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GATE Mock Test Computer Science Engineering (CSE) - 4 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - GATE Mock Test Computer Science Engineering (CSE) - 4

GATE Mock Test Computer Science Engineering (CSE) - 4 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The GATE Mock Test Computer Science Engineering (CSE) - 4 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 4 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Computer Science Engineering (CSE) - 4 below.
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GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 1

Find the number of digits in 176440, given that log⁡6 = 0.778 and log⁡7 = 0.845:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 1

Given,

176440

log⁡6 = 0.778

log⁡7 = 0.845

As we know that,

Power Rule: logb⁡(PQ) = q × logb⁡P

The above property of the product rule states that the logarithm of a positive number p to the power q is equivalent to the product of q and log of p.

⇒ log⁡176440 = 40log⁡ (62×72)

= 40 (2 log⁡6+2 log⁡7)

= 40 (2 × 0.778 + 2 × 0.845)

= 40 (1.556 + 1.69)

= 40 (3.246)

= 129.84

So, number of digits in 176440 = [(log⁡176440)+ integral part of 1]

= 129 + 1

= 130 digits

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 2

Direction: Choose the correct synonym (word with similar meaning) of the head word form the four options given and mark the right option.

DEFIANCE

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 2

The word 'Defiance' means refusal to obey.

The synonyms of the word 'Defiance' are "disobedience, rebellion, disrespect".

From the synonym of the given word, we can say that the word 'Disobedience' has the same meaning.

The word 'Disobedience' means failure or refusal to obey rules or someone in authority.

Hence, the correct option is (B).

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GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 3

If all the students of class VI shake hands with each other and the total number of the handshake is 780. Find the number of students in class VI:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 3

Given,

Number of shake hands = 780

Let the total number of students is y.

yC= 780
= 780
=780

⇒ y(y−1) = 1560

⇒ y2 − y − 1560 = 0

Using discriminant method:

D = b− 4ac

= (−1)2−4 (1) (−1560)

= 1 + 6240

= 6241
X = 


= 80/2

= 40

∴ Number of students is 40.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 4

The scores of 15 students in an examination were recorded as 10,5,8,16,18,20,8,10,16,20,18,11,16,14 and 12. After calculating the mean, median and mode, an error is found. One of the values is wrongly written as 16 instead of 18. Which of the following measures of central tendency will change?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 4

Recorded data are:

10, 5, 8, 16, 18, 20, 8, 10, 16, 20, 18, 11, 16, 14 and 12

Let us arrange the following data in ascending order to find out the median and mode, i.e.:

5, 8, 8, 10, 10, 11, 12, 14, 16, 16, 16, 18, 18, 20, 20

Total number of observation = 15

As we know,

Mean = ∑x/n

= 202/15

Median = value of  observation

∴ Median =8th observation = 14

From above observation 16 occurs maximum number of times.

∴ Mode =16

According to question,

One of the values is wrongly written as 16 instead of 18.

= 5, 8, 8, 10, 10, 11, 12, 14, 16, 16, 18, 18, 18, 20, 20

 Mean = ∑x/n

= 204/15

Median = value of  observation

Median = 8th  observation = 14

From the above observation 18 occurs the maximum number of times.

∴ Mode = 18

We can see that the mean and mode are changed.

Hence, the correct option is (D).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 5

Direction: Given below is an idiom/phrase which is followed by four alternative meanings to each. Choose the correct option which is the most appropriate meaning.

Gordian knot

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 5

The most appropriate meaning of the given idiom/phrase 'Gordian knot' is 'A difficult problem'.

Gordian knot: A complicated problem that can only be solved with creative or unorthodox thinking.

For example: The coding problem looked like a Gordian knot until we realized we could bypass it altogether with a different approach.​

By the given meaning and example we can say that 'A difficult problem' is the appropriate meaning.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 6

Look at this series: 2, 6, 12, 20, 30, 42, ...

