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Recent posts tagged iit-interviews

If you put "Null" as introducer it will work. But by self introduction I mean "john" introducing himself- introducer will be "john" for customer "john". But I guess this needn't be considered unless specified as by the meaning of "introduction" we can assume it is a different person.
posted Jul 18, 2020 in Interview Experience vijayp_ 674 views
A)The n^{th} statement in a list of 100 statements is: "Exactly n of the statements in this list are false." 1)Exactly 1 of the statements in this list are false. 2)Exactly 2 of the statements in this list are false. 3)Exactly 3 of the statements in this list are false. . . ... (100) will be false. B)Same way we can solve it. So we will get 1-50 are true and 51-100 are false. C)This cannot happen.
posted May 23, 2020 in Interview Experience GateAspirant2020 1,850 views
If $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$, then $\frac{g(h(x))}{h(g(x))}$ is: $\frac{h(x)}{g(x)}$ $\frac{-1}{x}$ $\frac{g(x)}{h(x)}$ $\frac{x}{(1-x)^{2}}$
posted Apr 25, 2019 in Interview Experience Pooja Khatri 974 views
Answer is D. $L_1$ is context-free and hence recursive also. Recursive set being closed under complement, $L_1$' will be recursive. $L_1$' being recursive it is also recursively enumerable and Recursively Enumerable set is closed under Union. So, $L_1' \cup L_2$ is recursively enumerable. ... $L_2$')' $= L_2$ is also recursive which is not the case here. So, $II$ is also false.
posted Apr 25, 2019 in Interview Experience Pooja Khatri 960 views
$\lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}}$ is $\infty $ 0 1 Not defined
posted Apr 25, 2019 in Interview Experience Pooja Khatri 352 views
2 is correct answer. fig 2, b and c have {f,g} as upperbound. but for the graph to be a lattice it should have a least upper bound. Since b and c have two upper bounds they cannot have a least upper bound<which is always unique for a pair for vertices>. In ... {f,g}. therefore it is not a join semilattice(every pair of element should have a least upper bound). henceforth it is also not a lattice
posted Mar 16, 2019 in Interview Experience Prashansa Mittal Agr 1,883 views
$(C012.25)_H - (10111001110.101)_B $ $= 1100\;0000\;0001\;0010.\;0010\;0101$ $- 0000\;0101\;1100\;1110.\;1010\;0000$ $= 1011\;1010\;0100\;0011.\;1000\;0101$ $= 1\;011\;101\;001\;000\;011 .\;100\;001\;010$ $= (135103.412)_o$ Binary subtraction is like decimal subtraction:$ 0-0 = 0, 1-1 = 0, 1-0 = 1, 0-1 = 1$ with $1$ borrow. Correct Answer: $A$
posted May 24, 2018 in Interview Experience VS 4,087 views
Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
posted May 4, 2018 in Preparation Advice Shikha Mallick 1,127 views
Solution 48: spills to m/m while execution of program how many times m/m will be hit, we have 2 regiters Program Execution c = a + b; R2 <- R1 + R2 d = c * a; [m/m spill] <- R2*R1(we need R & R1 further value cant be replaced) e = c + a; [m/m spill] <- R2+R1 x = c * c; R2 < ... } else { d = d * d; R3<- R3*R3 e = e * e; R4<- R4*R4 } If 4 registers are used, m/m spills will be 0, hence answer is (B)
posted Feb 16, 2018 in Preparation Advice Arjun 3,197 views
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