GATE 2025 DA (Data Science & AI) question paper with solutions

Courage and bravery are synonyms, so the second word of the pair must be a synonym of yearning. Yearning means a strong desire for something, which is exactly what longing means. Yelling (shouting), yawning (opening the mouth when tired) and glaring (staring angrily) only resemble yearning in sound or spelling. Answer: (A) Longing.

Concept: Synonym analogy. The trap is the distractors that start with Y or share letters with yearning; match the meaning relation, not the look of the word.

The interrupting event (it started to rain) is in the simple past. An action that was already in progress before and up to a past event takes the past perfect continuous: had been playing. 'Have been playing' is present perfect continuous and clashes with the past-tense clause; the modal forms change the meaning to something hypothetical. Answer: (B) had been playing.

Concept: Tests the past perfect continuous for an ongoing action interrupted by a past event. The usual trap is picking the present perfect continuous, which cannot pair with a past-tense time clause.

Count the entries that are at most 4, row by row. Row 1 (0, 1, 0, 2): all 4 qualify. Row 2 (4, 7, 3, 3): 4, 3, 3 qualify - 3 pixels. Row 3 (5, 5, 4, 4): the two 4s - 2 pixels. Row 4 (6, 7, 3, 2): 3 and 2 - 2 pixels. Total = 4 + 3 + 2 + 2 = 11. Answer: (C) 11.

Concept: Simple table reading with an inclusive inequality. The trap is treating ≤ 4 as < 4, which drops the four 4s and gives 7, or miscounting a row.

The labelled region 4 lies in the rectangle (1) and the ellipse (3): 1 + 3 = 4, so a region's number is the sum of the numbers of the shapes containing it. P is in all three: 1 + 2 + 3 = 6. Q is in the triangle and the ellipse: 2 + 3 = 5. R is in the rectangle and the triangle: 1 + 2 = 3. Answer: (A) P = 6; Q = 5; R = 3.

Concept: Pattern induction from the one labelled overlap region. The trap is guessing a product or a count rule; check the rule against the given overlap (4 = 1 + 3) before applying it.

New width = 1.1W, length unchanged. Area: L × 1.1W = 1.1LW, a 10% increase for every L and W. Perimeter: 2(L + 1.1W) = 2(L + W) + 0.2W, an increase of 0.2W/(2(L + W)), less than 10%. Diagonal: √(L² + 1.21W²) grows by less than 10% because L is unchanged. Square: only if 1.1W = L, which is not true in general. Answer: (C) Area increases by 10%.

Concept: Area is linear in each side, so scaling one side scales area by the same factor; perimeter and diagonal are not. The trap is assuming every measure scales by 10%.

Match each phrasal verb to its meaning. P (a messy house) - 'clear it up' means tidy it: P-3. Q (unhappy with marks) - 'take it up with your teacher' means raise the matter with someone in authority: Q-4. R (avoids politics) - 'won't bring it up' means won't mention the topic: R-1. S (doesn't know a word) - 'look it up' means find it in a dictionary: S-2. Answer: (B) P - 3; Q - 4; R - 1; S - 2.

Concept: Tests phrasal verbs with 'up': clear up, take up, bring up, look up. The trap is that 'clear it up' can also mean explain, which tempts matching it to S; only 'look it up' fits the dictionary sense.

Both brothers follow the same weight function w(a), with a₁ ≠ a₂ and w(a₁) = w(a₂). If both ages were at most 50, w would be strictly increasing there, so a₁ < a₂ would force w(a₁) < w(a₂). If both were at least 50, w would be strictly decreasing, giving w(a₁) > w(a₂). Either way the weights could not be equal. So the two ages must lie on opposite sides of the peak, and neither can be exactly 50 (the peak value is attained only once). Hence a₁ < 50 < a₂. Answer: (B).

Concept: A strictly monotone function is one-to-one, so equal values at different inputs require the inputs to straddle the turning point. The trap is option (D), thinking the peak must be involved.

