Proof says any number greater than or equal to 6, its square is greater than or equal to 25.
6×6>25. 7×7>25——-.
Proposition 1 Fails because it does not prove that number (6,7,8—) square is not greater than 25.
Proposition 2 fails because there is no requirement of assumption as numbers are already given.
Proposition 3 is correct.
Ans .
(D) DEFGH1000
Explanation :
S[0]=A at address 1000.
S[1]=B at address 1001.
S[2]=C at address 1002.
S[3]=D. at address 1003.
S[4]=E. at address 1004.
S[5]=F S[6]=G S[7]=H.
printf(“%c”,*(&S[3]));
prints character at subscript S[3].
printf(“%s”,S+4);
prints from base address 1000+4.
printf(“%u”,S);
prints base address.
Ans .
(C) data members
Explanation :
Ans .
(A) 0
Explanation :
–(x)+-1.0==7.5>1>=2.
5.5==7.5>0 (as 1>=2 is false).
6.5==1.(as 7.5>0).
6.5==1 is false.
1.An assertion is a predicate expressing a condition we wish the database to always satisfy.
2. Domain constraints, functional dependency and referential integrity are special forms of assertion.
3. Where a constraint cannot be expressed in these forms, we use an assertion.
e.g. Ensuring the sum of loan amounts for each branch is less than the sum of allaccount balances at the branch. Ensuring every loan customer keeps a minimum of $100 in an account.
4. An assertion in DQL-92 takes the form.
Ans .
(C) (b) only
Explanation :
Ans .
(C) (d)
Explanation :
Ans .
(D) (a), (b), (c) and (d)
Explanation :
Ans .
(C) Owner entity set
Explanation :
Ans .
(B)
Explanation :
Kruskal algo arrange weights in ascending order.
6-1->10
3-4->12
2-7->14
2-3->16
7-4->18
4-5->22
7-5->24
5-6->25
1-2->28.
put all the nodes in ascending order so that it does not form a circuit.untill all the nodes are traversed.
So nodes are.
6-1,3-4,2-7,2-3,4-5,5-6,1-2.
it avoids 7-4 and 7-5 to avoid the circuit.
Ans .
(D) debfgca
Explanation :
Ans .
(A) Breadth First Search
Explanation :
Ans .
(A) The item to be searched is in some where middle of the Array
Explanation :
let there are n elements to be searched linearly.
1st element will take 1 search.
2nd element will take 2 search.
—-------
—-------
n th element will take n search.
Average search will take. (1+2+3—–n)/n
= n(n+1)/2n hence approximately n/2.
when item is not there it will search whole list.
last element will take n search time.
Hence option A is the correct
Ans .
(B) Counting number of key operations
Explanation :
Ans .
(D) Session Initiation Protocol
Explanation :
Ans .
(D) (ii) (iv) (i) (iii)
Explanation :
Ans .
(C) Semantics of the information transmitted
Explanation :
Ans .
(C) 16 bits
Explanation :
Ans .
(D) All of the above
Explanation :
Ans .
(D) (iv) (iii) (ii) (i)
Explanation :
Ans .
(C) Macro pre-processor
Explanation :
Ans .
(A) context free grammar
Explanation :
Ans .
(B) A parsing algorithm which performs a left to right scanning and a right most deviation is RL (1).
Explanation :
Ans .
(A) Syntax analysis
Explanation :
Ans .
(D) 0.42 sec
Explanation :
current position of head is at cylinder 24.
24->26->12->8->4->42->50.
total cylinders traversed are.
(2+14+4+4+38+8)=70 cylinders.
1 cylinders is traversed in 6 ms.
70 cylinders are traveresed in 420ms.= 420s/1000=0.42sec
Ans .
(C) R(Pi)∩R(Pj)=Φ
Explanation :
Ans .
(B) 7
Explanation :
LRU
Whenever a page is used it is stamped with a counter value. When a page is to be replaced page with lowest stamp value is selected as the victim.
0172 3 2 7 1 0 3.
0(1 page fault,Counter=1),1(2 page fault,Counter=2),7(3 page fault,Counter=3),2 (4 page fault,Counter=4).
0 is selected as the victim .and replaced by.3. now frame is. 3(5 page fault,counter=5),1(2 page fault,Counter=2),7(3 page fault,Counter=3),2 (4 page fault,Counter=4)
2 is already in the frame so counter is increased. 3(5 page fault,counter=5),1(2 page fault,Counter=2),7(3 page fault,Counter=3),2 (4 page fault,Counter=6)
7 is already in the frame so counter is increased.3(5 page fault,counter=5),1(2 page fault,Counter=2),7(3 page fault,Counter=7),2 (4 page fault,Counter=6)
1 is already in the frame so counter is increased. 3(5 page fault,counter=5),1(2 page fault,Counter=8),7(3 page fault,Counter=7),2 (4 page fault,Counter=6)
0 is not in the frame so counter 5 is the least so 3 is replaced by 0.and page fault is increased.
0(6 page fault,counter=9),1(2 page fault,Counter=8),7(3 page fault,Counter=7),2 (4 page fault,Counter=6) 3 replaces 2
0(6 page fault,counter=9),1(2 page fault,Counter=8),7(3 page fault,Counter=7),3 (7 page fault,Counter=10)
Ans .
(A) It will print all of the lines in the file x that match the search string “abc”
Explanation :
Ans .
(C) Scheduling parameters
Explanation :
Ans .
(D) (iii) (i) (iv) (ii)
Explanation :
Ans .
(A) Validation testing
Explanation :
Ans .
(A) Waterfall model
Explanation :
Ans .
(C) Information hiding concept
Explanation :
Ans .
(C) Identify the activities, milestones and deliverables produced by a project
Explanation :
Ans .
(C) 1/((1-P)+P/N)
Explanation :
Ans .
(D) In PVM Model, the message order is not preserved.
Explanation :
Ans .
(C) Radix sort
Explanation :
Ans .
(D) Tuples in a weak entity set are not partitioned according to their relationship with tuples in a strong entity set.