Let `f(x)=sin x`
`bb"Statement I"` f is not a polynominal function.
`bb"Statement II"` nth derivative of f(x), w.r.t. x, is not a zero function for any positive integer n.

Answer

To solve the problem, we will analyze both statements regarding the function \( f(x) = \sin x \).

### Step 1: Analyze Statement I
**Statement I:** \( f \) is not a polynomial function.

- A polynomial function is defined as a function that can be expressed in the form:
\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]
where \( a_n, a_{n-1}, \ldots, a_0 \) are constants and \( n \) is a non-negative integer.
- The function \( f(x) = \sin x \) is a trigonometric function and cannot be expressed in the polynomial form mentioned above.
- Therefore, **Statement I is true.**

### Step 2: Analyze Statement II
**Statement II:** The \( n \)th derivative of \( f(x) \) with respect to \( x \) is not a zero function for any positive integer \( n \).

- Let's find the first few derivatives of \( f(x) = \sin x \):
1. First derivative:
\[ f'(x) = \cos x \]
2. Second derivative:
\[ f''(x) = -\sin x \]
3. Third derivative:
\[ f'''(x) = -\cos x \]
4. Fourth derivative:
\[ f^{(4)}(x) = \sin x \]
- We can see that the derivatives cycle every four derivatives:
- \( f^{(1)}(x) = \cos x \)
- \( f^{(2)}(x) = -\sin x \)
- \( f^{(3)}(x) = -\cos x \)
- \( f^{(4)}(x) = \sin x \)
- None of these derivatives are equal to zero for all \( x \). Thus, the \( n \)th derivative is not a zero function for any positive integer \( n \).
- Therefore, **Statement II is also true.**

### Conclusion
Both statements are true:
- **Statement I** is true because \( \sin x \) is not a polynomial function.
- **Statement II** is true because the \( n \)th derivative of \( f(x) \) is never the zero function.

### Final Answer
Both statements are correct, and Statement II provides a correct explanation for Statement I.

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