Question

Let $f:R\to R$ be a function defined by:
$f\left(x\right)=\{\begin{array}{ccc}\begin{array}{c}\text{max}\{t^3-3t\}\\ t\le x\end{array}& ;& x\le 2\\ x^2+2x-6& ;& 2<x<3\\ [x-3]+9& ;& 3\le x\le 5\\ 2x+1& ;& x>5\end{array}$
where $\left[t\right]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I={\int }_{-2}^{2}f\left(x\right)dx$. Then the ordered pair $(m,I)$ is equal to :

Answer 1

$\begin{array}{c}\text{max}\{t^3-3t\}\\ t\le x\end{array};x\le 2$
$g\left(t\right)=t^3-3t\Rightarrow g^{\prime }\left(t\right)=3t^2-3=3(t-1)(t+1)$


$f\left(x\right)=\{\begin{array}{cc}{x}^3-3x& x<-1\\ 2& -1\le x\le 2\\ x^2+2x-6& 2<x<3\\ 9& 3\le x<4\\ 10& 4\le x<5\\ 11& x=5\\ 2x+1& x>5\end{array}$


Points of non-differentiability $=\{2,3,4,5\}$
$\Rightarrow m=4$
$I={\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^{-1}(x^3-3x)dx+{\int }_{-1}^{2}2dx$
$={[\frac{x^4}{4}-\frac{3x^2}{2}]}_{-2}^{-1}+2(2+1)=(\frac{1}{4}-\frac{3}{2})-(4-6)+6$
$=\frac{27}{4}$