If `f(x) ={(x^3 sin (1/x), x ne 0), (0, x = 0):}`, then

Options

A. `f”(2/pi) = frac{12 – pi^2}{2 pi}`
B. `f”(2/pi) = frac{24 – pi^2}{2 pi}`
C. `f"(0) = 0`
D. `f"(0) = 1`

Answer

To solve the problem, we need to find the second derivative of the function defined as:

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

### Step 1: Find the First Derivative \( f'(x) \)

For \( x \neq 0 \), we can use the product rule to differentiate \( f(x) = x^3 \sin\left(\frac{1}{x}\right) \).

Using the product rule:
\[ f'(x) = u'v + uv' \]
where \( u = x^3 \) and \( v = \sin\left(\frac{1}{x}\right) \).

Calculating \( u' \) and \( v' \):
- \( u' = 3x^2 \)
- To find \( v' \), we apply the chain rule:
\[ v' = \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} \]

Now substituting back into the product rule:
\[ f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) + x^3 \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right) \]
\[ f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right) \]

### Step 2: Find the Second Derivative \( f''(x) \)

Now we differentiate \( f'(x) \):
\[ f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right) \]

Using the product rule again:
1. Differentiate \( 3x^2 \sin\left(\frac{1}{x}\right) \):
- \( u = 3x^2 \) and \( v = \sin\left(\frac{1}{x}\right) \)
- \( u' = 6x \)
- \( v' = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} \)

Thus,
\[ \frac{d}{dx}(3x^2 \sin\left(\frac{1}{x}\right)) = 6x \sin\left(\frac{1}{x}\right) + 3x^2 \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) \]

2. Differentiate \( -x \cos\left(\frac{1}{x}\right) \):
- \( u = -x \) and \( v = \cos\left(\frac{1}{x}\right) \)
- \( u' = -1 \)
- \( v' = \frac{\sin\left(\frac{1}{x}\right)}{x^2} \)

Thus,
\[ \frac{d}{dx}(-x \cos\left(\frac{1}{x}\right)) = -\cos\left(\frac{1}{x}\right) + x \cdot \frac{\sin\left(\frac{1}{x}\right)}{x^2} = -\cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} \]

Combining these results:
\[ f''(x) = \left(6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right)\right) + \left(-\cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x}\right) \]
\[ f''(x) = 6x \sin\left(\frac{1}{x}\right) - 4 \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} \]

### Step 3: Evaluate \( f''(0) \)

To find \( f''(0) \), we need to check the limit as \( x \to 0 \):
\[ f''(0) = \lim_{x \to 0} \left(6x \sin\left(\frac{1}{x}\right) - 4 \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x}\right) \]

As \( x \to 0 \):
- \( 6x \sin\left(\frac{1}{x}\right) \to 0 \)
- \( -4 \cos\left(\frac{1}{x}\right) \) oscillates between -4 and 4 (not defined)
- \( \frac{\sin\left(\frac{1}{x}\right)}{x} \) oscillates (not defined)

Thus, \( f''(0) \) is not defined.

### Conclusion

The second derivative \( f''(x) \) is given by:
\[ f''(x) = 6x \sin\left(\frac{1}{x}\right) - 4 \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} \]
And \( f''(0) \) is not defined.