Each triangle has two sides equal to r with an included angle of 360°/12 = 30°. Area of one triangle = (1/2) r² sin 30° = r²/4. Area of a square of side r = r², so each square needs r² / (r²/4) = 4 triangles. With 12 triangles, the number of squares = 12/4 = 3. Answer: (A) 3; 4.

Concept: Area bookkeeping with the triangle area formula (1/2)ab sin C. The trap is using the side d in the area, or dividing 12 by 3 because a dodecagon suggests multiples of 3.

Write everything in base 3: 3^(x²) = 3³ × 3^(2x) = 3^(3 + 2x), so x² = 2x + 3, i.e. x² - 2x = 3 (x = 3 or x = -1). The required value is 2^(x²) / 2^(2x) = 2^(x² - 2x) = 2³ for either root. Answer: (C) 2³.

Concept: Equate exponents after a common base, then notice the target only needs x² - 2x, not x itself. The trap is reading (2^x)² as 2^(x²), or solving for x and plugging in only one root.

Average X = (5×20 + 6×40 + 7×30 + 8×10)/100 = (100 + 240 + 210 + 80)/100 = 6.3. Earnings are linear in X, so average earnings = 1000(6.3 - 0.2) = ₹ 6,100. Answer: (A) 6,100.

Concept: Weighted mean from a frequency chart and linearity of the mean. The trap is forgetting to subtract 0.2 (giving 6,300) or averaging the four X values without weights.

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.

In a dependency-preserving decomposition, every functional dependency can be checked inside a single decomposed relation. If it is not preserved, some dependency X → Y spans attributes held in different relations, so to check it on every update the relations must be recombined - that is, joined - before the dependency can be verified. Hence joins are executed more often. Answer: (C) Join.

Concept: Dependency preservation means FDs can be enforced locally. The trap is choosing projection because decomposition itself is a projection; the extra cost of enforcement comes from rejoining.

Inside out: σ_maker="ABC"(Make) gives the models made by ABC. Joining with Car on model gives the cars of those models. σ_color="red" keeps only red ones, so we have red cars made by ABC. Joining with Own on serial attaches the owners of those cars, and π_owner returns the owners. Answer: (C) All owners of a red car made by ABC.

Concept: Reading nested selections and natural joins: selections are conjunctive (red AND ABC), not a union. Option (D) forgets the final projection is onto owner, and (B) invents a count condition.

h(1) = 3 -> slot 3. h(4) = 12 mod 10 = 2 -> slot 2. h(5) = 15 mod 10 = 5 -> slot 5. h(6) = 18 mod 10 = 8 -> slot 8. h(14) = 42 mod 10 = 2: slot 2 full, 3 full, 4 free -> slot 4. h(15) = 45 mod 10 = 5: slot 5 full, 6 free -> slot 6. Answer: (D) 4 and 6.

Concept: Linear probing moves to the next free slot on a collision. The trap is reporting the home slot (2 and 5) instead of where the key actually lands after probing.

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).

Here y_n = +1 and f < 0, so the sample is misclassified and an update happens: w_new = w_old + x_n, b_new = b_old + 1. f(x_n; w_new, b_new) = (w_old + x_n)ᵀx_n + b_old + 1 = f(x_n; w_old, b_old) + ‖x_n‖² + 1. Since ‖x_n‖² + 1 > 0, the new score is strictly larger. (D) is false because y_n f = f < 0. Answer: (A).

Concept: This is the perceptron update: each mistake pushes the score for that sample toward its correct sign by ‖x‖² + 1. The trap is thinking one update must fix the sign or leave f unchanged.

A.extend(B) appends each element of B to A in place, giving [1, 2, 3, 4, 5, 6]. A.append(B) adds B as a single element: [1, 2, 3, [4, 5, 6]]. Lists have no update method (that is for dict and set), so A.update(B) raises AttributeError. A.insert needs an index and a value; A.insert(B) raises TypeError. Answer: (A) A.extend(B).

Concept: List methods: extend concatenates, append nests. The trap is append, which produces a nested list, and update, which exists only on dict/set.

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).

