The least integral value of p for which f " (x) is everywhere continuous where f(x)=xpsin⁡1x+x|x|,    x≠00,    x=0 is ______ .

∵f(x)=xpsin⁡1x+x2,    x>0xpsin⁡1x−x2,    x<00,    x=0f′′(x)=−xp−4sin⁡1x−(p−2)xp−3cos⁡1x    −pxp−3cos⁡1x                                   x>0+p(p−1)xp−2sin⁡1x+2,    −xp−4sin⁡1x−(p−2)xp−3cos⁡1x    pxp−3cos⁡1x    +p(p−1))xp−2sin⁡1x−2,                       x<00,                                                             x=0RHL = LHL = f(0) = 0Since sin ∞ and cos ∞ lie between -1 to 1, for p ≥ 5, RHL = 2LHL = -2f(0) = 0For p∈(5,∞),f′′(x) is not continuous.