If $\underset{x\to 0}{\lim }{\left(1+px+q{x}^2\right)}^{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c} x}=2048$, then the value of $\frac{p}{11}$ is equal to (take $\ln 2=0.69$)

The given limit is of the form $1^{\infty }$

$\therefore e^{\underset{x\to 0}{\lim }\left(\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c} x\right)\left(1+px+q{x}^2-1\right)}=2048$

$\underset{x\to 0}{\lim }\left(\frac{px+q{x}^2}{\sin x}\right)=\ln 2048$

$\underset{x\to 0}{\lim }\left(\frac{px+q{x}^2}{x}\right)\cdot \left(\frac{x}{\sin x}\right)=11\ln 2$

$$\begin{gathered} \underset{x\to 0}{\lim }\left(p+qx\right)\left(1\right)=11\ln 2 \\ \Rightarrow p=11\ln 2 \end{gathered}$$

Hence, $\frac{p}{11}=\ln 2=0.69$