Let ( a ) and ( a ) be the roots of the equation ( 1 + a 3 - 1 ) x 2 + ( 1 + a - 1 ) x + ( 1 + a 6 - 1 ) = 0 , where a 1 . Then lim a 0 + ( a ) and lim a 0 + ( a ) are [2012]

Question

Let $α (a)$ and $β (a)$ 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 $\lim_{a \to 0^{+}} α (a) and \lim_{a \to 0^{+}} β (a)$ are [2012]

Smart Achievers · Accepted Answer

(2)

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

Let $a + 1 = y,$ then equation reduces to

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

On dividing both sides by $y - 1$ , we get

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

On taking limit as $y \to 1 i.e. a \to 0$ on both sides, we get

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

$\Rightarrow x = - 1, - \frac{1}{2}$ (roots of the equation)

$∴$ $\lim_{a \to 0^{+}} α (a) = - 1, \lim_{a \to 0^{+}} β (a) = - \frac{1}{2}$

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