---

Solution

Period $T=100\ ms=0.1\ sec$

Frequency:

$f=\frac{1}{T}=\frac{1}{0.1}=10\ Hz$

Now convert into kHz:

$10\ Hz=\frac{10}{1000}\ kHz=0.01\ kHz=10^{-2}\ kHz$

---

Solution

Abstraction means hiding or removing unnecessary details and keeping only important information.

So, removing details from a state representation is called abstraction.

---

Solution

AI problem solving commonly uses heuristic search.

In AI, solution steps may not always be explicit, and knowledge may be imprecise.

Repetition mechanism is mainly a feature of conventional algorithmic programming, not a special feature of AI problem solving.

---

Solution

“Some” means there exists at least one person.

“my friends” means $f(x)$.

“not perfect” means $\neg p(x)$.

So, the statement becomes:

$\exists x(f(x)\land \neg p(x))$

---

Solution

In CNF, formula is written as conjunction of clauses.

Each clause is a disjunction of literals.

Example:

$(p\vee \neg q)\wedge(r\vee s)$

Here, $(p\vee \neg q)$ and $(r\vee s)$ are clauses.

So, clauses in CNF are disjunction of literals.

---

Solution

For reflexive relation, every element must have pair $(x,x)$.

But $(c,c)$ and $(d,d)$ are missing.

So, relation is not reflexive.

For antisymmetric relation, if $(x,y)$ and $(y,x)$ both exist, then $x=y$.

Here, no reverse pair exists for $(c,d),(c,g),(d,g)$.

So, relation is antisymmetric.

For symmetric relation, if $(x,y)$ exists, then $(y,x)$ must also exist.

But $(c,d)$ exists and $(d,c)$ does not exist.

So, relation is not symmetric.

Therefore, relation satisfies B and not A.

---

Solution

From $\forall x(P(x)\vee Q(x))$, we can get:

$P(c)\vee Q(c)$

So, step (B) is correct.

But from $P(c)\vee Q(c)$, we cannot conclude $P(c)$ alone.

Also, we cannot conclude $Q(c)$ alone.

So, steps (C) and (E) are not correct inferences.

Step (G) would be correct only if (D) and (F) were validly obtained.

Hence, the main error is in steps (C) and (E).

---

Solution

Symbol table stores information about identifiers like name, type, scope, address and other attributes.

It helps in storage allocation.

It helps in checking type compatibility.

It also helps in avoiding or suppressing duplicate error messages for the same identifier.

So, all A, B and C are correct.

---

Solution

Given minterms are:

$0,1,2,3,6,12,13,14,15$

$A'B'$ covers minterms $0,1,2,3$

$AB$ covers minterms $12,13,14,15$

Remaining minterm is $6$

Option (B):

$A'CD'$ covers minterms $2,6$

Both are present in the function.

So, option (B) is correct.

Option (D):

$BCD'$ covers minterms $6,14$

Both are present in the function.

So, option (D) is also correct.

Options (A) and (C) include minterm $4$, which is not present in the function.

---

Solution

In a k-segment pipeline, theoretical maximum speedup is $k$.

So, statement A is true.

In practical cases, due to pipeline filling time, emptying time, delays and unequal stages, exact maximum speedup $k$ is not achieved.

So, statement B is true.

All pipeline segments need not take exactly the same computation time.

So, statement C is false.

---

Solution

A multiprocessor system has multiple processors controlled by one operating system.

So, statement A is true.

Multiple computers connected through communication lines describes a distributed system or multicomputer system, not a tightly coupled multiprocessor system.

So, statement B is false.

Multiprocessor systems are generally classified as MIMD systems, that means multiple instruction stream and multiple data stream.

So, statement C is true.

Therefore, A and C only are correct.

---

Solution

XML is used to store and transport data, not to display data like HTML.

So, statement A is false.

XML defines syntax for representing data, but meaning depends on application.

So, statement B is true.

XML tags are case-sensitive.

So, <Letter>, <LETTER>, and <letter> are different tags.

So, statement C is true.

---

Solution

In perspective projection, parallel lines generally meet at vanishing point.

So, statement A is true.

Perspective transformation involves perspective division, commonly by z-coordinate.

So, statement B is true.

Because division is involved, perspective transformation is not linear.

So, statement C is false.

---

Solution

Phong shading requires more computation than Gouraud shading.

So, statement A is false.

Gouraud shading computes color at vertices and interpolates color inside the polygon.

So, statement B is true.

Phong shading interpolates normal vectors and then computes illumination.

So, statement C is true.

---

Solution

Two operations conflict if:

They belong to different transactions.

They operate on the same data item.

At least one operation is write.

So, statements A, B and C are true.

Statement D is false because conflict needs same data item, not different data items.

---

Solution

Software reliability can be described with respect to different time measures such as execution time, calendar time and clock time.

So, all A, B and C are correct.

---

Solution

For object-behavioral modeling, analyst evaluates use-cases.

State transition diagram is also built to show object states and transitions.

The model is reviewed for accuracy and consistency.

Statement D is not correct as written, because events should drive the interaction sequence, not “do not derive” it.

---

Solution

Behavioral testing means black-box testing.

Equivalence partitioning is a black-box testing technique.

Graph-based testing method is also a behavioral testing technique.

Boundary value analysis is also a black-box testing technique.

Data flow testing and loop testing are white-box testing techniques.

Therefore, A, B and C only are correct.

---

Solution

For an algorithm to have tight running time $\Theta(g(n))$, worst-case should be $O(g(n))$ and best-case should be $\Omega(g(n))$.

This condition is given in statement A.

Statement B reverses the required bounds.

Statement C is false because $O(g(n))$ and $\Omega(g(n))$ are not always equal.

Statement D is false because a function cannot be both smaller order and larger order than the same $g(n)$.

---

Solution

(A) True, because every tree is bipartite, so it is 2-colorable.

(B) False, because a bipartite graph has no odd cycle, but it can have even cycles.

