Question

If \[f\left( x \right) = \left| \log_e |x| \right|\] 

विकल्प

• f (x) is continuous and differentiable for all x in its domain

• f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1

•  (x) is neither continuous nor differentiable at x = ± 1

• none of these

   

Answer 1

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1 

We have, 
\[f\left( x \right) = \left| \log_e |x| \right|\]
\[\text{We know that log function is defined for positive value} . \]
\[\text{Here,} \left| x \right| \text { is positive for all non zero x} . \]
\[\text{Therefore, domain of function is R} - \left\{ 0 \right\}\]

And we know that logarithmic function is continuous in its domain.

\[\text{Therefore, }\left| \log_e \left| x \right| \right| \text { is continuous in its domain} .\]

\[\text{We will check the differentiability at its critical points} . \]

`|log_e |x||| = {(log_e (-x) , -infty <x < -1),(-log_e (-x) ,-1 <x<0),(-log_e (x) ,0<x<1),(log_e(x) ,1<x< infty):}`

\[\left( \text { LHD at x } = - 1 \right) = \lim_{x \to - 1^-} \frac{f\left( x \right) - f\left( - 1 \right)}{x - \left( - 1 \right)}\]
\[ = \lim_{x \to - 1^-} \frac{\log_e \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left[ - \left( - 1 - h \right) \right]}{- 1 - h + 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 + h \right)}{- h}\]
\[ = - 1\]
\[\left( \text { RHD at x } = - 1 \right) = \lim_{x \to - 1^+} \frac{f\left( x \right) - f\left( - 1 \right)}{x - \left( - 1 \right)}\]
\[ = \lim_{x \to - 1^+} \frac{- \log_e \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left[ - \left( - 1 + h \right) \right]}{- 1 + h + 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left( 1 - h \right)}{h}\]
\[ = {- \lim}_{h \to 0} \frac{\log_e \left( 1 - h \right)}{h}\]
\[ = - 1 \times - 1 = 1\]
\[\text { Here, LHD }\neq \text { RHD }\]
\[\text{Therefore, the given function is not differentiable at x} = - 1 .\]

\[\left( \text { LHD at x } = 1 \right) = \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ = \lim_{x \to 1^-} \frac{- \log_e \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left[ \left( 1 - h \right) \right]}{1 - h - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 - h \right)}{h}\]
\[ = - 1\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - \left( 1 \right)}\]
\[ = \lim_{x \to 1^+} \frac{\log_e \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left[ \left( 1 + h \right) \right]}{1 + h - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 + h \right)}{h}\]
\[ = 1\]
\[\text { Here, LHD } \neq \text { RHD }\]
\[\text{Therefore, the given function is not differentiable at x} = 1 .\]

Therefore, given function is continuous for all x in its domain but not differentiable at x = ± 1