Question

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

• A. $c(\alpha -\beta )$
• B. $\left|\frac{c}{2\alpha }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$
• C. $\frac{1}{\beta }(\frac{1}{\alpha }-\frac{1}{\beta })$
• D. $\left|\left(\frac{1}{\alpha }-\frac{1}{\beta }\right)\frac{c}{\alpha }\right|$

Answer 1

The correct option is B $\left|\frac{c}{2\alpha }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$

As $\alpha ,\beta$ are roots of $a{x}^2+bx+c=0$
then the roots of $c{x}^2+bx+a=0$ are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$
$\underset{x\to \frac{1}{\alpha }}{lim}\sqrt{\frac{1-\cos (c{x}^2+bx+a)}{2(1-\alpha x{)}^2}}=\underset{x\to \frac{1}{\alpha }}{lim}{\left(\frac{{\sin }^2\left(\frac{c{x}^2+bx+a}{2}\right)}{{(1-\alpha x)}^2}\right)}^{1/2}$
$=\underset{x\to \frac{1}{\alpha }}{lim}\left|\frac{\sin \left(\frac{c{x}^2+bx+a}{2}\right)}{1-\alpha x}\right|=\underset{x\to \frac{1}{\alpha }}{lim}\left|\frac{\sin \left(\frac{c}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta })\right)}{-\alpha (x-\frac{1}{\alpha })}\right|$
$=|\underset{x\to \frac{1}{\alpha }}{lim}\frac{\sin \left(\frac{c}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta })\right)}{\left(\frac{c}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta })\right)}\times \underset{x\to \frac{1}{\alpha }}{lim}\frac{\frac{c}{2}(x-\frac{1}{\beta })}{-\alpha }|$
$=\left|\frac{1\times \frac{c}{2}(\frac{1}{\alpha }-\frac{1}{\beta })}{-\alpha }\right|=\left|\frac{c}{2\alpha }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$
Hence, option 'B' is correct.