(C) True, because every bipartite graph is 2-colorable.

(D) True, by greedy coloring, any graph with maximum degree $d$ can be colored using at most $d+1$ colors.

(E) False, because having $O(|V|)$ edges does not guarantee coloring with only $O(\log |V|)$ colors.

So, incorrect statements are (B) and (E).

---

Solution

A hash function takes an input message of arbitrary length and produces a fixed length hash value.

So, statement (A) is true.

Statement (B) is false because hash function output is fixed length, not variable length.

Statement (C) is true because two different messages may produce the same hash value. This is called collision.

Therefore, true statements are (A) and (C) only.

---

Solution

For context free languages, equivalence problem is undecidable.

So,

$L(G_1)=L(G_2)$ is undecidable.

Also, inclusion problem for context free languages is undecidable.

So,

$L(G_2)\subseteq L(G_1)$ is undecidable.

Now,

$L(G_1)=R$

is also undecidable because if $R=\Sigma^*$, then this becomes CFL universality problem:

$L(G_1)=\Sigma^*$

which is undecidable.

Therefore, all three problems are undecidable.

---

Solution

For $L_1$:

$L_1=\{a^{2^z}\mid z\text{ is an integer}\}$

The possible lengths are:

$1,2,4,8,16,32,\ldots$

So,

$L_1=\{a,a^2,a^4,a^8,a^{16},\ldots\}$

This is not regular because the gap between consecutive string lengths keeps increasing.

So, $L_1$ is not regular.

For $L_2$:

$L_2=\{a^{2z}\mid z\ge 0\}$

The possible lengths are:

$0,2,4,6,8,\ldots$

So,

$L_2=\{\lambda,a^2,a^4,a^6,\ldots\}$

This language contains all strings of even length over alphabet $\{a\}$.

It can be written as:

$L_2=(aa)^*$

Since $(aa)^*$ is a regular expression, $L_2$ is regular.

For $L_3$:

$L_3=\{\omega\omega\mid \omega\in\{a,b\}^*\}$

In this language, the second half of the string must be exactly the same as the first half.

A finite automaton cannot remember the complete first half and compare it with the second half.

So, $L_3$ is not regular.

Therefore, only $L_2$ is regular.

---

Solution

A grammar is ambiguous if at least one string has more than one parse tree or more than one leftmost derivation.

For grammar $(A)$:

$S\rightarrow SS\mid aSb\mid bSa\mid \lambda$

The string $abab$ can be generated in more than one way.

First derivation:

$S\Rightarrow SS$

$\Rightarrow aSbS$

$\Rightarrow abS$

$\Rightarrow abaSb$

$\Rightarrow abab$

Second derivation:

$S\Rightarrow aSb$

$\Rightarrow abSab$

$\Rightarrow abab$

So, grammar $(A)$ is ambiguous.

For grammar $(B)$:

$S\rightarrow aSbS\mid bSaS\mid \lambda$

The string $abab$ can be generated in more than one way.

First derivation:

$S\Rightarrow aSbS$

$\Rightarrow abS$

$\Rightarrow abaSbS$

$\Rightarrow abab$

Second derivation:

$S\Rightarrow aSbS$

$\Rightarrow a(bSaS)bS$

$\Rightarrow abab$

So, grammar $(B)$ is ambiguous.

For grammar $(C)$:

$S\rightarrow aAB$

$A\rightarrow bBb$

$B\rightarrow A\mid \lambda$

The string $abbbb$ can be generated in two ways.

First way:

$S\Rightarrow aAB$

$\Rightarrow a(bb)(bb)$

$\Rightarrow abbbb$

Second way:

$S\Rightarrow aAB$

$\Rightarrow a(bbbb)(\lambda)$

$\Rightarrow abbbb$

So, grammar $(C)$ is also ambiguous.

Therefore, grammars $(A)$, $(B)$ and $(C)$ all are ambiguous.

---

Solution

In cloud computing, pooling and sharing of resources means multiple users can use shared computing resources like CPU, memory, storage and network.

Virtualization is the main concept that helps in pooling and sharing of resources.

Using virtualization, one physical machine can be divided into multiple virtual machines.

So, many users can share the same physical resources.

Virtual Memory is an operating system concept, not mainly a cloud computing concept for resource pooling.

Service means providing facilities like $IaaS$, $PaaS$ and $SaaS$, but the direct concept related to pooling and sharing is virtualization.

Therefore, only $(C)$ Virtualization is correct.

---

Solution

POP is used by a client to download emails from the mail server.

POP has two modes:

Delete mode and keep mode.

In Delete mode, after the client retrieves the mail, the mail is deleted from the mailbox.

So, statement $(B)$ is true.

In Delete mode, mail is not deleted before retrieval.

So, statement $(C)$ is false.

In keep mode, mail is not deleted after retrieval.

The mail remains in the mailbox even after retrieval.

So, statement $(E)$ is true.

Statement $(D)$ is false because in keep mode mail is not deleted before retrieval.

Therefore, true statements are $(A)$, $(B)$ and $(E)$.

---

Solution

A firewall can filter packets using information present in the packet header.

It can filter packets on the basis of Source IP Address.

So, statement (A) is true.

It can filter packets on the basis of Destination IP Address.

So, statement (B) is true.

It can filter packets on the basis of TCP Source Port.

So, statement (C) is true.

It can filter packets on the basis of UDP Source Port.

So, statement (D) is true.

It can also filter packets on the basis of TCP Destination Port.

So, statement (E) is true.

Therefore, all statements (A), (B), (C), (D) and (E) are correct.

---

Solution

Given statement says that a radical person is electable if he or she is conservative, otherwise not electable.

This means, for a radical person, electability depends exactly on conservativeness.

So, the correct logical form is:

$R\Rightarrow(E\Leftrightarrow C)$

Therefore, statement (B) is true.

