Question

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

• A. $ 0 $
• B. $ \frac{1}{2}{{(\alpha -\beta )}^{2}} $
• C. $ \frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}} $
• D. $ (\alpha -\beta ) $

Answer 1

The Correct Option is C

Since, $ \alpha $ and $ \beta $ are the roots of $ a{{x}^{2}}+bx+c=0 $ .
$ \therefore $ $ a(x-\alpha )(x-\beta )=a{{x}^{2}}+bx+c $ Now, $ \underset{x\to \alpha }{\mathop{\lim }}\,=\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}} $
$=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\left( \frac{a{{x}^{2}}+bx+c}{2} \right)}{{{(x-\alpha )}^{2}}} $
$=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\frac{a(x-\alpha )(x-\beta )}{2}}{{{\left( \frac{a(x-\alpha )(x-\beta )}{2} \right)}^{2}}}\times \frac{{{a}^{2}}{{(x-\beta )}^{2}}}{4} $
$=\frac{{{a}^{2}}{{(\alpha -\beta )}^{2}}}{2} $