GATE Engineering Mathematics previous year questions

(A) Being a classmate is symmetric, so it is true. (B) ∀x P(x) is equivalent to ¬∃x ¬P(x), so the implication holds. (D) is the standard translation of a universal statement. (C) ∀x IsKing(x) ∧ IsPerson(x) says everything is a king and a person; the correct form is ∀x IsKing(x) ⇒ IsPerson(x). Answer: (C).

Concept: With ∀ the connective is ⇒, with ∃ it is ∧. The classic trap is writing ∀ with ∧.

There are 2²⁰²⁶ - 1 non-empty subsets, each equally likely. The product is odd exactly when every element is odd. S has 1013 odd numbers, giving 2¹⁰¹³ - 1 non-empty all-odd subsets. Even-product subsets: (2²⁰²⁶ - 1) - (2¹⁰¹³ - 1) = 2²⁰²⁶ - 2¹⁰¹³ = 2¹⁰¹³(2¹⁰¹³ - 1). Probability = 2¹⁰¹³(2¹⁰¹³ - 1)/(2²⁰²⁶ - 1). Answer: (C).

Concept: Complement counting. The trap is dividing by 2²⁰²⁶, forgetting that the empty set is excluded from the sample space.

Taking each solution as equally likely: by stars and bars, non-negative solutions of n₁ + n₂ + n₃ + n₄ = 20 number C(20 + 3, 3) = C(23,3). Positive solutions: substitute mᵢ = nᵢ - 1 ≥ 0, so m₁ + ... + m₄ = 16, giving C(16 + 3, 3) = C(19,3). Probability = C(19,3)/C(23,3). Answer: (A).

Concept: Stars and bars for non-negative vs positive solutions. The trap is using C(n,k) with k = 4 instead of k - 1 = 3 bars.

M is rotation by θ, so Mᵏ is rotation by kθ. M⁵ is rotation by 2π, which is I₂. 2026 = 5 · 405 + 1, so M²⁰²⁶ = (M⁵)⁴⁰⁵ · M = M. Answer: (B) M.

Concept: Powers of a rotation matrix are periodic. The trap is computing 2026 mod 5 wrongly or expecting I₂ because 2026 is even.

X entails Y means no model makes X true and Y false. That is exactly: X ⇒ Y is valid (A), X ∧ ¬Y is unsatisfiable, i.e. always FALSE (B), and 'if X then Y' (C). The converse 'if Y then X' (D) need not hold: Y can be true while X is false. Answer: (A), (B), (C).

Concept: Deduction theorem: entailment equals validity of the implication. The trap is accepting the converse.

f(1) = 1 - 3 + 2 = 0, so (D) holds; f(x) = (x - 1)(x² - 2x - 2), other roots 1 ± √3 ≈ -0.732 and 2.732. Only -0.732 lies in [-0.9, 0], so (A) is false. f'(x) = 3x² - 6x = 0 at x = 0, 2. f(0) = 2, f(2) = -2, f(3) = 2, and f(x) → -2 as x → -1 but x = -1 is excluded. So the minimum value -2 is attained only at x = 2 (B true), while the maximum value 2 is attained at both x = 0 and x = 3 (C false). Answer: (B), (D).

Concept: Extrema on a half-open interval: check endpoints and excluded points. The trap is forgetting the endpoint x = 3 ties with x = 0, and that x = -1 is not in the domain.

Both densities are symmetric about 0, so G(0) = H(0) = 0.5 (A true). g(0) = 1/√(2π) ≈ 0.399 and h(0) = 1/π ≈ 0.318 (t₁ is Cauchy), so (D) false. t₁ has heavier tails: h > g for |y| > c and h < g near 0. Hence P(Y > c) > P(Z > c), i.e. H(c) < G(c), so (B) false. By symmetry P(Y < -c) > P(Z < -c), i.e. G(-c) < H(-c) (C true). Answer: (A), (C).

Concept: Heavy tails of the t distribution versus the normal. The trap is assuming a heavier-tailed CDF is larger everywhere; it is smaller on the right tail.

f(f(n)) = n means f is an involution: a permutation whose cycles all have length 1 or 2. Count by number of 2-cycles: none: 1 (identity); one 2-cycle: C(4,2) = 6; two 2-cycles: 3 ways to pair four elements. Total 1 + 6 + 3 = 10. Answer: 10.

Concept: Involutions consist of fixed points and transpositions. The trap is forgetting the identity or the double-transposition cases.

P(X > 10 | X > 5) = P(X > 10)/P(X > 5) = e^(-10/λ)/e^(-5/λ) = e^(-5/λ) = P(X > 5) = 0.35, by the memoryless property. Answer: 0.35 (accepted range 0.34 to 0.36).

Concept: Memorylessness of the exponential distribution. The trap is computing λ first or squaring 0.35.

The rule: for a creative person, success holds exactly when discipline holds, i.e. C ⇒ (S ⇔ D) - statement (ii). (i) fails: a non-creative disciplined person makes D true and C ∧ S false, yet the rule says nothing about non-creative people. (iii): (D ⇒ S) ∨ ¬S = ¬D ∨ S ∨ ¬S, which is always true, so (iii) reduces to C ⇔ TRUE, i.e. just C - not the rule. Answer: (B) Only (ii).

Concept: Translating a conditional rule scoped to a condition. The trap is (i), which wrongly forces D to imply creativity and success.

Let W = 2X - 1, which is ±1, so W² = 1. By independence, E[WY] = E[W]E[Y] = E[W] · 0 = 0. Var(WY) = E[W²Y²] - (E[WY])² = E[W²]E[Y²] - 0 = 1 · 100 = 100. Answer: (A) 100.

Concept: Variance of a product of independent variables with a zero-mean factor. The trap is computing Var(2X - 1) = 0.84 and multiplying by 100 to get 84 or similar.

The sum is P(Nₙ ≤ n) where Nₙ ~ Poisson(n). Nₙ is a sum of n independent Poisson(1) variables, with mean n and variance n. By the central limit theorem, (Nₙ - n)/√n → N(0, 1), so P(Nₙ ≤ n) = P((Nₙ - n)/√n ≤ 0) → Φ(0) = 0.5. Answer: (A) 0.5.

Concept: Recognising a Poisson CDF and applying the CLT. The trap is answering 1 because the full series Σ n^k/k! equals e^n; here the sum stops at k = n.

The sum of eigenvalues equals the trace: 1 + cos t + cos t = 1 + 2cos t. Setting 1 + 2cos t = 1 + √2 gives cos t = 1/√2. In [-π, π], t = π/4 or t = -π/4. Answer: (C) {π/4, -π/4}.

Concept: Trace equals sum of eigenvalues, so no eigenvalue computation is needed. The trap is forgetting the negative solution of cos t = 1/√2.

In standard first-order semantics the domain is non-empty. (A) If P holds for every element, it holds for some element: valid. (B) Some element satisfying P does not make all satisfy it: not valid. (C) contains (B) as one direction: not valid. (D) If P holds for all x, then ¬P(x) holds for none, so the implication fails whenever ∀x P(x) is true: not valid. Answer: (B), (C), (D).

Concept: Universal implies existential (non-empty domain), not the reverse. The trap is thinking (D) is valid by misreading it as ∀x P(x) ⇒ ¬∃x ¬P(x).

M is the centering matrix. (A) Iₙ and 𝟏𝟏ᵀ are symmetric, so M is symmetric. M² = Iₙ - (2/n)𝟏𝟏ᵀ + (1/n²)𝟏(𝟏ᵀ𝟏)𝟏ᵀ = Iₙ - (2/n)𝟏𝟏ᵀ + (1/n)𝟏𝟏ᵀ = M, so M² = M, not Iₙ (B false). Trace = n - (1/n) · n = n - 1 (C false). Symmetric and idempotent means M is an (orthogonal) projection (D true). Answer: (A), (D).

Concept: Properties of the centering matrix. The trap is confusing idempotent (M² = M) with involutory (M² = I).

(A) A sum of n squared independent standard normals is χ²ₙ: true. (B) The centred sum of squares loses one degree of freedom (Cochran's theorem): χ²ₙ₋₁, true. (C) X₁² + Xₙ² is χ²₂, whose density is (1/2)e^(-x/2), i.e. exponential with mean 2: true. (D) X̄ ~ N(0, 1/n), so √n X̄ ~ N(0, 1) and its square is χ²₁, not χ²₂: false. Answer: (A), (B), (C).

Concept: Chi-square facts and χ²₂ = Exp(mean 2). The trap is (D), giving the wrong degrees of freedom for a single squared normal.

F(x) = P(X ≤ x). Below the smallest support point F(x) = 0, so it is non-negative but not always positive (A false). Any CDF is non-decreasing (B true). For a discrete variable F is a step function that jumps by P(X = a) at each support point a (C true). A CDF defined with ≤ is right continuous; at a jump it is not left continuous (D false). Answer: (B), (C).

Concept: CDF properties. The trap is (D): with F(x) = P(X ≤ x), continuity is from the right, not the left.

P(+) = P(+|D)P(D) + P(+|not D)P(not D) = 0.8 · 0.3 + 0.1 · 0.7 = 0.24 + 0.07 = 0.31. By Bayes, P(D|+) = 0.24/0.31 ≈ 0.774. Answer: 0.77 (accepted range 0.76 to 0.78).

Concept: Bayes' theorem with a base rate. The trap is answering the sensitivity 0.80 and ignoring false positives.

Write xᵢ - xⱼ = (xᵢ - x̄) - (xⱼ - x̄). Expanding, the cross term sums to zero, so Σᵢ Σⱼ (xᵢ - xⱼ)² = 2n Σᵢ (xᵢ - x̄)². With n = 100: 200 · SS / 2000 = 99, so SS/10 = 99 and SS = 990. Then (1/99) · 990 = 10. Answer: 10.

Concept: Pairwise squared differences equal 2n times the sum of squared deviations. The trap is forgetting the factor 2 (pairs are counted in both orders).

E[Y | X] = x², so E[XY] = E[X · E[Y | X]] = E[X³] = 0, since X is symmetric about 0. Also E[X] = 0. Cov(X, Y) = E[XY] - E[X]E[Y] = 0, and both variances are positive, so the correlation is 0, even though Y clearly depends on X. Answer: 0.

Concept: Zero correlation does not mean independence; correlation only captures linear dependence. The trap is expecting a non-zero value because Y is a function of X plus noise.

Row 2 and column 3 share the entry a₂₃; the other 4 entries in each are independent. If a₂₃ = 1 (prob 1/2), each needs 2 ones among 4: C(4,2)/16 = 6/16, so this case gives (1/2)(6/16)² = 36/512. If a₂₃ = 0, each needs 3 ones among 4: 4/16, giving (1/2)(4/16)² = 16/512. Total = 52/512 ≈ 0.1016. Answer: 0.10 (accepted range 0.08 to 0.12).

Concept: Conditioning on the shared cell. The trap is treating the row and column sums as independent and squaring C(5,3)/32.

A is symmetric and idempotent (the centering projection), so its eigenvalues are 0 (eigenvector 𝟏) and 1 (any vector orthogonal to 𝟏, multiplicity n - 1). The maximum of xᵀAx over unit vectors is the largest eigenvalue (Rayleigh quotient), which is 1, attained e.g. at x = (1, -1, 0, ..., 0)/√2. Answer: 1.

Concept: Rayleigh quotient maximum equals the largest eigenvalue; projections have eigenvalues 0 and 1. The trap is answering n - 1 (the trace) instead of the largest eigenvalue.

Start R=1, B=1. Step 1: red with prob 1/2 gives (2,1); blue gives (1,2). After step 2 there are 4 balls; equal means (2,2). From (2,1), pick blue with prob 1/3 -> (2,2). From (1,2), pick red with prob 1/3 -> (2,2). P = 1/2 x 1/3 + 1/2 x 1/3 = 1/3. Answer: (B) 1/3.

Concept: Polya urn with conditional probabilities along a two-step tree. The composition after k steps is uniform over {1,...,k+1} red balls, here 1, 2 or 3 each with probability 1/3. The trap is assuming each pick is 1/2.

Fill the top-left 3 × 3 block freely: 2⁹ ways. The last entry of each of the first three rows is forced to make the row parity even; the last entry of each of the first three columns is forced likewise. The bottom-right entry is then forced by row 4, and it automatically makes column 4 even too, because the total number of 1s is even from both views. So the count is 2⁹ = 512. Answer: (A) 512.

