Answer:
The answer is explained below
Explanation:
i) [tex]p(n) \ is\ an\ element\ of\ O(n^k)[/tex]
Therefore [tex]p(n)\leq c.n^k,therefore\ p(n) \ is\ an\ element\ of\ O(n^k)[/tex]
ii) [tex]p(n) \ is\ an\ element\ of\ \Omega(n^k)[/tex]
[tex]p(n)=a_kn^k+a_{k-1}n^{k-1}+.\ .\ .+a_o\\p(n)=a_kn^k(1+\frac{a_{k-1}}{a_k}\frac{1}{n} +.\ .\ .+\frac{a_o}{a_k}\frac{1}{n_k} )\\p(n)=1, as\ n\ tends\ to\ \infty, the \ limit\ in\ the \ parenthesis=1. For\ c=\frac{1}{2}a_k,n\geq n_o \\Therefore:\\p(n)\ge \frac{1}{2}a_kn^k=c.n^k[/tex]
[tex]p(n) \ is\ an\ element\ of\ \Omega(n^k)[/tex]
From i and ii, [tex]p(n) \ is\ an\ element\ of\ \Theta(n^k)[/tex]