Variable elimination sums out hidden variables factor by factor and returns the exact posterior P(query | evidence), so it is an exact algorithm used to compute conditional probabilities: (A) false, (C) true. Gibbs sampling and rejection sampling estimate probabilities from samples; the estimates converge only as the number of samples grows, so both are approximate: (B) false, (D) true. Answer: (C), (D).

Concept: Exact inference (enumeration, variable elimination) versus approximate inference (rejection sampling, likelihood weighting, Gibbs/MCMC). The trap is swapping the categories.

Binary search needs sorted data (otherwise discarding half the range is unjustified) and O(1) access to the middle element. An unsorted array or list fails the first requirement. A sorted linked list fails the second: reaching the middle takes O(n) steps, so the total is O(n). Only a sorted array gives O(log n). Answer: (C).

Concept: Binary search requires both order and random access. The trap is the sorted linked list: it has order but no random access, so it degrades to linear time.

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.

Insertion sort makes one adjacent swap per inversion. The pair (15, 13) is always one inversion, so x must create exactly one more. x = 10: 11 > 10 before it - 1 inversion; 10 < 15, 13. Total 2. Yes. x = 12: nothing before it is larger, and 12 < 15, 13. Total 1. No. x = 14: 14 > 13 after it - 1 inversion. Total 2. Yes. x = 16: 16 > 15 and 16 > 13 - 2 inversions. Total 3. No. Answer: (A) 10, (C) 14.

Concept: Swaps in insertion sort equal the number of inversions. The trap is counting only inversions involving x and forgetting the fixed (15, 13) pair.

Single linkage measures the distance between two clusters by their closest pair of points - the minimum, DIS-2. Complete linkage uses their farthest pair - the maximum, DIS-1. Answer: (B), (D).

Concept: Linkage criteria in hierarchical clustering: single = nearest neighbour (min), complete = farthest neighbour (max), average = mean. The trap is associating 'complete' with 'closest'.

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'.

Inner query: SR Nagar loans are 40000, 25000 and 65000, so MAX = 65000. Outer query keeps loans with amount > 65000: L11 (90000), L23 (80000), L25 (70000). L19 itself (65000) is not strictly greater. Rows returned = 3.

Concept: A scalar subquery with MAX, then a strict comparison. The trap is using >= (which would add L19) or comparing against the SR Nagar total instead of the maximum.

Minimise Σ(yᵢ - w xᵢ)². Setting the derivative to zero gives w = Σxᵢyᵢ / Σxᵢ². Σxy = (-1)(1) + (2)(-5) + (3)(5) = -1 - 10 + 15 = 4. Σx² = 1 + 4 + 9 = 14. w = 4/14 = 0.2857... ≈ 0.286 (key accepts 0.285 to 0.287).

Concept: Regression through the origin (no intercept): w = Σxy/Σx². The trap is using the intercept formula with means subtracted, which answers a different model.

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).

Forward pass: h₁ = R(2 + 3) = 5; h₂ = R(2 - 3) = R(-1) = 0; output pre-activation = 4·5 + (-1)·0 = 20 > 0, so y = 20. The output ReLU is active (derivative 1). ∂y/∂f = 1 · h₂ = 0. ∂y/∂a = 1 · e · R'(5) · u = 4 · 1 · 2 = 8 (h₁'s ReLU is active). Answer: (A) ∂y/∂a = 8, ∂y/∂f = 0.

Concept: Backpropagation through ReLUs: the gradient passes unchanged through active units and is zero through inactive ones. The trap is forgetting the chain factors e and u, or not noticing h₂ is zero so ∂y/∂f vanishes.

Tree-1: after the left leaf, A has α = 2. At MIN node B, after seeing x, β = x. Prune B's right child when β ≤ α, i.e. x ≤ 2: x ∈ (-∞, 2]. Tree-2: D = min(5, y), so C has α = min(5, y). At E, after the leaf 2, β = 2. Prune when 2 ≤ α = min(5, y), i.e. y ≥ 2 (and 2 ≤ 5 holds): y ∈ [2, ∞). Answer: (C).

