The least integral value of p for which f"(x) is everywhere continuous where `f(x)={x^p sin(1/x)+x|x|, x!=0 and 0, x=0` is _________'

Answer

To find the least integral value of \( p \) for which \( f''(x) \) is everywhere continuous, we start with the given function:

\[ f(x) = \begin{cases} x^p \sin\left(\frac{1}{x}\right) + x^2 & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 1: Determine \( f'(x) \)

We need to differentiate \( f(x) \) for \( x > 0 \) and \( x < 0 \).

**For \( x > 0 \):**
\[ f'(x) = \frac{d}{dx}(x^p \sin\left(\frac{1}{x}\right) + x^2) \]
Using the product rule:
\[ f'(x) = p x^{p-1} \sin\left(\frac{1}{x}\right) + x^p \cdot \frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right) + 2x \]
The derivative of \( \sin\left(\frac{1}{x}\right) \) is:
\[ \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} \]
Thus,
\[ f'(x) = p x^{p-1} \sin\left(\frac{1}{x}\right) - x^{p-2} \cos\left(\frac{1}{x}\right) + 2x \]

**For \( x < 0 \):**
\[ f'(x) = \frac{d}{dx}(x^p \sin\left(\frac{1}{x}\right) - x^2) \]
This will be similar to the above, but we will have a negative sign for \( x^2 \):
\[ f'(x) = p x^{p-1} \sin\left(\frac{1}{x}\right) - x^{p-2} \cos\left(\frac{1}{x}\right) - 2x \]

### Step 2: Determine \( f''(x) \)

Now we need to differentiate \( f'(x) \) to find \( f''(x) \).

**For \( x > 0 \):**
\[ f''(x) = \frac{d}{dx}(p x^{p-1} \sin\left(\frac{1}{x}\right) - x^{p-2} \cos\left(\frac{1}{x}\right) + 2x) \]
Using the product rule again, we will differentiate each term carefully.

**For \( x < 0 \):**
The same approach applies, but we will have to consider the negative sign from the \( -x^2 \) term.

### Step 3: Continuity of \( f''(x) \) at \( x = 0 \)

To ensure \( f''(x) \) is continuous at \( x = 0 \), we need to check the limits as \( x \) approaches \( 0 \) from both sides.

1. **Limit from the right (\( x \to 0^+ \))**:
\[ \lim_{x \to 0^+} f''(x) \]
This limit must yield a finite value.

2. **Limit from the left (\( x \to 0^- \))**:
\[ \lim_{x \to 0^-} f''(x) \]
This limit must also yield the same finite value.

### Step 4: Analyze the conditions for continuity

For \( f''(x) \) to be continuous at \( x = 0 \), the powers of \( x \) in \( f''(x) \) must be such that the limit exists.

- The term \( x^{p-2} \) must vanish as \( x \to 0 \), which requires \( p - 2 \geq 0 \) or \( p \geq 2 \).
- The term \( x^{p-3} \) must also vanish, which requires \( p - 3 \geq 0 \) or \( p \geq 3 \).
- Finally, for the term \( x^{p-4} \) to vanish, we need \( p - 4 \geq 0 \) or \( p \geq 4 \).

### Conclusion

The least integral value of \( p \) that satisfies all these conditions is \( p = 5 \).

Thus, the answer is:
\[ \boxed{5} \]