Here, we use the AND as a accumulator with immediate. F leaves the high nibble whatever it is, 0 clears the lower nibble.
→ The XOR's don't reliably clear random bits and ANI OF clears the upper nibble, not the lower nibble.

We know that stack works on the principle of LIFO.

A vectored interrupt is a processing technique in which the interrupting device directs the processor to the appropriate interrupt service routine vector.

X is remains unchanged. As the if condition is becomes false.
X = -10

Merge sort uses the divide and conquer strategy.

The locality of reference is also known as principle of locality.
→ The things which are used more frequently those are stored in locality of reference.
→ For this purpose we use the cache memory.

The segment table is often too large to fit in one page. This is true and the segment table can be divided into pages. Thus page table for each segment table, pages are created.
Segment paging is different from paged segmentation.

FIFO is suffers from Belady's Anamoly.

Which is to be letter followed by any number of letters (or) digits
L(L ∪ D)*

S ∝→ [violates context free]
Because LHS must be single non-terminal symbol.
S ∝→ b [violates CSG]
→ Length of RHS production must be atleast same as that of LHS.
Extra information is added to the state by redefining iteams to include a terminal symbol as second component in this type of grammar.
Ex: [A → αβa]
A → αβ is a production, a is a terminal (or) right end marker $, such an object is called LR(k).
So, answer is (D) i.e., LR(k).

Formal arguments (or) formal parameters is a special kind of variables used in subroutime to refer to one of pieces of data provided as input to the subroutine.

Distance = Minimum hamming distance = 2
010101 ⊕ 011001 = 001100

Note: Out of syllabus.

In object module it includes names and locations of all external symbols defined in the object module.
To link to external symbols it must know the location of external symbols.

In Round robin, we use the time quentum based on this execution can be done. Then it is said to be time shared operating system.

In best case, no. of comparisons is Min(m,n).
In worst case, no. of comparisons is m+n-1.
Then we require O(m+n) comparisons to merging two sorted lists.

A binary tree is a tree data structure in which each node has atmost two child nodes.
The no. of subtrees of a node is called the degree of the node. In a binary tree, all nodes have degree 0, 1 and 2.
The degree of a tree is the maximum degree of a node in the tree. A binary tree is of degree 2.
The number of nodes of degree 2 in T is "n - 1".

If a relation is equivalence then it must be
i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is said to be symmetric and transitive then we can't say the relation is reflexive and equivalence.

S = {(φ), 1, (2, 3)}
P(S) = {φ, {{φ}}, {1}, {{2, 3}}, {{φ}, 1}, {1, {2, 3}}, {{φ}, 1, {2, 3}}}
In P(S) it contains 8 elements.

x & cos(x) are intersecting at only one point.

Note: Out of syllabus.

Note: Out of syllabus.

In a normal graph number of edges required for n vertices is n-1, and in cyclic graph it is n.
In cyclic graph:
No. of edges = No. of vertices
⇒ n = n

ROM memory size = 2^m × n
m = no. of address lines
n = no. of data lines
Given, 4K × 16 = 2¹² × 16
Address lines = 12
Data lines = 16

Let X=3 and Y=5
1^st pass : X=3 and Y=2
2^nd pass : X=1 and Y=2
3^rd pass : X=1 and Y=1
Write(X), which writes 1. Which is nothing but GCD of 3 & 5.

Since nothing is said in question. So we will assume by default call by value.
X in the procedure FIND is a local variable. No change will be reflected in global variable X.

For verification, just put up the values and check for AND, OR operations and their outputs.

Here, in option (B) and (C) they have given CPU idle time is 0 which is not possible as per schedule (B) and (C).
So, (B) and (C) will be eliminated.
Now in (A) and (D):
For r(A),

So, idle time is between 0 to 1 which in case of option (A).
For option(D),

We can see there is no idle time at all, but in option given idle time is 5, which is not matching with our chart. So option (D) is eliminated.
Therefore, the correct sequence is option (A).

Page are fitted in frames, so first we need to determine the pages but given request are just the record request in decimal. We can assume that first page to address from 0000 to 0099 and page 2 contains records from 0100 to 0199 and so on (it is given in question that each page contains 100 records) and so on. So page request string is
01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.
Clearly 7 page faults.

Just put values of (C) in place of x. It will satisfy the equation.

concat (a, head (tail (tail (acbc))))
concat (a, head (tail (cbc)))
concat (a, head (bc))
concat (a, b)
ab

⇒ 23131
Note SR is bottom up parser.

Note: Out of syllabus.

Probability of first ball white and second one black is,

Probability of first ball black and second one white is,

Note: Out of syllabus.

Primary memory < Virtual memory < Secondary memory.

As the matrices are non-singular so their determinant ±0. Hence, the inverse matrix always exist. But for a group to be Abelian it should follow commutative property. As matrix multiplication is not commutative, is a group but not an abelian group.

Note: Out of syllabus.

From the axiom ¬p → q, we can conclude that p ∨ q.
So, either p or q must be True.
Now,
¬p ∨ (p → q)
= ¬p ∨ (¬p ∨ q)
= ¬p ∨ q
Since nothing c an be said about the truth values of p, it implies that ¬p ∨ q can also be True or False. Hence, the value cannot be determined.

(I) is the correct definition and the other two is wrong because the other two can have any no. of x and y. There is no such restriction over the number of both being equal.

The postfix expression will be,
A B C D + * F / + D E * +

Binary search using linked list is not efficient as it will not give O(log n), because we will not be able to find the mid in constant time. Fnding mid in linked list takes O(n) time.
Recursive program requires stack for storing the function state. Any recursive program can be rewritten as an iterative one. Advantage of recursion is "less programming effort", while it always lags behind ietrative one in terms of performance.

If A is the start state, shortest sequence is 10 'or' 00 to reach C.
If B is the start state, shortest sequence is 0 to reach C.
If C is the start state, shortest sequence is 10 or 00 to reach C.
If D is the start state, shortest sequence is 0 to reach C.
∴ (B) is correct.

L∪R is nothing but L itself. Because R is subset of L and hence regular. R is deterministic context free but not regular as we require a stack to keep the count of 0's to make that of 1's.

No. of sets = 4K/ (64×4) = 2¹²/ 2⁸ = 2⁴
So we need 4 set bits.
And,
64 words per block means WORD bit is 6-bits.
So, answer is option (D).