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The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}$ increases for all $\quad$
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The correct answer is:
$\mathrm{x}>1$ only
Given $f(x)=x^{2}-2 x$ On differentiating w.r.t $^{\circ} x^{\prime}$, we get $f^{\prime}(x)=2 x-2$ $f(x)$ is increasing, if $f^{\prime}(x)>0$ $\Rightarrow 2 x-2>0$
$\Rightarrow x>1$
$\Rightarrow x>1$
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