What number should come next?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 6

The terms are of type x2 + x. (Starting from x = 1)
So, next term will be = 72 + 7 = 56

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 7

One of the four words given in the four options does not fit the set of words. The odd word from the group is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 7

'Hiatus' means 'a pause or break in continuity of a sequence or an activity'. So, options 1, 2 and 3 refer to gap or rest. But, 'end' means 'a termination of a state or situation', thus the odd one out.

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 8

Directions: The question is based on the following bar graph giving the number of students in different branches of an engineering college for the years 1996-1997 and 1997-98.

The percentage decrease in electronics students of ABC Engineering College for the period 1996-97 to 1997-98 exceeds the percentage decrease in electronics students of XYZ Engineering College for the same period by:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 8

% decrease in 1997-98 for ABC Engineering College = 70/150 × 100 for electronics
% decrease in 1997-98 for XYZ Engineering College = 60/180 × 100 for electronics
Difference = (70/150 - 60/180) × 100 = 13.33%

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 9

Directions: Read the given statement carefully and answer the question that follows.

Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists and so on.

Which of the following is the best inference from the above statement?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 9

Hetero means different. It is clearly mentioned that the movement comprised of moderates, liberals, radicals, socialists and so on. Hence, option (4) is correct.

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 10

A bottle of whisky contains 3/4 th whisky and the rest is water. What quantity of the mixture must be taken away and substituted by equal quantity of water to have half whisky and half water?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 10

Percentage of whisky in mixture = 75%
Percentage of whisky in water = 0%

Therefore, mixture and water should be mixed in the ratio of 2 : 1.
or,
We need to replace 1/3rd or of the mixture with water.

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 11

A bolt is manufactured by 3 machines A, B and C. Machine A turns out twice as many items as B and machines B and C produce an equal number of items. 2% of bolts produced by A and B are defective and 4% of bolts produced by C are defective. All bolts put into 1 stockpile and 1 is chosen from this pile. What is the probability that it is defective?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 11

Let A be the event in which an item has been produced by machine A and so on.

D = The event of the item being defective.

P(A) = 1/2

P(B) = 1/4

P(D/A) = P = 2/100 (An item is defective, given that A has produced it)

P(D/B) = 2/100

P(D/C) = 4/100

P(C) = 1/4

By theorem of total productivity,

Required probability = Chosen a defective bolt from Machine A+ Chosen a defective bolt from Machine B+ Chosen a defective bolt from Machine C.

P(D) = P(A) × P(D/A) + P(B) × P(D/B) + P(C) × P(D/C)

=0.025

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 12

Which statement defines an Algorithm?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 12

According to the definition of the Algorithm,

  • An algorithm is a sequence combination of finite steps to solve a particular problem. 
  • It is a finite step-by-step procedure to achieve a required result.
  • An algorithm always produces at least one output.
  • It always terminates an infinite amount of time.
  • An algorithm is independent of the programming language.

Hence, the correct option is (B).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 13

The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 13

For every packet sent the window size increases by 2. So, when

Window size [WS=1] initially

⇒ After 1 RTT, window size = 2 and 1 segment is sent

⇒ After 2 RTT window size = 4 and 3 segment sent in total

⇒ After 3 RTT, window size = 8 and 7 segment sent in total

⇒ After ‘x’ RTTS window size 2x and 2x−1 segment are sent

Now, 2x−1 = 3999

⇒ 2x = 4000

⇒ x = log2⁡(4000)

⇒ x = 12 RTT's

Alternately we can say that,

1st RTT- 1 segment sent.

2nd RTT- 2 new segments sent.

3rd RTT- 4 new segment sent.

4th RTT- 8 new segments sent.

5th RTT- 16 new segments sent.

6th RTT- 32 new segments sent.

7th RTT- 64 new segments sent.

8th RTT- 128 new segments sent.

9th RTT- 256 new segments sent.

10th RTT- 512 new segments sent.

11th RTT- 1024 new segments sent.

12th RTT- 2048 new segments sent.

Total segments sent = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048

=4095 > 3999

Hence, the correct answer is 12.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 14

Which of the following statement is wrong?

