Question
Download Solution PDFमान लीजिए f : R → R, c ∈ R पर अवकलनीय है तथा f(c) = 0 है। यदि g(x) = |f(x)| है, तो x = c पर g है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF\({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{{\rm{g}}\left( {\rm{x}} \right)}}{{\rm{x}}} - \frac{{{\rm{g}}\left( {\rm{c}} \right)}}{{\rm{x}}}\)
\(\Rightarrow {\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\left| {f\left( x \right)} \right| - \left| {f\left( c \right)} \right|}}{{x - c}}\)
चूँकि, f(c) = 0
तब, \({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\left| {f\left( {\rm{x}} \right)} \right|}}{{x - c}}\)
\(\Rightarrow {\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{f\left( x \right)}}{{x - c}};\) यदि f(x) > 0
और \({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{ - f\left( {\rm{x}} \right)}}{{{\rm{x}} - {\rm{c}}}}\) ; यदि f(x) < 0
⇒ g'(c) = f'(c) = -f'(c)
= f'(c) + f'(c)
⇒ 2f'(c) = 0
⇒ f'(c) = 0
अतः, यदि f'(c) = 0 है तो g(x) अवकलनीय है।Last updated on Jul 11, 2025
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