Question

$\underset{x\to 1/\alpha }{lim}\frac{1-\cos (c{x}^2+bx+a)}{{(1-x\alpha )}^2}$, where $\alpha$ is a root of $a{x}^2+bx+c=0$, is equal to

• A. $\frac{b^2-4ac}{2{\alpha }^2}$
• B. $\frac{b^2-4ac}{{\alpha }^2}$
• C. $\frac{4ac-b^2}{2{\alpha }^2}$
• D. None of these

Answer 1

The correct option is A $\frac{b^2-4ac}{2{\alpha }^2}$

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

$=\underset{x\to \frac{1}{\alpha }}{lim}\frac{1-\cos ({cx}^2+bx+a)}{{({cx}^2+bx+a)}^2}.\frac{{({cx}^2+bx+a)}^2}{{(1-x\alpha )}^2}$

$=\underset{z\to 0}{lim}\frac{1-\cos z}{z^2}.\underset{y\to \alpha }{lim}\frac{{({ay}^2+by+c)}^2}{y^2{(y-\alpha )}^2}$

$[$if $\alpha ,\beta$ are roots of ${ax}^2+bx+c=0$ then ${ax}^2+bx+c=a(x-\alpha )(x-\beta )]$

$=\frac{a^2{(\alpha -\beta )}^2}{2{\alpha }^2}=\frac{a^2}{2{\alpha }^2}[{(\alpha +\beta )}^2-4\alpha \beta ]$

$=\frac{a^2}{2{\alpha }^2}(\frac{b^2}{a^2}-\frac{4c}{a})=\frac{b^2-4ac}{2{\alpha }^2}$