fn 2Inn and gn n In2 fn ogn fn Ogn fn thetagn fn ohmgn

f(n) = 2^In(n), and g(n) = n In(2). f(n) = o(g(n)) f(n) = O(g(n)) f(n) = theta(g(n)) f(n) = ohm(g(n)) f(n) = omega(g(n)) f(n) = 100 middot (log(n))^log(n), and g(n) = 1000 f(n) = o(g(n)) f(n) = O(g(n)) f(n) = theta(g(n)) f(n) = ohm(g(n)) f(n) = omega(g(n)) f(n) = n^Squareroot n, and g(n) = n^10 middot 2^n. f(n) = o(g(n)) f(n) = O(g(n)) f(n) = theta(g(n)) f(n) = ohm(g(n)) f(n) = omega(g(n))

Solution

1.) f(n) = 2^ln(n); and g(n) = n*ln(n)

f(n) = O(g(n)) and f(n) = Omega(g(n))

Therefore, we can say f(n) = theta(g(n)).

Hence, option (c) is correct.

2.) f(n) = 100 * (log(n))^log(n) and g(n) = 1000n

c.g(n) <=f(n)

therefore, f(n) = Big-Omega(g(n)) i.e., option(d) is correct.

3.) f(n) = n^sqrt(n) and g(n) = n¹⁰* 2ⁿ

since, c.g(n) >= f(n) for c=10 and n0>=1;

therefore, f(n) = O(g(n)).

Hence, option(b) is correct.