Note.x3Sgn x=x3, if x>00, if x=0−x3, if x<0Clearly f is continuous at x=0f'0−=ddx−x3x=0=0f'0+=ddxx3x=0=0∴f'(0)=0⇒f is derivable at x=0.

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$If f (x) = x^{3} Sgn, then$

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$f is derivable at x = 0$

$f is continuous but not derivable$

$f' (0^{+}) = 2$

$f' (0^{-}) = 1$

answer is A.

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Detailed Solution

$\begin{array}{l} Note. \\ x^{3} S g n x = \{\begin{cases} x^{3}, i f x > 0 \\ 0, i f x = 0 \\ - x^{3}, i f x < 0 \end{cases} \\ Clearly f is continuous a t x = 0 \\ f' (0^{-}) = {[\frac{d}{d x} (- x^{3})]}_{x = 0} = 0 \\ f' (0^{+}) = {[\frac{d}{d x} (x^{3})]}_{x = 0} = 0 \\ ∴ f' (0) = 0 \Rightarrow f is derivable at x = 0. \end{array}$