JEE Main 2022 (Online) 29th June Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be a function defined by : $f(x)=\begin{cases}\max_{t\le x}\{t^{3}-3t\}; & x\le2\\ x^{2}+2x-6; & 2<x<3\\ [x-3]+9; & 3\le x\le5\\ 2x+1; & x>5\end{cases}$ where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I=\int_{-2}^{2}f(x)\,dx$. Then the ordered pair $(m,I)$ is equal to :
Differentiation and its Application
Solution
Concept. Find $\max_{t\le x}(t^3-3t)$ piece by piece, then check every joining point. Formulas used. (1) $t^3-3t$ has a local max $2$ at $t=-1$, a local min $-2$ at $t=1$, and equals $2$ again at $t=2$Step 1 - First piece. By (1), the running max is $x^3-3x$ for $x\le-1$, and stays $2$ for $-1\le x\le2$. Step 2 - Joins. $x=-1$: slopes $0$ and $0$; smooth. $x=2$: value $2$ on both sides; slopes $0$ and $2x+2=6$. Corner. $x=3$: left limit $9+6-6=9$, right $[0]+9=9$; slopes $8$ and $0$. Corner. $x=4$: $[x-3]$ jumps from $9$ to $10$. Not differentiable. $x=5$: left limit $10$, $f(5)=11$. Not differentiable. So $m=4$. Step 3 - Integral. $$I=\int_{-2}^{-1}\left(x^3-3x\right)dx+\int_{-1}^{2}2\,dx=\left[\frac{x^4}{4}-\frac{3x^2}{2}\right]_{-2}^{-1}+6.$$ The bracket is $\left(\frac14-\frac32\right)-(4-6)=\frac34$, so $I=\frac{27}{4}$. Answer: (C)