JEE Main 202126 Feb, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The maximum slope of the curve y=12x4−5x3+18x2−19xy = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19xy=21x4−5x3+18x2−19x occurs at the point :AOption A. (3,212)\left( {3,{{21} \over 2}} \right)(3,221)BOption B. (0, 0)COption C. (2, 9)DOption D. (2, 2)CheckSkip · show solution
