Gate-2002
Question 1 |
4 | |
2 | |
1 | |
0 |
Question 2 |
The trapezoidal rule for integration gives exact result when the integrand is a polynomial of degree
0 but not 1 | |
1 but not 0 | |
0 or 1 | |
2 |
Question 3 |
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T(1)=1
k=0; T(1)=3T(2-1)+1
k=1; T(2)=3T(20)+1=3(1)+1=4
k=2; T(4)=3T(21)+1=3(4)+1=13
k=3; T(8)=3T(22)+1=3(13)+1=40
k=4; T(16)=3T(23)+1=3(40)+1=121
Simply go through the options.
Option B:
k=4 ⇒ (34+1-1)/2
⇒ (243-1)/2
⇒ 121
Question 4 |
2 | |
3 | |
4 | |
n - 2[n/2] + 2 |
Question 5 |
log n | |
n/2 | |
(log2)n - 1 | |
n |
Question 6 |
The set of all rational negative numbers forms a group under multiplication. | |
The set of all non-singular matrices forms a group under multiplication. | |
The set of all matrices forms a group under multiplication. | |
Both B and C are true. |
a. Closure: result of a * b must be in group G.
b. There must be an identity element i.e. e * a = a * e = a
c. There must be an inverse element b for every element a such that a * b = b * a = e
d. Associativity i.e. (a * b) * c = a * (b * c)
Rational negative numbers don't form a group under multiplication, as multiplying two negative numbers results into a positive number, so closure property is not satisfied.
Set of non-singular matrices forms a group under multiplication.
The set of all matrices doesn't form a group under multiplication, since there may not be an inverse for a matrix (in particular, for singular matrices).
Question 7 |
Context free | |
Regular | |
Deterministic Context free | |
Recursive |
Question 8 |
(X∧¬Z)→Y | |
(X∧Y)→¬Z | |
X→(Y∧¬Z) | |
(X→Y)∧¬Z |
⇒ Z ∨ ¬X ∨ Y
⇒ ¬X ∨ Z ∨ Y
Option A:
(X ∧ ¬Z) → Y = ¬(X ∧ ¬Z ) ∨ Y = ¬X ∨ Z ∨ Y Hence, option (A) is correct.
Question 9 |
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HOLD is active | |
READY is active | |
None of the above |
Question 10 |
Only PCHL instruction | |
Only ADD instructions | |
Only JMP and CALL instructions | |
All instructions |
ADD Instruction: increments the program counter.
JMP & CALL: Change the values of PC.
Question 11 |
Receiver is to be synchronized for byte reception | |
Receiver recovers lost ‘0’s and ‘1’s from these padded bits | |
Padded bits are useful in parity computation | |
None of the above |
Question 12 |
xz+y'z | |
xz'+zx' | |
x'y+zx' | |
None of the above |

