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

ધારોકે \([\bullet]\) મહત્તમ પૂર્ણાંક વિધેય દર્શાવે છે, તથા \(f(x)=\min \left\{\sqrt{2} x, x^2\right\}\). ધારોકે \(S =\left\{x \in(-2,2):\right.\) વિધેય \(g (x)=|x|\left[x^2\right]\) એ \(x\) પર અસતત છે \(\}\). તો \(\sum_{x \in S} f(x)=\) ___ .

  1. A \(2-\sqrt{2}\)
  2. B \(2 \sqrt{6}-3 \sqrt{2}\)
  3. C \(1-\sqrt{2}\)
  4. D \(\sqrt{6}-2 \sqrt{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1-\sqrt{2}\)

Step-by-step Solution

Detailed explanation

\(g ( x )=| x |\left[ x ^2\right]\) points of discontinuity of \(g(x)\) in \((-2,2)\) are \(( \pm 1, \pm \sqrt{2}, \pm \sqrt{3})\) \(\therefore S =\{-1,1,-\sqrt{2}, \sqrt{2},-\sqrt{3}, \sqrt{3}\}\) \(\because f(x)=\min \left\{\sqrt{2} x, x^2\right\}\)…
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