A) ?128
B) ?126
C) ?120
D) None of these
Correct Answer: A
Solution :
Given, \[f(x)=2{{x}^{3}}-21{{x}^{2}}+36x-20\] \[f'(x)=6{{x}^{2}}-42x+36\] Put \[f'(x)=0\]Þ \[6{{x}^{2}}-42x+36=0\]Þ \[{{x}^{2}}-7x+6=0\] Þ \[{{x}^{2}}-6x-x+6=0\]Þ \[(x-1)(x-6)=0\]Þ \[x=1,\,6\] Now, \[{f}''\,(x)=12x-42\] \[{f}''\,(1)=-30=-ve\] and \[{f}''\,(6)=30=+ve\] Hence \[x=6\]is the point of minima Minimum value = \[f(6)=2{{(6)}^{3}}-21{{(6)}^{2}}+36\times 6-20\] \[f(6)=-128\].
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