If f(x) = |cos(2x)cos(2x)sin(2x)−cosxcosx−sinxsinxsinxcosx|, then f'(x)=0 at exactly three points in (−π,π)? f'(x)=0 at more than three points in (−π,π)? f(x) attains its maximum at x=0? f(x) attains its minimum at x=0?

f(x) = |cos(2x)cos(2x)sin(2x)−cosxcosx−sinxsinxsinxcosx| = cos2x(cos2x+sin2x)−cos2x(−cos2x+sin2x)+sin2x(−cosxsinx−cosxsinx) = cos2x+cos22x−sin22x = cos2x+cos4x=2cos(2x+4x2).cos(2x−4x2)=2cos3xcosx=cos2x+cos4x
f'(x) = −2sin2x−4sin4x=0
−sin2x−2sin4x=0
−sin2x(1+4cos2x)=0
∴sin2x=0 or 4cos2x+1=0
x=0,−π2,π2 or cos2x=−1/4
Now, cos2x=0 has 4 solutions. So, option B is correct.
f'(x)=0 at more than 3 points
f''(x) = −4cos4x−8cos4x=−12cos4x
∴f''(0)=−12<0
∴ this is maximum at x=0. Hence, option C is also correct.