The transistor in the circuit shown is operating in

1. cut-off region
2. active region
3. saturation region
4. either in the active or saturation region

From the figure,

I_B =	5 - V_BE	=	4.2	= 4.2 mA
	100 K		100 K

I_c =	4.8	mA
	5

Since (I_B)_min =	I_c	≤ I_B ,
	h_FE

hence the transistor will operate in saturation region.