JEE Advance - Mathematics (2012 - Paper 2 Offline - No. 17)
Let $$\alpha$$(a) and $$\beta$$(a) be the roots of the equation $$(\root 3 \of {1 + a} - 1){x^2} + (\sqrt {1 + a} - 1)x + (\root 6 \of {1 + a} - 1) = 0$$ where $$a > - 1$$. Then $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a)$$ and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a)$$ are
$$ - {5 \over 2}$$
$$ - {1 \over 2}$$
$$ - {7 \over 2}$$
$$ - {9 \over 2}$$
Explanation
Let a + 1 = t6. Thus, when a $$\to$$ 0, t $$\to$$ 1.
$$\therefore$$ $$({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$$
$$ \Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$$,
as t $$\to$$ 1
$$2{x^2} + 3x + 1 = 0$$
$$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$$
$$ \Rightarrow (2x + 1)(x + 1) = 0$$
Thus, x = $$-$$1, $$-$$1/2
or, $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$$
and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$$
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