GATE CS 2025 Set 2

The sentence contrasts Rehman's 'initial hesitation' with a quality that 'never wavered' and contributed to the project's success.

• Ambivalence means having mixed feelings or contradictory ideas. While related to hesitation, saying his ambivalence never wavered would mean he remained undecided, which doesn't fit the context of contributing to success.
• Satisfaction refers to fulfillment, which doesn't fit the context of determination.
• Resolve means firm determination to do something. 'Rehman's resolve... never wavered' indicates that despite being hesitant at the start, his determination to succeed remained firm.
• Revolve is a verb meaning to move in a circle, which is grammatically incorrect here.

Thus, resolve is the correct word.

The analogy follows the relationship of Animal : Natural Habitat/Home.

• A Bird lives in a Nest.
• Similarly, a Bee lives in a Hive.

Explanation of other options:

• Kennel: A small shelter for a dog.
• Hammock: A type of bed made of canvas or rope mesh suspended from supports.
• Lair: A wild animal's resting place, such as a den.

Given the equation $Pe^x = Qe^{-x}$ holds for all real values of $x$.Method 1: Testing specific values of $x$

• For $x = 0$: $P e^0 = Q e^{-0} \implies P(1) = Q(1) \implies P = Q$.
• For $x = 1$: $P e^1 = Q e^{-1} \implies Pe = \frac{Q}{e} \implies Pe^2 = Q$.

Substituting $P = Q$ into $Pe^2 = Q$:
$Pe^2 = P \implies P(e^2 - 1) = 0$.
Since $e^2 - 1 \neq 0$ (as $e \approx 2.718$), we must have $P = 0$.
Since $P = Q$, it follows that $Q = 0$.Method 2: Using properties of functions
Rearrange the equation: $P e^{2x} = Q$.
For this to be true for all $x$, the left side must be a constant function. The function $f(x) = P e^{2x}$ is constant if and only if its derivative is zero everywhere:
$f'(x) = 2P e^{2x} = 0 \implies P = 0$.
If $P = 0$, then $0 \cdot e^{2x} = Q \implies Q = 0$.Thus, $P = Q = 0$ is the only correct statement.

The given figure is a net of a cube. Let's analyze the relative positions of the symbols with respect to the central face containing the Triangle ($\Delta$).From the net:

• The Triangle ($\Delta$) is in the center.
• The Solid Circle ($\bullet$) is directly above the Triangle.
• The Circle ($\circ$) is directly below the Triangle.
• The Empty Square is to the left of the Triangle.
• The Solid Triangle ($\blacktriangle$) is to the right of the Triangle.

When folded into a cube, if we position the Triangle on the Front face in its upright orientation (pointing up):

• The Top face must be the Solid Circle.
• The Bottom face must be the Circle.
• The Left face must be the Empty Square.
• The Right face must be the Solid Triangle.

Let's check the options:

• (A) Shows the Triangle on the Front. The Top face is the Solid Circle. The Left face is the Empty Square. This matches the net perfectly.
• (B) Shows the Triangle on the Front. The Top face is an Empty Square (incorrect, should be Solid Circle). The Right face is the Solid Triangle (correct position, but Top is wrong).
• (C) Shows the Triangle on the Front. The Top face is an Empty Square (incorrect).
• (D) Shows the Triangle on the Front. The Top face is the Solid Triangle (incorrect). The Right face is the Solid Circle (incorrect).

Thus, only option (A) correctly represents the folded cube.

We analyze each option using properties of prime numbers and counterexamples:

• Counterexample: Let $p_1 = 2$ and $p_2 = 3$. Both are prime. Their sum is $2 + 3 = 5$, which is a prime number. Thus, the statement is false.

2. Option (B): "$p_1 p_2$ is not a prime number."

• By definition, a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Since $p_1$ and $p_2$ are primes, $p_1 \ge 2$ and $p_2 \ge 2$. The product $N = p_1 p_2$ has divisors $1, p_1, p_2,$ and $p_1 p_2$. Since $p_1 > 1$, $N$ has at least three divisors (or four if $p_1 \neq p_2$). Therefore, the product of two primes is always composite (not prime). This statement is always true.

3. Option (C): "$p_1 + p_2 + 1$ is a prime number."

• Counterexample: Let $p_1 = 3$ and $p_2 = 5$. Their sum plus one is $3 + 5 + 1 = 9$, which is composite ($3 \times 3$). Thus, the statement is false.

