Here P-N junction diode rectifies half of the ac wave i.e., acts as half wave rectifier.
During + ve half cycle Diode is forward biased output across R_L will be

During –ve half cycle Diode is reverse biased output will not obtained.

The Boolean expression of this arrangement is

Y = $$\overline {\overline A + \overline B } $$ = $$\overline{\overline A} + \overline{\overline B} $$ = A.B

Thus, the combination represents AND gate.

Potential difference across resistance R

= 3.5 – 0.5 = 3.0 V

Current in circuit, I = V/R = 3/100 = 30 mA

Given, V_i = 2cos(15t + $${\pi \over 3}$$)

and Voltage gain A_V = 150

For CE transistor phase difference between input and output signal is $$\pi $$ = 180^o

Using formula, A_V = $${{{V_0}} \over {{V_i}}}$$

$$ \Rightarrow $$ V₀ = A_V $$ \times $$ V_i

= 150$$ \times $$2cos(15t + $${\pi \over 3}$$ + $$\pi $$)

= 300cos$$\left( {15t + {{4\pi } \over 3}} \right)$$