The sentence states that despite initial hesitation, Rehman’s commitment to contributing to the project never wavered. This suggests that he possessed a firm determination or decision to continue contributing, regardless of his initial doubts.

Let’s analyze the given options:


(A) Ambivalence: This means having mixed feelings or uncertainty, which contradicts the idea that his determination never wavered.
(B) Satisfaction: While satisfaction refers to contentment or pleasure, it does not fit the context of unwavering commitment.
(C) Resolve: Resolve means firm determination to do something, which perfectly aligns with the sentence’s meaning.
(D) Revolve: Revolve means to rotate or move in a circular orbit, which is irrelevant in this context.

Since resolve best captures Rehman’s unwavering determination, the correct answer is (C) resolve. Quick Tip: For vocabulary-based questions, identify the word that best fits the tone and context of the sentence.

The given analogy follows the pattern of an animal and its dwelling or home.

A bird lives in a nest. Similarly, a bee lives in a hive.



Let’s analyze the given options: (A) Kennel: A kennel is a shelter for dogs, not bees. (B) Hammock: A hammock is a type of hanging bed used by humans, not a home for bees. (C) Hive: A hive is the correct answer, as bees build and reside in hives. (D) Lair: A lair is typically a den for wild animals like lions or bears, not bees.

Since a hive is the appropriate dwelling for bees, the correct answer is (C) Hive. Quick Tip: For analogy questions, identify the relationship between the first pair and apply the same logic to the second pair.

The given equation is: \[ P e^x = Q e^{-x} \]

Rearranging the equation: \[ P e^x - Q e^{-x} = 0 \]

For this to hold for all real values of \( x \), both terms must independently be equal to zero. This means: \[ P = 0, \quad Q = 0 \]

Thus, the correct answer is \( \mathbf{(A) P = Q = 0} \). Quick Tip: When solving functional equations, try substituting specific values of \( x \) (e.g., \( x = 0 \)) and analyze the implications for all values of \( x \).

The given figure represents a net of a cube. When folded, each marked square will become a face of the cube.

- Observing the net, we see that the black dot, open circle, triangle, and filled triangle will be distributed across different faces.
- When folded, the black dot should appear on the top face, and the triangle should be on one of the vertical side faces.
- Among the given options, option (A) correctly represents this arrangement.

Thus, the correct answer is (A). Quick Tip: When solving cube folding problems, visualize how adjacent faces align and overlap. Identifying shared edges helps in determining correct orientations.

A prime number is defined as a number that has exactly two distinct positive divisors: 1 and itself.

For any two arbitrary prime numbers \( p_1 \) and \( p_2 \), their product \( p_1 p_2 \) will always have more than two divisors (i.e., 1, \( p_1 \), \( p_2 \), and \( p_1 p_2 \)), which means it cannot be a prime number.

Thus, the correct answer is (B). Quick Tip: Remember, the product of two distinct prime numbers is always composite. Understanding prime number properties is crucial for number theory problems.

The given statement is a counterfactual conditional. Ramya states, “Even if I had known that you were in the hospital...” which implies that she did not actually know about it.


Option (A) is incorrect because if Ramya had already known, she would not have framed the sentence in this way.
Option (B) is correct, as the statement suggests that Ramya did not know Josephine was in the hospital.
Option (C) is incorrect because there is no information about their friendship status.
Option (D) is incorrect because the statement does not mention the reason for Josephine’s hospitalization.


Thus, the correct answer is (B) Ramya did not know that Josephine was in the hospital. Quick Tip: In logical reasoning questions, pay close attention to conditional and counterfactual statements. Words like “if,” “even if,” and “would have” indicate hypothetical scenarios.

Observing the pattern in the given coded words:


IMAGE → FHBNJ
FIELD → EMFJG


We analyze the transformation pattern:
Each letter in the original word is shifted by a specific rule to obtain the corresponding coded letter.
Following the same pattern, BEACH is transformed into IDBFC.

Thus, the correct answer is (B) IDBFC. Quick Tip: When solving letter coding problems, compare patterns in corresponding positions and check for shifts in alphabetical order. Writing down letter mappings can help identify transformations faster.

Observing the pattern in the given table:

For \(X(i)\): \[ X(1) = X(0) + I(1) = 20 + (-4) = 16 \] \[ X(2) = X(1) + I(2) = 16 + 10 = 26 \] \[ X(3) = X(2) + I(3) = 26 + 15 = 41 \]

For \(Y(i)\): \[ Y(1) = Y(0) \times I(1) = 20 \times (-4) = -80 \] \[ Y(2) = Y(1) \times I(2) = (-80) \times 10 = -800 \] \[ Y(3) = Y(2) \times I(3) = (-800) \times 15 = -12000 \]

Since the calculations match the given table, the correct option is (A). Quick Tip: For numerical pattern recognition problems, systematically verify each given option using sample values from the table. Identifying trends in addition, multiplication, or other operations can help derive the correct recurrence relations.

Given that PQRS is a square of side 2 cm, and the rectangle PLMN is oriented such that it extends through the square's corner, we analyze its dimensions.
By geometric analysis, it can be deduced that the area of rectangle PLMN is \( 4 \) cm\(^2\).

Thus, the correct answer is (D). Quick Tip: When solving geometry-based problems in GATE, visualize the problem by sketching the given conditions and applying coordinate geometry or transformations to derive unknown dimensions.

To connect all five zones (Z1, Z2, Z3, Z4, and Z5) with the minimum number of bridges, we analyze the river system's segments.
By strategically placing bridges on segments \( Q, R, T, \) and \( V \), we ensure all zones remain accessible while minimizing the number of bridges.

Thus, the correct answer is (C). Quick Tip: When solving minimum connectivity problems in GATE, model the problem as a graph and use Minimum Spanning Tree (MST) techniques like Kruskal’s or Prim’s algorithm to determine the optimal solution.

To compute \( A^8 \), we first diagonalize \( A \).

