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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
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 $\times $ 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 $\times $ 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.
$|\log_\text{e}|\text{x}||=\begin{cases}\log_\text{e}(-\text{x}) & -\infty<\text{x<-1}\\-\log_\text{e}(-\text{x}) &-1<\text{x}<0\\-\log_\text{e}(\text{x})&0<\text{x}<1\\\log_\text{e}(\text{x})&1<\text{x}<\infty\end{cases}$
$(\text{LHL}$ at $x = -1) =\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\log_\text{e}(-\text{x})-0}{\text{x}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[-(-1-\text{h})]}{-1-\text{h}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{-\text{h}}$
$=-1$
$\text{(RHL}$ 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 $x = -1.$
$\text{(LHL}$ at $x = 1) =\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{-\log_\text{e}(\text{x})-0}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[(1-\text{h})]}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$
$=-1$
$\text{(RHL}$ 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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