Previous year questions

The complete June 2023 paper is solved and free to read — every question, with the reasoning behind each option.

Every PYQ solved, tagged by topic and trap type. Free samples are open; the full set needs the PYQ Pack.

2023 DecemberPart Brolle-counting-rootsshow ▾

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)?

A.F(x) has exactly one real root.
B.F(x) has exactly four real roots.
C.F(x) has at most two real roots.✓
D.F(x) has at most one real root.

Solution

F′ = f changes sign exactly once, so F is monotone on each side of one point: at most two real roots (and the constant of integration can realise 0, 1 or 2).

Topic: Continuity and Differentiation › Differentiability, mean value theorems, Taylor, L'Hôpital

2023 DecemberPart Bshow ▾

Locked — Continuity and Differentiation. Unlock the PYQ Pack

2023 DecemberPart Bratio-f-prime-over-fshow ▾

Locked — Continuity and Differentiation. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

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?

A.Y is an infinite set.
B.Y has $2^{∣ X ∣}$ elements.✓
C.There is a bijective function from X to Y.
D.There is a surjective function from X to Y.

Solution

Every subset of X is the zero set of its indicator-complement (f = 1 off the subset, 0 on it), so Y is the power set of X, with $2^{∣ X ∣}$ > |X| elements.

Topic: Lebesgue Measure and Integration › Measurable sets and functions

2023 DecemberPart Bschur-invariant-flagshow ▾

Let A be an n × n matrix with complex entries. If n ≥ 4, which one of the following statements is true?

A.A does not have any non-zero invariant subspace in $C^{n}$ .
B.A has an invariant subspace in $C^{n}$ of dimension n − 3.✓
C.All eigenvalues of A are real.
D. $A^{2}$ does not have any invariant subspace in $C^{n}$ of dimension n − 1.

Solution

Over $C$ every matrix is triangularisable (Schur), so the span of the first k basis vectors of a triangularising basis is an invariant subspace of every dimension k — in particular n − 3 ≥ 1.

Topic: Vector Spaces and Linear Maps › Linear transformations, matrix representation, change of basis

2023 DecemberPart Bsylvester-rank-inequalityshow ▾

Locked — Vector Spaces and Linear Maps. Unlock the PYQ Pack

2023 DecemberPart Binvertibility-over-ringsshow ▾

Locked — Vector Spaces and Linear Maps. Unlock the PYQ Pack

2023 DecemberPart Brank-one-eigenvalueshow ▾

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?

A.1
B.n
C.n(n+1)/2
D.n(n+1)(2n+1)/6✓

Solution

A = vvᵀ with v = (1, 2, …, n) has rank 1; its only non-zero eigenvalue is $v^{T} v = \sum i^{2} = n (n + 1) (2 n + 1) /6$ .

Topic: Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

2023 DecemberPart Bindefinite-implies-isotropicshow ▾

Locked — Inner Product Spaces and Forms. Unlock the PYQ Pack

2023 DecemberPart Bleading-principal-minorsshow ▾

For

a \in R

, let A_a = [[2, −1, 0], [−1, 2, −1], [0, −1, a]]. Which one of the following statements is true?

A.A_a is positive definite for all a < 3.
B.A_a is positive definite for all a > 3.✓
C.A_a is positive definite for all a ≥ −2.
D.A_a is positive definite only for finitely many values of a.

Solution

Leading principal minors: 2, 3 and det A_a = 3a − 2. Positive definite iff a > 2/3, which contains all a > 3.

Topic: Inner Product Spaces and Forms › Quadratic forms, positive definiteness, Sylvester's law

2023 DecemberPart Bunion-of-proper-subgroupsshow ▾

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

A.G is a union of proper subgroups.
B.G is a union of proper subgroups if |G| has at least two distinct prime divisors.
C.If G is abelian, then G is a union of proper subgroups.
D.G is a union of proper subgroups if and only if G is not cyclic.✓

Solution

A finite group is the union of its proper subgroups iff it is not cyclic: if it is cyclic a generator lies in no proper subgroup; if not cyclic every element generates a proper subgroup.

Topic: Groups › Subgroups, cosets, Lagrange, cyclic groups

2023 DecemberPart Bpower-sums-mod-pshow ▾

Which one of the following is equal to

1^{37} + 2^{37} + 3^{37} + \dots + 8 8^{37}

Z /89 Z

A.88
B.−88
C.−2
D.0✓

Solution

89 is prime and $(Z /89) *$ is cyclic of order 88. For a generator $g, \sum_{k}$ $k^{37} = \sum_{j}$ $g^{37 j}$ is a geometric sum equal to 0 because 88 ∤ 37 (so $g^{37} \neq = 1)$ .

