Need help finding the drain current of a jfet
The image (taken from wikipedia) below shows the JFET chara. It can be seen that the drain current reduces to zero as \$V_{GS}\$ approaches pinch off (\$V_{P}\$). And the channel is off for \$|V_{GS}|&gt;|V_{P}|\$ and no current flow happens. 



So the transistor is in saturation and the Shockley's equation

$$I_D = I_{DSS}\left [ 1-\frac{V_{GS}}{V_P} \right ]^{2} $$

is valid only if \$|V_{GS}| &lt; |V_P|\$ and \$V_{DG} &gt; V_P\$.



Now calculating \$V_{GS}\$ in your case,

case1: \$I_{D1} =1.16mA\$
$$V_{GS} = -2.76V$$ 
But \$V_{P}=-2V\$ so \$|V_{GS}| &gt; |V_P|\$ and hence transistor is in cut-off. 

case2: \$I_{D2} =1.0258mA\$
$$V_{GS} = -1.2838V$$ 
Here, \$|V_{GS}| &lt; |V_P|\$ and hence transistor is in saturation.

So \$I_{D} =1.0258mA\$ is the correct answer.



PS: You should have faced this problem while calculating the value of \$R_S\$ also. 

$$1mA = 4mA\left( 1 + \dfrac{10-1mA\times R_S}{2}\right)^2$$

\$R_S = 11k\Omega\$  and \$R_S = 13k\Omega\$ will satisfy this equation. The value \$R_S = 13k\Omega\$ can not be used because of the same reason discussed above.