JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha_\theta\) and \(\beta_\theta\) be the distinct roots of \(2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi)\). If m and M are the minimum and the maximum values of \(\alpha_\theta^4+\beta_\theta^4\), then \(16(M+m)\) equals :
- A 24
- B 25
- C 17
- D 27
Answer & Solution
Correct Answer
(B) 25
Step-by-step Solution
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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