For alpha belonging to an interval of length | vedclass

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Question

(D) Since $(\alpha, -\alpha)$ lies inside the ellipse $4x^2 + 5y^2 - 1 = 0$,we have: $4(\alpha)^2 + 5(-\alpha)^2 - 1 < 0$ $4\alpha^2 + 5\alpha^2 - 1 <

Vedclass · Accepted Answer

$201$

Vedclass · Answer

(D) Since $(\alpha, -\alpha)$ lies inside the ellipse $4x^2 + 5y^2 - 1 = 0$,we have: $4(\alpha)^2 + 5(-\alpha)^2 - 1 $4\alpha^2 + 5\alpha^2 - 1 $9\alpha^2 $\alpha^2 $-\frac{1}{3} Thus,the interval for $\alpha$ is $(-\frac{1}{3}, \frac{1}{3})$. The length of this interval is $\beta = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}$. Now,substitute $\beta = \frac{2}{3}$ into the expression: $(6\beta - 4)^{201} + 201 = (6 	imes \frac{2}{3} - 4)^{201} + 201$ $= (4 - 4)^{201} + 201$ $= 0^{201} + 201 = 201$.

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

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