# Solve this

Question:

If $f(x)=\left|\log _{e}\right| x||$, 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

Solution:

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

We have, $f(x)=\left|\log _{e}\right| x||$

We know that $\log$ function is defined for positive value.

Here, $|x|$ is positive for all non zero $x$.

Therefore, domain of function is $R-\{0\}$

And we know that logarithmic function is continuous in its domain.

Therefore, $\left|\log _{\mathrm{e}}\right| x||$ is continuous in its domain.

We will check the differentiability at its critical points.

$\left|\log _{\mathrm{e}}\right| x||=\left\{\begin{array}{lc}\log _{e}(-x) & -\infty$(\mathrm{LHD}$at$x=-1)=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x-(-1)}=\lim _{x \rightarrow-1^{-}} \frac{\log _{\mathrm{e}}(-x)-0}{x+1}=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}[-(-1-h)]}{-1-h+1}=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}(1+h)}{-h}=-1(\mathrm{RHD}$at$x=-1)=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x-(-1)}=\lim _{x \rightarrow-1^{+}} \frac{-\log _{e}(-x)-0}{x+1}=\lim _{h \rightarrow 0} \frac{-\log _{e}[-(-1+h)]}{-1+h+1}=\lim _{h \rightarrow 0} \frac{-\log _{e}(1-h)}{h}=-\lim _{h \rightarrow 0} \frac{\log _{e}(1-h)}{h}=-1 \times-1=1$Here,$\mathrm{LHD} \neq \mathrm{RHD}$Therefore, the given function is not differentiable at$x=-1$.$(\mathrm{LHD}$at$x=1)=\lim _{\mathrm{x} \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{-\log _{\mathrm{e}}(x)-0}{x-1}=\lim _{h \rightarrow 0} \frac{-\log _{\mathrm{e}}[(1-h)]}{1-h-1}=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}(1-h)}{h}=-1(\mathrm{RHD}$at$x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-(1)}=\lim _{x \rightarrow 1^{+}} \frac{\log _{e}(x)-0}{x-1}=\lim _{h \rightarrow 0} \frac{\log _{e}[(1+h)]}{1+h-1}=\lim _{h \rightarrow 0} \frac{\log _{e}(1+h)}{h}=1$Here, LHD$\neq$RHD Therefore, the given function is not differentiable at$x=1$. Therefore, given function is continuous for all$x$in its domain but not differentiable at$x=\pm 1\$

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