Let α. and β be the roots of ax2+bx+c=0, then limx→a 1−cos⁡ax2+bx+c(x−α)2 is equal to

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

1. A

0

2. B

$\frac{1}{2}\left(\alpha -\beta {\right)}^{2}$

3. C

$\frac{{a}^{2}}{2}\left(\alpha -\beta {\right)}^{2}$

4. D

$-\frac{{a}^{2}}{2}\left(\alpha -\beta {\right)}^{2}$

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### Solution:

It is given that ${\alpha }_{,}\beta$ are roots of $a{x}^{2}+bx+c$

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

$\begin{array}{l}=2\underset{x\to \alpha }{lim} \frac{{\mathrm{sin}}^{2}\left\{\frac{\left(a{x}^{2}+bx+c\right)}{2}\right\}}{\left(x-\alpha {\right)}^{2}}\\ =2\underset{x\to \alpha }{lim} \frac{{\mathrm{sin}}^{2}\left\{\frac{a\left(x-\alpha \right)\left(x-\beta \right)}{2}\right\}}{\left(x-\alpha {\right)}^{2}}}\\ =2\underset{x\to \alpha }{lim} \left[\frac{{\mathrm{sin}\left\{\frac{a\left(x-\alpha \right)\left(x-\beta \right)}{2}\right\}]}^{2}}{\frac{a\left(x-\alpha \right)\left(x-\beta \right)}{2}}\right]×\frac{{a}^{2}}{4}\left(x-\beta {\right)}^{2}\\ =2\left(1{\right)}^{2}×\frac{{a}^{2}}{4}\left(\alpha -\beta {\right)}^{2}=\frac{{a}^{2}}{2}\left(\alpha -\beta {\right)}^{2}\end{array}$  Register to Get Free Mock Test and Study Material

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