UGC NET Computer Science PYQ, November 2017 (Paper II & III, Selected Questions)

Explanation

Group the bits in threes from the right: 001 011 101 011. 001 = 1, 011 = 3, 101 = 5, 011 = 3. So (1011101011)₂ = (1353)₈.

Explanation

Convert both numbers to decimal, add, and convert back to base 4: (313)₄ = 3×4² + 1×4¹ + 3×4⁰ = 55; (322)₄ = 3×4² + 2×4¹ + 2×4⁰ = 58. 55 + 58 = 113, and 113 in base 4 is (1301)₄.

Explanation

Grouping the 1s: columns AB = 01 and 11 give B, the row CD = 11 together with the cell AB = 10, CD = 01 give CD and AD. SOP form = AD + CD + B = (A + C).D + B.

Explanation

In the Boolean product, multiplication is replaced by logical AND and addition by logical OR. Row i of A ⊙ B is the OR of the rows of B selected by the 1s in row i of A: row 1 = B1 = 1110, row 2 = B2 = 0110, row 3 = B1 ∨ B2 = 1110, row 4 = B3 = 1011. So option (a) is correct.

Explanation

P ↔ Q = P'Q' + PQ, so ¬(P ↔ Q) = P'Q + PQ', which is true exactly when P and Q differ. This is the same as P ↔ ¬Q (and ¬P ↔ Q).

Explanation

Proposition: ∃x H(x). Negation: ¬∃x H(x) ≡ ∀x ¬H(x).

Explanation

The clock repeats every 12 hours. 101 mod 12 = 5, so the time after 101 hours = 4 + 5 = 9 O'clock.

Explanation

The number of permutations of n objects with n₁, n₂, ..., nₖ identical objects of each type is n!/(n₁! n₂! ... nₖ!). BANANA has 6 letters with A three times, N twice and B once: 6!/(3! 2! 1!) = 60.

Explanation

cx = 0Ah = 10, so the loop body runs 10 times; loop decrements cx and stops when cx = 0h. ax starts at 0 and is decremented 10 times: 0 − 10 = FFF6h (16-bit two's complement). So ax = FFF6h and cx = 0h.

Explanation

After the rotate and shift instructions al = 00011110, and mul cl multiplies it by 2 (append a 0 bit at the LSB): ax = 0000000000111100 = 003CH. For mul, CF = 0 when the high byte of the result is 0 and CF = 1 otherwise; here the high byte is 0, so CF = 0. Hence ax = 003CH; CF = 0.

Explanation

Sequential part = 5% = 0.05, parallel part = 0.95. By Amdahl's law with an infinite number of processors, speed up S = 1 / (1 − 0.95) = 1 / 0.05 = 20.

Explanation

An in-order traversal of a binary search tree always gives the keys in ascending order, so the in-order sequence is 3, 4, 5, 7, 9, 14, 15, 16, 17, 18, 20.

Explanation

Inserting 10, 1, 3, 5, 15, 12, 16: 10 is the root; 1 is its left child, 3 is the right child of 1 and 5 is the right child of 3; 15 is the right child of 10 with children 12 and 16. The longest root-to-leaf path is 10-1-3-5 with 3 edges, so the height is 3.

Explanation

Merge sort performs O(n log n) comparisons, and comparing two strings of length n takes O(n) time in the worst case. So the worst-case running time is O(n · n log n) = O(n² log n).

Explanation

In a binary heap stored in an array from index 1 to n, the children of index i are at 2i and 2i + 1, so the parent of the element at index i is at floor(i/2).

Explanation

Kruskal's algorithm picks edges in increasing order of weight while avoiding cycles: GC = 2, AD = 2, GB = 3, GA = 3, then BF = 4 and AE = 4. So the order is GC, AD, GB, GA, BF, AE. Note: more than one order is possible since edge weights are repeated.