Concept: Parity-constrained counting: an n × n binary matrix with all row and column sums even has 2^((n-1)²) choices. Options near 2¹⁰ tempt those who think only rows constrain.

The characteristic polynomial of an n × n matrix has degree n, so the sum of algebraic multiplicities of all eigenvalues is n, and one eigenvalue can take all of it. Example: the identity matrix I has eigenvalue 1 with multiplicity n. Answer: (A) n.

Concept: Multiplicity is bounded by the degree of the characteristic polynomial. n + 1 is impossible; 1 confuses this with distinct-eigenvalue matrices.

M·e₂ = 0 with e₂ ≠ 0, so M has a non-trivial null space and is singular: det M = 0 (B true, A false). Rank is at most n - 1 but need not be 1, e.g. M with a zero second column and identity elsewhere has rank n - 1 (C false). The null space is a subspace containing e₂, so it also contains 2e₂, -e₂ and so on: at least two non-zero vectors (D true). Answer: (B), (D).

Concept: A non-zero null vector means singular, det 0. Any non-trivial subspace contains infinitely many vectors; (C) confuses 'rank deficient' with 'rank 1'.

For x > 0, f(x) = c₁eˣ - c₂ ln(1/x) = c₁eˣ + c₂ ln x. As x → 0⁺, ln x → -∞, so the right limit is finite only if c₂ = 0. Then the right limit is c₁e⁰ = c₁. Continuity at 0 needs c₁ = f(0) = 3. So c₁ + c₂ = 3 + 0 = 3. Answer: 3.

Concept: A one-sided limit must be finite and equal to f(0). The log term diverges, forcing its coefficient to zero; missing that leaves the problem looking underdetermined.

f(x) = -(x - |x|/2)². For x ≥ 0: x - x/2 = x/2, so f = -x²/4. For x < 0: x + x/2 = 3x/2, so f = -9x²/4. f(0) = 0 and f < 0 elsewhere: local (global) maximum at 0, no local minimum. f′(x) = -x/2 for x > 0, -9x/2 for x < 0, and both one-sided derivatives at 0 are 0, so f′ is continuous. f″ is -1/2 on the right and -9/2 on the left, so f′ is not differentiable at 0. Answer: (A), (C), (D).

Concept: Split |x| into cases. A function can be C¹ but not C²: matching slopes at the joint, different curvatures. Students often assume an |x| term breaks f′ itself.

(A) If v ∉ S, every edge at v must be covered by its other endpoint, so all of v's at least k + 1 neighbours are in S, giving |S| ≥ k + 1, a contradiction. So v ∈ S. (B) Path a-b-c: k = 1, S = {b}; a lies on the path of length 2 but a ∉ S. (C) Triangle: k = 2, S = {a, b}; c is on the 3-cycle but c ∉ S. (D) Single edge u-v: k = 1, S = {u}; v is a clique of size 1 but v ∉ S. Answer: (A).

Concept: Classic kernel rule from parameterised vertex cover (Buss): a vertex of degree > k must be in every cover of size k. Small counterexamples kill the other options.

2-colorable means bipartite, and a graph is bipartite iff it has no odd cycle: (A) false. Even cycles such as C₄ are bipartite: (B) true. BFS or DFS colouring alternately and checking every edge decides it in Θ(|V| + |E|) with adjacency lists, and every vertex and edge must be read, so this is optimal: (C) true, (D) false. Answer: (B), (C).

Concept: Bipartite iff no odd cycle; test by BFS layering in linear time. The log factor in (D) belongs to sorting or heap-based algorithms, not to this check.

Let M(n) be the number of matchings (including the empty one) in a path on n vertices. For the last vertex: either it is unmatched, giving M(n - 1), or it is matched with its neighbour, giving M(n - 2). So M(n) = M(n - 1) + M(n - 2) with M(1) = 1, M(2) = 2. Sequence: 1, 2, 3, 5, 8, 13, 21, 34. M(8) = 34. Answer: 34.

Concept: Matchings in a path follow the Fibonacci recurrence. The usual trap is forgetting the empty matching (giving 33) or counting only maximum matchings.

Probabilities sum to 4/6 + 4/12 = 1. E[X] = (1 + 2 + 5 + 7)/6 + (3 + 4 + 6 + 8)/12 = 15/6 + 21/12 = 2.5 + 1.75 = 4.25. Answer: 4.25.

Concept: Expectation of a discrete random variable is Σ x·p(x). Grouping values with equal probability keeps the arithmetic short.

Let y be the special person: 'there exists y'. For every other person x (for all x with x ≠ y): x knows y, M(x, y), and y does not know x, ¬M(y, x). This is (∃y)(∀x)((x ≠ y) → (M(x, y) ∧ ¬M(y, x))). Answer: (A).

Concept: Translating English to predicate logic: one specific person (∃) related in a fixed way to all others (∀), with the guard written as an implication. The trap is swapping the quantifier order, which changes the meaning, or using ∃x which makes the statement nearly vacuous.

The normal density is (1/(σ√(2π))) exp(-(x - μ)²/(2σ²)). With μ = 0 and σ = 3: 2σ² = 18 and the constant is 1/(3√(2π)), matching exactly. So X ~ N(0, 9). Poisson is discrete, exponential is supported on x ≥ 0, uniform has a constant density. Answer: (B).

Concept: Recognise the Gaussian form exp(-x²/(2σ²)) with matching normalisation. The trap is being misled by 'exp' into choosing exponential, whose density is λe^(-λx) on x ≥ 0 only.

Reflexive: x·x = x² is a square. Symmetric: xy = yx. Transitive: if xy = a² and yz = b², then xz·y² = (ab)², so xz = (ab/y)² is a rational square that is an integer, hence a perfect square. Antisymmetric fails: (1,4) and (4,1) are both in R but 1 ≠ 4. Answer: (A), (B) and (C).

Concept: R relates numbers with the same square-free part, so it is an equivalence relation. The trap is doubting transitivity, or assuming antisymmetry without testing a pair like (1,4).

∫₋₁¹ 3x² dx = [x³]₋₁¹ = 2. ∫₋₁¹ ax dx = 0 because x is odd over a symmetric interval. ∫₋₁¹ 1 dx = 2. So I(a) = 4 for every a. It is independent of a and always positive, never negative. Answer: (A) and (C).

Concept: Odd functions integrate to zero over symmetric limits, removing the a-dependence. The trap is assuming a parameter in the integrand must affect the value.

Infinitely many solutions require the two equations to be proportional: a/16 = 1/a = b/24. From a/16 = 1/a, a² = 16, a = ±4. Then b = 24/a: a = 4 gives b = 6, a = -4 gives b = -6. Either way ab = 24. Answer: 24.

Concept: Two lines coincide when coefficient ratios and constant ratio are all equal. The trap is setting only the determinant to zero (that also allows no solution) and forgetting the constant ratio.

The coefficient is 0 exactly when two spanning trees share no edge. Kₙ has n(n-1)/2 edges, and for n ≥ 4 it contains two edge-disjoint spanning trees (Nash-Williams; indeed Kₙ decomposes into floor(n/2) of them). Example for n = 5: T1 = path 1-2-3-4-5 = {12, 23, 34, 45} and T2 = path 1-3-5-2-4 = {13, 35, 25, 24}, which share no edge. So the minimum is 0. Answer: (C) 0.

Concept: Edge-disjoint spanning trees exist in dense graphs; Kₙ has enough edges (at least 2(n-1) for n ≥ 4). The trap is assuming all spanning trees must share some edge and computing 1/(2n-3), which is the value for exactly one shared edge.

Scaling an n x n matrix by k scales the determinant by kⁿ, since each of the n rows is multiplied by k. det(2A) = 2⁴ x 3 = 16 x 3 = 48. Answer: 48.

Concept: det(kA) = kⁿ det(A). The trap is answering 6 (multiplying once) or 24 (2³).

E2: exactly 2 heads among tosses 1, 2, 3, 5 (6 equally likely patterns). Toss 2 is the only overlap with E1; given E2, toss 2 is H in 3 of 6 patterns, so P(toss2 = H | E2) = 1/2. Tosses 4 and 6 are independent of E2. If toss 2 = H, E1 needs at least one H in tosses 4, 6: 3/4. If toss 2 = T, E1 needs both: 1/4. P(E1 | E2) = 1/2 x 3/4 + 1/2 x 1/4 = 1/2. Answer: 0.5.

Concept: Condition on the single shared toss and use independence for the rest. The trap is treating E1 and E2 as independent without checking, or miscounting the 6 patterns of E2.

f is constant on each interval [0.d₁d₂, 0.d₁d₂ + 0.01). Within each tenth, second digits 0-9 give values 0,0,1,1,0,0,1,1,0,0. The value changes at second digits 2 (0→1), 4 (1→0), 6 (0→1) and 8 (1→0): 4 jumps per tenth. Between tenths (second digit 9 to 0) the value stays 0, so no jump. Ten tenths give 4 x 10 = 40 discontinuities, all inside (0,1) (0.02 up to 0.98). Answer: 40.

Concept: A step function is discontinuous only where its value actually changes. The trap is counting every 0.01 boundary (99) or every boundary of an in-set digit block without noticing that 9→0 is not a jump.

E[X|Y] is a function of Y. Taking its expectation over Y (discrete case): E[E[X|Y]] = Σ_y E[X|Y=y] P(Y=y) = Σ_y Σ_x x P(X=x|Y=y) P(Y=y) = Σ_x x Σ_y P(X=x, Y=y) = Σ_x x P(X=x) = E[X]. This is the law of total expectation (tower property). Answer: (C) E[X].

Concept: Law of iterated expectations. The trap is treating E[X|Y] as a number (option A) rather than a random variable that still has to be averaged over Y.

The matrix is already upper triangular, so the forward-elimination stage (the O(n³) part) has nothing to do. What remains is back substitution: x_n needs 1 division, x_(n-1) needs about 2 operations, ..., x_1 needs about n operations. Total ≈ 1 + 2 + ... + n = n(n + 1)/2 multiplications and a similar number of additions, i.e. O(n²). Answer: (B) O(n²).

Concept: Separates the cost of forward elimination (O(n³) for a dense matrix) from back substitution (O(n²)). The trap is quoting the general O(n³) cost without noticing the matrix is already triangular.

Let e be the all-ones vector. Row sums equal 1 means Ae = e, so A²e = e and A³e = e. Then Be = e - 2e + e = 0 with e ≠ 0, so B has a non-trivial null space and is singular. A homogeneous system Bx = 0 always has x = 0; once it has one non-zero solution e, every multiple ce is also a solution, so there are infinitely many. Answer: (C).

Concept: Row-sum 1 means 1 is an eigenvalue with eigenvector of all ones; B = A(A - I)² kills that vector. Homogeneous systems are never 'no solution' and never have exactly two solutions.

f(x) = sinh x. Differentiating: f' = cosh x, f'' = sinh x, so derivatives alternate between cosh and sinh. Every even-order derivative equals sinh x, so f⁽¹⁰⁾(x) = sinh x and f⁽¹⁰⁾(0) = 0. Equivalently, sinh x = x + x³/3! + x⁵/5! + ... has only odd powers, so the x¹⁰ coefficient f⁽¹⁰⁾(0)/10! is 0. Answer: (A) 0.

Concept: sinh is odd, so its Taylor series at 0 has only odd powers and all even derivatives vanish at 0. The trap is confusing the derivative with the Taylor coefficient f⁽ᵏ⁾(0)/k!, which suggests the 1/10! options.

p → q is false only when p is true and q is false, which is exactly when ¬p ∨ q is false. So S1 ≡ S3. Check the others: S2 (¬p ∧ q) is false at p = F, q = F while S3 is true; S4 (¬p ∨ ¬q) is true at p = T, q = F while S1 is false; S2 is false at p = T, q = F while S4 is true. Answer: (A) S1 ≡ S3.

Concept: Material implication rewrite p → q ≡ ¬p ∨ q. The trap is mixing up ∧ and ∨, or confusing ¬p ∨ ¬q, which is ¬(p ∧ q), with the implication.

X is uniform on [t, 4]. The median m satisfies F_X(m) = 1/2: (3 - t)/(4 - t) = 1/2, so 6 - 2t = 4 - t, giving t = 2. Check: the median of uniform[2, 4] is the midpoint 3. Answer: (A) 2.

Concept: Median from a CDF: solve F(m) = 0.5. For a uniform distribution it is the midpoint, so t = 2·3 - 4 directly. The trap is setting F(3) = 3 or using the mean formula incorrectly.

