JEE Main 201911 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002The maximum value of the function f(x) = 3x³ – 18x² + 27x – 40 on the set S = {x ∈R:x2+30≤11x}\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}{x∈R:x2+30≤11x} is :AOption A. −-− 222BOption B. −-− 122COption C. 122122122DOption D. 222CheckSkip · show solution
