QJEEAdv12Maf022: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Maf022 Solution: Let (a) and (a) be the roots of the equation ( 1 + a 3 − 1 ) x…

Let a + 1 = t⁶. Thus, when a $\to$ 0, t $\to$ 1.

$\therefore$ $({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$

$\Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$ ,

as t $\to$ 1

$2{x^2} + 3x + 1 = 0$

$\Rightarrow 2{x^2} + 2x + x + 1 = 0$

$\Rightarrow (2x + 1)(x + 1) = 0$

Thus, x = $-$ 1, $-$ 1/2

or, $\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$

and $\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$