We are supposed to find \$R\$ such that \$I_S=0\$ and then find the maximum \$V_{SRC}\$ for operation in the linear region. I tried to use KCL at the nodes but it didn't work out eventually since I got a value of R that is negative. And for the second part of the question how does one determine the maximum possible value in the linear range: I am not sure how to go about this. I would appreciate any help.
Step by step:
The current, from left to right, through \$R\$ is
$$I_R = \frac{V_{SRC} - V_{O2}}{R} $$
The current, from left to right, through the left-most 10k resistor is
$$I_{10k} = \frac{V_{SRC}}{10k\Omega} $$
KCL at the input node yields
$$I_S = I_R + I_{10k}$$
Using the well-known inverting op-amp gain formula, the two op-amp cascade has a gain of
$$\frac{V_{O2}}{V_{SRC}} = (-\frac{40k}{10k}) \cdot (-\frac{20k}{10k}) = 8 $$
Now, set \$I_S = 0\$ and solve.
A rewarding exercise is to solve for the input resistance seen by the input voltage source:
$$R_{IN} = \frac{V_{SRC}}{I_S} = \frac{V_{SRC}}{I_R + I_{10k}} = \frac{1}{\frac{1}{10k\Omega} - \frac{7}{R}}$$
Note that the input resistance is positive for \$R > 70k\Omega\$, is negative for \$R < 70k\Omega\$ (the circuit supplies power to the voltage source), and is 'infinite' (open circuit) for \$R = 70k\Omega\$
Start by forgetting about Vsrc and R, and just analyze the the two-opamp circuit by itself. You should be able to see what each stage does from inspection, which then easily gives you what both stages together do. Not even a calculator is needed here.
Now consider the input impedance of this circuit. What is it? It should also be obvious from inspection. This gives you a certain current the two-opamp circuit will draw from its input for any particular voltage. Now consider what voltage is on the right side or R for any particular input voltage. This current thru R must be the same as that drawn by the inut for Vsrc to have no current thru it.
Now stop and think about what this circuit is really doing. Start out with R infinite and consider what happens as R is lowered. At inifinite R, this is just the basic two-opamp circuit that should be obvious. Can you see that as R decreases, it supplies ever more current to the input? The effect is that it increases the apparent input impedance of the overall circuit. Eventually R gets to the point where all the input bias current is exactly offset, which is what this question is asking about. At that point (ideally) you have a circuit with infinite input impedance. If you keep going and make R smaller, you get something called hysterisis if Vsrc has any positive resistance.
If you regard V\$_{SRC}\$ as an input, the output of the 2nd op-amp will be 8 x V\$_{SRC}\$ and ideally the resistor R and the 10k resistor should form a potential divider that would recreate V\$_{SRC}\$ as if V\$_{SRC}\$ wasn't there.
This would mean that when V\$_{SRC}\$ becomes connected, it would see an exact replica of itself and no current would flow i.e. I\$_S\$ would be zero.
For the 2nd part, if the op-amps were rail-to-rail perfect devices, the maximum magnitude of voltage seen would be on the output of the 2nd op-amp - this cannot be greater than 12V - knowing this means you can calculate what V\$_{SRC}\$ limits are.
