Question

The coefficient of \( (x - \frac{\pi}{2}) \) in the Taylor series expansion of the function

\[ f(x) = \begin{cases} \frac{4(1 - \sin(x))}{2x - \pi}, & x \neq \frac{\pi}{2} \\ 0, & x = \frac{\pi}{2} \end{cases} \]

about \( x = \frac{\pi}{2} \) is ............

Hint:

The coefficient of \( (x - a) \) in the Taylor series is the first derivative of the function evaluated at \( x = a \).

Answer 1

Correct Answer: 1

Step 1: Examine the function at \( x = \frac{\pi}{2} \).
We are asked to find the coefficient of \( (x - \frac{\pi}{2}) \) in the Taylor expansion around \( x = \frac{\pi}{2} \). To begin, evaluate the function at \( x = \frac{\pi}{2} \): \[ f\left(\frac{\pi}{2}\right) = 0. \]
Step 2: Derivative of \( f(x) \) near \( x = \frac{\pi}{2} \).
Next, we compute the first derivative of \( f(x) \). However, the function is defined piecewise, so we need to find the limit of \( f'(x) \) as \( x \to \frac{\pi}{2} \). For \( x \neq \frac{\pi}{2} \), we differentiate the expression: \[ f'(x) = \frac{d}{dx}\left( \frac{4(1 - \sin(x))}{2x - \pi} \right). \] By applying the quotient rule: \[ f'(x) = \frac{(2x - \pi)(-4 \cos(x)) - 4(1 - \sin(x))(2)}{(2x - \pi)^2}. \] At \( x = \frac{\pi}{2} \), both terms in the numerator vanish, which results in: \[ f'(x) = 0 \text{ at } x = \frac{\pi}{2}. \]
Step 3: Conclusion.
Since the first derivative is zero, the coefficient of \( (x - \frac{\pi}{2}) \) in the Taylor expansion is zero. Therefore, the correct answer is \( \boxed{0} \).