Let `f(x)={(x^(p)"sin"1/x,x!=0),(0,x=0):}`
then `f(x)` is continuous but not differentiable at `x=0`, if

Options

A. `0ltple1`
B. `1lepltoo`
C. `-ooltplto`
D. `p=0`

Answer

A

Explanation

Given `f(x)` is continuous at `x=0`
`:.` Limit must exist.
i.e. `lim_(xto0)x^(p)"sin"1/x=(0)^(@)sin oo=0`
when `0lt p lt oo`………..i
Now RHD `=lim_(hto0)(h^(p)"sin"1/h-0)/h`
`=lim_(hto0)h^(p-1)"sin"1/h`
`LHD=lim_(hto0)((-h)^(p)sin(-1/h)-0)/(-h)`
`=lim_(hto0)(-1)^(p)h^(p-1)"sin"1/h`
Since `f(x)` is not differentiable at `x=0`
`:.ple1`...........ii
From eqs (i) and (ii) `0 lt ple1`