Answer:
Option B
Explanation:
Concept Involved To make the quadratic into the simple form we should eliminate radical sign
Description of Situation As forgiven equation when $a\rightarrow 0$ equation reduces to identity in x.
i.e, $ax^{2}+bx+c=0$ for all $x \epsilon R $ or a=b=c $\rightarrow 0$
Thus, first we should make above equation indepenrlent from coefficients as 0.
Sol. Let $(a+1=t^{6}$ Thus, when $a \rightarrow 0,t \rightarrow 1$
$\therefore$ $(t^{2}-1)x^{2}+(t^{3}-1)x+(t-1)=0$
$\Rightarrow$ $(t-1){(t+1)x^{2}+(t^{2}+t+1)x+1}=0$
As $t \rightarrow 1$
$2x^{2}+3x+1=0$
$2x^{2}+2x+x+1=0$
$\Rightarrow$ $(2x+1) (x+1)=0$
Thus, x=-1, -1/2
or $\lim_{a \rightarrow {0^{+}}}\alpha(a)=-\frac{1}{2}$
and $\lim_{a \rightarrow {0^{+}}}\beta(a)=-1$