Now,

$E\Leftrightarrow C$

can be written as:

$(E\Rightarrow C)\land(C\Rightarrow E)$

So,

$R\Rightarrow[(E\Rightarrow C)\land(C\Rightarrow E)]$

Now convert implications:

$E\Rightarrow C\equiv \neg E\lor C$

$C\Rightarrow E\equiv \neg C\lor E$

Therefore,

$R\Rightarrow[(\neg E\lor C)\land(\neg C\lor E)]$

This is equivalent to:

$(\neg R\lor \neg E\lor C)\land(\neg R\lor \neg C\lor E)$

So, statement (D) is also true.

Statement (A) is not correct because it says that a person is conservative if and only if the person is radical and electable, which is not the meaning of the given sentence.

Statement (C) is not the exact logical assertion of the given sentence.

Therefore, correct statements are (B) and (D).

---

Solution

Statement (A) is false.

Entailment means if $\alpha$ is true, then $\beta$ must be true in all models where $\alpha$ is true.

Correct definition is:

$\alpha\models\beta$

means $\beta$ is true in every model in which $\alpha$ is true.

But statement (A) says few worlds, so it is false.

Statement (B) is true.

Forward chaining and backward chaining are natural reasoning algorithms for Horn form knowledge bases.

Statement (C) is false.

Sound inference algorithm derives only those sentences that are logically entailed.

But deriving all entailed sentences is the property of completeness, not soundness.

Statement (D) is true.

Propositional logic does not scale well to environments of unbounded size because it cannot represent general rules using variables and quantifiers.

Therefore, true statements are (B) and (D).

---

Solution

Minimax search is generally implemented using depth-first search, not breadth-first search.

Statement (B) is true.

Alpha-beta pruning depends highly on the order in which states are examined.

If the best moves are examined first, more pruning is possible.

Statement (C) is true.

Alpha-beta pruning does not change the final result of minimax.

It only reduces the number of nodes examined.

So, alpha-beta search computes the same optimal move as minimax.

Statement (D) is false.

In games of imperfect information, optimal play requires reasoning about current and future belief states of each player.

Therefore, true statements are (B) and (C).

---

Solution

Given,

$R_1=\{(1,1),(2,2),(3,3)\}$

$R_2=\{(1,1),(1,2),(1,3),(1,4)\}$

Now,

$R_1\cup R_2$

means all elements of $R_1$ and $R_2$.

So,

$R_1\cup R_2=\{(1,1),(1,2),(1,3),(1,4),(2,2),(3,3)\}$

Therefore, (A) matches with I.

Now,

$R_1-R_2$

means elements that are in $R_1$ but not in $R_2$.

Common element is $(1,1)$.

So,

$R_1-R_2=\{(2,2),(3,3)\}$

Therefore, (B) matches with IV.

Now,

$R_1\cap R_2$

means common elements of $R_1$ and $R_2$.

So,

$R_1\cap R_2=\{(1,1)\}$

Therefore, (C) matches with II.

Now,

$R_2-R_1$

means elements that are in $R_2$ but not in $R_1$.

So,

$R_2-R_1=\{(1,2),(1,3),(1,4)\}$

Therefore, (D) matches with III.

Final matching is:

A-I, B-IV, C-II, D-III

---

Solution

Handshaking requires two control signals working in opposite directions.

So, (A) matches with II.

Programmed I/O requires the CPU to continuously check the I/O flag and perform the transfer.

So, (B) matches with IV.

In interrupt-initiated I/O, the I/O interface informs the CPU that the device is ready for transfer.

So, (C) matches with I.

An I/O processor has local memory and controls a large set of I/O devices.

So, (D) matches with III.

Final matching is:

A-II, B-IV, C-I, D-III

---

Solution

For (A),

$\left[[2,0,0],[0,2,0],[0,0,1]\right]$

This is a scaling matrix.

Here, $S_x=2$ and $S_y=2$.

So, (A) matches with (II).

For (B),

$\left[[1,0,2],[0,1,2],[0,0,1]\right]$

This is a translation matrix.

Here, $T_x=2$ and $T_y=2$.

So, (B) matches with (IV).

For (C),

$\left[[0.5,-0.866,0],[0.866,0.5,0],[0,0,1]\right]$

The standard rotation matrix is:

$\left[[\cos\theta,-\sin\theta,0],[\sin\theta,\cos\theta,0],[0,0,1]\right]$

Here,

$\cos 60^\circ=0.5$

and

$\sin 60^\circ=0.866$

So, this matrix represents rotation by $60^\circ$.

Therefore, (C) matches with (III).

For (D),

$\left[[1,2,0],[0,1,0],[0,0,1]\right]$

This is shear in $x$ direction.

Here, shear factor is $2$.

So, (D) matches with (I).

Final matching is:

A-II, B-IV, C-III, D-I

---

Solution

In E-R diagram, double rectangle represents weak entity type.

So, (A) matches with (II).

Oval with underlined attribute represents key attribute type.

So, (B) matches with (I).

Double oval represents multivalue attribute type.

So, (C) matches with (IV).

Double line between entity and relationship represents total participation of entity in a relation.

So, (D) matches with (III).

Final matching is:

A-II, B-I, C-IV, D-III

---

Solution

CMM has different maturity levels.

INITIAL level means the process is ad hoc.

So, (A) matches with IV.

REPEATABLE level focuses on project management.

So, (B) matches with III.

DEFINED level focuses on process definition.

So, (C) matches with II.

MANAGED level focuses on process measurement.

So, (D) matches with I.

Final matching is:

A-IV, B-III, C-II, D-I

---

Solution

Branch-and-bound keeps track of partial paths which can be candidates for further exploration.

So, (A) matches with I.

Steepest-accent hill climbing considers all possible moves from the current state and selects the best move.

So, (B) matches with IV.

Constraint satisfaction means finding state or states that satisfy a given set of constraints.

So, (C) matches with III.

Means-end-analysis detects the difference between current state and goal state.

So, (D) matches with II.

Final matching is:

A-I, B-IV, C-III, D-II

---

Solution

Topological sorting of a DAG takes $\theta(V+E)$ time.

