1. Rewrite the trigonometric part

We have $f(x)=\log_{\sqrt m}\Big(\sqrt2(\sin x-\cos x)+m-2\Big).$

Use $\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right).$ Therefore, $\sqrt2(\sin x-\cos x)=2\sin\left(x-\frac\pi4\right).$

So the argument of the logarithm becomes $2\sin\left(x-\frac\pi4\right)+m-2.$

Since $\sin\theta\in[-1,1]$, this expression ranges over $[m-4,\,m].$

2. Use the given range of the logarithmic function

The range of $f$ is given as $[0,2]$.

Now the base is $\sqrt m$. For the logarithm to be valid, we need $\sqrt m>0, \qquad \sqrt m\ne 1.$ So $m>0$ and $m\ne 1$.

Also, for the logarithm with base $\sqrt m$ to be increasing, we need $\sqrt m>1 \iff m>1.$ All options satisfy this except no issue here.

If the argument ranges from $m-4$ to $m$, then because $\log_{\sqrt m}(t)$ is increasing, the range of $f$ is $\left[\log_{\sqrt m}(m-4),\,\log_{\sqrt m}(m)\right].$

This must equal $[0,2]$.

Hence, $\log_{\sqrt m}(m-4)=0$ and $\log_{\sqrt m}(m)=2.$

3. Solve the endpoint conditions

From $\log_{\sqrt m}(m-4)=0,$ we get $m-4=1$ because $\log_a 1=0$. Thus, $m=5.$

Check the second condition: $\log_{\sqrt5}(5)=2$ if and only if $(\sqrt5)^2=5,$ which is true.

So $m=5$ works.

4. Verify completely

For $m=5$, $f(x)=\log_{\sqrt5}\Big(\sqrt2(\sin x-\cos x)+3\Big).$ The argument ranges from $5-4=1 \quad \text{to} \quad 5.$ Therefore, $f(x)\in\left[\log_{\sqrt5}1,\,\log_{\sqrt5}5\right]=[0,2].$ So the given range is exactly satisfied.

5. Evaluate options

• A: $4$
  Argument range $[0,4]$ includes $0$, so logarithm not defined for all real $x$. Reject.

• B: $3$
  Argument range $[-1,3]$ includes negative values, not defined. Reject.

• C: $5$
  Gives range $[1,5]$, hence logarithmic range $[0,2]$. Correct.

• D: $2$
  Argument range $[-2,2]$ includes negative values, not defined. Reject.

Therefore the correct answer is Option C.