To analyze this question, let’s go through each option one by one for the function $f(x) = x^2 \sin(\sin(x^{-1}))$ with $f(0) = 0$.

1. Option (A): Differentiable Everywhere

   To check differentiability everywhere, we need to investigate the behavior of $f(x)$ at $x = 0$ and for other values of $x \neq 0$.

   • For $x \neq 0$, the function $f(x) = x^2 \sin(\sin(x^{-1}))$ is differentiable because it is the product of $x^2$ (a differentiable function) and $\sin(\sin(x^{-1}))$ (bounded for all $x$).

   • At $x = 0$, we must verify the differentiability. A function is differentiable at $x = 0$ if the limit $\lim_{x \to 0} \frac{f(x) - f(0)}{x}$ exists. Here, $\frac{f(x) - f(0)}{x} = \frac{x^2 \sin(\sin(x^{-1}))}{x} = x \sin(\sin(x^{-1})).$ Since $\sin(\sin(x^{-1}))$ oscillates between -1 and 1 as $x \to 0$, the limit does not converge to a single value. Therefore, $f(x)$ is not differentiable at $x = 0$.

   Hence, (A) is incorrect.

2. Option (B): $f'(0) \neq 0$

   From the previous calculation, we saw that the derivative $f'(0)$ does not exist because the limit $\lim_{x \to 0} \frac{f(x) - f(0)}{x}$ does not converge. Therefore, $f'(0)$ is undefined rather than nonzero. So, (B) is incorrect.

3. Option (C): $f'(0) = 0$

   Since $f'(0)$ does not exist, it cannot equal 0. Thus, (C) is also incorrect.

Conclusion: None of the options (A), (B), or (C) is correct because $f(x)$ is not differentiable at $x = 0$ and $f'(0)$ does not exist.

Would you like more details or have any questions?

Related Questions

1. How is differentiability at a point evaluated for functions with oscillating terms?
2. What conditions ensure the differentiability of a piecewise-defined function?
3. Why does the oscillatory behavior of $\sin(\sin(x^{-1}))$ affect differentiability at $x = 0$?
4. How do bounded functions behave when multiplied by terms approaching zero?
5. What techniques are useful for proving non-differentiability at specific points?

Tip

When analyzing differentiability at a specific point, checking the behavior of the function’s limit is crucial, especially with oscillatory or undefined terms.