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JEE 2012 Math: ${\lim _{a \to {0^ + }}}\alpha (a)$ & ${\lim _{a \to {0^ + }}}\beta (a)$

Let $\alpha (a)$ and $\beta (a)$ be the roots of the equation

$\left( {\sqrt[3]{{1 + a}} - 1} \right){x^2} + \left( {\sqrt {1 + a}  - 1} \right)x + \left( {\sqrt[6]{{1 + a}} - 1} \right) = 0$ where a>-1.

Then ${\lim _{a \to {0^ + }}}\alpha (a)$ and ${\lim _{a \to {0^ + }}}\beta (a)$ 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

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Solution

Looking at the options and noticing that while the sum of the limits is same the product of the limits is different and hence the following quick method can be employed:

Consider,

{\lim _{a \to {0^ + }}}\alpha (a).{\lim _{a \to {0^ + }}}\beta (a)

= {\lim _{a \to {0^ + }}}\alpha (a).\beta (a) assuming both limits exist

= {\lim _{a \to {0^ + }}}\frac{{\sqrt[6]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}}

= {\lim _{a \to {0^ + }}}\frac{{\sqrt[6]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}} \times \frac{{\sqrt[6]{{1 + a}} + 1}}{{\sqrt[6]{{1 + a}} + 1}}

= {\lim _{a \to {0^ + }}}\frac{{\sqrt[3]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}} \times \frac{1}{{\sqrt[6]{{1 + a}} + 1}}

= \frac{1}{2}

Hence, (B).