Prove or Disprove (4 Marks)a. ¼ n 2 - 2n + ½ Ɛ O(n 2 )b. 4n 2 + 25 n – 2010 =  (n2)c. 8n 2 + 2n - 3  O(n 2 )d. 2n 2 = 4n + O(22n)

a. ¼ n^2 - 2n + ½ ∈ O(n^2)

To prove this, we need to show that there exist constants c and n0 such that 0 ≤ ¼ n^2 - 2n + ½ ≤ c*n^2 for all n ≥ n0.

Let's choose c = 1 and n0 = 1. Then for all n ≥ 1, we have:

0 ≤ ¼ n^2 - 2n + ½ ≤ n^2

So, ¼ n^2 - 2n + ½ ∈ O(n^2) is true.

b. 4n^2 + 25n – 2010 = Ω(n^2)

To prove this, we need to show that there exist constants c and n0 such that 0 ≤ c*n^2 ≤ 4n^2 + 25n – 2010 for all n ≥ n0.

Let's choose c = 1 and n0 = 2010. Then for all n ≥ 2010, we have:

0 ≤ n^2 ≤ 4n^2 + 25n – 2010

So, 4n^2 + 25n – 2010 = Ω(n^2) is true.

c. 8n^2 + 2n - 3 ∈ O(n^2)

To prove this, we need to show that there exist constants c and n0 such that 0 ≤ 8n^2 + 2n - 3 ≤ c*n^2 for all n ≥ n0.

Let's choose c = 10 and n0 = 1. Then for all n ≥ 1, we have:

0 ≤ 8n^2 + 2n - 3 ≤ 10n^2

So, 8n^2 + 2n - 3 ∈ O(n^2) is true.

d. 2n^2 = 4n + O(22n)

This statement is false. The left side of the equation is a quadratic function, while the right side is an exponential function. As n grows, the right side will eventually become much larger than the left side, so there are no constants c and n0 that can satisfy the definition of Big O notation. Therefore, 2n^2 ≠ 4n + O(22n).