Question

$f(x) = (\cos x + \text{i}\sin x) \cdot (\cos 3x + \text{i}\sin 3x) \cdots [\cos(2\text{n} - 1)x + \text{i}\sin(2\text{n} - 1)x] \text{n} \in \mathbb{N}$ Then $f''(x) = $ ________, (Where $\text{i} = \sqrt{-1}$ )

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

Hint:

Sum of first n odd numbers = $n^2$.

Answer 1

The Correct Option is B

Concept:
Using Euler's form: \[ \cos\theta + i\sin\theta = e^{i\theta} \] Step 1: Convert product. \[ f(x) = e^{ix} \cdot e^{i3x} \cdots e^{i(2n-1)x} \] \[ = e^{i(1+3+5+\cdots+(2n-1))x} \] Sum of odd numbers: \[ 1+3+\cdots+(2n-1) = n^2 \] \[ f(x) = e^{i n^2 x} \]
Step 2: Differentiate twice. \[ f'(x) = i n^2 e^{i n^2 x} \] \[ f''(x) = (i n^2)^2 e^{i n^2 x} = -n^4 f(x) \]
Step 3: Conclusion. \[ f''(x) = -n^4 f(x) \]