E[X] = aE[Z] + b = b, so b = 1. X - E[X] = aZ, so E[(X - E[X])Z] = aE[Z²] = a = -2. Check: E[(X - E[X])²] = a² E[Z²] = 4. Consistent. Answer: (A) a = -2, b = 1.

Concept: Uses E[Z] = 0 and E[Z²] = 1. The variance alone gives a = ±2; the covariance with Z fixes the sign. The trap is stopping at the variance and picking a = 2.

For an exponential with rate λ, P(X ≥ x) = e^(-λx). So e^(-2λ) = 0.25, giving -2λ = ln 0.25 = -ln 4, λ = (ln 4)/2 = ln 2. Answer: (A) ln 2.

Concept: Exponential survival function e^(-λx). The trap is forgetting to divide by 2 (giving ln 4) or dropping the minus sign (ln 0.25, which is negative and not a valid rate).

Sum, difference and product of functions differentiable at c are differentiable at c. f/g: g > 1 so g(c) ≠ 0, and the quotient rule applies - differentiable. f ∘ g needs f differentiable at g(c), and g ∘ f needs g differentiable at f(c); we only know differentiability at c. Example: c = 0, g(x) = 2 + eˣ (always > 1), f(x) = |x - 3|; f is differentiable at 0, but f(g(x)) = |eˣ - 1| has a corner at 0, while g(f(x)) = 2 + e^|x - 3| is differentiable at 0, so the sum is not differentiable at 0. Answer: (A), (B), (C).

Concept: Algebraic combinations preserve differentiability at a point (quotient needs a non-zero denominator, guaranteed by g > 1). Composition needs differentiability at the image point, which is not given.

Any rotation of the standard basis is another orthonormal basis (e.g. in ℝ², (1/√2)(1, 1) and (1/√2)(1, -1)), so the orthonormal basis is not unique: (A) false, (B) true. (C) false: (1, 0) and (1, 1) are independent but not orthogonal. (D) true: if Σ cᵢvᵢ = 0, taking the dot product with v_j gives c_j = 0 for every j. Answer: (B), (D).

Concept: Orthonormal implies independent, not the converse; orthonormal bases are unique only up to an orthogonal transformation. Note for n = 1 there are still two (±1).

Ax = x + x(xᵀx) = 2x, so 2 is an eigenvalue. For any v ⊥ x, Av = v, giving eigenvalue 1 with multiplicity n - 1. All eigenvalues (2, 1, ..., 1) are non-zero, so A is invertible with rank n: (A), (B) true; (C) false. A⁻¹ has eigenvalues 1/2 and 1, all positive: (D) false. Answer: (A), (B).

Concept: A rank-one update of the identity: eigenvalues are 1 + xᵀx along x and 1 elsewhere. The trap is confusing it with I - xxᵀ, a projection that does have eigenvalue 0.

P(W) = (1/2)(1/3) + (1/6)(2/3) + (1/3)(3/6) = 1/6 + 1/9 + 1/6 = 3/18 + 2/18 + 3/18 = 8/18. P(Box-2 and W) = (1/6)(2/3) = 2/18. P(Box-2 | W) = (2/18)/(8/18) = 1/4 = 0.25. Answer: 0.25.

Concept: Bayes' theorem with the law of total probability in the denominator. The trap is ignoring the unequal box priors and using 2/3 or the plain fraction of white balls.

Multiply by the conjugate: √(t² + t) - t = (t² + t - t²)/(√(t² + t) + t) = t/(√(t² + t) + t) = 1/(√(1 + 1/t) + 1). As t → ∞ this tends to 1/(1 + 1) = 0.5. Answer: 0.5.

Concept: An ∞ - ∞ form resolved by rationalising. The trap is concluding the limit is 0 because √(t² + t) 'is about t'.

Joint scores: P(x|y₁)P(y₁) = (3/4)(1/3) = 1/4 and P(x|y₂)P(y₂) = (1/4)(2/3) = 1/6. Since 1/4 > 1/6 the classifier predicts y₁. It is wrong when the true class is y₂: P(y₂|x) = (1/6)/(1/4 + 1/6) = (2/12)/(5/12) = 0.4. Answer: 0.40 (key range 0.39 to 0.41).

Concept: Bayes decision rule picks the larger posterior; the error probability for that x is the smaller posterior. The trap is reporting the unnormalised 1/6 or the prior 2/3.

Z is standard normal, so E[Z²] = 1 and E[Z⁴] = 3. Var(Y) = E[Z⁴] - (E[Z²])² = 3 - 1 = 2. (Y is chi-squared with 1 degree of freedom, whose variance is 2k = 2.) Answer: (B) 2.

Concept: The fourth moment of a standard normal is 3. The trap is answering 3 (the fourth moment itself) or 1 (the mean of Y).

rank(A) = rank(A³) = rank(A · A²) ≤ rank(A²) = rank(A · A) ≤ rank(A). So rank(A²) = rank(A) always: (D) true. Counterexamples for the rest: A = 0 satisfies A³ = A but is not invertible and has trace 0, killing (A) and (C); A = I satisfies it with determinant 1, killing (B). Answer: (D).

Concept: Rank of a product never exceeds the rank of either factor; sandwiching gives equality. Eigenvalues are in {-1, 0, 1}, so determinant and trace vary. The trap is assuming A must be singular.

Let X be the matrix with columns x₁, ..., xₙ. Then A = XᵀX (a Gram matrix). For any z, zᵀAz = (Xz)ᵀ(Xz) = ‖Xz‖². Because the columns are independent, Xz ≠ 0 whenever z ≠ 0, so zᵀAz > 0: A is positive definite. A positive definite matrix has all eigenvalues (and singular values) positive, so det A > 0 and A is invertible. (B), (C), (D) are all false. Answer: (A).

Concept: Gram matrix of independent vectors is positive definite. The trap is option (D): zᵀAz = ‖Xz‖² is zero only if Xz = 0, impossible for independent columns.

X² ≤ 0.25 means -0.5 ≤ X ≤ 0.5. X is continuous, so P = F(0.5) - F(-0.5) = (1/4)(1.5)² - (1/4)(0.5)² = (2.25 - 0.25)/4 = 0.5. Answer: (C) 0.5.

Concept: Convert the event on X² into an interval on X, then difference the CDF. The trap is computing only F(0.5) = 0.5625 (option D), forgetting the lower end.

E[Y] = 300 × 0.25 = 75, Var(Y) = 300 × 0.25 × 0.75 = 56.25, SD = 7.5. Standardise: (60 - 75)/7.5 = -2 and (90 - 75)/7.5 = 2. P(60 ≤ Y ≤ 90) ≈ φ(2) - φ(-2). Answer: (A).

Concept: CLT approximation of a binomial sum by N(nθ, nθ(1 - θ)); here φ is the standard normal CDF. The trap is plugging raw values into φ (option D) or using the variance instead of the SD.

Mean 1/ln 10 means rate λ = ln 10, so P(X ≥ x) = e^(-x ln 10) = 10^(-x). P(Y = ℓ) = P(ℓ ≤ X < ℓ + 1) = 10^(-ℓ) - 10^(-(ℓ+1)) = (0.1)^ℓ (1 - 0.1). Matching q^ℓ(1 - q) gives q = 0.1. Answer: (A) 0.1.

Concept: The floor of an exponential variable is geometric with q = e^(-λ). The trap is option (D), 0.434 = 1/ln 10, which is the mean, not e^(-λ).

Complement: no die shows a 1. Each die avoids 1 with probability 5/6, independently, so P(no 1) = (5/6)¹⁰⁰. P(A) = 1 - (5/6)¹⁰⁰. Answer: (C).

Concept: 'At least one' is easiest via the complement 'none'. The trap is option (D), which is the complement itself, or (B), since 100 dice make a 1 very likely but not certain.

ReLU is 0 for x ≤ 0 and x for x ≥ 0; both pieces meet at 0, so it is continuous everywhere: (A) true. At 0 the left derivative is 0 and the right derivative is 1, so it is not differentiable there: (B) false, (C) true. (D) fails, e.g. x = 1, a = 2: ReLU(1) = 1 but ReLU(2) = 2. Answer: (A), (C).

Concept: A corner gives continuity without differentiability. The trap is (D): ReLU is positively homogeneous, ReLU(ax) = a ReLU(x) for a ≥ 0, not invariant to scaling.

f'(x) = x² + 7x + 10 = (x + 2)(x + 5); critical points x = -5 (local max, f'' = 2x + 7 = -3 < 0) and x = -2 (local min, f'' = 3 > 0). Values: f(-8) = -512/3 + 224 - 80 + 66.5 ≈ 39.83; f(-5) = -125/3 + 87.5 - 50 + 66.5 ≈ 62.33; f(-2) = -8/3 + 14 - 20 + 66.5 ≈ 57.83; f(0) = 66.5. Global max = 133/2 at x = 0, not at -5: (A) false, (C) true. Global min is at x = -8, not -2: (B) false. f' is a parabola opening upward with vertex at x = -7/2, inside [-8, 0]: (D) true. Answer: (C), (D).

Concept: On a closed interval compare critical points with the endpoints. The trap is equating local extrema with global ones: here both global extrema occur at the endpoints.

A is the orthogonal projection onto span{x₁, ..., x₅}: Axᵢ = xᵢ for i = 1..5, and Av = 0 for v orthogonal to all of them. So A is symmetric with eigenvalues 1 (five times) and 0 (five times). For a symmetric positive semidefinite matrix the singular values equal the eigenvalues: (A) true, and they are 0 or 1: (B) true. A has rank 5 < 10, so det A = 0 and A is not invertible: (C), (D) false. Answer: (A), (B).

Concept: A sum of outer products of orthonormal vectors is a projection matrix (A² = A, Aᵀ = A). The trap is thinking it is the identity; it is only the identity when the vectors span the whole space.

f'' > 0 means f' is strictly increasing, so f' can vanish at most once: (B) true. Any local minimum of a differentiable function has f' = 0, so there is at most one local minimum: (D) true, and hence at most one global minimum: (C) true. (A) is not always true: f(x) = eˣ has f'' > 0 but f' = eˣ is never 0, so f has no minimum. Answer: (B), (C), (D).

Concept: Strict convexity forbids two stationary points but does not guarantee one exists. The trap is (A): eˣ is strictly convex yet has no minimum.

‖Ax‖² = xᵀAᵀAx = xᵀx for all x. Since AᵀA - I is symmetric and its quadratic form is identically zero, AᵀA = I: A is orthogonal: (A) true. Orthogonal matrices are invertible, so A has full rank: (D) true. (B) false: -I or any rotation also works. (C) false: a 2-D rotation by 90° preserves length but has eigenvalues ±i; real eigenvalues of an orthogonal matrix are ±1, but eigenvalues can be complex with modulus 1. Answer: (A), (D).

Concept: Norm-preserving linear maps are exactly the orthogonal matrices. The trap is (C): eigenvalues have |λ| = 1, but they need not be real.

E[p̂] = (1/n) Σ E[Xᵢ] = (1/n)(np) = p: (A) true, (B) false. Var(p̂) = (1/n²) Σ Var(Xᵢ) = (1/n²) · n p(1 - p) = p(1 - p)/n, which decreases as n grows: (C) true, (D) false. Answer: (A), (C).

Concept: The sample proportion is unbiased and its variance shrinks like 1/n. The trap is dividing the mean by n twice (option B).

Divide by |x - y|: |(f(x) - f(y))/(x - y)| ≤ |x - y|. Letting y → x gives f'(x) = 0 for every x, so f is constant and f(1) - f(0) = 0. (Directly: split [0, 1] into n equal parts; the total change is at most n · (1/n)² = 1/n for every n, hence 0.) Answer: 0.

Concept: A Hölder condition with exponent greater than 1 forces the derivative to vanish everywhere. The trap is applying the bound once with x = 1, y = 0 and answering 1.

The data are centred, so the covariance matrix is C = (1/n) Σ x⁽ⁱ⁾x⁽ⁱ⁾ᵀ, and (1/n) Σ (uᵀx⁽ⁱ⁾)² = uᵀCu, the variance of the data along u. The direction of maximum variance is the top eigenvector of C, and then uᵀCu equals the largest eigenvalue. λᵢ = 100^(2-i) decreases with i, so the largest is λ₁ = 100¹ = 100. Answer: 100.

Concept: The first principal component captures variance equal to the top eigenvalue of the covariance matrix. The trap is reading λᵢ as increasing in i and taking λ₁₀₀, or summing the eigenvalues.

