Question

If $\alpha$ and $\beta$ be the roots of equation $a{x}^2+bx+c=0,$ then $\underset{x\to \alpha }{lim}{\left(1+a{x}^2+bx+c\right)}^{\frac{1}{x-\alpha }}$ is

• A. $a\left(\alpha -\beta \right)$
• B. $\ln \left|a\left(\alpha -\beta \right)\right|$
• C. $e^{a\left(\alpha -\beta \right)}$
• D. $e^{a\left|\alpha -\beta \right|}$

Answer 1

Answer: (D)

$e^{a\left|\alpha -\beta \right|}$

Given $\alpha$ & $\beta$ are roots of equation, $a{x}^2+bx+c=0\to 1$

To find $\underset{x\to \alpha }{lim}{(1+a{x}^2+bx+c)}^{\frac{1}{x-\alpha }}=l$ (say)

From equation $1,\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

& $\alpha$ satisfies equation $1\Rightarrow a{\alpha }^2+b\alpha +c=0\to 2$

Now, $l=\underset{x\to \alpha }{lim}{(1+(a{x}^2+bx+c\left)\right)}^{\frac{1}{(x-\alpha )}}$

$l=e^{\underset{x\to \alpha }{lim}\left(\frac{a{x}^2+bx+c}{x-\alpha }\right)}$ (As when we get $1^{\infty }$ on putting the limit $\underset{x\to \alpha }{lim}{(1+f\left(x\right))}^{\frac{1}{g\left(x\right)}}=e^{\underset{x\to \alpha }{lim}\frac{f\left(x\right)}{g\left(x\right)}}$)

$l=e^{\underset{x\to \alpha }{lim}\frac{2xa+b}{1}}$ (applying L hospital rule)

$l=e^{2\alpha a+b}\to 3$

As we know sum of roots$=\frac{-b}{a}\Rightarrow \alpha +\beta$

$\Rightarrow -b=a\alpha +a\beta$

$\Rightarrow b=-(a\alpha +a\beta )\to 4$

Put value of b from equation $4$ to equation $3$,

$\Rightarrow l=e^{2\alpha a-a\alpha -a\beta }$

$\Rightarrow l=e^{a\alpha -a\beta }$

$\Rightarrow l=e^{a(\alpha -\beta )}$