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

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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[JEE 2012]

Solution

The following is a short method making use of the options.

Consider the product of the limits,

${\lim _{a \to 0 + }}\alpha \times {\lim _{a \to 0 + }}\beta$

$= {\lim _{a \to 0 + }}\alpha \beta$

$= {\lim _{a \to 0 + }}\frac{{\sqrt[6]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}}{\rm{ }}\left[ {\frac{0}{0}} \right]$

$= {\lim _{a \to 0 + }}\frac{{\frac{1}{6}{{(1 + a)}^{ - \frac{5}{6}}}}}{{\frac{1}{3}{{(1 + a)}^{ - \frac{2}{3}}}}}$

$= \frac{1}{2}$

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Only option (B) gives us 1/2 as the product.

Hence, (B).

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