See figure attached.

I agree with part a) but in part b) isn't there a mistake?

\(v_{A} - v_{B} = I5k\)

\(v_{A} = v_{B} = -I_{1}5k + v_{B}\)

Isn't he off by a negative sign on VB?

 

The voltage at Vc will only be positive and will be one sixth of the input minus two diode drops.

 

The voltage at Vc will only be positive and will be one sixth of the input minus two diode drops.

This doesn't answer my question?

 

You are right - part (b) solution is incorrect as shown.

VA=VB-5000*I is the right answer.