Computer Science Previous Year Questions (page 2)

Explanation

$? : exit status of the last command executed. $# : number of arguments supplied to a script. $0 : file name of the current script. $* : all the arguments (list of arguments). So A-iv, B-iii, C-i, D-ii, i.e. option (c).

Explanation

In Linux the seq command prints a sequence of numbers from FIRST to LAST in steps of INCREMENT: seq FIRST INCREMENT LAST. If FIRST or INCREMENT is omitted, it defaults to 1. Here FIRST = 1, INCREMENT = 2 and LAST = 10, so the output is 1 3 5 7 9 (each number on its own line). Hence the answer is (c).

Explanation

Win·txt is a group of classes that is use for windows programming.

Explanation

R₁ ∩ R₂ ≠ φ. This makes the decomposition lossless join, as all the attributes are keys, R₁ ∩ R₂ will be a key of the decomposed relation Now, even the original relation R is in 3NF (Even BCNF) as all the attributes are prime attributes (in fact attribute is a candidate key) Hence, any decomposition will also be in 3NF and 2NF, So the answer is option (d).

Explanation

2NF says that all the non-key attributes should depend the keys, partial dependency (an attribute is defined by a part of a composite key) is not allowed 3NF says that any dependency between non-key attribute should be cleared. BCNF says that all determinants in a table is key for every non trivial (no overlapping attributes) function. In the table book, (Title Author) identifies catalog-no. Which if turn identifies publisher and year further (publisher, Title, Year) identifies price. Now here the dependents are partially dependent on the key attributes on left hand side of (R). So the table book is in 2NF. In the table collection, (Title, Author) determines catalog-no., the only non-key attribute. So it is in 3NF as well as BCNF.

Explanation

2-phase locking is a concurrency control method that guarantees serializability, but because transactions may block one another while holding locks, 2PL can lead to deadlocks. Timestamp-based (time phase) ordering is a non-lock concurrency method: it ensures conflict serializability and no deadlock can occur since no transaction ever waits. So the answer is (b) II only.

Explanation

The host-to-host layers main purpose is to shield the upper layer application from the complexity of network. So the answer is (a).

Explanation

Six 1200 bps terminals give 6 × 1200 = 7200 bps, and n 300 bps terminals give 300n bps. 7200 + 300n = 9600 ⇒ 300n = 2400 ⇒ n = 8. So the answer is (b).

Explanation

At 1 Mbps the sender sends 1,000,000 bits per second. The receiver clock is 0.1% faster, so the extra bits received = 1,000,000 × 0.1/100 = 1000 bits, i.e. the receiver receives 1,001,000 bits per second. Extra bits per second = 1000 bps. Hence option (c) is correct.

Explanation

With n = 4 bits for sequence numbers, the maximum sender window size (Go-Back-N) is 2ⁿ − 1 = 2⁴ − 1 = 15. So the answer is 15.

Explanation

IPv4 is the fourth version in the development of the IP. and IPv4 use loose source for an IP option can be used for address translation and IPv4 also use strict source routing for the contrast with loose source routing, in every step of the route is decided in advance where the packet is send and IPv4 also time stamp. So the correct answer is (d).

Explanation

Ping is not a client server application. Ping is a computer network administration unility used to test the reach ability of a host on an Internet protocol ping there is no server that provides a service.

Explanation

CMM (Capability maturity model)– It is framework that describes an evolutionary improvement path for software organizations from an immature process to a mature, discipline one. The software process is at CMM level 5, if there is continuous process improvement, if there is quantitative feedback from the process and from piloting innovative ideas and technologies. Software processes are the CMM level 5, if there are process change management, detect prevention and technology change management, So software product engineering in not a key process area in CMM level 5

Explanation

Return : returns a value associated with it after completion of the function body. goto : control jumps from one statement to another by following the label. break : exits only from the innermost loop in which it appears. continue : skips the rest of the loop body and passes control to the conditional test and increment portions of the loop. So A-ii, B-iii, C-iv, D-i, i.e. option (a).

Explanation

Error mathematical is the difference between the observed value and approximate value and true value of a quantity. Error is used to refer to human action that results in software containing a defector fault whereas failure refers to a condition that causes a system to fail.

