GPSC ICT Officer (Class 2) Technical Model Paper with Solutions

Question 1 • Database Management Systems (Normalization)

EN: A relational schema R(A, B, C, D) has functional dependencies F = {A -> B, B -> C, C -> D}. What is the highest normal form satisfied by relation R?
GU: રિલેશનલ સ્કીમા R(A, B, C, D) માટે ફંક્શનલ ડિપેન્ડન્સી F = {A -> B, B -> C, C -> D} છે. આ સંબંધ R કયું સર્વોચ્ચ નોર્મલ ફોર્મ (Highest Normal Form) સંતોષે છે?

Option (A): First Normal Form (1NF) only
✅ Option (B): Second Normal Form (2NF)
Option (C): Third Normal Form (3NF)
Option (D): Boyce-Codd Normal Form (BCNF)

💡 In-Depth Analysis: The candidate key for R is {A}. In 2NF, no non-prime attribute should be partially dependent on a candidate key. Here, all LHS are single attributes or superkeys, so 2NF holds. However, transitive dependencies exist (A -> B and B -> C where C is non-prime, and B is not a superkey), violating 3NF. Thus, R is in 2NF.

Question 2 • DBMS (Concurrency Control & ACID)

EN: Which concurrency control protocol guarantees conflict serializability as well as freedom from deadlocks?
GU: કયો કન્કરન્સી કંટ્રોલ પ્રોટોકોલ કોન્ફ્લિક્ટ સીરિયલાઈઝેબિલિટી (Conflict Serializability) ની સાથે સાથે ડેડલોક-મુક્ત (Deadlock Freedom) ની ખાતરી આપે છે?

Option (A): Basic Two-Phase Locking (2PL)
✅ Option (B): Timestamp Ordering Protocol (Thomas Write Rule)
Option (C): Strict Two-Phase Locking
Option (D): Rigorous Two-Phase Locking

💡 In-Depth Analysis: Timestamp ordering protocols process transactions strictly by chronological timestamp order; conflicting operations execute according to timestamp priorities. Because no transaction waits for locks held by another transaction, deadlock cannot occur, although cascading aborts can occur unless enhanced.

Question 3 • Data Structures & Algorithms (Trees)

EN: What is the maximum number of rotations required during a single deletion operation in an AVL tree containing n nodes?
GU: n નોડ ધરાવતા AVL ટ્રીમાંથી એક નોડ ડિલીટ (Deletion) કરતી વખતે વધુમાં વધુ કેટલા રોટેશન્સ (Rotations) ની જરૂર પડી શકે છે?

Option (A): At most 1 single or double rotation
Option (B): At most 2 rotations
✅ Option (C): O(log n) rotations
Option (D): O(n) rotations

💡 In-Depth Analysis: In an AVL tree insertion, at most 1 single or double rotation (O(1)) restores balance. However, in deletion, rebalancing one subtree may reduce its height and unbalance its parent all the way up to the root, requiring up to O(log n) rotations.

Question 4 • Algorithms (Graph Traversal & Shortest Path)

EN: What is the time complexity of Dijkstra's single-source shortest path algorithm when implemented using a Fibonacci Heap for a graph with V vertices and E edges?
GU: V વર્ટિસિસ અને E એજીસ ધરાવતા ગ્રાફ માટે ફિબોનાકી હીપ (Fibonacci Heap) નો ઉપયોગ કરીને ડાઈકસ્ટ્રા અલ્ગોરિધમની ટાઈમ કોમ્પ્લેક્સિટી શું થાય?

Option (A): O(V^2)
Option (B): O((V + E) log V)
✅ Option (C): O(E + V log V)
Option (D): O(V log E)

💡 In-Depth Analysis: Dijkstra performs V Extract-Min operations (each O(log V) amortized) and E Decrease-Key operations (each O(1) amortized in Fibonacci heap). Hence total time = V * O(log V) + E * O(1) = O(E + V log V).

