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

• A

  $(-\infty, 0) \cup(2, \infty)$
• B

  $(-\infty, 1)$
• Correct answer
  C

  (0,1)$\cup(2, \infty)$
• D

  (1,2)

Solution

$$ \text { } \begin{array}{l} \text { Here, } \mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2} \\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1) \end{array} $$ For $\mathrm{f}(\mathrm{x})$ as increasing, $\mathrm{f}^{\prime}(\mathrm{x})>0$ $$ \text { So, } 4 x(x-1)(x-2)>0 \Rightarrow x(x-1)(x-2)>0 $$
From the above figure required interval is, (0,1)$\cup(2, \infty)$

Hence, the correct answer is option: 'C'.