JEE Main 2021 (Online) 24th February Evening ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be defined as
$f(x)=\begin{cases}-55x, & \text{if } x<-5\\ 2x^{3}-3x^{2}-120x, & \text{if } -5\le x\le4\\ 2x^{3}-3x^{2}-36x-336, & \text{if } x>4\end{cases}$
Let $A=\{x\in\mathbf{R}:f$ is increasing$\}$. Then $A$ is equal to :
Differentiation and its Application
Solution
Concept. Check the sign of $f'$ piece by piece; $f$ is continuous at $-5$ and $4$. Formulas used. (1) $6x^2-6x-120=6(x-5)(x+4)$ and $6x^2-6x-36=6(x-3)(x+2)$Step 1 - $x<-5$. $f'=-55<0$: decreasing. Step 2 - $-5<x<4$. By (1), $f'>0$ only for $x<-4$: increasing on $(-5,-4)$. Step 3 - $x>4$. By (1), $f'>0$: increasing. Step 4 - Answer. $A=(-5,-4)\cup(4,\infty)$. Answer: (B)