4. Option (D): "$p_1 p_2 + 1$ is a prime number."

• Counterexample: Let $p_1 = 3$ and $p_2 = 5$. The product plus one is $3 \times 5 + 1 = 16$, which is composite ($4^2$). Thus, the statement is false.

Therefore, Option (B) is the correct statement.

The statement uses a counterfactual conditional structure: "Even if I had known...". In English grammar, this specific construction (even if + past perfect) is used to describe a hypothetical situation in the past that did not actually occur. Therefore, the use of "had known" implies that the speaker, Ramya, did not actually know that Josephine was in the hospital at that time.

• (A) is incorrect because the counterfactual implies the opposite.
• (C) might be a reasonable real-world assumption based on the cold tone, but it is not a logically certain inference based only on the provided text.
• (D) is completely unsupported by the text.

Thus, (B) is the only logically sound inference.

The coding pattern involves reversing the word and then shifting each letter forward by one position in the alphabet ($+1$).Step 1: Analyze the pattern for IMAGE $\to$ FHBNJ

• E $+ 1 \to$ F
• G $+ 1 \to$ H
• A $+ 1 \to$ B
• M $+ 1 \to$ N
• I $+ 1 \to$ J

Result: FHBNJ. This matches the given code.Step 2: Verify with FIELD $\to$ EMFJG

• D $+ 1 \to$ E
• L $+ 1 \to$ M
• E $+ 1 \to$ F
• I $+ 1 \to$ J
• F $+ 1 \to$ G

Result: EMFJG. This matches the given code.Step 3: Apply the pattern to BEACH

• H $+ 1 \to$ I
• C $+ 1 \to$ D
• A $+ 1 \to$ B
• E $+ 1 \to$ F
• B $+ 1 \to$ C

Result: IDBFCTherefore, the correct code for BEACH is IDBFC.

To find the correct recurrence relations, we test the given options with the data from the table for $i > 0$.1. Testing $X(i)$:

• For $i = 1$: $X(1) = 16$. Using $X(i) = X(i-1) + I(i)$, we get $X(0) + I(1) = 20 + (-4) = 16$. (Matches)
• For $i = 2$: $X(2) = 26$. Using $X(i) = X(i-1) + I(i)$, we get $X(1) + I(2) = 16 + 10 = 26$. (Matches)
• For $i = 3$: $X(3) = 41$. Using $X(i) = X(i-1) + I(i)$, we get $X(2) + I(3) = 26 + 15 = 41$. (Matches)

Thus, $X(i) = X(i-1) + I(i)$. This eliminates options (B) and (C).2. Testing $Y(i)$:

• For $i = 1$: $Y(1) = -80$.
• Option (A): $Y(0)I(1) = 20 \times (-4) = -80$. (Matches)
• Option (D): $Y(0)I(0) = 20 \times 20 = 400$. (Does not match)
• For $i = 2$: $Y(2) = -800$.
• Option (A): $Y(1)I(2) = -80 \times 10 = -800$. (Matches)
• For $i = 3$: $Y(3) = -12000$.
• Option (A): $Y(2)I(3) = -800 \times 15 = -12000$. (Matches)

Therefore, the correct relations are $X(i) = X(i - 1) + I(i)$ and $Y(i) = Y(i - 1)I(i)$.

Let the vertices of the square PQRS be represented in a Cartesian coordinate system. Let $P$ be at $(0, 2)$, $Q$ at $(0, 0)$, $R$ at $(2, 0)$, and $S$ at $(2, 2)$. The side length of the square is 2 cm.The point $L$ lies on the side $QR$, which is on the x-axis. Let the coordinates of $L$ be $(x, 0)$, where $0 \le x \le 2$.The rectangle is PLMN. The side $PL$ connects $P(0, 2)$ and $L(x, 0)$.
The length of side $PL$ is the distance between these two points:$$PL = \sqrt{(x - 0)^2 + (0 - 2)^2} = \sqrt{x^2 + 4}$$The side $MN$ is parallel to the side $PL$. The width of the rectangle, $PN$, is the perpendicular distance between the line segment $PL$ and the line segment $MN$.
Since the side $MN$ passes through the corner $S(2, 2)$, the width $PN$ is equal to the perpendicular distance from point $S(2, 2)$ to the line passing through $P$ and $L$.The equation of the line passing through $P(0, 2)$ and $L(x, 0)$ can be found using the intercept form $\frac{X}{a} + \frac{Y}{b} = 1$:$$\frac{X}{x} + \frac{Y}{2} = 1 \implies 2X + xY - 2x = 0$$The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here, point $S$ is $(2, 2)$, and the line is $2X + xY - 2x = 0$.$$PN = d = \frac{|2(2) + x(2) - 2x|}{\sqrt{2^2 + x^2}} = \frac{|4 + 2x - 2x|}{\sqrt{4 + x^2}} = \frac{4}{\sqrt{4 + x^2}}$$The area of the rectangle PLMN is the product of its length and width:$$\text{Area} = PL \times PN = \sqrt{x^2 + 4} \times \frac{4}{\sqrt{x^2 + 4}} = 4$$Thus, the area of the rectangle is 4 cm$^2$.

