A) \[-1\]and\[2\]
B) \[1\]
C) \[1\]and\[-\frac{1}{2}\]
D) \[1\]and\[2\]
Correct Answer: B
Solution :
We have, \[f(x)={{(x-2)}^{3/2}}(2x+1)\] \[\Rightarrow \] \[f'(x)=\frac{2}{3}{{(x-2)}^{-1/3}}(2x+1)\] \[+{{(x-2)}^{2/3}}\cdot 2\] \[=\frac{2(2x+1)}{3{{(x-2)}^{1/3}}}+2{{(x-2)}^{2/3}}\] \[=\frac{2(2x+1)+6(x-2)}{3{{(x-2)}^{1/3}}}\] \[=\frac{4x+2+6x-12}{3{{(x-2)}^{1/3}}}\] \[=\frac{10(x-1)}{3{{(x-2)}^{1/3}}}\] For critical points, \[f'(x)=0\] \[\Rightarrow \] \[x=1\] and \[f'(x)\] is not defined at\[x=2.\]
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