Options I am attaching below pls see.

The figure is attached.

The object is initially at rest.

The graph under the force-displacement curve = Work done

work done from x=0 to x=8 = Change in kinetic energy from x = 0 to x = 8

If velocity at x = 8 be

Then,$$\begin{gathered} Area under curve = \frac{1}{2}m{v_1}^2-0 \\ =\frac{1}{2}m{v_1}^2=20\times 5+10\times 3 \\ =v_1=22.8\approx 23m/s \\ Similarly work done from x=8 to x=12 m=change in kinetic energy \\ \frac{1}{2}m({v_2}^2-{v_1}^2)=-20.5\times 2-\frac{1}{2}\times 0.5\times 2+10\times 2 \\ =\frac{0.5}{2}({v_2}^2-{23}^2)=-41-0.5+20 \\ ={v_2}^2=433.84 \\ =v_2=20.82 \approx 20.6 m/s \end{gathered}$$

So option 4 is correct