# GATE-CS-2007

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Question 1 |

Consider the following two statements about the function f(x)=|x|

Which of the following is TRUE?P.f(x) is continuous for all real values of xQ.f(x) is differentiable for all real values of x

P is true and Q is false. | |

P is false and Qis true. | |

Both P and Q are true | |

Both P and Q are false. |

**Numerical Methods and Calculus**

**GATE-CS-2007**

**Discuss it**

Question 1 Explanation:

A function is continuous if for every value of 'x', we have a corresponding f(x). Here, for every x, we have f(x) which is actually the value of x itself, without the negative sign for x < 0.

But, the given function is not differentiable for x = 0 because for x < 0, the derivative is negative and for x > 0, the derivative is positive. So, the left hand derivative and right hand derivative do not match.

Hence, P is correct and Q is incorrect. Thus, A is the correct option.

Please comment below if you find anything wrong in the above post.

But, the given function is not differentiable for x = 0 because for x < 0, the derivative is negative and for x > 0, the derivative is positive. So, the left hand derivative and right hand derivative do not match.

Hence, P is correct and Q is incorrect. Thus, A is the correct option.

Please comment below if you find anything wrong in the above post.

Question 2 |

Let S be a set of nelements. The number of ordered pairs in the largest and the smallest equivalence relations on S are:

n and n | |

n ^{2} and n | |

n ^{2} and 0 | |

n and 1 |

**Set Theory & Algebra**

**GATE-CS-2007**

**Discuss it**

Question 2 Explanation:

Consider an example set, S = (1,2,3) Equivalence property follows, reflexive, symmetric and transitive Largest ordered set are s x s = { (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3) } which are 9 which equal to 3^2 = n^2 Smallest ordered set are { (1,1) (2,2) ( 3,3)} which are 3 and equals to n. number of elements.

Question 3 |

What is the maximum number of different Boolean functions involving n Boolean variables?

n ^{2 } | |

2 ^{n} | |

2 ^{2n} | |

2 ^{n2} |

**Combinatorics**

**GATE-CS-2007**

**Discuss it**

Question 3 Explanation:

No of inputs sequences possible for a n variable Boolean function = 2^{n}Each input sequence can give either T or F as output ( 2 possible values ) So, Total no of Boolean functions are - 2X2X2X2X2X2X.............X2X2X2X2X2X2 <-------------------- 2^{n}Times --------------> 2^{2n}

Question 4 |

Let G be the non-planar graph with the minimum possible number of edges. Then G has

9 edges and 5 vertices | |

9 edges and 6 vertices | |

10 edges and 5 vertices | |

10 edges and 6 vertices |

**Graph Theory**

**GATE-CS-2007**

**Discuss it**

Question 4 Explanation:

According to

**Kuratowski's Theorem**, a graph is planar if and only if it does not contain any subdivisions of the graphs K_{5}or K_{3,3}. That means K_{5}and K_{3,3}are minimum non-planar graphs. These graphs have 5 vertices with 10 edges in K_{5}and 6 vertices with 9 edges in K_{3,3}graph. So, graph K_{5}has minimum vertices and maximum edges than K_{3,3}. So, option (B) is correct.Question 5 |

Consider the DAG with Consider V = {1, 2, 3, 4, 5, 6}, shown below. Which of the following is NOT a topological ordering?

1 2 3 4 5 6 | |

1 3 2 4 5 6 | |

1 3 2 4 6 5 | |

3 2 4 1 6 5 |

**Graph Traversals**

**GATE-CS-2007**

**Discuss it**

Question 5 Explanation:

In option D, 1 appears after 2 and 3 which is not possible in Topological Sorting.
In the given DAG it is directly visible that there is an outgoing edge from vertex 1 to vertex 2 and 3 hence 2 and 3 cannot come before vertex 1 so clearly option D is incorrect topological sort.
But for questions in which it is not directly visible we should know how to find topological sort of a DAG.
This solution is contributed by

**Parul sharma.**Question 6 |

Which of the following problems is undecidable? [2007]

Membership problem for CFGs | |

Ambiguity problem for CFGs. | |

Finiteness problem for FSAs. | |

Equivalence problem for FSAs. |

**Undecidability**

**GATE-CS-2007**

**Discuss it**

Question 6 Explanation:

A set is closed under an operation means when we operate an element of that set with that operator we get an element from that set.

Here, CFG generates a CFL and set of all CFLs is the set. But ambiguity is not an operation and hence we can never say that CFG is closed under such operation.

Only ambiguity problem for CFGs are undecidable.

Thus, option (B) is correct.

Please comment below if you find anything wrong in the above post.

Question 7 |

Which of the following is TRUE?

Every subset of a regular set is regular. | |

Every finite subset of a non-regular set is regular. | |

The union of two non-regular sets is not regular. | |

Infinite union of finite sets is regular. |

**Regular languages and finite automata**

**GATE-CS-2007**

**Discuss it**

Question 7 Explanation:

Some points for Regular Sets:

This solution is contributed by

- A set is always regular if it is finite.
- A set is always regular if a DFA/NFA can be drawn for it.

**Option A:**Every subset of a regular set is regular is False. For input alphabets a and b, a*b* is regular. A DFA can be drawn for a*b* but a n b n for n≥0 which is a subset of a*b* is not regular as we cannot define a DFA for it.**Option B:**Every finite subset of a non-regular set is regular is True. Each and every set which is finite can have a well-defined DFA for it so whether it is a subset of a regular set or non-regular set it is always regular.**Option C:**The union of two non-regular sets is not regular is False. For input alphabets a and b, a^{n}b^{n}for all n≥0 is non-regular as well as a^{n}b^{m}for n≠m is also non- regular but their union is a*b* which is regular.**Option D:**TInfinite union of finite sets is regular is False. For input alphabets a and b sets {ab}, {aabb}, {aaabbb}…….. are regular but their union {ab} U {aabb} U {aaabbb} U …………………….. gives {a n b n for n>0} which is not regular.This solution is contributed by

**Yashika Arora**.Question 8 |

How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates?

7 | |

8 | |

9 | |

10 |

**Digital Logic & Number representation**

**GATE-CS-2007**

**Combinational Circuits**

**Discuss it**

Question 9 |

Consider the following Boolean function of four variables:
f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14)
The function is:

independent of one variables. | |

independent of two variables. | |

independent of three variables. | |

dependent on all the variables. |

**Digital Logic & Number representation**

**GATE-CS-2007**

**Logic functions and Minimization**

**Discuss it**

Question 10 |

Consider a 4-way set associative cache consisting of 128 lines with a line size of 64 words. The CPU generates a 20-bit address of a word in main memory. The number of bits in the TAG, LINE and WORD fields arerespectively:

9,6,5 | |

7, 7, 6 | |

7, 5, 8 | |

9, 5, 6 |

**Computer Organization and Architecture**

**GATE-CS-2007**

**Cache and main memory**

**Discuss it**

Question 10 Explanation:

Here the number of sets = 128/4 = 32 (as it is 4 say set associative) We have total 64 words then we need 6 bits to identify the word So the line offset is 5 bits and the word offset is 6 bits and the TAG = 20-(5+6) =9 bits so it should be 9,5,6

There are 85 questions to complete.