This truth table represents NAND gate.

$${{{I_C}} \over {{I_E}}}$$ = 0.98 = $$\alpha $$

$$ \therefore $$ Current gain, $$\beta $$ = $${{{I_C}} \over {{I_B}}}$$ = $${\alpha \over {1 - \alpha }}$$ = 49

$$\beta $$ = $${{{I_C}} \over {{I_B}}}$$ = $${{{I_C}} \over {{I_E} - {I_C}}}$$ = $${{{{{I_C}} \over {{I_E}}}} \over {1 - {{{I_C}} \over {{I_E}}}}}$$ = $${\alpha \over {1 - \alpha }}$$

A diode is said to be forward biased if p-type semiconductor of p-n junction is at positive potential with respect to n-type semiconductor of p-n junction. It is so for circuit (a).