This problem can be modeled as finding a spanning tree in a graph where the zones ($Z1$ to $Z5$) are nodes and the river segments ($P, Q, R, S, T, U, V$) represent potential edges (bridges) connecting adjacent zones.1. Identify Adjacencies (Edges):
Based on the topology of the diagram:

• P connects $Z1$ and $Z2$.
• S connects $Z1$ and $Z5$.
• Q connects $Z1$ and $Z3$ (separating the left zone from the top-right zone).
• R connects $Z2$ and $Z3$.
• T connects $Z3$ and $Z5$.
• V connects $Z3$ and $Z4$.
• U connects $Z4$ and $Z5$.

2. Goal:
Connect all 5 nodes ($Z1, Z2, Z3, Z4, Z5$) using the least number of bridges. For a graph with $n=5$ nodes, the minimum number of edges required to connect them is $n-1 = 4$ (a spanning tree).3. Evaluate Options:

• (A) Bridges on P, Q, and T:
• Edges: $(1,2), (1,3), (3,5)$.
• Connected components: $\{1, 2, 3, 5\}$. Node $4$ is isolated.
• Incorrect.
• (B) Bridges on P, Q, S, and T:
• Edges: $(1,2), (1,3), (1,5), (3,5)$.
• This creates a cycle $1-3-5-1$. Node $4$ is still isolated.
• Incorrect.
• (C) Bridges on Q, R, T, and V:
• Edges: $Q(1,3), R(2,3), T(3,5), V(3,4)$.
• All edges connect to node $3$ (forming a star-like structure centered at $Z3$).
• Connectivity: $1-3, 2-3, 5-3, 4-3$. All nodes $\{1, 2, 3, 4, 5\}$ are connected.
• Number of bridges: 4 (minimal).
• Correct.
• (D) Bridges on P, Q, S, U, and V:
• Number of bridges: 5. Since 4 is sufficient, 5 is not the least number.
• Incorrect.

Given the matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}$.Step 1: Calculate $A^2$.$$A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (2)(2) & (1)(2) + (2)(-1) \\ (2)(1) + (-1)(2) & (2)(2) + (-1)(-1) \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5I$$where $I$ is the $2 \times 2$ identity matrix.Step 2: Calculate $A^4$.$$A^4 = (A^2)^2 = (5I)^2 = 25I^2 = 25I$$Step 3: Calculate $A^8$.$$A^8 = (A^4)^2 = (25I)^2 = 625I^2 = 625I = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}$$Therefore, the correct option is (C).

To find the value of $x > 1$ satisfying the equation $\int_{1}^{x} t \ln t \, dt = \frac{1}{4}$, we evaluate the integral using integration by parts.Let $u = \ln t$ and $dv = t \, dt$. Then $du = \frac{1}{t} dt$ and $v = \frac{t^2}{2}$.The indefinite integral is:$$\int t \ln t \, dt = uv - \int v \, du = \frac{t^2}{2} \ln t - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt = \frac{t^2}{2} \ln t - \frac{1}{2} \int t \, dt = \frac{t^2}{2} \ln t - \frac{t^2}{4}$$Now, apply the limits from $1$ to $x$:$$\left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_{1}^{x} = \left( \frac{x^2}{2} \ln x - \frac{x^2}{4} \right) - \left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right)$$Since $\ln 1 = 0$, the expression simplifies to:$$\frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4}$$Setting this equal to the given value $\frac{1}{4}$:$$\frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} = \frac{1}{4}$$$$\frac{x^2}{2} \ln x - \frac{x^2}{4} = 0$$$$\frac{x^2}{4} (2 \ln x - 1) = 0$$Since $x > 1$, $x^2 \neq 0$. Therefore:$$2 \ln x - 1 = 0$$$$\ln x = \frac{1}{2}$$$$x = e^{1/2} = \sqrt{e}$$Thus, the correct option is (A).