Explanation

With distinct edge weights the MST is unique and always contains Eₘᵢₙ. Eₘₐₓ can be in the MST (when it is a bridge, its removal disconnects G), so the statement "No minimum spanning tree contains Eₘₐₓ" is false.

Explanation

Scalability is the ability of a process, network, software or organization to grow and manage increased demand. A scalable system can adapt to the changing needs of its users.

Explanation

To minimise access time on a sequential device, store files in increasing order of size: F4 (60 KB), F6 (65 KB), F3 (75 KB), F1 (150 KB), F2 (225 KB), F5 (275 KB).

Explanation

The dispatcher is the module that gives control of the CPU to the process selected by the short-term scheduler.

Explanation

Wait & signal operations are the two atomic operations performed on a semaphore.

Explanation

Relations are decomposed into normal forms (normalization) to prevent insertion, deletion and update anomalies.

Explanation

As keyed by the book, MODIFY is the SQL command that changes one or more fields in a record.

Explanation

A view is a virtual table through which a selective portion of the data from one or more tables can be seen; it does not contain data of its own. A stored procedure executed when an event occurs is a trigger, not a view. So only (B) is true.

Explanation

AVG returns the average of the values in a group, ignoring null values. It is an aggregate function. Syntax: AVG([ALL | DISTINCT] expression).

Explanation

Entity integrity - no duplicate rows in a table (iv); Domain integrity - enforces valid entries for a column (iii); Referential integrity - rows can't be deleted which are used by other records (ii); User defined integrity - enforces specific business rules that do not fall into entity or domain integrity (i).

Explanation

Non-adaptive (static) routing algorithms do not base their routing decisions on measurements or estimates of the current traffic and topology. Instead, the route from one node to another is computed in advance.

Explanation

The transport layer is responsible for end-to-end communication over a network; it provides logical communication between application processes running on different hosts.

Explanation

Switch takes data sent from one network device & forwards it to the destination node based on MAC address. A switch is a multi port bridge with a buffer and a design that can boost its efficiency and performance. Switch is data link layer device. It can perform error checking before forwarding data.

Explanation

Gigabit Ethernet uses the standard Ethernet frame, whose destination and source address (MAC address) fields are each 48 bits long.

Explanation

The address 0.0.0.0 is used by a host as its source address while it is booting (e.g. in a DHCP request) before it knows its own IP address.

Explanation

As software is used, it receives multiple change requests; each change can introduce errors in component interactions, so the software deteriorates over time.

Explanation

Software re-engineering is a process of software development which is done to improve the maintainability of a software system.

Explanation

Agile philosophy stresses self-organizing teams, collaboration with customers, welcoming change and rapid delivery of working software. Testing is done continuously along with development, not as a separate phase after the build phase.

Explanation

Traditional testing starts with unit testing, followed by integration testing, then system testing and finally validation (acceptance) testing: (C), (A), (B), (D).

Explanation

Regression testing verifies that previously developed and tested software still performs correctly after it has been changed, e.g. after enhancements or patches during maintenance.

Explanation

A virtual function cannot be static, because a virtual function is a non-static member function whose call is resolved through a specific object at run time.

Explanation

Function templates are instantiated implicitly when the function is called with arguments, whereas class templates are instantiated by defining an object with explicit template arguments. So class templates and function templates are not instantiated in the same way; statement (a) is not correct.

Explanation

Only array element access ([]) is left-to-right associative among the options listed. All others are unary operators and are right-to-left associative.

Explanation

A member function can always access all the data (public, private and protected) of the class of which it is a member.

Explanation

Postfix ++ has higher precedence than *, so *ptrdata++ is evaluated as *(ptrdata++): the value at ptrdata is used and then the pointer moves to the next location. (*ptrdata)++ would instead increment the value stored at that location.

Explanation

As keyed by the book, Bluetooth (in its low-energy form) consumes the least power among the options; Wi-Fi and GSM/CDMA consume much more.

