A) \[-2\]
B) \[0\]
C) \[2\]
D) \[1\]
Correct Answer: C
Solution :
Given, \[f(x)={{x}^{3}}-3x\] \[\therefore \] \[f'(x)=3{{x}^{2}}-3\] ... (i) and \[f'\,\,'(x)=6x\] ...(ii) For maxima, \[f'(x)=0\] \[\Rightarrow \] \[3{{x}^{2}}-3=0\Rightarrow x=\pm 1\,\,\,\,\because x=1\in [0,\,\,2]\] \[\therefore \]\[x=1\]is the only solution. At \[x=1,\,\,f'\,\,'(x)>0\] \[\therefore \] \[x=1\]is point of minima. \[f(0)=,\,\,f(1)=-2\]and\[f(2)=2\] \[\therefore \]\[f(x)\]attains the maximum at \[x=2\] and maximum value\[=2\]
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