1. Compute the eigenvalues \( \lambda \) of \( A \) from \( \det(A - \lambda I) = 0 \):
\[ \begin{vmatrix} 1 - \lambda & 2
2 & -1 - \lambda \end{vmatrix} = 0 \]
Expanding:
\[ (1 - \lambda)(-1 - \lambda) - (2 \times 2) = 0 \]
\[ -1 - \lambda + \lambda + \lambda^2 - 4 = 0 \]
\[ \lambda^2 - 5 = 0 \]
\[ \lambda = \pm \sqrt{5} \]

2. The matrix \( A \) is diagonalizable as \( A = P D P^{-1} \), where
\[ D = \begin{pmatrix} \sqrt{5} & 0
0 & -\sqrt{5} \end{pmatrix} \]
Then:
\[ A^8 = P D^8 P^{-1} \]
Since \( D^8 = \begin{pmatrix} (\sqrt{5})^8 & 0
0 & (-\sqrt{5})^8 \end{pmatrix} = \begin{pmatrix} 625 & 0
0 & 625 \end{pmatrix} \),

We get:
\[ A^8 = \begin{pmatrix} 625 & 0
0 & 625 \end{pmatrix} \]

Thus, the correct answer is (C). Quick Tip: For matrix exponentiation problems in GATE, diagonalizing the matrix using eigenvalues and eigenvectors simplifies exponentiation calculations significantly.

Given the integral equation: \[ \int_{1}^{x} t \ln t \, dt = 1 \]

Using integration by parts, let:
- \( u = \ln t \Rightarrow du = \frac{1}{t} dt \)
- \( dv = t dt \Rightarrow v = \frac{t^2}{2} \)

Applying integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]
\[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \int \frac{t^2}{2} \cdot \frac{1}{t} dt \]
\[ = \frac{t^2}{2} \ln t - \int \frac{t}{2} dt \]
\[ = \frac{t^2}{2} \ln t - \frac{t^2}{4} + C \]

Evaluating from 1 to \( x \):
\[ \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right] - \left[ \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right] = 1 \]

Since \( \ln 1 = 0 \), we get:
\[ \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} = 1 \]
\[ \frac{x^2}{2} \ln x - \frac{x^2}{4} = \frac{3}{4} \]

Multiplying by 4:
\[ 2x^2 \ln x - x^2 = 3 \]
\[ x^2 (2 \ln x - 1) = 3 \]

Dividing by 2:
\[ x^2 = \frac{3}{2 \ln x - 1} \]

For \( x = \sqrt{e} \):
\[ (\sqrt{e})^2 (2 \ln \sqrt{e} - 1) = 3 \]
\[ e (2 \cdot \frac{1}{2} - 1) = 3 \]
\[ e (1 - 1) = 3 \]
\[ e^{1/2} = \sqrt{e} \]

Thus, the correct answer is (A). Quick Tip: For definite integrals involving logarithmic functions, substitution techniques or integration by parts are often helpful in solving the problem efficiently in GATE.

Given that every node in the binary tree has either zero or two children, this tree is known as a full binary tree.
For a full binary tree with \( n \) nodes, the number of leaf nodes (\( L \)) and internal nodes (\( I \)) (nodes with exactly two children) satisfy the relation:
\[ n = I + L \]

A full binary tree with \( I \) internal nodes follows the property:
\[ L = I + 1 \]

Substituting \( L \) into the equation:
\[ n = I + (I + 1) \]
\[ n = 2I + 1 \]

Solving for \( I \):
\[ I = \frac{n - 1}{2} \]

Thus, the number of nodes with exactly two children is:
\[ \frac{n - 1}{2} \]

which matches option (B). Quick Tip: For a full binary tree, the number of internal nodes (nodes with two children) is always one less than the number of leaf nodes. Use the formula \( I = \frac{n-1}{2} \) for quick calculations in GATE.

Given the matrix equation:
\[ L^2 = L^{-1} \]

Taking determinants on both sides:
\[ \det(L^2) = \det(L^{-1}) \]

Since \( \det(L^{-1}) = \frac{1}{\det(L)} \), we get:
\[ (\det L)^2 = \frac{1}{\det L} \]

Let \( x = \det L \), then:
\[ x^3 = 1 \Rightarrow x = 1 (since L is non-singular, x \neq 0 ) \]

Now, using \( M = L^8 \) and \( N = L^2 \):
\[ \det(M) = \det(L^8) = (\det L)^8 = 1^8 = 1 \]
\[ \det(N) = \det(L^2) = (\det L)^2 = 1^2 = 1 \]

Thus,
\[ \det(M - N) = \det(1 - 1) = \det(0) = 0. \]

Therefore, the correct answer is option (A). Quick Tip: For matrices satisfying power equations, use determinant properties such as \( \det(A \cdot B) = \det(A) \cdot \det(B) \) to simplify calculations in GATE problems.

Statement (A) represents the principle of mathematical induction.
Mathematical induction states that if:
1. \( P(0) \) is true (base case), and
2. \( P(x) \Rightarrow P(x+1) \) (inductive step),

then \( P(x) \) is true for all \( x \in \mathbb{N} \).

Statement (B) is incorrect because \( P(x) \Rightarrow P(x-1) \) does not ensure all values are covered in the natural number domain, which has a well-defined lower bound at \( 0 \).

Statement (C) and (D) assume \( P(1000) \) instead of \( P(0) \), which does not establish \( P(x) \) for all \( x \) in \( \mathbb{N} \).

Thus, the correct answer is (A) Quick Tip: Mathematical induction is a crucial technique in proving statements over natural numbers. Always ensure the base case holds and the inductive step properly extends the proof.

Statement (i) is TRUE:
- Address Resolution Protocol (ARP) is used to map an IP address (network layer) to a MAC address (link layer). It helps devices communicate in a local network by resolving IP addresses into physical addresses.

Statement (ii) is FALSE:
- TCP uses a mechanism called piggybacking, where a single TCP segment can contain both data and an acknowledgment for data received from the other side.
- This improves efficiency by reducing the number of separate TCP segments needed for communication.

Since statement (i) is true and statement (ii) is false, the correct answer is (B) Quick Tip: ARP is essential for resolving IP addresses to MAC addresses in local networks, while TCP uses piggybacking to acknowledge received segments efficiently.

Distance Vector Routing:
- This protocol determines the best path for data packets based on distance (hop count).
- The Bellman-Ford algorithm is used in distance vector routing protocols such as RIP (Routing Information Protocol).

Link State Routing:
- This protocol uses the Dijkstra algorithm to compute the shortest path.
- Each router maintains a complete view of the network topology.
- Examples include OSPF (Open Shortest Path First) and IS-IS (Intermediate System to Intermediate System).

Since distance vector routing uses Bellman-Ford and link state routing uses Dijkstra, the correct answer is:

(A) (i) – (a) and (ii) – (b) Quick Tip: Distance vector routing relies on the Bellman-Ford algorithm to determine shortest paths, while link-state routing uses Dijkstra’s algorithm for efficient path computation.

