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# Choose the correct option.$f(x) = \left| {{{\log }_e}\left| x \right|} \right|,then f’(x): \\ A.\;\;\;\;\dfrac{1}{{\left| x \right|}},x \ne 0 \\ B.\;\;\;\;\dfrac{1}{x}for\left| x \right| > 1\;and\;\;\dfrac{{ - 1}}{x}for\left| x \right| < 1 \\ C.\;\;\;\;\; - \dfrac{1}{x}for\left| x \right| > 1\;and\;\;\dfrac{1}{x}for\left| x \right| < 1\; \\ D.\;\;\;\;\;\;\dfrac{1}{x}for\left| x \right| > 0\;and\;\; - \dfrac{1}{x}for\left| x \right| < 0\; \\$

Last updated date: 19th Sep 2024
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Hint: In order to solve this problem we need to try to find $f'(x)$. Knowing that if the value of x is negative the modulus will open with a negative sign and conversely. We also need to know that differentiation of log x is $\dfrac{1}{x}$. After that we have to consider the interval and get the right answer.

So it is given to us that $f(x) = \left| {{{\log }_e}\left| x \right|} \right|$ then,
For x>1, we have
$f(x) = \left| {{{\log }_e}\left| x \right|} \right|$=logx
And hence on doing the differentiation we have
$\Rightarrow f'(x) = \dfrac{1}{x}$
Now again for x<-1 we have
$f(x) = \left| {{{\log }_e}\left| x \right|} \right|$=-logx
And hence on doing the differentiation we have,
$\Rightarrow f'(x) = \dfrac{{ - 1}}{x}$
Now this time for 0$f(x) = \left| {{{\log }_e}\left| x \right|} \right|$= logx
And hence again on doing the differentiation, we have
$\Rightarrow f'(x) = \dfrac{1}{x}$
Now for -1$f(x) = \left| {{{\log }_e}\left| x \right|} \right|$ =-log(x)
And hence on doing the differentiation, we have
$\Rightarrow f'(x) = \dfrac{{ - 1}}{x}$
Hence f’(x)=$\left\{ \begin{gathered} \dfrac{1}{x},\left| x \right| > 1 \\ \dfrac{{ - 1}}{x},\left| x \right| < 1 \\ \end{gathered} \right.$

So, the correct answer is “Option B”.

Note: In this type of question we have to find that on solving the above expression which of the options will be suitable and hence for that we’ll try to differentiate the above expression on given conditions and proceeding like above we have done will take you to the right answer.