GATE Data Science & Artificial Intelligence Mock Test - 1 - GATE Architecture and Planning MCQ

The square root can be rewritten as:

Thus, the limit becomes:

Step 2: Factor x² out of the square root.
Factor x² inside the square root:

The limit becomes:

Step 3: Expandusing a binomial approximation.
For large x, use the approximation:

Thus:

Substitute this back:

Simplify:

Step 4: Subtract a: and take the limit.
The expression becomes:

Simplify:

As a: → ∞ n, the term 

Let A be the event that the sum obtained is a multiple of 4.
B be the event that the score of 4 has appeared at least once.
A = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}
B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}
(A ∩ B) = (4, 4)
n(A ∩ B) = 1
Required probability = P(B/A)
= P(A ∩ B)/P(A)
= 1/9
Hence option C is the answer.

First, we need to find all the candidate keys of relation R.
As, in the given FDs, all the attributes are present in RHS except X. Therefore X must be present in all candidate keys possible.
X+ = X
XY+ = XYV
XZ+ = XZWYZ (CK)
XW+ = XWVZY (CK)
XV+ = XV
So, Candidate Keys = {XZ, XW}
Prime attributes = {X, Z, W}
For a given relation to be in 2NF, there must not be any partial dependencies.
Partial dependency exists if a non–key attribute in R partially dependent on any key of R.
In the given functional dependencies, W → Y is partial dependency. Therefore, R is not in 2NF.
For a relation R to be in 3NF; first it must be in 2NF.
As, R is not in 2NF, it is not in 3NF.

Total number of families: Assume there are 100 families for simplicity.

• 50 families have one child each.
• 50 families have two children each.

Total number of children:

• From the 50 families with one child, we have 50 x 1 50 children.
• From the 50 families with two children, we have 50 X 2 100 children.

Therefore, the total number of children is 50 + 100 150.
Children with siblings: Only children from the families with two children have siblings. Since
there are 50 families with two children, there are 50 X 2 100 children with siblings.
Probability calculation: The probability that a randomly selected child has a sibling is the
number of children with siblings divided by the total number of children:

For x = 2: (5 × 2 + 2) mod 7 = (10 + 2) mod 7 = 12 mod 7 = 5
So, the number 2 will be inserted at index 5.

For x = 4: (5 × 4 + 2) mod 7 = (20 + 2) mod 7 = 22 mod 7 = 1
So, the number 4 will be inserted at index 1.

For x = 9: (5 × 9 + 2) mod 7 = (45 + 2) mod 7 = 47 mod 7 = 5
Since index 5 is already occupied (by 2), we apply closed hashing and move to the next available index, which is index 6. So, the number 9 will be inserted at index 6.

For x = 12: (5 × 12 + 2) mod 7 = (60 + 2) mod 7 = 62 mod 7 = 6
Since index 6 is already occupied (by 9), we move to the next available index, which is index 0. So, the number 12 will be inserted at index 0.

Now, we can write the final contents of the hash table:

Index 0: 12
Index 1: 4
Index 5: 2
Index 6: 9
So, the contents of the hash table are:
12, 4, -, -, -, 2, 9

A max heap must satisfy the heap property, where every parent node is greater than or equal to its children.
Option A: Incorrect. The root 2 is smaller than its children (4, 6, and 7).
Option B: Incorrect. The root 12 is valid, but 10 (child of 12) has a child 6 that violates the heap property.
Option C: Incorrect. The root 12 is valid, but 6 (child of 12) has a child 7 that violates the heap property.
Option D: Correct. The root 12 is greater than its children 7, 10, and 6. Similarly, all other nodes satisfy the max heap property.

SVM or Support Vector Machine is a linear model for classification and regression problems. It can solve linear and non-linear problems and work well for many practical problems. The algorithm creates a line or a hyperplane which separates the data into classes.
A hyperplane in an n-dimensional Euclidean space is a flat, n-1 dimensional subset of that space that divides the space into two disconnected parts.

- True, BFS explores all nodes at the present depth before moving on to the nodes at the next depth level. Since each step cost is the same, the first solution BFS finds is guaranteed to be the shortest path in terms of the number of steps, which translates to the lowest cost due to the uniform step cost.
- False, DFS is not guaranteed to find the optimal path in this scenario. It goes as deep as possible down one branch before backtracking. This approach can lead to finding a longer path first if the goal state is not on the deepest branch that DFS explores initially.

- True. UCS expands the node with the lowest path cost first. Since all step costs are a constant k, this approach will effectively behave like BFS, expanding nodes in a level-by-level manner. It will find the shortest path, which is also the least costly due to the uniform step cost.

- True, IDS combines the advantages of DFS's space efficiency and BFS's completeness and optimality for uniform step costs. It iteratively deepens the depth at which it searches, ensuring that it eventually explores all possible paths in a manner similar to BFS. Since all step costs are equal, IDS will find the optimal path, as it will find the shallowest solution, which is also the least costly.

L is the number of leaf nodes.
I is the number of internal nodes (nodes that are not leaves).
Additionally, we can use the formula that counts the total number of children in the tree:
21 - D₁ + 2D₂
Where:
D₁ is the number of internal nodes with one child.
D₂ is the number of internal nodes with two children.
From the question:
D₁ = 7 (internal nodes with one child).
D₂ = 8 (internal nodes with two children).
Step 1: Find the total number of internal nodes
The total number of internal nodes, I, is the sum of the nodes with one child and two children:
I = D₁ + D₂ = 7 + 8 = 15
Step 2: Use the formula to find the number of leaf nodes
Now, using the relationship between leaf nodes and internal nodes:
L -1+1- 15+1 =16

Step 1: Derivative of the loss function with respect to y
First, we calculate the derivative of the loss function with respect to y:

Since I, we substitute it into the equation:
Step 2: Substituting the value of y 0.8
Now, substitute the observed output y 0.8 into the derivative:

Step 3: The negative sign
The derivative is often considered negative to reflect the direction of the gradient descent process,
where we want to move towards minimizing the loss. Therefore, we have: