If function $f\left(x\right)=\left\{\begin{array}{ccc}a\sqrt{x+7}& ;& 0\le x<2\\ bx+1& ;& x\ge 2\end{array}\right.$ is differentiable $\forall x\ge 0$, then $2a+4b$ is equal to

$f\left(x\right)$ must be continuous and differentiable at $x=2$
For continuity at $x=2$
$f\left(2^{-}\right)=f\left(2^{+}\right)=f\left(2\right)$
$a\sqrt{2+7}=b\left(2\right)+1\Rightarrow 3a=2b+1\dots \left(\mathrm{i}\right)$
For differentiability at $x=2$
$f^{\prime }\left(x\right)=\left\{\begin{array}{ccc}\frac{a}{2\sqrt{x+7}}& ;& 0\le x<2\\ b& ;& x\ge 2\end{array}\right.$
$f^{\prime }\left(2^{-}\right)=f^{\prime }\left(2^{+}\right)$
$\frac{a}{2\sqrt{2+7}}=b\Rightarrow a=6b\dots \left(\mathrm{i}\mathrm{i}\right)$
From $\left(\mathrm{i}\right) \& \left(\mathrm{i}\right)$, we get, 

$18b=2b+1\Rightarrow b=\frac{1}{16}$
$2a+4b=12b+4b=16b=1$