GATE DA 2025

\(E[X|Y]\) is itself a random variable: a function of \(Y\) that gives the average of \(X\) for each value \(Y\) can take. Averaging it over the distribution of \(Y\) recovers the overall average of \(X\). For discrete variables,

\(E[E[X|Y]] = \sum_{y} E[X|Y = y]\,P(Y = y) = \sum_{y}\sum_{x} x\,P(X = x, Y = y) = \sum_{x} x\,P(X = x) = E[X]\).

This is the law of total expectation (the tower property). Option A is still random, not a fixed number, and B and D fail in general: if \(Y\) is independent of \(X\), then \(E[X|Y] = E[X]\), so the result is \(E[X]\) whatever \(E[Y]\) is.

Answer: C

Gaussian elimination first creates zeros below the diagonal. For a full matrix this costs about \(\frac{2}{3}n^{3}\) additions and multiplications, the source of the \(O(n^{3})\) option. An upper triangular matrix already has those zeros, so this phase has nothing to do, and only back substitution remains:

\(x_{i} = \dfrac{1}{a_{ii}}\Big(b_{i} - \sum_{j \gt i} a_{ij}x_{j}\Big)\) for \(i = n, n - 1, \dots, 1\).

Row \(i\) needs \(n - i\) multiplications and \(n - i\) additions, so the total is \(\frac{n(n-1)}{2}\) of each: \(O(n^{2})\). \(O(n)\) is too small, because each of the roughly \(\frac{n^{2}}{2}\) entries above the diagonal is used once.

Answer: B

Let \(e = (1, 1, \dots, 1)^{T}\). Each row of \(A\) sums to 1, so \(Ae = e\); then \(A^{2}e = e\) and \(A^{3}e = e\) as well. Hence

\[Be = A^{3}e - 2A^{2}e + Ae = e - 2e + e = 0.\]

(Equivalently, \(B = A(A - I)^{2}\) and \((A - I)e = 0\).) So \(Bx = 0\) has the non-zero solution \(e\), and every multiple \(ce\) solves it too: infinitely many solutions.

• A fails: \(x = 0\) always solves \(Bx = 0\).
• B is impossible: the solutions form a subspace, so there is exactly one solution or infinitely many.
• D fails: \(e \ne 0\) is a second solution.

Answer: C

Differentiating \(f\) once gives \(f'(x) = \frac{e^{x} + e^{-x}}{2}\), and differentiating again brings back \(f''(x) = \frac{e^{x} - e^{-x}}{2} = f(x)\). So every even-order derivative equals \(f\) itself: \(f^{(10)}(x) = f(x)\), and \(f^{(10)}(0) = \frac{1 - 1}{2} = 0\).

Another view: \(f\) (the hyperbolic sine) is an odd function, so its Maclaurin series \(x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdots\) has only odd powers. The coefficient of \(x^{10}\), which equals \(\frac{f^{(10)}(0)}{10!}\), is therefore 0. Options C and D look like series coefficients rather than derivatives.

Answer: A

\(p \to q\) is false only when \(p\) is true and \(q\) is false, and \(\neg p \vee q\) is false in exactly the same case, so \(S_{1} \equiv S_{3}\).

Every other pair differs on at least one row of the truth table:

• \(S_{2}\) and \(S_{3}\): with \(p\) and \(q\) both true, \(S_{2}\) is false but \(S_{3}\) is true.
• \(S_{2}\) and \(S_{4}\): with both false, \(S_{2}\) is false but \(S_{4}\) is true.
• \(S_{1}\) and \(S_{4}\): with both true, \(S_{1}\) is true but \(S_{4}\) is false.

Answer: A

A functional dependency is cheap to enforce when all its attributes sit in one relation of the decomposition: each insert or update is checked against that relation alone. In a decomposition that is not dependency-preserving, some dependency \(X \to Y\) cannot be derived from the dependencies holding within the individual relations, because its attributes are spread over several of them. To check it after every change, those relations must first be joined to bring \(X\) and \(Y\) back together, so joins are executed far more often. Selection, projection and set union cannot reunite attributes stored in different relations.

Answer: C

Evaluate the expression from the innermost part outwards.

1. \(\sigma_{\texttt{maker}=\text{“}\mathit{ABC}\text{”}}\texttt{Make}\) keeps the models that ABC makes.
2. Joining with Car (on model) keeps only cars of those models, and \(\sigma_{\texttt{color}=\text{“}\mathit{red}\text{”}}\) then keeps the red ones. Both conditions hold for the same car.
3. Joining with Own (on serial) pairs each such car with its owners, and \(\pi_{\texttt{owner}}\) keeps only the owners.

The result is every owner of a red car made by ABC. Option A would need a union of separate selections, B needs a count that the expression never takes, and D describes cars although the projection outputs owners.

Answer: C

Compute the home slot \(h(x) = 3x \bmod 10\) of each key in order, moving to the next index whenever a slot is taken.

• \(h(1) = 3\), \(h(4) = 12 \bmod 10 = 2\), \(h(5) = 15 \bmod 10 = 5\) and \(h(6) = 18 \bmod 10 = 8\). All four slots are free.
• \(h(14) = 42 \bmod 10 = 2\). Slot 2 holds 4 and slot 3 holds 1, so 14 goes to slot 4.
• \(h(15) = 45 \bmod 10 = 5\). Slot 5 holds 5, so 15 goes to slot 6.

Option A lists the home slots 2 and 5, which were already occupied when 14 and 15 arrived.

Answer: D

The median \(m\) is the point where the CDF reaches one half, \(F_{X}(m) = \frac{1}{2}\). Since 3 lies in the middle piece (\(t \leq 3 \leq 4\)),

\[\frac{3 - t}{4 - t} = \frac{1}{2} \;\Rightarrow\; 6 - 2t = 4 - t \;\Rightarrow\; t = 2.\]

Check: with \(t = 2\), \(X\) is uniform on \([2, 4]\), whose midpoint is 3. The other options fail; for example, \(t = 0\) gives \(F_{X}(3) = \frac{3}{4}\), so the median would be 2 instead.

Answer: A

Use \(E[Z] = 0\) and \(E[Z^{2}] = \operatorname{Var}(Z) = 1\).

• \(E[X] = aE[Z] + b = b\), so \(b = 1\).
• \(X - E[X] = aZ\), so \(E[(X - E[X])Z] = aE[Z^{2}] = a\), which gives \(a = -2\).
• Check: \(E[(X - E[X])^{2}] = a^{2}E[Z^{2}] = 4\) holds. This condition alone cannot fix the sign of \(a\); the second one does.

Option C has the right \(a\) but the wrong mean, while B and D fail the condition \(E[(X - E[X])Z] = -2\).

Answer: A