The least integral value of p for which f " (x) is everywhere continuous where $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{x}}^{\mathrm{p}}\sin \left(\frac{1}{\mathrm{x}}\right)+\mathrm{x}\left|\mathrm{x}\right|,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\ne 0\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=0\end{array}\right.$ is ______ .

$\because \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{x}}^{\mathrm{p}}\sin \left(\frac{1}{\mathrm{x}}\right)+{\mathrm{x}}^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}>0\\ {\mathrm{x}}^{\mathrm{p}}\sin \left(\frac{1}{\mathrm{x}}\right)-{\mathrm{x}}^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}<0\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=0\end{array}\right.$
${\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)=\left\{\begin{array}{l}-{\mathrm{x}}^{\mathrm{p}-4}\sin \left(\frac{1}{\mathrm{x}}\right)-(\mathrm{p}-2){\mathrm{x}}^{\mathrm{p}-3}\cos \left(\frac{1}{\mathrm{x}}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ -{\mathrm{px}}^{\mathrm{p}-3}\cos \left(\frac{1}{\mathrm{x}}\right)\hspace{0.25em} \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\mathrm{x}>0\\ +\mathrm{p}(\mathrm{p}-1){\mathrm{x}}^{\mathrm{p}-2}\sin \left(\frac{1}{\mathrm{x}}\right)+2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ -{\mathrm{x}}^{\mathrm{p}-4}\sin \left(\frac{1}{\mathrm{x}}\right)-(\mathrm{p}-2){\mathrm{x}}^{\mathrm{p}-3}\cos \left(\frac{1}{\mathrm{x}}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ {\mathrm{px}}^{\mathrm{p}-3}\cos \left(\frac{1}{\mathrm{x}}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ +\mathrm{p}(\mathrm{p}-1)){\mathrm{x}}^{\mathrm{p}-2}\sin \left(\frac{1}{\mathrm{x}}\right)-2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{x}<0\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{x}=0\end{array}\right.$
RHL = LHL = f(0) = 0
Since sin ∞ and cos ∞ lie between -1 to 1, for p ≥ 5, RHL = 2
LHL = -2
f(0) = 0
For $\mathrm{p}\in (5,\mathrm{\infty }),{\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)$ is not continuous.