Consider the set S = {1, ω, ω2}, where ω and ω2 are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) forms
Option A. A group Option B. A ring Option C. An integral domain Option D. A field True Answer A
A group should satisfy the relation (i) Closure (ii) Inverse (iii) Identity (iv) Associative in We can say all are elements of group than it is closure Hence, we think about inverse, then this should satisfy where e is identity element means G have identity element 1 means identities Hence it should be inverse means means b is the inverse element a, b, e 64 here 1 * 1 = 1, inverse So, it also hold Associative for this means Associative means is a group.
Option A. TRUE Option B. FALSE Option C. Option D. True Answer B
Option A. A Option B. B Option C. C Option D. None of the above True Answer A
Option A. residing in many computers Option B. created using HTML Option C. with links to other web pages Option D. residing in many computers linked together using HTML True Answer B
Option A. TRUE Option B. FALSE Option C. Option D. True Answer A
Fetch_And_Add(X, i) is an atomic Read-Modify-Write instruction that reads the value of memory location X, increments it by the value i, and returns the old value of X. It is used in the pseudocode shown below to implement a busy-wait lock. L is an unsigned integer shared variable initialized to 0. The value of 0 corresponds to lock being available, while any non-zero value corresponds to the lock being not available.
AcquireLock (L) { while (Fetch_And_Add (L,1)) L = 1; } ReleaseLock (L) { L = 0; } This implementation
Option A. fails as L can overflow Option B. fails as L can take on a non-zero value when the lock is actually available Option C. works correctly but may starve some processes Option D. works correctly without starvation True Answer B
1) Acquire lock (L) { 2) While (Fetch_And_Add (L,1)) 3) L = 1 } 4) Release Lock (L) { 5) L = 0; 6) } Let P and Q be two concurrent processes in the system currently executing as follows P executes 1, 2, 3 then Q executes 1 and 2 then P executes 4, 5, 6 then L = 0 now Q executes 3 by which L will be set to 1 and thereafter no process can set L to zero, by which all the processes could starve.
Option A. Output device Option B. Virus Option C. Threat Option D. None of These True Answer D
Option A. To catch two birds with your hands Option B. To be dissatisfied with what you have Option C. It is better to be satisfied with what you have than to try to get what is not yours Option D. To trap birds in bushes True Answer C
Option A. note outlines for teachers Option B. project presentations by students Option C. communication of planning Option D. All of above True Answer D
The surface area of a cube is 96 cm2. What is the length of its side?
Option A. 4 cm Option B. 5 cm Option C. 6 cm Option D. 8 cm True Answer A
We have, Surface area of a cube = Therefore,
Option A. F5 Option B. F6 Option C. F7 Option D. F8 True Answer C
Newton-Raphson method is used to compute a root of the equation x2 − 13 = 0 with 3.5 as the initial value. The approximation after one iteration is
Option A. 3.575 Option B. 4.21 Option C. 3.97 Option D. 3.607 True Answer D
Since in Tayor's series Since h is small, so we neglect higher power of h. So as first iteration Since
Option A. ARPANET Option B. USENET Option C. INFRARED Option D. DATANET True Answer A
Option A. Deterministic finite automata (DFA) and Non-deterministic finite automata (NFA) Option B. Deterministic push down automata (DPDA) and Non-deterministic push down automata (NPDA) Option C. Deterministic single-tape Turing machine and Non-deterministic single-tape Turing machine Option D. Single-tape Turing machine and multi-tape Turing machine True Answer B
NPDA is more powerful than DPDA.
Hence, Answer is B.
Option A. Logic gate Option B. Bus Option C. Modem Option D. Serial port True Answer B
Option A. Financial statements, Business forecasting Option B. Transaction registers, inventory control Option C. Accounts receivable, accounts payable Option D. Any of the above True Answer D