Given that

So, the function will be defined for those values of x for which

1-x² ≥0

⇒ x² ≤ 1

⇒ |x| ≤ 1

⇒ -1 ≤ x≤ 1

∴ Function is continuous in [-1, 1].

Now, we will check the differentiability at x =0

LHD at x =0,

∵ LHD does not exist, so f(x) is not differentiable at x =0.

∴ f(x) is not differentiable at x =0.

Hence, option B is correct.