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  • If  <img alt="\mathrm{ f(x)=\left|\log _{10} x\right|~, ~then ~at ~x=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20f%28x%29%3D%5Cleft%7C%5Clog%20_%7B10%7D%20x%5C

If  \mathrm{ f(x)=\left|\log _{10} x\right|~, ~then ~at ~x=1} 

Option: 1

f(x) is continuous and \mathrm{f^{\prime}\left(1^{+}\right)=\log_{10}}


Option: 2

f(x) is continuous and  f^{\prime}\left(1^{-}\right)=\log _e 10


Option: 3

f(x) is continuous and \mathrm{f^{\prime}\left(1^{-}\right)=\log _c 10}


Option: 4

f(x) is continuous and  \mathrm{f^{\prime}\left(1^{-}\right)=-\log e_{10}^e}


Answers (1)

best_answer

As is evident from the graph of f(x) that it is continuous but not differentiable at x=1.

\mathrm{\text { Now, } f^{\prime}\left(1^*\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} }
\mathrm{ =\lim _{h \rightarrow 0} \frac{\log \ln (1+h)-0}{h}=\lim _{h \rightarrow 0} \frac{\log (1+h)}{h \cdot \log _e 10}=\frac{1}{\log _e 10}=\log _{10} e }
\mathrm{ f\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} }
\mathrm{=\lim _{h \rightarrow 0} \frac{\log 10(1-h)}{h}=\lim _{h \rightarrow 0} \frac{\log (1-h)}{h \log _6 10}=-\log _h 10 }


 

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chirag

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