With replacement, each draw is black with probability 10/15 = 2/3, independently. S₁₀₀ ~ Binomial(100, 2/3), so E[S₁₀₀] = 100 × 2/3 = 66.67 ≈ 66.7 (key accepts 66.6 to 66.7, depending on truncating or rounding 66.666...).

Concept: Expectation of a binomial count is np, by linearity of expectation. The trap is using 10/5 as the probability or worrying about the bag emptying, which cannot happen with replacement.

Counterexample for onto: A = {1}, B = {1, 2}, C = {c}, g(1) = 1, f(1) = f(2) = c. f and f∘g are onto, but g misses 2. Counterexample for one-to-one: A = {1, 2}, B = {1}, C = {c}, g(1) = g(2) = 1, f(1) = c. f and f∘g are onto, but g is not injective. So g need be neither. Answer: (D).

Concept: If f∘g is onto then f must be onto; nothing follows about g. The trap is reversing the rule (that holds for injectivity: f∘g injective implies g injective).

Determinant = 1 - k². If k ≠ ±1 there is a unique solution, so infinitely many k give exactly one solution: (C) false. k = 1: x + y = 1 and x + y = -1 are inconsistent: no solution. k = -1: x - y = 1 and -x + y = -1 are the same line: infinitely many solutions. So exactly one k (k = 1) gives no solution: (A) true, (B) false; exactly one k (k = -1) gives infinitely many: (D) true. Answer: (A), (D).

Concept: Singular coefficient matrix means no solution or infinitely many, decided by consistency of the right-hand side. Check each singular k separately rather than assuming both behave alike.

Complement count. Total strings of length 5: 3⁵ = 243. Strings with no two equal adjacent symbols: 3 choices for the first, then 2 for each of the next 4 (anything but the previous symbol): 3 x 2⁴ = 48. Strings with at least one equal adjacent pair: 243 - 48 = 195. Answer: 195.

Concept: "At least one" counts are easiest via the complement. Direct counting of aa/bb/cc occurrences double-counts strings with several pairs.

Continuity at x = 1: a + b = 1 + 1 + 1 = 3. Differentiability at x = 1: left derivative a equals right derivative 3x² + 2x at x = 1, which is 5. So a = 5 and b = 3 - 5 = -2. Answer: -2.0.

Concept: Differentiable at a join needs both continuity (values match) and equal one-sided derivatives. Using only one condition leaves b undetermined.

P(fake) = 1/5, P(HH | fake) = 1. P(regular) = 4/5, P(HH | regular) = 0.25. P(HH) = 0.2 x 1 + 0.8 x 0.25 = 0.2 + 0.2 = 0.4. By Bayes, P(fake | HH) = 0.2 / 0.4 = 0.5. Answer: 0.50.

Concept: Bayes' theorem: posterior = likelihood x prior / evidence. The trap is answering the prior 0.2, ignoring that two heads is much likelier with the fake coin.

Characteristic equation of A: (1 - λ)(-1 - λ) - 1 = λ² - 2 = 0, so λ = ±√2. Eigenvalues of A¹³ are λ¹³ = (±√2)¹³ = ±2⁶√2 = ±64√2. (Also A² = 2I, so A¹³ = 2⁶A = 64A.) Answer: 64√2, -64√2 (D).

Concept: Eigenvalues of Aᵏ are λᵏ. The odd power keeps the signs; the trap is dropping the extra √2 factor (answering ±64).

"Exactly one" = existence plus uniqueness: for each x, some y is x's mother and no z different from y is x's mother. (B) states exactly this. (D) is the same with ∀z(p → ¬q) rewritten as ¬∃z(p ∧ q). (A) only says some y is a mother and some z is not, which allows two mothers. (C) is trivially true (take z = y) and asserts neither existence nor uniqueness. Answer: (B), (D).

Concept: Uniqueness pattern ∃y[P(y) ∧ ∀z(z ≠ y → ¬P(z))]. Recognise the equivalent ¬∃ form; reject formulas satisfied by a trivial witness.

Closure: a sum of non-negative integers is a non-negative integer. Associativity and commutativity follow pointwise from + on integers. Identity: the zero function z(n) = 0. Inverses: f(n) = 1 would need g(n) = -1, which is not in A, so not a group. Hence a commutative (Abelian) monoid. Answer: (B).

Concept: Pointwise operations inherit the properties of the underlying operation on the codomain; (N, +) is a commutative monoid without inverses, so the function space is too.

Each bit arrives correctly with probability 0.99, independently. All 5 correct: 0.99⁵ = 0.95099..., which rounds to 0.951. Answer: 0.951.

Concept: Independent events multiply. The trap is 1 - 5 x 0.01 = 0.95, a first-order approximation that is slightly off at three decimals.

Normalise: ∫ from 1 to 4 of Cx² dx = C(64 - 1)/3 = 21C = 1, so C = 1/21. P(2 ≤ x ≤ 3) = (1/21) ∫ from 2 to 3 of x² dx = (1/21)(27 - 8)/3 = 19/63 ≈ 0.302. Answer: 0.302 (accepted range 0.300 to 0.302).

Concept: Find the constant from total probability 1, then integrate over the interval. Trap: forgetting to normalise.

A² = [[1+4, 2-2], [2-2, 4+1]] = [[5, 0], [0, 5]] = 5I. Then A⁸ = (A²)⁴ = 5⁴ I = 625 I. Answer: (C) [[625, 0], [0, 625]].

Concept: Squaring first exposes A² as a scalar matrix, so high powers are immediate. Equivalently the eigenvalues are ±√5. The trap is stopping at A⁴ = 25I and picking (A).

By parts: ∫ t ln t dt = (t²/2) ln t - t²/4. From 1 to x: (x²/2) ln x - x²/4 + 1/4. Setting this equal to 1/4 gives x²(2 ln x - 1)/4 = 0, so ln x = 1/2 and x = √e. Answer: (A) √e.

Concept: Integration by parts with u = ln t. Forgetting the lower-limit term +1/4 is the usual slip and leads to a wrong equation.

L² = L⁻¹ gives L³ = I. Then M = L⁸ = L⁶ L² = L², so M = N and M - N is the zero matrix. det(M - N) = 0. Answer: (A) 0.

Concept: Reduce powers using L³ = I. The trap is computing det L instead of noticing M and N are the same matrix.

(A) is the principle of mathematical induction, valid for every predicate. (B) take P(x) true only at 0: the step P(x) ⇒ P(x-1) holds, yet not all P(x). (C) take P(x) true for x ≤ 1000: premises hold but P(1001) is false. (D) take P(x) true for x ≥ 1000: premises hold but P(0) is false. Answer: (A).

Concept: Induction must start at the least element and step upward. Each wrong option covers only a ray of naturals, shown by a predicate true exactly on that ray.

≼ is the componentwise order on n-bit vectors. Reflexive (f(x) ≤ f(x)), antisymmetric (f ≤ g and g ≤ f pointwise gives f = g) and transitive: a partial order. Meet and join are the pointwise min (AND) and max (OR), which are in ℱ, so it is a lattice (isomorphic to the subset lattice of {1..n}). It is not symmetric: the all-0 function ≼ the all-1 function but not conversely, hence also not an equivalence relation. Answer: (B) and (C).

Concept: Functions to {0,1} are characteristic vectors of subsets, and ≼ is set inclusion, the Boolean lattice. A partial order is symmetric only on a trivial set.

(A) LUX = Q: put Y = UX, solve LY = Q by forward substitution, then UX = Y by back substitution: true. det L = 1 and det U = u11 u22 u33, so det P = u11 u22 u33. (B) P invertible gives det U ≠ 0, and L is always invertible: true. (C) P singular gives u11 u22 u33 = 0, so some uii = 0: true. (D) false: triangular matrices with nonzero off-diagonal entries are not symmetric; for symmetric P one gets P = LDLᵀ, not symmetric L and U. Answer: (A), (B), (C).

Concept: The determinant of a triangular matrix is the product of its diagonal, which links P's invertibility to U's pivots. Symmetry of P gives U = DLᵀ, not symmetric factors.

The root multisets must agree: {α, β} = {α², β²}. Case 1: α² = α and β² = β, so α, β ∈ {0, 1}: polynomials x², x(x - 1), (x - 1)². Case 2: α² = β and β² = α, so α⁴ = α, giving α = 0 (β = 0), α = 1 (β = 1), or α a primitive cube root of unity ω with β = ω², giving (x - ω)(x - ω²) = x² + x + 1. Distinct polynomials: x², x² - x, x² - 2x + 1, x² + x + 1. Equal roots in x² and (x - 1)²: probability 2/4 = 0.5. Answer: 0.5.

Concept: Count distinct polynomials, not ordered root pairs. The trap is missing the complex cube-root case or double-counting (ω, ω²) and (ω², ω).

Let the cut be U ~ Uniform(0,1). If U < 0.4, 0.4 lies in (U, 1) of length 1 - U; if U > 0.4, it lies in (0, U) of length U. E = ∫ from 0 to 0.4 of (1 - u) du + ∫ from 0.4 to 1 of u du = (0.4 - 0.08) + (0.5 - 0.08) = 0.32 + 0.42 = 0.74. This lies in the accepted range 0.7 to 0.8. Answer: 0.74.

Concept: Condition on where the cut falls relative to the fixed point. The trap is answering 0.5: the piece containing a given point is size-biased, so it is longer on average.

(i) For a Poisson random variable with parameter λ, E[X] = λ and Var(X) = λ, so they are equal: true. (ii) The standard normal N(0, 1) has mean 0 and variance 1 by definition: true. Answer: (A) Both (i) and (ii) are true.

Concept: Basic moments of standard distributions. The trap is confusing the Poisson with the exponential, whose standard deviation (not variance) equals its mean.

T ∩ S would need at least 2 heads and at least 2 tails, i.e. at least 4 tosses. With only 3 tosses this is impossible, so T and S are mutually exclusive and P(T ∩ S) = 0. Answer: (A) 0.

Concept: Mutually exclusive events. The trap is multiplying P(T) × P(S) = 0.25 as if the events were independent.

Characteristic equation: λ² - (trace)λ + det = 0, with trace = 2 + 1 = 3 and det = 2·1 - (-1)·3 = 5. So λ² - 3λ + 5 = 0. Discriminant = 9 - 20 = -11 < 0, so λ = (3 ± i√11)/2, a complex conjugate pair. Answer: (B).

Concept: Eigenvalues of a 2 × 2 real matrix from trace and determinant. The trap is the sign of the off-diagonal product: det = 2 + 3, not 2 - 3.

This is the second derivative test. f′(x*) = 0 makes x* a stationary point; f″(x*) > 0 means f is locally convex there, so f(x) ≥ f(x*) in some neighbourhood of x*: a local minimum. It need not be global, e.g. f(x) = x³ - 3x has f′(1) = 0, f″(1) = 6 > 0, yet f(x) → -∞ as x → -∞. Answer: (A) local minimum.

Concept: Second derivative test gives only local information. The trap is choosing 'global minimum', which holds only for convex functions.

An implication fails only when the premise is true and the conclusion false. (A) Premise true means x = F, y = T; then y ⇒ x is T ⇒ F = F. Not a tautology. (B) Premise true means x = T, y = F; ¬x ⇒ y has false antecedent, so true. Tautology. (C) Premise true means x = F, y = T; ¬x ⇒ y is T ⇒ T = T. Tautology. (D) Premise true means x = T, y = F; y ⇒ x has false antecedent, so true. Tautology. Answer: (B), (C), (D).

Concept: Check only the rows where the premise is true. The trap is building full truth tables and making sign errors with the negations.

For the sigmoid, f′(x) = e^(-x)/(1 + e^(-x))² = f(x)(1 - f(x)). With f(x) = 0.4: f′(x) = 0.4 × 0.6 = 0.24. Answer: 0.24.

Concept: The sigmoid's derivative identity σ′ = σ(1 - σ), used throughout logistic regression and neural networks. The trap is solving for x first, which is unnecessary.

Sum of the 50 points = 50 × 40 = 2000. Adding 142 gives 2142 over 51 points. New average = 2142/51 = 42. Answer: 42.

Concept: Updating a mean via the running total. The trap is averaging 40 and 142 directly, ignoring that the old mean carries weight 50.

M² + 12M = M(M + 12I), so det(M² + 12M) = det(M) · det(M + 12I). det(M) = 1(1·6 - 3·3) - 2(3·6 - 3·4) + 3(3·3 - 1·4) = -3 - 12 + 15 = 0. (Indeed row 3 = row 1 + row 2.) Hence the determinant is 0. Answer: 0.