Question 5 • Operating Systems (Deadlock Avoidance)

EN: In Dijkstra's Banker's Algorithm for deadlock avoidance, if a system state is neither deadlocked nor in an unsafe state, which statement is strictly true?
GU: ડેડલોક નિવારણ માટે બેન્કર્સ અલ્ગોરિધમમાં, જો સિસ્ટમ સેફ સ્ટેટ (Safe State) માં હોય, તો કયું વિધાન સાચું છે?

✅ Option (A): There exists at least one safe sequence of processes such that all processes can finish execution
Option (B): No process can ever request further resources
Option (C): Every future sequence of resource allocations is guaranteed not to deadlock
Option (D): Paging thrashing is completely eliminated

💡 In-Depth Analysis: By definition, a state is safe if there exists an ordering (a safe sequence) such that for each Pi, the resources that Pi can still request can be satisfied by currently available resources plus resources held by all Pj (j < i).

Question 6 • Operating Systems (Virtual Memory & Belady's Anomaly)

EN: Belady's anomaly demonstrates that increasing the number of page frames can increase the number of page faults. Which of the following page replacement algorithms CANNOT suffer from Belady's anomaly?
GU: બેલાડી એનોમલી દર્શાવે છે કે પેજ ફ્રેમ વધારવા છતાં પેજ ફોલ્ટ વધી શકે છે. નીચેનામાંથી કયો અલ્ગોરિધમ બેલાડી એનોમલીથી સંપૂર્ણ મુક્ત છે?

Option (A): First-In First-Out (FIFO)
✅ Option (B): Least Recently Used (LRU)
Option (C): Second Chance Page Replacement
Option (D): Clock Algorithm

💡 In-Depth Analysis: Stack algorithms (such as LRU and Optimal) satisfy the inclusion property: the set of pages in memory for n frames is always a subset of the pages in memory for n+1 frames. Therefore, stack algorithms never suffer from Belady's Anomaly, unlike FIFO.

Question 7 • Computer Networks (Subnetting & CIDR)

EN: An IP network is assigned the CIDR block 192.168.10.0/27. What is the broadcast address and the total number of usable host IP addresses in this subnet?
GU: એક IP નેટવર્કને CIDR બ્લોક 192.168.10.0/27 ફાળવવામાં આવ્યો છે. આ સબનેટનું બ્રોડકાસ્ટ એડ્રેસ અને કુલ વાપરી શકાય તેવા (Usable) હોસ્ટ એડ્રેસ કેટલા થશે?

✅ Option (A): Broadcast: 192.168.10.31, Usable hosts: 30
Option (B): Broadcast: 192.168.10.32, Usable hosts: 32
Option (C): Broadcast: 192.168.10.255, Usable hosts: 27
Option (D): Broadcast: 192.168.10.63, Usable hosts: 62

💡 In-Depth Analysis: A /27 subnet mask has 32 - 27 = 5 host bits. Total addresses = 2^5 = 32 (range: 192.168.10.0 to 192.168.10.31). First address (.0) is Network ID, last address (.31) is Directed Broadcast Address. Usable hosts = 32 - 2 = 30.

Question 8 • Computer Networks (Transport Layer & TCP)

EN: In TCP Congestion Control, what action does TCP take when a packet loss is detected via 3 duplicate ACKs (Triple Duplicate ACK)?
GU: TCP કન્જેશન કંટ્રોલમાં, જ્યારે 3 ડુપ્લિકેટ ACK (Triple Duplicate ACK) મળે ત્યારે TCP કઈ પ્રક્રિયા કરે છે?

Option (A): Slow Start phase is restarted with Congestion Window (cwnd) = 1 MSS
✅ Option (B): Fast Retransmit and Fast Recovery are triggered; cwnd is halved
Option (C): The connection is terminated immediately with RST flag
Option (D): Window size is doubled exponentially

💡 In-Depth Analysis: 3 duplicate ACKs indicate mild packet loss (Fast Retransmit & Fast Recovery). The lost segment is retransmitted immediately without waiting for retransmission timeout (RTO), ssthresh is set to cwnd/2, and cwnd is set to ssthresh + 3 MSS, avoiding a return to Slow Start (cwnd = 1).

