CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Write the points where $f(x) = |\log_e x|$ is not differentiable.
✓
Answer
Given: $\text{f(x)}=|\log_\text{e}\text{x}|=\begin{cases}-\log_\text{e}\text{x}, & 0<\text{x}<1\\\log_\text{e}\text{x}, & \text{x}\geq1\end{cases}$
Clearly $f(x)$ is differentiable for all $x > 1$ and for all $x < 1.$
So, we have to check the differentiability at $x = 1$.
$(\text{LHL}$ at $x = 1)$
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=-\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
$(\text{RHL}$ at $x = 1)$
$\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{+}}\frac{\log(1+\text{h})}{1+\text{h}-1}$
$=\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1+\text{h})}{\text{h}}$
$=1$
Thus$, (\text{LHL}$ at $x = 1) \neq (\text{RHL}$ at $x = 1)$
So$, f(x)$ is not differentiable at $x = 1.$
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