Ta có:

$\begin{align}  & f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x \\  & \Rightarrow f'\left( x \right)=4\cos x{{\sin }^{3}}x-4\sin x{{\cos }^{3}}x \\ \end{align}$

$\begin{align}  & \Rightarrow f'\left( x \right)=4\sin x\cos \left( {{\sin }^{2}}x-{{\cos }^{2}}x \right) \\  & \Rightarrow f'\left( x \right)=2\sin 2x.\left( -\cos 2x \right) \\ \end{align}$

$\begin{align}  & \Rightarrow f'\left( x \right)=-\sin 4x\Rightarrow f''\left( x \right)=-4\cos 4x \\  & \Rightarrow S=f'\left( \frac{\pi }{4} \right)+\frac{1}{4}f''\left( \frac{\pi }{4} \right)=-\sin \pi -\cos \pi =1 \\ \end{align}$

Vậy đáp án đúng là B.