1

Answer: B f(3) = [(3-2)(3-4)*2]/[(1-2)(1-4)] + [(3-1)(3-4)*4]/[(2-1)(2-4)] + [(3-1)(3-2)*16]/[(4-1)(4-2)] = -2/3 + 8/2 + 16/3 = 26/3 = 8(1/3). Ref: https://mat.iitm.ac.in/home/sryedida/public_html/caimna/interpolation/lagrange.html

3

For answering there is no need to execute the query, we can directly answer this as $2$ How? Group by Student_Names It means all name that are same should be kept in one row. There are $3$ names. But in that there is a duplicate with Raj being repeated $\implies$ Raj produces ...

5

Let $G = (V, E)$ be a simple undirected graph, and $s$ be a particular vertex in it called the source. For $x \in V$, let $d(x)$ denote the shortest distance in $G$ from $s$ to $x$. A breadth first search (BFS) is performed starting at $s$. Let $T$ ... of $G$ that is not in $T$, then which one of the following CANNOT be the value of $d(u) - d(v)$? $-1$ $0$ $1$ $2$

6

Clustered- this is the definition of clustered indexing and for the same reason a table can have only one clustered index. http://www.ece.rutgers.edu/~yyzhang/spring03/notes/7-B+tree.ppt Correct Answer: C

7

*This is my first post, so forgive me if its written badly*

After preparing twice (Gate 17 & 18) and failing twice I was frustrated and disheartened. I left the hope of getting into premium institutions and joined NIT Delhi with no other option. But still that dream was awake in some corner of my heart. I decided to appear for Gate 19. I started stealing time from academics to prepare for gate 19. I had this in mind that I know the concepts well as I have had prepared earlier so I will revise the syllabus and practice gate questions. I joined GO_classroom and started reading discrete maths and doing programming assignments given but I couldn't continue...

After preparing twice (Gate 17 & 18) and failing twice I was frustrated and disheartened. I left the hope of getting into premium institutions and joined NIT Delhi with no other option. But still that dream was awake in some corner of my heart. I decided to appear for Gate 19. I started stealing time from academics to prepare for gate 19. I had this in mind that I know the concepts well as I have had prepared earlier so I will revise the syllabus and practice gate questions. I joined GO_classroom and started reading discrete maths and doing programming assignments given but I couldn't continue...

10

Before you read any further, you should judge me, because I am not a topper! My rank(558) and gate score(729), marks (63.67). It is important for you to decide if my experience is worth your time. Also, the usual disclaimer, below is My reasoned routine for GATE 2019.

First things first, there is "NO SHORT CUT TO THE GATE JOURNEY". Quite trivial a line it is and you must indeed learn to value the meaning that it holds. For those preparing for GATE 2020, the biggest challenge you have right now is time. Not that you have less time in hand, but the fact that you have way way more than whats required to kill GATE, calls for immense dedication to keep the...

12

I am Shreyansh Jain, a B.Tech CSE 2018 graduate from a tier-3 college, managed AIR-86 in GATE-19 with two months of dedicated preparation.

This post might help people who messed up their exam or could not held their nerves on the D-day. I will try to pin point everything wrong I did while preparing last year. Before that I would like to thank everyone who has contributed in the making of GO book and Arjun sir especially for all his selfless efforts to make this platform and bringing all aspirants under one roof.

After GATE 2018, I was looking for an answer to the question “*Should I drop completely / or continue preparation...*

13

x* x=e, x is its own inverse y *y= e, y is its own inverse (x *y)*( x* y)= e, x *y is its own inverse (y* x)* (y* x)= e, y* x is its own inverse also x* x* e= e*e can be rewritten as follows x* y* y *x= e *y* y* e= e, (Since y *y ... group any element should have only one inverse element (unique) This implies x *y= y* x (is one element) So the elements of such group are 4 which are {x, y,e,x *y }.

14

M=14.25=1110.01= 1.11001*2^3 M=11001 MSB 1 is for sign bit , since exponent is 8 bit biased so, 2^8 -1= 127.. E= 127 +3= 130=10000010 So , 1 foe sign bit 10000010(8 bits) for exponent and 1100100000....0(23bits)= C1640000H

15

T(n)=n/n+1 so T(n)=O(n) and yes thank you for posting such an awesome question regards grv.

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