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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 $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{x \rightarrow 0^{+}} \beta(a)$ are
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$-\frac{1}{2}$ and $-1$
$(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$
Let $a+1=y$, then equation reduces to $\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0$
On dividing both sides by $y-1$, we get $\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0$
On taking limit as $y \rightarrow 1$ i.e. $a \rightarrow 0$ 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)
$\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}$
Let $a+1=y$, then equation reduces to $\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0$
On dividing both sides by $y-1$, we get $\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0$
On taking limit as $y \rightarrow 1$ i.e. $a \rightarrow 0$ 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)
$\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}$
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