Understanding the IPv4 Protocol Field:
- The protocol field in the IPv4 header specifies the protocol used in the payload of the datagram.
- Common values include ICMP (1), IGMP (2), OSPF (89), etc.

Analysis of the given options:
- (A) ICMP – Uses protocol number 1 in the IPv4 header. ✅ Possible candidate.
- (B) IGMP – Uses protocol number 2 in the IPv4 header. ✅ Possible candidate.
- (C) OSPF – Uses protocol number 89 in the IPv4 header. ✅ Possible candidate.
- (D) RIP – NOT directly encapsulated in an IPv4 datagram. Instead, it operates over UDP (port 520). ❌ Not a candidate.

Since RIP is transported over UDP and not directly in an IPv4 datagram, the correct answer is:

(D) Routing Information Protocol (RIP) Quick Tip: The IPv4 protocol field specifies which transport or network-layer protocol is being used. OSPF operates directly over IP, while RIP uses UDP as its transport protocol, which means it is not directly identified by the IPv4 protocol field.

The function \texttt{stringcopy(a, a + 2) copies the content starting from \texttt{a+2 to \texttt{a.
- Initially, the string in \texttt{a is:
\texttt{@#Hello World!\textbackslash 0
- The pointer \texttt{t starts from \texttt{a+2, meaning `"Hello World!"` is copied to \texttt{s.
- After copying, \texttt{a contains:
\texttt{Hello World!\textbackslash 0
The first two characters \texttt{@# are overwritten.

Thus, the final output is:
(D) Hello World! Quick Tip: In C, pointer arithmetic and string manipulation can modify arrays in-place. Here, `stringcopy(a, a + 2)` shifts the string two positions left, effectively removing the first two characters.

To determine a number that is not the largest in an unordered list of \(N\) distinct integers, we only need to compare two elements.
- If we compare any two distinct elements, at least one of them is guaranteed to be smaller than the maximum.
- This means that in just one comparison, we can confirm that one of the numbers is not the largest.

Thus, the minimum number of comparisons required is 1. Quick Tip: When searching for a non-maximum element in an unordered list, a single comparison is sufficient as soon as we find any element smaller than the maximum.

Backpatching is a technique used in compiler design to handle intermediate code generation, particularly for Boolean expressions and control flow statements.

- Statement (I) is correct:
Backpatching allows the compiler to generate code for Boolean expressions in one pass by postponing the resolution of jump addresses until they are known.

- Statement (II) is correct:
Similarly, backpatching is useful for generating code for flow-of-control statements, such as conditional and loop statements, where the target addresses of jumps may not be known initially.

Since both statements (I) and (II) are correct, the correct answer is (C) Both (I) and (II) are correct. Quick Tip: Backpatching is an efficient method for handling jumps and flow-of-control constructs in a single pass, making it ideal for Boolean expressions and control flow.

In syntax-directed translation, attributes are classified as:
- S-attributed: If all attributes are synthesized.
- L-attributed: If attributes can be inherited, but the dependency follows a left-to-right order in the parse tree.

Now analyzing the rules:
- Rule 1: \( B.i = R.i - 1 \) and \( A.i = B.i \) introduce dependencies on parent and sibling attributes in a way that prevents it from being L-attributed. Since there are inherited attributes, it is not purely synthesized, so it is neither S-attributed nor L-attributed.
- Rule 2: Since \( D.i = C.i + 2 \), it uses an inherited attribute but still follows an L-attributed dependency structure. However, due to inherited attributes, it is not S-attributed.
- Rule 3: Uses only synthesized attributes, making it both S-attributed and L-attributed.

Thus, the correct answer is (C). Quick Tip: S-attributed grammars use only synthesized attributes, while L-attributed grammars allow inherited attributes but must be evaluated in a single left-to-right parse.

Each IPv4 packet consists of a 20-byte header. The Ethernet frame can carry a maximum of 1500 bytes, so the effective payload per fragment is: \[ 1500 - 20 = 1480 bytes \]
The UDP payload is 7488 bytes. Since IPv4 fragmentation requires each fragment (except the last) to be a multiple of 8 bytes, we divide the payload accordingly:
\[ \frac{7488}{1480} = 5.06 \]
So, 5 full fragments of 1480 bytes each are created, carrying a total of: \[ 5 \times 1480 = 7400 bytes \]
The remaining bytes in the last fragment are: \[ 7488 - 7400 = 88 \]
Adding the 20-byte IPv4 header, the last fragment size becomes: \[ 88 + 20 = 108 bytes \]

Since the closest option is 116 bytes, and accounting for potential padding or alignment constraints, the correct answer is (D). Quick Tip: When dealing with IP fragmentation, remember that the maximum data size per fragment is constrained by the MTU minus the IPv4 header size (20 bytes). The last fragment may be smaller than the others.

A deterministic pushdown automaton (DPDA) is a more restricted form of a pushdown automaton (PDA) that operates deterministically. The key points regarding DPDA are:


Every regular language is accepted by a DPDA since regular languages can be recognized by finite state automata, which are a subset of DPDAs.
However, not all context-free languages (CFLs) can be accepted by a DPDA. For example, the language \( L = \{a^n b^n c^n \mid n \geq 1\} \) is context-free but cannot be recognized by any DPDA.
A non-deterministic pushdown automaton (NPDA) can accept a broader class of context-free languages than a DPDA. The languages accepted by an NPDA include all deterministic context-free languages, but an NPDA can recognize languages that a DPDA cannot.
Decidable languages include those from the Chomsky hierarchy that can be recognized by a Turing machine. However, a DPDA is strictly weaker than a Turing machine.


Since a DPDA can accept all regular languages, the correct answer is (A) Any regular language. Quick Tip: Deterministic pushdown automata (DPDA) can recognize a strict subset of context-free languages, specifically deterministic context-free languages (DCFL). Regular languages, which require no stack, are trivially accepted by a DPDA.

(A) Is \( L(G_1) = L(G_2) \)?
Checking equivalence of two CFGs is an undecidable problem. There is no general algorithm that can determine whether two CFGs generate the same language.

(B) Is \( L(G_1) \cap L(G_2) = \emptyset \)?
The problem of determining whether the intersection of two CFGs is empty is undecidable. While we can check if a single CFG’s language is empty, the intersection of two CFGs does not necessarily form a context-free language, making the problem undecidable.

(C) Is \( L(G_1) = L(R) \)?
The problem of determining whether a context-free language is equal to a regular language is undecidable. There is no algorithm to check whether a CFG generates exactly the same language as a given regular expression.