Concept: Factorising a matrix polynomial and using det(AB) = det(A)det(B). The trap is computing M² + 12M entry by entry instead of noticing M is singular.

Each throw is even with probability p = 1/2. Let E be the expected throws from the start and E₁ the expected remaining throws after one even. From the start: one throw, then with 1/2 we are in state 1, with 1/2 back to start: E = 1 + (1/2)E₁ + (1/2)E. From state 1: one throw, with 1/2 we finish, with 1/2 back to start: E₁ = 1 + (1/2)E. Substituting: E = 1 + (1/2)(1 + E/2) + E/2, so E/4 = 3/2 and E = 6. (General formula 1/p + 1/p² = 2 + 4.) Answer: (C) 6.

Concept: First-step analysis (conditioning on the next throw) for waiting times of patterns. The trap is answering 4 = 1/p², which ignores that a failure after one success resets the run.

Continuity at x = 2: 4a + 2b + c = 2. Continuity at x = -2: 4a - 2b + c = 2. Subtracting gives b = 0, so 4a + c = 2. Differentiability at x = 2: 2a(2) + b = 1, so 4a = 1 and a = 1/4. (At x = -2: -4a + b = -1, consistent.) Then c = 2 - 1 = 1. Answer: (A) a = 1/4, b = 0, c = 1.

Concept: Matching values and slopes at the join points of a piecewise function. The trap is enforcing only continuity, which (B) and (D) partly satisfy, without matching derivatives.

A solution, when one exists, is unique exactly when the null space of M is {0}. (A) A unique solution for a square M means M is invertible, so Mx = q is also unique: false. (B) Take a singular M, e.g. diag(1, 1, 0): p = (0, 0, 1) is outside the column space (no solution) and q = 0 gives infinitely many: true. (C) A 2 × 3 matrix has rank ≤ 2, so its null space has dimension ≥ 1; any consistent system has infinitely many solutions, never a unique one: false. (D) A 3 × 2 matrix of rank 2 has trivial null space; its column space is a plane in ℝ³, so p in the plane gives a unique solution and q outside gives none: true. Answer: (B), (D).

Concept: Existence depends on whether the right side is in the column space; uniqueness depends only on the null space. The trap is thinking uniqueness can vary with the right-hand side.

The range of a projection onto U is U, so rank(M) = dim U and, by rank-nullity, nullity = 3 - dim U. (A) dim U = 1 gives a 2-dimensional null space: false. (B) dim U = 2 gives a 1-dimensional null space: true. (C) Projecting a vector already in U leaves it unchanged, so M(Mx) = Mx for all x: M² = M (idempotent): true. (D) M³ = M·M² = M·M = M: true. Answer: (B), (C), (D).

Concept: Projection matrices are idempotent and their rank equals the dimension of the target subspace. The trap is (A): the null space is the complement, of dimension 3 - dim U.

f′(x) = x³ - 2x² - 3x = x(x - 3)(x + 1), so the stationary points are x = -1, 0, 3. f″(x) = 3x² - 4x - 3. f″(0) = -3 < 0: local maximum. f″(3) = 27 - 12 - 3 = 12 > 0: local minimum. f″(-1) = 3 + 4 - 3 = 4 > 0: local minimum, not maximum. Answer: (A), (B).

Concept: Second derivative test at each stationary point. The trap is assuming the stationary points simply alternate max/min in the order listed in the options; check each one.

'All balls are round except rugby balls' says two things: a ball not used in rugby is round (s1), and a rugby ball is not round (s3). s2 is the contrapositive of s1 (¬round ⇒ rugby is equivalent to ¬rugby ⇒ round), and s4 is the contrapositive of s3. So s1 ∧ s3 and s2 ∧ s3 both express the assertion. s1 ∧ s2 is just s1 twice (says nothing about rugby balls), and s3 ∧ s4 is just s3 twice (says nothing about non-rugby balls). Answer: (A), (C).

Concept: Reading 'except' as a two-way statement and recognising contrapositives as equivalent. The trap is choosing (B) or (D), which pair a sentence with its own contrapositive.

(X, Y) is uniform on the square [1, 3] × [2, 4], of area 4, with density 1/4. X ≥ Y needs X ≥ 2 and Y between 2 and X: the triangle with vertices (2, 2), (3, 2), (3, 3), area 1/2. P(X ≥ Y) = (1/2)/4 = 1/8 = 0.125. Answer: 0.125.

Concept: Probabilities for two independent uniforms as areas in the joint rectangle. The trap is using the unit-square intuition P = 1/2, which only applies when both intervals coincide.

For the exponential distribution, E(X) = 1/λ and Var(X) = 1/λ². The condition 5/λ = 1/λ² gives λ = 1/5 = 0.2. Answer: 0.2.

Concept: Moments of the exponential distribution. The trap is mixing up the variance 1/λ² with the standard deviation 1/λ, which makes the condition 5/λ = 1/λ impossible.

P(T) = 1 - 0.6 = 0.4. Total probability: P(S) = P(S|T)P(T) + P(S|T̄)P(T̄) = 0.3 × 0.4 + 0.6 × 0.6 = 0.12 + 0.36 = 0.48. Bayes: P(T|S) = 0.12/0.48 = 0.25. Answer: 0.25.

Concept: Bayes' theorem with the law of total probability. The trap is using P(T̄) = 0.6 as P(T).

ln((x² + 1) cos x) = ln(1 + x²) + ln(cos x). Near 0, ln(1 + x²) = x² + O(x⁴) and cos x = 1 - x²/2 + O(x⁴), so ln(cos x) = -x²/2 + O(x⁴). The numerator is x²/2 + O(x⁴); dividing by x² gives the limit 1/2 = 0.5. Answer: 0.5.

Concept: Series expansion (or L'Hospital twice) for a 0/0 limit. The trap is dropping the cos x factor and answering 1.

M = uuᵀ is symmetric positive semidefinite of rank 1, so its singular values equal its eigenvalues. Mu = u(uᵀu) = ‖u‖²u, so the non-zero eigenvalue is ‖u‖² = 1 + 4 + 9 + 16 + 25 = 55, and the other four are 0. Sum = 55. Answer: 55.

Concept: A rank-one outer product has a single non-zero singular value ‖u‖². The trap is answering √55 (the norm) or 15 (the sum of entries).

E[X] = P(HH) = 1/4. E[Y] = P(at least one head) = 3/4. If X = 1 then Y = 1, so XY = X and E[XY] = 1/4. Cov(X, Y) = E[XY] - E[X]E[Y] = 1/4 - 3/16 = 1/16 = 0.0625, i.e. 0.062 or 0.063 depending on rounding. Answer: 0.062 to 0.063.

Concept: Covariance of indicator variables via E[XY] - E[X]E[Y]. The trap is treating X and Y as independent (covariance 0), when X = 1 forces Y = 1.

x³ − x = x(x − 1)(x + 1), so x³ ≥ x on [−1, 0] and [1, ∞), and x ≥ x³ elsewhere. The pieces meet at x = −1, 0, 1. Slopes: at −1, x³ has slope 3 and x has slope 1; at 0, slope 0 vs 1; at 1, slope 1 vs 3. The slopes differ at every crossing, so f has a corner at each. Answer: (D) {−1, 0, 1}.

Concept: A max of two smooth functions can fail to be differentiable only where they cross with different slopes. The trap is forgetting the crossing at x = −1 or 0.

The product of eigenvalues equals the determinant. Row 3 − Row 2 = Row 2 − Row 1 = (3, 3, 3), so the rows are linearly dependent and the determinant is 0. Answer: (B) 0.

Concept: Product of eigenvalues = determinant; sum = trace. The trap is trying to compute the eigenvalues directly instead of spotting the dependent rows.

By symmetry P(X) = P(Y) = 1/2. The relative orders of {1, 2} and {3, 4} are independent: of the 4 equally likely combined orderings, exactly one has 1 before 2 and 3 before 4, so P(X ∩ Y) = 1/4 = P(X)P(Y). Hence independent. They can co-occur (not exclusive), both can fail (2 before 1 and 4 before 3), and they are equally likely. Answer: (B).

Concept: Events about disjoint sets of elements in a random permutation are independent. The trap is confusing independent with mutually exclusive.

(A) P(A)P(B) = 0.15 ≠ 0.1, so not independent. (B) P(A ∪ B) = 0.3 + 0.5 − 0.1 = 0.7, true. (C) P(A ∩ Bᶜ) = P(A) − P(A ∩ B) = 0.2, true. (D) P(Aᶜ ∩ Bᶜ) = 1 − P(A ∪ B) = 0.3, not 0.4. Answer: (B), (C).

Concept: Inclusion-exclusion and De Morgan's law. The trap in (D) is computing (1 − 0.3)(1 − 0.5) = 0.35 or similar as if independent.

A bijection A → B gives |A| = |B| = n. A bijection A × A → A ∪ B gives n² = |A ∪ B|, and n ≤ |A ∪ B| ≤ 2n. So n² ≤ 2n, i.e., n ≤ 2. n = 1: take A = B, |A ∪ B| = 1 = 1². n = 2: take A, B disjoint, |A ∪ B| = 4 = 2². Both work, so 2 values. Answer: 2.

Concept: Bijection means equal cardinality; then bound |A ∪ B| between n and 2n. The trap is forgetting that A and B may overlap or be disjoint as needed.

By Cayley's formula the complete graph Kₙ on n labelled vertices has nⁿ⁻² spanning trees. For n = 4: 4² = 16. Answer: 16.

Concept: Cayley's formula nⁿ⁻² counts labelled spanning trees. The trap is counting non-isomorphic trees (only 2) instead of labelled ones.

rank(A) ≤ n, so the null space has dimension m − rank ≥ m − n: (A) true. (B) needs nullity exactly m − n; for A = 0 the nullity is m, so false. (C) fails for n = 1, m = 2, A = [1 1]: solutions (t, −t) have no zero entry when non-zero. (D) fails for A = [[1,0,0],[0,1,0]] (n = 2, m = 3): solutions (0, 0, t) have at most one non-zero entry. Answer: (A).

Concept: Rank-nullity gives a lower bound on the solution space, not an exact size. The trap is (B), which treats the bound as equality.

(A) False: the 5-cycle has k = 3 but no triangle. (B) True: an optimal colouring splits n vertices into k independent colour classes, so one has at least n/k. (C) True: in an optimal colouring every pair of colour classes must have an edge between them (else merge them), giving at least C(k,2) edges. (D) False: K_k has chromatic number k but maximum degree k − 1. Answer: (B), (C).

Concept: Colour classes are independent sets, and optimality forces an edge between every pair of classes. The trap is (A): high chromatic number does not require a clique.

(A) (a◊b)◊c = a + 2b + 2c but a◊(b◊c) = a + 2b + 4c: not associative. (B) Multiplication is associative: true. (C) a◊(b□c) = a + 2bc, while (a◊b)□(a◊c) = (a + 2b)(a + 2c): false. (D) a□(b◊c) = a(b + 2c) = ab + 2ac = (a□b)◊(a□c), and (b◊c)□a = ba + 2ca = (b□a)◊(c□a): true. Answer: (B), (D).

Concept: Check each law symbolically; one counterexample disproves it. The trap is assuming ◊ is associative because it is built from +.

Given the first ball is red, 9 red remain among 24 balls. P(both red | first red) = 9/24 = 0.375. (Equivalently P(both red)/P(first red) = (10/25 × 9/24)/(10/25).) The key accepts 0.37 to 0.38. Answer: 0.375.

Concept: Conditioning on the first draw just changes the second draw's sample space. The trap is giving the unconditional P(both red) = 0.15.

"A when B" means "if B then A". Here B = student scores more than 50% marks (q) and A = fail grade cannot be given (¬p). So the statement is q → ¬p. Answer: (A) q → ¬p.

Concept: Translating 'when'/'if' clauses: the condition after 'when' is the antecedent. The trap is reversing the implication or dropping the negation.

Characteristic equation r² − 5r + 6 = 0 gives r = 2, 3, so T(n) = a·2ⁿ + b·3ⁿ. From T(0) = 1: a + b = 1. From T(1) = 2: 2a + 3b = 2. Subtracting twice the first equation: b = 0, so a = 1. Hence T(n) = 2ⁿ exactly (check: T(2) = 10 − 6 = 4) and T(n) = Θ(2ⁿ). Answer: (A).

