MathematicsMedium179×since 2002Q4125Let f(x)=1−tanx4x−πf(x) = {{1 - \tan x} \over {4x - \pi }}f(x)=4x−π1−tanx, x≠π4x \ne {\pi \over 4}x=4π, x∈[0,π2]x \in \left[ {0,{\pi \over 2}} \right]x∈[0,2π]. If f(x)f(x)f(x) is continuous in [0,π2]\left[ {0,{\pi \over 2}} \right][0,2π], then f(π4)f\left( {{\pi \over 4}} \right)f(4π) isA−1-1−1B12{1 \over 2}21C−12-{1 \over 2}−21D111Check answerSkip