Question

If $f\left(x\right)=\left|{\log }_e\left|x\right|\right|$, then
(a) f (x) is continuous and differentiable for all x in its domain
(b) f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1
(c) f (x) is neither continuous nor differentiable at x = ± 1
(d) none of these

Answer 1

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\left|{\log }_e\left|x\right|\right| \\ \mathrm{We}\mathrm{know}\mathrm{that}\log \mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{positive}\mathrm{value}. \\ \mathrm{Here},\left|x\right|\mathrm{is}\mathrm{positive}\mathrm{for}\mathrm{all}\mathrm{non}\mathrm{zero}x. \\ \mathrm{Therefore},\mathrm{domain}\mathrm{of}\mathrm{function}\mathrm{is}R-\left\{0\right\} \end{gathered}$$
And we know that logarithmic function is continuous in its domain.
$\mathrm{Therefore},\left|{\log }_{\mathrm{e}}\left|x\right|\right|\mathrm{is}\mathrm{continuous}\mathrm{in}\mathrm{its}\mathrm{domain}.$
$$\begin{gathered} \mathrm{We}\mathrm{will}\mathrm{check}\mathrm{the}\mathrm{differentiability}\mathrm{at}\mathrm{its}\mathrm{critical}\mathrm{points}. \\ \left|{\log }_{\mathrm{e}}\left|x\right|\right|=\left\{\begin{array}{l}\begin{array}{l}{\log }_e\left(-x\right)-\infty <x<-1\\ -{\log }_e\left(-x\right)-1<x<0\end{array}\\ \begin{array}{l}-{\log }_e\left(x\right)0<x<1\\ {\log }_e\left(x\right)1<x<\infty \end{array}\end{array}\right. \\ \left(\mathrm{LHD}\mathrm{at}x=-1\right)=\underset{x\to -1^{-}}{\lim }\frac{f\left(x\right)-f\left(-1\right)}{x-\left(-1\right)} \\ =\underset{x\to -1^{-}}{\lim }\frac{{\log }_{\mathrm{e}}\left(-x\right)-0}{x+1} \\ =\underset{h\to 0}{\lim }\frac{{\log }_{\mathrm{e}}\left[-\left(-1-h\right)\right]}{-1-h+1} \\ =\underset{h\to 0}{\lim }\frac{{\log }_{\mathrm{e}}\left(1+h\right)}{-h} \\ =-1 \\ \left(\mathrm{RHD}\mathrm{at}x=-1\right)=\underset{x\to -1^{+}}{\lim }\frac{f\left(x\right)-f\left(-1\right)}{x-\left(-1\right)} \\ =\underset{x\to -1^{+}}{\lim }\frac{-{\log }_{\mathrm{e}}\left(\mathit{-}x\right)-0}{x+1} \\ =\underset{h\mathit{\to }0}{\lim }\frac{-{\log }_{\mathrm{e}}\left[-\left(-1+h\right)\right]}{-1+h+1} \\ =\underset{h\to 0}{\lim }\frac{-{\log }_{\mathrm{e}}\left(\mathit{1}\mathit{-}h\right)}{h} \\ =\underset{h\to 0}{-\lim }\frac{{\log }_{\mathrm{e}}\left(1-h\right)}{h} \\ =-1\times -1=1 \\ \mathrm{Here},\mathrm{LHD}\ne \mathrm{RHD} \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=-1. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{\mathrm{x}\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ =\underset{x\to 1^{-}}{\lim }\frac{-{\log }_{\mathrm{e}}\left(x\right)-0}{x-1} \\ =\underset{h\to 0}{\lim }\frac{-{\log }_{\mathrm{e}}\left[\left(1-h\right)\right]}{1-h-1} \\ =\underset{h\to 0}{\lim }\frac{{\log }_{\mathrm{e}}\left(1-h\right)}{h} \\ =-1 \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-\left(1\right)} \\ =\underset{x\to 1^{+}}{\lim }\frac{{\log }_{\mathrm{e}}\left(x\right)-0}{x-1} \\ =\underset{h\to 0}{\lim }\frac{{\log }_{\mathrm{e}}\left[\left(1+h\right)\right]}{1+h-1} \\ =\underset{h\to 0}{\lim }\frac{{\log }_{\mathrm{e}}\left(1+h\right)}{h} \\ =1 \\ \mathrm{Here},\mathrm{LHD}\ne \mathrm{RHD} \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=1. \end{gathered}$$
Therefore, given function is continuous for all x in its domain but not differentiable at x = ± 1