So, (A) matches with III.

Bellman-Ford single-source shortest path algorithm takes $O(VE)$ time.

So, (C) matches with II.

Floyd-Warshall all pair shortest path algorithm takes $\theta(V^3)$ time.

So, (D) matches with IV.

Now, from the given matching list, Kruskal's MST algorithm is matched with $O(V+E)$.

So, (B) matches with I.

Final matching is:

A-III, B-I, C-II, D-IV

---

Solution

For (I),

$\overline{L_{REC}}\cup L_{RE}$

Recursive languages are closed under complement.

So, $\overline{L_{REC}}$ is also recursive.

Every recursive language is recursively enumerable.

Union of a recursively enumerable language with a recursive language is recursively enumerable.

So, (I) represents recursively enumerable language.

Therefore, (A) matches with I.

For (II),

$\overline{L_{CF}}\cup L_{REC}$

Every context free language is recursive.

Recursive languages are closed under complement.

So, $\overline{L_{CF}}$ is recursive.

Union of two recursive languages is recursive.

So, (II) represents recursive language.

Therefore, (B) matches with II.

For (III),

$L_R^*\cap L_{CF}$

Regular languages are closed under Kleene star.

So, $L_R^*$ is regular.

Intersection of a regular language with a context free language is context free.

So, (III) represents context free language.

Therefore, (C) matches with III.

Final matching is:

A-I, B-II, C-III

---

Solution

Serial Line IP $(SLIP)$ is used for transmitting IP packets over serial links.

It works at the Data Link layer.

So, (A) matches with III.

Border Gateway Protocol $(BGP)$ is used for routing between autonomous systems.

So, it is associated with the Network layer.

So, (B) matches with IV.

User Data Protocol $(UDP)$ is a Transport layer protocol.

So, (C) matches with II.

Simple Network Management Protocol $(SNMP)$ is an Application layer protocol.

So, (D) matches with I.

Final matching is:

A-III, B-IV, C-II, D-I

---

Solution

Greedy Best-First Search selects the node that appears closest to the goal according to heuristic value.

It may get trapped and can be incomplete even if the search space is finite.

So, (A) matches with II.

$A^*$ search is optimal and complete under proper conditions, but its computation and space complexity are very high because it stores many generated nodes.

So, (B) matches with IV.

Recursive Best-First Search uses limited memory and its space complexity is linear in depth.

So, (C) matches with I.

$SMA^*$ means Simplified Memory-Bounded $A^*$.

It is optimal if the optimal solution is reachable within available memory; otherwise, it returns the best reachable solution.

So, (D) matches with III.

Final matching is:

A-II, B-IV, C-I, D-III

---

Solution

Greedy Best-First Search selects the node that appears closest to the goal according to heuristic value.

It may get trapped and can be incomplete even if the search space is finite.

So, (A) matches with II.

$A^*$ search is optimal and complete under proper conditions, but its computation and space complexity are very high because it stores many generated nodes.

So, (B) matches with IV.

Recursive Best-First Search uses limited memory and its space complexity is linear in depth.

So, (C) matches with I.

$SMA^*$ means Simplified Memory-Bounded $A^*$.

It is optimal if the optimal solution is reachable within available memory; otherwise, it returns the best reachable solution.

So, (D) matches with III.

Final matching is:

A-II, B-IV, C-I, D-III

---

Solution

SISD means Single Instruction Single Data.

It is the simplest type of machine.

SIMD means Single Instruction Multiple Data.

It is more complex than SISD.

MIMD means Multiple Instruction Multiple Data.

It is the most complex because multiple processors can execute different instructions on different data.

So, descending order of complexity is:

$MIMD>SIMD>SISD$

According to given symbols:

$B>C>A$

---

Solution

For perspective projection only:

Perspective projection needs homogeneous coordinates to represent it using matrix form.

So, $(i)$ matches with D.

For orthographic projection only:

In orthographic projection, far away objects appear the same size as closer objects.

So, $(ii)$ matches with C.

For both orthographic and projective transformations:

Both transformations map straight lines to straight lines.

So, $(iii)$ matches with A.

For neither orthographic nor projective transformation:

Distances and angles are not preserved in general by these projections.

So, $(iv)$ matches with B.

Final order is:

D, C, A, B

---

Solution

The strength of cohesion from best to worst is generally:

Functional Cohesion

Sequential Cohesion

Communication Cohesion

Procedural Cohesion

Temporal Cohesion

Logical Cohesion

Coincidental Cohesion

Here, Functional Cohesion and Coincidental Cohesion are not given.

So, from the given options, best to worst order is:

Sequential Cohesion $\rightarrow$ Communication Cohesion $\rightarrow$ Procedural Cohesion $\rightarrow$ Temporal Cohesion $\rightarrow$ Logical Cohesion

According to given symbols:

$B\rightarrow C\rightarrow E\rightarrow D\rightarrow A$

---

Solution

---

Solution

For $r=a(b+a)^*$:

This represents all strings over $\{a,b\}$ that start with $a$.

So,

$L(r)=\{a,aa,ab,aaa,aab,aba,\ldots\}$

For $s=a(a+b)^+$:

Here, after first $a$, at least one more symbol from $\{a,b\}$ must be present.

So, $s$ represents all strings over $\{a,b\}$ that start with $a$ and have length at least $2$.

Therefore,

$L(s)\subseteq L(r)$

For $t=aa^*b$:

This represents strings of the form:

$ab,aab,aaab,aaaab,\ldots$

So, every string of $L(t)$ starts with $a$ and has length at least $2$.

Therefore,

$L(t)\subseteq L(s)$

Hence,

$L(r)\supseteq L(s)\supseteq L(t)$

---

Solution

Statement I:

We need to check whether $5$ divides $n^5-n$.

By Fermat's theorem,

$n^5\equiv n\pmod 5$

So,

$n^5-n\equiv 0\pmod 5$

Therefore, $5$ divides $n^5-n$.

So, Statement I is correct.

