Let $$f : \mathbb{R} \to \mathbb{R}$$ be a function defined by $$f(x) = \log_{\sqrt{m}} \{\sqrt{2}(\sin x - \cos x) + m - 2\}$$, for some $$m$$, such that the range of $$f$$ is $$[0, 2]$$. Then the value of $$m$$ is _____.

We want to find $$m$$ such that $$f(x) = \log_{\sqrt{m}}\{\sqrt{2}(\sin x - \cos x) + m - 2\}$$ has range $$[0, 2]$$. Define $$u = \sqrt{2}(\sin x - \cos x) = 2\sin(x - \pi/4)$$. Since $$\sin(x - \pi/4)\in[-1,1]$$, the range of $$u$$ is $$[-2,2]$$. Therefore the argument of the logarithm is $$g(x) = u + m - 2\in[m-4,m]$$.

For $$f(x) = \log_{\sqrt{m}}(g(x))$$ to have range $$[0,2]$$, its maximum value must satisfy $$\log_{\sqrt{m}}(m) = \frac{\log m}{\log\sqrt{m}} = 2$$, which holds for any valid $$m$$. Its minimum value must satisfy $$\log_{\sqrt{m}}(m-4) = 0$$, implying $$m-4 = 1$$ and hence $$m = 5$$.

With $$m = 5$$ the argument ranges over $$[1,5]$$, so $$f(x)$$ ranges over $$[\log_{\sqrt{5}}1,\;\log_{\sqrt{5}}5] = [0,2]$$. Thus the required value is $$m = 5$$, corresponding to Option 1.