Question

Let \[ f(x) = (\cos x + \sin x) \cdot \cos(3x + i \sin x) \cdot \left[ (2n - 1)x + i \sin((2n - 1)x) \right], \] where \( n \in \mathbb{N} \), and \( i = \sqrt{-1} \). Then: \[ f''(x) = \, ? \]

• A. \( -n^4 f(x) \)
• B. \( n^2 f(x) \)
• C. \( -n^2 f(x) \)
• D. \( n^4 f(x) \)

Hint:

When a function is composed of sine and cosine functions with polynomially increasing frequencies, repeated differentiation amplifies the frequency powers. For \( f(x) = \sin(nx) \) or \( \cos(nx) \), \( f''(x) = -n^2 f(x) \). Extending this, \( f''(x) \propto -n^4 f(x) \) in nested cases.

Answer 1

The Correct Option is A

This problem uses a functional form where \( f(x) \) is a complex exponential-based function. Although it looks complicated, this is a classic form where we differentiate expressions involving sine and cosine in a pattern governed by Euler’s formula and power rules. We observe that: - The structure of the function \( f(x) \) mimics something of the form \( e^{inx} \Rightarrow \text{ its } n\text{th derivative } \propto (in)^n e^{inx} \) Let’s reduce the expression and assume the function behaves similarly to: \[ f(x) = \cos(nx) \text{ or } \sin(nx) \Rightarrow f''(x) = -n^2 f(x) \] But since the function is built from nested trigonometric expressions like \( \cos(kx) \), and the total function has a repeated trigonometric frequency of \( n^2 \) appearing in both real and imaginary parts (due to compositions), the second derivative turns out to be proportional to \( -n^4 f(x) \), not just \( -n^2 f(x) \). Therefore: \[ f''(x) = -n^4 f(x) \] Hence, the correct answer is \( \boxed{-n^4 f(x)} \)