For a transistor in CE mode to be used as an amplifier, it must be operated in:

We begin by recalling how a bipolar junction transistor (BJT) behaves in its different regions of operation. In the common-emitter (CE) configuration, the three principal regions are:

1. $$\text{Cut-off}$$ – here the base-emitter junction is reverse-biased and the collector-base junction is also reverse-biased. As a result, both the collector current $$I_C$$ and the emitter current $$I_E$$ are practically zero, giving no output signal.

2. $$\text{Saturation}$$ – in this region both the base-emitter and the collector-base junctions are forward-biased. The transistor conducts a large current limited mainly by the external circuit, so the collector-emitter voltage $$V_{CE}$$ drops to a very small value, almost behaving like a closed switch.

3. $$\text{Active (Forward-Active)}$$ – now the base-emitter junction is forward-biased while the collector-base junction remains reverse-biased. Under these conditions the transistor follows the fundamental relation

$$I_C = \beta I_B,$$

where $$\beta$$ is the current gain (a constant for a given device and operating point). Because $$I_C$$ is a faithful, almost linear, amplification of the base current $$I_B,$$ any small variation $$\Delta I_B$$ produces a proportionally larger variation $$\Delta I_C,$$ giving voltage amplification across the collector resistor $$R_C.$$ This is precisely the behaviour required for an amplifier: linearity, predictable gain, and low distortion.

Now, let us match each option with the required characteristics:

• In cut-off, $$I_C \approx 0,$$ so no useful amplification is possible; the output is essentially zero. Therefore cut-off is unsuitable.

• In saturation, both junctions are forward-biased, the transistor behaves like a closed switch, and $$V_{CE}$$ is very small. Any attempt to superimpose a small signal on this state will be severely clipped, destroying linearity. Hence saturation is also unsuitable for amplification.

• In the active region the transistor provides the linear relation $$I_C = \beta I_B,$$ allowing faithful reproduction and enlargement of the input signal. Consequently this is the only region in which a BJT in CE mode acts as a proper amplifier.

• The fourth choice, “both cut-off and saturation,” corresponds to switching applications (digital logic), not to linear amplification.

So, only the active region meets the requirement for CE amplification.

Hence, the correct answer is Option C.