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3
There is no limit on a large value being considered. f(n)= n^0.0000001 = n^(10^-7) g(n)=lg n(base 2) n= 10^8 f(n)= 1.000001842 , g(n)=26.565 Now, I take n = 10^(10^8) f(n) = n^(10^-7) = (10^(10^8))^(10^-7) = 10^(10^(8-7)) = 10^10 g(n) = 10^8 log10 2 So, f(n) becomes larger. Thus for any x, we can have an n, where nx becomes larger than log n and stays larger from there on wards for any higher n.
posted Jun 1, 2018 in Preparation Advice Shikha Mallick 1,448 views
4
The function terminates for all powers of $2$ (which is infinite), hence (i) is false and (ii) is TRUE. Let $n = 5. $ Now, recursive calls will go like $5 - 14 - 7 - 20 - 10 - 5 -$ And this goes into infinite recursion. And if we multiply $5$ with ... possible, there are infinite recursions possible (even considering this case only). So, (iv) is TRUE and (iii) is false. So, correct answer is (D).
posted May 19, 2018 in Preparation Advice Shikha Mallick 2,239 views
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