If f(x)=| _e|x||, then:

MCQ

If $\text{f(x)}=|\log_\text{e}|\text{x}||,$ then:

A
f(x) is continuous and differentiable for all x in its domain.
✓
f(x) is continuous for all for all × in its domain but not differentiable at $\text{x}=\pm1$
C
f(x) is neither continuous nor differentiable at $\text{x}=\pm1$
D
None of these.

✓

Answer

Correct option: B.

f(x) is continuous for all for all × in its domain but not differentiable at $\text{x}=\pm1$

We have,

$\text{f(x)}=|\log_\text{e}|\text{x}||$

We know that log function is defined for posirive value.

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

Therefore, domian of function is R - {0}

And we know that logarithmic function continuous in its domain.

Therefore, $|\log_\text{e}|\text{x}||$ is continuous in its domain.

We will check the differentiability at its critical points.

$|\log_\text{e}|\text{x}||=\begin{cases}\log_\text{e}(-\text{x}) & -\infty<\text{x<-1}\\-\log_\text{e}(-\text{x}) &-1<\text{x}<0\\-\log_\text{e}(\text{x})&0<\text{x}<1\\\log_\text{e}(\text{x})&1<\text{x}<\infty\end{cases}$

(LHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$

$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\log_\text{e}(-\text{x})-0}{\text{x}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[-(-1-\text{h})]}{-1-\text{h}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{-\text{h}}$

$=-1$

(RHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$

$=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{-\log_\text{e}(-\text{x})-0}{\text{x}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[-(-1+\text{h})]}{-1+\text{h}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}(1-\text{h})}{\text{h}}$

$=-\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$

$=-1\times-1=1$

Here, $\text{LHL}\neq\text{RHL}$

Therefore, the given function is not differentiable at x = -1.

(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{-\log_\text{e}(\text{x})-0}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[(1-\text{h})]}{1-\text{h}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$

$=-1$

(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-(1)}$

$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log_\text{e}(\text{x})-0}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[(1+\text{h})]}{1+\text{h}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{\text{h}}$

$=1$

Here, $\text{LHL}\neq\text{RHL}$

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 $\text{x}=\pm1.$

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