Question

If \( f(x) = \begin{cases} x^3 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} \), then:

• A. \( f''(0) = 1 \)
• B. \( f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi} \)
• C. \( f''\left(\frac{2}{\pi}\right) = \frac{12 - \pi^2}{2\pi} \)
• D. \( f''(0) = 0 \)

Answer 1

The Correct Option is B

The function provided is:

\( f(x) = x^3 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right). \)

The second derivative of \( f(x) \) is:

\( f''(x) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right) \left(-\cos\left(\frac{1}{x}\right)\right). \)

Evaluating at \( x = \frac{2}{\pi} \):

\( f''\left(\frac{2}{\pi}\right) = 6\left(\frac{2}{\pi}\right) \sin\left(\frac{\pi}{2}\right) - 3 \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right). \)

Simplification yields:

\( f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} - \frac{\pi^2}{2\pi} = \frac{24 - \pi^2}{2\pi}. \)

The resulting value is:

\( f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi}. \)