could you explain what "eoc" and" eos" (edge of conduction and edge of saturation) is?
The problem says "Vbe = 0.7 at edge of conduction" , is that in cutoff mode?
BJT Saturation depends on collector voltage or the Δ(Vcb-Vbe)=Vce .
Also hFE drops towards 10% of it's max hFE when Vce=Vce(sat) @ If rated. The actual current ratio varies from device to device.
In datasheets, we define Vce(sat) at some currents Ic ( 1 or more levels) although 10:1 or Ic/Ib=10 is most common, sometimes it defined by OEMs at 20:1 and in super-beta transistors >> 500) it may be defined at 50:1.
As Vbe approaches Vbe(on), the BJT reaches the edge of conduction (EOC), and past that it enters the forward active (FA) region, where it becomes fully conductive. Henceforth, we have Ve = Vb – VBE(on) ≅ Vb – 0.7 V, that is, the emitter will follow the base, albeit with an offset of about –0.7 V ..aka the emitter follower mode.