Let `f:[-1,1] to R` be a function defined by `f(x)={x^(2)|cos((pi)/(x))| "for" x ne 0, "for "x=0`, The set of points where f is not differentiable is

Options

A. `{x in [-1,1], x ne 0}`
B. `[x in 0-1,1]: x=0 " or " x=(2)/(2n+1), n in Z}`
C. `{ x in [-1,1]: x=(2)/(2n+1),n in Z}`
D. `[-1,1]`

Answer

C

Explanation

`f(x)={{:(x^(2)|cos(pi)/(x)|xne0),(0" "x=0):}}`
The possible point of non differential of f(x) are
`X=0,(2)/(2n+1)` where `n in I`
When x =0 f(0)=0 and ` underset(x to 0)(lim) x^(2)|cos(pi)/(x)|=0`
Hence f(x) is continueous at x=0
Now f(0) `= underset(x to 0)(lim)(f(0-h)-0)/(-h)`
`=underset(x to 0)(h^(2)cos (pi)/(h))/(-h)=0`
Similarly Rf'(0)=0 hence differentiable at x=0
Clearly non differentiable at `x=(2)/(2n+1),n in I`