Let `f(x)={{:(,x^(n)"sin "(1)/(x),x ne 0),(,0,x=0):}` Then f(x) is continuous but not differentiable at x=0. If

Options

A. `0 lt p le 1`
B. `1 le p lt oo`
C. `-oo lt p lt 0`
D. p=0

Answer

A

Explanation

`f(x)=x^(p) sin""1/x, x gt 0 and f(x)=0, x=0`
Since at x=0, f(x) is a continuous function.
`therefore underset(x to 0^(+))lim f(x)=f(0)= 0 rArr underset(x to 0^(+))lim x^(p) sin""1/x=0 rArr p gt 0`
f(x) is differentiable at x=0, if `underset(x to 0^(+))lim (f(x)-f(0))/(x-0)"exists"`
`rArr underset(x to 0^(+))lim (x^(p)sin""1/x-0)/(x-0)"exists" rArr underset(x to 0^(+))lim x^(p-1) sin""1/x" exists"`
`rArr p-1 gt or p gt 1`
If `p lt 1", then "underset(x to 0)lim x^(p-1) sin(1/x)` does not exist and at x=0, f(x) is not differentiable.
For `0 lt p le 1`, f(x) is a continuous function at x=0 but not differentiable