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

Answer: (C)

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

Answer 2

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, $\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}$
$=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\left( \frac{a{{x}^{2}}+bx+c}{2} \right)}{{{(x-\alpha )}^{2}}}$
$=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\frac{a(x-\alpha )(x-\beta )}{2}}{{{\left( \frac{a(x-\alpha )(x-\beta )}{2} \right)}^{2}}}\times \frac{{{a}^{2}}{{(x-\beta )}^{2}}}{4}$
$=\frac{{{a}^{2}}{{(\alpha -\beta )}^{2}}}{2}$