If `f:[0,1]rarrR` is defined as `f(x)={(x^(3)(1-x)"sin"1/(x^(2)) 0ltxle1),(0 x=0):}`, then

Options

A. `f` is continuous in `[0,1]`
B. `f` is differentiable in `[0,1]`
C. `f` is discontinuous in `[0,1]`
D. `f` is not differentiable in `[0,1]`

Explanation

To solve the problem, we need to analyze the function defined as:

\[ f(x) = \begin{cases} x^3(1-x)\sin\left(\frac{1}{x^2}\right) & \text{if } 0 < x \leq 1 \\ 0 & \text{if } x = 0 \end{cases} \]

We will check the continuity and differentiability of the function \( f \) on the interval \([0, 1]\).

### Step 1: Check Continuity at \( x = 0 \)

To check the continuity at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( 0 \):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^3(1-x)\sin\left(\frac{1}{x^2}\right) \]

Since \( \sin\left(\frac{1}{x^2}\right) \) oscillates between -1 and 1, we can bound the function:

\[ -x^3(1-x) \leq f(x) \leq x^3(1-x) \]

As \( x \to 0^+ \), both bounds approach \( 0 \):

\[ \lim_{x \to 0^+} -x^3(1-x) = 0 \quad \text{and} \quad \lim_{x \to 0^+} x^3(1-x) = 0 \]

By the Squeeze Theorem:

\[ \lim_{x \to 0^+} f(x) = 0 \]

Since \( f(0) = 0 \), we conclude that \( f \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to evaluate the derivative:

\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x} \]

Substituting \( f(x) \):

\[ f'(0) = \lim_{x \to 0} \frac{x^3(1-x)\sin\left(\frac{1}{x^2}\right)}{x} = \lim_{x \to 0} x^2(1-x)\sin\left(\frac{1}{x^2}\right) \]

Again, using the bounds for \( \sin\left(\frac{1}{x^2}\right) \):

\[ -x^2(1-x) \leq f'(0) \leq x^2(1-x) \]

As \( x \to 0 \), both bounds approach \( 0 \):

\[ \lim_{x \to 0} -x^2(1-x) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2(1-x) = 0 \]

Thus, by the Squeeze Theorem:

\[ f'(0) = 0 \]

### Conclusion

The function \( f(x) \) is continuous on the interval \([0, 1]\) and differentiable at \( x = 0 \) with \( f'(0) = 0 \).