$f\left(x\right)=\underset{n\to \infty }{\lim }{\sin }^{4}x+\frac{1}{4}{\sin }^{4}2x+\dots \dots .+\frac{1}{4^n}{\sin }^4\left(2^{n}x\right)\text{ and }g\left(x\right)\text{ is a differentiable function }$$\text{ satisfying }g\left(x\right)+f\left(x\right)=1,\text{ then the value of }g\left(\frac{\pi }{4}\right)\text{ is }$

$\begin{array}{l}\text{We know that }{\sin }^{4}x={\sin }^{2}x-\frac{1}{4}{\sin }^{2}2x\\ \left({\sin }^{2}x-\frac{1}{4}{\sin }^{2}2x\right)+\left(\frac{1}{4}{\sin }^{2}2x-\frac{1}{4^2}{\sin }^2{2}^{2}x\right)+\mathrm{....}+\left(\frac{1}{4^n}{\sin }^2{2}^{n}x-\frac{1}{4^{n+1}}{\sin }^2{2}^{n+1}x\right)\\ {\sin }^{2}x-\frac{1}{4^{n+1}}{\sin }^2{2}^{n+1}x\\ \underset{n\to \infty }{\lim }{S}_n={\sin }^{2}x=f\left(x\right)\end{array}$

$\text{ And }g\left(x\right)+f\left(x\right)=1,\text{ so that }g\left(x\right)={\cos }^{2}x$

$\text{ Hence, }g\left(\frac{\pi }{4}\right)={\cos }^2\left(\frac{\pi }{4}\right)=\frac{1}{2}=0.5$