GATE-CS-2001
Question 1 |
S1: The sum of two singular n × n matrices may be non-singular S2: The sum of two n × n non-singular matrices may be singular.Which of the following statements is correct?
S1 and S2 are both true | |
S1 is true, S2 is false | |
S1 is false, S2 is true | |
S1 and S2 are both false |
Discuss it
M1 and M2 are singular M1 = 1 1 1 1 M2 = 1 -1 -1 1 But M1+M2 is non-singular M1+M2 = 2 0 0 2S2 is True: The sum of two n × n non-singular matrices may be singular
M1 and M2 are non-singular M1 = 1 0 0 1 M2 = -1 0 0 -1 But M1+M2 is singular M1+M2 = 0 0 0 0
Question 2 |
R1(a,b) iff (a+b) is even over the set of integers R2(a,b) iff (a+b) is odd over the set of integers R3(a,b) iff a.b > 0 over the set of non-zero rational numbers R4(a,b) iff |a - b| <= 2 over the set of natural numbersWhich of the following statements is correct?
R1 and R2 are equivalence relations, R3 and R4 are not | |
R1 and R3 are equivalence relations, R2 and R4 are not | |
R1 and R4 are equivalence relations, R2 and R3 are not | |
R1, R2, R3 and R4 are all equivalence relations |
Discuss it
- R1(a,b)
- Reflexive : Yes, because (a+a) is even.
- Symmetrix : Yes, (a+b) is even ⟹ (b+a) is even.
- Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.
- R2(a,b)
- Reflexive : No, because (a+a) is even.
- R3(a,b)
- Reflexive : Yes, because a.a > 0.
- Symmetrix : Yes, a.b > 0 ⟹ b.a > 0.
- Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.
- R4(a,b)
- Reflexive : Yes, because |a-a| ≤ 2.
- Symmetrix : Yes, |a-b| ≤ 2 ⟹ |b-a| ≤ 2.
- Transitive : No, because |a-b| ≤ 2 and |b-c| ≤ 2 ⇏ (a-c) is even.
Question 3 |
Consider two well-formed formulas in prepositional logic.
F1 : P ⇒ ¬P
F2 : ( P⇒¬P)∨(¬P⇒P)
Which of the following statements is correct?
F1 and F2 are both satisfiable | |
F1 unsatisfiable, F2 is satisfiable | |
F1 is unsatisfiable, F2 is valid | |
F1 is satisfiable, F2 is valid |
Discuss it
The concept behind this solution is:
a) Satisfiable
If there is an assignment of truth values which makes that expression true.
b) UnSatisfiable
If there is no such assignment which makes the expression true
c) Valid
If the expression is Tautology
Here, P => Q is nothing but –P v Q
F1: P => -P = -P v –P = -P
F1 will be true if P is false and F1 will be false when P is true so F1 is Satisfiable
F2: (P => -P) v (-P => P) which is equals to (-P v-P) v (-(-P) v P) = (-P) v (P) =
Tautology
So, F1 is Satisfiable and F2 is valid
Option (a) is correct.
Question 4 |
Only S1 is correct | |
Only S2 is correct | |
Both S1 and S2 are correct | |
None of S1 and S2 is correct |
Discuss it
Question 5 |
If a language is context free it can always be accepted by a deterministic push-down automaton | |
The union of two context free languages is context free | |
The intersection of two context free languages is context free | |
The complement of a context free language is context free |
Discuss it
Question 6 |
N2 | |
2N | |
2N | |
N! |
Discuss it
Question 7 |
exploit the temporal locality of reference in a program | |
exploit the spatial locality of reference in a program | |
reduce the miss penalty | |
none of the above |
Discuss it
Question 8 |
Virtual memory implements the translation of a program‘s address space into physical memory address space | |
Virtual memory allows each program to exceed the size of the primary memory | |
Virtual memory increases the degree of multiprogramming | |
Virtual memory reduces the context switching overhead |
Discuss it
Question 9 |
INTER | |
RESET IN | |
HOLD | |
READY |
Discuss it
Question 10 |
It cannot have subroutine call instruction | |
It can have subroutine call instruction, but no nested subroutine calls | |
Nested subroutine calls are possible, but interrupts are not | |
All sequences of subroutine calls and also interrupts are possible |
Discuss it