A) 1
B) 2
C) 0
D) \[\frac{4}{27}\]
Correct Answer: D
Solution :
Given \[f(x)=(x-1){{(x-2)}^{2}}\] \[f(x)=(x-1)({{x}^{2}}+4-4x)\]; \[f(x)=({{x}^{3}}-5{{x}^{2}}+8x-4)\] Now \[f'(x)=3{{x}^{2}}-10x+8\], \[f'(x)=0\] Þ \[3{{x}^{2}}-10x+8=0\]Þ \[(3x-4)(x-2)=0\]Þ \[x=\frac{4}{3}\], 2 Now \[{f}''(x)=6x-10\] \[{f}''(4/3)=6\times 4/3-10<0\] \[{f}''(2)=12-10>0\] Hence at \[x=\frac{4}{3}\] the function will occupy maximum value. \[\therefore \] Maximum value =\[f(4/3)=4/27\].
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