Concept: Alpha-beta prunes a MIN node's remaining children once its upper bound β drops to or below the MAX ancestor's lower bound α. The trap is reversing the inequality or using 5 instead of min(5, y) as α.

f = g + h. Expand S: A (g 4, f 6), E (g 1, f 7). Expand A (6): B (g 6, f 8). Frontier: E 7, B 8. Expand E (7): C (g 3, f 9). Frontier: B 8, C 9. Expand B (8): C via B has g 8, worse than 3, ignored. Expand C (9): D (g 6, f 8). Expand D (8): G (g 9, f 9). Expand G. Order: S, A, E, B, C, D, G. Answer: (C).

Concept: A* orders by g + h, not by g or h alone. The trap is expanding E first because its edge cost is smallest (that is uniform-cost thinking), or expanding C before B.

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.

CUBE(D1, D2) produces groupings for every subset of {D1, D2}: (D1, D2): 3 × 2 = 6 rows (D1): 3 rows (D2 shown as NULL) (D2): 2 rows (D1 shown as NULL) (): 1 grand-total row Total = 6 + 3 + 2 + 1 = 12. Answer: (D) 12.

Concept: CUBE over k attributes gives all 2ᵏ grouping sets; ROLLUP gives only the k + 1 hierarchical ones (here 10). The trap is counting just the 6 base groups.

Tuple assignment evaluates all right-hand sides with the old values first. Iteration 1: "other" in C. "this" in A, so A = A - B = {this}, B = B - C = {that}, C = C - A = {other}. "that" in B, so A = C | A = {other, this}, B = A | B = {this, that}, C = B | C = {that, other}. Iteration 2: "other" in C. "this" in A: A = A - B = {other}, B = B - C = {this}, C = C - A = {that}. "that" is not in B, so no second update. Iteration 3: "other" not in C, loop ends with A = {other}, B = {this}, C = {that}. Answer: (B) Only B.

Concept: Set difference and union with simultaneous tuple assignment. The trap is updating A first and then using the new A when computing B and C.

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.

The closest point of Class-1's convex hull to the origin is (1, 1), the midpoint of (2, 0)-(0, 2), at distance √2. The maximum-margin boundary is the perpendicular bisector: x₁ + x₂ = 1. Scale so the support vectors satisfy |wᵀx + b| = 1: with w = (1, 1)ᵀ, b = -1, we get (2, 0) and (0, 2) at +1 and (0, 0) at -1, while (2, 2) gives 3. So (A) is false. Support vectors: (2, 0), (0, 2), (0, 0) - three: (B) true. Margin = 2/‖w‖ = 2/√2 = √2: (C) true. The data are separable, so hard-margin SVM gets 100% training accuracy: (D) false. Answer: (B), (C).

Concept: Hard-margin SVM: the margin width is 2/‖w‖ under the canonical scaling. The trap is (A): w = (4, 4), b = 1 does not even separate the data, since (0, 0) would score +1.

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).

Expand: ‖μ_red - x‖² = ‖μ_red‖² - 2μ_redᵀx + ‖x‖², and similarly for green. The ‖x‖² terms cancel: f(x) = 2(μ_green - μ_red)ᵀx + (‖μ_red‖² - ‖μ_green‖²). So f(x) = wᵀx + b with w = 2(μ_green - μ_red) and b = ‖μ_red‖² - ‖μ_green‖²: (B) and (C) true; it is not quadratic: (D) false. At x = 0, f = ‖μ_red‖² - ‖μ_green‖² < 0 when ‖μ_red‖ < ‖μ_green‖, so x = 0 is labelled red, not green: (A) false. Answer: (B), (C).

Concept: A nearest-mean classifier has a linear (hyperplane) decision boundary because the quadratic terms cancel. The trap is (D), reading the squared distances as making f quadratic.

In Customer, aadhaar is only a foreign key with no UNIQUE constraint, so several customers (emails) may share one aadhaar; it is not a candidate key: (A) false. phone has no NOT NULL constraint, so it can be NULL: (B) true. In Person, aadhaar is the PRIMARY KEY, hence a candidate key: (C) true, and primary keys cannot be NULL: (D) false. Answer: (B), (C).

