Let m and M respectively be the minimum and the maximum values of $f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$. Then m + M is equal to :

Given,
\( f(x)=\sin^{-1}(2x)+\sin 2x+\cos^{-1}(2x)+\cos 2x \)

Using identity,
\( \sin^{-1}a+\cos^{-1}a=\frac{\pi}{2} \)

So,
\( f(x)=\frac{\pi}{2}+\sin 2x+\cos 2x \)

Let \( t=2x \).
Since \( x\in\left[0,\frac{\pi}{8}\right] \),
\( t\in\left[0,\frac{\pi}{4}\right] \)

Now,
\( \sin t+\cos t \) is increasing in \( \left[0,\frac{\pi}{4}\right] \).

Minimum at \( t=0 \):
\( m=\frac{\pi}{2}+\sin 0+\cos 0 \)
\( m=\frac{\pi}{2}+1 \)

Maximum at \( t=\frac{\pi}{4} \):
\( M=\frac{\pi}{2}+\sin\frac{\pi}{4}+\cos\frac{\pi}{4} \)
\( M=\frac{\pi}{2}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \)
\( M=\frac{\pi}{2}+\sqrt{2} \)

Therefore,
\( m+M=\left(\frac{\pi}{2}+1\right)+\left(\frac{\pi}{2}+\sqrt{2}\right) \)

\( \boxed{m+M=\pi+1+\sqrt{2}} \)