Question 9 • Cyber Security & IT Act 2000

EN: Under the Information Technology Act, 2000 of India, which section penalizes cheating by personation by using computer resources (identity theft/impersonation)?
GU: ભારતના ઇન્ફોર્મેશન ટેકનોલોજી એક્ટ, ૨૦૦૦ હેઠળ કમ્પ્યુટર સંસાધનો દ્વારા ખોટી ઓળખ ધારણ કરીને છેતરપિંડી (Cheating by personation) કરવા બદલ કઈ કલમ હેઠળ સજાની જોગવાઈ છે?

Option (A): Section 65
Option (B): Section 66C
✅ Option (C): Section 66D
Option (D): Section 70

💡 In-Depth Analysis: Section 66D prescribes punishment for cheating by personation by using computer resource (up to 3 years imprisonment and fine up to Rs 1 lakh). Section 66C deals specifically with identity theft (fraudulent use of electronic signature/password).

Question 10 • Cryptography (Public Key Infrastructure)

EN: In asymmetric public key cryptography, when Alice wants to send a digitally signed confidential message to Bob, which keys are used for signing and encryption respectively?
GU: અસમાન પબ્લિક કી ક્રિપ્ટોગ્રાફીમાં જ્યારે એલિસ બોબને ડિજિટલ સહી કરેલો ગુપ્ત સંદેશ મોકલવા માંગે ત્યારે સહી (Signing) અને એન્ક્રિપ્શન (Encryption) માટે અનુક્રમે કઈ કી વપરાય છે?

✅ Option (A): Signed with Alice's Private Key, Encrypted with Bob's Public Key
Option (B): Signed with Alice's Public Key, Encrypted with Bob's Private Key
Option (C): Signed with Bob's Public Key, Encrypted with Alice's Private Key
Option (D): Both signing and encryption use Bob's Public Key

💡 In-Depth Analysis: Digital signature guarantees non-repudiation and authenticity; hence Alice signs with her own Private Key (only she has it). Confidentiality ensures only Bob can read it; hence Alice encrypts the payload using Bob's Public Key (which only Bob's Private Key can decrypt).

Question 11 • Software Engineering (Design Patterns)

EN: Which design pattern is classified under Gang of Four (GoF) 'Structural' design patterns?
GU: ગેંગ ઓફ ફોર (GoF) ડિઝાઇન પેટર્ન્સમાં નીચેનામાંથી કઈ 'સ્ટ્રક્ચરલ' (Structural) ડિઝાઇન પેટર્ન છે?

Option (A): Abstract Factory
✅ Option (B): Adapter
Option (C): Singleton
Option (D): Observer

💡 In-Depth Analysis: GoF patterns are classified into: Creational (Singleton, Factory, Builder, Prototype), Structural (Adapter, Bridge, Composite, Decorator, Facade, Flyweight, Proxy), and Behavioral (Observer, Strategy, Command, Iterator, State).

Question 12 • Web Technologies (RESTful Architecture)

EN: According to HTTP/1.1 specifications (RFC 7231), which of the following HTTP methods is idempotent?
GU: HTTP/1.1 વિશિષ્ટતાઓ મુજબ, નીચેનામાંથી કઈ HTTP પદ્ધતિ આઇડેમ્પોટેન્ટ (Idempotent) છે?

Option (A): POST
✅ Option (B): PUT
Option (C): PATCH (in general non-idempotent updates)
Option (D): CONNECT

💡 In-Depth Analysis: An HTTP method is idempotent if the side-effects of N > 0 identical requests is the same as for a single request. GET, HEAD, PUT, and DELETE are idempotent. POST is NOT idempotent because repeating it creates multiple new resources.