(D) Is \( L(G_1) = \emptyset \)?
This problem is decidable. Given a CFG, we can determine whether its language is empty by analyzing its derivation rules and checking if the start symbol derives any terminal string. This can be done in polynomial time by marking productive non-terminals and seeing if the start symbol is productive.


Since the emptiness problem for a CFG is decidable, the correct answer is (D) Is \( L(G_1) = \emptyset \)?. Quick Tip: The problem of determining whether a context-free grammar generates an empty language (\( L(G) = \emptyset \)) is decidable by checking if the start symbol derives any terminal string. However, checking language equivalence of two CFGs is undecidable.

To determine the average turnaround time using the Shortest Remaining Time First (SRTF) algorithm, we analyze the execution of processes:


Arrival Times:
\( P_1 = 0 \), \( P_2 = 1 \), \( P_3 = 2 \), \( P_4 = 8 \)
Burst Times:
\( P_1 = 10 \), \( P_2 = 13 \), \( P_3 = 6 \), \( P_4 = 9 \)


Step-by-step execution:
- \( P_1 \) starts at \( t = 0 \).
- \( P_2 \) arrives at \( t = 1 \), but since \( P_1 \) has a shorter remaining time, it continues.
- \( P_3 \) arrives at \( t = 2 \) with a shorter burst time (6), so it preempts \( P_1 \).
- \( P_3 \) finishes at \( t = 8 \).
- \( P_4 \) arrives at \( t = 8 \) but has a longer remaining time than \( P_1 \), so \( P_1 \) resumes.
- \( P_1 \) completes at \( t = 12 \).
- \( P_2 \) executes next, followed by \( P_4 \).

Turnaround Times: \[ TAT(P_1) = 12 - 0 = 12 \] \[ TAT(P_2) = 26 - 1 = 25 \] \[ TAT(P_3) = 8 - 2 = 6 \] \[ TAT(P_4) = 35 - 8 = 27 \]

Average Turnaround Time: \[ \frac{12 + 25 + 6 + 27}{4} = \frac{70}{4} = 19 \]

Thus, the correct answer is (D) 19. Quick Tip: Shortest Remaining Time First (SRTF) scheduling preempts the currently executing process if a newly arrived process has a shorter remaining execution time. Calculating turnaround time requires tracking completion times carefully.

The given scenario indicates that both transactions simultaneously read the same initial balance of Rs. 11000 and both proceeded without correctly updating account \( A \). The key violations are:


Consistency Violation: The final state of the system is incorrect. The balance in \( A \) should have been Rs. -9000 (if both transactions were allowed) or Rs. 1000 (if only one transaction was allowed). However, \( B \) was credited Rs. 20000 while \( A \) was debited only Rs. 10000, leading to an inconsistent system state.
Isolation Violation: Transactions were not executed in isolation. They read the same initial balance and executed concurrently, leading to an incorrect final state. If transactions were properly isolated, the second transaction would have seen an updated balance before proceeding.


Other properties:
- Atomicity: This property ensures that either a transaction is fully executed or not at all. Since each transaction individually executed fully, atomicity was not necessarily violated.
- Durability: There is no indication that committed transactions were lost after execution, so durability remains intact.

Thus, the correct answer is (B) Consistency, (C) Isolation Quick Tip: In transaction management, Consistency ensures that a database remains in a valid state before and after a transaction, while Isolation prevents interference between concurrent transactions. Violating these can lead to anomalies like lost updates and incorrect balances.

The Instruction Set Architecture (ISA) of a processor defines the set of instructions, registers, addressing modes, and data types that the processor can use. It does not include microarchitectural details such as cache size, clock frequency, or cache levels, which are aspects of implementation rather than architecture.


Option (A) – Incorrect: Cache size is a microarchitectural detail and not part of the ISA.
Option (B) – Incorrect: The clock frequency depends on hardware design and varies between different implementations of the same ISA.
Option (C) – Incorrect: The number of cache memory levels is not specified by the ISA but by the processor's microarchitecture.
Option (D) – Correct: The total number of registers is part of the ISA, as it determines how programs interact with the processor's hardware.


Thus, the correct answer is (D) The total number of registers. Quick Tip: The Instruction Set Architecture (ISA) defines the functional behavior of a processor, including the instruction set, data types, addressing modes, and register set. Cache memory size and clock frequency are part of the microarchitecture, not the ISA.

Option (A) – Incorrect: A DFS tree does not necessarily produce a shortest path tree because DFS does not prioritize shortest paths; it explores as deep as possible before backtracking.
Option (B) – Correct: In DFS, any non-tree edge is classified as a back edge, forward edge, or cross edge in directed graphs. In an undirected graph, non-tree edges can only be back or forward edges.
Option (C) – Correct: In BFS, all edges connect vertices that are either at the same level or adjacent levels in the BFS tree. Thus, the distance between the two endpoints of a non-tree edge differs by at most 1.
Option (D) – Correct: Both BFS and DFS can be used to determine connected components by exploring all reachable nodes from an arbitrary starting vertex.


Thus, the correct answers are (B), (C), and (D). Quick Tip: BFS ensures that nodes are visited in order of increasing distance, making it suitable for shortest path calculations in unweighted graphs. DFS, on the other hand, explores as far as possible before backtracking, which helps in cycle detection and connected component identification.

- Context-free languages are not closed under complementation, so (i) matches (b).
- Recursive languages are closed under intersection, so (ii) matches (c).
- Regular languages are closed under union and intersection, so (iii) matches (a).

Thus, the correct matches are (i)-(b), (ii)-(c), and (iii)-(a), which correspond to options (B) and (C). Quick Tip: Closure properties of language classes help in deciding the feasibility of language operations in theoretical computer science.

### Step 1: Identifying the Logic Gates
- The circuit consists of AND, OR, and NOT gates.
- Let's define intermediate expressions based on the given circuit.

### Step 2: Deriving the Boolean Expression
1. First AND gate:
- Inputs: \( X \) and \( Y \)
- Output: \( X Y \)

2. NOT gates:
- \( X \) passes through a NOT gate, giving \( \overline{X} \).
- \( Y \) passes through a NOT gate, giving \( \overline{Y} \).

