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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{f(x)}=|\log_{10}\text{x}|\text{fx}=\log_{10}\text{x},$ then at $x = 1:$
  • A
    $f(x)$ is continuous and $\text{f}\ '(1^+)=\log_{10}\text{e}$
  • B
    $f(x)$ is continuous and $\text{f}\ '(1^-)=-\log_{10}\text{e}$
  • ✓
    both $a$ and $b$
  • D
    None of these
✓

Answer

Correct option: C.
both $a$ and $b$
Given,
$\text{f(x)}=|\log_{10}\text{x}|=\bigg|\frac{\log_{\text{e}}\text{x}}{\log_{\text{e}}10}\bigg|$
$=|(\log_{\text{e}}\text{x})\times(\log_{10}\text{e})|$
$=(\log_{10}\text{e})|\log_{10}\text{x}|$
$\text{f}\ '(1^+)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{{10}}\text{e})|\log_{\text{e}}(1+\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1|}{\text{h}}$
$=(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1$
$=(\log_{10}\text{e})$
Also,
$f\ '(1^-)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{10}\text{e})|\log_{\text{e}}(1-\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1}{\text{h}}$
$=-(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$
$=(\log_{10}\text{e})\times1=-(\log_{10}\text{e})$

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