If $\alpha, \beta$ are the distinct roots of $x^{2}+b x+c=0$, then $\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$ is equal to:
Correct Option: , 3
$\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$
$\Rightarrow \lim _{x \rightarrow \beta} \frac{1\left(1+\frac{2\left(x^{2}+b x+c\right)}{1 !}+\frac{2^{2}\left(x^{2}+b x+c\right)^{2}}{2 !}+\ldots\right)-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$
$\Rightarrow \lim _{x \rightarrow \beta} \frac{2\left(x^{2}+b x+1\right)^{2}}{(x-\beta)^{2}}$
$\Rightarrow \lim _{x \rightarrow \beta} \frac{2(x-\alpha)^{2}(x-\beta)^{2}}{(x-\beta)^{2}}$
$\Rightarrow 2(\beta-\alpha)^{2}=2\left(b^{2}-4 c\right)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.