Question

Assertion :If sum of the coefficient in the expansion of ${({\alpha }^2{x}^2-2ax+1)}^{51}$ , as a polynomial in x vanishes, then the points $(\alpha ,2{\alpha }^2)$ lies out side the circle $x^2+y^2=4$. Reason: The point $(\alpha ,\beta )$ lies out side the circle $x^2+y^2=r^2$ if ${\alpha }^2+{\beta }^2-r^2>0$.

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,
• B. Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion,
• C. Assertion is true but Reason is false,
• D. Assertion is false but Reason is true.

Answer 1

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,

Let $x=1$
Hence, we get
$({\alpha }^2-2\alpha +1{)}^{51}$
$=\left[\right(\alpha -1{)}^2{]}^{51}$
$=[\alpha -1{]}^{102}$
Now sum of the coefficients is $0$.
Hence
$\alpha -1=0$
$\alpha =1$
Therefore the point $(\alpha ,2{\alpha }^2)$ becomes $(1,2)$
$x^2+y^2-4=0$ is the equation of the given circle.
Substituting the acquired point, we get
$1+4-4$
$1>0$
Hence the point lies outside the circle.