A) \[\frac{2}{27}\]
B) \[\frac{4}{27}\]
C) 5
D) 0
Correct Answer: B
Solution :
Given \[f(x)=x{{(1-x)}^{2}}\], \[f(x)={{x}^{3}}-2{{x}^{2}}+x\] Now \[f'(x)=3{{x}^{2}}-4x+1\] Put \[f'(x)=0\] i.e., \[3{{x}^{2}}-4x+1=0\] \[3{{x}^{2}}-3x-x+1=0\] Þ \[x=1,\,\,1/3\] \[{f}''(x)=6x-4\] \[\therefore {f}''\,(1)=2=\] positive and \[f''(1/3)=-2=\]?ve Hence maximum value will be at \[x=\frac{1}{3}\] Maximum value \[f\,\left( \frac{1}{3} \right)=\frac{4}{27}\].
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