Let $f(n) = n$ and $g(n) = n^{(1 + \sin n)}$, where $n$ is a positive integer. Which of the following statements is/are correct? I. $f(n) = O(g(n))$ II. $f(n) = \Omega(g(n))$

Here $f(n) = n$ and $g(n) = n^{1 + \sin n}$. The function $\sin n$ oscillates continuously between $-1$ and $1$ as $n \to \infty$.

1. Statement I ($f(n) = O(g(n))$): $f(n) = O(g(n))$ requires that there exist constants $c > 0$ and $n_0$ such that $n \le c \cdot n^{1 + \sin n}$ for all $n \ge n_0$. However, whenever $\sin n < 0$ (for infinitely many values of $n$, $\sin n \approx -1$), $g(n) = n^{1 + \sin n} \approx n^0 = 1$, whereas $f(n) = n \to \infty$. Thus, no such constant $c$ exists, and statement I is false.

2. Statement II ($f(n) = \Omega(g(n))$): $f(n) = \Omega(g(n))$ requires that there exist constants $c > 0$ and $n_0$ such that $n \ge c \cdot n^{1 + \sin n}$ for all $n \ge n_0$. However, whenever $\sin n > 0$ (for infinitely many values of $n$, $\sin n \approx 1$), $g(n) = n^{1 + \sin n} \approx n^2$, which grows much faster than $f(n) = n$. Thus, no such constant $c$ exists, and statement II is false.

Hence, Neither I nor II is correct (D).