Concept: A foreign key need not be unique or non-null; a primary key is always unique and NOT NULL. The trap is assuming the referencing column inherits the key property of the referenced one.

A⁺ = A, B, C, E, F, then D, G from E: all of ABCDEFG, so A is a key. BC⁺ = B, C, A, then everything A gives: BC is a superkey; B⁺ = B and C⁺ = C, so BC is minimal - a candidate key. E⁺ = E, D, G only: not a key. D, F, G appear only on right sides, so any key must reach A; the candidate keys are exactly A and BC: (A), (C) false; (B) true. E → DG has a left side that is not a superkey, so R violates BCNF: (D) true. Answer: (B), (D).

Concept: Find keys with attribute closures and check minimality; BCNF requires every non-trivial FD's left side to be a superkey. The trap is missing BC as a key because BC → A is listed last.

Each given shortest path has length 3, so no edge may create a shorter route between its endpoints. (b, d): a - b - d would have length 2 < 3. Not an edge. (b, h): a - b - h would have length 2 < 3. Not an edge. (e, g): e - g - h would have length 2 < 3. Not an edge. (b, g): it creates a - b - g - h (length 3, not shorter), and gives no shorter a-d or e-h path, so it may exist. Answer: (A), (C), (D).

Concept: An edge between two vertices of a shortest path that skips ahead would shorten it, so such edges cannot exist. The trap is ruling out (b, g), which only creates a path of equal length.

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.

The division returns customers who have a loan at every branch. Branches: Banjara Hills, Kondapur, SR Nagar, Balanagar. Anand: Banjara Hills only. Ankita: SR Nagar. Gopal: SR Nagar. Sunil: Balanagar (L23), Kondapur (L25). Karteek: Banjara Hills (L11), Kondapur (L14), SR Nagar (L22), Balanagar (L23) - all four. Result: {Karteek}, so 1 tuple.

Concept: Relational division answers 'for all' queries. The trap is counting customers with loans at many branches (Sunil) or counting loans rather than distinct branches.

b = 10 throughout. (a-1)/2 is a float in Python 3, but the comparisons still work (7.0 == 7). f(15) = 2 f(7); f(7) = 2 f(3); f(3) = 2 f(1); f(1) = 2 f(0); f(0) = 10. So f(1) = 20, f(3) = 40, f(7) = 80, f(15) = 160. The even branch b + f(a-1, b) is never reached. Printed value: 160 (as 160.0 in Python 3; the integer answer is 160).

Concept: Recursion tracing: unwind the calls to the base case and multiply back up. The trap is assuming the even branch is used, or worrying that / produces a float, which does not change the value.

While flag = 0 the loop pushes 1, 2, ..., 8, so S = [0, 1, ..., 8]. In the iteration that pushes 8, flag becomes 1, and from then on each iteration pops. Pop 8 (even): sum = 8. Pop 7 (odd), also pop 6: sum = 15. Pop 5, pop 4: sum = 20. Pop 3, pop 2: sum = 23. Pop 1, pop 0: sum = 24. S is empty, the loop ends. Answer: 24.

Concept: Stack trace with a conditional extra pop: odd values are added and the even value below them is discarded. The trap is adding every popped value (giving 36).

Each vertex i has out-edges to i + 2 and i + 1, visited in that order. DFS from 0 always takes the +2 edge first: 0, 2, 4, ..., 100 are discovered in that order. 100 has no out-edges. Backtracking: 98's next neighbour 99 is new, discovered (its neighbour 100 is done). Then 96 discovers 97, and so on down to 0 discovering 1. Discovery order: 0, 2, ..., 100, then 99, 97, ..., 1. After 50: the evens 52, ..., 100 (25 vertices) and all 50 odd vertices. Total = 25 + 50 = 75.

Concept: Adjacency order decides the DFS path: the larger neighbour first means a dive along even vertices, and odd vertices are found only during backtracking. The trap is assuming 51 is found right after 50.