Correct Answer:

$\frac{1}{\sqrt{2}}$

Solution : Given: $\sin (4\alpha -\beta )=1$ and $\cos (2\alpha +\beta)=\frac{1}{2}$
$\sin (4\alpha -\beta )=1$
⇒ $\sin (4\alpha -\beta )=\sin 90^{\circ}$
⇒ $4\alpha -\beta= 90^{\circ}$...................................... (1)
Now, $\cos (2\alpha +\beta)=\frac{1}{2}$
⇒ $\cos (2\alpha +\beta)=\cos 60^{\circ}$
⇒ $2\alpha +\beta=60^{\circ}$....................................(2)
Solving equations 1 and 2, we get:
$6\alpha=150^{\circ}$
⇒ $\alpha=25^{\circ}$
Putting the value of $\alpha$ in equation 1, we get,
$4×25^{\circ} -\beta= 90^{\circ}$
⇒ $\beta=10^{\circ}$
Thus, $\sin (\alpha +2\beta)=\sin (25^{\circ} +2×10^{\circ})$
$=\sin (25^{\circ} +20^{\circ})$
$=\sin45^{\circ}$
$=\frac{1}{\sqrt{2}}$
Hence, the correct answer is $\frac{1}{\sqrt{2}}$.