1.For α\alphaα belonging to an interval of length β\betaβ, suppose (α,−α)(\alpha, -\alpha)(α,−α) is an interior point of the ellipse 4x2+5y2=14x^2 + 5y^2 = 14x2+5y2=1. Then (6β−4)201+201=(6\beta - 4)^{201} + 201 =(6β−4)201+201=a.202202202b.000c.402402402d.201201201Login to continueOnly logged in users canattempt or see the solution.