Let $I$ be the number of internal nodes (nodes with 2 children) and $L$ be the number of leaf nodes (nodes with 0 children).
In a binary tree where every node has either 0 or 2 children (a full or strict binary tree), the relationship between the number of leaf nodes and internal nodes is given by:$$L = I + 1$$The total number of nodes $n$ is the sum of leaf nodes and internal nodes:$$n = L + I$$Substituting $L = I + 1$ into the equation for $n$:$$n = (I + 1) + I$$$$n = 2I + 1$$We need to find the number of nodes with exactly two children, which is $I$. Rearranging the equation to solve for $I$:$$2I = n - 1$$$$I = \frac{n - 1}{2}$$Thus, the number of nodes with exactly two children is $\frac{n-1}{2}$.

Given the equation $L^2 = L^{-1}$. Since $L$ is a non-singular matrix, we can multiply both sides by $L$ to obtain:$$L^3 = L \cdot L^{-1} = I$$where $I$ is the identity matrix of order 3.Now, let's evaluate the matrix $M$:$$M = L^8 = L^6 \cdot L^2 = (L^3)^2 \cdot L^2 = I^2 \cdot L^2 = I \cdot L^2 = L^2$$We are given that $N = L^2$. Therefore, the matrix $(M - N)$ is:$$M - N = L^2 - L^2 = O$$where $O$ is the $3 \times 3$ zero matrix.The determinant of a zero matrix is always 0:$$\det(M - N) = \det(O) = 0$$Hence, the correct option is (A).

The statement in option (A) corresponds to the Principle of Mathematical Induction.For a predicate $P(x)$ over the natural numbers (typically starting from 0):If both conditions hold, then $P(x)$ is true for all natural numbers $x$. This is exactly what option (A) states: $(P(0) \wedge (\forall x [P(x) \Rightarrow P(x+1)])) \Rightarrow (\forall x P(x))$.Analysis of other options:

• (B) suggests backward induction from 0, which would imply properties for negative integers, not covering the natural numbers $x > 0$.
• (C) suggests backward induction starting from 1000. This would prove $P(x)$ for $0 \le x \le 1000$, but not for $x > 1000$.
• (D) suggests forward induction starting from 1000. This would prove $P(x)$ for all $x \ge 1000$, but not for $0 \le x < 1000$.

Statement (i) is TRUE. The Address Resolution Protocol (ARP) is used to map a network layer address (like an IPv4 address) to a link layer address (like a MAC address).Statement (ii) is FALSE. TCP supports piggybacking, where an acknowledgment for a received packet can be included in the header of an outgoing data packet. Therefore, a single TCP segment can carry both data (payload) and an acknowledgment (in the ACK field of the header).

Distance Vector Routing protocols (e.g., RIP) typically use the Bellman-Ford algorithm (or a variant like split horizon with poison reverse) to determine the shortest path.Link State Routing protocols (e.g., OSPF, IS-IS) typically use Dijkstra's algorithm (Shortest Path First) to compute the routing table.Therefore, the correct matching is:
(i) Distance vector routing $\rightarrow$ (a) Bellman-Ford
(ii) Link state routing $\rightarrow$ (b) Dijkstra

The Protocol field in the IPv4 header identifies the protocol used in the data portion of the IP datagram.Thus, RIP is not a possible candidate for X.

The program defines a function stringcopy that copies characters from the source pointer t to the destination pointer s as long as *t is not the null terminator (\0).Inside main:

• The array contents (indices 0-14) are: @, #, H, e, l, l, o, , W, o, r, l, d, !, \0.

2. stringcopy(a, a + 2) is called.

• s points to a[0] ('@').
• t points to a[2] ('H').

3. The while(*t) loop executes:

• It copies characters from t to s and increments both pointers.
• It copies the substring "Hello World!" (length 12) from indices 2-13 to indices 0-11.
• a[0] becomes 'H', a[1] becomes 'e', ..., a[11] becomes '!'.

4. Crucially, the loop condition while(*t) checks the value before copying. When t reaches index 14 (the null terminator \0), the condition fails, and the loop terminates. The assignment *s++ = *t++ is NOT executed for the null terminator.

• Original a[12] was 'd'.
• Original a[13] was '!'.
• Original a[14] was \0.

6. The final string in a is:

• Indices 0-11: Hello World!
• Index 12: d (unchanged)
• Index 13: ! (unchanged)
• Index 14: \0 (unchanged)
• Result: "Hello World!d!"

7. printf("%s\n", a) prints the string up to the first null terminator.

• Output: Hello World!d!

