A) \[0\]
B) \[1/2\]
C) \[-1\]
D) \[-1/2\]
Correct Answer: A
Solution :
Let\[f(x)=|x{{|}^{3}}=\left\{ \begin{matrix} 0, & x=0 \\ {{x}^{3}}, & x>0 \\ -{{x}^{3}}, & x<0 \\ \end{matrix} \right.\] Now,\[R\,\,f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}-0}{h}=0\] \[\therefore \] \[Rf'(0)=Lf'(0)=0\] \[\therefore \] \[f'(0)=0\] Alternative Solution: Let \[f(x)=|x{{|}^{3}}\] It is clear from the figure that derivative (tangent) at \[x=0\] is\[=0\].You need to login to perform this action.
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