Concept: Linear homogeneous recurrence: the initial conditions decide which root survives. The trap is taking the dominant root 3 without solving for the constants, which here make the 3ⁿ coefficient zero.

Let I = ∫₀² f(x) dx. Substituting x → 2 − x gives I = ∫₀² f(2 − x) dx. Adding the two forms: 2I = ∫₀² [f(x) + f(2 − x)] dx = ∫₀² 1 dx = 2. So I = 1. Answer: (B) 1.

Concept: The property ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx turns the functional equation into a constant integrand.

A is its own inverse means A² = I. Since A is a symmetric 0/1 matrix, the diagonal entry (A²)ᵢᵢ equals the degree of vertex i, so every vertex has degree exactly 1. A graph in which every vertex has degree 1 is a set of disjoint edges covering all vertices - a perfect matching. Conversely, for a perfect matching no two edges share a vertex, so A² = I holds. Answer: (B).

Concept: (A²)ᵢⱼ counts walks of length 2 from i to j; diagonal entries are degrees. A cycle has degree 2 everywhere, so it fails.

Total outcomes = 6⁶ = 46656. Favourable outcomes: the six dice show 1 to 6 in some order, 6! = 720 ways. Probability = 720/46656 = 5/324. Answer: (B) 5/324.

Concept: The dice are distinguishable, so favourable outcomes are permutations (6!), not a single arrangement. Using 1/6⁶ is the trap.

A total order containing P is a linear arrangement with 1 before 2, 3 before 2 and 3 before 4. The first element must be minimal: 1 or 3. If 1 is first, the rest {2,3,4} must have 3 first (it precedes both 2 and 4), then 2 and 4 in either order: 2 orders (1324, 1342). If 3 is first, the rest {1,2,4} only needs 1 before 2: 3!/2 = 3 orders (3124, 3142, 3412). Total = 5. Answer: 5.

Concept: Counting linear extensions of a small poset. Fix the first element among the minimal elements and recurse; forgetting that 3 must also precede 4 overcounts.

Since 0 ≤ y ≤ 1 and x ≥ 0, z = xy ≤ x always, so E[z] ≤ E[x], i.e. z̄ ≤ x̄ - (D) always holds. The others fail without independence: if x = y uniform on [0,1], z̄ = E[x²] = 1/3 > 1/4 = x̄ȳ, breaking (A) and (B); if y = 1 − x, z̄ = E[x − x²] = 1/6 < 1/4, breaking (C). Answer: (D).

Concept: E[xy] = E[x]E[y] needs independence (zero covariance). Only the pointwise bound xy ≤ x survives for dependent variables.

(A) A row swap multiplies the determinant by −1 - true. (B) det B = −det A ≠ 0, so B is invertible - true. (C) False: A = [[1,2],[2,3]] is symmetric, but swapping its rows gives B = [[2,3],[1,2]], which is not. (D) False: A = [[0,1],[1,0]] has trace 0, swapping rows gives I with trace 2. Answer: (A), (B).

Concept: Elementary row operations: a swap flips the determinant's sign but keeps it non-zero. Symmetry and trace depend on positions of entries, so they are not preserved.

Colour odd-numbered vertices red and even-numbered vertices blue. Every edge joins vertices whose labels differ by an odd amount (sides differ by 1 or 7, chords by 3 or 5), so every edge joins an odd and an even vertex: the graph is bipartite and 2 colours suffice. It has edges, so 1 colour is not enough. Answer: 2.

Concept: A graph is 2-colourable exactly when it is bipartite (no odd cycle). The crossing chords make the picture look dense, which tempts answers of 3 or 4.

An element is its own inverse when g + g = 0, componentwise. In Z2 both 0 and 1 satisfy 2g = 0 (2 choices). In Z3 only 0 (2g ≡ 0 mod 3 forces g = 0). In Z4, 0 and 2 (2 choices). Total = 2 × 1 × 2 = 4. Answer: 4.

Concept: Self-inverse elements in a direct product are tuples of self-inverse components. Remember the identity counts as its own inverse.

The characteristic equation r² = r + 1 has roots φ = (1 + √5)/2 and ψ = (1 - √5)/2, so Lₙ = aφⁿ + bψⁿ; options with (1 - √5)/3 are not solutions of the recurrence. Test (A): n = 1: φ + ψ = 1 = L1. n = 2: φ² + ψ² = (φ + ψ)² - 2φψ = 1 - 2(-1) = 3 = L2. Test (D): n = 1 gives φ - ψ = √5 ≠ 1. Answer: (A).

Concept: Linear recurrences are solved by the roots of the characteristic equation, then fitted to initial values. The trap is (D), the Fibonacci form without the 1/√5 factor; Lucas numbers use the sum.

B is obtained from A by swapping row 1 and row 3 (rows 2 and 4 are unchanged). A single row interchange multiplies the determinant by -1, so det(B) = -det(A). det(A) is not 0 (A is a circulant whose eigenvalues 10, -2 ± 2i, -2 are all non-zero; det(A) = -160), so (A) and (C) fail, and det(AB) = det(A)det(B) = -det(A)², not a sum. Answer: (B).

Concept: Row operations and determinants: swapping two rows flips the sign. Spot the swap rather than expanding a 4×4 determinant; also recall det(AB) is a product, not a sum.

(B): if Q(x, y) holds for all x, y, then for every x, ∃y Q(x, y) is true, so P(x) ⇒ ∃y Q(x, y) holds for all x. (C): take the witness y0; for every x, P(x) ⇒ Q(x, y0), hence P(x) ⇒ ∃y Q(x, y). So ∃y∀x implies ∀x∃y. (A) and (D) speak about only one particular x; another x with P(x) true and no suitable y would still falsify the conjecture. Answer: (B) and (C).

Concept: Quantifier strength: ∃y∀x implies ∀x∃y, and a universal statement implies its weaker forms. Statements about a single x (existentials) never imply a universal claim.

f'(x) = 3x² + 30x - 33 = 3(x² + 10x - 11) = 3(x + 11)(x - 1), so critical points are x = -11 and x = 1. f''(x) = 6x + 30: f''(-11) = -36 < 0, a local maximum; f''(1) = 36 > 0, a local minimum. Answer: (B) f(x) has a local maximum and (D) f(x) has a local minimum.

Concept: A cubic with two distinct real critical points has one local maximum and one local minimum (but no global extrema). The trap is confusing "no global maximum" with "no local maximum".

Split the integrand. ∫∫∫ 4x²y: the y-integral ∫ from -2 to 2 of y dy = 0 (odd function over a symmetric interval), so this term is 0. ∫∫∫ z³: the z-integral ∫ from -1 to 1 of z³ dz = 0, also odd. Hence the whole integral is 0. Answer: 0.

Concept: Odd-function symmetry: integrating an odd power of a variable over a symmetric interval gives zero. Spotting it avoids a long computation.

Choose the 2k positions occupied by A ∪ B: C(n, 2k). Fill the other n - 2k positions with the remaining elements: (n - 2k)!. Within the chosen positions, either A takes the first k or B does: 2 ways; then arrange A among its k positions (k!) and B among its k (k!). Total = 2 C(n, 2k) (n - 2k)! (k!)². The two cases are disjoint since k ≥ 1. Answer: (D).

Concept: Positions-first counting. The trap is (C), which forgets that either A or B may come first.

Well-defined: if [x] = [y] then x ∼ y, i.e. f(x) = f(y), so F([x]) does not depend on the representative; (A) is false. Onto: for b ∈ B, f onto gives x with f(x) = b, so F([x]) = b. One-to-one: F([x]) = F([y]) means f(x) = f(y), i.e. x ∼ y, so [x] = [y]. Being injective and surjective, F is bijective. Answer: (B), (C), (D).

Concept: The quotient of a function by its kernel relation is always injective; surjectivity is inherited from f. This is the set-level first isomorphism theorem.

Symmetric difference is closed on subsets of X and associative (it is XOR of characteristic functions). Identity: AΔ∅ = A. Inverse: AΔA = ∅, so each A is its own inverse. Hence H is a group (an abelian one), (A) true, (B) false. The complement is not the inverse: AΔA' = X ≠ ∅ unless X is empty. Answer: (A) and (D).

Concept: The power set under symmetric difference is a group with identity ∅ in which every element is self-inverse. The trap is guessing the complement as inverse, which gives X, not the identity.

P(A) = 1/4, P(B) = P(C) = 1/2. A ⊆ B, so P(A ∩ B) = 1/4 ≠ P(A)P(B) = 1/8: not independent. Likewise A and C. B ∩ C = A, P = 1/4 = (1/2)(1/2): B and C are independent, which is the same as P(B|C) = P(B ∩ C)/P(C) = (1/4)/(1/2) = 1/2 = P(B). Answer: (C) and (D).

Concept: Independence means P(X ∩ Y) = P(X)P(Y). An event contained in another (A ⊆ B) with probabilities strictly between 0 and 1 is never independent of it.

Each vertex receives a color not used by any already-colored neighbour, and later neighbours avoid it in turn, so the coloring is proper: (A). A vertex has at most Δ(G) neighbours, so among the colors 1..Δ(G)+1 at least one is free: (B). (C) fails for a complete graph K_n, where Δ = n - 1 but n colors are needed. (D) fails because greedy depends on the order; e.g. a path a-b-c-d ordered a, d, b, c uses 3 colors though it is 2-colorable. Answer: (A) and (B).

Concept: Greedy coloring is always proper and uses at most Δ + 1 colors, but it is not optimal in general. Complete graphs and odd cycles show Δ colors can be insufficient.

tr(AB) = Σᵢ Σⱼ aᵢⱼ bⱼᵢ (AB is m×m) and tr(BA) = Σⱼ Σᵢ bⱼᵢ aᵢⱼ (BA is n×n). Both are the same double sum, so tr(AB) = tr(BA) even though AB and BA have different sizes. Statement 2 is the special case m = n. Answer: (C) Both Statement 1 and Statement 2 are correct.

Concept: The cyclic property of trace holds whenever both products are defined. The trap is thinking different sizes of AB and BA break the identity.

(A) xyxy = xxyy; cancel x on the left and y on the right: yx = xy. True. (B) Every element is its own inverse, so xy = (xy)⁻¹ = y⁻¹x⁻¹ = yx. True. (C) A group of order 2 is cyclic, hence abelian. True. (D) False: elements of a subgroup are elements of G, so they commute. Answer: (A), (B), (C).

Concept: Standard abelian criteria via cancellation laws. Every subgroup of an abelian group is abelian, and every group of prime order is cyclic.

A disconnected graph splits into at least two components. With parts of size k and 10 − k the edge maximum is C(k,2) + C(10−k,2), largest when one part is a single vertex: K₉ plus an isolated vertex gives C(9,2) = 36 edges. Answer: 36.

Concept: Maximum edges in a disconnected simple graph on n vertices is C(n−1, 2). The trap is splitting the vertices evenly, which minimises the count.

Identical balls and identical bins means we count partitions of 6 into at most 3 parts (empty bins allowed): 6; 5+1; 4+2; 3+3; 4+1+1; 3+2+1; 2+2+2. That is 7 arrangements. Answer: 7.

Concept: Identical objects into identical boxes is a partition count, not stars and bars. The trap is using C(8,2) = 28, which assumes distinct bins.

At x = 0 both numerator and denominator are 0, so apply L'Hôpital's rule: derivative of x is 1; derivative of 1 − e²ˣ is −2e²ˣ. The limit is 1 / (−2e⁰) = −1/2. Equivalently e²ˣ ≈ 1 + 2x, so the ratio ≈ x / (−2x). Answer: −0.5.

Concept: 0/0 form handled by L'Hôpital or the series e^u ≈ 1 + u. The usual slip is dropping the minus sign.

Write aₙ = 1 + (n if n odd, else 0). Σ xⁿ = 1/(1 − x). For the odd part, let g(x) = Σ over odd n of xⁿ = x/(1 − x²). Then Σ over odd n of n·xⁿ = x·g'(x). g'(x) = [(1 − x²) + 2x²]/(1 − x²)² = (1 + x²)/(1 − x²)², so x·g'(x) = x(1 + x²)/(1 − x²)². Total = x(1 + x²)/(1 − x²)² + 1/(1 − x). Check: coefficient of x is 1 + 1 = 2 = a₁. Answer: (A).

Concept: Split the sequence into simple pieces and use x·d/dx to bring down the factor n. Checking a₁ = 2 and a₃ = 4 quickly eliminates the other options.