Question 13 • Web Security (OWASP Top 10)

EN: What defense mechanism is specifically designed to prevent Cross-Site Request Forgery (CSRF) attacks in modern web applications?
GU: આધુનિક વેબ એપ્લિકેશનમાં ક્રોસ-સાઇટ રિક્વેસ્ટ ફોર્જરી (CSRF) હુમલાને રોકવા માટે મુખ્યત્વે કઈ પદ્ધતિ વપરાય છે?

✅ Option (A): Anti-CSRF Synchronizer Tokens and SameSite Cookie Attribute
Option (B): Parameterized SQL Prepared Statements
Option (C): Content Encoding gzip compression
Option (D): Strict Content-Length validation

💡 In-Depth Analysis: CSRF forces an authenticated browser to submit forged state-changing requests. Defenses include: Anti-CSRF Synchronizer Token Pattern (unpredictable, unique per-session/request tokens) and SameSite=Strict/Lax cookies.

Question 14 • Operating Systems (CPU Scheduling)

EN: In a pre-emptive Shortest Job First (SRTF) scheduling algorithm, what problem can arise if short processes continually enter the ready queue?
GU: SRTF (Shortest Remaining Time First) શેડ્યુલિંગ અલ્ગોરિધમમાં જો સતત ટૂંકી પ્રક્રિયાઓ આવતી રહે તો કઈ સમસ્યા સર્જાય છે?

Option (A): Deadlock
✅ Option (B): Starvation of long processes
Option (C): Priority Inversion
Option (D): Race Condition

💡 In-Depth Analysis: Shortest Remaining Time First (SRTF) favors jobs with minimal remaining burst time. If short bursts arrive incessantly, long CPU-bound processes are indefinitely preempted, causing Starvation (indefinite postponement).

Question 15 • Cloud Computing (Service Models)

EN: In the Cloud Shared Responsibility Model, which component is the customer responsible for securing in an Infrastructure as a Service (IaaS) deployment?
GU: ક્લાઉડ શેર્ડ રિસ્પોન્સિબિલિટી મોડેલમાં, IaaS (ઇન્ફ્રાસ્ટ્રક્ચર એઝ અ સર્વિસ) હેઠળ ગ્રાહક (Customer) ની સુરક્ષા જવાબદારી નીચેનામાંથી કઈ છે?

Option (A): Physical Data Center Security
Option (B): Virtualization Hypervisor Patching
✅ Option (C): Operating System, Network Configuration & Application Data
Option (D): Physical server cooling and power

💡 In-Depth Analysis: In IaaS (e.g. AWS EC2), the cloud provider manages physical hardware, hypervisor, facilities, and physical networking. The customer is responsible for guest OS patching, firewall/security groups, application software, and data protection.

Question 16 • Object Oriented Programming (Polymorphism)

EN: In C++, how is runtime polymorphism (dynamic method dispatch) implemented internally by the compiler for classes containing virtual functions?
GU: C++ માં વર્ચ્યુઅલ ફંક્શન ધરાવતા ક્લાસ માટે રનટાઇમ પોલીમોર્ફિઝમ (Dynamic Dispatch) કમ્પાઈલર દ્વારા આંતરિક રીતે કેવી રીતે અમલમાં મુકાય છે?

✅ Option (A): Through Virtual Method Table (vtable) and virtual pointer (vptr)
Option (B): By static inline code expansion at compile time
Option (C): Using memory heap compaction
Option (D): By function overloading resolution during linking

💡 In-Depth Analysis: When a class has virtual functions, the compiler generates a Virtual Table (vtable) containing function pointers. Each object instance contains an invisible virtual pointer (vptr) pointing to the class vtable, resolving calls at runtime.

Question 17 • Data Structures (Hashing & Collisions)

EN: What is the worst-case time complexity of searching an element in a Hash Table that uses separate chaining collision resolution with n keys inserted?
GU: n કી ધરાવતા અને સેપરેટ ચેઇનિંગ (Separate Chaining) વાપરતા હેશ ટેબલમાં કોઈ એલિમેન્ટ શોધવાની સૌથી ખરાબ સ્થિતિ (Worst-case) ની સમય જટિલતા શું છે?

