The correct answer is (d) 2.5 and 2.5 V
To explain: I1 = \(\frac{V_1-V_2}{2}\)
Applying KCL at node 1, 5 = \(\frac{V_1}{1} + \frac{V_1-V_2}{2} + V_1 \)
10 = 2V1 + V1 – V2 + 2V1
Or, 10 = 5V1 – V2
KCL at node 2, \(\frac{V_1-V_2}{2}\) + V1 + 2I1 = V2
∴ 1.5 V1 – V2 = 0
∴ V1 = V2 = 2.5 V.