Explanation

The parser (syntax analyzer) recognizes syntax (A-iii); the scanner (lexical analyzer) turns the character stream into words/tokens (B-iv); the semantic analyzer checks meanings and consistent use of names (C-ii); the optimizer is an IR-to-IR transformer that improves the IR program (D-i).

Explanation

Sunway TaihuLight was named the world's fastest supercomputer in the TOP500 list at that time. It was developed by China's National Research Center of Parallel Computer Engineering & Technology (NRCPC).

Explanation

ERP software packages are themselves combinations of separate applications for manufacturing, materials resource planning, general ledger, human resources, procurement and order entry.

Explanation

K-means, self-organizing feature maps and agglomerative methods are clustering methods. The k-nearest neighbour (K-NN) algorithm is a non-parametric method used for classification and regression, not clustering.

Explanation

Odd binary numbers end with 1 in the LSB and even numbers end with 0. Only 0*1*0*1+ always ends with at least one 1, so its language contains no even values.

Explanation

F1, F2, F3 each with 7 micro-operations need 3 bits each = 9 bits; CD for 4 status bits needs 2 bits; BR with 4 options needs 2 bits; ADF for 128 locations needs 7 bits. Total = 9 + 2 + 2 + 7 = 20 bits.

Explanation

The isolated I/O method isolates memory and I/O addresses so that the memory address range is not affected by interface address assignment. Memory-mapped I/O uses the same address space for memory and I/O devices. In asynchronous serial transfer the two units do not share a common clock; in synchronous serial transfer they do.

Explanation

AC is the auxiliary carry flag, CY is the carry flag and Z is the zero flag. All these flags are affected by arithmetic operations.

Explanation

The 8085 microprocessor has a 16-bit address bus, so 2¹⁶ = 65,536 different memory locations can be addressed.

Explanation

The ALU forms the processing unit (A-iii); timing and control belongs to the instruction unit (B-iv); the general-purpose registers are in the storage and interface unit (C-ii).

Explanation

Indexed addressing mode is best suited for accessing an array in contiguous memory location.

Explanation

For a full binary tree with n internal nodes, e = i + 2n. Example: a full binary tree with 7 internal nodes of depth up to 2 has i = 0 + 1×2 + 2×4 = 10 and 8 leaves at depth 3, so e = 24 = 10 + 2×7.

Explanation

Standard planning algorithms assume the environment to be both deterministic and fully observable.

Explanation

In the worst case greedy (best-first) search keeps all generated nodes in memory. With branching factor b and maximum depth m, the space complexity is O(bᵐ).

Explanation

Heap allocation is required for languages that support dynamic data structures, whose storage is allocated and freed at run time in an arbitrary order.

Explanation

Scaling by 3 units in x' and 4 units in y' gives x = x'/3 and y = y'/4. Substituting in x² + y² = 1 gives (x'/3)² + (y'/4)² = 1.

Explanation

Window: xwmin = 1, ywmin = 1, xwmax = 3, ywmax = 5. Viewport (normalized screen): xvmin = 0, yvmin = 0, xvmax = 1, yvmax = 1. Scale factors sx = (1 − 0)/(3 − 1) = 1/2 and sy = (1 − 0)/(5 − 1) = 1/4. N = T(xvmin, yvmin) · S(sx, sy) · T(−xwmin, −ywmin), which gives the matrix [1/2 0 −1/2], [0 1/4 −1/4], [0 0 1].

Explanation

The producer-consumer (bounded buffer) problem describes two processes, the producer and the consumer, which share a common fixed-size buffer used as a queue. The producer puts items into the buffer and waits if it is full; the consumer removes items and waits if it is empty.

Explanation

In statement set (A), the many-to-one model maps many user threads to one kernel thread, the one-to-one model maps one user thread to one kernel thread, and the many-to-many model maps many user threads to a smaller or equal number of kernel threads; all are true. Set (B) reverses the roles of user and kernel threads, so it is false.