Explanation

Software testing is a process of executing a program or application with the intent of finding errors (the software bugs) and the main objective of software testing is to find any bug present in the software program. Besides, through the process of testing. One can determine the strengths and weaknesses of the program in other criteria such as path testing, dependency testing etc.

Explanation

Name attribute specifies the name of an anchor and it is used to create a bookmark inside a document. Syntax : <a name = "value"> ⇒ href attribute specifies the URL of the page the link goes to, and if the href attribute is not present, the <a> tag is not a hyperlink. Syntax : <a href = "URL">

Explanation

Each pixel needs 1 byte (8 bits) and there are 1024 × 1024 pixels. Total storage space = 1024 × 1024 × 1 = 2¹⁰ × 2¹⁰ = 2²⁰ bytes = 1 MB. So the answer is (b).

Explanation

RGB (Red, Green, Blue) and CMY (Cyan, Magenta, Yellow) are each defined by three primary colors. HSV (Hue, Saturation, Value), also known as HSB (Hue, Saturation, Brightness), and HLS (Hue, Lightness, Saturation) are not defined by primary colors. Hence the answer is (d).

Explanation

Butt Cap : Chops off the stroke at the ends. Round Cap : Extends the stroke past the ends with semicircular ends. Projecting Cap : Extends the stroke past the ends with squared ends.

Explanation

Line : attributes like type, width and color. Fill Area : a region can be filled with some color in different styles and patterns. Text : can be written with different font, size, color etc. Marker : the attributes are marker type, marker size and marker color. So a-iv, b-iii, c-ii, d-i, i.e. option (c).

Explanation

In programming language we use unique keywords to store data, just like that, the control string is use to access the different type of keyword, and control strings are the characters that use with the % (percent) and it is a classification of format specifier, white space and non-white space characters.

Explanation

STRIPs was devised SRI in the early 1970s to control a robot called SHAKEY strips is not related with NLP (Natural Language Processing).

Explanation

Frames : Frames are structured objects that represent knowledge and contain labeled slots with their values. Slots : Slots are almost similar to attributes as in ERD or object-oriented systems in the sense that they may contain declarative as well as procedural information. Facets : Facets are almost similar to description to the slot values that may function as criteria or validity of the values inserted into a slop.

Explanation

Chomsky Normal Form : production rules are of the form A → BC or A → a (and S → ε) — grammar (iii). Greibach Normal Form : production rules are of the form A → aA₁A₂...Aₙ or A → a — grammar (iv). S-grammar : every production's right-hand side begins with a terminal, and this terminal is different for any two productions with the same left-hand side — grammar (i). LL grammar : two production rules for the same nonterminal can begin with the same terminal only if their select sets differ — grammar (ii). So A-iii, B-iv, C-i, D-ii, i.e. option (c).

Explanation

Prolog variables begin with an upper case letter, while predicate names, function names and the names for objects must begin with a lower case letter; rules and facts are terminated by a full stop (.). So the right answer is option (c).

Explanation

The Excess-3 decimal code is a self-complementing code because the binary sum of a code and its 9's complement is equal to 9 and the complement can be generated by inverting each bit pattern.

Explanation

To add two BCD digits we need two 4-bit adders: the first adds the two BCD numbers and the second corrects the result to a valid BCD number. The first 4-bit adder needs 3 full adders and 1 half adder (giving A₃ A₂ A₁ A₀). In the correction step 0110 is added to A₃ A₂ A₁ A₀. A₀ is always added to 0, so no adder is needed; A₁ is added to 1 with no incoming carry, so a half adder is used; A₂ is added to 1 plus a carry, so a full adder is used; A₃ is added only to the carry, so a half adder is used. Total = 3 + 1 = 4 full adders and 1 + 2 = 3 half adders.

Explanation

Total inputs to the ROM = 8 + 8 (two 8-bit numbers) + 1 (mode) + 1 (carry in) = 18, so the ROM has 2¹⁸ words. Outputs = 8 result bits + 1 carry out + 1 overflow = 10 bits. Hence the ROM size is 2¹⁸ × 10.

Explanation

The postfix form of the expression is AB*CD*E/+. Counting a PUSH for each of the 5 operands and a POP for each of the 4 operators (ignoring the pushes of intermediate results), the evaluation needs 5 PUSH and 4 POP instructions.

