For $\alpha$ belonging to an interval of length $\beta$, suppose $(\alpha,-\alpha)$ is an interior point of the ellipse $4 x^2+5 y^2=1$. Then,$(6 \beta-4)^{201}+201=$

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Aicharya · Accepted Answer

$\because(\alpha,-\alpha)$ lies inside $4 x^2+5 y^2-1=0$ $\begin{aligned} & \Rightarrow \quad 4 \alpha^2+5 \alpha^2-1 < 0 \ & \Rightarrow \quad 9 \alpha^2-1 < 0 \Rightarrow 9 \alpha^2 < 1 \ & \Rightarrow \quad \alpha^2 < \frac{1}{9} \Rightarrow-\frac{1}{3} < \alpha < \frac{1}{3} \ & \Rightarrow \quad \alpha \in\left(-\frac{1}{3}, \frac{1}{3}ight)\end{aligned}$ Thus, $\beta=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ Then, $(6 \beta-4)^{201}+201$ $=\left(6 	imes \frac{2}{3}-4ight)^{201}+201=201$

For $\alpha$ belonging to an interval of length $\beta$, suppose $(\alpha,-\alpha)$ is an interior point of…

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