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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

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detailed solution

Correct option is A

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)=x3 Sgn, then

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