GATE DA 2025

An exponential random variable with mean \(\frac{1}{\lambda}\) has rate \(\lambda\) and tail probability \(P(X \geq x) = e^{-\lambda x}\) for \(x \geq 0\). So

\[e^{-2\lambda} = 0.25 = \frac{1}{4} \;\Rightarrow\; 2\lambda = \ln 4 = 2\ln 2 \;\Rightarrow\; \lambda = \ln 2.\]

Check: \(e^{-2\ln 2} = 2^{-2} = 0.25\). Option B is the value of \(2\lambda\), not \(\lambda\), and option D is negative (\(\ln 0.25 = -\ln 4\)), which is impossible for a rate.

Answer: A

Here \(y_{n} = +1\), so the update adds the sample: \(w_{new} = w_{old} + x_{n}\) and \(b_{new} = b_{old} + 1\). Evaluate the new score on the same point:

\[f(x_{n}; w_{new}, b_{new}) = (w_{old} + x_{n})^{\top}x_{n} + b_{old} + 1 = f(x_{n}; w_{old}, b_{old}) + \lVert x_{n} \rVert^{2} + 1.\]

Since \(\lVert x_{n} \rVert^{2} + 1 \geq 1\), the score strictly increases, moving towards the correct positive side, so B and C are false. Option D fails because \(y_{n}f(x_{n}; w_{old}, b_{old}) = f(x_{n}; w_{old}, b_{old}) < 0\).

Answer: A

A.extend(B) adds each element of B to the end of A in place, so A becomes [1, 2, 3, 4, 5, 6].

• A.append(B) adds the whole list as a single element, giving [1, 2, 3, [4, 5, 6]].
• Lists have no update method (dictionaries and sets do), so option C raises AttributeError.
• insert needs two arguments, an index and a value, so A.insert(B) raises TypeError.

Answer: A

Options A, B and C use only the values and derivatives of \(f\) and \(g\) at \(c\):

• sums, differences and products of functions differentiable at \(c\) are differentiable at \(c\);
• the quotient rule for \(\frac{f}{g}\) also needs \(g(c) \neq 0\), which holds because \(g\) takes values in \((1, \infty)\).

Option D is different: \(f \circ g\) needs \(f\) to be differentiable at \(g(c)\), and \(g \circ f\) needs \(g\) to be differentiable at \(f(c)\). Neither is given. For example, take \(c = 0\), \(f(x) = |x - 2|\) and \(g(x) = 1 + e^{x}\) (both differentiable at 0, and \(g > 1\)). Then \(f(g(x)) = |e^{x} - 1|\) has a corner at 0 while \(g(f(x))\) is smooth there, so the sum is not differentiable at 0.

Answer: A, B, C

• A and B: orthonormal bases are not unique. Rotating the standard basis gives another one; in \(\mathbb{R}^{2}\), \(\frac{1}{\sqrt{2}}(1, 1)\) and \(\frac{1}{\sqrt{2}}(1, -1)\) also form an orthonormal basis. Even changing the sign of one vector gives a new basis. So A is false and B is true.
• C is false: \((1, 0)\) and \((1, 1)\) are linearly independent, but they are not orthogonal and \((1, 1)\) is not a unit vector.
• D is true: if \(\sum_{i} c_{i}v_{i} = 0\) for orthonormal \(v_{i}\), taking the dot product with \(v_{j}\) leaves \(c_{j} = 0\), so every coefficient is zero.

Answer: B, D

• Variable elimination is an exact method. It multiplies the relevant conditional probability tables and sums out the hidden variables one at a time, so A is false.
• What it returns is a conditional probability such as \(P(Q \mid e)\) for a query \(Q\) given evidence \(e\), so C is true.
• Gibbs sampling and rejection sampling both estimate probabilities from random samples. Their answers are approximate and improve as more samples are drawn, so B is false and D is true.

Answer: C, D

Binary search compares the target with the middle element and discards half of the remaining range. It needs two things to be correct and \(O(\log n)\):

• Sorted data. Discarding a half is justified only when the elements are in order. On unsorted input the search can throw away the half that holds the target, so binary search does not work for A or B.
• Constant-time access to the middle. A linked list must be walked node by node to reach its middle, costing \(\frac{n}{2} + \frac{n}{4} + \cdots = O(n)\) in total, so D is not \(O(\log n)\) even though the list is sorted.

Only the sorted array (C) meets both conditions.

Answer: C

Find the eigenvalues of \(A\) using \(x^{\top}x = 1\):

• \(Ax = x + x(x^{\top}x) = 2x\), so 2 is an eigenvalue.
• Any \(v\) orthogonal to \(x\) has \(x^{\top}v = 0\), so \(Av = v\). These vectors fill an \((n - 1)\)-dimensional subspace, so 1 is an eigenvalue of multiplicity \(n - 1\).

No eigenvalue is 0, so \(A\) is invertible with rank \(n\): A and B are true and C is false. The eigenvalues of \(A^{-1}\) are the reciprocals \(\frac{1}{2}\) and 1, all positive, so D is false.

Answer: A, B

Each swap in insertion sort exchanges two adjacent elements that are out of order, removing exactly one inversion. So the number of swaps equals the number of inversions.

Without \(x\), the only inversion is the pair (15, 13), so \(x\) must add exactly one more:

• \(x = 10\): only 11, placed before it, is larger. Total 2.
• \(x = 12\): no element is out of order with it. Total 1.
• \(x = 14\): only 13, placed after it, is smaller. Total 2.
• \(x = 16\): both 15 and 13, placed after it, are smaller. Total 3.

Answer: A, C

In hierarchical clustering, the linkage rule sets the distance between two clusters from the distances between their members.

• Single linkage uses the closest pair, one object from each cluster. That is the minimum of \(D(x, y)\), which is DIS-2.
• Complete linkage uses the farthest pair. That is the maximum of \(D(x, y)\), which is DIS-1.

Options A and C swap the two rules.

Answer: B, D