JEE Main 202116 Mar, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The maximum value of f(x)=∣sin2x1+cos2xcos2x1+sin2xcos2xcos2xsin2xcos2xsin2x∣,x∈Rf(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|,x \in Rf(x)=sin2x1+sin2xsin2x1+cos2xcos2xcos2xcos2xcos2xsin2x,x∈R is :AOption A. 5\sqrt 55BOption B. 34{3 \over 4}43COption C. 5DOption D. 7\sqrt 77CheckSkip · show solution
