Let $\alpha_ heta$ and $\beta_ heta$ be the distinct roots of $2 x^2+(\cos heta) x-1=0, heta \in(0,2 \pi)$. If m and M are the minimum and the maximum values of $\alpha_ heta^4+\beta_ heta^4$, then $16(M+m)$ equals :

Question

Let $\alpha_	heta$ and $\beta_	heta$ be the distinct roots of $2 x^2+(\cos 	heta) x-1=0, 	heta \in(0,2 \pi)$. If m and M are the minimum and the maximum values of $\alpha_	heta^4+\beta_	heta^4$, then $16(M+m)$ equals :

Aicharya · Accepted Answer

$\begin{aligned} & 2 x^2+(\cos 	heta) x-1=0 \ & \alpha_	heta+\beta_	heta=\frac{-\cos 	heta}{2} \ & \alpha_	heta \cdot \beta_	heta=\frac{-1}{2} \ & \alpha_	heta^2+\beta_	heta^2=\left(\alpha_	heta+\beta_	hetaight)^2-2 \alpha_	heta \beta_	heta \frac{\cos ^2 	heta}{4}+1\end{aligned}$ $\begin{aligned} & \alpha_	heta^4+\beta_	heta^4=\left(a_	heta^2+\beta_	heta^2ight)^2-2 \alpha_	heta^2 \beta_	heta^2=\left(\frac{\cos ^2 	heta}{4}+1ight)^2-\frac{2}{4} \ & =\left(\frac{\cos ^2 	heta}{4}+1ight)^2-\frac{1}{2} \end{aligned}$ Maximum when $\cos 	heta=1$ $\begin{aligned} & M=\left(\frac{1}{4}+1ight)^2-\frac{1}{2} \ & M=\frac{17}{16} \end{aligned}$ Minimum when $\cos 	heta=0$ $\begin{aligned} & m=1-\frac{1}{2}=\frac{1}{2} \ & 16(M+m)=16\left(\frac{17}{16}+\frac{1}{2}ight)=25 \end{aligned}$ *

Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2 x^2+(\cos \theta) x-1=0, \theta \in(0,2…

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