Question

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

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

Answer 1

Answer: (D)

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

We have, $\underset{x\to \alpha }{lim}\frac{1-\cos (a{x}^2+bx+c)}{{(x-\alpha )}^2}$

$=\underset{x\to \alpha }{lim}\frac{2{\sin }^2\left(\frac{a{x}^2+bx+c}{2}\right)}{{(x-\alpha )}^2}$
[Since, $\alpha$ and $\beta$ are the roots of $a{x}^2+bx+c=0$, so it can be written as $a(x-\alpha )(x-\beta )=0$]

$=\underset{x\to \alpha }{lim}\frac{2{\sin }^2\left(\frac{a(x-\alpha )(x-\beta )}{2}\right)}{{(x-\alpha )}^2}$

$=\underset{x\to \alpha }{lim}\frac{2{\sin }^2\left(\frac{a}{2}(x-\alpha )(x-\beta )\right){\left(\frac{a}{2}\right)}^2{(x-\beta )}^2}{{\left[\left(\frac{a}{2}\right)(x-\alpha )(x-\beta )\right]}^2}$

$=\underset{x\to \alpha }{lim}2{\left(\frac{a}{2}\right)}^2{(x-\beta )}^2$

$=\underset{x\to \alpha }{lim}\frac{a^2}{2}\cdot {(x-\beta )}^2=\frac{a^2}{2}{(\alpha -\beta )}^2$