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IIT M FOUNDATION DIPLOMA ENDTERM QPB2 07 Aug 2022 IBA NS - 07 Aug 2022

MLF End Term22 questions50 marks12 SA · 8 MCQ · 2 MSQFoundation / Diploma / QPB2

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Q1MCQ0 marks

Q2SA2 marks

Suppose that eigenvalues of matrix A are 1, 0.5, 0.25. The determinant of (A−1)T is NOTE: Enter your answer to the nearest integer.

Your numeric answer:

Q3SA2 marks

NOTE: Enter your answer to the nearest integer. question figure

Your numeric answer:

Comprehension passage - the questions below it refer to this

Based on the above data, answer the given subquestions. question figure

Q4SA1 mark

Compute the equation of the function’s linear approximation at point (2,3). If the equation of the linear approximation is L = ax + by + c, then provide the value of a. NOTE: Enter your answer to the nearest integer.

Your numeric answer:

Q5SA1 mark

Compute the equation of the function’s linear approximation at point (2,3). If the equation of the linear approximation is L = ax + by + c, then provide the value of b. NOTE: Enter your answer upto two decimal.

Your numeric answer:

Q6SA1 mark

Compute the equation of the function’s linear approximation at point (2,3). If the equation of the linear approximation is L = ax + by + c, then provide the value of c. NOTE: Enter your answer upto two decimal.

Your numeric answer:

Comprehension passage - the questions below it refer to this

Based on the above data, answer the given subquestions. question figure

Q7SA1 mark

Calculate the value of a NOTE: Enter your answer upto two decimal.

Your numeric answer:

Q8SA2 marks

NOTE: Enter your answer upto two decimal. question figure

Your numeric answer:

Q9MCQ3 marks

If two complex vectors v1 and v2 have lengths l1 and l2 and v1.v2 = l1l2, then which of the following statements is/are true?

v1 and v2 vectors are parallel to each other.
v1 and v2 vectors are conjugate to each other.
v1 and v2 vectors are perpendicular to each other.
v1 and v2 can be any two vectors.

Q10MCQ3 marks

Q11MCQ3 marks

Q12MCQ2 marks

positive definite
positive semi-definite
negative definite
negative semi-definite
None of these

Q13MSQ3 marksselect all that apply

Comprehension passage - the questions below it refer to this

A manufacturing company makes two types of television sets; one is smart led and the other is normal led. The company has resources to make at most 300 sets a week. It takes Rs 1800 to make a smart led set and Rs 2700 to make a normal led set. The company can spend not more than Rs 648000 a week to make television sets. If it makes a profit of Rs 510 per smart led and Rs 675 per normal led. Based on the above data, answer the given subquestions.

Q14MCQ2 marks

The objective function to maximise the profit for the given problem is

510x + 675y
410x + 675y
510x + 575y
102x + 135y

Q15MSQ3 marksselect all that apply

The constraints for the given problem is

x + y ≤ 300
2x + 3y ≤ 720
x + 3y ≤ 120
x ≥ 0, y ≥ 0

Q16SA3 marks

NOTE: Enter your answer to the nearest integer. question figure

Your numeric answer:

Q17SA3 marks

NOTE: Enter your answer in one decimal place. question figure

Your numeric answer:

Q18SA3 marks

NOTE: Enter your answer to the nearest integer. question figure

Your numeric answer:

Q19SA3 marks

NOTE: Enter your answer upto one decimal. question figure

Your numeric answer:

Q20SA3 marks

NOTE: Enter your answer upto two decimal question figure

Your numeric answer:

Comprehension passage - the questions below it refer to this

Based on the above data, answer the given subquestions. question figure

Q21MCQ3 marks

Q22MCQ3 marks

Answer key

Q1 (MCQ, 0 marks): A
Q2 (SA, 2 marks): 8
Q3 (SA, 2 marks): 45
Q4 (SA, 1 marks): -2
Q5 (SA, 1 marks): -0.8 to -0.7
Q6 (SA, 1 marks): 10.0 to 10.5
Q7 (SA, 1 marks): 0.23 to 0.27
Q8 (SA, 2 marks): 0.42 to 0.48
Q9 (MCQ, 3 marks): A
Q10 (MCQ, 3 marks): D
Q11 (MCQ, 3 marks): B
Q12 (MCQ, 2 marks): A
Q13 (MSQ, 3 marks): A, B, C
Q14 (MCQ, 2 marks): A
Q15 (MSQ, 3 marks): A, B, D
Q16 (SA, 3 marks): 4
Q17 (SA, 3 marks): 3.5 to 4.0
Q18 (SA, 3 marks): 5 to 7
Q19 (SA, 3 marks): 7.1 to 7.7
Q20 (SA, 3 marks): 0.23 to 0.27
Q21 (MCQ, 3 marks): A
Q22 (MCQ, 3 marks): A

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 the core topics, concepts, principles, and equations essential for the "Diploma Level: Machine Learning Foundations" exam, along with a Mermaid knowledge graph to visualize relationships.

Core Topics & Key Concepts

1. Linear Algebra

Eigenvalues & Determinants
- If eigenvalues of matrix $A$ are $\lambda_1, \lambda_2, \lambda_3$ , then:
  - Determinant of $A$ : $\det(A) = \lambda_1 \lambda_2 \lambda_3$ .
  - Determinant of $A^{-1}$ : $\det(A^{-1}) = \frac{1}{\det(A)}$ .
  - Determinant of $(A^{-1})^T$ : Same as $\det(A^{-1})$ (since $\det(A^T) = \det(A)$ ).
- Equation:

\det((A^{-1})^T) = \frac{1}{\lambda_1 \lambda_2 \lambda_3}

Example: For eigenvalues $1, 0.5, 0.25$ , $\det((A^{-1})^T) = \frac{1}{1 \times 0.5 \times 0.25} = 8$ .

