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

ધારો કે \(2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi)\) સમીકરણના ભિન્ન બીજ \(\alpha_\theta\) અને \(\beta_\theta\) છે. જો m અને M એ \(\alpha_\theta^4+\beta_\theta^4\) ના ન્યૂનતમ અને મહત્તમ મૂલ્યો હોય, તો \(16(M+m)\) = __________

  1. A 24
  2. B 25
  3. C 17
  4. D 27
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Answer & Solution

Correct Answer

(B) 25

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Detailed explanation

\begin{aligned} & 2 x^2+(\cos \theta) x-1=0 \\ & \alpha_\theta+\beta_\theta=\frac{-\cos \theta}{2} \\ & \alpha_\theta \cdot \beta_\theta=\frac{-1}{2} \\ & \alpha_\theta^2+\beta_\theta^2=\left(\alpha_\theta+\beta_\theta\right)^2-2 \alpha_\theta \beta_\theta \frac{\cos ^2…

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