We are given the function $f(x) = x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right)$ for $x \neq 0$ and $f(0) = 0$. We are asked which of the following is true:

(A) $f(x)$ is differentiable everywhere.

(B) $f'(0) \neq 0$.

(C) $f'(0) = 0$.

1. Behavior of the Function at $x = 0$:

First, let's consider the behavior of $f(x)$ around $x = 0$. The term $\frac{1}{x}$ causes potential oscillations in the sine function as $x$ approaches zero. Therefore, we need to check whether the function $f(x)$ behaves smoothly and whether it is differentiable at $x = 0$.

2. Limit of $f(x)$ as $x \to 0$:

When $x = 0$, we are given that $f(0) = 0$. Now, we need to check whether the limit of $f(x)$ as $x \to 0$ exists and is continuous.

• For small $x$, $\frac{1}{x}$ becomes large, and $\sin\left( \frac{1}{x} \right)$ oscillates between $-1$ and $1$. However, since $\sin \left( \sin \left( \frac{1}{x} \right) \right)$ is bounded between $-1$ and $1$, the term $x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right)$ approaches zero as $x \to 0$, because $x^2$ dominates the oscillations of the sine function.

Therefore, the limit of $f(x)$ as $x \to 0$ is zero, and the function is continuous at $x = 0$.

3. Differentiability at $x = 0$:

To check whether $f(x)$ is differentiable at $x = 0$, we need to compute $f'(0)$ using the definition of the derivative: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ Since $f(0) = 0$, this simplifies to: $f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left( \sin\left( \frac{1}{h} \right) \right)}{h} = \lim_{h \to 0} h \sin\left( \sin\left( \frac{1}{h} \right) \right)$ Again, since $\sin\left( \sin\left( \frac{1}{h} \right) \right)$ is bounded between $-1$ and $1$, this limit becomes: $f'(0) = \lim_{h \to 0} h \cdot \text{(bounded term)} = 0$ Thus, $f'(0) = 0$, meaning the function is differentiable at $x = 0$ and the derivative at 0 is zero.

4. Differentiability Everywhere Else:

For $x \neq 0$, the function $f(x) = x^2 \sin\left( \sin\left( \frac{1}{x} \right) \right)$ is clearly differentiable because it's a composition of differentiable functions. The problem arises only at $x = 0$.

Conclusion:

• (A) True: The function is differentiable everywhere, including at $x = 0$.
• (B) False: $f'(0) = 0$, so this statement is false.
• (C) True: $f'(0) = 0$, so this statement is true.

The correct answers are (A) and (C).

Do you have any questions, or would you like further clarification? Here are some related questions to explore:

1. What are the conditions for a function to be differentiable at a point?
2. How can you check the differentiability of a piecewise function at a boundary point?
3. Why do oscillating terms like $\sin\left( \frac{1}{x} \right)$ cause trouble near zero?
4. How does L'Hopital's Rule help with limits involving oscillating functions?
5. What are some common techniques for finding derivatives of functions with oscillations?

Tip: When dealing with oscillating functions near singularities, always focus on the behavior of the dominant term, which can often simplify the analysis!