Question

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 \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

Answer 1

The Correct Option is B

PLAN To make the quadratic into simple form we should eliminate radical sign. Description of Situation As for given equation, when a $\rightarrow$ 0 the equation reduces to identity in x. i, e., $ ax^2 + bx + c = 0, \forall \, x \, \in R $ or a = b = c $ \rightarrow $ 0 Thus, first we should make above equation independent from coefficients as 0 . Let $ a + 1 = t^6> $ Thus when $ a \rightarrow 0, t \rightarrow 1 $. $ (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 $ \rightarrow 1 $ $\hspace25mm$ $ 2x^2 + 3x + 1 = 0 $ $\Rightarrow $ $\hspace25mm$ $ 2x^2 + 2x + x + 1 = 0 $ $\Rightarrow $ $\hspace25mm$ ( 2x + 1) (x + 1) = 0 Thus, $\hspace25mm$ x = - 1, - 1 / 2 or $\hspace25mm$ $ lim_{a \to 0^+ } \alpha (a) = -1 / 2 $ and $\hspace25mm$ $ lim_{a \to 0^+} \beta (a) = - 1 $ .