Question

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

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

Answer 1

Answer: (A)

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

As $\alpha$ and $\beta$ are roots of $a{x}^2+bx+c=0$, then

sum of roots $\alpha +\beta =-\frac{b}{a}$

and product of roots $\alpha \beta =\frac{c}{a}$

Given limit = $\underset{x\to a}{lim}$ $\frac{1-cosa(x-\alpha )(x-\beta )}{(x-\alpha {)}^2}$
$=\underset{x\to a}{lim}\frac{2si{n}^2\left(a\frac{(x-a)(x-\beta )}{2}\right)}{(x-\alpha {)}^2}$
$=\underset{x\to \alpha }{lim}\frac{2}{(x-\alpha {)}^2}\times \frac{si{n}^2\left(a\right)\frac{(x-\alpha )(x-\beta )}{2}}{\frac{a^2(x-\alpha {)}^2(x-\beta {)}^2}{4}}\times \frac{a^2(x-\alpha {)}^2(x-\beta {)}^2}{4}=\frac{a^2(\alpha -\beta {)}^2}{2}$