Question

Let $\alpha \left(a\right)$ and $\beta \left(a\right)$ be the roots of the equation $(\sqrt[3]{31+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{61+a}-1)=0$ where $a>-1.$ Then $\underset{a\to 0^{+}}{lim}\alpha \left(a\right)$ and $\underset{a\to 0^{+}}{lim}\beta \left(a\right)$ are

• A. $-\frac{5}{2}$ and $1$
• B. $-\frac{1}{2}$ and $-1$
• C. $-\frac{7}{2}$ and $2$
• D. $-\frac{9}{2}$ and $3$

Answer 1

The correct option is B $-\frac{1}{2}$ and $-1$

Given $\left(\right(1+a{)}^{1/3}-1)x^2+\left(\right(a+1{)}^{1/2}-1)x+\left(\right(a+1{)}^{1/6}-1)=0$

Let $a+1=t^6$
$\therefore (t^2-1)x^2+(t^3-1)x+(t-1)=0$
$\Rightarrow (t+1)x^2+(t^2+t+1)x+1=0$

As $a\to 0,t\to 1$
Hence $2x^2+3x+1=0$
$\therefore x=-1$ and $x=-\frac{1}{2}$