GATE DA Machine Learning Previous Year Questions

GATE DA 2025 | Question 22

MCQ · 1 mark · −0.33

Consider designing a linear classifier

\[y = \operatorname{sign}(f(x; w, b)), \quad f(x; w, b) = w^{\top}x + b\]

on a dataset \(D = \{(x_{1}, y_{1}), (x_{2}, y_{2}), \ldots, (x_{N}, y_{N})\}\), \(x_{i} \in \mathbb{R}^{d}, y_{i} \in \{+1, -1\}, i = 1, 2, \ldots, N\). Recall that the sign function outputs +1 if the argument is positive, and \(-1\) if the argument is non-positive. The parameters \(w\) and \(b\) are updated as per the following training algorithm:

\[w_{new} = w_{old} + y_{n}x_{n}, \quad b_{new} = b_{old} + y_{n}\]

whenever \(\operatorname{sign}(f(x_{n}; w_{old}, b_{old})) \neq y_{n}\). In other words, whenever the classifier wrongly predicts a sample \((x_{n}, y_{n})\) from the dataset, \(w_{old}\) gets updated to \(w_{new}\), and likewise \(b_{old}\) gets updated to \(b_{new}\). Consider the case \((x_{n}, +1)\), \(f(x_{n}; w_{old}, b_{old}) < 0\). Then

\(f(x_{n}; w_{new}, b_{new}) > f(x_{n}; w_{old}, b_{old})\) \(f(x_{n}; w_{new}, b_{new}) < f(x_{n}; w_{old}, b_{old})\) \(f(x_{n}; w_{new}, b_{new}) = f(x_{n}; w_{old}, b_{old})\) \(y_{n}f(x_{n}; w_{old}, b_{old}) > 1\)

Explanation

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

GATE DA 2025 | Question 30

MSQ · 1 mark

Let \(C_{1}\) and \(C_{2}\) be two sets of objects. Let \(D(x, y)\) be a measure of dissimilarity between two objects \(x\) and \(y\). Consider the following definitions of dissimilarity between \(C_{1}\) and \(C_{2}\).

\[\begin{aligned} \textbf{DIS-1}(C_{1}, C_{2}) &= \max_{x \in C_{1},\, y \in C_{2}} D(x, y) \\ \textbf{DIS-2}(C_{1}, C_{2}) &= \min_{x \in C_{1},\, y \in C_{2}} D(x, y) \end{aligned}\]

Which of the following statements is/are correct?

Single Linkage Clustering uses DIS-1 Single Linkage Clustering uses DIS-2 Complete Linkage Clustering uses DIS-2 Complete Linkage Clustering uses DIS-1

Explanation

Answer: (B), (D)

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

GATE DA 2025 | Question 34

NAT · 1 mark

Given data \(\{(-1, 1), (2, -5), (3, 5)\}\) of the form \((x, y)\), we fit a model \(y = wx\) using linear least-squares regression. The optimal value of \(w\) is __________
(Round off to three decimal places)

Your answer

Explanation

Answer: 0.29

The model has no intercept, so least squares picks \(w\) to minimise \(L(w) = \sum_i (y_i - w x_i)^{2}\). Setting \(L'(w) = -2\sum_i x_i (y_i - w x_i) = 0\) gives

\[w = \frac{\sum_i x_i y_i}{\sum_i x_i^{2}}.\]

Here \(\sum x_i y_i = (-1)(1) + (2)(-5) + (3)(5) = 4\) and \(\sum x_i^{2} = 1 + 4 + 9 = 14\), so \(w = \frac{4}{14} = \frac{2}{7} \approx 0.286\).

Fitting a line with an intercept would give a different slope (about 0.308), but this model is forced through the origin.

