Question

Let $\alpha \left(a\right)$ and $\beta \left(a\right)$ be the roots of the equation $(\sqrt[3]{31+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{61+a}-1)=0$ where $a>-1$. Then $\underset{a\to 0^{+}}{lim}\alpha \left(a\right)$ and $\underset{a\to 0^{+}}{lim}\beta \left(a\right)$ 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

Answer: (B)

$-\frac{1}{2}$ and $-1$

Let $1+a=y$.

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

Dividing by $y-1$, we get

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

When $a\to 0,y\to 1$.

Hence, taking $\underset{y\to 1}{lim}$ on both the sides,

$\Rightarrow \underset{y\to 1}{lim}\left(\frac{y^{\frac{1}{3}}-1}{y-1}\right)x^2+\left(\frac{y^{\frac{1}{2}}-1}{y-1}\right)x+\left(\frac{y^{\frac{1}{6}}-1}{y-1}\right)=0$

$\Rightarrow \frac{1}{3}{x}^2+\frac{1}{2}x+\frac{1}{6}=0$ ...[Using $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}]$

$\Rightarrow 2x^2+3x+1=0$

Solving the quadratic equation, we get

$\Rightarrow x=-1$ or $x=\frac{-1}{2}$.

Since $\alpha \left(a\right)$ and $\beta \left(a\right)$ are roots of the given equation, we get

$\underset{a\to 0^{+}}{lim}\alpha \left(a\right)=-1$ and $\underset{a\to 0^{+}}{lim}\beta \left(a\right)=\frac{-1}{2}$.