Topic: Rings and Fields › Ideals, quotient rings, prime & maximal ideals, CRT

2023 DecemberPart Bdense-subfieldsshow ▾

Locked — Rings and Fields. Unlock the PYQ Pack

2023 DecemberPart Bcr-orthogonalityshow ▾

Let

f : C \to 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?

A.For $c_{1}, c_{2} \in C$ , the level curves $u = c_{1}$ and $v = c_{2}$ are orthogonal wherever they intersect.
B. $\nabla u \cdot \nabla v = 0$ at every point.
C.If f is an entire function, then $\nabla u \cdot \nabla v = 0$ at every point.✓
D.If $\nabla u \cdot \nabla v = 0$ at every point, then f is an entire function.

Solution

For entire f the Cauchy–Riemann equations give $\nabla u \cdot \nabla v = u_{x} v_{x} + u_{y} v_{y} = u_{x} (- u_{y}) + u_{y} u_{x} = 0$ . Without holomorphy there is no such relation, and orthogonal gradients alone do not force CR (e.g. f = z̄ has $\nabla u \cdot \nabla v = 0)$ .

Topic: Analytic Functions › Cauchy–Riemann equations, harmonic functions

2023 DecemberPart Bshow ▾

Locked — Analytic Functions. Unlock the PYQ Pack

2023 DecemberPart Brouche-dominant-termshow ▾

How many roots does the polynomial

z^{100} - 50 z^{30} + 40 z^{10} + 6 z + 1

have in the open disc {

z \in C

: |z| < 1}?

A.100
B.50
C.30✓
D.0

Solution

On |z| = 1, | $- 50 z^{30}$ | = 50 > 1 + 40 + 6 + 1 = 48 ≥ | $z^{100} + 40 z^{10} + 6 z + 1$ |, so by Rouché the polynomial has as many zeros inside as $- 50 z^{30}$ : thirty.

Topic: Zeros and Mappings › Argument principle, Rouché's theorem, open mapping

2023 DecemberPart Bargument-principle-extra-factorshow ▾

Locked — Zeros and Mappings. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

Locked — Ordinary Differential Equations. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

The partial differential equation

x^{2} u_{xx} - 2

u_{x}

ᵧ

- 3 y^{2} u_{y} y + u_{x} - u_{y} = 0

A.elliptic in {(x, y) : y > 0}
B.parabolic in {(x, y) : x > 0, y > 0}
C.hyperbolic in {(x, y) : xy ≠ 0}✓
D.parabolic in {(x, y) : xy ≠ 0}

Solution

$B^{2} - 4$ AC $= 4 x^{2} y^{2} + 12 x^{2} y^{2} = 16 x^{2} y^{2} > 0$ wherever xy ≠ 0: hyperbolic.

Topic: Partial Differential Equations › Classification and canonical forms

2023 DecemberPart Bdalembert-limitshow ▾

Locked — Partial Differential Equations. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx

= 3 x + 2 y + 1, y (1) = 1

, at x = 1.1 (rounded to two decimal places) is

A.1.50
B.1.65
C.1.25✓
D.1.15

Solution

$y_{1} = 1 + 0.05 6 \approx 1.1225; y_{2} = 1.1225 + 0.05 3.15 + 2.245 + 1 \approx 1.1225 + 0.05 \cdot 2.529 \approx 1.249 \approx 1.25$ .

Topic: Numerical Analysis › Numerical ODE: Euler, Runge–Kutta

2023 DecemberPart Bshow ▾

Locked — Calculus of Variations. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

The value of

λ

for which the integral equation

y (x) = λ \int_{0}^{1} x^{2}

e^{x + t}

y(t) dt has a non-zero solution is

A. $4/ (1 + e^{2})$
B. $2/ (1 + e^{2})$
C. $4/ (e^{2} - 1)$ ✓
D. $2/ (e^{2} - 1)$

Solution

Separable kernel: $y = c x^{2} e$ ˣ with $c = λ c \int_{0}^{1} t^{2}$ $e^{2 t}$ dt $= λ c (e^{2} - 1) /4$ , so $λ = 4/ (e^{2} - 1)$ .

Topic: Linear Integral Equations › Fredholm and Volterra equations

2023 DecemberPart Bshow ▾

Locked — Classical Mechanics. Unlock the PYQ Pack

2023 DecemberPart Bshow ▾

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

A.P(X > 9) ≥ 11/12
B.P(X < 9) ≥ 11/12✓
C. $E (X - 3/4)^{2} \geq 11/12$
D.(11/9)X ~ Poisson(11/12)

Solution

Markov: P(X ≥ 9) ≤ E[X]/9 = 1/12, so P(X < 9) ≥ 11/12. The variance is 3/4 < 11/12, and a scaled Poisson is not Poisson.

Topic: Probability › Random variables, distributions, moments, MGF