Let $\cos \mathrm{\theta },\cos \frac{\mathrm{\theta }}{2},\cos \frac{\mathrm{\theta }}{4}\mathrm{a}$ are the roots of the equation ${\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=0\text{. If }\underset{\mathrm{\theta }\to 0}{lim} \frac{\mathrm{p}+\mathrm{q}+\mathrm{r}+1}{{\mathrm{\theta }}^6}=\mathrm{K}$ then the value of the reciprocal of K is

${\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=(\mathrm{x}-\cos \mathrm{\theta })\left(\mathrm{x}-\cos \frac{\mathrm{\theta }}{2}\right)\left(\mathrm{x}-\cos \frac{\mathrm{\theta }}{4}\right)$ Consider
$\begin{array}{l}\underset{\mathrm{\theta }\to 0}{lim} \frac{\mathrm{p}+\mathrm{q}+\mathrm{r}+1}{{\mathrm{\theta }}^6}=\underset{\mathrm{\theta }\to 0}{lim} \frac{(1-\cos \mathrm{\theta })\left(1-\cos \frac{\mathrm{\theta }}{2}\right)\left(1-\cos \frac{\mathrm{\theta }}{4}\right)}{{\mathrm{\theta }}^6}\\ \Rightarrow \underset{\mathrm{\theta }\to 0}{lim} \frac{\left(2{\sin }^2\frac{\mathrm{\theta }}{2}\right)\left(2{\sin }^2\left(\frac{\mathrm{\theta }}{4}\right)\right)\left(2{\sin }^2\left(\frac{\mathrm{\theta }}{8}\right)\right)}{{\mathrm{\theta }}^6}\\ =8\underset{\mathrm{\theta }\to 0}{lim} \left(\frac{{\sin }^2\frac{\mathrm{\theta }}{2}}{{\mathrm{\theta }}^2}\right)\left(\frac{{\sin }^2\left(\frac{\mathrm{\theta }}{4}\right)}{{\mathrm{\theta }}^2}\right)\left(\frac{{\sin }^2\left(\frac{\mathrm{\theta }}{8}\right)}{{\mathrm{\theta }}^2}\right)=8\times \frac{1}{4}\times \frac{1}{16}\times \frac{1}{64}=\frac{1}{512}=\mathrm{K}\end{array}$
Therefore reciprocal of  k is $\frac{1}{\mathrm{K}}=512$