⇒ xz' + zx'
Question 13 |
instruction cache | |
instruction register | |
instruction opcode | |
translation look-a-side buffer |
Question 14 |
is equivalent to the binary value 0.1 | |
is equivalent to the binary value 0.01 | |
is equivalent to the binary value 0.00111… | |
cannot be represented precisely in binary |
Multiply 0.25 by 2.
0.25×2 = 0.50 (product)
Fractional part = 0.50
Carry = 0
2nd Multiplication iteration:
Multiply 0.50 by 2.
0.50×2 = 1.00 (product)
Fractional part = 0.00
Carry = 1
The fractional part in the 2nd iteration becomes zero and so we stop the multiplication iteration.
Carry from 1st multiplication iteration becomes MSB and carry from 2nd iteration becomes LSB. So the result is 0.01.
Question 15 |
1111 | |
11111 | |
111111 | |
10001 |
-15 = 11111
1's complement = 10000
2's complement = 10001
Question 16 |
floating point multiplication | |
signed 16 bit integer addition | |
arithmetic left shift | |
converting a signed integer from one size to another |
Question 17 |
At most one activation record exists between the current activation record and the activation record for the main | |
The number of activation records between the current activation record and the activation record for the main depends on the actual function calling sequence. | |
The visibility of global variables depends on the actual function calling sequence. | |
Recursion requires the activation record for the recursive function to be saved on a different stack before the recursive fraction can be called. |
Question 18 |
Do not differ | |
Differ in the presence of loops | |
Differ in all cases | |
May differ in the presence of exception |
Question 19 |
Zero | |
More than zero but less than that of an equivalent 3NF decomposition | |
Proportional to the size of F+ | |
Indetermine |
Question 20 |
Relational algebra is more powerful than relational calculus. | |
Relational algebra has the same power as relational calculus. | |
Relational algebra has the same power as safe relational calculus. | |
None of the above. |
A query can be formulated in safe Relational Calculus if and only if it can be formulated in Relational Algebra.
Question 21 |
is flagged whenever there is carry from sign bit addition | |
cannot occur when a positive value is added to a negative value | |
is flagged when the carries from sign bit and previous bit match | |
None of the above |
Question 22 |
Round Robin | |
First-In First-Out | |
Multilevel Queue Scheduling | |
Multilevel Queue Scheduling with Feedback |
Question 23 |
Has not been used for the longest time in the past. | |
Will not be used for the longest time in the future. | |
Has been used least number of times. | |
Has been used most number of times. |
Question 24 |
the operand is inside the instruction | |
the address of the operand is inside the instruction | |
the register containing the address of the operand is specified inside the instruction | |
the location of the operand is implicit |
The operand is inside the instruction --- absolute addressing.
The register containing the address of the operand is specified inside the instruction --- Register addressing.
The location of the operand is implicit --- Implicit addressing.
Question 25 |
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Question 26 |
Σ(1,4,5) | |
Σ(6,7) | |
Σ(0,1,3,5) | |
None of the above |
= (Σ(0,1,3,5) ⋅ Σ(6,7) + Σ(1,4,5))
[Σ(0,1,3,5) and Σ(6,7) ⇒ No common terms]
= (Σ(1,4,5))
Question 27 |
xyz' | |
xy+z | |
x+y | |
None of the above |
F = (xyz' + xyz + y'zy + zy')z'
= (xyz' + xyz + y'z(y+1))z'
= (xyz' + xyz + y'z)z'
= (xy(z+z') + y'z)z'
= (xy + y'z)z'
= (xyz' + y'zz')
= (xyz')
Question 28 |
x'+z | |
xyz | |
xy'+z | |
None of the above |
⇒ f(f((x+y), y), z)
⇒ f(((x+y)' + y), z)
⇒ f(((x'⋅y') + y), z)
⇒ f((x'⋅y') + y), z)
⇒ ((x'⋅y') + y)' + z
⇒ (x'⋅y')⋅y' + z
⇒ (x+y)⋅y' + z
⇒ (xy'+yy') + z
⇒ xy' + z
Question 29 |
AC = 0 and CY =0 | |
AC = 1 and CY =1 | |
AC = 1 and CY =0 | |
AC = 0 and CY =1 |
⇒ H = 0101 1101
MOV L, 6BH
⇒ L = 0110 1011
MOV A, H
A = 0101 1101
ADD L ⇒ A+L =

Here, AC=1; CY=0
Question 30 |
Outputs the sum of the present and the previous bits of the input. | |
Outputs 01 whenever the input sequence contains 11 | |
Outputs 00 whenever the input sequence contains 10 | |
None of the above |
(A,1) = (B, 01)
Previous input + Present input = 0+1 = 01
(B,0) = (A, 01)
Previous input + Present input = 1+0 = 01
(A,0) = (A, 00)
Previous input + Present input = 0+0 = 00
(A,1) = (B, 01)
Previous input + Present input = 0+1 = 01
(B,1) = (C, 10)
Previous input + Present input = 1+1 = 10
(C,1) = (C, 10)
Previous input + Present input = 1+1 = 10
Question 31 |
the pipeline stages have different delays | |
consecutive instructions are dependent on each other | |
the pipeline stages share hardware resources | |
All of the above |
If pipeline stages can’t have different delays, no dependency among consecutive instructions and sharing of hardware resources should not be there.
Question 32 |
does not require use of signal decoders | |
results in larger sized microinstructions than vertical microprogramming | |
uses one bit for each control signal | |
all of the above |
Question 33 |
4040 | |
4050 | |
5040 | |
5050 |
= 0 + [40 * 100 * 1] + [50 * 1]
= 4000 + 50
= 4050
Question 34 |
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Question 36 |
O(n) | |
O(log n) | |
O(log log n) | |
O(1) |
Let n=2m
So, T(2m) = T(2m/2)+1
We substitute T(2m) = S(m),
So now the equation becomes,
S(m) = S(m/2)+1
Now after applying master's theorem,
S(m) = O(log m)
T(n) = O(log log n)
Question 37 |
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No. of nodes in left sub tree = 2 right sub tree
No. of nodes in left sub tree = (n-1/3)
No. of nodes in right sub tree = 2(n-1/3)

Height of the tree = log3/2 n
Question 38 |
2 states | |
3 states | |
4 states | |
5 states |

Minimum no. of states that we require is "3".
Question 39 |
The complement of a recursive language is recursive. | |
The complement of a recursively enumerable language is recursively enumerable. | |
The complement of a recursive language is either recursive or recursively enumerable. | |
The complement of a context-free language is context-free. |
Question 40 |
X2 = 3 | |
X3 = 3 | |
X2 = 2 | |
X3 = 2 |