Statement II:

$n^3-n=n(n-1)(n+1)$

This is the product of three consecutive integers.

In any three consecutive integers, one number is divisible by $3$ and at least one number is divisible by $2$.

So, the product is divisible by $6$.

Therefore, $6$ divides $n^3-n$.

So, Statement II is also correct.

---

Solution

Statement I is true.

Hardwired control unit uses fixed logic circuits.

So, it is faster than microprogrammed control unit and can be optimized for fast operation.

Statement II is also true.

In indirect addressing mode, the address field of the instruction contains the address of the memory location where the effective address is stored.

So, first memory reference is used to get the effective address.

Second memory reference is used to fetch the actual operand.

Therefore, indirect addressing needs two memory references to fetch the operand.

---

Solution

Solution:
Statement I is false.

Bezier curves do not interpolate all of their control points.

A Bezier curve passes through the first and last control points, but the intermediate control points generally control the shape of the curve.

So, it does not pass through all control points.

Statement II is true.

A cubic Bezier curve has degree $3$.

For a Bezier curve of degree $n$, number of control points is $n+1$.

So, for cubic Bezier curve:

$n+1=3+1=4$

Therefore, a cubic Bezier curve has four control points.

---

Solution

Statement I is true.

If $V_1$ and $V_2$ have correlation $1$, then they have perfect positive linear relationship.

So, all data points lie on a straight line.

In $k$ means clustering, centroids are calculated as the mean of points in each cluster.

Mean of points lying on a straight line also lies on the same straight line.

Therefore, the cluster centroids will be in a straight line.

Statement II is false.

If $V_1$ and $V_2$ have correlation $0$, then there is no linear relationship between them.

So, the data points are not necessarily on a straight line.

Therefore, the cluster centroids will not necessarily be in a straight line.

/* P1 */int main() {    fork();    fork();    fork();    print("Happy\n");}

/* P2 */int main() {    fork();    print("Happy\n");    fork();    print("Happy\n");    fork();    print("Happy\n");}

---

Solution

For program $P1$:

There are $3$ fork calls before the print statement.

Each $fork()$ doubles the number of processes.

So, total number of processes after $3$ fork calls is:

$2^3=8$

Each process executes the print statement.

Therefore, $P1$ displays "Happy" $8$ times.

So, Statement I is true.

For program $P2$:

Initially, there is $1$ process.

After first $fork()$:

Number of processes $=2$

So, first print statement executes $2$ times.

After second $fork()$:

Number of processes $=4$

So, second print statement executes $4$ times.

After third $fork()$:

Number of processes $=8$

So, third print statement executes $8$ times.

Total number of times "Happy" is printed:

$2+4+8=14$

So, $P2$ displays "Happy" $14$ times, not $12$ times.

Therefore, Statement II is false.

---

Solution

Statement I is true.

Quality control is related to inspections, reviews and testing of software products.

Its main purpose is to check whether the work product satisfies the given requirements or not.

Statement II is also true.

Quality assurance is a management activity.

It includes auditing and reporting functions to ensure that proper software processes are being followed.

Therefore, both statements are true.

---

Solution

Statement I is true.

In DFS of an undirected graph, there are no cross edges.

Edges are generally tree edges or back edges.

So, no edge of $G$ is a cross edge with respect to DFS tree $T_D$.

Statement II is false.

In BFS tree, for every edge $(u,v)$, the depths of $u$ and $v$ differ by at most $1$.

So, the correct condition is:

$|i-j|\le 1$

But the statement says:

$|i-j|=1$

This is not always true because two adjacent vertices can also be at the same level in a BFS tree.

Therefore, Statement I is true and Statement II is false.

---

Solution

Statement I is true.

For context sensitive languages, the emptiness problem is undecidable.

Since context sensitive languages are closed under intersection, checking whether:

$L_1\cap L_2=\phi$

is also undecidable.

So, Statement I is true.

Statement II is also true.

Membership problem for context sensitive languages is decidable.

This is because context sensitive languages are accepted by Linear Bounded Automata, and for a fixed input there are only finitely many configurations.

Therefore, we can decide whether $W\in L$ or not.

So, both statements are true.

---

Solution

Statement I is true.

Nyquist theorem gives the limit for a noiseless channel.

It says that for a channel of bandwidth $H$, the maximum number of independent samples per second is $2H$.

So, Statement I is true.

Statement II is also true.

Shannon capacity theorem gives the maximum data rate of a noisy channel as:

$C=H\log_2\left(1+\frac{S}{N}\right)$

where $H$ is bandwidth and $\frac{S}{N}$ is signal-to-noise ratio.

Therefore, both statements are true.

---

Solution

Statement I is true.

Caesar Cipher is a substitution cipher.

In Caesar Cipher, each letter of the plain text is replaced by another letter by shifting it by a fixed number of positions.

So, Statement I is true.

Statement II is also true.

Diffie-Hellman algorithm is used for secure key exchange between two parties.

It helps two users to agree on a common secret key over an insecure channel.

So, Statement II is true.

Therefore, both statements are true.

---

Solution

Statement I is incorrect.

A genetic algorithm maintains a population of states, but it is generally considered a variant of stochastic beam search, not simple stochastic hill-climbing.

Hill-climbing usually works with one current state, while genetic algorithm works with a population of states and uses selection, crossover and mutation.

So, Statement I is false.

Statement II is true.

In nondeterministic environments, actions may have more than one possible outcome.

AND-OR search is used to handle such situations.

It generates contingent plans, where the next action depends on the outcome that actually occurs.

So, Statement II is true.

Therefore, Statement I is incorrect but Statement II is true.

---

Solution

Sector size $=512$ bytes

Number of sectors per track $=50$

So, capacity of one track is:

$T=512\times 50$

$T=25600$ bytes

Since,

$1K=1024$ bytes

So,

$T=\frac{25600}{1024}=25K$

Now, each surface has $2000$ tracks.

