Question

Let f(n)= 32n⁴+ 12 n² log n . Then f(n)= (?), f(n)= Ω (?), f(n)= θ (?)

Replace question marks with the correct answer.
a) None of the cases are correct.
b) f(n)= (n⁴), f(n)= Ω(n² log n²), θ is not applicable
c) f(n)= (n), f(n)= Ω(n⁴), f(n)= θ(n⁴)
d) f(n)= (n² log n), f(n)= Ω(n² log n), f(n)= θ (n² log n)
e) f(n)= (n⁴), f(n)= Ω(n²), θ is not applicable
f) f(n)= (n²), f(n)= Ω(n²), f(n)= θ(n²)
g) f(n)= (log n), f(n)= Ω(log n), f(n)= θ(log n)

Answer 1

The function f(n) can be expressed as 32n⁴ + 12n² log n. It is Ο(n⁴), Ω(n² log n), and θ(n² log n).

Step-by-step explanation:

In this case, f(n) can be expressed as f(n) = 32n⁴ + 12n²logn.

Therefore, the correct answers are:

• f(n) = Ο(n⁴) because the highest power of n in the equation is n⁴.
• f(n) = Ω(n² log n) because the term 12n²logn grows faster than any constant times n² log n.
• f(n) = θ(n² log n) because both f(n) = Ο(n⁴) and f(n) = Ω(n² log n) are true.