Solutions. 9. Ans. C. Memory size. 11. Ans. D. a b c d e f a d e b c f a b d c e f a d b c e f a b d e c f a d b e c f. 13. Ans. D. Merge sort n log n

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1 Solutions. Ans. B. ~ p q ~ r ~ s ~ p q r s ~ ~ p q r s p ~ q r s ~ p q ~ r v ~ s For x only the above compound preposition is true.. Ans. B. Case I First bit is 0 Case II First bit is 6. Ans. A. # has the property of xor operation. So x#y = xy +x y 7. Ans. D. 00 s complement and st bit is. So result is - 8. Ans. C. 0->->0->->0-> 0000->000->000->000->000->00 There are 6 states and of them correspond to same state. To differentiate between 0,,, we need bits. To differentiate between 0's we need bits. So Total bits required = FF. Ans. C. Memory size GB bytes Word size = bytes No. of Address bits Memory size bytes t bits Word size bytes a a a n n n. Ans. B. sinx lim x x sin x lim x 0 x sin y lim By taking y x x y y. Ans. A. f x x a, Given x 0 other wise We know that ' f x dx a ' dx a x x a a a Ans. A. Both ENQUEUE and DEQUEUE can be performed in O() time using circular array.. Ans. D. a b c d e f a d e b c f a b d c e f a d b c e f a b d e c f a d b e c f. Ans. D. Here function f takes two arguments one is int and the other is short and its return type is void. So, in main function P is a pointer to short and when we call f (i,*p) there won t be any type checking error.. Ans. D. Merge sort n log n in all the cases Quick sort n log n best case and n Insertion sort n best case R n. Ans. A. worst cases worst case 5. Ans. B. Given that and are two Eigen values of real matrix is, i and are Eigen values. But -i also Eigen values (complex roots occurs in pair only) det = Product of Eigen values i i 5 5 P a g e

2 MST of G does not change but shortest path between some pair may change when every edge weight is increased by same value as shown in above. 5. Ans. B. Output is not affected by the function mystery () as it is just taking the address of a&b into ptra & ptrb and contents of ptra & ptrb are swapped leaving a&b as it is. 6. Ans. D. Given grammar generates all strings of a s and b s including null string L a b * 7. Ans. C. There is no known algorithm to check whether the language accepted by TM is empty. Similarly there is no algorithm to check whether language CFG s are equivalent. 8. Ans. B. A) contains 00 & consecutively which is not the required condition. C) Doesn t guaranty that both 00 & will be present in the string. D) Says string should start with & ends with 00 or vice versa.. Ans. A. x = u - t; y = x * v; x = y + w; y = t - z; y = x * y; So, we require total 0 variables in SSA form of the given code. 0. Ans. A. SRTF is pre-emptive SJF which produces less average waiting time.. Ans. B. Any superset of VY is a super key.. Ans. D. D means durability not deadlock freedom.. Ans. A. candidate key is (volume, number, start page, end page) (Volume number) year is a partial dependency. So original table is in NF but not in NF. Ans. C. Except DHCP, remaining all the protocols are used to resolve one form of address nother one. 5. Ans. C. FTP and POP are stateful application layer protocols 6. Ans. A. x x x 5 x 6... x x x... x x n n x x x n n 0 For coefficient of x x put 5 n 0 7. Ans. B. The recurrence relation can be written as a a 6n n (i) n n Characteristic equation is m 0,m Complementary solution Let the particular solution be n b n a C C n a An Bn c n () ( RHS is second deg ree polynomial and is root) By substuting n A,B, C General solution is a An Bn C n in () and solving c b a a a C n n n given a B B c B c 0 Given a k K 0 K 8 8. Ans. A. n Given f n f is n is even f n 5 if n is odd We can observe that f f f f f 6 f 7... and f 5 f 0 f 5... n n n Clearly, the range of f(x) will contain two distinct elements only.. Ans. A. From the given steps we can observe that probabilities of y are,,,... Required probability P a g e

3 Ans. D. Output of first multiplexer is Y P0 PR PR Output of second multiplexer is X QR QY QR QPR QR PQR 8. Ans. A. If x then the shortest path between d & c will contain edge with lable x.. Ans. D.. Ans. B. DMA controller needs 5kB byte. Ans. B. Old design t p 800 New design t p 600 Throughput %.% 600. Ans. B. Look ahead carry generator gives output in constant time if fan in = number of inputs. If we have 8 inputs, and OR gate with inputs, to build an OR gate with 8 inputs, we will need gates in level-, in level- and in level-. Hence gate delays, for each level. Similarly an n-input gate constructed with -input gates, total delay will be O(log n).. Ans. D. When a=b then P[a] will have the maximum value of the array. 5. Ans. A. 0. Ans. B. Statement II is correct. If e is the heaviest edge in cycle every mst excludes it. Statement is incorrect. Complete graph with vertices and edge weights,,5,6 in non diagonal and diagonal edges and.,5,6 will create a cycle and we can exclude the lighest edge e () from it, in a MST. So every MST of G need not include e.. Ans. B. Exp. Maximum number of iterations will be n 56 n 6. Ans. D. Lagrange s generated by G a * b * Lagrange s generated by G a b b. Ans. D. Given PDA can accept any number of a s by staying at initial state. This PDA can also accept the strings a s followed by b s but equal number of a s and b s by reaching other final state. So n n n L a n 0 a b n 0 and is deterministic contextfree.. Ans. C. Y reduces to W and we know Y is Recursively enumerable but not recursive, so W is not recursively enumerable. Z reduces to X and we know X is recursive language, then Z is also recursive language. Output is 6. Ans. D. Dynamic scoping looks for the definition of free variable in the reverse order of calling sequence. 7. Ans. B. Time complexity of heapification is O (height) = O(d) 5. Ans. C. 5 7 * 5 7 * 6 7 * 6 7 * * * * P a g e

4 6. Ans. C. 7. Ans. A. 0 Given LA 0 bit LAS Page size = 6 KB Page table Entry size (e) = 8 bits (or) 6 bytes Page table size =? Size of the page table n e 0 LAS 6 No. of pages (n) 6M PS Page table size 6 6B 8MB 8. Ans. A. C-Look disc Scheduling For first a 0 fault 0 Next a Hit again a 0 replace any location from 0 to for a 0 Hit. So total 0 page fault Difference = 0 = 50. Ans. A. It can guarantee that at most one process can be in critical section at any time. But other conditions will fail in some cases. 5. Ans. A. PL ensures serializability and here as we are following linear order in acquiring the locks there will not be any deadlock. 5. Ans. B. Total Head movements Ans. B. a,a,...a,a,a,...a Ans. B. So, no. of fragments that are transferred in this scenario is. For first a 0 0 page fault Now a Hit again a 0 0 replace only th position, so page fault. So total page fault 5. Ans. B. Given C Mb Max Output rate = 0 Mbps Arrival rate = 0 Mbps The minimum time required to transmit the data is S c m p P a g e

5 Mb S 0.sec 0 0Mpbs 0 For Mb of data, S value becomes. seconds 55. Ans. B. Frame size (L) =000bytes Sender side bandwidth (B S) = 80kbps Acknowledgement (L A) =00bytes Receiver side bandwidth (B R) =8kbps T 00ms p Tx n T T T p msg T T Ack T x ack p L 000Bytes 00ms B 0 0 BPS x S L 00 Bytes 00ms B 0 BPS A A R 00ms Channel Utilization Tn 00ms T T T 00ms 00ms 00ms n ack p Throughput B 0 0.5Kbps or 500Bps *** 5 P a g e