Explanation

Each cell needs 4 words. In row-major order, rows 0 to 10 contain 11 × 10 = 110 elements, and A[11][0] to A[11][4] add 5 more, so 115 elements precede A[11][5]. Address = 100 + 4 × 115 = 560.

Explanation

^ has the highest precedence, so B^D becomes BD^, and A + B^D becomes ABD^+. E – F becomes EF–. Dividing gives ABD^+EF–/, and adding G gives ABD^+EF–/G+.

Explanation

A full binary tree with n leaves contains n – 1 internal nodes, so total nodes = internal nodes + leaves = (n – 1) + n = 2n – 1.

Explanation

For a sorted list followed by a few random elements: bubble sort and selection sort take O(n²) time, quick sort degrades to its O(n²) worst case on nearly sorted input, while insertion sort takes close to O(n) time. Hence insertion sort is most suitable.

Explanation

The DES algorithm works with a 64-bit block size, employs 16 processing rounds, and is driven by a 56-bit key (with 8 additional parity bits), making up a total of 19 distinct stages.

Explanation

Probability of k = 1 in one trial = 1/1000000 = 10⁻⁶, so the probability of k ≠ 1 is 1 – 10⁻⁶. Probability that 1 is never printed in 10⁶ printouts = (1 – 10⁻⁶)^(10⁶) ≈ 1/e. Probability that 1 is printed at least once = 1 – (1 – 10⁻⁶)^(10⁶) ≈ 1 – 1/e = 0.632121.

Explanation

gof(x) = √(x⁸ + 1), so ho(gof)(x) = (x⁸ + 1) + 72 = x⁸ + 73. hog(x) = (x² + 1) + 72 = x² + 73, so (hog)of(x) = x⁸ + 73. Hence both are x⁸ + 73.

Explanation

A friend function of a class is a non-member function that is given the same access as member methods to the class's private and protected data. It is declared inside the class that grants access, but it is not a member of the class.

Explanation

x++ is a post-increment, so the old value 128 is used in the expression and printf prints 1 + 128 = 129; x becomes 129 only afterwards.

Explanation

Average internal fragmentation per process is y/2. A process has x/y pages and each page-table entry needs z bytes, so the page table takes xz/y bytes. Total overhead = xz/y + y/2. Setting d/dy (xz/y + y/2) = –xz/y² + 1/2 = 0 gives y = √(2xz).

Explanation

Average time to service a page fault = 0.7 × 20 + 0.3 × 8 = 14 + 2.4 = 16.4 m. sec.

Explanation

Two-level scheduling is used to schedule efficiently when some processes are swapped out. A lower-level scheduler (e.g. round robin) selects only among the processes in memory, while a higher-level scheduler swaps processes in and out of memory. Hence I and III are correct.

Explanation

The program text is shared, so it is not multiplied by 5. As given in the book, total memory = 200 + 15 + 50 + 70 = 335 K.

Explanation

A correct solution must not let a process running outside its critical section block other processes. So condition (c) does not hold good for a solution to the critical section problem.

Explanation

Only I is correct: a self join is implemented on a single table using alias names. Outer join is not a basic relational algebra operation, and natural join and outer join are not equivalent.

Explanation

A weak entity does not have sufficient attributes to form a primary key; it is drawn as a double rectangle in an ER diagram. A weak entity set can be converted into a strong entity set simply by adding suitable attributes.

Explanation

The candidate keys are AB, CD, BC and AD, so A, B, C and D are all prime attributes and the relation is in 3NF. It is not in BCNF because in C → A and D → B the left side is not a superkey. Hence the best normal form is 3NF.

Explanation

{A}⁺ = {A, B, C, D}, which does not include E, so A alone is not a key. {AE}⁺ = {A, B, C, D, E}, and nothing can be removed from AE, so AE is the minimal key.

Explanation

Correct match: A. Call control protocol – iii. Connection management; B. A-bis – i. Interface between BTS and BSC; C. BSMAP – iv. Works between MSC and BSS; D. CDMA – ii. Spread spectrum.

Explanation

The signal sends 2000 signal elements (baud) per second and 6000 bits per second, so each signal element carries 6000/2000 = 3 bits/baud.