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JEE Mains · Maths · STD 12 - 6. Application of derivatives

माना \(\lambda\) के सभी धनात्मक मानों का समुच्चय, जिसके लिए फलन \(\left(1+x\left(\lambda^2-x^2\right)\right)\) का स्थानीय निम्नतम बिंदु \(\frac{x^2+x+2}{x^2+5 x+6}<0\) को संतुष्ट करता है, \((\alpha, \beta)\) हो। तो \(\alpha^2+\beta^2\) = ...........

  1. A \(13\)
  2. B \(40\)
  3. C \(39\)
  4. D \(50\)
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Answer & Solution

Correct Answer

(C) \(39\)

Step-by-step Solution

Detailed explanation

\( \frac{x^2+x+2}{x^2+5 x+6}<0 \) \( \Rightarrow \frac{1}{(x+2)(x+3)}<0\) \( \mathrm{x} \in(-3,-2) \ldots \ldots \ldots . .(1) \) \( \mathrm{f}(\mathrm{x})=1+\mathrm{x}\left(\lambda^2-\mathrm{x}^2\right)\) Finding local minima…
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