  1. Timestamp protocol suffers from more aborts.
  2. A block hole in a DFD is a data store with only inbound flows.
  3. Multivalued dependency among attributes is checked at 3 NF level.
  4. An entity-relationship diagram is a tool to represent the event model.
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 14

STATEMENT- (I): TRUE

Timestamp protocol suffers from more aborts (suffer from cascading rollback).

STATEMENT- (II): TRUE

A block hole in a DFD is a data store with only inbound flows.

STATEMENT- (III): FALSE

Multivalued dependency among attributes is checked at 3 NF level because it is checked at 4 NF level.

STATEMENT- (IV): FALSE

An entity-relationship diagram is a tool to represent the event model because it is a tool to represent the data model.

Hence, the correct options are (C) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 15

Consider the following grammar G:
S → X|a|YXZ
X → Y∣b
Y → S∣c
Z → d
After removing the left recursion how many Production Rules exist in the grammar.
(In G there are 8 Production Rules)

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 15

Given:

S → X|a|YXZ

X → Y∣b

Y → S∣c

Z → d

Substitute to X production in S.

S → Y|b|a|YYZ|YbZ

Y → S∣c

Z → d

Substitute to Y production in S.

S → c|b|a|SZ|ccZ|ScZ|cSZ|SbZ|cbZ

Z → d

Left recursion of above productions is,

S → SZ ⇒ S → pX′,X′ → ZX′/ε

S → ScZ ⇒ S → pY′,Y′ → cZY′/ε

S → SbZ ⇒ S → pZ′,Y′ → bZZ′/ε

Total productions are:

S → pY′,Y′→cZY′/ε

S → pX,X→ZX′/ε

S → pZ′,Y′→bZZ′/ε

S → c|b|a|ccZ|cSZ|cbZ

Z → d

So, the total productions are 13.

Hence, the correct answer is (A)

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 16

The standard ordered basis of R3 is {e1,e2,e3}. Let T:R3 → R3 be the linear transformation such that,

T(e1)=7e1−5e3, T(e2)=−2e2+9e3,T(e3)=e1+e2+e3

The standard matrix of T is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 16

Given,

linear transformation are:

T(e1)=7e1−5e3

T(e2)=−2e2+9e3

T(e3)=e1+e2+e3

Let the standard matrix be A with respect to the basis e1,e2,e3,

Now, T(e1)=7e1+0e2−5e3

T(e2)=0e1−2e2+9e3

T(e3)=e1+e2+e3

The standard matrix will be (transpose of linear combinations).

Hence, the correct option is (A).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 17

The output of a multiplexer shown below is:
f (w1, w2, w3, w4) = 


The boxes 1,2 and 3 represents 2 input logic gates.

Which of the following statements is/are true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 17

It is a combinational circuit that selects binary information from one of many input lines and directs it is to single output lines. The selection of particular input lines is controlled by a set of selection lines.


Y = S¯10I0 + S¯1S0I1 + S10I2 + S1S0I3

Now,

From the given multiplexer,
f = 

Given that the output is,
f = 

Now, we need to express every term in terms of w1 and w4 as these are the inputs of selection lines. 


  represents NOR gate.

 represents AND gate.

represents OR gate.

Hence, the correct options are (A), (B) and (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 18

Consider the following program written in C:

#include<stdio.h>
f(int a, int b)
{
  if(b != 0)
   {  
       if(a != 0)
       {
          printf("#");
                 f(a/2, b);
        }
  }
        else
         {
              b = b - 1;
              f(1024, b);
         }
}
int main()
  {
        f(2048,2048);
  }

The number of times # will be printed is __________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 18

Only if part should be consider because if part contains the statement printf(“#”);

12 times # is printed which can be seen from the above given tree

Important points:

# will be printed if a ≠ 0 and b ≠ 0

log2 22048+1

=11+1

=12

Hence, the correct answer is 12.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 19

Processes P1,P2,P3 and P4 run on a single processor. Arrival time of processes P1,P2,P3 and P4 are 1,3,3 and 5 ms respectively while computation time of P1,P2,P3 and P4 are 2,X,5 and 7 ms respectively. Algorithm used by processor for processing is Shortest Job First. If the average time turnaround time is 
13/2 ms then the value of X is_____________ ms (computation time of P2 is less than P4).


