A) \[0\]
B) \[\frac{4}{27}\]
C) \[-4\]
D) None of these
Correct Answer: B
Solution :
We have, the function, \[y=x{{(x-1)}^{2}}\] \[\frac{dy}{dx}={{(x-1)}^{2}}+2(x-1)x\] \[\Rightarrow \] \[\frac{dy}{dx}=3{{x}^{2}}-4x+1\] \[\Rightarrow \] \[\frac{dy}{dx}=(x-1)(3x-1)\] Put\[\frac{dy}{dx}=0\], (for maximum and minimum) \[(x-1)(3x-1)=0\] \[\Rightarrow \] \[x=1,\,\,\frac{1}{3}\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=6x-4\] \[{{\left[ \frac{{{d}^{2}}y}{d{{x}^{2}}} \right]}_{x=1}}=2>0\] \[{{\left[ \frac{{{d}^{2}}y}{d{{x}^{2}}} \right]}_{x=1/3}}=2-4=-2<0\] Thus, the function is maximum at\[x=\frac{1}{3}\] and maximum value of the function \[y\left( \frac{1}{3} \right)=\frac{1}{3}{{\left( \frac{1}{3}-1 \right)}^{2}}\] \[=\frac{1}{3}\cdot \frac{4}{9}=\frac{4}{27}\]
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