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 $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Limits XII

Let

α (a)

and

β (a)

be the roots of the equation

(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0

where

a > - 1

.
Then

lim_{a \to 0^{+}} α (a)

and

lim_{a \to 0^{+}} β (a)

are

Answer: (B)

Let

y = 1 + a

. The equation becomes:

(y^{1/3} - 1) x^{2} + (y^{1/2} - 1) x + (y^{1/6} - 1) = 0

Dividing by

y - 1

\frac{y ^{1/3} - 1}{y - 1} x^{2} + \frac{y ^{1/2} - 1}{y - 1} x + \frac{y ^{1/6} - 1}{y - 1} = 0

a \to 0^{+}

y \to 1

. Using L'Hôpital's rule or the limit

lim_{y \to 1} \frac{y ^{r} - 1}{y - 1} = r

\frac{1}{3} x^{2} + \frac{1}{2} x + \frac{1}{6} = 0

Multiplying by 6:

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

, which factors as

(2 x + 1) (x + 1) = 0

.

Thus

x = - \frac{1}{2}

x = - 1

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option