If $\alpha > \beta > 0$ are the roots of the equation $a x^{2}+b x+1=0$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equal to }$$ :

Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$ <br/><br/>Replace $x \rightarrow \frac{1}{x}$ <br/><br/>$\frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0$ <br/><br/>So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots. <br/><br/>Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}$ <br/><br/>$$ =\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} $$ <br/><br/>$$ \begin{aligned} & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\ & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\ & =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\ & \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\ & \Rightarrow k=2 \alpha \end{aligned} $$