question_answer

The function\[f(x)={{(x(x-2))}^{2}}\]is increasing in the set

A)  \[(-\infty ,0)\cup (2,\infty )\]    

B)  \[(-\infty ,1)\]

C)  \[(0,1)\cup (2,\infty )\]

D)         \[(1,2)\]

E)  \[(0,2)\]

Correct Answer: C

Solution :

\[f(x)={{x}^{2}}{{(x-2)}^{2}}\] \[f(x)={{x}^{2}}.2(x-2)+2x.{{(x-2)}^{2}}\]                 \[=2x(x-2)(x+x-2)\]                 \[=4x(x-1)(x-2)\] The function is increasing. \[\Rightarrow \]               \[f(x)>0\]                 \[4x(x-1)(x-2)>0\] \[\Rightarrow \]               \[x\in (0,1)\cup (2,\infty )\]