Question

If $\alpha \left(a\right)$ and $\beta \left(a\right)$ are the roots of the equation $(\sqrt{1+a}-1)x^2+(i-i\sqrt{1+a}-\sqrt[6]{61+a}+1)x+i(\sqrt[6]{61+a}-1)=0,$
then the equations whose roots are $\underset{a\to 0}{lim}(\alpha +\beta )$ and $\underset{a\to 0}{lim}(\alpha -\beta )$ is/are

• A. $9x^2-18ix-10=0$
• B. $9x^2+18ix-10=0$
• C. $9x^2+6x+10=0$
• D. $9x^2-6x+10=0$

Answer 1

Answer: (A), (D)

The correct options are
A $9x^2-18ix-10=0$
D $9x^2-6x+10=0$

Let $\sqrt[6]{61+a}=k$
$(k^3-1)x^2+i(1-k^3)x-x(k-1)+i(k-1)=0\Rightarrow (k^3-1)x(x-i)-(k-1)(x-i)=0$
$\Rightarrow (x-i)\left(\right(k^3-1)x-(k-1\left)\right)=0\Rightarrow x=i,\frac{k-1}{k^3-1}$

Let $p$ and $q$ be the roots of required equation.
Case 1:
$\alpha =i,\beta =\frac{k-1}{k^3-1}$
$p=\underset{a\to 0}{lim}(\alpha +\beta )=i+\frac{1}{3}$
$q=\underset{a\to 0}{lim}(\alpha -\beta )=i-\frac{1}{3}$
$p+q=2i,pq=\frac{-10}{9}$
$\therefore$ Required equation is
$x^2-2ix-\frac{10}{9}=0$
$\Rightarrow 9x^2-18ix-10=0$

Case 2:
$\alpha =\frac{k-1}{k^3-1},\beta =i$
$p=\underset{a\to 0}{lim}(\alpha +\beta )=\frac{1}{3}+i$
$q=\underset{a\to 0}{lim}(\alpha -\beta )=\frac{1}{3}-i$
$p+q=\frac{2}{3},pq=\frac{10}{9}$
$\therefore$ Required equation is
$x^2-\frac{2}{3}x+\frac{10}{9}=0$
$\Rightarrow 9x^2-6x+10=0$