Sedra and Smith Microelectronics 7th edition Example 6.12
First of all 'B' or beta, which is the measure of a bjt transistors gain in a known circuit, is not a factor here based on your question. The emitters are tied together and referenced to ground via a 1K ohm resistor. A little research into NPN and PNP bjt transistors show they have a base-emitter turn-on voltage (forward) of about .55 to .65 vdc for the NPN and about -.55 to -.65 volts for the PNP. Based on your diagram if the base of both transistors was zero volts compared to ground, neither one would be on. If the base voltage rises above .55 volts or so Q1 begins to turn on,but Q2 is reverse biased so it is OFF. If we swap things around and input a -.55 volts to -.65 volts at the bases, then Q2 is ON and Q1 is reverse biased so it is OFF. This forward bias voltage needed creates a ~1.20 volt gap (centered around zero volts,or ground) inside of which neither transistor is ON. So logically, both transistors cannot be on at the same time.