Question

If $\alpha$, $\beta$ are the root of quadratic equation $a{x}^2+bx+c=0$

then ${lim}_{x\to \alpha }\frac{1-\cos (a{x}^2+bx+c)}{(x-\alpha {)}^2}$ equals?

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

Answer 1

Answer: (C)

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

Substituting gives us indeterminate form.

Applying L'Hospitals Rule,

$={lim}_{x\to \alpha }\frac{\sin (a{x}^2+bx+c).(2ax+b)}{2(x-\alpha )}$

Substituting gives indeterminate form, applying L'Hospitals Rule,

$={lim}_{x\to \alpha }\frac{\cos (a{x}^2+bx+c).(2ax+b{)}^2+\sin (a{x}^2+bx+c).(2a)}{2}$

$=\frac{(2a\alpha +b{)}^2}{2}$

We know,

$(\alpha +\beta )=\frac{-b}{a}and(\alpha .\beta )=\frac{c}{a}$

$=\frac{(2a\alpha +b{)}^2}{2}$

$=\frac{a^2(2\alpha +\frac{b}{a}{)}^2}{2}$

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

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

Hence, option 'C' is correct.