Capacity of each surface is:

$S=25K\times 2000$

$S=50000K$

Therefore,

$(T,S)=(25K,50000K)$

---

Solution

Given,

Sector size $=512$ bytes

Sectors per track $=50$

Tracks per surface $=2000$

Number of double-sided platters $=5$

So, number of surfaces:

$5\times 2=10$

Capacity of one track:

$512\times 50=25600$ bytes

Since,

$1K=1024$ bytes

So,

$\frac{25600}{1024}=25K$

Capacity of one surface:

$25K\times 2000=50000K$

Capacity of total disk:

$50000K\times 10=500000K$

---

Solution

Statement I is true.

Number of cylinders is equal to the number of tracks per surface.

Given,

Tracks per surface $=2000$

So, total number of cylinders $=2000$.

Therefore, Statement I is true.

Statement II is false.

Given sector size is:

$512$ bytes

A valid block size should be a multiple of sector size.

Now,

$51200=512\times 100$

So, $51200$ bytes is a valid block size.

But Statement II says that $51200$ bytes is not a valid block size.

Therefore, Statement II is false.

---

Solution

Given,

Rotational speed $=5400$ rpm

It means:

$5400$ revolutions in $60$ seconds

So, revolutions per second:

$\frac{5400}{60}=90$

Time taken for one revolution:

$\frac{1}{90}$ seconds

$=0.0111$ seconds

Maximum rotational delay is equal to time for one full revolution.

So, maximum rotational delay is approximately:

$0.011$ seconds

---

Solution

From previous calculation, capacity of one track is:

$25K$ bytes

Given rotational speed:

$5400$ rpm

Revolutions per second:

$\frac{5400}{60}=90$

If one track of data is transferred per revolution, then data transferred per second is:

$25K\times 90$

$=2250K$ bytes/second

So, data transfer rate is:

$2250$ KBytes/second

---

Solution

We need to retrieve distinct employee IDs.

So, we must use:

SELECT DISTINCT EMPID

Now, the project number can be $20$, $30$ or $40$.

For checking multiple possible values in SQL, we use IN operator.

So, correct condition is:

WHERE PROJNO IN (20,30,40)

Therefore, correct SQL query is:

SELECT DISTINCT EMPID
FROM PROJECTWORK
WHERE PROJNO IN (20,30,40);

Option (A) is incorrect because PROJNO=(20,30,40) is not valid SQL syntax.

Option (B) is incorrect because it does not use DISTINCT.

Option (D) is incorrect because PROJNO=20,30,40 is not valid SQL syntax.

Correct Answer: (C)
SELECT DISTINCT EMPID
FROM PROJECTWORK
WHERE PROJNO IN (20,30,40);

---

Solution

We need to find salary details of employees who belong to the English department.

For this, EMPLOYEE table and DEPARTMENT table must be joined.

In Statement I:

DEPTNO=DID

This condition joins EMPLOYEE table with DEPARTMENT table.

Also,

DNAME='ENGLISH'

filters only English department employees.

So, Statement I is correct.

In Statement II:

Only condition is:

DNAME='ENGLISH'

There is no join condition between EMPLOYEE and DEPARTMENT.

So, it will not correctly connect employees with their department.

It may produce incorrect result due to Cartesian product.

Therefore, Statement II is false.

---

Solution

We need employees whose salary is greater than the salary of all employees in department number $20$.

For comparing one value with all values returned by a subquery, we use:

SALARY > ALL (subquery)

So, the correct condition is:

SALARY > ALL (SELECT SALARY FROM EMPLOYEE WHERE DEPTNO = 20)

Also, the question asks to order results by employee ID.

So, ORDER BY EID is also required.

Option (A) is incorrect because the subquery may return more than one salary value. So, direct comparison using $>$ is not correct.

Option (B) is incorrect because it also uses direct comparison with a subquery that may return multiple values, and it does not use ORDER BY EID.

Option (C) is correct because it uses ALL and also orders the result by EID.

Therefore, only query (C) is correct.

---

Solution

To drop a column from a table, correct SQL syntax is:

ALTER TABLE table_name DROP column_name;

In Statement (A):

ALTER TABLE EMPLOYEE DROP SALARY CASCADE;

This query is valid because it uses ALTER TABLE and drops the SALARY column.

CASCADE means if there are dependent objects, they may also be removed.

So, Statement (A) is correct.

In Statement (B):

ALTER TABLE EMPLOYEE DROP SALARY RESTRICT;

This query is also valid because it uses ALTER TABLE and drops the SALARY column.

RESTRICT means the column will be dropped only if there is no dependent object.

Here, SALARY is not used as a primary key or foreign key.

So, Statement (B) is also correct.

In Statement (C):

ALTER EMPLOYEE DROP SALARY;

This query is incorrect because the keyword TABLE is missing.

Correct syntax should have ALTER TABLE, not only ALTER.

So, Statement (C) is incorrect.

Therefore, correct queries are (A) and (B) only.

---

Solution

We need to increase the salary of all employees who belong to the 'SCIENCE' department.

The salary is stored in EMPLOYEE table.

The department name is stored in DEPARTMENT table.

So, first we need to find the DID of the 'SCIENCE' department.

This can be done by:

SELECT DID FROM DEPARTMENT WHERE DNAME='SCIENCE'

Then update salary of employees whose DEPTNO matches that DID.

So, correct query is:

UPDATE EMPLOYEE
SET SALARY = SALARY * 1.20
WHERE DEPTNO IN (SELECT DID FROM DEPARTMENT WHERE DNAME='SCIENCE');

Therefore, query (A) is correct.

Query (B) is incorrect because UPDATE TABLE is not valid syntax.

Also, DNAME is not present in EMPLOYEE table.

Query (C) is incorrect because ALTER TABLE is used to change table structure, not to update data.

Also, it compares DEPTNO with DNAME, which is wrong.

Therefore, only query (A) is correct.