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 19

Assume: Computation time of P2 is less than or equal to computation time of  P3.

Gantt chart:

Process table:

Average Turnaround time =13/2
 =13/2
3X + 17 = 26
∴ X = 3 ms
As we know that,
TAT = CT − AT
ms → milliseconds
Explore: If computation time of P2 is more than computation time of P3.
Hence, the correct answer is 3.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 20

Foreign key is a set of attributes references to primary key or alternative key of same relation or some other relation. Select the correct option regarding the foreign key:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 20

A foreign key is a set of attributes references to the primary key or alternative key of the same relation or some other relation. The correct statement regarding the foreign key is that a referencing relation record whose foreign key value is NULL is out of referential integrity constraints. Between referenced and referencing is (1:M) mapping. Each referenced record can be related to many [0 or more] referencing records and each referencing record can relate to at most 1 referenced records.

Referential integrity is the state of a database in which all values of all foreign keys are valid. A foreign key is a column or a set of columns in a table whose values are required to match at least one primary key or unique key value of a row in its parent table. A referential constraint is a rule that the values of the foreign key are valid only if one of the following conditions is true:

  • They appear as values of a parent key.
  • Some component of the foreign key is null.

To map (1:M) relationships, the primary key on the `one side' of the relationship is added to the `many sides' as a foreign key.

Hence, the correct options are (B), (C) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 21

Consider the following schedule S of transactions T1,T2,T3,T4:

Which one of the following statement/s is correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 21

We know that:

Using precedence graph method we can find out Schedule is conflict serializable or not.

If precedence graph has any loop then that schedule is going to be not conflict-serializable.

If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.

If one transaction  (T1) is writing some value × and that value is read by some other transaction (T2) then T1 should commit first then only  (T2) should read that value and commit in order to become recoverable schedule.

There is no loop in this graph so schedule is conflict serializable.

T2: T3: T1: T4 is the only serial schedule that is equivalent to given schedule. (Using Topological Sorting)

There are two write read dependency.

1. T1 write (x) and T4 read (x)

2. T2 write (y) and T4 read (y)

In first case T1 is committed first then read by T4 then T4 is committed so this is recoverable.

In second case T2 is committed first then read by T4 then T4 is committed so this is also recoverable so schedule S is recoverable.

Hence, the correct answer is (C).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 22

Consider the two processors P1 and P2 with intermediate register gateway is 0.

P1: Has four stage pipeline with stage latency 1.5 n/sec, 2 n/sec, 1 n/sec and 0.5 n/sec.

P2: Has five stage pipeline with stage latency 1 n/sec, 2.5 n/sec, 1.5 n/sec, 2 n/sec and 1 n/sec.

If each processor has infinite number of instruction to execute, then which of the following is true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 22

The clock rate or clock speed typically refers to the frequency at which the clock generator of a processor can generate pulses, which are used to synchronize the operations of its components, and is used as an indicator of the processor's speed.

For P1:

Pipeline time =max(1.5,2,1 and 0.5) n\sec

=2 n\sec

Clock rate =1/2n\sec

=500 MHz

For P2:

Pipeline time =max(1, 2.5, 1.5, 2 and 1) n\sec

= 2.5 n\sec

Clock rate = 1/2.5 n\sec

= 400 MHz

So, processor P1 faster than processor P2.

Hence, the correct option is (A).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 23

Consider the following multiplexer where 0,1,2,3 are four data input lines selected by two address line combinations of CB =00, 01,10,11 respectively and f is the output of the multiplexer:

The function f(S2, S1, A) implemented by the above circuit is?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 23

As we know that,

For a general 4 to 1 multiplier as shown, with S1 as MSB and S2 as LSB, the output expression is written as: 
F= 

i.e. I0 will be the output when the select inputs are 00,I1 will be the output when the select Inputs are 01, and so on

We get it as,

OR

Hence, the correct options are (A) and (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 24

L = { anbmcmd∣ n, m > = 1} which of the following is grammar of L:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 24

If we see b,c grammar by generating different string we can generate above language L. But if we take a,d grammar by generating different string we cannot generate above language.

