\sqrt{3},23/3)$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  f has a simple pole at z = 0.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  $T_{2,n}$ → 0 in probability as $n\to \infty$ under $H_0$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $T_{2,n}$ converges in distribution to a degenerate random variable under $H_0$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  a is divisible by $3^6$.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  y(x) exists on $(0,\infty )$, but not bounded above.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  J = (−1, 1)

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  −1/3

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  5/6

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  $y(1)=(1+\pi /4)e$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $y(\sqrt{3})=(1+$ $3\pi /4)e^{\sqrt{3}}$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  1875/3456

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  E(Y) = 0

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $P(W_2$ˣ $<W_3$ʸ) = 13/16

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $P(W_1$ˣ $<W_1$ʸ) = 1/4

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  $W_p(n-1,\sum )$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  A is positive definite.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  3

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  There exists a prime number p such that g(X) mod p is irreducible in $(\mathbb{Z}/p\mathbb{Z})[X]$.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $E(T_2)=1/35$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $(1/n_2){\sum }_j(Y_{2j}-$ Ȳ${}_2)$ is an unbiased estimator of ${\alpha }_2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  T ≥ ½ ${\chi }_{48,0.05}^2$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  T ≥ ½ ${\chi }_{24,0.05}^2$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  T ≤ ½ ${\chi }_{24,0.95}^2$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  0.83

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  0.77

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  0.84

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

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

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

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

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  $S^2=77$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  0

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  2

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  3z(x) + y(x) = 2 for x ∈ [0,1].

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option D

  y(x) + z(x) = x for x ∈ [0,1].

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option B

  $li{m}_n\exp (lo{g}_e((n{\pi }^2+(-1{)}^n{e}^2)/(7n)))$ does not exist.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  Expectation of $X_{(2)}$ is 1/2.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option A

  $P_{H_0}(R$ = 6) = 28/286, $P_{H_0}(R$ = 9) = 28/143.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  $P_{H_0}(R$ = 6) = 21/286, $P_{H_0}(R$ = 9) = 15/286.

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  The test in (B) is the uniformly most powerful test of size $\alpha$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  P(X > 4) = P(X > 1)P(X > 2)

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  6X ~ ${\chi }_3^2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  $W_1(1,2)$

  The analysis of this one comes with the PYQ pack.

  Part B · june 2023

• Option B

  Size of the test $\psi$ is 0.05.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option D

  a > b.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  If ${\alpha }_0=2$, then ${\beta }_0=2,P=1/4,Q=3/4$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  If ${\beta }_0=3$, then ${\alpha }_0=3,P=1/6,Q=5/6$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

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

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option D

  $(\sum Y_i,\sum$ iY${}_i)$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  X̄ is the method of moments estimator of $\theta$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  $a_{22}$ ~ ${\chi }_3^2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $(1/33)(a_{11}-4a_{13}+4a_{33})$ ~ ${\chi }_3^2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  $(1/9)(a_{11}-4a_{13}+4a_{33})$ ~ ${\chi }_5^2$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  J[y] ≤ J[0] for all $y\in B_0(0,\epsilon )$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  If ${\psi }_1$ rejects $H_0$, then ${\psi }_2$ also rejects $H_0$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  If ${\psi }_2$ does not reject $H_0$, then ${\psi }_1$ also does not reject $H_0$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  $(1-2cos\theta )\theta$̈ $+\sin \theta (\theta$̇${)}^2-(g/a)\sin \theta =0$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option C

  $m(1-2cos\theta )\theta$̈ $+\sin \theta (\theta$̇${)}^2+(g/a)\sin \theta =0$

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  n⁻${}^1{\sum }_{i=1}^{2n}$ $X_i$

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option C

  a + b = 2

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  a = 1, b = 1

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

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