Answer: 0.286

GATE DA 2025 | Question 35

NAT · 1 mark

The naive Bayes classifier is used to solve a two-class classification problem with class-labels \(y_1, y_2\). Suppose the prior probabilities are \(P(y_1) = \frac{1}{3}\) and \(P(y_2) = \frac{2}{3}\). Assuming a discrete feature space with

\[P(x|y_1) = \frac{3}{4} \quad \text{and} \quad P(x|y_2) = \frac{1}{4}\]

for a specific feature vector \(x\). The probability of misclassifying \(x\) is __________
(Round off to two decimal places)

Your answer

Explanation

Answer: 0.39 to 0.41

Naive Bayes compares prior times likelihood for the two classes:

\(P(y_1)P(x|y_1) = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\)
\(P(y_2)P(x|y_2) = \frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}\)

So \(P(x) = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}\), giving posteriors \(P(y_1|x) = \frac{3}{5}\) and \(P(y_2|x) = \frac{2}{5}\). The classifier picks the larger posterior, \(y_1\), and it is wrong exactly when \(x\) really belongs to \(y_2\), which has probability \(P(y_2|x) = 0.40\). The larger prior of \(y_2\) is outweighed by the much higher likelihood under \(y_1\).

Answer: 0.40

GATE DA 2025 | Question 42

MCQ · 2 marks · −0.66

Consider the neural network shown in the figure with
inputs: \(u, v\)
weights: \(a, b, c, d, e, f\)
output: \(y\)
\(R\) denotes the ReLU function, \(R(x) = \max(0, x)\).

Network diagram: input u connects to the upper ReLU node R with weight a and to the lower ReLU node R with weight c, and input v connects to the upper node with weight b and to the lower node with weight d. The upper and lower nodes feed an output ReLU node R with weights e and f, and that node outputs y.

Given \(u = 2, v = 3\),
\(a = 1, b = 1, c = 1, d = -1, e = 4, f = -1\),
which one of the following is correct?

\(\frac{\partial y}{\partial a} = 8, \frac{\partial y}{\partial f} = 0\) \(\frac{\partial y}{\partial a} = 1, \frac{\partial y}{\partial f} = 0\) \(\frac{\partial y}{\partial a} = 1, \frac{\partial y}{\partial f} = -1\) \(\frac{\partial y}{\partial a} = 2, \frac{\partial y}{\partial f} = -1\)

Explanation

Answer: (A)

Forward pass, with \(h_1\) and \(h_2\) the outputs of the upper and lower hidden nodes:

\(h_1 = R(au + bv) = R(2 + 3) = 5\)
\(h_2 = R(cu + dv) = R(2 - 3) = R(-1) = 0\)
\(y = R(eh_1 + fh_2) = R(20 + 0) = 20\)

The output node and the upper node have positive inputs, so their ReLU slopes are 1. By the chain rule,

\[\frac{\partial y}{\partial a} = 1 \cdot e \cdot 1 \cdot u = 4 \times 2 = 8, \qquad \frac{\partial y}{\partial f} = 1 \cdot h_2 = 0.\]

The gradient for \(f\) is 0 because \(f\) only multiplies \(h_2\), and the lower ReLU outputs 0 here.

Answer: A

GATE DA 2025 | Question 53

MSQ · 2 marks

Consider designing a linear binary classifier \(f(x) = \operatorname{sign}(w^{\top}x + b)\), \(x \in \mathbb{R}^{2}\) on the following training data:

\[ \text{Class-1: } \left\{ \begin{pmatrix} 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \end{pmatrix} \right\}, \ \text{Class-2: } \left\{ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right\} \]

Hard-margin support vector machine (SVM) formulation is solved to obtain \(w\) and \(b\). Which of the following options is/are correct?

\(w = \begin{pmatrix} 4 \\ 4 \end{pmatrix}\) and \(b = 1\) The number of support vectors is 3 The margin is \(\sqrt{2}\) Training accuracy is 98%

Explanation

Answer: (B), (C)

The triangle formed by the Class-1 points comes closest to the origin at \((1, 1)\), the midpoint of its edge from \((2, 0)\) to \((0, 2)\). The maximum-margin boundary is the perpendicular bisector of the segment from \((0, 0)\) to \((1, 1)\), the line \(x_{1} + x_{2} = 1\). In canonical form (Class-1 as \(+1\)), \(w = (1, 1)^{\top}\) and \(b = -1\): \(w^{\top}x + b\) is 1 at \((2, 0)\) and \((0, 2)\), \(-1\) at \((0, 0)\) and 3 at \((2, 2)\).

A fails: \(w = (4, 4)^{\top}\), \(b = 1\) gives every point, \((0, 0)\) included, a positive score, so it does not separate the classes.
B holds: \((2, 0)\), \((0, 2)\) and \((0, 0)\) lie on the margin lines; \((2, 2)\) lies beyond them.
C holds: the margin lines \(x_{1} + x_{2} = 0\) and \(x_{1} + x_{2} = 2\) are \(\frac{2}{\|w\|} = \sqrt{2}\) apart.
D fails: the data are separable, so training accuracy is 100%.