Question 41 |
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Atleast one head = 1/16
Atleast one tail = 1/16
Probability of getting one head and one tail is = 1 - 1/16 - 1/16 = 16-1-1/16 = 14/16 = 7/8
Question 42 |
Neither reflexive nor symmetric | |
Symmetric and reflexive | |
Transitive and reflexive | |
Transitive and symmetric |
A×S = {(ɸ,1), (ɸ,2), (ɸ,3), (1,ɸ), (2,ɸ), (3,ɸ)}
Not reflexive = (1,1), (2,2), (3,3) not there
Symmetric: if (a,b) is present then (b,a) is also present
Transitive: True; if (x,y), (y,z) then (z,x) is also present
Question 43 |
A context free language | |
A context sensitive language | |
A regular language | |
Parsable fully only by a Turing machine |
Question 44 |
One stack is enough | |
Two stacks are needed | |
As many stacks as the height of the expression tree are needed | |
A Turning machine is needed in the general case |
Question 45 |
Security is dynamic | |
The path for searching dynamic libraries is not known till runtime | |
Linking is insecure | |
Cryptographic procedures are not available for dynamic linking |
Question 46 |
A | |
A and B | |
A and C | |
A, B and C |
Question 47 |
the size of the blocks, and the size of the address of the blocks. | |
the number of blocks used for the index, and the size of the blocks. | |
the size of the blocks, the number of blocks used for the index, and the size of the address of the blocks. | |
None of the above |
No. of block addresses available for addressing one file(B) = No. of Maximum blocks we can use for the Index * No. of addressable blocks using one Index block(A)
Size of File = B * Size of Block
Question 48 |
A B+- tree index is to be built on the Name attribute of the relation STUDENT. Assume that all student names are of length 8 bytes, disk blocks are of size 512 bytes, and index pointers are of size 4 bytes. Given this scenario, what would be the best choice of the degree (i.e. the number of pointers per node) of the B+-tree?
16 | |
42 | |
43 | |
44 |
Then,
8(P-1) + 4P ≤ 512
12P - 8 ≤ 512
12P ≤ 520
P ≤ 43.33
P = 43
Question 49 |
Relation R is decomposed using a set of functional dependencies, F, and relation S is decomposed using another set of functional dependencies, G. One decomposition is definitely BCNF, the other is definitely 3NF, but it is not known which is which. To make a quaranteed identification, which one of the following tests should be used on the decompositions? (Assume that the closures of F and G are available).
Dependency-preservation | |
Lossless-join | |
BCNF definition | |
3NF definition |
Question 50 |
A functionally determines B and B functionally determines C | |
A functionally determines B and B does not functionally determines C | |
B does not functionally determines C | |
A does not functionally determines B and B does not functionally determines C |
But for the given instance it can be seen that B does not functionally determines C, and it can be concluded for entire relation
Question 52 |
Theory explanation is given below. |
The given relation S is irreflexive because no diagonal elements are present in the relation i.e., (x,x)∉R, ∀x∈A
If a relation is equivalence then it is
Reflexive
Symmetric
Transitive
Given relation S is not Reflexive & Transitive.
Reflexive closure on S = {(1,1), (2,2), (3,3), (1,2), (2,1)}
Transitive closure on S is does not change after performing reflexive closure on S.
S = {(1,1), (2,2), (3,3), (1,2), (2,1)}
(b) Given S = {a,b}
Powerset of S i.e., P(S) = {(ɸ), (a), (b), (a,b)}
Hasse diagram:

Question 53 |
Theory Explanation is given below. |

Eigen value of upper/ lower triangular matrix are the diagonal elements of matrix.

(b) (i) A↔(A∨A): This can tells that if A then (A or A)and if (A or A) then A. It represents result as a tautology.
(ii) (A∨B)→B: This is neither tautology nor contradiction.
(iii) A∧(¬(A∨B)): here when A is true then (¬(A∨B)) is false, then it results contradiction.
Question 54 |
Theory Explanation is given below. |

Total 5 binary trees are possible with the preorder C, B, A.
Question 55 |
Theory Explanation is given below. |
Question 59 |
Theory Explanation is given below. |
move disk from source to dest //Step-2
move (disk-1, aux, dest, source) //Step-3
Recurrence: 2T(n - 1) + 1
T(n) = 2T (n - 1) + 1
= 2[2T(n - 2) + 1] + 1
= 22T(n - 2) + 3
⁞
2k T(n - k) + (2k - 1)
= 2n-1T(1) + (2n-1 - 1)
= 2n-1 + 2n-1 - 1
= 2n - 1
≅ O(2n)
void move (int n, char A, char B, char C) {
if (n>0)
move(n-1, A, C, B);
printf("Move disk%d from pole%c to pole%c\n", n,A,C);
move(n-1, B, A, C);
}
}
Question 60 |
Theory Explanation is given below. |
{ for j = 1 ... n
{ if a[i,j] = 0 then P[i,j] = infinite;
else P[i,j] = a[i,j];
}
}
(b) ALGORITHM: For i = 1 ... n
{ for j = 1 ... n;
{ for k = 1 ... n
{
P[i,j] = min (P[i,j], P[i,k]+P[k,j])
}
}
}
(c) Actual graph:

MST: There are 2 possible MST's

