f(x) = f(x)=(sin−1x)2cos(1x) if x≠0; f(x)=0, f(x) is

cont. everywhere in − 1 ≤ x ≤ 1

cont. everywhere in − 1 ≤ x ≤ 1

differentiable everywhere in − 1 ≤ x ≤ 1

cont. everywhere in − 1 ≤ x ≤ 1

differentiable nowhere in − 1 ≤ x ≤ 1

cont. nowhere in − 1 ≤ x ≤ 1

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f(x) = $f (x) = {(\sin^{- 1} x)}^{2} \cos (\frac{1}{x}) if x \neq 0; f (x) = 0, f (x)$ is

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differentiable everywhere in $- 1 \leq x \leq 1$

cont. everywhere in $- 1 \leq x \leq 1$

differentiable nowhere in $- 1 \leq x \leq 1$

cont. nowhere in $- 1 \leq x \leq 1$

answer is B.

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$\lim_{x \to 0} {(\sin^{- 1} x)}^{2} \cos (\frac{1}{x}) = 0 \cos (\frac{1}{x}) i s o s c i l a t i n g b e t w e e n [- 1, - 1]$

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