Limit of series

$\displaystyle f(x)=\sum _{r=1}^n \tan \left(\frac x{ 2^r} \right) \sec \left( \frac x{2^{r-1}}\right); \quad r,n \in \mathbb{N} \\ g(x)=\begin{cases} \displaystyle \lim _{ n \to \infty} \frac {\ln(f(x)+\tan(\frac { x }{ { 2 }^{ n } } ))-(f(x)+\tan(\frac { x }{ { 2 }^{ n } } ))^{ n }\lfloor \sin(\tan(\frac { x }{ 2 } )) \rfloor}{ 1+{ (f(x)+\tan(\frac { x }{ { 2 }^{ n } } )) }^{ n } } & x\ne \frac \pi 4 \\ K & x= \frac \pi 4 \end{cases}$

Given the above and that the domain of $g(x)$ is $\left(0, \frac \pi 2 \right)$ .

• Find the value of $K$ such that $g(x)$ is continuous at $x=\frac \pi 4$ .
• Also find $M$ the number of points of discontinuity of $g(x)$ in $\left(0, \frac \pi 4 \right)$ .
• Enter the answer as value of $K^2+M^2$ . If you think $K$ does not exist enter the answer as $111$ .

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .