Solve the Following Questions

Question:

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:

  1. $b^{2}+4 c$

  2. $2\left(b^{2}+4 c\right)$

  3. $2\left(b^{2}-4 c\right)$

  4. $b^{2}-4 c$


Correct Option: , 3

Solution:

$\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{1\left(1+\frac{2\left(x^{2}+b x+c\right)}{1 !}+\frac{2^{2}\left(x^{2}+b x+c\right)^{2}}{2 !}+\ldots\right)-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2\left(x^{2}+b x+1\right)^{2}}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2(x-\alpha)^{2}(x-\beta)^{2}}{(x-\beta)^{2}}$

$\Rightarrow 2(\beta-\alpha)^{2}=2\left(b^{2}-4 c\right)$

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