The correct option is C [0,98]
Given : f(x)=sin4x+cos4x−12sin2x
⇒f(x)=1−2sin2xcos2x−12sin2x⇒f(x)=1−sin22x2−sin2x2⇒f(x)=−sin22x+sin2x−22⇒f(x)=−(sin2x+12)2−942
We know that,
−1≤sin2x≤1⇒−12≤sin2x+12≤32⇒0≤(sin2x+12)2≤94⇒−98≤(sin2x+12)2−942≤0⇒0≤−(sin2x+12)2+942≤98∴f(x)∈[0,98]