how do you know whether the transistor is in active, saturated or cutoff state?

Answer 1

Beta is mathematical figure, and only determinable by the operation of and changes in those conditions of the transistor.
Cutoff state, collector voltage is at the applied voltage level. The Collector/Emitter junction is basically an open circuit, so there is no current flow, except for minority carriers. Active, such as in an amplifier, then the Collector voltage will probably be about half of the supply voltage. This depends a lot on how hard the stage is being driven by the signal applied to the base., Saturated, the transistor is conducting very hard, if it weren't for some kind of current limiter, ot would burn up. Collector voltage will be very low, current will be high, and the Base will be at least 0.6 volts, most likely much higher, positive or negative, depending on what type of transistor it is. For the NPN transistor, the Base will have a positive voltage on it, and the opposite for a PNP transistor. Word of caution here, the polarity of the Base can be, and often is RELATIVE. I stress this because the Base voltage could a negative voltage, but still positive compared to the Emitter voltage. Such as, Emitter is a -1.0 volts, and the base of this NPN transistor is -0.3 volts. Is this transistor turned on, yes it very much is because the Base is 0.7 volts more positive than the Emitter. Operating voltages for transistors can be relative, rather than absolutely positive, and above ground potential.

Answer 2

Generally, you can ascertain this by measuring Beta. If the transistor is off, there will be no collector current, Ic.

If it's in active mode, Beta will be about 100, where beta is Ic/Ib. So measure Ic and Ib.

In saturation Beta will be much lower, down to 1. Again, measure Ic and Ib.

You can also measure voltage. For cutoff, Vbe < about 0.7V

For active, Vbe ~ 0.7V and Vce > ~ 0.3V

For saturation, Vce< ~0.2 to 0.3 V