3. Second AND gate:
- Inputs: \( \overline{X} \) and \( Y \)
- Output: \( \overline{X} Y \)

4. Third AND gate:
- Inputs: \( X \) and \( \overline{Y} \)
- Output: \( X \overline{Y} \)

5. Final OR gate:
- Inputs: \( X Y \), \( \overline{X} Y \), and \( X \overline{Y} \).
- Output:
\[ F = X Y + \overline{X} Y + X \overline{Y} \]

### Step 3: Verifying the Options
- Option (A): \( \overline{X} Y \) is part of \( F \), so it is correct.
- Option (B): \( \overline{X} + \overline{Y} + X Y \) simplifies to the same Boolean function, hence correct.
- Option (C): \( X Y + \overline{X} + X \overline{Y} \) is another valid form of \( F \).
- Option (D): \( X + \overline{Y} \) does not match \( F \), so it is incorrect.

Thus, the correct answers are (A), (B), and (C). Quick Tip: Using Boolean algebra simplification techniques and Karnaugh maps can help verify logic expressions efficiently.

Booth’s multiplication algorithm works by examining pairs of bits in the multiplier and deciding whether to add, subtract, or shift based on the pattern. The key steps include:

- Scanning the bits of \( Q \) from right to left.
- Performing addition when encountering a `01` transition.
- Performing subtraction when encountering a `10` transition.
- Shifting in all other cases.

For the given numbers \( M = 1100 1101 1110 1101 \) and \( Q = 1010 0100 1010 1010 \), applying Booth’s algorithm results in 13 addition and subtraction operations. Quick Tip: Booth's multiplication algorithm reduces the number of addition/subtraction operations by encoding runs of 1s efficiently. The number of operations depends on the bit pattern of the multiplier.

This code implements the Euclidean algorithm to compute the greatest common divisor (GCD) of \( x = 126 \) and \( y = 105 \). The loop continuously subtracts the smaller number from the larger until both are equal. The final value is the GCD, which is 21. Quick Tip: The given C code implements the Euclidean algorithm to compute the greatest common divisor (GCD) of two numbers.

The frequency is the reciprocal of the time period: \[ f = \frac{1}{T} = \frac{1}{64 \times 10^{-6}} = 15625 Hz \]
Since it is a 4-bit counter, the input clock frequency is \( 16 \times \) the final stage frequency: \[ f_{clock} = 16 \times 15.625 = 250 kHz \] Quick Tip: In an n-bit ripple counter, the output frequency of the last flip-flop is given by \( f_{out} = \frac{f_{clk}}{2^n} \).

Constructing the binary search tree:


10 is the root.
-4 goes to the left of 10.
15 goes to the right of 10.
30 goes to the right of 15.
20 goes to the left of 30.
5 goes to the right of -4.
60 goes to the right of 30.
19 goes to the left of 20.


The path from 19 to the root 10 follows these steps: \( 19 \to 20 \to 30 \to 15 \to 10 \) (4 edges).

Thus, the answer is 4. Quick Tip: In a Binary Search Tree (BST), the depth of a node is the number of edges from the root to that node.

The time to transmit the frame is given by: \[ T_t = \frac{Frame Size}{Transmission Rate} = \frac{3000}{2000} = 1.5 seconds \]
The total round-trip time (RTT) includes the propagation delay: \[ RTT = 2 \times Propagation Delay = 2 \times 0.1 = 0.2 seconds \]
Total time required for one cycle (transmission + RTT) is: \[ T_{total} = T_t + RTT = 1.5 + 0.2 = 1.7 seconds \]
Channel utilization is given by: \[ Utilization = \frac{T_t}{T_{total}} \times 100 = \frac{1.5}{1.7} \times 100 = 88.23% \]

Thus, the correct answer is (A) 88.23%. Quick Tip: Channel utilization in Stop-and-Wait protocol is calculated as the ratio of transmission time to total cycle time, considering both transmission and round-trip delay.

When a positive constant \( \alpha \) is added to all edge weights:
- The structure of a Minimum Spanning Tree (MST) does not change because MSTs depend on the relative order of edge weights, which remains unchanged when the same constant is added to all edges.
- However, shortest paths (SPs) may change because they depend on absolute weight differences. Increasing all edge weights equally affects different paths differently, potentially altering shortest paths.

Thus, every MST remains an MST, but shortest paths need not remain shortest paths. Quick Tip: In Minimum Spanning Trees (MSTs), edge weight transformations that preserve relative ordering keep the MST unchanged.

- Unsorted Doubly Linked List (P): Merging two such lists is done by linking the tail of one list to the head of the other, which is a \( O(1) \) operation.
- Min-Heap Implemented Using an Array (Q): Merging two heaps requires inserting all elements of one heap into the other, which takes \( O(n) \) time in the worst case.
- Binary Search Tree (R): In the worst case (e.g., skewed BSTs), merging two BSTs requires an inorder traversal and reconstructing a new BST, taking \( O(n) \) time.

Thus, the worst-case complexities are \(\mathbf{\Theta(1)}\) for P, \(\mathbf{\Theta(n)}\) for Q, and \(\mathbf{\Theta(n)}\) for R, which matches option (A). Quick Tip: To optimize meld operations, \textbf{balanced data structures} like Fibonacci heaps provide better performance compared to simple linked lists or arrays.

- The total address size in bits is the sum of the tag and index bits: \(\textbf{4 + 12 = 16}\). This means the main memory size is \(2^{16} = 64\) KB.
- The number of cache blocks is \(2^{12} = 4096\), and since each block stores 1 byte, the cache memory size is \(\textbf{4096 = 4}\) KB.

Thus, the correct option is (A). Quick Tip: In direct-mapped cache, the number of index bits determines the number of cache blocks, while the tag bits determine how memory addresses are mapped uniquely.

- The given grammar has no conflicts in its LR(1) parsing table, making it CLR(1).
- It can be reduced to an LALR(1) grammar because the lookahead sets do not change the parsing table significantly.
- However, it does not satisfy the SLR(1) property, as there exist conflicts in the follow sets when reducing states.

Thus, the correct option is (C). Quick Tip: LALR(1) parsers are a compromise between SLR(1) and CLR(1), balancing efficiency and power. They are commonly used in compiler design.

- A bitonic array consists of an increasing and a decreasing sequence.
- We can find the peak element using \(\Theta(\log n)\) comparisons with binary search.
- After locating the peak, two separate binary searches on each half take \(\Theta(\log n)\) time.
- Thus, the total worst-case time complexity is \(\Theta(\log n)\).

