Question
Let $$\alpha \left( a \right)$$ and $$\beta \left( a \right)$$ be the roots of the equation $$\left( {\root 3 \of {1 + a} - 1} \right){x^2} + \left( {\sqrt {1 + a} - 1} \right)x + \left( {\root 6 \of {1 + a} - 1} \right) = 0$$ where $$a > - 1.$$ Then $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha \left( a \right)$$ and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta \left( a \right)$$ are-
A.
$$ - \frac{5}{2}\,\,{\text{and}}\,\,1$$
B.
$$ - \frac{1}{2}\,\,{\text{and}}\,\, - 1$$
C.
$$ - \frac{7}{2}\,\,{\text{and}}\,\,2$$
D.
$$ - \frac{9}{2}\,\,{\text{and}}\,\,3$$
Answer :
$$ - \frac{1}{2}\,\,{\text{and}}\,\, - 1$$
Solution :
$$\eqalign{
& {\text{Given equation,}} \cr
& \left( {\root 3 \of {1 + a} - 1} \right){x^2} + \left( {\sqrt {1 + a} - 1} \right)x + \left( {\root 6 \of {1 + a} - 1} \right) = 0 \cr
& \left( {{{\left( {1 + a} \right)}^{\frac{1}{3}}} - 1} \right){x^2} + \left( {{{\left( {1 + a} \right)}^{\frac{1}{2}}} - 1} \right)x + \left( {{{\left( {1 + a} \right)}^{\frac{1}{6}}} - 1} \right) = 0.....\left( 1 \right) \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \alpha \left( a \right),\,\mathop {\lim }\limits_{a \to {0^ + }} \beta \left( a \right){\text{ and }}\alpha \left( a \right),\,\beta \left( a \right){\text{ are the roots}} \cr
& {\text{Put}}\,1 + a = y,{\text{ equation }}\left( 1 \right){\text{ can be re - written as:}} \cr
& \left( {{y^{\frac{1}{3}}} - 1} \right){x^2} + \left( {{y^{\frac{1}{2}}} - 1} \right)x + \left( {{y^{\frac{1}{6}}} - 1} \right) = 0 \cr
& \Rightarrow \frac{{\left( {{y^{\frac{1}{3}}} - 1} \right){x^2}}}{{\left( {y - 1} \right)}} + \frac{{\left( {{y^{\frac{1}{2}}} - 1} \right)x}}{{\left( {y - 1} \right)}} + \frac{{\left( {{y^{\frac{1}{6}}} - 1} \right)}}{{\left( {y - 1} \right)}} = 0 \cr
& {\text{Taking }}y \to 1,{\text{ as }}a \to 0 \cr
& \Rightarrow \frac{{\left( {{y^{\frac{1}{3}}} - 1} \right){x^2}}}{{\left( {{y^{\frac{1}{3}}} - 1} \right)\left( {{y^{\frac{2}{3}}} + 1 + 2{y^{\frac{1}{3}}}} \right)}} + \frac{{\left( {{y^{\frac{1}{2}}} - 1} \right)x}}{{\left( {{y^{\frac{1}{2}}} - 1} \right)\left( {{y^{\frac{1}{2}}} + 1} \right)}} + \frac{{\left( {{y^{\frac{1}{6}}} - 1} \right){x^2}}}{{\left( {{y^{\frac{1}{6}}} - 1} \right)\left( {{y^{\frac{5}{6}}} + 2{y^{\frac{1}{6}}} + 2{y^{\frac{3}{6}}} + 1} \right)}} = 0 \cr
& {\text{As putting }}y \to 1,{\text{ we get}} \cr
& \Rightarrow \frac{{{x^2}}}{3} + \frac{x}{2} + \frac{1}{6} = 0 \cr
& \Rightarrow 2{x^2} + 3x + 1 = 0 \cr
& {\text{we get, }}x = - 1,\, - \frac{1}{2} \cr
& {\text{Hence, option B is correct}}{\text{.}} \cr} $$