(A³)ᵢᵢ counts closed walks of length 3 starting and ending at vertex i. In a simple graph such a walk must trace a triangle. Each triangle {u, v, w} yields 6 closed walks: 3 choices of starting vertex times 2 directions. So the number of triangles = tr(A³)/6. Answer: (D) A³ divided by 6.

Concept: Powers of the adjacency matrix count walks. The trap is dividing by 3 (forgetting direction) or by 2 (forgetting the starting vertex).

Doolittle (unit diagonal L): U row 1 = [1, 1, −2]; L₂₁ = 1, L₃₁ = 2. Row 2: U₂₂ = 3 − 1 = 2, U₂₃ = −1 − (−2) = 1. L₃₂ = (1 − 2·1)/2 = −1/2. U₃₃ = −5 − (2)(−2) − (−1/2)(1) = −5 + 4 + 1/2 = −1/2. Solve: eq2 − eq1 gives 2x₂ + x₃ = 3; eq3 − 2·eq1 gives −x₂ − x₃ = −1. Adding: x₂ = 2, x₃ = −1, then x₁ = 4 − 2 − 2 = 0. Answer: (D) L₃₂ = −1/2, U₃₃ = −1/2, x₁ = 0.

Concept: LU factors are the Gaussian elimination multipliers (L) and the reduced rows (U). The trap is a sign slip in L₃₂, whose multiplier is negative here.

(B) f(2) = 2·1 − 1 = 1, and if f(2ᵏ) = 1 then f(2ᵏ⁺¹) = 2·1 − 1 = 1: true. (A) f(1) = 1, f(3) = 2f(1) + 1 = 3, f(7) = 7; 2ⁿ⁺¹ − 1 = 2(2ⁿ − 1) + 1 gives f = 2(2ⁿ − 1) + 1 = 2ⁿ⁺¹ − 1: true. (C) f(5) = 2f(2) + 1 = 3 = 2¹ + 1; f(5·2ⁿ⁺¹) = 2f(5·2ⁿ) − 1 = 2(2ⁿ⁺¹ + 1) − 1 = 2ⁿ⁺² + 1: true. (D) f(5) = 2f(2) + 1 = 3, but 2² + 1 = 5: false. Answer: (A), (B), (C).

Concept: Prove each closed form by induction along the recurrence's own step (n to 2n or 2n + 1). One counterexample such as f(5) = 3 kills (D).

(A) (A²)ᵢᵢ = Σⱼ aᵢⱼaⱼᵢ = Σⱼ aᵢⱼ = degree of i: true. (B) False: path 1-2-3 (n = 3). A² + I has entry (1,2) = 0, since there is no walk of length exactly 2 from 1 to 2. (C) Sum of entries = 2|E| ≤ 2(n − 1) means |E| ≤ n − 1, but a triangle plus an isolated vertex (n = 4, 3 edges) has a cycle: false. (D) No isolated vertex does not imply connected (two disjoint edges): false. Answer: (A).

Concept: Aᵏ counts walks of length exactly k, so bipartite parity can leave zeros; (I + A)ⁿ⁻¹ would be the correct test. Few edges does not guarantee a forest unless the graph is connected.

Multiply the matrix by each vector. (A) (−1,1,0,1): rows give 9−6−4 = −1, 8−6−1 = 1, −20+15+5 = 0, −32+21+12 = 1, so Av = v (λ = 1). (B) (1,0,−1,0): first two rows give −9+2 = −7 and −8+3 = −5, not proportional to (1,0,...): not an eigenvector. (C) (−1,0,2,2): 9−4−8 = −3, 8−6−2 = 0, −20+16+10 = 6, −32+14+24 = 6, giving 3v (λ = 3). (D) (0,1,−3,0): −6+6 = 0, −6+9 = 3, 15−24 = −9, 21−21 = 0, giving 3v (λ = 3). Answer: (A), (C), (D).

Concept: Check Av = λv directly; no need for the characteristic polynomial. A zero entry in v must map to zero in Av, which quickly rules out (B).

Edges go from i to j only when j ≤ i, so every edge points to a smaller (or the same) vertex. In a spanning tree rooted at 5, each other vertex v has exactly one incoming tree edge from a parent u with u > v. Any such choice is automatically acyclic because parents are strictly larger, and every vertex then reaches back to 5. Choices: vertex 4 - parent 5 (1); vertex 3 - 4 or 5 (2); vertex 2 - 3, 4 or 5 (3); vertex 1 - 2, 3, 4 or 5 (4). Total = 1 × 2 × 3 × 4 = 24. Answer: 24.

Concept: An arborescence is fixed by choosing one parent per non-root vertex, provided no cycle forms; here edge direction rules out cycles. Self-loops (i = j) are never used.

Σ aₙxⁿ = 2 Σ n xⁿ + 3 Σ xⁿ. Known forms: Σ xⁿ = 1/(1 - x), Σ n xⁿ = x/(1 - x)². So G(x) = 2x/(1 - x)² + 3/(1 - x) = [2x + 3(1 - x)]/(1 - x)² = (3 - x)/(1 - x)². Check: at x = 0 this gives 3 = a₀. Answer: (D).

Concept: Uses the standard generating functions for 1 and n and linearity. A quick check is G(0) = a₀ = 3, which eliminates (B) and (C). The trap is forgetting to bring 3/(1 - x) to the common denominator.

A trial is a tie when both show the same number: P(tie) = 6/36 = 1/6, so P(someone wins) = 5/6. Winning on the third trial needs ties on trials 1 and 2 and no tie on trial 3: (1/6)(1/6)(5/6) = 5/216 ≈ 0.0231. Answer: 0.023 (accepted range 0.021 to 0.024).

Concept: A geometric distribution: first success on trial k has probability (1 - p)^(k-1) p with p = 5/6. The trap is computing the probability that a particular person (say P) wins, which would halve the answer.

Substitute u = x², du = 2x dx. The limits become 0 to (π/4)² = π²/16. Integral = (1/2)∫₀^(π²/16) cos u du = (1/2) sin(π²/16). With π = 3.14: π² = 9.8596, π²/16 = 0.61623 rad, sin(0.61623) ≈ 0.5780. Value ≈ 0.289. Answer: 0.289 (accepted range 0.27 to 0.30).

Concept: Simple substitution; the derivative of x² appears in the integrand. Traps: forgetting to change the upper limit to (π/4)², and evaluating the sine in degrees instead of radians.

A = uvᵀ = [[1, 1], [2, 2]]. It has rank 1, so one eigenvalue is 0. The other equals the trace, 1 + 2 = 3 (equivalently vᵀu = 1 + 2 = 3, since A u = u(vᵀu) = 3u). Eigenvalues are 0 and 3. Answer: 3.

Concept: An outer product uvᵀ has rank 1; its only nonzero eigenvalue is vᵀu with eigenvector u. The trap is confusing uvᵀ (a 2×2 matrix) with vᵀu (a scalar) when forming A.

a, b, c form a triangle (a-b, b-c, a-c), so at least 3 colours are needed. A 3-colouring exists: a = 1, b = 2, c = 3; e (adjacent to a, c) = 2; d (adjacent to b, c) = 1; f (adjacent to b, d, e, which use 2, 1, 2) = 3. Every edge joins different colours. Chromatic number = 3.

Concept: Lower bound from a clique (here a triangle), upper bound from an explicit colouring. The trap is stopping at the lower bound without checking that 3 colours suffice, or assuming 4 because the graph looks dense.

By Lagrange's theorem the order of a subgroup divides 84. The proper divisors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and the largest is 42. It is attained, for example the cyclic group Z₈₄ has a subgroup of order 42 (generated by 2). Answer: 42.

Concept: Lagrange's theorem limits subgroup orders to divisors of the group order; a proper subgroup excludes the whole group. The trap is answering 84 (not proper) or thinking the bound may not be reachable.

P is 2×2 with only one independent eigenvector. Two distinct eigenvalues would give two independent eigenvectors, so the eigenvalue is repeated (II true). Diagonalization needs 2 independent eigenvectors, so P cannot be diagonalized (III true). Invertibility depends on whether the repeated eigenvalue is 0. With N = [[4, -1], [16, -4]] (N² = 0, only eigenvectors multiples of [1, 4]ᵀ), P = N is singular but P = 2I + N is invertible, and both satisfy the condition, so (I) is not necessarily true. Answer: (D).

Concept: Geometric multiplicity 1 in a 2×2 matrix forces algebraic multiplicity 2 and a defective (non-diagonalizable) matrix. The trap is assuming singularity; a matrix is singular only if 0 is an eigenvalue.

P: rationals are countable (enumerate p/q by |p| + q). Q: a function from {0, 1} to N is a pair (f(0), f(1)), so Q is in bijection with N × N, which is countable. R: functions N → {0, 1} correspond to subsets of N (characteristic functions), i.e. the power set of N, which is uncountable by Cantor's diagonal argument. S: finite subsets of N can be listed by size and largest element (a countable union of finite sets), so countable. Answer: (D) P, Q and S only.

Concept: Finite products and countable unions of countable sets are countable; the power set of N (or infinite binary sequences) is not. The trap is confusing functions from {0, 1} to N with functions from N to {0, 1}.

In the 7-element model pick witnesses a, b, c for s, t, u. Restrict the model to the universe {a, b, c} (at most 3 elements), keeping the predicates' truth values on these elements. With no function symbols this restriction is a valid structure. ψ(a, b, c, v, w, x, y) held for all v, w, x, y in the large universe, so it holds in particular for all choices from {a, b, c}, since ψ is quantifier-free and its truth depends only on the chosen elements. Hence the restriction is a model of φ of size at most 3. Answer: (A).

Concept: Universal sentences are preserved under substructures, and an ∃³∀* sentence (Bernays-Schönfinkel class) has a model of size at most the number of existential variables. Options (B), (C), (D) cannot be guaranteed. The trap is thinking the 7-element model fixes the size.

A permutation of 100 numbers has 99 adjacent positions, and each adjacent swap gives a different permutation, so every vertex has degree y = 99. Any permutation can be sorted by adjacent swaps (bubble sort), so every vertex is connected to the identity permutation: the graph is connected, z = 1. y + 10z = 99 + 10 = 109. Answer: 109.

Concept: Adjacent transpositions generate all permutations, which is why bubble sort always works. The trap is counting all C(100,2) swaps for the degree, or thinking parity splits the graph into two components (that happens only if moves preserve parity, and adjacent swaps flip it).

Total probability: P(H_D) = 0.2 × 0.40 + 0.5 × 0.10 + 0.3 × 0.01 = 0.08 + 0.05 + 0.003 = 0.133. Bayes: P(H_G | H_D) = P(H_D | H_G) P(H_G) / P(H_D) = 0.08 / 0.133 ≈ 0.6015. Answer: 0.60 (accepted range 0.60 to 0.62).

Concept: Bayes' theorem with the law of total probability for the denominator. The trap is answering 0.40, which is P(H_D | H_G), the reverse conditional.

¬((p → q) ∧ (¬r ∨ ¬s)) = ¬(p → q) ∨ ¬(¬r ∨ ¬s) = (p ∧ ¬q) ∨ (r ∧ s). r ∧ s needs x to be both a perfect square and a prime, which is impossible. So we need p ∧ ¬q: x ∈ {8, 9, 10, 11, 12} and x not composite. Since x ≥ 2, not composite means prime; the only prime in the set is 11. Answer: 11.

Concept: Applying De Morgan's law and ¬(p → q) ≡ p ∧ ¬q. The trap is mis-negating the implication as ¬p → ¬q or forgetting that r ∧ s is unsatisfiable.

Classify valid n-bit strings by their last bit. If it ends in 0, the first n-1 bits form any valid string: aₙ₋₁ ways. If it ends in 1, the bit before must be 0, and the first n-2 bits form any valid string: aₙ₋₂ ways. So aₙ = aₙ₋₁ + aₙ₋₂ (with a₁ = 2, a₂ = 3). Answer: (B).

Concept: Standard Fibonacci-type recurrence from case analysis on the last bit. The trap is counting the '1' ending as aₙ₋₁ without forcing the preceding 0.

Substitute t = x - 4; as x → 4, t → 0. The limit becomes lim (t→0) sin t / t = 1. Answer: 1.

Concept: The standard limit sin t / t → 1 after a shift of variable. The trap is substituting x = 4 directly and stopping at the indeterminate form 0/0.

A pdf must integrate to 1: ∫ from a to 1 of x⁻² dx = [-1/x] from a to 1 = -1 + 1/a = 1. So 1/a = 2 and a = 0.5. Answer: 0.5.