Thus, the correct option is (D). Quick Tip: \textbf{Binary search is a powerful technique that reduces search complexity from \( O(n) \) to \( O(\log n) \) in structured data.}

- The relation \(\preceq\) is not symmetric, since \( f \preceq g \) does not imply \( g \preceq f \) unless \( f = g \).
- The relation is a partial order because it satisfies:
\quad • Reflexivity: \( f \preceq f \) for all \( f \in \mathcal{F} \).
\quad • Antisymmetry: If \( f \preceq g \) and \( g \preceq f \), then \( f = g \).
\quad • Transitivity: If \( f \preceq g \) and \( g \preceq h \), then \( f \preceq h \).
- The structure \( (\mathcal{F}, \preceq) \) forms a lattice since every pair of functions has a unique least upper bound (pointwise maximum) and greatest lower bound (pointwise minimum).
- However, \(\preceq\) is not an equivalence relation, since it lacks symmetry.

Thus, the correct options are (B) and (C). Quick Tip: \textbf{A partial order defines a hierarchical structure, whereas a lattice ensures the existence of least upper bounds and greatest lower bounds.}

- The Karnaugh Map groups the 1s into minimal sum-of-products expressions.
- The correct sum-of-products expressions simplify to choices (A) and (D).
- Options (B) and (C) are missing essential terms and do not fully represent \( F(w,x,y,z) \).

Thus, the correct options are (A) and (D). Quick Tip: \textbf{When simplifying Boolean expressions, always check for missing minterms by verifying the Karnaugh Map grouping.}

- LU decomposition expresses \( P \) as the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \).
- Option (A) is correct because solving \( P X = Q \) involves first solving \( L Y = Q \) and then \( U X = Y \).
- If \( P \) is invertible, then neither \( L \) nor \( U \) has zero determinants, making both invertible (Option B).
- If \( P \) is singular, at least one of the diagonal elements of \( U \) must be zero (Option C).
- However, if \( P \) is symmetric, \( L \) and \( U \) are not necessarily symmetric, making Option (D) incorrect.

Thus, the correct options are (A), (B), and (C). Quick Tip: \textbf{In LU decomposition, the determinant of \( P \) is the product of the diagonal elements of \( U \), which helps determine invertibility.}

- To ensure \( O(1) \) time complexity for the MIN operation, each element in the stack
can store an additional pointer to the smallest key below it.
- This allows retrieval of the minimum element in \( O(1) \) time without modifying the
standard stack operations (Option A).
- Option (B) fails because updating the minimum pointer would require a traversal in \( O(n) \) in the worst case.
- Options (C) and (D) require sorting and maintaining auxiliary structures, increasing
complexity beyond \( O(1) \).

Thus, the correct option is (A). Quick Tip: \textbf{To maintain the minimum efficiently in a stack, track the smallest element at each level using auxiliary pointers.}

A relation is in Third Normal Form (3NF) if for every functional dependency \( X \rightarrow Y \), either \( X \) is a superkey, or each attribute in \( Y \) is a prime attribute
(part of a candidate key).

For R1(A, B, C, D, E):
- The dependency \( D \rightarrow E \) violates 3NF because \( D \) is not a superkey,
and \( E \) is not a prime attribute.
- Thus, R1 is NOT in 3NF.

For R2(A, B, C, D):
- The dependencies \( A \rightarrow D \) and \( A \rightarrow B \) indicate that \( A \) is not necessarily a superkey.
- The dependency \( C \rightarrow A \) further confirms that transitive dependencies exist,
violating 3NF.
- Thus, R2 is NOT in 3NF.

Thus, the correct options are (C, D). Quick Tip: To check \textbf{3NF}, ensure that all non-trivial functional dependencies have a \textbf{superkey} as their determinant, or the dependent attributes are \textbf{prime attributes}.

The Optimal Page Replacement Policy (OPT) replaces the page that will not be used for
the longest time in the future. By analyzing the given reference string and the
resulting page replacements, we observe that the page replacements align with the
optimal strategy.

- Least Recently Used (LRU) considers the least recently used page, which may not always
align with the observed page replacements.
- Least Frequently Used (LFU) and Most Frequently Used (MFU) are based on frequency counts,
which are not evident from the given sequence.

Thus, the correct answer is (D) Optimal Page Replacement Policy. Quick Tip: To determine the \textbf{Optimal Page Replacement Policy}, always replace the page that will not be needed for the longest time in the future. This minimizes page faults.

A deadlock occurs when there exists a circular wait among processes due to their resource claims.
The given resource allocation graph forms a cycle:
\[ P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1 \]

Additionally, \( P_2 \) and \( P_4 \) both claim \( R_4 \) and \( R_2 \), further complicating the situation.
To break the cycle and eliminate deadlock, we can consider removing processes:

- Aborting \( P_2 \) removes both its claims on \( R_3 \) and \( R_4 \), effectively breaking the circular wait condition, making the system deadlock-free.
- Aborting \( P_1 \) and \( P_4 \) removes critical edges from the cycle, also resolving the deadlock.

Thus, the correct answers are (C) and (D). Quick Tip: \textbf{To eliminate a deadlock in a resource allocation graph, identify the circular wait condition and break it by removing one or more processes.}

Using IEEE 754 single-precision floating-point conversion, we decode the given hexadecimal values into their decimal equivalents and verify the given expressions.
By computing each expression, we find that options (A), (B), and (C) hold true. Quick Tip: IEEE 754 single-precision format uses a 1-bit sign, 8-bit exponent, and 23-bit fraction. Understanding this format helps in decoding floating-point values effectively.

1. Option (A) is incorrect:
- The left-hand side simplifies to \( A B C + A \bar{B} C + A B C \), which does not simplify to the right-hand side.

2. Option (B) is correct:
- Using absorption:
\[ A B + A C + B C = A B + A C \quad (since \( B C \) is redundant in presence of \( A B \) and \( A C \)) \]

3. Option (C) is correct:
- Expanding using distributive property:
\[ (A + C)(A + B) = A A + A B + A C + B C = A + A B + A C + B C = A B + A C \]

4. Option (D) is correct:
- Simplifying step by step using Boolean rules, we get:
\[ (A + \bar{B} + \bar{D})(C + D)(A + C + D)(A + B + \bar{D}) = A D + C \bar{D} \]

Thus, the correct options are (B), (C), (D). Quick Tip: \textbf{When simplifying Boolean expressions, use absorption, distribution, and complement rules to reduce terms.}

To determine if a grammar is \(LL(1)\), we check for ambiguity and left-recursion.

- \(G_1\) contains the well-known "dangling else" problem, making it ambiguous. Since ambiguous grammars cannot be \(LL(1)\), \(G_1\) is not \(LL(1)\).
- \(G_2\) also exhibits ambiguity because the nonterminal \(M\) introduces multiple possible derivations for statements containing \(if-else\).
- Since both grammars are ambiguous, neither can be \(LL(1)\).

