Cutoff leaves the collector output at the supply rail. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A low input leaves the transistor in cutoff

The input equals the stated drop, so base current is 0 A and collector current is 0 A.

IB=0 A,IC=0 AI_B=0\ \text{A},\quad I_C=0\ \text{A}
Input lowThe inverter result cites the cutoff region.base 0 Adrive 0 Acollector 0 Ainput 1 Vdrop 1 Vload 3 Aoutput 12 VcutoffHIGH

With no collector current, the output stays high

The collector resistor pulls the output to 12 V, so the checked output state is HIGH.

Vout=12 VHIGHV_\text{out}=12\ \text{V}\Rightarrow \text{HIGH}
Output highHIGH comes from the checked output voltage.base 0 Adrive 0 Acollector 0 Ainput 1 Vdrop 1 Vload 3 Aoutput 12 VcutoffHIGH