Question

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

• A. $\left|\frac{c}{2\alpha }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$
• B. $\left|\frac{c}{2\beta }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$
• C. $\left|\frac{c}{\alpha \beta }(\frac{1}{\alpha }-\frac{1}{\beta })\right|$
• D. None of these

Answer 1

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

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

Then $c{x}^2+bx+a=c(x-\frac{1}{\alpha })(x-\frac{1}{\beta })$

$\underset{x\to \frac{1}{\alpha }}{lim}\sqrt{\frac{1-cos\left(c\right)(c-\frac{1}{\alpha })(x-\frac{1}{\beta })}{2{\alpha }^2(x-\frac{1}{\alpha }{)}^2}}$

$={lim}_{x\to \frac{1}{\alpha }}\sqrt{\frac{2si{n}^2\frac{c}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta })}{2{\alpha }^2(x-\frac{1}{{\alpha }^2})}}$
$=\underset{x\to \frac{1}{\alpha }}{lim}\sqrt{\frac{1}{{\alpha }^2}\left[\frac{si{n}^2\left(\right(x-\frac{1}{\alpha }\left)\right(x-\frac{1}{\beta }\left)\frac{c}{2}\right)}{{\left[\right(x-\frac{1}{\alpha }\left)\right(x-\frac{1}{\beta }\left)\frac{c}{2}\right]}^2}\right]}\times (x-\frac{1}{\beta }{)}^2\frac{c^2}{4}$
$=\underset{x\to \frac{1}{\alpha }}{lim}\sqrt{\frac{1}{{\alpha }^2}\times \frac{c^2}{4}(\frac{1}{\alpha }-{\frac{1}{\beta }}^2)}$

$=\left|\frac{c}{2\alpha }(\frac{1}{2\alpha }-\frac{1}{\beta })\right|$

Hence, option 'A' is correct.