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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો \(S\) એ બિંદુઓનો ગણ છે કે જ્યાં વિધેય \(f(\mathrm{x})=|2-| \mathrm{x}-3 \|, \mathrm{x} \in \mathrm{R},\) એ વિકલનીય ન હોય તો \(\sum\limits_{\mathrm{x\in s}} f(f(\mathrm{x}))\) મેળવો.

  1. A \(5\)
  2. B \(2\)
  3. C \(3\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(f(x)=|2-| x-||3\) \(f\) is not differentiable at \(x=1,3,5\) \(\sum_{\mathrm{xes}} \mathrm{f}(\mathrm{f}(\mathrm{x}))=\) \( \mathrm{f}(\mathrm{f}(1))+\mathrm{f}(\mathrm{f}(3))+\mathrm{f}(\mathrm{f}(5)) \) \(=\mathrm{f}(0)+\mathrm{f}(2)+\mathrm{f}(0) \) \(=1+1+1=3 \)
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