Question

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1$. Then $\lim _{a \rightarrow o^{+}} \alpha(a)$ and $\lim _{a \rightarrow o^{+}} \beta(a)$ 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

Answer: (B)

Let $1+a=y$
$\Rightarrow \left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+y^{1 / 6}-1=0$
$\Rightarrow \left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\frac{y^{1 / 6}-1}{y-1}=0$
Now taking $\displaystyle\lim _{y \to 1}$ on both the sides
$\Rightarrow \frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0$
$\Rightarrow 2 x^{2}+3 x+1=0$
$x=-1,-\frac{1}{2}$