Windows: Wrong sample rates list and wrong nearestFormat decision

On Windows 7 64bit, using QAudioOutput, trying to play the attached WAV file.
It's a crappy audio: 6000Hz, 1 channel.
The playback works fine on Linux with my code.
Also, it plays fine on Windows with the 'player' example which uses the QMediaPlayer class.

Here's the issue: QAudioDeviceInfo::isFormatSupported returns false on the file, cause on my audio card (M-Audio Fast Track Pro) the accepted sample rates go from 8000 to 96000.
After that, I call QAudioDeviceInfo::nearestFormat to adjust the sample rate and it returns a sample rate of 48000 (instead of eventually 8000) !!

In any case, this is all wrong. When I used my native implementation on Qt4, I went directly on WAVEOUT and I could play that file.
This code didn't give any error on a sample rate == 6000:

HWAVEOUT dev;
WAVEFORMATEX fmt;
fmt.wFormatTag = WAVE_FORMAT_PCM;
fmt.wBitsPerSample = 16;
fmt.nChannels = 1;
fmt.nSamplesPerSec = (unsigned long)(6000);
fmt.nBlockAlign = fmt.nChannels * fmt.wBitsPerSample/8;
fmt.nAvgBytesPerSec = fmt.nSamplesPerSec * fmt.nChannels * fmt.wBitsPerSample/8;

waveOutOpen (&dev, WAVE_MAPPER, &fmt, (DWORD)wave_callback, 0, CALLBACK_FUNCTION))

So there are 2 issues here:

• QAudioDeviceInfo stores wrong information of the actually supported sample rates. The last lines of code of QWindowsAudioDeviceInfo::updateLists() is this:
  if (sampleRatez.count() > 0)
  sampleRatez.prepend(8000);
  Guys, seriously ? This is blind shooting code !

• QAudioDeviceInfo::nearestFormat should return the nearest format, and not a random one