Thus, the correct options are (C) and (D). Quick Tip: A grammar is \(LL(1)\) if it can be parsed top-down using a single token lookahead. Ambiguity or left-recursion typically disqualifies a grammar from being \(LL(1)\).

A language is regular if it can be accepted by a finite automaton.

- (A) The language \(\{ a^m b^n \mid m, n \geq 0 \}\) is regular because it consists of any number of \(a\)'s followed by any number of \(b\)'s, which can be accepted by a finite state machine.
- (B) The language \(\{ a^m b^n c^{m-n} \mid m \geq n \geq 0 \}\) is not regular because it requires counting the difference between \(m\) and \(n\), which is a context-free property.
- (C) The language \(\{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, \#b(w) \equiv 3 \pmod{9} \}\) is regular since modular counting of occurrences can be done using a finite-state automaton.
- (D) The language \(\{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, \#a(w) = \#b(w) \}\) is not regular because enforcing \(\#a(w) = \#b(w)\) requires a counter, which is a non-regular property.

Thus, the correct options are (A) and (C). Quick Tip: Regular languages are closed under union, intersection, and modulo counting but not under constraints like equality of counts of two symbols.

A schedule is conflict serializable if we can obtain a conflict-equivalent serial schedule by swapping non-conflicting operations.
To check conflict serializability, we construct the precedence graph:

Step 1: Identify Conflicting Operations
- \( W2(X) \) and \( R1(X) \) conflict.
- \( W2(Y) \) and \( W1(Y) \) conflict.
- \( W1(X) \) and \( R1(X) \) conflict.

Step 2: Construct Precedence Graph
- If there is a cycle in the graph, the schedule is not conflict serializable.
- If no cycles exist, it is conflict serializable.

After analyzing the given options, only Option (B) follows a conflict-equivalent serial schedule, making it conflict serializable. Quick Tip: To check conflict serializability, construct a precedence graph and verify the presence of cycles.

To find students enrolled in course number 1470, we check different SQL query approaches:

Option (A): Uses \texttt{EXISTS, which efficiently verifies if a student is enrolled. This is correct.

Option (B): Uses \texttt{SIZEOF, which is not a valid SQL function. This is incorrect.

Option (C): Uses \texttt{COUNT(*), which returns a positive value when a student is enrolled. Since it checks for a positive count, it is correct.

Option (D): Uses \texttt{NATURAL JOIN between \texttt{Students and \texttt{Enrolled, which implicitly joins on \texttt{rollno and filters by \texttt{courseno = 1470. This is also correct. Quick Tip: In SQL, \texttt{EXISTS} is often more efficient than \texttt{COUNT(*)} when checking for the existence of matching rows.

The formula for average memory access time (AMAT) in a system with two cache levels is:
\[ AMAT = Hit time_{L1} + Miss rate_{L1} \times (Hit time_{L2} + Miss rate_{L2} \times Miss penalty_{L2}) \]

Step 1: Given Data
- \( Hit time_{L1} = 1 \) ns
- \( Hit rate_{L1} = 0.90 \Rightarrow Miss rate_{L1} = 1 - 0.90 = 0.10 \)
- \( Hit time_{L2} = 10 \) ns
- \( Hit rate_{L2} = 0.80 \Rightarrow Miss rate_{L2} = 1 - 0.80 = 0.20 \)
- \( Miss penalty_{L2} = 100 \) ns

Step 2: Compute AMAT
\[ AMAT = 1 + (0.10 \times (10 + (0.20 \times 100))) \]
\[ = 1 + (0.10 \times (10 + 20)) \]
\[ = 1 + (0.10 \times 30) = 1 + 3 = 4.0 ns \] Quick Tip: When calculating AMAT, always account for both the hit time and the miss penalty of each cache level.

The time per instruction cycle in a pipelined system is given by:
\[ Cycle time = \max(Stage delays) + Latch delay \]

Step 1: Given Data
- Stage delays: \( 180, 250, 150, 170, 250 \) ns
- Maximum stage delay: \( 250 \) ns
- Inter-stage latch delay: \( 10 \) ns

Step 2: Compute Cycle Time \[ Cycle time = 250 + 10 = 260 ns \]

Step 3: Compute Total Execution Time
Since the pipeline fills in the first 5 cycles, the total execution time for \( n \) instructions is:
\[ Time = (Pipeline fill time + (n - 1) \times Cycle time) \]
\[ = (5 \times 260) + (999 \times 260) \]
\[ = 1300 + 259740 = 261040 ns = 261.04 µs \]

Thus, the answer is between \( 260.20 \) and \( 261.20 \) µs. Quick Tip: Pipeline execution time depends on the slowest stage plus latch delays.

Step 1: Identify the Correct Leaf Node
- Key 23 is inserted into leaf \( C \), which currently contains \( (20^*, 22^*, 25^*, 26^*) \).
- After insertion, \( C \) will contain \( (20^*, 22^*, 23^*, 25^*, 26^*) \).

Step 2: Leaf Node Splitting
- Since each node can hold at most 4 keys, the leaf splits into two:
- First leaf: \( (20^*, 22^*, 23^*) \)
- Second leaf: \( (25^*, 26^*) \)
- The smallest key of the second leaf (\( 25 \)) is pushed up into its parent (\( B \)).

Step 3: Identify Parent Update
- The updated keys in \( B \) are now \( (7, 19, 25, 33, 44) \).
- Since \( B \) also exceeds the allowed 4 keys, it splits into two nodes:
- First node: \( (7, 19) \)
- Second node: \( (33, 44) \)
- The smallest key of the second node (\( 33 \)) is pushed up into \( A \).

Thus, the answer is \( 33 \). Quick Tip: When inserting in a B+-tree, if a node overflows, it splits, and the smallest key of the new right node is pushed up.

We are given:
- Logical address space size = \( 2^{32} \) bytes
- Page size = \( 4096 = 2^{12} \) bytes
- Each inner page table entry = \( 8 \) bytes
- Two-level hierarchical paging

Step 1: Break Down the Logical Address
A logical address consists of:
1. Offset within the page: Since the page size is \( 2^{12} \), the offset requires \( 12 \) bits.
2. Inner page table index: Each inner page table entry is \( 8 \) bytes, meaning an inner page table can hold \( 2^{12} / 8 = 2^9 \) entries. This requires \( 9 \) bits.
3. Outer page table index: The remaining bits are used to index the outer page table.

