JEE Main 20204 Sep, Evening shiftMathematicsEasySequence and Series179× since 2002The minimum value of 2^sinx + 2^cosx is :AOption A. 2−1+2{2^{-1 + \sqrt 2 }}2−1+2BOption B. 21−12{2^{1 - {1 \over {\sqrt 2 }}}}21−21COption C. 21−2{2^{1 - \sqrt 2 }}21−2DOption D. 2−1+12{2^{-1 + {1 \over {\sqrt 2 }}}}2−1+21CheckSkip · show solution