---

Solution

Positive odd integers not exceeding $100$ are:

$1,3,5,\ldots,99$

Number of odd integers $=50$

Perfect squares not exceeding $100$ are:

$1^2,2^2,3^2,\ldots,10^2$

Number of perfect squares $=10$

Odd perfect squares are:

$1^2,3^2,5^2,7^2,9^2$

Number of odd perfect squares $=5$

Required number:

$50+10-5=55$

---

Solution

Books are identical and boxes are also identical.

So, we need to find the number of partitions of $6$ into at most $4$ parts.

Possible distributions are:

$6$

$5+1$

$4+2$

$4+1+1$

$3+3$

$3+2+1$

$3+1+1+1$

$2+2+2$

$2+2+1+1$

Total ways $=9$

---

Solution

Check option (a):

$(p\to r)\land(q\to r)$

$=(\neg p\lor r)\land(\neg q\lor r)$

$=(\neg p\land\neg q)\lor r$

$=\neg(p\lor q)\lor r$

$=(p\lor q)\to r$

So, option (a) is logically equivalent.

Option (b) is also equivalent because if $p$ and $q$ have same truth value, then $\neg p$ and $\neg q$ also have same truth value.

Option (c) is the standard form of biconditional:

$p\leftrightarrow q\equiv(p\land q)\lor(\neg p\land\neg q)$

So, option (c) is logically equivalent.

Now check option (d):

$((p\land q)\to r)=\neg(p\land q)\lor r=\neg p\lor\neg q\lor r$

But,

$(p\to r)\land(q\to r)=(\neg p\lor r)\land(\neg q\lor r)$

Both are not always same.

---

Solution

Corner points are checked for maximum value.

At $(0,0)$:

$z=2(0)+3(0)=0$

At $(2,0)$:

$z=2(2)+3(0)=4$

At $(0,2.5)$:

$z=2(0)+3(2.5)=7.5$

Now solve intersection:

$2x_1+x_2=4$

$x_1+2x_2=5$

From first equation,

$x_2=4-2x_1$

Put in second equation:

$x_1+2(4-2x_1)=5$

$x_1+8-4x_1=5$

$-3x_1=-3$

$x_1=1$

Then,

$x_2=4-2(1)=2$

At $(1,2)$:

$z=2(1)+3(2)=2+6=8$

Maximum value is $8$.

---

Solution

Main memory size $=2^{16}$ bytes

So, address size $=16$ bits.

Block size $=8$ bytes $=2^3$ bytes

So, word field bits $=3$

Cache has $32$ lines $=2^5$ lines

So, line field bits $=5$

Now,

Tag bits $=16-5-3=8$

Therefore, Tag, line and word bits are:

$8,5,3$

---

Solution

Initially,

$AC=0000H$

After $CLA$:

$AC=0000H$

After $ADD\ 217$:

$AC=0000H+1234H=1234H$

After $INC$:

$AC=1235H$

After $STA\ 217$:

$M[217]=1235H$

After $LDA\ 218$:

$AC=9CE2H$

After $CMA$:

$AC=\overline{9CE2H}=631DH$

Now execute $AND\ 217$:

$AC=631DH\ AND\ 1235H$

$631DH=0110\ 0011\ 0001\ 1101$

$1235H=0001\ 0010\ 0011\ 0101$

So,

$AC=0000\ 0010\ 0001\ 0101=0215H$

---

Solution

Non-pipeline time for $500$ tasks:

$500\times50=25000\ ns$

For pipeline:

Number of stages $=6$

Clock cycle $=10ns$

Pipeline time:

$(k+n-1)\times clock$

$=(6+500-1)\times10$

$=505\times10=5050\ ns$

Speedup:

$\frac{25000}{5050}=4.95$

---

Solution

Let radix be $r$.

In base $r$,

$10=r$

$26=2r+6$

Given roots are $4$ and $7$.

Sum of roots:

$4+7=11$

For equation $x^2-10x+26=0$, sum of roots $=10$ in base $r$.

So,

$r=11$

Also, product of roots:

$4\times7=28$

And,

$26=2r+6=2(11)+6=28$

So, radix is $11$.

---

Solution

Initially,

$p=1=a^0$

After $1$ iteration:

$p=a^1$

After $2$ iterations:

$p=a^2$

After $x$ iterations:

$p=a^x$

Loop stops when:

$p\ge b$

So,

$a^x\ge b$

Minimum such value of $x$ is:

$x=\lceil \log_a b\rceil$

---

Solution

The $<map>$ tag is used with the $<area>$ tag to create an image map.

Image map means different clickable regions can be defined inside an image.

---

Solution

$short\ s$: starts at offset $0$, size $2$, next offset $2$

$char\ c$: starts at offset $2$, size $1$, next offset $3$

$short\ t$: needs multiple of $2$, so $1$ byte padding, starts at offset $4$, next offset $6$

$char\ d$: starts at offset $6$, size $1$, next offset $7$

$double\ r$: needs multiple of $8$, so $1$ byte padding, starts at offset $8$, next offset $16$

$int\ i$: needs multiple of $4$, starts at offset $16$, next offset $20$

Struct size must be multiple of maximum alignment $8$.

So, $20$ is padded to $24$ bytes.

For $A[10]$:

$10\times24=240$ bytes

---

Solution

$f(1,3)=\frac{1}{3}-f(2,9)$

$=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+\cdots$

This is a geometric series with first term $\frac{1}{3}$ and common ratio $-\frac{2}{3}$.

Sum:

$\frac{\frac{1}{3}}{1-(-\frac{2}{3})}=\frac{\frac{1}{3}}{\frac{5}{3}}=\frac{1}{5}=0.2$

---

Solution

In Hadoop, DataNode stores actual data blocks, so it works as worker node.

NameNode manages metadata of files and blocks, so it works as master node.

Hence, DataNode and NameNode are Worker Node and Master Node respectively.

---

Solution

Attributes $Y$ and $Z$ are not present on RHS of any FD.

So, every candidate key must contain $Y$ and $Z$.

Now check:

$(UYZ)^+=\{U,Y,Z,V,W,X\}$, so $UYZ$ is a key.

