ধরা যাক, f : R → R-কে c ∈ R এবং f(c) = 0-তে পার্থক্য করা যায়। যদি g(x) = |f(x)| হয়, তাহলে x = c, g কত?

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JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 1)
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  1. f'(c) = 0 হলে পার্থক্যযোগ্য নয়
  2. f'(c) ≠ 0 হলে পার্থক্যযোগ্য
  3. f'(c) = 0 হলে পার্থক্যযোগ্য
  4. পার্থক্যযোগ্য নয়

Answer (Detailed Solution Below)

Option 3 : f'(c) = 0 হলে পার্থক্যযোগ্য
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JEE Main 04 April 2024 Shift 1
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\({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{{\rm{g}}\left( {\rm{x}} \right)}}{{\rm{x}}} - \frac{{{\rm{g}}\left( {\rm{c}} \right)}}{{\rm{x}}}\)

\(\Rightarrow {\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\left| {f\left( x \right)} \right| - \left| {f\left( c \right)} \right|}}{{x - c}}\)

যেহেতু, f(c) = 0

অতএব, \({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\left| {f\left( {\rm{x}} \right)} \right|}}{{x - c}}\)

\(\Rightarrow {\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{f\left( x \right)}}{{x - c}};\) যদি f(x) > 0

এবং \({\rm{g'}}\left( {\rm{c}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{c}}} \frac{{ - f\left( {\rm{x}} \right)}}{{{\rm{x}} - {\rm{c}}}}\) ; যদি f(x) < 0 হয়

⇒ g'(c) = f'(c) = -f'(c)

= f'(c) + f'(c)

⇒ 2f'(c) = 0

⇒ f'(c) = 0

সুতরাং, f'(c) = 0 হলে g(x) পার্থক্যযোগ্য
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