# GATE CS 1996

Question 1 |

Let

*A*and*B*be sets and let*A*and^{c}*B*denote the complements of the sets^{c}*A*and*B*. The set*(A−B)*∪*(B−A)*∪*(A*∩*B*) is equal to a). A ∪ B b). A^{c }∪ B^{c}c). A ∩ B d). A^{c }∩ B^{c}a | |

b | |

c | |

d |

**Set Theory & Algebra**

**GATE CS 1996**

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Question 2 |

Let X= {2, 3, 6, 12, 24}, Let ≤ be the partial order defined by X ≤ Y if x divides y. Number of edges in the Hasse diagram of (X,≤) is

3 | |

4 | |

9 | |

None of the above |

**Set Theory & Algebra**

**GATE CS 1996**

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Question 3 |

Suppose X and Y are sets and |X| and |Y| are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. From this one can conclude that

|X|=1,|Y|=97 | |

|X|=97,|Y|=1 | |

|X|=97,|Y|=97 | |

None of the above |

**Set Theory & Algebra**

**GATE CS 1996**

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Question 4 |

Which of the following statement is false?

The set of rational numbers is an abelian group under addition | |

The set of integers in an abelian group under addition | |

The set of rational numbers form an abelian group under multiplication | |

The set of real numbers excluding zero is an abelian group under multiplication |

**Set Theory & Algebra**

**GATE CS 1996**

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Question 5 |

Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is

1/36 | |

1/3 | |

25/36 | |

11/36 |

**Probability**

**GATE CS 1996**

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Question 5 Explanation:

There can be two cases:

This explanation is contributed by

**Case 1:**Exactly one dice has 6 facing up and other dice can have any number from 1,2,3,4,5 facing up. There will be 5*2=10 such occurrences.**Case 2:**Both of the dices having 6 coming up. Only one possible case exists for the same.

This explanation is contributed by

**Pradeep Pandey**.Question 6 |

The formula used to compute an approximation for the second derivative of a function

*f*at a point X_{0}isa | |

b | |

c | |

d |

**Numerical Methods and Calculus**

**GATE CS 1996**

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Question 7 |

Let Ax=b be a system of linear equations where A is an

*m×n*matrix and b is a*m×1*column vector and X is an*n×1*column vector of unknowns. Which of the following is false?The system has a solution if and only if, both A and the augmented matrix [Ab] have the same rank | |

If m | |

If m=n and b is a non-zero vector, then the system has a unique solution | |

The system will have only a trivial solution when m=n, b is the zero vector and rank(A) = n |

**Linear Algebra**

**GATE CS 1996**

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Question 8 |

Which two of the following four regular expressions are equivalent? (

*ε*is the empty string). (i). (00)*(ε+0) (ii). (00)* (iii). 0* (iv). 0(00)*(i) and (ii) | |

(ii) and (iii) | |

(i) and (iii) | |

(iii) and (iv) |

**Regular languages and finite automata**

**GATE CS 1996**

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Question 8 Explanation:

Here,

(00)*(ε+0) = (00)*.ε+ (00)*.0 = (00)* + (00)*0 = 0*It is equal to (iii) [ using regular expression properties ]. Here, we see that (00)* generates strings of even length and (00)*0 generated the strings of odd length. Option (C) is correct.

Question 9 |

Which of the following statements is false?

The Halting Problem of Turing machines is undecidable | |

Determining whether a context-free grammar is ambiguous is undecidable | |

Given two arbitrary context-free grammars G1 and G2 it is undecidable whether L(G1)=L(G2) | |

Given two regular grammars G1 and G2 it is undecidable whether L(G1)=L(G2) |

**Undecidability**

**GATE CS 1996**

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Question 9 Explanation:

Halting problem of Turing Machine is undecidable because there is no algorithm exists for it.
Determining a context free language is ambiguous is undecidable because no algorithm exists to decide CFG is ambiguous.
Equivalence problem of CFG is undecidable.
Everything about regular language is decidable.
Option (D) is correct.

Question 10 |

Let L ⊆ ∑* where ∑ = {a, b}. Which of the following is true ?

L = { x | x has an equal number of a's and b's } is regular | |

L = { a | n ≥ 1 } is regular^{n}b^{n} | |

L = { x | x has more a's than b's } is regular | |

L = { a | m ? 1, n ? 1 } is regular^{m}b^{n} |

**Regular languages and finite automata**

**GATE CS 1996**

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Question 10 Explanation:

For option (A) :-
L = { x | x has an equal number of

*a*'s and*b*'s } is regular for equal number of a’s and b’s we need s stack for to store the number of a’s and we push all a’s into the stack and pop all b’s for each a’s hence, a cannot be regular language. For option (B) :- L = {*a*| n ≥ 1 } is regular is also not regular , it is same as above language. This language also says that equal number of a’s followed by equal number of b’s so it also need a stack to push all a’s and pop all b’s for each a’s. For option (C) :- L= { x | x has more^{n}b^{n}*a*'s than*b*'s } is also not regular it is also CFL we need stack here for a should be greater than b. For option (D) :- L = {*a*| m ≥ 1, n ≥ 1 } is regular language because there is no restriction that equal number of a’s and b’s this language only says that a should be followed by b therefore we can draw a DFA for it. Option (D) is correct.^{m}b^{n}
There are 75 questions to complete.