Grammar in the theory of computation is a finite set of formal rules that are generating syntactically correct sentences. The formal definition of grammar is that it is defined as four tuples- G=(V, T, P, S) G is a grammar, which consists of a set of production rules. It is used to generate the strings of a language.

Hence, the correct options are (B) and (C).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 25

Consider the following problems, which among them can be implemented on a Turing machine :

S1: Calculating GCD of two numbers.

S2: Copying string.

S3: Predicting the result of tomorrow's cricket match.

S4: Calculating surface area of given cone dimensions.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 25

According to Church Turing Thesis, every problem which can be logically solved using some algorithm can be implemented on a Turing machine.

So, now these statements can be analyzed:

S1: For calculating GCD, we have an algorithm, so it can be implemented on a Turing machine.

S2: Copying string have algorithm, so it can be implemented on a Turing machine.

S3: Predicting a tomorrow's match result is undecidable and it has no algorithm. So, it cannot be implemented on a Turing machine.

S4: Calculating surface area of a cone has also logic behind it, a formula has to be used, so it can be implemented on a Turing machine.

Hence, the correct option is (B).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 26

Which of the given below options is/are FALSE?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 26

Option (A): TRUE

In the polling method, the CPU polls each device to check status bits to find out if the device has raised any interrupt. It is a software method.

Option (B): FALSE

In the vectored interrupt, a vector address is given to the CPU to identify the source of the interrupt. For example, in 8085μP, RST 4.5 is a vectored interrupt. Here, 4.5 is the interrupt vector which gives the vector address as 4.5 × 8 = (36)10 = (0024)16.

Option (C): FALSE

In DMA mode, either the CPU or the DMA controller would gain control of the system bus time, but not both. Accordingly, the CPU would be either in a busy state or a Hold state. It would be in a busy state until the I/O device prepares the data and would go to Hold state when the I/O device starts transferring data to the main memory via the DMA controller.

Option (D): TRUE

In the daisy-chaining method of interrupt handling, the devices are connected serially in such a manner that the nearest device to the CPU has the highest priority, followed by the next device and so on.

Hence, the correct options are (B) and (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 27

Consider the following augmented grammar:
S' ->.S
S ->.Aa/.bAc
A ->.d
How many rows are present in the parsing table of the LALR(1) parser?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 27

The DFA shown for the above grammar using LALR(1) parser is:

Hence, the correct answer is 9.

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 28

Find derivative of the f(x) = 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 28

Given,

Simplify the function, we get

We know that,

( a + b )2 = a2 + 2ab + b2

As we know,

Differentiating with respect to x, we get

f′(x) = 24x × 4 × ln⁡2 + 23x × 3 × ln⁡2

f′(x) = ( 4 × 24x + 3 × 23x) ln⁡2

Hence, the correct option is (D).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 29

The simplified SOP (Sum of Product) form of the Boolean expression 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 29

The simplified SOP (Sum of Product) form of the boolean expression by the two methods:

Method ( 1 ): By Minimization

Method ( 2 ): By Truth Table

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 30

Let S be a set of n elements. The number of ordered pairs in the largest and the smallest equivalence relations on S is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 4 - Question 30

The largest equivalence relation on S is a relation that contains all pairs (x, y), where x and y are elements of S. The number of such (ordered) pairs is n x n = n2
The smallest equivalence relation on S is such a relation that every element x of S is only equivalent to itself. Thus, this relation will have n ordered pairs.
Note that any equivalence relation must be reflexive (i.e. each element must be equivalent to itself), so we cannot have 0 or 1 pair in this case.
So, the smallest equivalence relation for set of n elements has n ordered pair and largest equivalence relation has n2 ordered pairs.

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