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The sum of the coefficients in the expansion or \small \left(\alpha^2 x^2-2 \alpha x+1\right)^{51}, as a polynomial in \small x vanishes. Then the position of the point \small \left(\alpha, 2 \alpha^2\right) with respect to the circle  \small x^2+y^2=4 is ....

Option: 1

On the circle


Option: 2

Inside the circle


Option: 3

Outside the circle 


Option: 4

None of these


Answers (1)

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\text { Sum of coefficients }=\left(\alpha^2-2 \alpha+1\right)^{51}=(\alpha-1)^{102}=0

\therefore \alpha=1 \\ \therefore\ Point\ is\ (1,2)

\text { Let circle } S=x^2+y^2-4=0

\therefore S_1=(1)^2+(2)^2-4=1>0

\text { Hence, point }\left(\alpha, 2 \alpha^2\right) \text { lies out side the circle } x^2+y^2=4 \text {. }

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Divya Prakash Singh

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