Let f(x) = frac{sqrt{x - 2sqrt{x - 1}}}{sqrt{ | vedclass

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(D) We have $f(x) = \frac{\sqrt{(\sqrt{x-1})^2 + 1 - 2\sqrt{x-1}}}{\sqrt{x-1} - 1} = \frac{\sqrt{(\sqrt{x-1} - 1)^2}}{\sqrt{x-1} - 1} = \frac{|\sqrt{x

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$f(x)$ is a constant function for $x > 2$

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(D) We have $f(x) = \frac{\sqrt{(\sqrt{x-1})^2 + 1 - 2\sqrt{x-1}}}{\sqrt{x-1} - 1} = \frac{\sqrt{(\sqrt{x-1} - 1)^2}}{\sqrt{x-1} - 1} = \frac{|\sqrt{x-1} - 1|}{\sqrt{x-1} - 1}$. For $x \in [1, 2)$,$\sqrt{x-1} < 1$,so $\sqrt{x-1} - 1 < 0$,which gives $f(x) = -1$. Thus $f'(x) = 0$ for $x \in (1, 2)$. For $x > 2$,$\sqrt{x-1} > 1$,so $\sqrt{x-1} - 1 > 0$,which gives $f(x) = 1$. Thus $f'(x) = 0$ for $x > 2$. The function is undefined at $x=1$ (division by zero) and $x=2$ (derivative does not exist). Since $f(x)$ is constant on $(2, \infty)$,the correct statement is that $f(x)$ is a constant function for $x > 2$.

Let $f(x) = \frac{\sqrt{x - 2\sqrt{x - 1}}}{\sqrt{x - 1} - 1}$. Then:

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