Why is the MOSFET not in the cutoff region?

Question

I'm struggling to understand an author's remark in the solving of a circuit including a MOSFET.

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In the following circuit, the threshold voltage \$V_T\$ is of \$1V\$ and -while I don't think it is necessary for my question- \$R=0.6k\Omega\$, with \$\text{Kn W/L}=1mA/V^2\$.

First we realize that it cannot be in the cutoff region since \$V_\text{GS}\$ would be \$5V\$, which is greater than \$V_T\$. So we assume it is in saturation...

As far as I know, a MOSFET is in the cutoff region if and only if \$V_\text{GS}<V_T\$, so I fail to understand the author's remark that if the MOSFET was in the cutoff region we would have \$V_\text{GS} = 5V > V_T\$.

Answer 1

If the MOSFET were in the cutoff (sub-threshold) there would be no current in the top resistor. It's voltage drop would therefore be 0, and the gate would be at 5V. But since the threshold voltage is less than that we arrive at a contradiction. We must reject the assumption that the MOSFET is in cutoff.

Answer 2

1. Let’s assume first that the MOSFET is in saturation and that in saturation its VDS is 0V.
2. In this case, we have 5V across 4*R, which makes 1.25V across each resistor.
3. The Gate voltage is 5 – 1.25 = 3.75V
4. The Source voltage is 5 – 3*R = 1.25V
5. The Gate to Source Voltage is 3.75V – 1.25V = 2.5V > 1V
6. Conclusion: The MOSFET is in saturation.

Let’s do a simulation: