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The function $\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}$ is increasing in the set
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Verified Answer
The correct answer is:
(0,1)$\cup(2, \infty)$
$$
\text { } \begin{array}{l}
\text { Here, } \mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2} \\
\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)
\end{array}
$$
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)$\cup(2, \infty)$
\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)$\cup(2, \infty)$
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