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प्रश्न
Discuss the continuity and differentiability of f (x) = |log |x||.
उत्तर
We have,
f (x) = |log |x||
`|x| = {(-x, ,-∞ <x<-1),(-x, ,-1<x<0),(x, ,0<x<1),(x, ,1<x<∞):}`
log `|x| = {(log(-x), ,-∞ <x<-1),(log(-x), ,-1<x<0),(log(x), ,0<x<1),(log(x), ,1<x<∞):}`
`|log |x|| = {(log(-x), ,-∞ <x<-1),(-log(-x), ,-1<x<0),(-log(x), ,0<x<1),(log(x), ,1<x<∞):}`
\[\left( \text { LHD at x } = - 1 \right) = \lim_{x \to - 1^-} \frac{f\left( x \right) - f\left( - 1 \right)}{x + 1}\]
\[ = \lim_{x \to - 1^-} \frac{\log \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{- 1 - h + 1}\]
\[ = - \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{h} = - 1\]
\[\left( \text { RHD at x } = - 1 \right) = \lim_{x \to - 1^+} \frac{f\left( x \right) - f\left( - 1 \right)}{x + 1}\]
\[ = \lim_{x \to - 1^+} \frac{- \log \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{- 1 + h + 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{h} = 1\]
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1
At 0 function is not defined.
\[\left( \text { LHD at x } = 1 \right) = \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ = \lim_{x \to 1^-} \frac{- \log \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{1 - h - 1}\]
\[ = - \lim_{h \to 0} \frac{\log \left( 1 - h \right)}{h} = - 1\]
\[\left( \text { RHD at x } = 1 \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ = \lim_{x \to 1^+} \frac{\log \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{1 + h - 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{h} = 1\]
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
At 0 function is not defined.
So, at 0 function is not continuous.
\[\left(\text { LHL at x } = - 1 \right) = \lim_{x \to - 1^-} f\left( x \right)\]
\[ = \lim_{x \to - 1^-} \log \left( - x \right)\]
\[ = \log \left( 1 \right) = 0\]
\[\left( \text { RHL at x } = - 1 \right) = \lim_{x \to - 1^+} f\left( x \right)\]
\[ = \lim_{x \to - 1^+} - \log \left( - x \right)\]
\[ = - \log 1 = 0\]
\[f\left( - 1 \right) = 0\]
\[\text { Therefore,} f\left( x \right) = \left| \log \left| x \right| \right| \text{is continuous at x} = - 1\]
\[\left( \text { LHL at x } = 1 \right) = \lim_{x \to 1^-} f\left( x \right)\]
\[ = \lim_{x \to 1^-} - \log \left( x \right)\]
\[ = - \log \left( 1 \right) = 0\]
\[\left( \text { RHL at x } = 1 \right) = \lim_{x \to 1^+} f\left( x \right)\]
\[ = \lim_{x \to 1^+} \log \left( x \right)\]
\[ = \log 1 = 0\]
\[f\left( 1 \right) = 0\]
\[\text { Therefore, at x } = 1, f\left( x \right) = \left| \log \left| x \right| \right|\text { is continuous .}\]
Hence, function f (x) = |log |x|| is not continuous at x = 0
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