A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\frac{1}{\sqrt{2}}\]
D) None of these
Correct Answer: B
Solution :
Let Number = x, then cube = \[{{x}^{3}}\] Now \[f(x)=x-{{x}^{3}}\] (Maximum) Þ \[f'(x)=1-3{{x}^{2}}\] Put \[{f}'(x)=0\] Þ \[1-3{{x}^{2}}=0\]Þ \[x=\pm \frac{1}{\sqrt{3}}\] Because \[{f}''(x)=-6x=-ve.\] when \[x=+\frac{1}{\sqrt{3}}\].
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