1) An NPN transistor in cut off mode:
"To get a transistor into cutoff mode, the base voltage must be less than both the emitter and collector voltages. VBC and VBE must both be negative. In reality, VBE can be anywhere between 0V and Vth (~0.7V for silicon) to achieve cutoff mode."
a) I've also read that both the base- emitter and base- collector junction will be reverse biased in cut off mode. But how? If the emitter is earthed, then even if Vbe (=Vb?) is less than 0.7V, the base is at a more positive voltage than the collector and consequently it is forward biased, right?
b) Also, why do we need to ground the input and the base if Vbe is already less than 0.7V? (How is the base grounded in the figure? I can see only the input being grounded)
2) In Saturation mode
a) This, I can understand because the collector voltage becomes close to zero during saturation and thus both Vc and Ve are less than 0.7V (and are thus forward biased)
b) But, why do we connect the input and base to Vcc?