Consider the following statements:
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
Question 1 Explanation:
Singular Matrix: A square matrix is singular if and only if its determinant value is 0. S1 is True: The sum of two singular n × n matrices may be non-singular It can be seen be taking following example. The following two matrices are singular, but their sum is non-singular.
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
Consider the following relations:
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
Question 2 Explanation:
So basically, we have to tell whether these relations are equivalence or not.
- 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.
- Reflexive : No, because (a+a) is even.
- 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.
- 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.
Consider two well-formed formulas in prepositional logic. Which of the following statements is correct?
F1 is satisfiable, F2 is valid
F1 unsatisfiable, F2 is satisfiable
F1 is unsatisfiable, F2 is valid
F1 and F2 are both satisfiable
Question 3 Explanation:
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. https://en.wikipedia.org/wiki/Tautology_(logic) This solution is contributed by Anil Saikrishna Devarasetty.
Only S1 is correct
Only S2 is correct
Both S1 and S2 are correct
None of S1 and S2 is correct
Question 4 Explanation:
We can easily build a DFA for S1. All we need to check is whether input string has even number of 0's. Therefore S1 is regular. We can't make a DFA for S2. For S2, we need a stack. Therefore S2 is not regular.
Which of the following statements is true?
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
Question 5 Explanation:
Given an arbitary non-deterministic finite automaton (NFA) with N states, the maximum number of states in an equivalent minimized DFA is at least
Question 6 Explanation:
More than one word are put in one cache block to
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
Question 7 Explanation:
See question 3 of http://www.geeksforgeeks.org/operating-systems-set-2/
Which of the following statements is false?
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
A low memory can be connected to 8085 by using
Question 9 Explanation:
A low memory can be connected to 8085 by using READY signal, Communication is only possible when READY signal is set .So (D) is correct option
Suppose a processor does not have any stack pointer register. Which of the following statements is true?
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
Question 10 Explanation:
Stack pointer register hold the address of top of stack, which is the location of memory at which CPU should resume its execution after servicing some interrupt or subroutine call. So if SP register is not available then no subroutine call instructions are possible. So (A) is correct option.
There are 50 questions to complete.
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