Answer: B, C

GATE DA 2025 | Question 55

MSQ · 2 marks

Consider a two-class problem in \(\mathbb{R}^{d}\) with class labels red and green. Let \(\mu_{\textit{red}}\) and \(\mu_{\textit{green}}\) be the means of the two classes. Given test sample \(x \in \mathbb{R}^{d}\), a classifier calculates the squared Euclidean distance (denoted by \(\| \cdot \|^{2}\)) between \(x\) and the means of the two classes and assigns the class label that the sample \(x\) is closest to. That is, the classifier computes

\[ f(x) = \|\mu_{\textit{red}} - x\|^{2} - \|\mu_{\textit{green}} - x\|^{2} \]

and assigns the label red to \(x\) if \(f(x) < 0\), and green otherwise. Which of the following statements is/are correct?

The sample \(x = 0\) is assigned the label green if \(\|\mu_{\textit{red}}\| < \|\mu_{\textit{green}}\|\) \(f\) is a linear function of \(x\) \(f(x) = w^{\top}x + b\), where \(w\) and \(b\) are functions of \(\mu_{\textit{red}}\) and \(\mu_{\textit{green}}\) \(f\) is a quadratic polynomial in \(x\)

Explanation

Answer: (B), (C)

Expand each squared distance as \(\|\mu - x\|^{2} = \|\mu\|^{2} - 2\mu^{\top}x + \|x\|^{2}\). The \(\|x\|^{2}\) terms cancel in the difference:

\[ f(x) = 2(\mu_{\textit{green}} - \mu_{\textit{red}})^{\top}x + \|\mu_{\textit{red}}\|^{2} - \|\mu_{\textit{green}}\|^{2} \]

B and C hold: \(f(x) = w^{\top}x + b\) with \(w = 2(\mu_{\textit{green}} - \mu_{\textit{red}})\) and \(b = \|\mu_{\textit{red}}\|^{2} - \|\mu_{\textit{green}}\|^{2}\), so the decision boundary is a hyperplane.
D fails: the quadratic term cancels.
A fails: \(f(0) = \|\mu_{\textit{red}}\|^{2} - \|\mu_{\textit{green}}\|^{2} < 0\) when \(\|\mu_{\textit{red}}\| < \|\mu_{\textit{green}}\|\), so the origin gets the label red, the nearer mean.

Answer: B, C

GATE DA 2025 | Question 60

NAT · 2 marks

Let \(D = \{x^{(1)}, \ldots, x^{(n)}\}\) be a dataset of \(n\) observations where each \(x^{(i)} \in \mathbb{R}^{100}\). It is given that \(\sum_{i=1}^{n} x^{(i)} = 0\). The covariance matrix computed from \(D\) has eigenvalues \(\lambda_{i} = 100^{2-i}\), \(1 \le i \le 100\). Let \(u \in \mathbb{R}^{100}\) be the direction of maximum variance with \(u^{\top}u = 1\).

The value of

\[ \frac{1}{n}\sum_{i=1}^{n} \left(u^{\top}x^{(i)}\right)^{2} = \]

________ (Answer in integer)

Your answer

Explanation

Answer: 100

Since \(\sum_{i} x^{(i)} = 0\), the data are already centred, so the covariance matrix is \(C = \frac{1}{n}\sum_{i=1}^{n} x^{(i)}x^{(i)\top}\). Then

\[ \frac{1}{n}\sum_{i=1}^{n} \left(u^{\top}x^{(i)}\right)^{2} = \frac{1}{n}\sum_{i=1}^{n} u^{\top}x^{(i)}x^{(i)\top}u = u^{\top}Cu, \]

the variance of the data along \(u\). Over unit vectors, \(u^{\top}Cu\) is largest for the top eigenvector of \(C\) (the first principal component), and that largest value is the top eigenvalue.

The eigenvalues \(100^{2-i}\) decrease with \(i\): \(100, 1, 0.01, \ldots\). So the value is \(\lambda_{1} = 100^{1} = 100\).