Option (A): O(1)
Option (B): O(log n)
✅ Option (C): O(n)
Option (D): O(n log n)

💡 In-Depth Analysis: In average case with a uniform hash function, search takes O(1 + alpha). However, in the worst case (where all n keys hash to the very same bucket), the bucket becomes a linked list of length n, resulting in O(n) search time.

Question 18 • Gujarat E-Governance & Infrastructure

EN: What is the primary role of GSWAN (Gujarat State Wide Area Network), established under the Department of Science and Technology, Government of Gujarat?
GU: ગુજરાત સરકારના વિજ્ઞાન અને પ્રૌદ્યોગિકી વિભાગ હેઠળ સ્થપાયેલ GSWAN (ગુજરાત સ્ટેટ વાઇડ એરિયા નેટવર્ક) નું મુખ્ય કાર્ય શું છે?

✅ Option (A): To provide high-speed secure C2G and G2G data, voice and video backbone connecting state Secretariat to all districts and talukas
Option (B): Solely to issue driving licenses online
Option (C): To run commercial cellular mobile services in rural areas
Option (D): To regulate private internet service provider tariffs

💡 In-Depth Analysis: GSWAN is one of the largest IP-based government wide-area communication networks in the world. It provides a secure intra-governmental digital backbone connecting Sachivalaya (Gandhinagar) across 33 district headquarters and 250+ taluka offices.

Question 19 • Artificial Intelligence & Machine Learning

EN: Which technique is specifically used to prevent overfitting in Deep Neural Networks by randomly deactivating a subset of neurons during forward-backward propagation?
GU: ડીપ ન્યુરલ નેટવર્ક્સમાં ફોરવર્ડ-બેકવર્ડ પ્રોપાગેશન દરમિયાન ન્યુરોન્સને અનિશ્ચિત રીતે નિષ્ક્રિય કરીને ઓવરફિટિંગ (Overfitting) અટકાવવા કઈ ટેકનિક વપરાય છે?

Option (A): Batch Normalization
✅ Option (B): Dropout Regularization
Option (C): Gradient Clipping
Option (D): Softmax Activation

💡 In-Depth Analysis: Dropout (introduced by Srivastava et al.) randomly drops units along with their connections during training with probability p. This prevents complex co-adaptations between neurons, acting as an effective ensemble regularizer against overfitting.

Question 20 • Computer Architecture & Pipelining

EN: In a 5-stage RISC processor instruction pipeline (IF, ID, EX, MEM, WB), what hazard occurs when an instruction depends on the result of a previous instruction that is still in the pipeline?
GU: ૫-સ્ટેજ RISC પ્રોસેસર પાઇપલાઇનમાં જ્યારે કોઈ સૂચના તેના અગાઉના નિર્દેશના પરિણામ પર નિર્ભર હોય જે હજુ સુધી પાઇપલાઇનમાં પૂર્ણ ન થયું હોય, ત્યારે કયું હેઝાર્ડ સર્જાય છે?

Option (A): Structural Hazard
✅ Option (B): Data Hazard (Read-After-Write - RAW)
Option (C): Control Hazard (Branch penalty)
Option (D): Cache Miss Hazard

💡 In-Depth Analysis: A Data Hazard (specifically Read-After-Write or RAW) occurs when instruction j tries to read a source register before instruction i writes to it. It is resolved using Operand Forwarding (bypassing) or pipeline stalls (bubbles).

GPSC ICT ઓફિસર (વર્ગ ૨) ટેકનિકલ મોડેલ પેપર & વિશ્લેષણGPSC ICT Officer (Class 2) Technical Model Paper with Solutions

📜 GPSC ICT Officer (Class 2) — Solved Model Technical Paper

💻 GPSC ICT Officer Exam Pattern

📝 20 Solved Model Questions with Explanations

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📜 GPSC ICT Officer (Class 2) — Solved Model Technical Paper

💻 GPSC ICT Officer Exam Pattern

📝 20 Solved Model Questions with Explanations

હવે પોતાની કસોટી કરોTest yourself next

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