Positive Definite Matrices
- A matrix $A$ is positive definite if for all $x \neq 0$ , $x^T A x > 0$ .
- Check via eigenvalues (all $> 0$ ) or principal minors (all $> 0$ ).
- Example: The matrix $A = \begin{bmatrix} 4 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is positive definite.

Eigenvectors
- For a matrix $A$ , eigenvectors $v$ satisfy $Av = \lambda v$ .
- Example: For $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{bmatrix}$ , possible eigenvectors are column vectors like $\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ .

Hessian Matrix
- For a function $f(x, y, z)$ , the Hessian $H$ is:

H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}

Example: For $f(x, y, z) = e^{-x} \cos(yz)$ , the Hessian at $(0, 1, \pi)$ is:

H = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\pi^2 & -\pi \\ 0 & -\pi & 1 \end{bmatrix}

2. Probability & Statistics

Expectation & Linearity
- For random variables $X$ and $Y = aX + b$ :

E[Y] = aE[X] + b

Example: Given $Y = 8X + 4$ and $E[Y] = 44$ , solve for $E[X] = 5$ , then $E[7X + 10] = 45$ .

Probability Density Functions (PDF)
- For a PDF $f(x)$ , the total probability integrates to 1:

\int_{-\infty}^{\infty} f(x) \, dx = 1

Example: For $f(x) = a|x|$ where $|x| \leq 2$ , solve for $a$ :

\int_{-2}^{2} a|x| \, dx = 1 \implies a = \frac{1}{4}

Probability Calculation:

P(X \geq \frac{1}{2}) = \int_{1/2}^{2} a x \, dx + \int_{-2}^{-1/2} a (-x) \, dx \approx 0.4375

Joint PDF & Transformations
- For joint PDF $f(x, y)$ , transform to $U = g_1(X, Y)$ , $V = g_2(X, Y)$ using the Jacobian $J$ :

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}

Example: For $U = X + Y$ , $V = X$ , $|J| = 1$ .

Maximum Likelihood Estimation (MLE)
- For exponential distribution $f(x; \lambda) = \lambda e^{-\lambda x}$ , the MLE for $\lambda$ given data $x_1, \dots, x_n$ is:

\hat{\lambda} = \frac{n}{\sum_{i=1}^n x_i}

Example: For lifetimes $3, 4, 2, 5, 6$ , $\hat{\lambda} = \frac{5}{20} = 0.25$ .

3. Calculus & Optimization

Linear Approximation
- For $f(x, y)$ , the linear approximation at $(a, b)$ is:

L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

Example: For $f(x, y) = \sqrt{41 - 4x^2 - y^2}$ at $(2, 3)$ :

f_x(2, 3) = -2, \quad f_y(2, 3) \approx -0.77, \quad f(2, 3) = 3

L(x, y) = 3 - 2(x - 2) - 0.77(y - 3) \implies a = -2, b \approx -0.77, c \approx 10.31

Gradient Descent
- Update rule for weights $w$ :

w^{t+1} = w^t - \eta \nabla J(w^t)

Example: For $y = [1, 2, 3]^T$ , $X = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 1 & 2 \end{bmatrix}$ , and $w^t = [1, 2]^T$ , compute $w^{t+1}$ .

Constrained Optimization
- Use Lagrange multipliers for constraints like $x^2 + y^2 = 2$ .
- Example: Maximize $f(x, y) = e^{2xy}$ under $x^2 + y^2 = 2$ . The maximum occurs at $x = y = 1$ , yielding $e^{2} \approx 7.39$ .

4. Machine Learning

Mean Squared Error (MSE)
- For model $f(x) = x + 3$ and true $Y = X + 5$ where $X \sim N(5, 5)$ :

\text{MSE} = E[(Y - f(X))^2] = E[(X + 5 - (X + 3))^2] = E[4] = 4

Linear Regression
- Gradient descent updates for $w$ in $y = Xw$ :

w^{t+1} = w^t - \eta X^T (Xw^t - y)

5. Geometry

Distance from Point to Plane
- For plane $r \cdot n = d$ and point $p$ , the distance is:

\text{Distance} = |p \cdot n - d|

Example: For $n = \frac{6i + 4j - 2k}{\sqrt{56}}$ , $d = \frac{40}{\sqrt{56}}$ , and $p = (1, 2, 1)$ , the distance is $\approx 3.71$ .

Graph 1: Linear Algebra Concepts

Graph 2: Probability & Statistics

Graph 3: Calculus & Optimization

Graph 4: Machine Learning

Graph 5: Geometry

Key Equations Summary

Topic	Equation
Determinant of $A^{-1}$	$\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{\lambda_1 \lambda_2 \lambda_3}$
Expectation Linearity	$E[aX + b] = aE[X] + b$
Linear Approximation	$L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$
MLE for Exponential	$\hat{\lambda} = \frac{n}{\sum x_i}$
Hessian Matrix	$H_{ij} = \frac{\partial^2 f}{\partial x_i \partial x_j}$
Gradient Descent	$w^{t+1} = w^t - \eta \nabla J(w^t)$
Plane Distance	$\text{Distance} = \|p \cdot n - d\|$
MSE	$\text{MSE} = E[(Y - \hat{Y})^2]$