If α >β >0 are the roots of the equation ax^2+bx+1=0, and undersetx→ 1 α lim ( 1- cos(x^2+bx+a) 2(1-α x)^2)^ 1 2= 1 k( 1 β - 1 α ), then k is equal to

Since, $\alpha$α and $\beta$β are the roots of the equation $a{x}^{2}+bx+1=0$ax2+bx+1=0, therefore $\frac{1}{\alpha }$α1​ and $\frac{1}{\beta }$β1​ would be the roots of ${x}^{2}+bx+a=0$x2+bx+a=0.

Let

$$L=\underset{x\rightarrow \frac{1}{\alpha }}{\mathrm{lim}}{[\frac{1-\mathrm{cos}({x}^{2}+bx+a)}{2{(1-\alpha x)}^{2}}]}^{\frac{1}{2}}\rightarrow \frac{0}{0}$$L=x→α1​lim​[2(1−αx)21−cos(x2+bx+a)​]21​→00​

$$\Rightarrow L=\underset{x\rightarrow \frac{1}{\alpha }}{\mathrm{lim}}{[\frac{2{\mathrm{sin}}^{2}(\frac{{x}^{2}+bx+a}{2})}{2{(1-\alpha x)}^{2}}]}^{\frac{1}{2}}$$⇒L=x→α1​lim​[2(1−αx)22sin2(2x2+bx+a​)​]21​

$$\Rightarrow L=\underset{x\rightarrow \frac{1}{\alpha }}{\mathrm{lim}}{[\frac{{\mathrm{sin}}^{2}(\frac{1}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta }))}{{\alpha }^{2}{(x-\frac{1}{\alpha })}^{2}}]}^{\frac{1}{2}}$$⇒L=x→α1​lim​[α2(x−α1​)2sin2(21​(x−α1​)(x−β1​))​]21​

$$\Rightarrow L=\underset{x\rightarrow \frac{1}{\alpha }}{\mathrm{lim}}{[\frac{{(x-\frac{1}{\beta })}^{2}}{4{\alpha }^{2}}\times \frac{{{\mathrm{sin}(\frac{1}{2}(x-\frac{1}{\alpha })(x-\frac{1}{\beta }))}}^{2}}{{(\frac{1}{2})}^{2}{(x-\frac{1}{\alpha })}^{2}{(x-\frac{1}{\beta })}^{2}}]}^{\frac{1}{2}}$$⇒L=x→α1​lim​[4α2(x−β1​)2​×(21​)2(x−α1​)2(x−β1​)2sin(21​(x−α1​)(x−β1​))2​]21​

$$\Rightarrow L=\frac{(\frac{1}{\alpha }-\frac{1}{\beta })}{2\alpha }$$⇒L=2α(α1​−β1​)​

On comparing this value with the value given in the question we get, $k=2\alpha$k=2α.

Hence this is the correct option.