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

ધારોકે વિધેય \(\left(1+x\left(\lambda^2-x^2\right)\right)\) નું સ્થાનીય ન્યૂનતમ બિંદુ \(\frac{x^2+x+2}{x^2+5 x+6}<0\) નું સમાધાન કરે તેવી \(\lambda\) ની તમામ ધન કિંમતોનો ગણ \((\alpha, \beta)\) છે. તો \(\alpha^2+\beta^2=\) .............

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

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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