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IIT M DIPLOMA FN EXAM QDD1 13 Apr 2025 - 13 Apr 2025

MLF End Term14 questions40 marks8 MSQ · 5 SA · 1 MCQDiploma / forenoon / QDD1
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Q1MCQ3 marks
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Q2MSQ3 marksselect all that apply
Which of the following is true about the singular value decomposition (SVD)?
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In the context of Principal Component Analysis (PCA), which of the following statements is/are correct?
  1. PCA projects the original data onto a lower-dimensional subspace thatcaptures the maximum variance.
  2. The eigenvectors of the covariance matrix correspond to the directions ofmaximum variance.
  3. The principal components form a basis that is not necessarily orthogonal.
  4. The principal components obtained through PCA are always uncorrelated.
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Which of the following statements about convex sets are true?
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Which of the following options is/are true?
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Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
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Which of the following statements correctly represents the Karush-Kuhn-Tucker (KKT) conditions for the given optimization problem?
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Which of the following statements are correct based on the KKT conditions and the solution to the optimization problem?
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Q14SA3 marks
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Answer key

  1. Q1 (MCQ, 3 marks): A
  2. Q2 (MSQ, 3 marks): A, B, C
  3. Q3 (MSQ, 3 marks): A, B, D
  4. Q4 (MSQ, 3 marks): C, D
  5. Q5 (MSQ, 3 marks): A, B, C
  6. Q6 (MSQ, 3 marks): A, C
  7. Q7 (MSQ, 4 marks): A, B
  8. Q8 (SA, 3 marks): 4
  9. Q9 (SA, 3 marks): 0.5
  10. Q10 (SA, 2 marks): 0.5
  11. Q11 (SA, 2 marks): 1
  12. Q12 (MSQ, 2 marks): A, C
  13. Q13 (MSQ, 3 marks): A, C
  14. Q14 (SA, 3 marks): 0.25

Concepts this paper tests

Auto-generated summary from the source site. Handy as a checklist; the lessons above explain each idea from scratch.

Here’s a structured summary of core topics, concepts, principles, and equations essential for the exam, followed by Mermaid knowledge graphs to visualize relationships.


1. Core Topics & Key Concepts

A. Probability & Statistics

1. Joint Probability Density Functions (PDFs)

fU,V(u,v)=fX,Y(x(u,v),y(u,v))⋅∣J∣f_{U,V}(u,v) = f_{X,Y}(x(u,v), y(u,v)) \cdot |J|

where JJ is the Jacobian matrix of the inverse transformation.

J=∣∂x∂u∂x∂v∂y∂u∂y∂v∣J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}

2. Expected Value & Linearity

E[∑i=1nXi]=∑i=1nE[Xi]E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n E[X_i]
E[X1+X2+X3]=0.5+3+0.5=4E[X_1 + X_2 + X_3] = 0.5 + 3 + 0.5 = 4

3. Markov’s Inequality

P(X≥a)≤μaP(X \geq a) \leq \frac{\mu}{a}

4. Cumulative Distribution Function (CDF)

FX(x)=x−ab−aF_X(x) = \frac{x - a}{b - a}

B. Linear Algebra & Matrix Decompositions

1. Singular Value Decomposition (SVD)

A=UΣVTA = U \Sigma V^T

where:

2. Principal Component Analysis (PCA)


C. Optimization

1. Convexity

f(θx+(1−θ)y)≤θf(x)+(1−θ)f(y)f(\theta x + (1-\theta)y) \leq \theta f(x) + (1-\theta) f(y)
θx+(1−θ)y∈S\theta x + (1-\theta)y \in S

2. Gradient Descent

xk+1=xk−η∇f(xk)x_{k+1} = x_k - \eta \nabla f(x_k)

where η\eta is the learning rate.

∇f=(2x−4,6y+6)\nabla f = (2x -4, 6y +6)

Starting at (2,−1)(2, -1) with η=0.1\eta = 0.1:

x1=2−0.1(0)=2,y1=−1−0.1(0)=−1x_1 = 2 - 0.1(0) = 2, \quad y_1 = -1 - 0.1(0) = -1

3. Karush-Kuhn-Tucker (KKT) Conditions

min⁡f(x)subject togi(x)≤0,hj(x)=0\min f(x) \quad \text{subject to} \quad g_i(x) \leq 0, h_j(x) = 0

The KKT conditions are:

  1. Stationarity: ∇f(x)+∑λi∇gi(x)+∑μj∇hj(x)=0\nabla f(x) + \sum \lambda_i \nabla g_i(x) + \sum \mu_j \nabla h_j(x) = 0.
  2. Primal Feasibility: gi(x)≤0g_i(x) \leq 0, hj(x)=0h_j(x) = 0.
  3. Dual Feasibility: λi≥0\lambda_i \geq 0.
  4. Complementary Slackness: λigi(x)=0\lambda_i g_i(x) = 0.

    • Example: For the problem:
min⁡x2+y2+z2+w2+1s.t.x+y+z+w=1,w≤δ\min x^2 + y^2 + z^2 + w^2 + 1 \quad \text{s.t.} \quad x + y + z + w = 1, w \leq \delta

The Lagrangian is:

L=x2+y2+z2+w2+1+α(1−x−y−z−w)+β(w−δ)L = x^2 + y^2 + z^2 + w^2 + 1 + \alpha(1 - x - y - z - w) + \beta(w - \delta)

KKT conditions include:


D. Function Analysis

1. Critical Points & Extrema

f′(x)=4x3−8x=0  ⟹  x=0,±2f'(x) = 4x^3 - 8x = 0 \implies x = 0, \pm \sqrt{2}
f′′(x)=12x2−8f''(x) = 12x^2 - 8

Graph 2: Linear Algebra & Matrix Decompositions

Graph 3: Optimization

Graph 4: Function Analysis


3. Key Equations Summary

TopicEquation
Joint PDF TransformationfU,V(u,v)=fX,Y(x(u,v),y(u,v))⋅∣J∣f_{U,V}(u,v) = f_{X,Y}(x(u,v), y(u,v)) \cdot |J|
Expected ValueE[∑Xi]=∑E[Xi]E[\sum X_i] = \sum E[X_i]
Markov’s InequalityP(X≥a)≤E[X]aP(X \geq a) \leq \frac{E[X]}{a}
Uniform CDFFX(x)=x−ab−aF_X(x) = \frac{x - a}{b - a}
SVDA=UΣVTA = U \Sigma V^T
PCAΣvi=λivi\Sigma v_i = \lambda_i v_i (Eigenvalue problem)
Convex Functionf(θx+(1−θ)y)≤θf(x)+(1−θ)f(y)f(\theta x + (1-\theta)y) \leq \theta f(x) + (1-\theta) f(y)
Gradient Descentxk+1=xk−η∇f(xk)x_{k+1} = x_k - \eta \nabla f(x_k)
KKT Conditions∇L=0\nabla L = 0, λigi(x)=0\lambda_i g_i(x) = 0, λi≥0\lambda_i \geq 0
Critical Pointsf′(x)=0f'(x) = 0, f′′(x)f''(x) determines concavity/convexity