Question

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

The correct option is B $-\frac{1}{2}\text{and}-1$

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 ${lim}_{a\to 0^{+}}\alpha \left(a\right)\text{and}{lim}_{a\to 0^{+}}\beta \left(a\right)$
Let $1+a=k^6$
as $a\to 0\Rightarrow k\to 1$
$\Rightarrow (k^2-1)x^2+(k^3-1)x+(k-1)=0\Rightarrow (k-1)\left(\right(k+1)x^2+(k^2+k+1)x+1)=0\left(\right(k+1)x^2+(k^2+k+1)x+1)=0{lim}_{k\to 1}\left(\right(k+1)x^2+(k^2+k+1)x+1)=02x^2+3x+1=0(2x+1)(x+1)=0x=-\frac{1}{2},x=1$
${lim}_{a\to 0^{+}}\alpha \left(a\right)=-\frac{1}{2}$ ${lim}_{a\to 0^{+}}\beta \left(a\right)=1$