Concept: Normalisation condition of a probability density function. The trap is a sign slip in the antiderivative -1/x.

For a real matrix, complex eigenvalues occur in conjugate pairs, so the third eigenvalue is 2 - i. The determinant is the product of eigenvalues: (2 + i)(2 - i) × 3 = (4 + 1) × 3 = 15. Answer: 15.

Concept: Real matrices have conjugate-paired complex eigenvalues, and det = product of eigenvalues. The trap is treating the given two as all the information and multiplying (2 + i) × 3.

(x³ + x⁴ + ···)³ = x⁹ (1 + x + x² + ···)³ = x⁹ (1 - x)⁻³. We need the coefficient of x³ in (1 - x)⁻³, which is C(3 + 2, 2) = C(5, 2) = 10. Check by counting: solutions of a + b + c = 12 with each ≥ 3 equal solutions of a' + b' + c' = 3 with each ≥ 0, which is C(5, 2) = 10. Answer: 10.

Concept: Generating functions / stars and bars with lower bounds. The trap is forgetting to factor out x⁹ or using C(n + k, k) with the wrong n.

Note that 6(1)² + 2(1) = 8 = a₁, so aₙ = Σ (k = 1 to n) (6k² + 2k). Σ 6k² = 6 · n(n + 1)(2n + 1)/6 = n(n + 1)(2n + 1); Σ 2k = n(n + 1). aₙ = n(n + 1)(2n + 1 + 1) = 2n(n + 1)². a₉₉ = 2 × 99 × 100² = 1,980,000 = 198 × 10⁴. K = 198 (key range 197.9 to 198.1). Answer: 198.

Concept: Unrolling a recurrence into standard power sums. The trick is noticing a₁ fits the same formula, so the sum can start at k = 1; otherwise an off-by-8 slip is easy.

Both rules preserve divisibility by 5 (halving an even multiple of 5 keeps it a multiple of 5; adding 5 too), so multiples and non-multiples of 5 never mix. Non-multiples: 1 → 6 → 3 → 8 → 4 → 2 → 1, so f(1) = f(2) = f(3) = f(4) = f(6) = f(8). Any other non-multiple n descends into this cycle: an odd n > 5 gives (n + 5)/2 < n after halving, and an even n halves, so the value strictly decreases until it lands in {1, 2, 3, 4, 6, 8}. All non-multiples share one value. Multiples of 5: 5 → 10 → 5, and every 5m reduces the same way to 5. All share one value. So f takes at most 2 distinct values. Answer: 2.

Concept: Find the invariant (residue class mod 5) and show each class collapses to one cycle. The trap is assuming one value per residue mod 5, giving 5.

Each round: P(Y) = 1/4, P(N) = 2/4, P(repeat) = 1/4. Let p = P(output Y). Then p = 1/4 + (1/4)p, so (3/4)p = 1/4 and p = 1/3 ≈ 0.33. Equivalently, conditioned on stopping, Y has 1 of the 3 stopping outcomes. Answer: 0.33 (key 0.33 to 0.34).

Concept: Repeated trials with a restart: condition on the process stopping. The trap is answering 0.25 by ignoring the restarts.

P ∧ (P ⇒ Q) ≡ P ∧ Q, so P and Q are both true whenever the premise holds. false - not implied; Q - implied; true - implied (everything implies true); P ∨ Q - implied; ¬Q ∨ P - implied since P is true. Count = 4. Answer: 4.

Concept: A implies B when B is true in every model of A. P ∧ (P ⇒ Q) is modus ponens, giving P ∧ Q. Forgetting that "true" is implied by everything is the usual slip.

f(x) + f(-x) keeps only the even-power terms of f (doubled), so the highest even power in f is x¹⁰. g(x) - g(-x) keeps only the odd-power terms of g = f′. Odd powers of f′ come from even powers of f: x¹⁰ gives 10x⁹, and a higher odd term in g would need a higher even term in f, which does not exist. Degree = 9. Answer: 9.

Concept: f(x)+f(-x) isolates even terms, g(x)-g(-x) isolates odd terms, and differentiation maps even powers to odd ones. The trap is assuming f itself has degree 10.

By the Four Colour Theorem every planar graph is 4-colourable, and K4 is planar and needs 4 colours, so 4 is sufficient for all planar graphs and cannot be lowered. Answer: 4.

Concept: Four Colour Theorem. Confusing it with the Five Colour Theorem, or with 3 colours that suffice only for special classes, is the usual mistake.

I is false: with m < n a system can still be inconsistent, e.g. x + y + z = 1 and x + y + z = 2. II is false: an overdetermined system can be consistent, e.g. x = 1, 2x = 2, 3x = 3. III is true: the homogeneous system with m = n always has the zero solution. Only III holds. Answer: (C).

Concept: Consistency depends on ranks, not on comparing m and n. Underdetermined does not mean solvable; overdetermined does not mean unsolvable.

Equal numbers of each type, so P(T1) = P(T2) = 0.5. By total probability, P(lasts > 100 h) = 0.5 × 0.7 + 0.5 × 0.4 = 0.35 + 0.20 = 0.55. Answer: 0.55.

Concept: Law of total probability over a partition. With an unequal mix a plain average would be wrong; here the mix is equal, so the weighted and plain averages coincide.

det A = product of eigenvalues = 1 × 2 × 4 = 8. Transposing does not change a determinant, and det(A⁻¹) = 1/det A. So det((A⁻¹)ᵀ) = 1/8 = 0.125. Answer: 0.125 (accepted range 0.124 to 0.126).

Concept: Determinant = product of eigenvalues; det(Aᵀ) = det A; det(A⁻¹) = 1/det A. The trap is answering 8 by forgetting the inverse.

Reflexive: (a, b)R(a, b) since a ≤ a. So P is true. Transitive: take (5, 5)R(1, 6) because 5 ≤ 6, and (1, 6)R(2, 1) because 1 ≤ 2. But (5, 5)R(2, 1) fails since 5 > 2 and 5 > 1. So Q is false. Answer: (B) P is true and Q is false.

Concept: An "or" condition breaks transitivity: the two steps can be justified by different coordinates. One concrete counterexample is enough; the trap is assuming it behaves like a product order (which uses "and").

(A): ∀x p(x) ⇒ ∀x q(x) ≡ ¬∀x p(x) ∨ ∀x q(x) ≡ ∃x ¬p(x) ∨ ∀x q(x), so it is a tautology. (B): ∃ distributes over ∨, valid. (C): a witness for p ∧ q is a witness for each, valid. (D): over the integers let p = "even", q = "odd". Every x is even or odd, but neither "all even" nor "all odd" holds, so (D) is not valid. Answer: (D).

Concept: ∀ distributes over ∧ and ∃ over ∨; the reverse directions (∀ over ∨, ∃ over ∧) hold only one way. A two-predicate counterexample such as even/odd settles it.

Model compounds as vertices and reactions as edges. The degree sum is even (handshaking lemma). The 9 compounds in S contribute 9 × 3 = 27, an odd number, so the degrees in U \ S must sum to an odd number, which forces at least one vertex in U \ S to have odd degree. That proves II. I and III are not forced: the 14 vertices in U \ S can have a mix of odd and even degrees. Answer: (B) Only II.

Concept: Handshaking lemma: the number of odd-degree vertices is even. Here S already has 9 odd vertices, so U \ S must hold an odd number of odd vertices, at least one.

13 ≡ -4 (mod 17). (-4)² = 16 ≡ -1, so (-4)⁴ ≡ 1. 99 = 4 × 24 + 3, so 13⁹⁹ ≡ (-4)³ = -64 ≡ -64 + 68 = 4 (mod 17). Answer: 4.

Concept: Modular exponentiation: find a small power that is ±1 (order of the element), then reduce the exponent. Fermat (exponent mod 16) also works: 99 mod 16 = 3.

"Not all that glitters is gold" is ¬∀x (glitters(x) → gold(x)). Push the negation in: ∃x ¬(glitters(x) → gold(x)) = ∃x (glitters(x) ∧ ¬gold(x)). So some thing glitters and is not gold. Option A says nothing that glitters is gold (too strong), B and C talk about gold things. Answer: (D) ∃x: glitters(x) ∧ ¬gold(x).

Concept: Negating a universal conditional gives an existential conjunction: ¬∀x(P→Q) ≡ ∃x(P ∧ ¬Q). The trap is option A, which reads "no glittering thing is gold", a much stronger claim.

Let the break point X be uniform on [0, 1]. The shorter piece has length min(X, 1 - X). For X in [0, 0.5] the shorter piece is X; for X in [0.5, 1] it is 1 - X. By symmetry, E = 2 ∫ from 0 to 0.5 of x dx = 2 × (0.5²/2) = 0.25. Answer: 0.25 (official range 0.24 to 0.27).

Concept: Expectation of a function of a uniform random variable. The trap is answering 0.5 (the expected break point) instead of the expected minimum of the two pieces.

At θ = π/6 the first row equals the second row, so f(π/6) = 0. At θ = π/3 the first row equals the third, so f(π/3) = 0. f is continuous on [π/6, π/3] and differentiable inside (tan is smooth there). By Rolle's theorem some θ in (π/6, π/3) has f′(θ) = 0, so I is true. Expanding along row 1: f(θ) = A sin θ - B cos θ + C tan θ with constant A = cos(π/6)tan(π/3) - tan(π/6)cos(π/3) = 3/2 - 1/(2√3) ≠ 0, so f is not identically zero; a non-constant differentiable function with equal end values must have f′ ≠ 0 somewhere, so II is true. Answer: (C) Both I and II.

Concept: Rolle's theorem applied to a determinant whose value vanishes at the endpoints because two rows coincide. The trap is assuming f is constant (zero) on the interval, which would make II false.

D: x ∈ f⁻¹(S ∩ T) ⇔ f(x) ∈ S and f(x) ∈ T ⇔ x ∈ f⁻¹(S) and x ∈ f⁻¹(T). So preimage always preserves intersection - true. Counter-example for the rest: X = {1, 2}, Y = {y}, f(1) = f(2) = y, A = {1}, B = {2}. Then A ∩ B = ∅, so f(A ∩ B) = ∅ but f(A) ∩ f(B) = {y} (B and C fail), and |f(A ∪ B)| = 1 ≠ 1 + 1 (A fails). Answer: (D).

Concept: Preimages commute with unions and intersections; images commute only with unions, and only when the function is injective for intersections. The trap is B, which holds only for one-to-one functions.

dim(V₁ + V₂) = dim V₁ + dim V₂ - dim(V₁ ∩ V₂). Since V₁ + V₂ ⊆ V, dim(V₁ + V₂) ≤ 6, so 4 + 4 - dim(V₁ ∩ V₂) ≤ 6, giving dim(V₁ ∩ V₂) ≥ 2. The value 2 is attained, e.g. V₁ = span(e₁..e₄), V₂ = span(e₃..e₆). Answer: 2.

Concept: Dimension formula for the sum of subspaces. The trap is answering 0, forgetting that two large subspaces of a small space must overlap.

On [0, 2π], x ≥ 0, so |x sin x| = x|sin x|. sin x ≥ 0 on [0, π] and ≤ 0 on [π, 2π]. An antiderivative of x sin x is sin x - x cos x. ∫ from 0 to π of x sin x dx = [sin x - x cos x] from 0 to π = π. ∫ from π to 2π of -x sin x dx = -[sin x - x cos x] from π to 2π = -(-2π - π) = 3π. Total = 4π, so k = 4. Answer: 4.

Concept: Split an absolute-value integrand at the sign changes, then integrate by parts. The trap is integrating x sin x over [0, 2π] directly, which gives -2π.

Mutually exclusive and exhaustive: P(A) + P(B) = P(S) = 1. Let P(A) = p, so P(B) = 1 - p and the product is p(1 - p). This is maximised at p = 1/2 (derivative 1 - 2p = 0), giving 1/4. Answer: 0.25.

Concept: Complementary events and maximising p(1 - p). The trap is treating A and B as independent, which contradicts their being mutually exclusive with positive probabilities.

"All rainy days are cold" is ∀d (Rainy(d) → Cold(d)). Negating it: ¬∀d (Rainy(d) → Cold(d)) ≡ ∃d ¬(Rainy(d) → Cold(d)) ≡ ∃d (Rainy(d) ∧ ¬Cold(d)): there is a rainy day that is not cold. Answer: (D).

Concept: Negation of a universally quantified implication. The trap is option A, which says every day is rainy and not cold.