Explanation

For this reference string with 3 frames, the optimal algorithm gives 9 page faults, fewer than FIFO, LIFO or LRU. So the optimal algorithm has the minimum number of page faults.

Explanation

UNIX provides read, write and execute permissions on files, and three sets of permissions: for the owner, for the group and for others. So both (A) and (B) are true.

Explanation

The query takes the Cartesian product R₁ × R₂ and projects A, B with duplicates removed. If R₂ is empty the product is empty, so R₂ must be non-empty; and since DISTINCT removes duplicates, the result equals R₁ only if R₁ has no duplicates.

Explanation

A → B is preserved in R1, B → C in R2, and B → D, D → B in R3; C → D follows from C → B (in R2, since C → D → B) and B → D. So all dependencies are preserved. R1 ∩ R2 = B and R2 ∩ R3 = B, and B is a key of R2 and R3, so the decomposition is also lossless.

Explanation

A relation is in 2NF if it is in 1NF and every non-key attribute is fully functionally dependent on the primary key. If every non-key attribute is functionally dependent on the primary key (no partial dependency), the relation is in second normal form.

Explanation

A⁺ contains all attributes except D, and D cannot be derived from any attribute, so D must be part of every key (e.g. AD, BD, ED, FD are keys). Then A → BC is a partial dependency of non-key attribute C on part of key AD, so R is in 1NF but not in 2NF.

Explanation

As keyed by the book, distributed systems use fixed routing; virtual and dynamic routing are not the technique keyed here.

Explanation

Quadrature amplitude modulation means changing both amplitude and phase of the Carrier.

Explanation

50,000 characters × 8 bits = 400,000 bits sent in 40 seconds, so the data rate = 400,000/40 = 10,000 bits/sec = 10 kbps.

Explanation

The three aspects of quantization programmers are concerned with are: (1) sampling rate - the number of samples per second; (2) aperture time - the time over which each sample is taken; (3) coding error - the difference between the conditioned signal and the quantized signal.

Explanation

Data link layer - framing (iv); Network layer - routing (iii); Transport layer - connection control (ii); Presentation layer - encryption (i).

Explanation

Class B has 16 host bits. Using 6 bits for the subnet leaves 10 host bits. Number of subnets = 2⁶ − 2 = 62; hosts per subnet = 2¹⁰ − 2 = 1022.

Explanation

In IPv6 the fragmentation fields of the IPv4 base header have moved to the fragmentation extension header, the authentication extension header is new in IPv6, and the record route option is not implemented in IPv6. So all the given statements are correct.

Explanation

Statistical software quality assurance in software engineering involves tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.

Explanation

Availability = MTBF / (MTBF + MTTR). MTBF = 91.25 × 24 × 60 = 131,400 minutes and MTTR = 5 minutes, so availability = 131,400 / 131,405 = 99.9962%.

Explanation

S1 and S2 interchange white-box and black-box testing: black-box tests are based on specifications and white-box tests are based on code, so both are false. S3 correctly describes alpha testing. So only S1 and S2 are FALSE.

Explanation

Components of a memory tube (direct-view storage tube) display include the flooding gun, primary (writing) gun, collector, storage grid, screen, ground (base) and focusing and deflection system. Liquid crystal is not a component of a memory tube display.

Explanation

In oblique projection, the parallel projection rays are not perpendicular to the viewing plane but strike it at an angle other than 90°; parallel lines in space appear parallel on the projected image; and it is used for pictorial purposes rather than formal working drawings. Projectors perpendicular to the plane of projection describe orthographic projection, so (d) is not true.

Explanation

In computer graphics, horizontal retrace is the path the electron beam takes when returning to the left side of the CRT after completing a scan line.

Explanation

Context-free languages are closed under union, concatenation and Kleene star, but not under complementation (or intersection).

Explanation

The ambiguity problem for context-free grammars is undecidable: no Turing machine can always halt and decide whether an arbitrary CFG is ambiguous. Equivalence of finite automata, CFG membership and finiteness for finite automata are decidable.