$(VYZ)^+=\{V,Y,Z,W,X,U\}$, so $VYZ$ is a key.

$(XYZ)^+=\{X,Y,Z,U,V,W\}$, so $XYZ$ is a key.

Thus, candidate keys are:

$UYZ,\ VYZ,\ XYZ$

---

Solution

There are $3$ page frames.

For FIFO page replacement, page faults are:

$P=11$

For LRU page replacement, page faults are:

$Q=12$

So,

$(P,Q)=(11,12)$

---

Solution

Using SSTF, from cylinder $20$ the order is approximately:

$20\to20\to22\to10\to6\to2\to38\to40$

Total head movement:

$0+2+12+4+4+36+2=60$ cylinders

Time for one cylinder movement $=2$ ms

Total time:

$60\times2=120$ milliseconds

---

Solution

Block size $=2K=2048$ bytes

Disk address size $=32$ bits $=4$ bytes

Number of addresses in one block:

$\frac{2048}{4}=512$

Direct blocks $=12$

Single indirect blocks $=512$

Double indirect blocks:

$512\times512=262144$

Total data blocks:

$12+512+262144=262668$

Total file size:

$262668\times2048=537944064$ bytes

In MB:

$\frac{537944064}{1024\times1024}\approx513$ MB

---

Solution

TLB hit time:

$15+150=165$ ns

TLB miss time:

$15+150+150=315$ ns

Hit ratio $=0.8$, miss ratio $=0.2$

Effective memory access time:

$0.8(165)+0.2(315)$

$=132+63=195$ ns

---

Solution

Static view shows the fixed structure or high-level view of the system.

Use-Case diagram represents actors and use-cases of the system.

Collaboration, state chart and activity diagrams mainly show dynamic behavior.

Therefore, Use-Case diagram has a static view.

---

Solution

Restructuring means improving the internal structure of software without changing its external behavior.

It is done to prevent future problems and improve maintainability.

So, it comes under preventive maintenance.

---

Solution

Let LOC for $L_1=x$

Then LOC for $L_2=3x$

For $L_1$:

Development cost $=\frac{x}{1000}\times70000=70x$

Maintenance cost for $10$ years:

$10\times100000=1000000$

Total cost for $L_1$:

$70x+1000000$

For $L_2$:

Development cost $=\frac{3x}{1000}\times90000=270x$

Maintenance cost for $10$ years:

$10\times40000=400000$

Total cost for $L_2$:

$270x+400000$

Now both costs are same:

$70x+1000000=270x+400000$

$600000=200x$

$x=3000$

---

Solution

Total effort required:

$Effort=\frac{864}{12}=72$ person-months

Team size $=6$

Project duration:

$\frac{72}{6}=12$ months

Monthly team cost:

$50000+70000+2(60000)+2(60000)$

$=50000+70000+120000+120000=360000$

Total project cost:

$360000\times12=4320000$

So, projected cost is ₹43,20,000.

---

Solution

For a complete n-ary tree:

$L=(n-1)I+1$

Given:

$L=41,\ I=10$

So,

$41=(n-1)10+1$

$40=10(n-1)$

$n-1=4$

$n=5$

---

Solution

In a max heap, the largest element is at the root.

The smallest element must be among the leaf nodes.

There are approximately $\frac{n}{2}$ leaf nodes.

To find the smallest element, we may need to scan all leaf nodes.

So, time complexity is $O(n)$.

---

Solution

One second $=10^6$ microseconds.

For algorithm A:

$\log_2 n \le 10^6$

$n \le 2^{10^6}$

For algorithm B:

$\sqrt n \le 10^6$

$n \le (10^6)^2=10^{12}$

So, the largest sizes are $2^{10^6}$ and $10^{12}$.

---

Solution

We need shortest path from 1 to 100 using operations:

$i \to i+1$ or $i \to 3i$

A shortest path is:

$1 \to 3 \to 9 \to 10 \to 11 \to 33 \to 99 \to 100$

Number of edges $=7$

---

Solution

In $L_1$, after symbol $2$, we have $0^n1^n$.

But in $L_2$, after symbol $2$, no $0$ or $1$ can appear.

So, for intersection, $n=0$ in $L_1$.

Thus,

$L=\{0^m1^m2\mid m\ge0\}$

This language is context free but not regular.

---

Solution

$(a+b)^*$ generates all strings over $\{a,b\}$.

Now, $(a^*+b)^*$ also generates all strings over $\{a,b\}$ because $a^*$ can generate any number of $a$'s and $b$ is also available.

So,

$L_1=L_2=\{a,b\}^*$

---

Solution

Context free languages are closed under union.

So, statement (a) is true.

CFLs are not closed under intersection and complement in general.

Also, every CFL is not accepted by a deterministic PDA.

Therefore, only statement (a) is true.

---

Solution

Syntax tree, three address code and postfix notation are standard intermediate code forms.

Quadruple is mainly a representation method of three address code, not a separate intermediate code form.

---

Solution

Find Hamming distance from received word $0000000111$.

Distance from $0000011111$ is $2$.

Distance from $0000000000$ is $3$.

Distance from $1111100000$ is $8$.

Distance from $1111111111$ is $7$.

Minimum distance is with $0000011111$.

Hence, the original code word must be $0000011111$.

---

Solution

$n=pq=3\times11=33$

$\phi(n)=(p-1)(q-1)=2\times10=20$

Now,

$d\times e\equiv1\pmod{\phi(n)}$

$7e\equiv1\pmod{20}$

$e=3$, because $7\times3=21\equiv1\pmod{20}$

Now encrypt $M=19$:

$C=M^e\mod n$

$C=19^3\mod33$

$19^2=361\equiv31\pmod{33}$

$19^3=31\times19=589\equiv28\pmod{33}$

So, $e=3$ and encrypted value $=28$.

---

Solution

In stop and wait protocol, sender sends one frame and waits for acknowledgement.

After receiving acknowledgement, it sends the next frame.