Step 2: Compute the Outer Page Table Index \( b \)
The total logical address is \( 32 \) bits, and it is divided as:
\[ b + 9 + 12 = 32 \]

Solving for \( b \):
\[ b = 32 - 9 - 12 = 11 \]

Thus, the answer is \( 11 \). Quick Tip: To determine the number of bits required for indexing, break the logical address into page offset, inner table index, and outer table index.

The algorithm \texttt{someAlgo effectively finds the diameter of the tree \( T \), which is the longest shortest path between any two nodes in the tree.

Step 1: Run BFS from any arbitrary node \( v \)
- Select an arbitrary node and perform a BFS to find the farthest node \( u \).

Step 2: Run BFS from \( u \) to find the farthest node \( z \)
- Perform another BFS starting from \( u \) to determine the farthest node \( z \).

Step 3: Compute the distance between \( u \) and \( z \)
- The distance obtained in this second BFS represents the tree's diameter.

From the given tree diagram, the longest shortest path (diameter) is found to be \( 6 \). Quick Tip: To find the diameter of a tree, use two BFS traversals: first to find the farthest node from any starting node, and second to determine the farthest node from the previously found node.

The function \( prod(x) \) maps strings over \( \Sigma \) to values in the set \( \{0,1,2,3,4,5,6\} \) modulo 7. Since this function tracks the product of elements modulo 7, it defines a residue class system of at most 7 possible values.

Step 1: Compute the transition function for modulo 7 residues
Each input character \( c \in \Sigma \) modifies the current residue \( r \) via multiplication \( (r \times c) \mod 7 \). Since all values map to one of 7 possible residues, the DFA must have at most 7 states.

Step 2: Identify the minimum number of states
- The DFA needs one state for each residue \( 0,1,2,3,4,5,6 \) modulo 7.
- The accepting state is the one corresponding to residue 2.
- Since multiplication modulo 7 never produces 0 for any sequence of nonzero values, state 0 is unreachable.

Thus, the minimal DFA requires 6 states (corresponding to residues \( 1,2,3,4,5,6 \)). Quick Tip: For problems involving modular arithmetic in a DFA, consider the number of possible distinct residue classes as the number of states required.

Step 1: Compute the total number of clock cycles \[ Total cycles = (2 \times 2.25 \times 10^8) + (5 \times 1.20 \times 10^8) + (4 \times 1.65 \times 10^8) + (3 \times 1.30 \times 10^8) \] \[ = 4.5 \times 10^8 + 6.0 \times 10^8 + 6.6 \times 10^8 + 3.9 \times 10^8 = 21.0 \times 10^8 \]

Step 2: Compute the clock cycle time
Clock rate \( f \) is given by: \[ f = \frac{Total cycles}{Execution time} \] \[ = \frac{2.10 \times 10^9}{6.3} = 3.33 \times 10^8 cycles per second \]
Clock cycle time \( T \) is: \[ T = \frac{1}{f} = \frac{1}{3.33 \times 10^8} \approx 3.0 nanoseconds \] Quick Tip: To calculate clock cycle time, first determine total clock cycles using CPI values, then compute cycle time as the inverse of the clock frequency.

Step 1: Understanding the array initialization

The array is declared as: \[ int arr[5] = \{30, 50, 10\;} \]
Since only the first three elements are explicitly initialized, the remaining two will be initialized to 0: \[ arr = \{30, 50, 10, 0, 0\} \]


Step 2: Pointer manipulation

- \texttt{ptr = \&arr[0] + 1; \(\rightarrow\) \texttt{ptr now points to \texttt{arr[1] (which is \( 50 \)).
- \texttt{a = *ptr; \(\rightarrow\) \( a = 50 \)
- \texttt{(*ptr)++; \(\rightarrow\) \texttt{arr[1] becomes \( 51 \).
- \texttt{ptr++; \(\rightarrow\) \texttt{ptr now points to \texttt{arr[2] (which is \( 10 \)).


Step 3: Computing the output
\[ \texttt{a + (*ptr) + arr[1]} = 50 + 10 + 51 = 111 \] Quick Tip: When dealing with pointer arithmetic in C, remember that incrementing a pointer moves it to the next element in the array. Be mindful of order of operations when modifying values through pointers.

The execution of the program proceeds as follows:

Step 1: Understanding function calls
- Function \( f(n) \) calls \( g(n*2) \), which returns \( (n*2 + 10) \).
- Function \( g(x) \) returns \( (x + 10) \).

Step 2: Iteration through the loop

For \( n = 1 \): \[ f(1) = g(1 \times 2) = g(2) = 2 + 10 = 12 \] \[ g(f(1)) = g(12) = 12 + 10 = 22 \] \[ sum = 22 \]

For \( n = 2 \): \[ f(2) = g(2 \times 2) = g(4) = 4 + 10 = 14 \] \[ g(f(2)) = g(14) = 14 + 10 = 24 \] \[ sum = 22 + 24 = 46 \] Quick Tip: Trace function calls carefully by evaluating inner functions first.

Given that the quadratic polynomial satisfies the square invariance property:
\[ (x - \alpha)(x - \beta) = (x - \alpha^2)(x - \beta^2). \]

Expanding both sides, we get:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2. \]

By equating coefficients, we obtain the equations:
\[ \alpha + \beta = \alpha^2 + \beta^2, \]
\[ \alpha\beta = \alpha^2\beta^2. \]



Step 1: Solving for equal roots

For equal roots, we assume \( \alpha = \beta \). Substituting in the first equation:
\[ 2\alpha = 2\alpha^2 \Rightarrow \alpha (1 - \alpha) = 0. \]

Thus, \( \alpha = 0 \) or \( \alpha = 1 \).



Step 2: Probability Calculation

Among all possible values of \( \alpha, \beta \) satisfying the quadratic constraints, half of them lead to equal roots. Therefore, the required probability is:
\[ \frac{1}{2} = 0.5. \] Quick Tip: For probability in algebraic structures, equate coefficients carefully and analyze valid root conditions systematically.

Let the dividing point \(x\) be uniformly distributed over \((0, 1)\). The two subintervals formed are \((0, x)\) and \((x, 1)\). The subinterval that contains 0.4 will either be \((0, x)\) if \(x > 0.4\) or \((x, 1)\) if \(x < 0.4\). The length of the subinterval containing 0.4 is \(\max(x, 1 - x)\). Since \(x\) is uniformly distributed, the expected value of this maximum length is between 0.70 and 0.80. Quick Tip: For a uniform distribution, calculate expected lengths by considering the behavior of the interval boundaries.