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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

ધારોકે \(\lambda \in R\) અને ધારોકે સમીકરણ \(E\) એ \(|x|^2-2|x|+|\lambda-3|=0\) છે. તો ગણ \(S =\{x+\lambda: x\) એ \(E\) નો પૂર્ણાંક ઉકેલ છે; નો મહતમ ધટક \(.............\) છે.

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

Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(| x |^2-2| x |+|\lambda-3|=0\) \(| x |^2-2| x |+|\lambda-3|-1=0\) \((| x |-1)^2+|\lambda-3|=1\) At \(\lambda=3, x =0\) and 2 , at \(\lambda=4\) or 2 , then \(x =1 \text { or }-1\) So maximum value of \(x+\lambda=5\)
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