JEE Main 2019 (Online) 11th January Morning ShiftQuestion 1
The maximum value of the function $f(x)=3x^{3}-18x^{2}+27x-40$ on the set $S=\left\{x\in R:x^{2}+30\le11x\right\}$ is :
Differentiation and its Application
Solution
Concept. First find the set $S$; then check whether $f$ is monotonic on it. Formulas used. (1) $f'>0$ on $[a,b]$ $\Rightarrow$ $\max f=f(b)$Step 1 - The set. $x^2-11x+30\le0$, i.e. $(x-5)(x-6)\le0$, so $S=[5,6]$. Step 2 - Monotonic. $f'(x)=9x^2-36x+27=9(x-1)(x-3)>0$ on $[5,6]$. Step 3 - Maximum. By (1), $$f(6)=3\cdot216-18\cdot36+27\cdot6-40=648-648+162-40=122.$$ Answer: (C)