To find an element that is NOT the largest in a list of $N$ distinct integers, we simply need to find an element that is smaller than at least one other element in the list.Algorithm:

• If $a < b$, then $a$ is definitely not the largest (since $b$ is larger).
• If $b < a$, then $b$ is definitely not the largest (since $a$ is larger).

In either case, we have identified an element that is not the largest using exactly 1 comparison.

Backpatching is a technique used in the intermediate code generation phase of a compiler. It is primarily used to generate code for boolean expressions and flow-of-control statements in a single pass.When generating code for boolean expressions (like a < b or A || B) and control flow statements (like if, while), the compiler often needs to generate jump instructions. In a single-pass compiler, the target address (label) of a forward jump is not known at the moment the jump instruction is generated.Backpatching solves this by generating the jump instruction with an empty or placeholder target and adding the location of this instruction to a list. Later, when the target label is determined, the compiler goes back and fills in (backpatches) the correct address in all the instructions on the list.Therefore:
(I) is TRUE: Backpatching allows single-pass generation for Boolean expressions (short-circuit evaluation involves jumps).
(II) is TRUE: Backpatching allows single-pass generation for flow-of-control statements.Hence, both statements are correct.

To determine if the rules are S-attributed or L-attributed, we analyze the dependencies of the attributes:Definitions:

• S-attributed: Only uses synthesized attributes (attributes of the LHS depend on attributes of the RHS symbols).
• L-attributed: Attributes can be synthesized or inherited. Inherited attributes of a symbol on the RHS can only depend on attributes of the LHS (parent) or attributes of symbols to its left.

Analysis of Rule 1: $R \rightarrow AB \{B.i = R.i - 1; A.i = B.i; R.i = A.i + 1;\}$

• $B.i = R.i - 1$: $B$ inherits from parent $R$. Allowed in L-attributed.
• $A.i = B.i$: $A$ inherits from $B$. In the production R → AB, $B$ is to the right of $A$. An inherited attribute depending on a right sibling is NOT allowed in L-attributed definitions.
• Since it uses inherited attributes, it is NOT S-attributed.
• Conclusion: Rule 1 is neither S-attributed nor L-attributed.

Analysis of Rule 2: $P \rightarrow CD \{P.i = C.i + D.i; D.i = C.i + 2;\}$

• $P.i = C.i + D.i$: Synthesized attribute. Allowed in both.
• $D.i = C.i + 2$: $D$ inherits from $C$. In P → CD, $C$ is to the left of $D$. This dependency is allowed in L-attributed definitions.
• Since $D.i$ is an inherited attribute (computed before the node is fully processed), it is NOT S-attributed.
• Conclusion: Rule 2 is L-attributed but not S-attributed.

Analysis of Rule 3: $Q \rightarrow EF \{Q.i = E.i + F.i;\}$

• $Q.i = E.i + F.i$: Synthesized attribute. No inherited attributes are present.
• Conclusion: Rule 3 is S-attributed (and by definition, all S-attributed definitions are also L-attributed).

Matching these conclusions with the options:

• Rule 1: Neither S nor L.
• Rule 2: Not S, is L.
• Rule 3: S and L.

This corresponds to Option (C).

• Ethernet MTU: The maximum data field in an Ethernet frame is 1500 bytes. This is the Maximum Transmission Unit (MTU) for the IP packet.
• IPv4 Header: The problem states no options are used, so the standard header size is $20$ bytes.
• Max IP Payload: The maximum amount of data an IP packet can carry per fragment is $MTU - IP\_Header = 1500 - 20 = 1480$ bytes.
• UDP Header: A standard UDP header is $8$ bytes.
• UDP Payload (Data): Given as $7488$ bytes.

2. Calculate Total Data for IP Layer:

• The IP layer encapsulates the entire UDP segment (Header + Data).
• Total IP Payload = UDP Header + UDP Data = $8 + 7488 = 7496$ bytes.

3. Calculate Number of Fragments:

• Total data to be fragmented = $7496$ bytes.
• Max data per fragment = $1480$ bytes.
• Number of fragments = $\lceil \frac{7496}{1480} \rceil = \lceil 5.065 \rceil = 6$ fragments.

4. Calculate Size of the Last Fragment:

• Data carried in the first 5 fragments = $5 \times 1480 = 7400$ bytes.
• Remaining data for the 6th fragment = $7496 - 7400 = 96$ bytes.
• Total size of the last fragment = IP Header + Remaining Data = $20 + 96 = 116$ bytes.

Thus, there are 6 fragments and the last fragment size is 116 bytes.

A Deterministic Pushdown Automaton (DPDA) accepts the class of Deterministic Context-Free Languages (DCFLs).