Question

If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
(a) continuous and differentiable at x = 3
(b) continuous but not differentiable at x = 3
(c) differentiable nut not continuous at x = 3
(d) neither differentiable nor continuous at x = 3

Answer 1

(d) neither differentiable nor continuous at x = 3

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\left|3-x\right|+\left(3+x\right),\mathrm{where}\left(x\right)\mathrm{denotes}\mathrm{the}\mathrm{least}\mathrm{integer}\mathrm{greater}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}x. \\ f\left(x\right)=\left\{\begin{array}{l}3-x+3+3,2<x<3\\ -3+x+3+4,3<x<4\end{array}\right. \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{l}-x+92<x<3\\ x+43<x<4\end{array}\right. \\ \mathrm{Here}, \\ \left(\mathrm{LHL}\mathrm{at}x=3\right)=\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }\left(-x+9\right)=-3+9=6 \\ \left(\mathrm{RHL}\mathrm{at}x=3\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }\left(x+4\right)=3+4=7 \\ \mathrm{Since},\left(\mathrm{LHL}\mathrm{at}x=3\right)\ne \left(\mathrm{RHL}\mathrm{at}x=3\right) \\ \mathrm{Hence},\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{continuous}\mathrm{at}x=3 \\ \mathrm{Therefore},\mathrm{the}\mathrm{function}\mathrm{will}\mathrm{also}\mathrm{not}\mathrm{be}\mathrm{differentiable}\mathrm{at}x=3 \end{gathered}$$