GATE CS 2010
Question 1 |
|S| = 2|T| | |
|S| = |T|-1 | |
|S| = |T| | |
|S| = |T|+1 |
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a / | \ b c dNow the questions is, if sum of degrees in trees are same, then what is the relationship between number of vertices present in both trees? The answer is, ξ(G) and ξ(T) is same for two trees, then the trees have same number of vertices. It can be proved by induction. Let it be true for n vertices. If we add a vertex, then the new vertex (if it is not the first node) increases degree by 2, it doesn't matter where we add it. For example, try to add a new vertex say 'e' at different places in above example tee.
Question 2 |
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
3.667 | |
3.607 | |
3.676 | |
3.575 |
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In Newton-Raphson's method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).
f(x) = x2-13 f'(x) = 2x Applying the above formula, we get Next x = 3.5 - (3.5*3.5 - 13)/2*3.5 Next x = 3.607
Question 3 |
What is the possible number of reflexive relations on a set of 5 elements?
220 | |
225 | |
215 | |
210 |
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Number of reflexive relations is 2n2-n which is 220 for n = 5
Question 4 |
A group | |
A ring | |
An integral domain | |
A field |
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Question 5 |
What is the value of Limn->∞(1-1/n)2n ?
1 | |
e-2 | |
e-1/2 | |
0 |
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Question 6 |
m2 + m4 + m6 + m7 | |
m0 + m1 + m3 + m5 | |
m0 + m1 + m6 + m7 | |
m2 + m3 + m4 + m5 |
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= PQ + QR’ + PR’ = PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’ = PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ = PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) = m2 + m4 + m6 + m7Option (A) is correct.
Question 7 |
A.100 nanoseconds
B.100×210 nanoseconds
C.100×220 nanoseconds
D.3200×220 nanoseconds
A | |
B | |
C | |
D |
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Question 8 |
P is a 16-bit signed integer. The 2's complement representation of P is (F87B)16.The 2's complement representation of 8*P
(C3D8)16 | |
(187B)16 | |
(F878)16 | |
(987B)16 |
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P = (F87B)16 is -1111 1000 0111 1011 in binary
Note that most significant bit in the binary representation is 1, which implies that the number is negative. To get the value of the number performs the 2's complement of the number. We get P as -1925 and 8P as -15400
Since 8P is also negative, we need to find 2's complement of it (-15400)
Binary of 15400 = 0011 1100 0010 1000
2's Complement = 1100 0011 1101 1000 = (C3D8)16
Question 9 |
(P(XOR)Q(XOR)R)' | |
P(XOR)Q(XOR)R | |
(P+Q+R)' | |
P+Q+R |
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Q | P | R | R’ | R’ | R | F |
0 | 0 | X | X | X | 1 | 1 |
0 | 1 | X | X | 1 | X | 1 |
1 | 0 | X | 1 | X | X | 1 |
1 | 1 | 1 | X | X | X | 1 |
Question 10 |
What does the following program print?
C
#include void f(int *p, int *q) { p = q; *p = 2; } int i = 0, j = 1; int main() { f(&i, &j); printf("%d %d n", i, j); getchar(); return 0; }
2 2 | |
2 1 | |
0 1 | |
0 2 |
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See below comments for explanation.
/* p points to i and q points to j */ void f(int *p, int *q) { p = q; /* p also points to j now */ *p = 2; /* Value of j is changed to 2 now */ }