Answer: 100

GATE DA 2024 | Question 17

MCQ · 1 mark · −0.33

Consider the dataset with six datapoints: \(\{(\boldsymbol{x}_{1}, \boldsymbol{y}_{1}), (\boldsymbol{x}_{2}, \boldsymbol{y}_{2}), \ldots, (\boldsymbol{x}_{6}, \boldsymbol{y}_{6})\}\), where \(\boldsymbol{x}_{1} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\), \(\boldsymbol{x}_{2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\), \(\boldsymbol{x}_{3} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}\), \(\boldsymbol{x}_{4} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}\), \(\boldsymbol{x}_{5} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}\), \(\boldsymbol{x}_{6} = \begin{bmatrix} -2 \\ -2 \end{bmatrix}\) and the labels are given by \(\boldsymbol{y}_{1} = \boldsymbol{y}_{2} = \boldsymbol{y}_{5} = 1\), and \(\boldsymbol{y}_{3} = \boldsymbol{y}_{4} = \boldsymbol{y}_{6} = -1\). A hard margin linear support vector machine is trained on the above dataset.

Which ONE of the following sets is a possible set of support vectors?

\(\{\boldsymbol{x}_{1}, \boldsymbol{x}_{2}, \boldsymbol{x}_{5}\}\) \(\{\boldsymbol{x}_{3}, \boldsymbol{x}_{4}, \boldsymbol{x}_{5}\}\) \(\{\boldsymbol{x}_{4}, \boldsymbol{x}_{5}\}\) \(\{\boldsymbol{x}_{1}, \boldsymbol{x}_{2}, \boldsymbol{x}_{3}, \boldsymbol{x}_{4}\}\)

Explanation

Answer: (D)

Each negative point is the mirror image of a positive one through the origin. Write the separator as \(\boldsymbol{w}^{\top}\boldsymbol{x} + b\) with \(\boldsymbol{w} = (w_{1}, w_{2})\). The hard-margin constraints \(y_{i}(\boldsymbol{w}^{\top}\boldsymbol{x}_{i} + b) \geq 1\) for \(\boldsymbol{x}_{1}\) and \(\boldsymbol{x}_{4}\) read \(w_{1} + b \geq 1\) and \(w_{1} - b \geq 1\), so \(w_{1} \geq 1\); \(\boldsymbol{x}_{2}\) and \(\boldsymbol{x}_{3}\) give \(w_{2} \geq 1\) in the same way. So \(\lVert\boldsymbol{w}\rVert^{2} \geq 2\), and \(\boldsymbol{w} = (1, 1)\), \(b = 0\) reaches this while meeting every constraint, so it is the maximum-margin solution.

For this hyperplane, \(\boldsymbol{x}_{1}\) to \(\boldsymbol{x}_{4}\) give \(y_{i}(\boldsymbol{w}^{\top}\boldsymbol{x}_{i} + b) = 1\): they lie on the margin and are the support vectors. \(\boldsymbol{x}_{5}\) and \(\boldsymbol{x}_{6}\) give 4, well outside the margin, which rules out A, B and C (A even holds points of one class only).

Answer: D

GATE DA 2024 | Question 18

MCQ · 1 mark · −0.33

Match the items in Column 1 with the items in Column 2 in the following table:

Column 1	Column 2
(p) Principal Component Analysis	(i) Discriminative Model
(q) Naïve Bayes Classification	(ii) Dimensionality Reduction
(r) Logistic Regression	(iii) Generative Model

(p) − (iii), (q) − (i), (r) − (ii) (p) − (ii), (q) − (i), (r) − (iii) (p) − (ii), (q) − (iii), (r) − (i) (p) − (iii), (q) − (ii), (r) − (i)

Explanation

Answer: (C)

Match each method by what it learns.

(p) Principal Component Analysis projects the data onto the few directions of largest variance, keeping most of the information in fewer features: dimensionality reduction, (ii).
(q) Naïve Bayes models how the data are generated, through \(P(\text{class})\) and \(P(\text{features} \mid \text{class})\), and gets the class from Bayes' theorem: a generative model, (iii).
(r) Logistic regression models \(P(\text{class} \mid \text{features})\) directly and never models the features themselves: a discriminative model, (i).

Options A and D call PCA a generative model, and B swaps the two classifiers.

Answer: C

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